Esteem 8 Technical Documentation.pdf

ESTEEM INTEGRATED TOTAL SOLUTION STRUCTURAL PACKAGE

Technical Documentation Copyright © 1994-2011

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Table of Contents Technical Manual Welcome Notations Feature Comparison Support reaction from secondary beam Basic Concepts Definitions Project properties setting Grid Lines Alignment of grid mark after grid line rotation Parameters Material properties Poisson's ratio Loading Live load reduction (BS6399) In walls In Foundations Slab load distribution Roof line method Region closer to node Small span beam in L-Shaped slab panel One straight free edge Two parallel free edges with two boundaries Wind Loading Wind load input comply to rectangular load distribution Limits of wind load application Wind loading validation Example calculations Brickwall Brickwall input on element with drop Finite Element Method Element Types Classification Frame element Wall frame element Shell Wall element Deep beam element Beam element with opening(s) Plate Plane stress Meshing Common Mesh quality control Bridging beam Intepretation of Results Different mesh sizes in slabs Residual forces in upper sub-frame column in 2D Analysis Wall Frame beam Unsymmetrical SFD for structural with symmetrical shape and p2

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loadings 2D Mesh Column meshing in flat slab configuration Continuous column in wall Meshing beam element Modelling a couple beam as a frame element PWallPlateNodes Concept Two beams connecting to a wall 2D corner slab mesh Length determination for beam loading Pinned condition of slab element 3D Mesh Alternative model Couple beam meshing 3D Slab Meshing Slab diaphragm effect Transfer wall seated perpendicularly to a wide beam Transfer wall seated at an eccentricity on a transfer beam Analysis 2D Analysis 2D wall sub-frame analysis 2D wall sub-frame design Auto Support Cantilever beam behaviour in FEM analysis Study of support reactions in statically determinate models Subframe and continuous beam analysis Transfer Wall Beam Stiffness increase Couple Beam connection to wall FE mesh Optimised pattern loading Two-way Slabs : Uniformly loaded rectangular panels 3D Analysis P-Delta Analysis Analyse 3D frame failed Seismic loading and analysis References Formulation of seismic parameters Site Classification, Table 1613.5.2 ref. (3) Fa, Table 15.2a ref. (1) Fv, Table 15.2b ref. (1) R, Table 1617.6 ref. (3) I, Table 15.5 ref. (1) Examples CT, Building Period Coefficient, ref. (1) Vertical Distribution of Forces Unstable Structure What is a closed form solution? Slab Mesh in 3D Analysis Diagnosing Results Trouble-shooting large displacement in model Alternative Load Assignment Design Modified basic span/depth ratio in cantilever spans Tensile strength of concrete Detailing - General p3

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Link diameter when fy not equal fyy Slab Slab Deflection check and control Slab Deflection Auto Pass Design Irregular slab span length Determination of Basic Ratio for Irregular Shape Slab Design Slab Top Bar (Support) Analysis Result Hogging Moment Selection Normal bar over Fabric Mesh Application of Clause 3.12.11.2.7 in slab bar spacing selection Skip Node Within Column Profile Detailing Hogging Moment at Slab mid span Limitation of cut section Cantilever Slab with drop Slab sagging moment at support Slab top bars curtailment Beam Beam fails in torsion design Beam deflection check failed Detailing Bar lapping issues at beam ends Cut sections Shear and torsional link detailing rules Link details in sections Beam detailing - Parallel Wall Beam sections of different width Links at intersecting beam Maximum distances between bars in tension Clause 3.12.11.2.2 (BS8110-1:1997) Transfer Beam Detailing Transfer Beam Detailing Specification Top rebar in cut section not drawn to the actual detailing Top Bar detailing rules Bottom Bar Detailing Use tension bar lapping for sagging moment at support Nib Beam Design Ultimate Moment Resistance of Parabolic Stress Block (BS8110) Design of multisection beam Deflection check and control Framing beam Fully supported transfer beam design Beam Deflection Auto Pass Design Example Design for torsion and shear reinforcement Behaviour of Deep Beam using the ‘Strut and Tie’ model Number or external loop in shear reinforcement Shear force diagram known issues Point load application Shear force diagram in beam-frame element vs. couple beam-shell element Study on the qualitative effects of the tension force in the transfer beam due to the surrounding elements p4

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Alternative deflection check (for cantilever beam only) Node Value in Full Report Not Tally with Summary Report Column Column Clear Height, lo Shear in Columns Computation of column beta value Couple column Column End Conditions (Out of plane cases) Curved beam framed into column Column dimension for shear design Column Nominal Moment Follow Top Column detailing Slender Column Checking of Column Slenderness Limit (BS8110) Column deflection under ultimate conditions (BS8110) K value Example for column slender moment calculation Sway Effects No column design when stump height is zero Wall Effective Length of Wall in Design of RC Wall Wall Detailing Wall Detailing with small openings Wall Not Design Not Design due to short transfer wall Wall bypass intermediate beam within two support Links for Containment of Large Amounts of Compression Reinforcement of Walls Wall Design and Detailing using Strut and Tie Model Support Width for Strut and Tie Model Shear Contour Centroid Wall Region sign for forces Wall Region local axis for Vx,Vy Wall Region local axis for Mx and My How RC Wall Self Weight in 3D Mesh Differs from Self Weight in 2D Mesh Wall Restrain Condition Wall Effective Height Example 1: Wall Connected to Foundation Example 2: Continuous Wall Across Multiple Floors Example 3: Wall Connected to Non-Orthogonal Members Example 4: Wall with Opening Example 5: Wall Connected to Cantilever Beam Example 6: Wall Connected to a Beam of Several Sizes Example 7: Wall Connected to Beam that is Not Attached to Lateral Restraint Element Example 8: Walls Above and Below Not Considered in Wall Relative Stiffness Calculation Example 9: Restraint condition for elements at wall end Example 10: Intermediate transverse wall or beams Foundation Pad Unequal bar distribution Pile Auto Pile Selection p5

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Shear check for Pile Cap Design (BS8110) Example 1 Pile group analysis calculation Pile cap size dependency on pile or column size Quantity Takeoff Quantity Take Off - Beam Quantity Take Off - Column Quantity Take Off - Wall Reinforcement (Normal Bar) Fabric Framing Beam Anti-Crack Bar Reinforcement (Wall Group Bar) Quantity Take Off - Slab Reinforcement (Normal Bar) Reinforcement (Support Normal Bar) Reinforcement (Fabric) Reinforcement (Support Fabric) Benchmarking Alternative Model Equivalent Lateral Force Method for Seismic Analysis 5x5m Slab Slab FEM result : EsteemPlus and Esteem 8 Beam FEM result : EsteemPlus and Esteem 8 Single Panel Shear Wall Flat Slab Verification Column Drop regions overlap or <500mm to each other Column Supporting Flat Slab Range of Column Drop B and Drop H values Slab Opening in Flat Slab Analysis Meshing of column drop area Design & Detailing Flat Slab not designed Punching Shear Calculator Average Moment Over Support Raft Input Column Drop Column Drop Value Outside Acceptable Range Slab Opening on Raft Slab Rectangular Opening Circular Opening Polygonal Opening Column Point Range of Subgrade Reaction Verification Column profile sitting outside of raft slab edge Column not fully input within raft slab Curve beam and curved slab not allowed in raft foundation Openings in Raft 3D Analysis p6

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Columns on Raft Raft Foundation Contour Cut Section on Raft Contour Raft slab cut section display Reset of 3D Mesh Analysis if input object data is edited on raft floor Benchmark 3D raft analysis with 2D frame analysis External Loads on Raft Slab 3D Mesh Band strip contour cut section Full Band Width Use Rectangular Mesh for Raft Slab Column Point as fixed node Analysis Failed due to Raft Foundation Uplift Rigid Zone Meshing of wall on raft edge View Raft support nodes reaction display Design Raft Floor Parameter By-pass Batch Design Parameter Edge Spring Multiplier Constant Design Result Punching Shear Perimeter Average Moment Detailing Limitation of band strip cut section Transfer Slab Verification Column is not supporting free edge of slab Column or wall sitting across two transfer slab Analysis Column sitting on a transfer slab Transfer Slab Mesh Design & Detailing Design Punching Shear Checking for Upper Column Punching Shear Checking for Upper Columns Multi-level Foundation Mutli-level Column Footing Multi-level Wall Footing View Design Beam with Opening Analysis 2D Mesh 3D Mesh Design To obtain 3D Analysis Result for Top Chord and Bottom Chord of a Beam Opening To obtain 3D beam result diagram for Beam with Opening Elaboration of 2D and 3D Opening Node Value Additional 2D Load Case for Reverse Moment Case Cracking Bar Calculation p7

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Detailing User-defined beam opening rebar lapping length Limitation Slenderness moment check Code of Practice Exception Code of Practice Safety factors BS8110 and CP65 Ec (Elastic Modulus) Maximum Shear Link Spacing Tension anchorage Enhanced Shear Multiplier BS8110: 1985 only vc (Design concrete shear stress) Shear & Torsion Design Calculations Distribution of Torsion Longitudinal Perimeter Bar BS8110: 1997 only vc (Design concrete shear stress) Column Eccentricity only checked on single axis Minimum Diameter of Longitudinal Reinforcement in Columns Enhanced Shear Multiplier CP65: 1999 only Ec (Elastic Modulus) ACI 318: 2008 Material Ec (Elastic Modulus) used in ACI 318 : 2008 Loading Load factor used in ACI 318: 2008 Strength Reduction Factor for ACI 318 : 2008 Design Design Concept Design ultimate moment resistance of parabolic stress block for ACI 318 : 2008 Modified span/depth ratio for cantilever spans Slab Design Flexural reinforcement for slab in ACI 318 : 2008 Slab deflection check in ACI 318 : 2008 Determination of Modified Span Depth Ratio for Irregular Shape Slab Beam Design Flexural reinforcement for rectangular beam in ACI 318 : 2008 Flexural reinforcement for flanged beam in ACI 318 : 2008 Minimum area of tension reinforcement for ACI 318 : 2008 Beam shear reinforcement in ACI 318 : 2008 Design concrete shear stress, vc for ACI 318 : 2008 Maximum shear link spacing for ACI 318 : 2008 Beam torsion design in ACI 318 : 2008 Maximum torsion link spacing for ACI 318 : 2008 Beam deflection check for ACI 318 : 2008 Column Design Column classification (short or slender column) for ACI 318 : 2008 Example Calculation of Relative Stiffness, Ψ Short and slender column design in ACI 318 : 2008 p8

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Pad Footing Design Flexural reinforcement design for pad footing in ACI 318 : 2008 Punching shear checking for pad footing in ACI 318 : 2008 Bearing strength of column and footing checking for pad footing in ACI 318 : 2008 Flexural shear checking for pad footing in ACI 318 : 2008 Pile Footing Design Flexural reinforcement design for pile footing in ACI 318 : 2008 Punching shear checking for pile footing in ACI 318 : 2008 Bearing strength of column and footing checking for pile footing in ACI 318 : 2008 Flexural shear checking for pile footing in ACI 318 : 2008 Detailing Beam Detailing Distribution of Torsion Longitudinal Bar used in ACI 318 : 2008 FAQ Does the program carry out shear check for slab design? Why the longitudinal steel area required for torsion at compression face is zero? How hanger bar is designed for ACI? How does Esteem 8 calculate beam shear and torsion link reinforcement required for the beam section in ACI : 08 ? IS456: 2000 Maximum Shear Link Spacing AS3600: 2001 vc (Design concrete shear stress) Maximum Shear Link Spacing NZ3101: 1995 Maximum Shear Link Spacing EC2: 2004 Material Poisson's ratio used in EC2 : 04 Stress Strain Diagram for reinforcing steel used in EC2 : 04 Mean tensile strength of concrete Ecd (Elastic Modulus) used in EC2 : 04 Loading Wind Loads for EC1 Pattern loading for EC2 : 04 Load factor used in EC2 : 04 Live load reduction for column, wall and foundation used in Euro Code Geometric Imperfections (Notional Horizontal Load) Design Design Concept Procedure for determining flexural reinforcement in EC2 : 04 Design ultimate moment resistance of parabolic stress block for EC2 design Modified Structural System for cantilever spans Slab Design Flexural reinforcement for slab in EC2 : 04 Slab deflection check in EC2 : 04 Slab Auto Deflection Control for EC2 : 04 Determination of Structural System, K, for Irregular Shape Slab Beam Design p9

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Flexural reinforcement for rectangular beam in EC2 : 04 Flexural reinforcement for flanged beam in EC2 : 04 Minimum and maximum area of tension reinforcement for EC2 : 04 Beam shear reinforcement in EC2 : 04 Maximum shear link spacing for EC2 : 04 Beam torsion design in EC2 : 04 Maximum torsion link spacing for EC2 : 04 Beam deflection check for EC2 : 04 Beam auto deflection control for EC2 : 04 Column Design Concrete Shear Capacity, VRd,c Column classification (short or slender column) for EC2 : 04 Example calculation of Relative Stiffness, k Method to Determine Braced/Unbraced Frames Short and slender column design in EC2 : 04 Pad Footing Design Flexural reinforcement design for pad footing in EC2 : 04 Punching shear checking for pad footing in EC2 : 04 Ultimate shear checking for pad footing in EC2 : 04 Flexural shear checking for pad footing in EC2 : 04 Pile Footing Design Flexural reinforcement design for pile footing design in EC2 : 04 Punching shear checking for pile footing in EC2 : 04 Ultimate shear checking for pile footing in EC2 : 04 Flexural shear checking for pile footing in EC2 : 04 Detailing Beam Side Bar Detailing FAQ Does the program carry out shear check for slab design? Does Esteem 7 do the shear between web and flanges checking for the flanged beam design in EC2 ? How does Esteem 7 calculate beam shear and torsion link reinforcement required for the beam section in EC2 : 04 ? Does the program do the shear check based on EC2 : 04 Part 1 Clause 6.2.1 (8) ? How hanger bar is designed for EC2? EC8: 2003 Parameters Ductility Classes and Material Requirements Input Beam Geometrical Constraints Column Geometrical Constraints Wall Geometrical Constraints Verification Design and Detailing Beam Minimum Reinforcement Ratio of the Tension Zone Maximum Link Distance from Support for DCM and DCH Class Beam Critical Regions for DCM Class Minimum Link Diameter and Maximum Link Spacing for DCM Class Beam Critical Regions for DCH Class p10

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Minimum Link Diameter and Maximum Link Spacing for DCH

832

Column Minimum and Maximum Compressive Steel Percentage Column Normalised Axial Force Check Column Critical Regions for DCM Class Column Maximum Link Spacing for DCM Class Column Critical Regions for DCH Class Column Maximum Link Spacing for DCH Class Wall Wall Normalised Axial Force Check Wall Critical Regions for DCM Class Minimum Link Diameter and Maximum Link Spacing for DCM

833 833 834 835 836 837 838 839 839 840 841

Class

Class Maximum Distance Between Longitudinal Bars for DCM Class Wall Critical Regions for DCH Class Minimum Link Diameter and Maximum Link Spacing for DCH

842 843 844

Maximum Distance Between Longitudinal Bars for DCH Class EC2: Singapore National Annex Wind Loads to Singapore National Annex Beam End Condition Setting Design Compressive and Tensile Strengths Maximum Concrete Strength Class Maximum Characteristic Yield Strength of Reinforcement Minimum Diameter of Longitudinal Reinforcement in Columns Slenderness Limit Calculation Foundation Pad Pile Technical Support

845 846 846 848 849 850 851 852 853 854 854 855 855

Class

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Technical Manual Welcome Welcome to Esteem 8 Technical Documentations. The information provided here will provide users with up-to-date and valuable information to Esteem 8. This compilation will complement Esteem User Manual. Modeling techniques, tips and good engineering practice are shared. We also hope that this will be helpful to you and become the de facto source of information whenever you have a technical query with regards to Esteem 8. One of the quickest way to search up your query is to type in the keyword in the search dialogue box. Support Team Malaysia Esteem Innovation, 7A-C, Jalan Kenari 10, Bandar Puchong Jaya, 47100 Puchong, Selangor, Malaysia. Telephone : +60-03-8076 2788 Fax : +60-03-8076 2677 Singapore Esteem Innovation Pte Ltd (200008042C) 20 Maxwell Road #09-17, Maxwell House, Singapore 069113. Telephone : +65 6408 9667 Fax : +65 6399 3699 World Wide Web : www.esteemsoft.com Email : [email protected] Office Hours : Monday to Friday 8:30 A.M. to 5:30 P.M. GMT

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Notations Ec

Static secant modulus of elasticity of concrete

fcu

Characteristic cube strength of concrete

fc

Characteristic cyclinder strength of concrete

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Feature Comparison Support reaction from secondary beam Loading diagram for load transfer to main beam from secondary beam. See topic "Auto Support" for further clarifications. Esteem Plus

Esteem 8

[$X$IF_CON2]No such point load is displayed.

Displayed as explicit point load assigned to main beam.

At the beam intersection, there is full displacement compatibility based on the global stiffness assembled from all the finite elements of the floor.

Reaction from secondary beam gb3.

Reaction from beam gb5.

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Esteem Plus

Esteem 8

[$X$IF_CON2]Shear at left side of beam gb5.

Beam loading diagram showing reaction from secondary beam.

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Intersecting beam point load now shown in beam loading diagram (after the analysis with slab node loadings shown).

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Esteem Plus

Esteem 8

[$X$IF_CON2]Beam loading diagram does not show point load as per Esteem Plus.

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Basic Concepts Definitions Bridging beams

Rigid elements in the model with very high stiffness (usually used in modelling elements with offsets to physically represent the element which is only a line representation in a computer model)

Deep Beam

1. [C lear span]/[overall depth] <= 4 2. C oncentrated loads less than twice the overall depth from the face of support. Notes : Any transfer beam, i.e. supporting either wall or column can be classified as deep beam.

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Project properties setting This menu can be used to perform the concrete grade verification of the whole structure.

Click on the cell you want to edit and type in the new value. Then click "OK". The necessity to have this concrete grade verification arises from the fact that in Esteem 8 each individual floor have its own set of parameters, one of which is the concrete grade. Therefore, it is possible to have defined different concrete grades for different floors. The verification process will produce a table so that the user can use this concrete grade information in the preparation of the design drawings. This table can also be used to change the concrete grade for a particular floor.

1. Project Properties Setting table can be called out within an opened project. 2. Click Verification->Project Properties Table. 3. Internally the program will search through every floor’s parameter to obtain the, a. Concrete Grade for every floors and, b. Bar list for column and wall 4. The program also searches the foundation concrete grade of the structure. Program 5. 6. 7. 8.

9.

will obtain the main bar and link bar list form the parameter. This list will be used in the bar list to control the maximum and minimum bar size to design the column. In the concrete grade list, concrete grades will be represented by differing colours. Text box will use to list the concrete grade. User is allowed to change the concrete grade which must be an integer. Column and wall bar list will be represented using combo box. User can only select the bar size in the list provided. User can change the value of the concrete but cannot delete the concrete grade. There are options to apply the changes on the particular flow to the other floors. Right click the cell and select apply to either, a. all floor, or b. Up i. All ii. Specify, or c. Down i. All ii. Specify User can change the maximum and minimum bar diameter for every floor. p18

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10. Click OK after assigning all values. 11. Project Properties Setting should be able to validate the concrete grade and bar list. Program will loop through every floor validating the concrete grade and bar list. 12. Bar list for every floor will fulfill the following condition

13. Program will check concrete grade floor by floor. 14. Concrete grade for every floor will fulfill the following condition. f = concrete grade

15. After checking through the above, program will check through following condition where f = concrete grade

16. A warning message will pop up if any of the conditions are not met. 17. After checking, the message, “Do you still want to proceed with these data at your own risk?” together with the report in the dialog box will appear to the user. Users are allowed to proceed changes to the values mentioned or retained the old values. 18. If user agrees with the changes by clicking OK, all the value in the Project Properties Setting will be assigned to the floor parameter but not project (global) parameter. 19. If user chooses Cancel, the dialog box will close and return to the Project Properties Setting. 20. If there are no changes, just click “OK” to exit the table. 21. The concrete grade for the foundation must always bigger or equal than lowers floor c olumn’s concrete grade. Program will always prompt warning message to the user if this condition is not fulfilled but still allowed user to revert the value to the floor parameter.

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Flow chart for showing the concrete grade.

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Grid Lines Alignment of grid mark after grid line rotation 1. Rotated grid will either trim or extend the group of grids in the other axis. 2. If there is a free grid amongst the group of grids, it will not be extended or trimmed. Example: When there is grid line rotated, the grid mark will be shorten as shown as below. In this case, user are not encourage to copy or mirror a rotated grid.

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Parameters

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Material properties Poisson's ratio The following are some extracts from "Finite Element Design of concrete structures, Practical problems and their solutions ", by G.A. Rombach, Thomas Telford 2004. pages 157 & 158, "Poisson's ratio  is not an exact value for reinforced concrete. Values varying from = 0.0 to  = 0.2 are generally used. The Poisson's ratio  is defined for an elastic member as the ratio between the lateral strain and the axial strain." "Bittner [Platten und Behalter. Springer Verlag, Wien, 1965] has conducted theoretical investigations to determine the correct value for Poisson's ratio. He proposed that a value of = 0.0 should be used in the design of reinforced concrete slabs. With this value the compressive stresses are usually not relevant in the design".

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Loading Live load reduction (BS6399) The allowance of live load reduction is as per Table 2 of BS6399:Pt.1-1996. For conveninece, the table is reproduced below:Number of floors with loads qualifying Reduction in total distributed imposed load for reduction carried by member under on all floors carried by the member under consideration consideration, % 1 0 2 10 3 20 4 30 5 to 10 40 over 10 50 max Application in Esteem 8

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The element in design consideration would collect the live loads from each floor. The load path should be continuous to establish the number of floors the element is carrying. As such, the live load reduction application is effective on individual continuous line of elements of columns or walls. On each floor of the set, the live load % reduction is applied to the particular floor as shown on the left starting from 0% to 50% for over 10 floors supported. Example: Floor

Live load column reaction from plan analysis (kN)

% LL Live Load collected reduction by column

By floor

11F 10F 9F 8F 7F 6F 5F 4F 3F 2F 1F

∑

100 100 100 100 100 100 100 100 100 100 100 1100

0 10 20 30 40 40 40 40 40 40 50

Cumulative from all floors above 100 100 90 190 80 270 70 340 60 400 60 460 60 520 60 580 60 640 60 700 50 750

Reminder

: While the specification of this live load reduction is conservative to the fact of having a water tank at the top most floor, design engineers must use judgment when a live load such as water tanks or swimming pools are located on an intermediate floor. In such cases, the engineer can opt not to have live load reduction.

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In walls The deducted live load is collected on "Panel" basis. For the Panel Considering Opening, Percentage of Total Live Load deducted from upper floor is distributed base on stress distribution in panel. From the Value of P and M, and we have the Panel Length L, Start and End Stress per thickness = P / L +- 6 M / L2; and we use triangle distribution to determine the percentage of deducted live load assigned to the "Design Zone".

The following live load reduction situation is not applied :-

1) One of the wall is not transferred to the foundation.

2) Part of the wall is not transferred to the foundation.

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3) One of the wall does not stop at the same level.

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4) Length of the wall panel not aligned from bottom to top.

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Notes

: No live load reduction is applied if uplift force occurs in the panel .

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In Foundations Live Load Reduction in Foundation Live load reduction in foundation will be enabled when the following criteria is fulfilled:a) Live load reduction in wall is enabled (for foundation supporting wall only) b) Live load reduction in column is enabled (for foundation supporting column only) c) Live load reduction in both column and wall are enabled (for foundation supporting wall and column)

If live load reduction for wall is applied (Refer to the condition in Live Load Reduction in Walls), live load reduction in foundation is taken from the total live load reduced from each floor for all walls or columns supported by the foundation. Live load reduction in foundation = Live load reduction in wall (all floors) + Live load reduction in column (all floors)

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Slab load distribution Methods for slab load distribution onto beams and walls:A. Reactions from finite element mesh analysis. B. Conventional "roof line method"

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Roof line method Introduction Traditionally beam analysis and design follow the basic process of taking off loadings from slabs and applying it to the beams. The analysis of the beam is carried out with these applied loadings. These applied loadings from slabs consist of various types of distributed loads. The pattern and intensity of these loads can be derived from “yield line” of the slab. In Esteem, we developed a similar approach called the “Roof Line Method”. The structural analysis is by the finite element method employing the stiffness matrix. Unlike conventional continuous beam analysis where the loads are taken off to a one dimensional model, the "roof line" loadings are applied to the beams in the 2D analysis. What is essential to take note here is that the stiffness of the "weightless" slab is taken into account.

Qualifications for the Roof Line Method The panel of slab must be fully bounded by either beams or walls or combination of both. Let’s consider a rectangular slab panel to illustrate the method. Assume a 6m by 3m panel shown below:-

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3D View Assumption in the program:1. The roof line boundary bisects the angle between beam/wall boundaries equally (the "alpha" angle as shown above). 2. "beta" angle is 45 degrees to formula the roof line planes. 3. The triangular and trapezoidal shapes shown in the plan view above is the projection of the 3D roof line onto the slab panel.

Differences between Conventional Way, Roof Line Method & FEM Loading Assignment to Beam from Slab In the conventional method:  The loading depends on whether it is a one-way or two-way slab. If it is one-way, the shorter edge will not assign any load to the beam. In the roof line method:  both long and short edges will assign loading from the slab. The load distribution pattern will be like a roof shape, and the beam is assumed to be stiffer than the column, therefore the load will transfer to the mid-beam (like the conventional way) FEM method:  load will transfer towards the columns p35

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Note: 1. When the program finds that one slab load cannot be distributed via roof line method, the whole project will switch to FEM as a combination of both is not logically consistent. 2. Then the only way to make the load to be distributed like the conventional way, is to increase the beam stiffness in the Project Parameter  2D analysis to 99. However, setting it this way may not be 100% making all the loads transferred to the beam as in the conventional way, because it will only increase the beam stiffness so that more loads will be distributed to mid-span of the beam.

Limitation

:

When there is a gap in the boundary in a single slab of multiple sub-slab as shown below,

the roofline method will not be possible. However, it will still be possible if we break the slab into individual slabs as shown below,

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Region closer to node Loads on areas that are closer to a node than to any beam /wall boundaries are applied as an equivalent UDL to the nearest adjacent boundary. The following example illustrates the issue:-

Loading on GB1

Total load in area “a” above shall be converted to equivalent UDL in shown Diagram 2 hence, b = Fa / w where Fa = total load of area “a”

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Small span beam in L-Shaped slab panel A special case when the application of "Region Closer to Node" does not hold is when the span of the beam where the load converted to an equivalent UDL is not long enough to accomodate the total load. The "extra" load is then distributed to the following bounding beam in the panel. The illustration below would clarify the above textual explanation:The example from plan has beam 3-2 as the short span.

Plotting the distribution of the slab load on beam 3-2-4 gives us a triangle load from node 3 to 2,

and the load from slab area A indicated on the plan is converted to an equivalent UDL load on the bounding beam starting from node 3 with the intensity as w . The total length of this equivalent UDL would be s1 + s2 if the require length if longer that s1.

Here the portion of equivalent UDL u2 is applied to the bounding beam 2-1.

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One straight free edge The same rules apply to the boundaries but special treatise to the free edge. The free edge is divided to two equal distances from the boundaries forming its begin and end. This midpoint shall be the ridge point between the two boundaries defining the free edge. The following example shall clarify this situation:-

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Two parallel free edges with two boundaries The definition of free edges in slab means the edge that is not bounded by supporting element such beams or walls. Assume the following setting in Project Parameter->2D Analysis->Analysis Option->Use Roof Line Method->"True". Consider the following slab panel and beam boundaries:A.

With one slab bounded as shown above, the program would not be able to perform the "Roof Line Method", loads will be taken to the beams from the nodes of the finite element mesh. B.

With two slab bounded as shown above, the program would perform the "Roof Line Method". UDL, GVL or Triangular Load will be defined as beam loadings. p40

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Wind Loading Wind Load input Reference: BS 6399 Part 2: 1997 For the wind load input, users have to manually define the maximum design wind speed at the top floor and minimum design wind speed at ground floor (reference floor)and the height from the ground surface (default = 0m). In this situation, it is assume that the user already manually calculate the design wind speed (clause 2.2.2.1) from the basic wind speed (clause 2.2.1).

Notes

: All lateral loads will be applied at the intersection of the element. Wind loads that are applied on elements that do not intersect with each other will be removed during 3D analysis. Data required: Maximum design wind speed Vs max (m/s) Reference floor Minimum design wind speed Vs min (m/s) Reference floor Height from the reference floor, href (m) Calculation Figure 1 General wind load act on the structure

Dynamic pressure of the wind, q (clause 2.1.2.1) (N/m2) Values of k are as stated in the clause 6: k = 0.613 in SI units Vs = design wind speed (m/s) * hmax = height of the top reference floor hmin = height of the reference floor (no need to be lowest floor) 

Vs will vary depend to the height from the ground level.

Therefore, to calculation

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Figure 1: General Diagram of Wind Load Calculation. General rules

Vabove for each floor,

Vbelow for each floor, Vaveg for each floor,

Vbottom for every project hcur = height of the node from the foundation point. Vbotttom refer to wind speed at the most bottom of the structure, the node between columns with foundation (pad or pile section).

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Figure 2 : Vbottom diagram At the boundary if user provide minimum wind load with reference height from the floor he stated the following calculation should carry out

Vabove for each floor, Vbelow for lowers floor, Vaveg for each floor, hcur = height of the node from the foundation point.

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Figure 3: Diagram above showing the node where wind force act to. After calculate the vaveg for each floor, the loading apply on to each node at each floor will be as shown in the figure 4 Obtaining Area that subject to the wing forces:

Area, A = H x l

Point load on to the node,

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Figure 4: Diagram showing symbol use in the calculation of wind load, F1 For transfer Column

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Area,

Point load on to the column c1, Wind Load Case Wind load will act on the building from difference direction. However the wind load will not act on the building from difference direction at one time. Therefore wind load have to separate into difference cases.

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Wind load Case input Users have to first define the wind load case. Program will define the wind load case name for the user. Example  Wind load case 1  Wind load case 2  Wind load case 3 To create a wind load case, user must key in the description of each wind load. Example Wind load case 1 Direction, Angle of wind speed Description: This wind load is act from north or 90 degree. Calculation

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To calculate the loading for the column c1

Area,

Point load on to the column c1, To calculate the loading for the column c2

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Area,

Point load on to the column c2, Note: All angles must in degree. Example Vmax = 30 m/s, Top floor height = 40m Vmin = 10 m/s, Bottom floor height = 13m, Reference height = -1.0 m Floor to floor height = 3 m

Find load at 22m height, Column to column (both side) distance = 3 m

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Find load at 13m height (Reference Floor), Column to column (both side) distance = 3.0 m

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Find load at 13m height (Reference floor’s corner), Column to column (both side) distance = 3, α = 30°

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Example 2

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From the example above, the wind load a will only apply onto column c1. Wind loads a will not being applied onto column c2. Calculation for c3a is as follow

l2 is equal to 0 because the area are hided from the wind force by the part of the building.

Area, Point load on to the column c1, Example 3

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From the example above, the wind load a will only apply onto column c3. Wind loads a will not being applied onto column c4. Calculation for c3b for corner is as follow

Area, *c3a will hide from the wind load because it is at same grid line of c3b.

Point load on to the column c1,

Example 4 p55

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From the example 4, the wind loads a will only act onto column c1, c1a, c1b, c1c, c1d and c1e. Wind loads a will not being applied onto column c2, and c3. Calculation for c1, c1a, c1b, c1c, cl d and cl e is as follow First project all the columns down

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C1

C1a

C1b

C1c

C1d

C1e

Example 5

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From the example 4, 1. The wind loads A will only act at node N1, N13, N12 and N11. 2. The wind loads B will only act at node N1, N2, N3, N4, and N5. 3. The wind loads C will only act at node N2, N3, N5, N6, N7 and N8. 4. The wind loads D will only act node N8, N9, N10, N14and N11. Basically all the wind loads that act on the wall will act on both end of the wall, wall-beam connection and column in the wall. The calculations of the wind loads on the wall are same as the calculation of the wind load on the column.

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Wind load input comply to rectangular load distribution Wind Load input (Rectangular Load Distribution) Reference: CP65 Example: Wind Load for HDB Residential Design o Rectangular load distribution

Data required: Model height (m) = Program automatically calculates the sum of storey height to the reference floor Height limit (m) = Refer to Table 1 for tabulation of the default values Reference floor = Always the lowest floor Basic wind speed, V (m/s) = 35 (default), value range = 0 - 100 m/s Force Coefficient, Cf = 1.0 (default), value range = 0.60 - 2.00 Calculation (automatically calculated by the program): S2 = 0.50 - 2.00 (according to Height Limit) Design Wind Speed, Vs (m/s) = S 2*V Dynamic pressure of the wind, q (N/m2) = 0.613*Vs^2 Table 1: Default values for wind load input complied to CP65 Stories Model Height S2 Vs height limit (m/s) (m) (m) 13 and below 40 40 0.89 31.15 13 to 25 68.9 70 1.00 35.00 25 to 50 113.7 170 1.15 40.25

q 2 (N/m )

Cf*q 2 (kN/m )

595 751 993

0.59 0.75 1.00

Notes : 1. If total storey height is less than the height limit, say model height  40, then only one row is shown. 2. User is allowed to edit the values in the boxes. 3. User is allowed to Delete any intermediate row. 4. Clicking Default button will load all the default values and it will overwrite all the existing values. Refer to sample calculations below for further details. Calculation of magnitude of point load to be assigned to node intersections: For rectangular load distribution, calculation of point load to be applied on a transition floor is shown below. Example 1: p60

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Plan View

Elevation View along grid 1

Height from lower to upper transition floor = h1 Height from lower floor to current floor = h2 Design wind speed at upper zone = VsB Design wind speed at lower zone = VsA At lower zone A, wind stress = qA At upper zone B, wind stress = qB

Design Wind Speed at transition floor =[VsA*h1 + VsB*h2]/(h1+h2) Point load at transition floor ={qA*[(h1+h2)/2] + qB*[(h1+h2)/2]}*(L1/2 + L2/2)

Sample calculation for Design Wind Speed at transition floor above. Total storey height = 150 m Floor height for every floor = 3 m Wind Load Description: Wind Load Case 2 = WL2 Acting at angle 180 degree Reference floor = Floor GF Basic Wind Speed, V = 35 m/s Force Coefficient, Cf = 1.00

Floor

Model Height (m)

Height Limit (m)

S2

1

13F

=height from GF to 13F =13*3 =39 < 40

40

0.89

=0.89*35 =31.15 m/s

q 2 (N/m ) =0.613*31.15^2 =595

2

23F

70

1.00

=1.00*35 =35.00 m/s

=0.613*35^2 =751

=1*751/1000 =0.751

3

50F

=height from GF to 23F =23*3 =69 < 70 =height from GF to 50F =50*3

170

1.15

=1.15*35 =40.25

=0.613*40.25^2 =993

=1*993/1000 =0.99

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Vs (m/s)

Cf*q 2 (kN/m ) =1*595/1000 =0.595

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=150 < 170

1a. Design Wind Speed for transition floor Find Design Wind Speed for transition floor at floor 13F:

Height from 12th to 13th floor = 3m Height from 11th to 12th floor = 3m Eg: Vs at floor 13F = (31.15*3 + 35*3)/6 = 33.075 m/s Sample calculation for point load assigned to node at the transition floor above (13th floor) 1b. Find point load assigned to node (2,D) at 13th floor = [0.59*3/2 + 0.751*3/2]*(3/2 + 3/2) = 6.0354 kN

Example 2: Plan View

Elevation View along Grid 1

Total storey height = 15.5m Wind Load Description: Wind Load Case 1 = WL1 Angle of attack = 0 degree. Reference floor = Floor gb Basic Wind Speed, V = 35 m/s Force Coefficient, Cf = 1.00

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Tabulated values based on the Wind Load data:

Sample calculation for Wind Load data tabulated above: Floor

1

5b

Model Height (m)

=height from 5b to gb =3*3 + 3.5 + 3 =15.5 m

Height Limit (m) 40

S2

Vs (m/s)

0.89 =0.89*35 =31.15 m/s

q 2 (N/m ) =0.613*31.15^2 =595

Cf*q 2 (kN/m ) =1*595/1000 2 =0.595 kN/m

Magnitudes of point loads applied to the node intersections:

Sample calculation for point loads assigned to node intersections above using the formula: Design wind speed, Vs = 31.15 m/s Height of floor 2b = 3.5 m Height of top floor 3b = 3 m Point load at node (2,A) at floor 2b Point load assigned to node (2,A) = (0.59*3.5/2 + 0.59*3/2)*(2/2 + 4/2) = 5.7525 kN (value differ due to decimal places). Point load at node (1,A) at floor 2b Point load assigned to node (1,A) = (0.59*3.5/2 + 0.59*3/2)*(4/2) = 3.835 kN (value differ due to decimal places).

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Limits of wind load application Wind load can only be added to any intersection point in condition that there is element, ie column, wall end, beam-beam intersection & beam- wall intersection. However, the engine will determine those points that can be applied, and purge others which will not be used during 3D analysis.

Refer to 2 different frame for cantilever beam frame as follows:1) Cantilever beam supporting secondary beam. Node will be detected & wind load applied as per figure below.

2) Cantilever beam which is not connected to other beam . No wind load will be apllied at the cantilever end.

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Wind loading validation 1. Wind load will be act at some node (figure 2) at a floor and from a floor to another 2. 3.

floor (figure 2). In order to make sure the wind loads are assign to the correct node, the following calculation are use to calculate the total external load and internal load. External loads are obtained multiply the wind load at the node to the total height of the node to the foundation. Internal loads are obtained from the analysis result. The internal loads are consist of Reaction load and moment in x or y direction. For example to load in y direction are sum of total moment of the reaction load (reaction multiply with x distance) and total moment of y

Figure 2: Wind force act at each floor Figure 1: Wind force act from elevation view

My at lowest floor

Reaction at lowest floor

Mx at lowest floor Wind force validation Calculation External force = Internal force

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Calculation of External Force, in y direction For each floor

Calculation of Internal Force in y direction At Lowest floor Reaction

Where w = wind load at the node at the current floor h = current floor height to the foundation

Moment y

Sum of External force

The validation should be for force in y direction

Similar The validation should be for force in x direction

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Example calculations 100 kN wind load was act at every node at 1b floor at 2 floors model with 5m x 5m length as in the drawing. Analysis result was shown as in the figure. Validate the wind force.

Gb floor plan, no wind load

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1b floor with 100 kN wind load

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Calculation Set origin at node (A,1) Y - Direction

X- Direction

External Load in y direction

External Load in x direction

Moment in y direction due to reaction

Moment in x direction due to reaction

Moment in y direction

Moment in x direction

Validation

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Difference are = 0% Difference are = 0.001666%

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Brickwall Brickwall input on element with drop You can input brickwall on a beam or a slab. By default, all the brickwall heights are set to default floor height. The default height in Esteem 8 are referred to the height of the floor level to upper floor or if there is no upper floor it will set to 4.0 m. Example:

When you applied a brickwall on the beam as per shown in figure above using default height, the total load applied on the beam as uniform distributed load are as follow. Given following data Brickwall Area Density: 5kN/m Floor height = 3.0 m

2

UDL from brick wall at Sec A

UDL from brick wall at Sec B

UDL from brick wall at Sec C

In this case, if users assume that the brickwall load is calculated from the surface level of the element to the floor above, for a) Sec A will applied enough loading on the section b) Sec B will miss 1kN/m load due to the drop of 0.2m c) Sec C will have extra load of 0.5 kN/m Therefore for the element with drop, users are advised to input line load to represent the brickwall (with default floor height) at the drop area.

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Finite Element Method

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Element Types Classification Beams and columns are classified under frame elements. Slabs in 2D floor analysis: a.) Are classified as Shell Elements if i.) option shell is turned on ii.) 2D wall subframe is turned on iii.) slab sitting on column with beam edge support > 180 degree iv.) slab sitting on wall edge support > 180 degree v.) slab sitting on column directly b.) Are classify as Plate Elements if option plate is turned on AND all of the sub-items a.) above is false Slabs in 3D floor analysis: a.) Are classified as Shell Elements if Slab is partly or wholly supported by wall. b.) Are classified as Plane Stress Elements if Slab is partly or wholly supported by beam AND not supported by wall or/and column alone. Walls in: a.) 2D floor analysis are classify as Shell Elements if 2D wall subframe is turned on b.) 3D floor analysis are classify as Shell Elements

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Frame element P,kN

Sz,kN

Sy,kN

Tx,kNm

My,kNm

Mz,kNm

Where P = Axial force to the element (kN) Sz = Shear along z axis (kN) Sy = Shear along y axis (kN) Tx = Torsional moment of the element (kNm) My = Bending moment about y axis (kNm) Mz = Bending moment about z axis (kNm)

Sy My

P

Y

X Tx

Sz

Z Mz

Esteem 8's beam elements do include the shear contribution in the beam stiffness matrix (see pp.256-258 in 'Introduction to Finite Elements in Engineering" by Tirupathani R. Chandrupatla & Ashok D. Belegundu).

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Wall frame element Frame elements will be defined at the floor level connecting all the wall nodes. Its width would equal the wall thickness and depth taken as twice the wall thickness. If the user has created a framing beam (conventional beams running along these wall nodes), the wall frame element would not be created. Wall frame element help in the transfer of forces between floor elements onto the wall elements. Note the following properties of wall framing element: No element nodal loads  No element stiffness except for torsional stiffness

Application in 3D Mesh/Analysis, if 1. the beam input/wall opening is meshed as couple beam (couple beam parameter = true) 2. For any wall opening, then apply wall framing beam (t x D), where D = 2t or actual depth whichever is less) Design considerations: 1. To combine into the couple beam the torsion from the 'Wall framing beam' for full design of moment, shear and torsion. 2. 'Wall framing beam' not sitting on opening (of either wall opening or beam input meshed as couple beam), design only for torsion and provide rebar (torsional link and longitudinal rebar for torsion) if the torsion design requires it. No shear and moment design is required.

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Shell Mx,kNm

My,kNm

Mxy,kNm

Sxz,kN

Syz,kN

Nx,kN

Ny,kN

Nxy,kN

Where M x = Moment in X direction (kN) M y = Moment in y direction (kN) M xy = Shear moment induce by different x-y plane shear stress along z axis (kN) S xz = Shear of the element in x-z plane (kN) S yz = Shear of the element in y-z plane (kN) N x = Stress in X direction (kN) =

N y = Stress in y direction (kN) = N xy = Shear stress in x-y plane (kN) = Simple explanation about shell Shell analysis is combine both plate and plane stress (membrane force) analysis. The Nx , Ny ,Nxy of membrane element is same of the plane stress interpretation (please refer to simple explanation of “plane stress”) and it’s shown as below,

The Syz, Sxz, Mx, My and Mxy of plate (please refer to simple explanation of “plate”) is shown as below,

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The shear (Syz and Sxz) and moment per unit length (Mx and My) is derived as below;

eq.1

eq.2 Similarly,

eq.3 So the analysis of plane stress combine of analysis of plate will give the following force and the element analysis with these 8 forces is termed as shell element.

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So the Mx , My , Mxy ,Sxz , Syz , Nx , Ny and Nxy in the 3D frame element in Esteem is related to Mx , My , Mxy ,Sxz , Syz x ,y ,xy. Their relationships are analogous to equation 1, 2 and 3.

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Wall element Sign conventions and definitions of axes for vertical walls

1. Global coordinate system: x-y-z 2. Wall face reference system: xref-yref-zref (FEM output)

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Deep beam element

When a beam with "Use Deep Beam" option is turned to TRUE in the Object Viewer, the beam will be meshed as Shell in 3D analysis as shown below.

Otherwise, the beam will be meshed by default as Line element.

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Beam element with opening(s)

When there is opening(s) in a beam, the beam will be meshed as Shell in 3D Analysis as shown below.

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Plate Mx,kNm

My,kNm

Mxy,kNm

Sx,kN

Sy,kN

Where M x = Moment in X direction (kN) M y = Moment in y direction (kN) M xy = Shear moment induce by different x-y plane shear stress along z axis (kN) S x = Shear of the element in x plane (kN) S y = Shear of the element in y plane (kN) Let us take an infinitesimal plate element and the state of stress is shown as below diagram. The Sxz and Syz in shear of the element in respective plane and it’s illustrated as below

In a manner similar to simple beam theory, we will relate the internal moments to the stresses. Figure below shows the net moments and shear forces per unit length on the t dx and t dy faces.

the moment per unit length Mx and My is related to x and y respectively by

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The following diagram shows an example of shell element. xy may vary over the thickness t . These variations will cause a moment M xy.

So the Mx , My , Mxy ,Sx , Sy in the slab plate element analysis in Esteem is related to Mx , My , Mxy ,Sxz , Syz

x ,y ,xy. Their relationships are analogous to equation 2 and equation 3.

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Plane stress Node

Nx,kN

Ny,kN

Nxy,kN

Where N x = Stress in X direction (kN) N y = Stress in y direction (kN) N xy = Shear stress in x-y plane (kN)

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Meshing

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Common Mesh quality control Generally, the software does have automatic adaptive capability to reduce the labor of preparing meshes and altering them to make the analysis more accurate. In Esteem 8, mesh generator will generate the meshing as good as possible for the model created by user. Two properties are checked to ensure the mesh quality. Firstly, Element Aspect Ratio The aspect ratio of a two or three dim ensional element is the ratio between its largest and smallest dimension. During generating the mesh, we should avoid finite elements which have the high aspect ratios, elongated or “skinny” elements.

Figure : Elements with good and bad aspect ratios As a rough guideline, elements with aspect ratios exceeding 3 should be viewed with caution and those exceeding 10 with alarm. However, such elements will not necessarily produce a bad result — that depends on the loading and boundary conditions of the problem—but may introduce potential trouble. In Esteem 8, the aspect ratio had been controlled for not more than 4. Secondly, Angle between two mesh lines In Esteem 7, the angle between two mesh lines is between 30° ~ 120°. By doing this, sharp corner element can be avoided in mesh generation.

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Example of Esteem 8 in 2D mesh

Example of Esteem 8 in 3D mesh

Refernce: http://www.colorado.edu/engineering/CAS/courses.d/IFEM.d/IFEM.Ch06.d/IFEM.Ch06.pdf

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Bridging beam Bridging beam is defined as a "very rigid" beam member. It is used to provide the stiff connection between elements. A common area, when this occurs is when there are offsets of members from their connected nodes.

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Intepretation of Results The accuracy of Finite Element Analysis is one of the most asked questions. The answer to these questions invariably depends on the type of finite element model, meshing and load application. Take for example a simply supported beam with an application of UDL. The m idspan moment would be (wl2 / 8) where w is the UDL, l the span between the supports. If we were to model this beam into discrete elements and so happen none of the nodes of the elements fall at the midspan location, the moment obtain from the analysis will not tally with the closed form solution above. Hence from the FEA point of view, the meshing aspect of the model is an important consideration. Other than this simply supported beam case, there are other examples of FEA intepretative requiring the user to have good knowledge of FEA.

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Different mesh sizes in slabs Different mesh sizes in Finite element will affect the accuracy of the analysis result. Figure 1 showing 2 models with same dimension and same properties. However model A come with one single mesh but model B come with 4 mesh.

After analysis, from cut section A through model A will give 3 values (due to only 3 nodes) to shown the analysis result but model B will have 5 values (5 nodes) along section A. As a result, the result plotted may not been smooth as model B. As the meshes getting more fine, the results obtain will be more smooth and accurate. In order to obtain the value in between 2 nodes, interpolation of the value need to done. Therefore the value use for design at the same point in between 2 node sometimes may be difference due to the interpolation result. Example below will explain the behavior of the difference size of the mesh.

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Figure 1: Model To Shown the effect of the m esh onto slab result and design, a simple project as above was create. 3 design at cut section 1 was created base on 3 types of m esh: 1000 mm mesh, 800mm mesh and 200mm mesh. Base on mesh 1000, we found that there are top bar provided at grid B to C. As a designer these top bar may not be required. However in computer simulation, because as it check the moment at the middle zone of the middle span of the slab, an hogging moment was found and this will lead the program to design top bar for the middle span of the slab. Looking into slab using mesh 800mm, now the mesh is getting smaller and the moment result graph obtained are more smooth and the result will be more accurate (From figure 5 to 7). As the mesh getting smaller, the moment result graph obtained will getting more fine and smooth (From 6 to 8)

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Figure 2: Slab mesh using 1000mm

Figure 3: Moment X accross cut section 1 base on m esh 1000mm

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Figure 4: Detailing at section1-1 base on mesh 1000mm

Figure 5: Slab mesh using 800mm Mesh

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Figure 6: Moment X accross cut section 1 base on m esh 800mm

Figure 7: Detailing at section1-1 base on mesh 800mm

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Figure 8: Slab mesh using 200mm Mesh

Figure 9: Moment X accross cut section 1 base on m esh 200mm

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Figure 10: Detailing at section1-1 base on mesh 200mm

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Residual forces in upper sub-frame column in 2D Analysis Suppose We have a model as shown below. 1. Create 5000mm x direction structure as shown in the figure below. 2. The project contain 2 floors. 3. Beam size is set to 200 x 500 and column size is set to 300 x 300 4. 3m height Brick wall added for 1st and 2nd floor. 5. Floor height is set to 3m.

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6. Click on Plan Loading Result View after run the plan (2D) analysis. 7. The user able to view there was a balance column axial forces as highlighted below.

8. The balance column axial force was correct and it was due to Residual forces in upper sub-frame column in 2D Analysis

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Wall Frame beam 1 Objective: This benchmark result on wall frame beam under shell and plate option for Esteem 8(Release 0.13.34.0) Project Data For Example:

Figure 1 Project sample

2 Project Data: 2.1 Create a project for the following data for the Benchmark. 1. 2. 3. 4. 5.

Create 5000mm x and y direction structure as show as the figure above Add 200 thickness slab . The project contain only 1 floor. The project is shown in figure above. Floor height is set to 3m.

2.2 Change the parameter of the FEM analysis in Esteem 7 1. Change the 2D Poisson ratio,V slab Poisson ratio Esteem 8 to 0.2. 2. Change the mesh size to 500 for Esteem 8 3. Change the pin support in Esteem7 parameter setting>3D Analysis>Pin foundation(column) to false. 4. Type of mesh change to mix in Esteem 8 p103

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5. Change the concrete as grade 30.

Figure 2 Esteem 8 Parameter

3 Result Benchmark Esteem 8 3.1 Wall in plane moment Comparison

Figure 3 Wall out of plane moment (Plate) p104

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Figure 4 Wall out of plane moment (shell)

3.2 Hogging moment Comparison

Figure 5 Slab Hogging moment(Plate)

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Figure 6 Slab Hogging Moment (shell)

4.0 Conclusion The example project used for the above, show that the benchmark result Esteem 8 is similar for both plate and shell option for wall frame beam.

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Unsymmetrical SFD for structural with symmetrical shape and loadings Although the structure is symmetrical in both shape and loadings, but the mesh generated on slab will unlikely to be symmetrical due to the program adopts random meshing with combination of triangular and quadrilateral mesh. Unsymmetrical slab mesh will cause unsymmetrical load transferring to beam hence created unsymmetrical shear force diagram. Below is an example for clarification: Unsymmetrical slab mesh:

Unsymmetrical slab mesh caused unsymmetrical slab loads transferring to beam:

Hence unsymmetrical SFD is produced on beam:

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2D Mesh Column meshing in flat slab configuration 1. Columns in consideration are: a. Columns that are within the boundary of the slab.

b. Columns which are fixed to the edge of the slab that is not supported by a beam or RC wall.

2. Integrity checks for columns in Flat Slab design a. A column is not allowed to support two flat slabs at the same time (see below)

b. A column is not allowed to cut across the slab it supports. For example, column (1000mm height) equals the slab dimension in y-direction. This is not allowed because such column input will divide the slab it supports.

3. Meshing for internal columns are as follow a. Rectangular Column i. If both the vertical and horizontal length of the column is less than or equals 1.5 x mesh size, a region for the

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boundary of the column will be created. The column element will connect to the center point of the region (which will consist of a single quad element).

Picture above: Plan view and Sub-frame view of internal column where height = 300mm, width = 300mm and mesh size = 1000mm. ii .If both the vertical and horizontal length of the column is more than 1.5 x mesh size, create a region for the boundary of the column and add a fixed point at the position of the column. The column element will connect to the fixed point.

Picture above: Plan view and Sub-frame view of internal column where height = 1600mm, width = 1600mm and mesh size = 1000mm. iii. If one of the column’s edge length is less than or equals 1.5 x mesh size, If the ratio of the longer edge to shorter edge is less than 2, do as (a – i). If the ratio of the longer edge to the shorter edge is more than 2, do as (a – ii). b. Circular Column i. The perimeter of the column boundary will be divided to the mesh size. 1. If the value is less than or equals 6, program will create a hexagonal boundary to represent the circular bound. 2. If the value is more than 6, program will create a circular boundary that has the number of edges equivalent to that value. ii. Program will create a region using the new boundary and, iii. Add a fixed point to the center of the boundary.

Picture above: Circular column with diameter 500mm and mesh size 500mm.

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Picture above: Circular column with diameter 2000mm and mesh size 300mm. c. L-Shape Column i. Program will create the region according to the L-Shape column. ii. If the center point’s distance to point d is less than 100mm, program will move point d to satisfy that requirement.

iii. Program will add a fixed point to the column point and connect the column mesh to the fixed point.

Picture above: L-shape column (leg length = 600mm and leg width = 200mm) with mesh size 500mm. d. T-Shape Column i. Program will create a region based on the boundary of the T-shape column ii. If the perpendicular distance of the center point to the side (d) is less than 100mm, extend the boundary to satisfy the requirement (see below).

iii. Program adds a fixed point to the center point. Connect the column mesh to the fixed point.

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e. Plus-Shape Column i. Program will create a region based on the boundary of the Plus-shape column. ii. If the distance, d, is less than 100mm, move the points to get 100mm (see below).

iii. Program adds a fixed point to the center point. Connect the column mesh to the fixed point.

f. U-Shape Column i. Progam will change the boundary of the U-shape column to a rectangular boundary.

ii. Program creates a region based on the new boundary. iii. Program will add a fixed point at the center point of the U-shape column. Connect the column element to the fixed point

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g. V-Shape Column Similar to L-Shape column.

h. Polygonal-Shape Column i. If the polygon is concave, program makes it convex. ii. Program creates a region for the convex polygon. iii. Program adds a fixed point to the centroid. iv. Program then connects the column mesh to the fixed point.

4. Meshing for columns on slab edge are as follow a. Rectangular Column i. For a column located and flushed to the corner of the slab, the meshing will be as below. The column element will be connected to the centroid of the slab quad element.

ii. For a column located at but intersects with the corner of the slab, the column’s boundary will be the one overlapping with the slab. The meshing will be as below. The column element will be connected to the nearest point on the quad element using a bridge beam.

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iii. For a column located at the corner of the slab but is offset to be within the slab, the column’s boundary will extend to the edge of the slab. The column element will be connected to the nearest point on the quad element using a bridge beam.

iv. For a column located at the corner of the slab that is rotated, a rectangle of equivalent area of the overlapped area between the column and slab that is parallel to the edge of the slab’s edge will be used for the meshing. The rectangular area will also follow the x-y ratio of the original column. See below.

v. For a column that is located along a slab edge…

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b. Other Column Types – Follow Rectangular style number (iv) i. Circular Column

ii. L-Shape Column

iii. Plus-Shape Column

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iv. T-Shape Column

v. U-Shape Column

vi. V-Shape Column

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vii. Polygonal-Shape Column

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Continuous column in wall A continuous column or wall in a 2D Subframe mesh is defined as the colum n or wall going beyond its floor to floor height. This situation arise when there is not any restraint at the floor level. A restraint could either be slabs, intersecting beams or walls. However if a column (column (1,B) and (1,C) shown below) is embedded in the wall or inside a wall (wall end to end, A to D), the column will not be continuous even if there is the wall and column in the same location in the lower floor.

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The resulting 2D sub-frame will be as follows:-

To ensure continuity, the column must be defined at the end points of the walls. In the example above, we need to defined multiple walls (1/A-B, 1/B-C, 1/C-D) with their ends to the columns (and at all the other floors as well to effect the contiuity):-

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and obtain the following 2D sub-frame:-

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Meshing beam element If the beam does not touch a slab, the element that make up the beam will always be about 500mm long. This is irregardless of the mesh size setting in the param eter. Example:

As shown above, the beam is divided into 10 discrete elements of 500mm each.

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Modelling a couple beam as a frame element In the situation where the user models the coupling beam as a frame element, the program will mesh it as a shell element if the depth of the beam defined is greater than 500mm. Mesh when beam depth < 500 mm

Mesh when beam depth greater or equal to 500mm

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PWallPlateNodes Concept Specification for PWallPlateNodes: For slab analysis using the plate option, when the slab is supported by the wall nodes at the floor level are > 2000 mm of any beam sitting on the wall, then it is assigned as PWallPlateNodes.

Function: To simulate the slab-wall subframe interaction effect with the plate-shell finite elem ent interface.

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Two beams connecting to a wall Two beams gb1 and gb2 are connected to wall B/1-2 at grid B/2 as illustrated below.

In this case where 2 beams of different depths meet, the shallower beam is meshed as frame(line) element while the deeper beam is meshed as shell(area) element. The mesh outcome would be as follows.

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2D corner slab mesh The circle mark in the 2D corner slab mesh as shown in figure below indicates that the slab element stiffness is divided by a factor of 10. This condition is m et when the slab element is connected to a beam end that is pinned.

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Length determination for beam loading

Beam 1b10 The mesh length for beam 1b10 is C. If beam 1b108 does not exist, the length of beam 1b10 will be C+E. The branching of the parallel bridge beams E and F in the above figure will remove E from the length consideration for beam 1b10. Beam 1b103 The mesh length for beam 1b103 is A+D. Because the branching bridge beams of D and F are not parallel to each other, it will not remove D from the length consideration. Beam 1b108 The mesh length for beam 1b108 is B. D is not taken into account because it has already be used for the length consideration of beam 1b103. Every bridge beam is only considered once in the determination of the beam mesh length. From 2D Analysis load validation, we have the total beam self weight = 158.4 kN

Manual calculations:-

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Self weight of beam 1b10 = 1 x 1 x 24 x 2 Self weight of beam 1b103 = 1 x 1 x 24 x 2.6 Self weight of beam 1b108 = 1 x 1 x 24 x 2 Total beam self weight

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= 48 kN = 62.4 kN = 48 kN = 158.4 kN

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Pinned condition of slab element Under the following conditions, the slab corners will be pinned:1. One end of beam at the respective corner is pinned.

2. Where slab edge is not connected to adjacent edge [where beam passed] AND it sits on c olumn with a) Single beam passing the column or b) Single beam sitting on the c olumn or c) Two beams ["broken" or ends at the same column AND along the same grid] sitting on the column.

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3D Mesh Alternative model The alternative model specify by the 3D Analysis parameter employs the "closed form" solution to calculated the results between nodes. Hence in a beam, the number of nodes for the finite element model is greatly reduced. Definition: Closed form solution is the option to obtain the same analysis results as non-closed form solution but by reducing numbers of nodes used in the 3D Analysis. This option can be used by setting from Project Parameter -> 3D Analysis -> Alternative Model and Alternative Model 4 Node to TRUE.

The purpose of application for the Alternative Model options are:



Reduced number of nodes for beam and slab in 3D Analysis Beam and slab nodes used in 3D Analysis are reduced. For example; when Alternative Model and Alternative Model 4 Node are set to TRUE, the 3D Analysis results will show single beam mesh with two node instead of ten nodes, while for slab, four nodes are used instead of nine nodes quadrilateral slab (depending on mesh size).

Table 1: Comparison of 3D Analysis results between Closed Form and Non-Closed Form solution Setting of parameter for 3D analysis

Expected Outcomes

Alternative Model = FALSE Alternative Model 4 Nodes = FALSE

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Alternative Model = TRUE Alternative Model 4 Nodes = FALSE

Alternative Model = TRUE Alternative Model 4 Nodes = TRUE



Increase memory efficiency and hence the software performance In FEM analysis when lesser nodes are used, the lesser the amount of time to run a whole large project model.

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Couple beam meshing The couple beam (from wall opening OR beam connecting wall with depth > 500 mm) meshing will use the rectangular (vertical) shape to the minimum mesh size setting provided that the following conditions are met:1. if the upper floor opening is touching 2. if no upper floor wall, i.e. only slab sitting on it 3. if the upper floor opening is partly overlap with it, then this special meshing only applies to the overlapping part. 4. If there is a cross beam sitting on it, that location should be a node. 5. If the mesh of the horizontal edge (distance) is too small use slanting 'vertical', but using the limit of 0.25 of vertical edge. * For a full model Couple Beam Column modeling, all the openings within a panel in the RC Wall must touch the bottom edge of the panel. Figure 1 shows the conditions for a full model modeling. The generated mesh should break the panels into appropriate regions as seen in Figure 2. Figure 3 shows the mesh result in Esteem.

Figure 1. All Openings must touch the bottom edge of the panel for a full couple beam column model

Figure 2. The panel broken into the appropriate regions

Figure 3. The 3D generated Mesh Average Node on Edge If two door openings on neighbouring panels do not share the same height but the difference in height is small (Figure 4), the generated node at that particular edge will use an average node (Figure 5).

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Figure 4

Figure 5 Partial Model * Condition for partial models is that no opening is obstructing the region that is above an opening that touches the bottom edge of the wall panel. Therefore the area where the openings that touches the bottom edge with no obstructing upper opening (window) will be model as Couple Beam Column model, while the rest will me modeled normally.

Figure 6

Figure 7 Bottom Window Region If a window (opening in the middle of the wall) is next to a door (opening that touches bottom edge of the wall), a region will be introduced under the window

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Figure 8

Figure 9

The "couple beam" shell element will be designed based on the actual forces (moment, shear, torsion) all along the span. p135

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Notes

:

1. A beam supporting a transfer wall should always be meshed as a couple beam. 2. If two beams that touches each other can be m eshed as couple beams, but the depth of the two beams are different, the one that supports the transfer wall will take priority and be meshed as couple beam while the other beam will be meshed as frame element. If no transfer wall exist, the beam which with deeper depth will be adopted. 3. If a beam has its ends pinned, it will not be meshed as a couple beam but to mesh as line element.

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3D Slab Meshing By default 3D slab mesh will be assigned the "Plane Stress" element. The 3D slab mesh will be assigned the "Shell" element if one of nodes touches a wall or c olumn.

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Planestress element has 2 degrees of freedom(x, y). As shown below only x and y has displacement value.

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Slab diaphragm effect When the width/length ratio of a slab is greater than 5, it will not be meshed in the 3D Mesh structure. Example: Input a slab with length = 20m, width = 3m Ratio = 20/3 = 6.67 Using three possible methods to input such slab are shown in Figure 1 below: 1. Input sub-slabs to produce the total length and width of slab, each sub-slab with length = 4m (portion A). In Figure 1, the sub-slabs created are FS3, FS4, FS9, FS7 and FS8. 2. Input the slab in the bounded area in portion B, and then cutting the edges to produce sub-slabs, each sub-slab with length = 4m (portion B). In Figure 1, these sub-slabs are FS5-1, FS5-2, FS5-3, FS5-3, FS5-5 and FS5-4. 3. Input the slab with length = 20m, and width = 3m (portion C). In Figure 1, the slab is FS6.

Figure 1: Plan view Results of Analysis:

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After the 3D Analysis is performed, the three slabs will be meshed as followed:

Figure 2: 3D mesh results

The different 3D mesh as shown above will produce different results of 3D analysis. In the Plan Loading and Result View, results of 3D analysis will produce different reactions for the beam's minor moment and minor shear as shown in Figure 3 and 4 below:

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Figure 3: Beam minor moment results for 3D Analysis

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Figure 4: Beam minor shear results for 3D Analysis

Next, the different 3D mesh will also produce differences in column reactions. For example, resulting column torsion as shown in Figure 5 below:

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Figure 5: Column torsion results for 3D Analysis To see the difference in 3D displacements of the elements, view the 3D Analysis Display for 3D Displacement of this projec as shown in figure 6 below. In 3D Analysis Display, select 3D Displacement. For Load Type of Individual Load, choose Notional Load Case 1 or NL Case 1. To see clearly the difference of the displacements between slabs in Portion A, B and C, increase the Scale to 10000.

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Figure 6: 3D Displacement in 3D Analysis Display for slabs in Portion A, B and C

Comparison of 3D mesh for three portions of slabs using three methods described above between: a) Slabs having ratio length/width or width/length more than 5 (ratio = 20/3 = 6.67) and, b) Slabs with ratio less than 5 (ratio = 8/3 = 2.67)

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Figure 6: a) Ratio of slab length/width more than 5, b) Ratio of slab length/width less than 5

Conclusion: From the results of analysis obtained above, it is shown that when user want to input a slab with length to width ratio more than 5, if the input is done directly by clicking on a bounded area, the results of analysis for that particular slab can produce larger reactions in the elements. Thus, if user wants to get larger reactions in the elements for a slab with length to width ratio more than 5, user can input a slab directly by clicking onto the bounded area. But the results of analysis in 3D might not be accurate. If user wants to get sm aller reactions and more accurate results, therefore it is recommended that user model any slab with length to width ratio more than 5 by using the input of sub-slabs to form the whole slab as shown in Figure 1; Portion A above.

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Transfer wall seated perpendicularly to a wide beam Consider the following model.

Floor Plan View

3D View

The 3D mesh would be as follows.

3D Mesh

Detail "A"

Elements No. 22 to 25 (supported bottom side of wall) would be meshed as "bridging beam elements" of the b=150 (thickness of wall) and h=300 (twice the thickness of the wall) with E value multiplied by 100.

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Transfer wall seated at an eccentricity on a transfer beam Consider the following model. Couple beam 1b1 is supporting an upper transfer wall which is seated at certain eccentricity on it.

Detail of "A", upper transfer wall sitting with an eccentricity on the transfer beam below. Floor Plan View The 3D mesh would be as follows.

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3D Mesh

Detail "A"

Elements on the supported bottom side of wall would be meshed as "bridging beam elements", except the first and the last nodes.

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Analysis

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2D Analysis 2D wall sub-frame analysis Wall Design for Moment Envelop from patterned load in 3D subframe analysis In Esteem 8, the wall can be designed to the full moment envelop of patterned load in 3D sub frame analysis. From the full moment envelop, the wall can be designed accurately for slenderness m oment effect at the mid-length location. The average of the moments at the mid 1/3 (default parameter = 1/3, range = 1/2 to 1/5 of mid-height range) height of the wall are used in combination to the slenderness moment in comparison with the wall end moments for the design. Both the upper and lower wall moment envelops for the same wall location are compared to ensure that the wall is design for the largest possible moment. The characteristic of wall mesh: 1. Restraints can be due to slab, beam, cross wall 2. Upper wall & column – upper overlapping wall (mesh full wall, i.e. both the overlapping and overhanging part, if any) the lower wall considered, only column within the wall (both lower & upper) considered, all independent upper column sitting on the lower wall is not considered. 3. Wall connected with different offsets – considered as 2 separate walls without connection(i.e. no bridging beam), If a single upper wall sits on the wall described in a. above, it will be split as 2 separate different offset walls without connection following the lower wall below. 4. Stress collection - moment: max, min, full 5. Design – from moment envelop in 4. above, store 2 pairs (upper & lower wall) of 2 values (wall end and wall mid-height) in every wall panel for 3D design. 6. If wall height is by-passed (due to no restraints) at any particular floor, that particular floor 2D sub-frame meshing will by-pass the wall element for that partic ular wall. 7. Support restraints nodal list: all nodes at lower end of lower wall and all nodes at upper end of upper wall that touches slab and beam.

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2D wall sub-frame design If the slab option = shell, there's no lateral restraints for wall nodes at floor but if the slab option = plate, there's lateral restraints for wall nodes at floor. Then the centroid of the wall gravity point load will be slightly distort. Since, the wall is only designed for out of wall plane moment in 2D(not the axial load and in-plane moment), it's acceptable for the sake of m uch faster solution time in the plate option.

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Auto Support

Figure showing 2 beams cross over each other at the m id span and both end support by two c olumns ll = longer dimension ls = Shorter dimension Cs = Shorter Beam Span Column Cl = Longer Beam Span Column Bs = Shorter Beam Span Bl = Longer Beam Span Conventional Analysis: User Defined Support According to the user defined support in conventional structure analysis, say Bl supported by Bs. Therefore Bl will be analyses as 2 span beams with one support at the mid span. Half of the loading from span one will go to column and the other half load from the beam will transfer to the mid span of Bs. This is what happens at the other half span.

Figure: Analysis will start at Span Bl

Figure: Analysis at Bs base on the load transfer from Bl p153

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When analyse beam Bs, a point load will transfer from Bl and the applied load will fully carry by the Bs. E7 analysis : Auto Support However in beam auto support, the both beams are integrated. The point load that applied at the center of the beam will be assign to the integrated structure according to the stiffness. Both beams will deflect equally at the point of intersection (displacement compatibility).

Figure: Behavior of the structure in auto support Let examine the example below: Objective: To create a model where 90kN live load applied at the center of the beam at beam intersection and to have the loads distributed according to ratio 1:2 (longer span: shorter span). In the “Auto Support” intersection, the displacement compatibility applies. This mean the deflection of both beams at the intersection equal and similarly column shall displace equally in order to maintain the ratio of 1:2 loading distribution. Beam Displacement Displacement formulae at mid span (beam), a

When 2 symmetrical beams was cross over each other at the center of each span, the reaction of both end each beam should be the same

When l1 = l2

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If l1 = 10000mm Then in order to get Fl : FS = 1:2

Column Displacement Column will displace when there is axial load applied on it. When more load applied on the column, the displacement will increase.The forces developed at the end of the m embers are

From the equation above to make sure all the loading transfer according to the ratio as above, the displacement of all columns must be the same, hence

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When column at longer span having dimension 300 x 300, F l : FS = 1:2

Column for shorter beam span should be

By having this combination the load will be aspect transfer as per design. When this case was model in Esteem 8, it returns the result as above.

Figure: Model of auto support

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Figure: Result of the analysis Result above shown that the loading transfer to shorter beam’s column and longer beam’s column are in ratio 1:2 symmetry This example proves the distribution of the point load according to the displacement compatibility in consideration of element stiffness. The elements of GB1 (with column (1, D) and (2A, D)) is twice the stiffness of beam GB2 (with column (1A, C) and (1A, E)).

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Cantilever beam behaviour in FEM analysis In FEM analysis, all the elements (slabs, beams, columns and walls) are integrated as one whole structure. The compatibility of displacement applies at all nodal connections. For example, consider a beam (presumably supporting another beam) intersect by another beam. At this point of support (intersection of nodes) we will have some displacement. Hence, the support beam at that node has displaced unlike conventional manual calculation where the designer would assume this particular beam to be the primary beam providing full support (no displacement) to the secondary beam. In a 2D sub-frame model where the elements are integrated the larger displacement usually occurs where the structure is less stiff eg. areas of large spanning slabs or beams. Let us study the following example.

Could beam gb3 be supporting beam gb1 at grid line (2)?

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Study the displacement contour below.

At the point of intersection of beam gb1 and gb3, there is displacement of approximately 3.8mm. This is the displacement plot along beam gb1.

And here is the moment envelope for the same beam.

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And moments for gb3 as follows.

So, as you can see, in this particular situation, beam gb1 is not fully supported by gb3 (as may be assumed in a conventional manual design). The moment diagram exhibits cantilever moment pattern. This is the true behavior of the 2D floor structure. There are exceptions when a model is incorrectly done. Esteem 8 software has the intelligence to provide a warning message shown in the detailing when it detects/suspects a beam is cantilevered when it should not be. The following write up would explain this feature. When a beam has 2 span and both spans identified as cantilevered spans, there exist the possibility of an "unstable/cantilever beam" situation. A warning message will be given under the beam details. This message will be discounted is there are sagging moments in the envelope of moments in either of the spans. This message would assist the engineer to identify genuine modeling error. Consider the following example.

Beam gb1 would seemed to be supported by column (1,A) and column (2,A). As beam gb2 would be expected to be cantilevered at span gb2A. But when the beam is designed the warning message would alert the engineer to a possible "unstable/cantilever beam" situation.

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There is a mistake in the model when creating beam gb1. Examine the beam end at intersection of grid line (A, 1A). Zoom in closer and you will see that the end of beam gb1 was not locked to grid line 2 where the column was locked to. Furthermore, it is not located in the profile of the column (if so, a bridging beam will automatically connect it to the colum n).

If it was a beam instead of a column and the beam's width is wide enough to overlap on to beam gb3's profile (see image below) we have a similar situation.

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The end of beam gb3 is not locked on to grid line 2 on which beam gb5 is locked to. Users have to check on the integrity of the input of the m embers in key plan to make sure all members are physically connected to each other.

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Study of support reactions in statically determinate models This report originated from the study of support reactions on to statically determ inate models. Example application of this study includes the analysis of pilegroups which consist of 2 or 3 piles in the group. The fact lies in the understanding that in a statically determ inate model the distribution of loads does not depend of the stiffness of the supporting m ember. Some examples will illustrate this principle:1. The following model is modeled in Esteem 7. The left hand side c olumn is dimensioned 1000 square while the right hand side column 300 square. The beam is arbritarily set to 200mm x 500mm

A live load of 100kN is placed at the midspan. The beam end moment release is set to False (i.e. Fixed),

and Analysis Option for Pin Foundation (Column) set to True (i.e. support is pinned)

The live load reactions reads 50kN to column at (1,A) and 50kN to column at (1,B) p163

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but when the option for Pin Foundation (Column) set to False (i.e. support is fixed), the live load reactions now reads 58.64kN to column at (1,A) and 41.36kN to column at (1,B)

2. We repeat this idea into a two span model, and setting the Pin Foundation (Column) to True, we obtained a symmetrical reactions.

However, setting the Pin Foundation (Column) to False, we have the following,

as expected more loads are attracted to the stiffer column. 3. Now, when we go further to four supports (indeterminacy comes in) and setting the Pin Foundation (Column) to True (to pinned the support), we have some results as follows,

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The support reactions are no longer equal or rather symmetrical. Conclusion: In the statically determinate model shown above, the size of column section does not attract more loads to itself. In the straight support geometry shown above, the static determinacy remain up to 3 supports.

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Subframe and continuous beam analysis According to BS8110:Part 1: 1997 1. Clause 3.2.1.2.1 subframe analysis need to be done for both column and beam design. 2. Clause 3.2.1.2.4 'continuous beam' analysis applies for beam design ONLY. Continuous Beam Simplification (BS8110: Clause 3.2.1.2.4.): "As a more conservative analysis of subframes, the moments and shear forces in the beams at one level may be obtained by considering the beams as a continuous beam over supports providing no restraint to rotation." Where the continuous beam simplification is used, the column moments may be calc ulated by simple moment distribution procedure, on the assumption that the c olumn and beam ends remote from the junction under consideration are fixed and that the beams possess half their actual stiffnesses. The simple moment distribution procedure = one of estimated subframe method BUT the exact subframe method is the full number of spans considered as done in all commercial softwares including Esteem.

Warning

:

To design the column with moments derived from minimum ecc entricities, i.e. WITHOUT subframe analysis when only 'continuous beam' analysis are carried out, is NOT best practice. This gross assumption that column moments are nominal is contrary to conscientious practice. Taking this shortcut in pursuit of economy will result in not only weak column design BUT most likely under-design ('very weak') column (i.e., NOT designed complying to code of practice, see item 1 above). The ignorance of the Clause 3.2.1.2.1 causes many buildings to be subjected to grave danger of 'under-designed by design'.

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Transfer Wall Beam Stiffness increase Transfer wall beam stiffness parameter is used to set the elastic m odulus of the beam elements that supports an RC wall. If a beam supports a transfer wall, all the elements on the beam that supports the wall will have their elastic modulus times with the factor set in the parameter making the element very stiff. During the 2D transfer beam design, the actual transfer wall is not modeled. In order to simulate the transfer wall stiffness sitting on the beam, the portion of beam supporting the wall is stiffened. However, the load from the transfer wall is not modeled. Therefore, the span for the beam supporting the transfer wall is not designed in 2D beam design. An example model have been done and increase the stiffness and the moment of the transfer beam has been recorded.

Figure 1: Example Below is the table of moment after user increase the stiffness from 1 to 100. Stiffness Increment 1 20 40 60 80 100

Mid span Moment (kNm) 3.28 1.54 1.40 1.34 1.32 1.30

From the table above, we can notice that the m id span moment will decrease as the stiffness increase.

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Couple Beam connection to wall FE mesh In Esteem 8, when a beam is connected in parallel to walls (couple beam) as shown as below, the beam result may vary due to finite element meshing of the walls:

A tabulation of the results according to mesh sizes is shown below:

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Optimised pattern loading Pattern Load

Figure 1 In pattern loading (Optimise pattern loading set to false), the pattern load are depending to number of beam and number of slab. In normal case the structure above, the total num bers of pattern loads are 9 where 9 beam contribute 9 pattern loads and slab contribute 1 pattern loads. This may take longer time to analysis and will slow down the process. When pattern loading are set to true, 1. If beam did not support any slab, this beam will contribute 1 pattern loads 2. Those beams support a slab will be combining with slab become 1 pattern loads. Therefore structure in figure 1 will produce 5 pattern loads where 1. 1 pattern from S1, b2,b8,b4, and b6 2. 4 pattern loads from each b1, b3, b7 and b5

Figure 2 In figure the total pattern loads will be 6 (5 from beam b1, b3, b7, b5 and b10 and 1 from slab, b2, b4, b6, b8, and b9)

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Case study

Figure 3 In figure 3, the total numbers of the pattern loads are 4: 1. Pattern 1: S1, b1, b2, b3 and b4 2. Pattern 2: S2, b5, b7, and b6 3. Pattern 3: S3, b10, b9 and b8 4. Pattern 4: S4, b12 and b11 The sequence of the analysis will start from Pattern 1 to Pattern 2 to Pattern 3 then Pattern 4. Analysis sequence:

The analysis will start from Pattern 1

After complete pattern 1, pattern 2 will start

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After complete pattern 2, pattern 3 will start

After complete pattern 3, pattern 4 will start Because of the sequence a phenomena will happen where the following elements will have same result compare with other element (when the meshes are equal): 1) b1 and b2 2) b3 and b4 3) b5 and b8 4) b6 and b10 5) b7 and b9 6) b12 and b11

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Create a project as below

Beam dimension 200 x 500 Slab thickness = 125 with DL = 1.2 and LL = 1.5 All parameter use default value except in Analysis Option: Optimised Pattern Loading = True

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Run 2D batch process. The Result should be as follow Beam

Span

Moment Value

GB1

Span 1 (Max)

34.36

Span 2 (Max)

34.42

Span 1 (Max)

34.42

Span 2 (Max)

34.46

Span 1 (Max)

34.42

Span 2 (Max)

34.36

Span 1 (Max)

34.36

Span 2 (Max)

34.42

Span 1 (Max)

88.72

Span 2 (Max)

88.72

Span 1 (Max)

88.72

Span 2 (Max)

88.72

GB2

GB3

GB4

GB5

GB6

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Two-way Slabs : Uniformly loaded rectangular panels Esteem 8 uses full FEM approach in slab analysis. Therefore, Elastic Analysis method is adopted in analyzing uniformly loaded rectangular slab panels instead of the c onventional BS8110:1997 Table 3.14 method. Below shows the table of Elastic Analysis for uniformly loaded rectangular slab panels that is extracted from Reynolds's Reinforced Concrete Designer's Handbook 11th edition, page 132.

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Example : Compare slab FS1 moment result between Esteem 8 graph and manual calculation.

Given : Slab thickness = 500mm 2 2 Slab unfactored DL = 0 kN/m , unfactored LL = 137.5 kN/m Slab area = 7m x 7m Result From Esteem 8 Before running the analysis in Esteem 8, to achieve the almost exact results as compared to the Elastic Analysis method, please take note of the parameter values that needed to be changed in the Project Parameter Settings:

Run 2D analysis, then open the Mx contour by clicking on the icon

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Result by Manual Calculation based on Elastic Analysis Table, Reynolds's Reinforced Concrete Designer's Handbook 11th edition: By referring to the Elastic Analysis Table, the coefficient that fulfills the slab FS1 condition is as highlighted below,

2

1) Negative moment at fixed edge = αwl 2 = 0.084 x (137.5 x 1.6) x 7.5 = 1039.5 kNm 2) Mid Span Mx

= 0.037 x (137.5 x 1.6) x 7.5 = 457.875 kNm

2

3) Mid Span My

= 0.031 x (137.5 x 1.6) x 7.5 = 383.625 kNm

2

Comparison : Type Negative moment at fixed edge Mid Span Mx Mid Span My

Esteem 8 1023.62 kNm

Manual Calculation 1039.5 kNm

% Difference 1.5

446.64 kNm 375.98 kNm

457.875 kNm 383.625 kNm

2.5 2.0

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3D Analysis P-Delta Analysis

The analysis of frame element without P-Delta  Background theory

Most computer programs adopt stiffness matrix in the analysis of frame elements subjected to various loads. Esteem is of no exception. This section briefly recounts the theory behind the stiffness matrix method.

If a frame element is under axial load only, then the Hooke’s law applies (see figure 1)

If a frame element is under distributed load (Figure 1), the equation 2 applies

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By representing the element as a two node element, a matrix between force and displacement can be form. The relationship between Force expressed by the following relationship

The arrow on the top of a symbol implies matrix.

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 The matrix (Chandrupatla et al. 1997)

The is given as follow, and are the moments of inertia, respectively. shear modulus and is the torsional constant:

is the

(4) is the transformation matrix that transforms the stiffness matrix from local coordinate to global coordinate. For the force matrix

, it is given as

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 Notes on the matrix

The obvious thing regarding equation 3 is that the relationship between force and the displacement is linear, as can be seen from the fact that the matrix

contains no term

dependant on or . This means that the equation 3 can be solved without iteration. It should be noted that this equation is strictly valid only when the element deflects linearly with the applied force (material linearity assumption) and when the element undergoes small displacements (geometric linearity assumption). When any of these two assumptions are violated then the matrix will have to be modified, usually becomes dependant on the applied force or the displacement. A numerical iterative solution thus has to be used. In Esteem, the inclusion of nonlinearity effect will be limited to geometric nonlinearity; even so, it only includes the nonlinearity involving axial load effect (P-delta effect). It doesn’t address other geometric nonlinearity such as chord shortening, large displacement effect etc.

The solution algorithm and internal forces calculation  The matrix asse mbly

Equation 3 is applicable to an element with two nodes, with the x axis always oriented in the element direction. Normal buildings usually consist of more than just one single element. p183

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Thus different elements have to be transformed to a global axis and they have to be assembled together before they are solved as one matrix. The displacement solution one obtains will be in global coordinates.  The numerical solve r

The equation 3 can be solved completely when the constraints and the external load vectors are specified. The solution algorithm normally takes advantage of the fact that the stiffness matrix is symmetric. The existing numerical solvers are fairly standard and easy to obtain. To read a lucid account of one of the solvers, readers are advised to consult text by (Chandrupatla et al. 1997).  Internal forces calculation

Once the global displacements are found for each node, the internal forces for each element can be calculated easily (Balfour 1992). The calculation recipe is given as below:

The subscript implies that all the force, stiffness matrix and displacement components have to be in local coordinates. is the force that is distributed on that particular element. Usually this is the weight of the element that causes axial loads and shear forces and moments on the element. The below two figures show the

The

Whereas

for a beam and a column.

for a column (figure 3) is:

for a beam (figure 4) is:

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P delta explanation and algorithm  The theory behind P delta effect

For a frame element, under a low axial load situation, the application of axial load doesn’t affect the deflection in lateral direction, at least not to a significant degree. However, if the compressive (tension) axial load is sufficiently high, the lateral stiffness (i.e., the resistance of the element towards external force) will be reduced (increased), increasing (decreasing) the deflection of the element. To see the whole theory in a complete picture, it is helpful to consider the following a frame subjected to both lateral and axial loads as shown below:

The differential equation governing the deflection of the frame is given below:

Where is the lateral load on the frame element. In the normal beam model the second term on the L.H.S is omitted (see equation 2). The very existence of in the equation above describes the interaction of the axial load and the lateral deflection. This effect is the P delta effect. It can be seen that equation 9 is a natural extension of equation 2, in the sense that equation 9 reduces to equation 2 when the axial load approaches 0. The modification to the stiffness matrix in equation 2 is shown as below. The rows and the columns that need to be modified are labeled on the top and on the right respectively.

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Figure 4: Equation

: The second order matrix (Ghali et al.2003)

By inserting terms in equation 10 at the appropriate places in equation 3, the effect of P delta can be captured.  Simplification

It is found that if the axial load is sufficiently small, then the equation 10 can be broken into two parts,

and

via Taylor series expansion:

is the first order matrix and is the familiar matrix we see in equation 3. due to geometric nonlinearity and is defined as:

is the matrix

Another common representation of P delta effect is shown below:

The difference between equation 12a and 12b is due to the fact that equation 12a takes into the consideration of axial load effect on the rotation of the element and 12b doesn’t. However, Equation 10 should be preferred over equation 12 in a computer implementation of P delta p187

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effect.

 The determination of axial load,

In the solution of first order stiffness matrix, the determination of can be deferred until the final solutions for displacements are found. However, for P delta consideration, the axial load changes each time the displacements are updated. The calculation of is done by subtracting the internal force at one node with the internal force at another and divide by two. The equation is given as follow:

Once the is determined, it can be plugged into equation 10 for the calculation of the real stiffness matrix. The solution must be iterative in nature.  The solution algorithm

First the displacements for the analysis without P delta are computed. Then the internal forces are calculated. Next these internal forces are used to compute the stiffness matrix with geometrical terms. Then the displacements are computed and are compared with the displacement obtained in the previous iteration. If the number of iteration exceeds a prescribed value or if the convergence is reached, then the program terminates, else the looping continues. Diagrammatically it can be represented as below:

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Figure 5: a general algorithm for determining the P delta effects on the performances of structures

Reference:  Ghali, A., Neville, A.M., and Brown, T.G., 2003, “Structural Analysis: A unified classical and matrix approach.” 5th ed., Spon Press, Great Britain.  Chandrupatla, T.R, and Belegundu, A.D., 1997, “Introduction to Finite Elements in Engineering.” 2nd ed., Prentice Hall, USA.  Balfour, J.A.D., 1992, “Computer Analysis of Structural Frameworks.”2 nd ed., Blackwell scientific publications, Oxford.

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Analyse 3D frame failed If a P-Delta Analysis failed to converge, an error message will be displayed.

In this situation, check for any input mistakes like a coulmn point load of a high magnitude or possiblity of slender column geometry.

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Seismic loading and analysis References 1. Nawy, Edward G. Reinforced concrete: a fundamental approach / Edward G. Nawy. -4th ed., 2000 Chap. 15 2. Farzad Naeim The Seismic Design Handbook 2nd Ed.,2001 pp. 140ff 3. International Code Council International Building Code 2006 (IBC)

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Formulation of seismic parameters The following table outlines the steps in obtaining the coefficient Cs (Seismic Load Factor) and k (Seisimic Distribution Exponent) required for the seismic load application. The " Equivalent lateral force method" is used in Esteem 7 for earthquake loadings. Steps Calculations

Description

1.

V = seismic base shear Cs = seismic response

2.

V = C s.W

Cs = S DS/(R/I)

Dependent Parameter

coefficient or seismic load factor W = The effective seismic weight fo the structure, inclusive or 20% live loading. SDS = Design spectral SS=Mapped Spectral 5% damped response acceleration at acceleration for short short period. periods, eg. 0.075g where design g=9.81 N/kg. SDS=2/3.S MS Fa=Site coefficent from SMS=Fa.SS

3.

Condition 1: Cs =calculated in

4.

Condition 2: Cs =calculated in

Table 15.2a of Ref. (1).

SD1 = Design spectral

S1=Mapped Spectral response acceleration at 1 acceleration for 1 second (2) above cannot second period. period, eg. 0.025g where exceed the g=9.81 N/kg. following, SD1=2/3.S M1 Cs = S D1/((R/I).T) Fv=Site coefficent from SM1=Fv.S1 Table 15.2b of Ref. (1).

(2) above cannot be taken less than following, Cs=0.044 SDS In lieu of an analysis, an R = Response modification approximate fundamental factor (see Table 1617.6 period can be used, ref. (3)) (3/4) Ta=CT.h I=Occupancy importance factor h = Program will automatically compute CT=Building Period the total height of Coefficient building. 5.

Assumptio n

Condition 3: For building and structures in seismic design categories E or F and in buildings and structures for which the 1-sec spectral response, S1 is equal to or greater than 0.6g, Cs =calculated in (2) above should not be taken less than following, p192

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6.

Cs=0.5 S1 /(R/I) Seismic Distribution Exponent, k

Based on building period, T calculated above. k=1 for T <=0.5 sec k=2 for T >=2.5 sec Interpolate linearly k value for T between 0.5 and 2.5 sec.

Notes : 1. Mapped Spectral accelerations, SS and S1 (Spectral Response Acceleration) are to be obtained from published regional geographic charts. In absence of such charts, engineers must use good judgment accordingly. See Figure 1613.5ff of the "International Building Code (IBC) 2006" for US charts as example.

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Site Classification, Table 1613.5.2 ref. (3)

Notes : 1. As far as Esteem 7 is concerned, site classification affects the calculation of Cs when Site Class E & F are selected. It invites the consideration of Condition 3 as outline in the topic above.

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Fa, Table 15.2a ref. (1) Value of Site Coefficient Fa as a function of site class and mapped spectral response acceleration at short periods (SS). Site Class

SS =< 0.25

SS =< 0.50

SS =< 0.75

SS =< 1.00

SS =< 1.25

A B C D E F

0.8 1.0 1.2 1.6 2.5

0.8 1.0 1.2 1.4 1.7

0.8 1.0 1.1 1.2 1.2

0.8 1.0 1.0 1.1 0.9

0.8 1.0 1.0 1.0

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Fv, Table 15.2b ref. (1) Value of Site Coefficient Fv as a function of site class and mapped spectral response acceleration at 1.0 sec periods (S1). Site Class

S1 =< 0.1

S1 =< 0.2

S1 =< 0.3

S1 =< 0.4

S1 =< 0.5

A B C D E F

0.8 1.0 1.7 2.4 3.5

0.8 1.0 1.6 2.0 3.2

0.8 1.0 1.5 1.8 2.8

0.8 1.0 1.4 1.6 2.4

0.8 1.0 1.3 1.5

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R, Table 1617.6 ref. (3) extract of Table 1617.6 ref. (3) for Response modification coefficient R . Basic Seismic-Force-Resisting System

Response System Limitations and Building Height Limitations (m) Modification by Seismic Deisgn Categoryc as determined in IBC C oefficient, R Section 1616.1 A&B C Dd Ee Ff

Bearing Wall System Special reinforced concrete shear wall Ordinary reinforced concrete shear wall Detailed plain concrete shear walls Ordinary plain concrete shear walls Building Frame System Ordinary reinforced concrete shear wall Detailed plain concrete shear walls Ordinary plain concrete shear walls Moment Resistant Frames Special reinforced concrete moment frames Intermediate reinforced concrete moment frames Ordinary reinforced concrete moment frames Dual System with Special Moment Frames Special reinforced concrete shear walls Ordinary reinforced concrete shear walls Dual System with Intermediate Moment Frames Special reinforced concrete shear walls Ordinary reinforced concrete shear walls Shear wall-frame Interactive system with ordinary reinforced concrete moement frames and ordinary reinforced concrete shear walls

5.5 4.5 2.5 1.5

NL NL NL NL

NL NL NL NP NL NL NP

48.8 NP NP NP NP NP NP NP

48.8 NP NP NP NP NP NP NP

48.8 NP NP NP NP NP NP NP

5 3 2

NL NL NL

8 5

NL NL

NL NL

NL NP

NL NP

NL NP

3

NLh

NP

NP

NP

NP

8 7

NL NL

NL NL

NL NP

NL NP

NL NP

6 5.5 5.5

NL NL NL

NL NL NP

48.8 NP NP

30.5 NP NP

30.5 NP NP

c NL = not limited and NP = not permitted d limited to buildings with a height of 73.2m or less. e limited to buildings with a height of 48.8m or less. f Ordinary moment frame is permitted to be used in lieu of Intermediate moment frame in seismic design categories B, and C. h Ordinary moment frames of reinforced concrete are not permitted as a part of the seismic-force-resisting system in seismic design category B structures fonded on Site Class E or F soils

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I, Table 15.5 ref. (1)

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Examples Example to Ref. 1 (Nawy) page 693

   

A building of 5 storey of 2.9m height per floor (total height = 14.5m). Building category II Site Class B Seismic group II (use in spectral response method)

Response modification method, R = 3.0 Occupancy importance factor, I = 1.25 From ground motion maps, S 1=0.42, Ss=0.85 (Site B with 5% damping)



Adjusted spectral response accelerations for site class effects: o for S1=0.42, Fv=1.0

o o o o o 

 

for SS=0.85, Fa=1.0 SMS=Fa.S S=1.0 x 0.85=0.85 SM1=Fv.S 1=1.0 x 0.42=0.42 SDS=2/3 . S MS=2/3 x 0.85 = 0.567 SD1=2/3 . S M1=2/3 x 0.42 = 0.278

CS=SDS /(R/I)=0.567/(3/1.25)=0.236 but cannot exceed CS=SD1 /(R/I) . T ; Ta = C T h(3/4) = 0.085 {for moment resisting frames} . 14.5 (3/4) = 0.63 sec Hence, CS=SD1 /(R/I) . T = 0.278 /((3/1.25) x 0.63) = 0.184 k = 1+ (0.63-0.5)/(2.5-0.5)=1.065

Example 2

   

A building of 5 storey of 3m height per floor (total height = 15m). Building category II Site Class B Seismic group II (use in spectral response method)

Response modification method, R = 3.0 Occupancy importance factor, I = 1.25 From ground motion maps, S 1=0.03, Ss=0.075 (Site B with 5% damping)



Adjusted spectral response accelerations for site class effects: o for S1=0.03, Fv=0.8

o o o o o 

for SS=0.075, Fa=0.8 SMS=Fa.S S=0.8 x 0.075=0.06 SM1=Fv.S 1=0.8 x 0.03=0.024 SDS=2/3 . S MS=2/3 x 0.06 = 0.04 SD1=2/3 . S M1=2/3 x 0.024 = 0.016

CS=SDS /(R/I)=0.04/(3/1.25)=0.0167 but cannot exceed CS=SD1 /(R/I) . T p199

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; Ta = C T h(3/4) = 0.085 {for moment resisting frames} . 15 (3/4) = 0.648 sec

 

Hence, CS=SD1 /(R/I) . T = 0.016 /((3/1.25) x 0.648) = 0.01 k = 1+ (0.648-0.5)/(2.5-0.5)=1.074

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CT, Building Period Coefficient, ref. (1)

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Vertical Distribution of Forces The lateral force Fx induced at any level can be determined from the following expressions illustrated by Nawy.

Cvx = vertical distribution factor V = total design lateral force or shear at the base of the building. Wi and Wx = the portion of the total gravity load of the building, W, located or assigned to Level i or x hi and hx = the height from the base to level i or x k = seismic distribution exponent

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Unstable Structure To ensure the stability of a structure, the m embers must be properly held or constrained by their supports. The model shown in Figure 1 with beams are connected in Pin condition without a support c olumn. This kind of connection will cause the unstability in the model.

Figure 1.0

Figure 1.1: Structure Model

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Figure 1.2: Free body diagram

Determinacy The equilibrium equations provide the necessary and suffic ient conditions for equilibrium. When all the forces in a structure can be determined from these equations, the structure is referred to as statically determinate. Structures having more unknown forces than available equilibrium equations are called statically indeterminate. The equlibrium equations are as follow: r = 3n, statically determinate r > 3n, statically indeterminate Where r is total force and moment reaction components in each part of the free body diagram n is total number for each part in free body diagram and Stability To ensure the equilibrium of a structure, besides satifying the equations of equilibrium, but the members must also be properly held or constrained by their supports. This is to ensure the stability of the structure. In general, a structure will be geometrically unstable -that is, it will move slightly or collapse - if there are fewer reactive forces than equations of equilibrium. If the structure is unstable, it does not matter if it is statically determinate or indeterminate. In all cases such types of structures must be avoided in practice. The equations are as follow: r < 3n, unstable r ≥ 3n, unstable if member reactions are concurrent or parallel or some of the components form a collapsible mechanism From the free body diagram above Figure 1.2, r = 10 n=4 r < 3n 10 < 12 Therefore, the structure is unstable.

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What is a closed form solution? Definition: Closed form solution is the option to obtain the same analysis results as non-closed form solution but by reducing numbers of nodes used in the 3D Analysis. This option can be used by setting from Project Parameter -> 3D Analysis -> Alternative Model and Alternative Model 4 Node to TRUE.

The purpose of application for the Alternative Model options are:



Reduced number of nodes for beam and slab in 3D Analysis Beam and slab nodes used in 3D Analysis are reduced. For example; when Alternative Model and Alternative Model 4 Node are set to TRUE, the 3D Analysis results will show single beam mesh with two node instead of ten nodes, while for slab, four nodes are used instead of nine nodes quadrilateral slab (depending on mesh size).

Table 1: Comparison of 3D Analysis results between Closed Form and Non-Closed Form solution Setting of parameter for 3D analysis

Expected Outcomes

Alternative Model = FALSE Alternative Model 4 Nodes = FALSE

Alternative Model = TRUE Alternative Model 4 Nodes = FALSE

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Alternative Model = TRUE Alternative Model 4 Nodes = TRUE



Increase memory efficiency and hence the software performance In FEM analysis when lesser nodes are used, the lesser the amount of time to run a whole large project model.

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Slab Mesh in 3D Analysis Release all slab element nodes that touches beams. (1) If no nodes are released (meaning no nodes touches beams), do not release the remaining nodes. The element will then be analyzed as SHELL. (2) If any of the remaining node touches a wall/column node, do not release the remaining nodes. The element will then be analyzed as SHELL. (3) If condition (1) and (2) don't exist (at least a node touches beam and no nodes touches wall/column), then release the remaining nodes. The element will then be analyzed as PLANE STRESS.

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Diagnosing Results Trouble-shooting large displacement in model Here we examine an example on how to trouble-shoot the problem area in a model: Suppose we have a model as shown below. We want beam "gb5" and "gb6" to be cantilevered on span 2/B to 2/C.

Select the Beam End Condition Command 1. CTRL + A to select all beam ends 2. Then press button P to ensure all the beam end condition was pinned as shown in the p208

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diagram below.

Model was wrongly done if the pinned condition applied to the end of the cantilever beam " gb5" and "gb6" By doing this, it will create large displacement nodes at the end of "gb5" and "gb6" as shown in the diagram below.

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To solve the large displacement problem, make the end of beam "gb5" and "gb6" supported by column (2,B) and (1,B) to be fixed as shown in the diagram below.

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Alternative Load Assignment This step is to by-pass the "Slab Equivalent Nodal Load" check and internally mesh 'freely' the slab load so that to enable the mesh to run successfully in hybrid mode if "Slab Equivalent Nodal Load" = False. In other words, the mesh will be assigned in the loadings of the slab (also called Alternative Load Assignment) rather than create the mesh according to the loads. The conditions of the mesh is as follow. A) Alternative load assignment will be used in 2D mesh for the floor when: 1. "Slab Equivalent Nodal Load' parameter is set to TRUE in Project Parameter < Analysis Setting < Misc.

2. At least a flat slab column in the plan has the "CustomizedDrop" property = TRUE.

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3. At least a slab load (eg. Point Load, Line Load or Patch Load) touches a flat slab column.

B) Alternative load assignment will be used in 3D mesh for the lowest floor when Raft Foundation is set to TRUE in Project Parameter < 3D Analysis < Raft.

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Design Modified basic span/depth ratio in cantilever spans A span is deemed to be a "cantilever span" when the % Difference of Moment defined above is less than 10%. BSDR ("Modified" basic span / depth ratio) = 7 + 16 (Diff% / 10)

Pure cantilever Maximum difference

Percentage difference, % 0 10

BSDR 7 23

% Difference of Moment = [ Mc - M ] / [ Mt - M ] If the Diff% > 10%, BSDR will use 20 or 26 based on the span condition (continuous/simply supported). 1. When MBSDR is TRUE: o The left and right side moment, Mc and Mt will be taken from the moment envelop. o The sagging moment, M will be taken from the Full combination. 2. When MBSDR is FALSE: o If there is a sagging moment detected for the cantilever span, the basic ratio will either be 20 or 26. Refer to example below (deflection check for Span 1): The design for moment and shear is based on support face:

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Moment diagram Moment envelop:

Remark The left and right side moment, Mc and Mt will be taken from moment envelop. The sagging moment,M is taken from Full load combination.

Full load combination moment:

DEFLECTION CHECKING FOR SPAN Left Side Moment Value = 0.0kNm Right Side Moment Value = -273.1kNm Maximum Sagging Moment Value = 3.0kNm Percentage Difference (%) = (3.0 / 276.1) × 100% = 1.10 % Modified Basic Span Depth Ratio = 7 + 16 × (1.10 / 10) = 8.8 Basic Span / Depth Ratio, Br = 8.8

Parameter is set to TRUE, calculate percentage difference of moment and get modified basic span to depth ratio. Percentage difference of moment calculated for Span 1 based on Full Moment. BSDR = 8.8

DEFLECTION CHECKING FOR SPAN Parameter is set to FALSE, there is sagging moment detected from 2D moment envelop (49.95kNm). Basic span/depth ratio (BSDR) = 26 for continuous beam

Basic Span / Depth Ratio, Br = 26.0 Span Length, l = 4300.0 mm

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Tensile strength of concrete Tension force is induced in situation like a beam supporting a column or wall (transfer beam / transfer column / transfer wall). According to BS8110-1:85 & 97 Clause 3.4.5.12, concrete shear capacity of a section subjected to shear and axial load (tension) can be calculated from equation below: BS8110-1:97

BS8110-1:85

Hence, the ultimate shear capacity of the concrete will reduce with the increase in tension force. Please note that only axial tension is considered in Esteem 8 as a more conservative approach.

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Detailing - General Link diameter when fy not equal fyy Link Diameter when fy not equal fyy During design stage, example in link design for column, user may set the main bar to use high yield steel and mild steel for link design. In BS there is no any clear statement on characteristic strength of shear reinforcement of column to be use in design (when shear design is not required, refer to BS8110-1:1997 cl3.8.4.6. In most of the case, characteristic of the main reinforcement will always use to apply for the shear reinforcement.

However users may wish to use different reinforcement characteristic strength for main reinforcement and shear reinforcement. Therefore in order to make the both reinforcement having equivalent strength a ratio was use to get the shear reinforcement when the characteristic strength for main reinforcement and shear reinforcement are different. This is to make sure the bar to provide always at the conservative side. According to BS8110-2:1985, Area of steel required can be calculated using parabolic stress block and the formula are as followed.

k1bx are depending to the concrete properties, fcu and ε0. Therefore

To obtain the ratio to converting high yield steel to mild steel or mild steel to high yield steel without changing the element properties and concrete properties

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Where High yield steel strength, fh Mild steel strength, fm Area of steel (high yield steel), Ah Area of steel (mild steel), Am Diameter of steel (high yield steel), d h Diameter of steel (mild steel), d m

For link design, when the main bar set to using high yield strength steel and link set to m ild steel the calculation of the link bar should be as follow

The following table will be use in determine the diameter of the link 2 When Main bar characteristic strength = 460 N/mm and shear reinforcement characteristic 2 strength = 250 N/mm

6

10

12

16

20

25

32

40

60

1.5

2.5

3

4

5

6.25

8

10

15

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,

,

6

6

6

6

6

10

10

10

16

6

6

6

6

10

10

12

12

25

2 When Main bar characteristic strength = 250 N/mm and shear reinforcement characteristic 2 strength = 460 N/mm , no conversion will be done because this may lead to bar dimension getting smaller than required base on

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Slab Slab Deflection check and control see "Deflection check and control" under "Beam"

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Slab Deflection Auto Pass Design In Esteem 8, User can let the slab deflection check autopass by enable param eter in parameters> Design > Floor > Slab > Special: 1. Auto deflection control = true(default) 2. As,prov / As,reqd = 2(default) (1.5 – 3) Steps of how to make the Slab Deflection Automatically Passed in Design: 1. Check the Slab Deflection. 2. If passed, stop here. Else, if failed continue the following steps. 3. Get the minimum new modification factor, MF to make it passed 4. If new MF > 2.0, new MF = 2.0 5. Back calculate using the new MF to find the As, prov 6. Report the calculation by using normal way of substituting the new As,prov into the eqn. 7 in BS8110:1985 Table 3.11. Modification factor =

By controling the As,prov / As,reqd, the program will find a As provided that will pass for the deflection check.

Notes

: The Auto-Deflection Control is NOT applicable when the slab is divided into multiple sub-slabs even though the option is set to TRUE in the parameter setting. Example: Below show the Auto Deflection Control option is set to TRUE, the report calculation will be as below:

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If the Auto Deflection Control is set to FALSE, the deflection check for the slab will fail.

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Irregular slab span length The way the software to determine FEM slab (irregular shape) span length is as following

For instances, input a irregular slab as follow,

Ab

= =

6*5 30m2

Aa

= = =

3 * 6 + 2 (4*1) 18 + 8 26m2

Notes

:

1. The generic method above is useful for handling irregular slab span length. In case the calculated irregular slab span length is lesser than the expected, user may overwrite the span length in Object Viewer > Span/Depth Options > Span Length.

Notes

:

1. Curve Slab fully bounded by an adjacent slab at straight edge and a curve beam at curve edge, Span Length for Deflection check = min ( 2 x Real Middle Offset , Arc Length)

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2. Straight Edge of curve slab not supported by beam/wall or not connected to adjacent slab, Span Length for Deflection check = Arc Length

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Determination of Basic Ratio for Irregular Shape Slab For irregular shape slabs, the Basic Span/Depth ratio, BR is determined using the following formula: Basic Span/Depth Ratio, Where,

br SE

= 7 + 13*(2*SE - 1) = Supported Edge Ratio

For example: 1. If the Supported Edge Ratio is 0.5, br = 7 + 13*[2*(0.5) - 1] =7 2. If the Supported Edge Ratio = 0.3, br = 7 + 13*[2*(0.3) - 1] = 7 (minimum Basic Span/Depth Ratio = 7) 3. If the Supported Edge Ratio > 0.8, say 1.0, use the formula in item [3a] to get the Final Basic Span/Depth ratio, BR. 3a. From the ratio of Continuous Edge Length to the Supported Edge Ratio, get the Final Basic Span/Depth. Final Basic Span/Depth Ratio, BR = br*(1 + 0.3*csr) Where, csr =

Example: The Basic Ratio, br = 7 + 13*[2*(1) - 1] = 20  7 Continuous Edge Length, c el = 18400mm Supported Edge Ratio, sel = 18400mm Continuous Edge on Supported Edge Ratio, csr = cel/sel = 18400/18400 = 1.0 > 0.8 Final Basic Span/Depth Ratio, BR = br*(1 + 0.3*csr) = 20*(1 + 0.3*1) = 26 For polygonal and L-shape slab, refer to Example 1 and 2 below. Example 1: Polygonal Slab Shape Refer to the following example on the determination of the Basic Span/Depth Ratio for slab FS10. Plan View

Remarks To show that the BR for polygonal slab FS10 is determined using the formulae for BR above. The length of the supported edges;

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1 2 3 4

= = = =

4800mm 5600mm 6000mm 2000mm

Total supported edge lengths = (1+2+3+4) = 18400mm

Slab Design > Full Report

Data and Result of Slab Mark : gb - FS10 Location : - 9/C - 9/D - 11/D - 1/C Slab Shape : Irregular-Shape

Remarks C ompute the length of the supported edges

SubSlab : FS10:1 Location : - 9/C - 9/D - 11/D - 1/C SubSlab Shape : Polygon Dimension Length of Edge 1 = 4800.0 mm Length of Edge 2 = 5600.0 mm Length of Edge 3 = 6000.0 mm Length of Edge 4 = 2000.0 mm Sub-Slab Thickness, h = 175 mm Sub-Slab Drop = 0 mm Supported edge ratio, SE = ratio of supported edge/total edge length = 18400/18400 = 1

DEFLECTION CHECKING Shorter span in X Direction (3954.0 mm)

Supported Edge Length, sel = 18400.0 mm Continuous Edge Length on Supported Edge, cel = 18400.0 mm Use formula above to get Total Edge Length, tel = 18400.0 mm Final Basic Span/Depth Ratio = 20 * (1 + 0.3 * 1.00) Supported Edge Ratio, SE = sel / tel = 18400.0 / 18400.0 = 1.00 = 26.0 Basic Span / Depth Ratio, BR = 7 + 13 × (2 × SE - 1) = 20.0 >= 7 Continuous Edge on Supported Edge Ratio, csr = cel / sel = 18400.0 / 18400.0 = 1.00 > 0.8 Final Basic Span / Depth Ratio, BR = 20 * (1 + 0.3 * 1.00) = 26.0 Basic Span / Depth Ratio, Br = 26.0 Span Length, l = 3954.0 mm Effective Depth, d = 145.0 mm Actual Span / Depth Ratio, Ar = 27.3 Ultimate Design Moment, Mu = 3.7 kNm Design Steel Strength, fy = 460.0 N/mm² Area of Tension Steel Required, AsReq = 228 mm² Area of Tension Steel Provided, AsProv = 314 mm² - Checking for deflection is based on BS8110: 1997 - Table 3.9: Basic span / effective depth ratio for rectangular or flange beams - Table 3.10: Modification factor for tension reinforcement Design Service Stress in Tension Reinforcement, Equation 8 p227

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fs = {(2 × fy × AsReq) / (3 × AsProv)} × (1 / ßb) = {(2 × 460.0 × 228) / (3 × 314)} × (1 / 1.00) = 222.1 N/mm² Modification Factor for Tension Reinforcement, Equation 7 MFt = 0.55 + {(477 - fs) / (120 × (0.9 + (M/bd²)))} = 0.55 + {(477 - 222.1) / (120 × (0.9 + (3.7 × 1000000 / (1000 × 145.0²)))} = 2.52 > 2.0 MFt taken as 2.0

The slab passed in Deflection checking.

Deflection Ratio = (Br × MFt) / Ar = (26.0 × 2.00) / 27.3 = 1.91 Ratio >= 1.0 : Deflection Checked PASSED

Example 2: L-Shape Slab Refer to the following example on the determination of the Basic Span/Depth Ratio for Sub-slabs FS29-1, FS29-2 and FS29-3. Plan View

Remarks The 1= 2= 3= 4= 5= 6=

length of the supported edges; 6401mm 2286mm 1829mm 1067m 4572mm 3353mm

Total edge lengths = (1+2+3+4+5+6) = 19508mm Total supported edge lengths = (1+2+3+4+5+6) = 19508mm

Slab Design > Full Report

Area of SubSlab (i): 4878324.0 mm² (1), 10451592.0 mm² (2), 4181094.0 mm² (3) Effective Depth of SubSlab (i): 70.5 mm (1), 70.5 mm (2), 70.5 mm (3) ∑ (Ai × di) = 1375526205.00 ∑ Ai = 19511010.00 Average Effective Depth, d = ∑ (Ai × di) / ∑ Ai = 70.50 Shorter span in X Direction (3196.9 mm) Supported Edge Length, sel = 19508.0 mm Continuous Edge Length on Supported Edge, cel = 16155.0 mm Total Edge Length, tel = 19508.0 mm Supported Edge Ratio, SE = sel / tel = 19508.0 / 19508.0 = 1.00 p228

Remarks

Supported edge ratio, SE = ratio of supported edge/total edge length = 19508/19508 = 1

Esteem Innovation Sdn Bhd, 2011

Use formula above to get Basic Span / Depth Ratio, BR = 7 + 13 × (2 × SE - 1) = 20.0 >= 7 Continuous Edge on Supported Edge Ratio, csr = cel / sel = 16155.0 / 19508.0 = 0.83 > Final Basic Span/Depth Ratio = 20 * (1 + 0.3 * 0.83) 0.8 = 25 Final Basic Span / Depth Ratio, BR = 20 * (1 + 0.3 * 0.83) = 25.0

Basic Span / Depth Ratio, Br = 25.0 Span Length, l = 3196.9 mm Effective Depth, d = 70.5 mm Actual Span / Depth Ratio, Ar = 45.3 Ultimate Design Moment, Mu = 1.2 kNm Design Steel Strength, fy = 460.0 N/mm² Area of Tension Steel Required, AsReq = 130 mm² Area of Tension Steel Provided, AsProv = 318 mm² - Checking for deflection is based on BS8110: 1997 - Table 3.9: Basic span / effective depth ratio for rectangular or flange beams - Table 3.10: Modification factor for tension reinforcement Design Service Stress in Tension Reinforcement, Equation 8 fs = {(2 × fy × AsReq) / (3 × AsProv)} × (1 / ßb) = {(2 × 460.0 × 130) / (3 × 318)} × (1 / 1.00) = 125.3 N/mm² Modification Factor for Tension Reinforcement, Equation 7 MFt = 0.55 + {(477 - fs) / (120 × (0.9 + (M/bd²)))} = 0.55 + {(477 - 125.3) / (120 × (0.9 + (1.2 × 1000000 / (1000 × 70.5²)))} = 3.12 > 2.0 MFt taken as 2.0

The slab passed in Deflection checking.

Deflection Ratio = (Br × MFt) / Ar = (25.0 × 2.00) / 45.3 = 1.10 Ratio >= 1.0 : Deflection Checked PASSED

Notes

:

1. User can overwrite the Basic Span/Depth ratio manually through Object Viewer after selecting the slab.

2. The Basic Span/Depth Ratio for irregular cantilever slab may be increased due to the ratio of Continuous Edge to the Supported Edge Ratio. Refer to following example:

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Shorter span in Y Direction (1870.8 mm) Supported Edge Length, sel = 4000.0 mm Continuous Edge Length on Supported Edge, cel = 4000.0 mm Total Edge Length, tel = 11236.1 mm Supported Edge Ratio, SE = sel / tel = 4000.0 / 11236.1 = 0.36 Basic Span / Depth Ratio, BR = 7 + 13 × (2 × SE - 1) = 3.3 < 7 Use Basic Span / Depth Ratio = 7 Continuous Edge on Supported Edge Ratio, csr = cel / sel = 4000.0 / 4000.0 = 1.00 > 0.8 Final Basic Span / Depth Ratio, BR = 7 * (1 + 0.3 * 1.00) = 9.1

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Design Slab Top Bar (Support) Analysis Result Moment values in moment contour and Full Report: The Mx and My moment values shown in the contour are based on their converted values to follow the local axis of the first slab edge as shown in the moment contour. This is to convert the moment values on the non-orthogonal edges (or support) of the slab to be based on the local axis of the first slab edge. In the Full Report (Support), the Mx and My moment values are mainly based on the local axis of the respective slab edges. Mx Contour

My Contour

The converted values (global) and moment values based on local axis of the slab support or slab edge are tabulated in the Slab Support's Analysis Result:

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Hogging Moment Selection The following example would illustrate how the slab hogging moment is calculated for "edge 1" of slab panel FS5. Slab edge 1 is the bottom horizontal edge of slab panel FS5 as shown below.

The nodal My moment are shown in the following figure.

The maximum node moment for edge 1 = -3.48 kNm. Average nodes moment for edge 1 = (-0.311 - 1.535 - 2.954 - 3.48 - 3.242 - 1.63 - 0.28) / 7 = -1.919 kNm. The slab design hogging moment will be calculated base on the parameter set by user in Design->Floor->Slab->Design Option->Hogging Moment Selec tion. User can set the value to be between 0 to 100%. Nodes 3

Mx Local (kNm) -0.27

My Local (kNm) -0.31

Vx (kN) 0.00

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Vy (kN) 0.00

d (mm) 0.329

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6 29 30 126 130 144

-0.28 0.13 0.00 -0.15 -0.09 0.29

-0.28 -3.24 -2.95 -1.53 -1.63 -3.48

0.00 0.00 0.00 0.00 0.00 0.00

Percentage used for Hogging Moment, P = 50.0% Bending Moment Bending Moment Maximum, MyMax = -3.48kNm Bending Moment Average, MyAvr = -1.92kNm Design Bending Moment, My = MyAvr + (MyMax - MyAvr) * P = -1.92 + (-3.48 - (-1.92)) * 50.0% = -2.70 kNm

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0.00 0.00 0.00 0.00 0.00 0.00

0.329 0.350 0.350 0.338 0.338 0.356

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Normal bar over Fabric Mesh Bar selection from fabric steel mesh to normal steel Bar 1. In Esteem 8, if user defined a slab to be designed using fabric mesh, the slab will be designed using fabric steel mesh. However, if all fabric mesh fail in the design, the program will automatically switch to design using normal steel bar. Example:

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Running above model, will give the following result Bar Required y – Direction bar required = 700mm (3000mm length) x – Direction bar required = 382mm (5000mm length) Bar selection BRC Design

Type A

Name

Main Area, 2 mm/m

Cross Area, 2 mm/m

Design Status

A5

98

98

Fail

A6

142

142

Fail

A7

193

193

Fail

A8

252

252

Fail

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A10

393

393

Fail

Name

Main Area, 2 mm/m

Cross Area, 2 mm/m

Design Status

B5

196

192

Fail

B6

283

192

Fail

B7

385

192

Fail

B8

503

251

Fail

B10

785

251

Fail

B12

1131

251

Fail

BRC Design

Type B

Referring to the list, no BRC bar type can be use to design this slab. Therefore the program will automatically switch to normal steel bar to design the slab. In this case Shorter direction (Y) provide = T10 – 75 Longer direction (X) provide = T10 – 200 2. The top bar fabric for the support will automatically switched to normal bar if the two slabs are not in the same level. Example: Project data: Slab FS1 is dropped at 25mm Slab FS8 has no drop.

Plan View

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Slab Top Bar Detailing

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Application of Clause 3.12.11.2.7 in slab bar spacing selection In Esteem 8, at times user may come across slab designs where bar spacings provided for steel reinforcement are found to be over-designed compared to the steel area required. This is due to the application of clause 3.12.11.2.7 being considered in Esteem 8 slab design. For instance:

Bottom Bar Spanning in Direction Parallel (X) to Sub-Slab Local Axis Bar Diameter, dia = 10 mm Effective Depth, d = 100 - 25 - 10 / 2 = 70 mm

From the slab reinforcement 2 (mm /m) table, 2 T10-300 (262mm Design as Singly Reinforced Rectangular Beam /m) is sufficient Concrete Neutral Axis, x = 4.2 mm Concrete Compression Force, Fc = k1 × b × x / 1000 = 12.12 × 1000 × 4.2 / 1000 = 50.42 kN enough as it is larger than Steel Area Required, AsReq = Fc × 1000 / (fy / γs) = 50.42 × 1000 / (460 / 1.05) = 116 mm² 130mm2/m. Moment Capacity = Fc × (d - k2 × x ) / 1000 = 50.42 × (70.0 - 0.4518 × 4.2) / 1000 = 3.4 kNm Maximum Depth of Section = 100.0 mm Minimum Tension Steel Area Required = 0.13% × 1000.0 × 100.0 = 130 mm²

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However, according to BS8110: Part 1, Cl 3.12.11.2.7, the clear spacing between bars should not exceed the lesser of 3d or 750mm. In this case, 3d = 3*70 = 210 mm If provide T10-300, Sv = 300mm > 210mm. Not OK! Therefore, Esteem 8 will reduce further the bar spacing to fulfill the clause requirement. In this case, bar spacing provided is reduced to; Sv = 200mm < 210mm ; OK!

Notes : This topic does not apply to slab design using welded steel fabric because it is not affected by BS 8110-1 Clause 3.12.11.2.7.

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Skip Node Within Column Profile User can set to TRUE for skip collecting the node values within column profile for slab support design

Figure 1: Turn to TRUE for skip node within column profile

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Figure 2: Support 27 for the slab design in the model above

Figure 3: Design report for Support 27 (If consider node within column profile)

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Figure 4: Analysis report for Support 27 (If consider node within column profile)

Figure 5: Design report for Support 27 (If Not consider node within column profile)

Figure 6: Analysis report for Support 27 (If Not consider node within column p242

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profile)

Notes

:

Skip Node Within Column Profile is Not applicable for wall support for slab

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Detailing Hogging Moment at Slab mid span In the case where there are nodes with hogging moment at the zone of the mid span of the slab, top bar at the mid span is provided as top reinforcement resisting Hogging Moment. This condition is normally happen when a slab in adjacent span has a heavy load. Definition of Mid Span Zone. The Mid span zone is area not covered by the Top Bar of slab support according to the curtailment of slab top bar. However, in some cases it is not exactly fall on the end of curtailment because the form of boundary of the mid span zone is not depending on the end support bar diameter but the curtailment of end support does. This Slab Mid Top Bar is NOT applicable to cantilever slab, where the top bar is extended from the support of the slab.

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Limitation of cut section The following condition does not constitute as a “bug” in the slab cut section detailing but rather a limitation. a) Cantilever slab

The top reinforcement is detailed for the whole cantilever span which can be seen in slab cut section 2-2 .

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But no top bar is detailed for slab cut section 1-1. b) Three-sided supported slab

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Similar to cantilever slab case, the top reinforcement also not detailed in full cantilever span. Hence, user is advised not to cut the cantilever slab/three sided supported slab in the transverse direction with the main bar.

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Cantilever Slab with drop Cantilever Slab Top Bar will be anchored at least 45 diameter into the beam from support face.

Calculation for bent length: (Anchorage by bend in the reinforcement) 45 bar diameter = Bent Length + Beam width - Design Cover - (bar diameter / 2) - r + 4r - r

Notes

:

Subjected that the bent length calculated is sm aller than the value in Project Parameter -> Detailing -> Slab -> Bent Length

Tips: For mild steel bars minimum r = 2 x bar diameter For high yield steel bars minimum r = 3 x bar diameter For 25 mm and above high yield steel bars minimum r = 4 x bar diameter e.g.

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45 x 10 [45 diameter] = Bent Length + 200 [Beam Width] - 25 [Cover] - 10/2 [diameter / 2] - (3 x 10) [r] + (4 x 3 x 10) [4r] - (3 x 10) [r] Bent Length = 220 mm

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Slab sagging moment at support In the case where there moment (vector perpendicular to the direction of slab support), the bar at both side of the slab should lap each other with a tension lap rather than anchorage lap.

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Slab top bars curtailment How does the software calculate the top support bar curtailment length in slabs? Refer to the example below for various top bar curtailment lengths:-

Parameters Setting: Ratio of top support bar curtailment = 30% Minimum anchorage length at discontinues edge = 600mm Calculation: A=30% of span 1 = 1800mm B=30% of span 3 = 1500mm < A (use 1800mm) C=20% of span 1 = 1200mm or 15% of span 2 = 3000mm but not more than 30% of span 1 (use 1800mm) and not less than MCCL (see note 1 below). D=20% of span 3 = 1000mm (parallel to span 3, no need to c onsider other span direction dimension) and not less than MCCL (see note 1 below) E=20% of span 1 = 1200mm (parallel to span 1, no need to consider other span direction dimension) and not less than MCCL (see note 1 below) F=20% of span 3 = 1000mm or 15% of span 4 = 900mm (use 1000mm); and not more than 30% of span 3 and not less than MCCL (see note 1 below)

Notes

:

1. MCCL (Mininum calculated c urtailment length) = Not less than minimum anchorage of 600mm or 40 (which ever is greater) 2. Some slabs top bar are designed as full length. This may happen in some cases where the slab hogging moment covers further into the slab panel. See the topic " Hogging moment at slab mid span" next.

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Beam Beam fails in torsion design A Beam may fail under several criteria. One of the failure types is to torsion design. In Esteem 8, torsion design is integrated into beam design in accordance to the code. The following method may help to pass the beam in torsion design. Beam Design Parameter

A) Link Diameter In the project parameter, under the Design field, Category “General”, there is “Link Diameter” parameter. If all beams use 6mm diameter as link bar, this value can be set to 6mm. Meanwhile change the minimum and maximum bar diameter in the beam design parameter to 6mm. By changing this link diameter, the x1 and y1 value will be increase, hence vt may reduce. (The x1 and y1 value calculated in the torsion design use this “Link Diameter” parameter and NOT the outcome of the design) Caution: The link for the design must be 6mm else it will cause the element to be “under design”

B) Design at Torsion Face In the project parameter, under “Design > Floor > Beam” field, Category “Design For”, there is “Torsion at Support Face” parameter. Be default this value will be set to false, meaning all beam torsion design value will be taken at the center of column instead of column face. Changing this value may help the torsion design when torsion value was very large at the center of the column but toward the column face the torsion value is getting less. This setting will be helpful when the beam support by a large column. Caution: The link for the design must be 6mm else it will cause the element under design

C) Beam End Release.

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In modeling, there may be a lot of beam supported by other beam. Esteem 8 by default will set all the beam end as fixed condition. Fixed end condition will cause big moment value in the supporting beam or column. This torsion moment can be reduced by changing the beam end condition to “Pin”. The Beam being support will not introduce any moment to the supporting beam. Caution: Do not change the cantilever beam end condition. This will cause large displacement in the analysis Example below shown 2 beams support to different End Condition

Highlighted beam use to shown the Effect of “Pin” and “Fixed”

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Beam support a “pin” Beam

Beam support a “Fixed” Beam

D) Torsion Reduction

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In esteem 8, user can reduce beam torsion value by applying torsion reduction. In the project parameter, under “2D anlaysis” field, Category “2D Analysis”, there is “Torsion Reduction” and under “3D anlaysis” field, Category “Analysis Option”, there is “Torsion Reduction” . This torsion reduction value by default is set to 50. With this default whether or not you set the beam torsion to true or false, in the analysis it will use 50. However changing the value to other value from 51– 99, this value will affect the torsion value for all beams that has the setting “Torsion Reduction” to “True”. For all beam that use “Torsion Reduction” to “False”, will use default value 50 for the analysis. To set the “Torsion Reduction” on the selected beam, users have to select a beam and change the value at the object viewer. Caution: Do not applied this on the “non straight beam” (beam connect beam using connector but not in the straight line). This may cause unstable condition. Example below shown 2 beams with different torsion reduction set to false and 99

Torsion Reduction = 99

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Torsion Reduction = False (Default = 50)

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Beam deflection check failed Title: Beam fails in deflection check due to no support at midspan location. Refer to example below. 1. The slab nodes are pinned. Therefore instead of being supported by beam B/1-6 at grid (3,B), beam 3/A2-A2' deflect together with beam 1/-6 due to following the hinged condition (where Auto Support also applies). Beam 3/A2-A2' then fails in deflection check due to no support at grid (3,B). To prove this, the deflection of beams B/1-6 and 3/A2-A2' should be the same at the grid location (3,B).

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2. Besides that, due to another column support at grid (3a,B), the stiffness of the slabs FS5, FS6, FS33 and FS34 tend to 'pull up' the deflected beam 3/A2-A2'. Therefore, the loadings acting on beam 3/A2-A2' cause the moment diagram shown in figure below:

3. In order to adjust the 'pulling effect' from the slabs, the slab's stiffness can be adjusted by going back to Input mode, selecting these four slabs (FS5, FS6, FS33 and FS34) and in the Object Viewer, input 1000 in Slab Stiffness Adjustment. This setting will make the slab's Static Secant Modulus (Ec=22454N/mm^2) to be divided by 1000 to reduce the slab's stiffness effect to the surrounding beam.

Notes

: Take note that this slab stiffness adjustment may cause the loadings to be different transferred to the adjacent beams. 4. The reduction of the 'pulling effect' from the slabs above will make the moment diagram for beam 3/A2-A2' to be reversed back as shown below. Thus making beam 3/A2-A2' supported solely by beam B/1-6 at grid (3,B). The beam then will not fail in deflection.

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Detailing Bar lapping issues at beam ends When a beam has been 'Broken' into different sections, each section is designed and detailed separately. Hence in the example below, the program details the beam (between Gridlines A and A1) below as in Figure 1 and not as Figure 2. Crank bar will be drawn between 3T12 and 2T12 instead of straight bar with tick.

Figure 1

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Figure 2

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Cut sections The program will produce cut sections between supports (columns, walls or possibly beam intersections). The number of cut sections defined is controlled via the beam parameter "Cutline" of the "Beam Design".

For the cut line fall exactly at the section change part, the cut line will be shifted slightly to smaller section.

Cutli Expected Output ne Setti ng One For beam with one cut section, this setting will show the cut section following the beam mark. Secti on (Bea m Mark) -Defa ult

Auto C ondition 1: Three cut section will be made for each beam span

C ondition 2: Only ONE cut section for each beam section

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C ondition 3: Only ONE cut section at cantilever span

None There will be no cut section in the beam detailing

One This will make only ONE cut section for every beam span. The cut section labelling will not follow the beam Secti mark. on C ondition 1: Only ONE cut section for every beam span

C ondition 2: Only ONE cut section for cantilever span

3 C ondition 1: Three cut section at cantilever span Secti on

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C ondition 2: Three cut section for every beam span

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Shear and torsional link detailing rules Esteem 8 provides links according to the following rules. Consider a beam span. A span here is defined as the center to center distance between primary supports such as columns, walls or "primary" beams. The basic areas for shear reinforcement placement would be at quarter spans from supports and the remainder half of the span. Sections are defined as parts of the beam that is divided by either a transfer column, secondary beam or an external point load application in the span. Zones are the sub-division of basic areas by sections.

If the "external point load" falls within distance d from the critical shear zone boundary (close to support), the shear zone boundary line will be extended to the distance of d from the location of the "point load". Apply to the above example, Zone 2 may extend beyond the quarter span distance to fulfill this rule should there be a point load somewhere within Zone 2.

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Link details in sections The following detail does not constitute as a "bug" in the program but rather a limitation. Link "a" and "b" are both to resist vertical shear. The torsional requirement are taken into account by link " a". (Notice that for link designed for torsion the lapping covers the full distance of the top bars.) Issues:  One of the top bar mark "2" is placed at the mid-section position and is seemingly placed out of the hook position of link "b".  This is not an error in the detailing of the program but rather a limitation. The program does not include discretive placement of the bars with respect to the links.

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Beam detailing - Parallel Wall

The beam shown above will be detailed as follows:-

1. Top Bar anchored into the wall with tension anchorage measured from the surface of wall. For example if diameter of bar = 16, and the factor for tension anchorage is 40 X bar diameter, then anc horage length = 40 x 16 = 640 mm 2. Bottom Bar anchored into the wall with compression anchorage measured from the surface of wall, If bar Diameter = 32 mm and factor for compression anchorage is 27, then the anchorage length is 27 x 32 = 864 mm 3. Curtailment for bottom bar is measured from the center of support which has a triangular mark if user shows it and the length taken as continuous 0.15l. The center of support is half of effective depth measured from the face of parallel wall (BS-8110 1997 cl3.12.9.4 b). but for convenience and bearing on conservative side, distance of half of beam depth is adopted. 4. Curtailment for Top Bar is measured from face of support and taken as 0.25l. Length of span l is taken from support to support.

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Beam sections of different width When the width of a section in a beam different with another section more than 100mm (total), the bars in the larger width beam section will be bent down.

Figure 1: Multisection beam with 2 different width where the total different beam width more than 100mm

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Figure 2: Detailing In detailing when the width of 2 section differ more than 100mm, it will bent down the bar before the end of the largest section. In the Figure 2 the bar will bent down at section 500 x 500 before section 200 x 500

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Links at intersecting beam If the intersecting beam is a support, then the stirrup will stop at the sides of the intersecting beam (with small offset). In the beam detail shown below the intersecting beam is not a support for current beam, then current beam shall resist the shear (perhaps due to heavy point load) introduced by this intersecting beam, hence the stirrup from 2 sides will not stop at the sides.

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Maximum distances between bars in tension Clause 3.12.11.2.2 (BS8110-1:1997) This clause is meant for maximum distance between bars in tension, where the diam eter smaller than 0.45 x largest bar should be ignored when checking for maximum spacing, it does not imply any allowable different of bar diameters in different layer.

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Transfer Beam Detailing Detailing notes on transfer beam supporting wall with the structural analysis based on the strut and tie model. For the transfer beam main bar detailing, there must be no curtailment of main rebar, i.e. full anchorage of main rebar is needed. This is due to the rebar being designed for full tension force in the strut and tie model. Therefore all the longitudinal rebars are required to be tied into the anchorage joints at the support.

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Transfer Beam Detailing Specification The transfer beam detailing for strut and tie design follows the specification below: For final tension reinforcement required, the additional longitudinal area of reinforcement from tie force will be added to the existing area of reinforcement required for tension. For final compression reinforcement required, the area of reinforcement required will be subtracted with the distributed area of longitudinal bar area from tie force, to find the m aximum area of compression reinforcement required. Example: Location: Midspan Top Compression Steel Area Required = 2106 mm² Bottom Tension Steel Area Required = 19379 mm² Area of Steel Required for Middle Bar, AsMid = 786 mm² Portion of Longitudinal Torsion Steel Area Distributed to Tension Reinforcement, Aslt = (Asl - AsMid) / 2 = (1442 - 786) / 2 = 328 mm² Portion of Longitudinal Torsion Steel Area Distributed to Compression Reinforcement, Aslc = (Asl - AsMid) / 2 = (1442 - 786) / 2 = 328 mm² Additional Tension Steel Required along beam span, Ast = Ft / (fyy × fy) = 2,970.9 × 10³ / (0.9524 × 460) = 6782 mm² Area of Longitudinal Bar Area Required by Top Reinforcement, AstTop = Ast / 4 = 1695 mm² Area of Longitudinal Bar Area Required by Bottom Reinforcement, AstBot = Ast = 6782 mm² Final Top Compression Steel Area Required (3D) = 2434 mm² Final Bottom Tension Steel Area Required (3D) = 26489 mm² Top Reinforcement Provided = 6T32 (4825 mm²) Bottom Reinforcement Provided = 26T32, 12T25 (26801 mm²) Final top compression steel area required = Max (AscReq, |AscReq - AstTop|) + Aslc = Max (2106, |2106 - 1695|) + 328 = 2106 + 328 = 2434 mm2 Final bottom tension steel area required = AstReq + AstBot + Aslt = 19379 + 6782 + 328 = 26489 mm2

Location: Left/Right Support Design to minimum steel percentage specified by code, Maximum Depth of Section = 1800.0 mm Minimum Tension Steel Area Required = 0.13% × 900.0 × 1800.0 = 2106 mm² Top Tension Steel Area Required = 2106 mm² Area of Steel Required for Middle Bar, AsMid = 1709 mm² Portion of Longitudinal Torsion Steel Area Distributed to Tension Reinforcement, Aslt = (Asl - AsMid) / 2 = (3137 - 1709) / 2 = 714 mm² Portion of Longitudinal Torsion Steel Area Distributed to Compression Reinforcement, Aslc = (Asl - AsMid) / 2 = (3137 - 1709) / 2 = 714 mm²

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Additional Tension Steel Required along beam span, Ast = Ft / (fyy × fy) = 4,525.4 × 10³ / (0.9524 × 460) = 10330 mm² Area of Longitudinal Bar Area Required by Top Reinforcement, AstTop = Ast / 4 = 2582 mm² Area of Longitudinal Bar Area Required by Bottom Reinforcement, AstBot = Ast = 10330 mm² Final Top Compression Steel Area Required (3D) = 5402 mm² Final Bottom Tension Steel Area Required (3D) = 8938 mm² Final top tension steel area required = AstReq + AstTop + Aslt = 2106 + 2582 + 714 = 5402 mm2 Final bottom compression steel area required = Max (AscReq, |AscReq - AstBot|) + Aslc = Max (2106, |2106 - 10330|) + 714 = 8938 mm2

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Top rebar in cut section not drawn to the actual detailing Basically, this problem occurs when one of the beam span length has a big difference with the adjacent span length. For instance, a beam (300x900mm) consists of two spans (spans length are 11m and 4.5m respectively), with uniform dead load, 75kN/m2.

From the beam detailing, the beam section for E should show 3T32+2T32+3T20 for the top reinforcement. However, section E only shows 2 layer of reinforcement 3T32+2T32. This is because beam section is drawn based on actual design result and did not reflect what is finally detailed with all things considered (curtailment lenght in this case). In fact, according to moment diagram, 4T25 is sufficient to resist the moment at the centre of span B-C and the actual curtailment for span BC should start at 1125mm (4500/4, quarter of the span). Since span AB has longer curtailment length, the program will select the longer curtailment length and assign the length for both side curtailments. Therefore, in this case, the discrepancy of the detailing result will be encountered between the beam section and the beam detailing. Engineers are advised that they should check all the p274

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detailing drawing before issue to third party.

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Top Bar detailing rules Condition 1: Single span with single section Condition 1a: Symmetry with total length < user set bar length (11m) - No Curtailment

Condition 1a: Symmetry with total length > user set bar length (9m) - With Curtailment (0.25L from face of support)

Condition 2: Multiple Spans with Single Section

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Condition 3: Single Span with Multiple Sections

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Bottom Bar Detailing Condition 1: Bottom bar at short section cantilever span (less than 500 mm) When there is a short cantilever span of beam which having lesser steel area provided compared to the inner span, the steel area provided from the inner span will be extended to the cantilever span. Refer to example below. The inner span is a transfer span (supporting transfer column). Therefore curtailment is not allowed. Although the short cantilever span requires 2T10 only, steel area provided from the inner span is used in the short cantilever span. (Figure 2).

Figure 1

Figure 2

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Use tension bar lapping for sagging moment at support 1. When there is a sagging moment detected at the support, tension lap will be used. Refer to example below, the program used tension lap for beam gb1 at the highlighted support (Figure 1 and Figure 3) because there is a sagging moment there (Figure 2). Moreover, the 2D Full Moment (Figure 2) will be used to decide whether tension or compression lap should be used.

Figure 1

Figure 2

Figure 3

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Nib Beam A "nib beam" is a type of bridging beam in the vicinity of column support. Summary: This specification explains the design and detailing of “nib beam ”. This occurs when there is a bridging beam from a beam to the column node in the condition where the profile of the beam does not sit fully within the column section profile. The figure below shows a typical example where a nib beam can occur.

Design and Detailing considerations: A. Application of nib beams 1. The nib beam will come into being when a main beam profile is not fully supported by a column. 2. When there is another beam “tied” to this column and provided that the angle is not greater than 135° then the nib beam would not be required. Nib beam required

Nib beam not required

Angle between the beams > 135°

Angle between the beams = 90° < 135°

3. Where there is a wall connection, the application of nib beam depends on the side of the beam where the wall is located. The nib beam is not required when the wall is connected at the side where the beam profile is not seated on the column support. Example: Nib beam required

Nib beam not required

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Angle between wall and beam = 90° < 135°

In addition to above if the wall connection angle in [3] above is also not greater than 135°, the nib beam is not required . Apart from this condition, the nib beam is applied for wall connection situation. B. Orientation of nib beam 1. The nib beam will be aligned parallel to the bridging beam. 2. The plan area of the nib will be bounded by the parallel lines which begin at the intersections of the beam and column profiles to the intersections at the other side of the column section. C. Design of the nib beam 1. Design as per conventional beam. 2. Beam width dimension is taken from column beam profile dimension intersection which is perpendicular to the beam, and the depth is taken as from of the supported beam. 3. If the nib beam width is more than 300 mm, the width is taken as 300m m 4. If the nib beam width is lesser than 200 mm, the width is taken as 200 m m. Refer to example below:

5. The concrete shear capacity is enhanced as per Clause 3.4.5.8 in BS8110: 1997 Part 1. D. Detailing of the nib beam 1. Anchorage length of reinforcement in nib beam must follow the conventional beam detailing rules. 2. The cantilever span length of the nib beam measured from surface of support to column centre must be larger or equal to: i. Lesser of (Minimum link spacing or 100mm) ii. Beam Side Cover dimension Otherwise, the nib beam detailing is unable to be reproduced and will be marked as Not Designed. E. Special considerations o In a situation where the bridging beam is not a nib beam but part of the beam profile that is without support of any kind (column, other beams, slabs); usually p283

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occurs in corner and T-Junction of beam column connections, the moment from the bridging beam is checked against the top bar resolved moment capacity calculated from the other beam connected to this intersection.

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Design Ultimate Moment Resistance of Parabolic Stress Block (BS8110) Expressions for the value and location of the compressive force C in the section are given in BS8110: Part 3, Appendix A.

= the concrete strain at the end of the parabolic section w = the distance from the neutral axis to strain x = depth to neutral axis k1 = mean concrete stress k2 = depth to the centroid of the stress block e.q 1 (a) To determine the mean concrete stress, k1 From the strain diagram

Therefore

e.q 2 For the stress block

; for area rst, see parabolic properties in item(b). Substituting for w from the equation 2 gives

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Same as equations given by BS8110:1985 Part 3 Appendix a. (b) To determine the depth of the centroid k2x

k2 is determined for a rectangular section by taking area moments of the stress block about the neutral axis – see figure above.

Substituting for w from the equation 2 gives

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Substituting for k1 from the equation 3 gives

Assume

as y and replace in the equations.

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Same as equations given by BS8110:1985 Part 3 Appendix a. According to BS8110:Part 1 Cl. 3.4.4.4 stated that x should not exceed 0.5d where the redistribution does not exceed 10%. In Esteem, we use it as a limiting effective depth factor Cb which is equivalent to 0.5.

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Divide the kk1 factor in Esteem calculation with fcu show at below:

Similar with k’ define in BS8110: Part 1 Cl.3.4.4.4 where the redistribution does not exceed 10% This is the value to design as a singly reinforced rectangular beam if it does not exceed the value. To calculate the concrete neutral axis, x: As we know the Concrete compression Force, Moment of resistance with respect to the concrete,

Since we have the value of moment, M, k1, k2, width, b and depth, d, replace it in the e.q 4 to get the concrete neutral axis, x. After getting the x value, substitute inside the e.q.4 to get the Concrete Compression Force, Fc. Using the equations, we can get the steel area required,

Example: Beam data: fcu= 30 N/mm2 width, b = 250 mm depth, d=663mm Step1: Calculate the mean concrete stress above neutral axis,k1 = 10.2 N/mm 2 using the

Step 2: Calculate the average concrete stress above neutral axis,k2 = 0.4557 using the

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Calculate the Limiting Concrete Moment capacity factor,kk1 by limiting the x/d should not over than 0.5. Kk1 = cb*k1(1-cb*k2)=3.9379 N/mm2 Step 4: Calculate the Concrete neutral axis, x using the M = k1*b*d*x – k1*k2*b*x2 X = 103.8 mm Step 5: Use the x = 103.8mm replace it into the equation 4, Fc= 264.91kN.

.

Step 6: Use the Fc to calculate the Steel area required. Ref: 1. 2. 3.

BS 8110:1985 Part1 BS 8110:1985: Part 3 Appendix A “Reinforced Concrete Design” 5th. Edition, Mosley,Bungey,Hulse. Pg 84.

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Design of multisection beam The following model consists of one span of beam with three sections. The first and third sections are 200x500mm and length is 1000mm. The sec ond section will be a 200x1200mm beam with length 4000mm. 15kN/m of UDL dead load will be imposed on the gb1.

Basically, the moment design will be done in three parts : 1. Span moment (middle span of beam)

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2. Left/right support moment (column face)

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3. The point which the section changed

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4. From the calculation above, o Span moment = 113.17 kNm , d = 1159mm, steel provided = 3T12 o Support moment = 5.8 kNm , d = 459mm, steel provided = 2T12 o Section moment = 55.04kNm, d = min (1159,459) = 459mm, steel provided = 3T12

5. Shear design also similar considerations.

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Deflection check and control Beam deflection checks are based on FE Analysis results (service load) BS8110 compliance Clause 3.4.6.3, the span / d ratio can be obtained by refer to the table 3.10. Deflection is influenced by the amount of tension reinforcement and its stress. Values obtained from table 3.10 should be multiplied by the appropriate factor using the equation 7 and 8 as modification factor tension reinforcement (cl. 6.4.6.5) and equation 9 for the modificaiton factor compression reinforcment. For beams: The allowable span depth ratio = basic span depth ratio x modification factor tension reinforcement x modification factor compression reinforcement.This value should be greater than the actual span/ d ratio. For slabs: The allowable span depth ratio = basic span depth ratio x modification factor tension reinforcement (the modification factor compression reinforcement is not considered as slab design does not design for compression reinforcement). This value should be greater than the actual span/ d ratio. Span definition: The span is taken as the distance between two ‘support’ exhibiting hogging moments. Effective depth, d' Esteem 8 employs the ‘weighted’ d for the multi-section span.

(  Mi bi di ) / (  Mi bi ) where M is the moment by FEA , b the breadth of section and d the effective depth at the i i i node(s) in the span. If the deflection cannot satisfy the codal criteria, the program will then check base on the actual FEA deflection, delta d, delta/span ratio, dr against the allowable deflection for a. short term, b. long term How to measure the actual FEA beam deflection? The actual FEA deflection = relative deflection measured from support = FEA’s maximum deflection – (support1+support2)’s deflection/2 How to measure the actual FEA slab deflection? The actual FEA deflection = relative deflection measured from support = FEA’s maximum deflection in span – SUM(supports)’s deflection/supportNos * Only deflection on the node at the support area (beam or wall) will be collected for the free edge is not consider a support. Slab deflection: For rectangular shape bounded by 4 sides, Esteem 8 will automatically determine the span, d and span/d ratio. Failing these Esteem 8 will use the modification factor to improve on the deflection check: Slab: Increase the slab tension modification factor using the Equation 7 and 8 by increasing the area of reinforcement provided. Beam: Increase the beam compression modification factor using the equation 9 by increasing the p296

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area of reinforcement provided.

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Framing beam A framing beam is a beam that sits along the top of an RC wall. It cannot be a taper beam . It must have the same length with the supporting RC Wall sits physically on the RC wall. If a framing beam is connected to a neighbouring beam at the end point, the connection at the endpoint must have the same width, depth and offset. Framing beam will be designed and detailed if torsion stress from section, v st is less than half 2

of the concrete torsional strength (0.067 SQRT (f cu) or 0.4 N/mm ) throughout the framing section.

Therefore for a framing beam sitting on top of an RC Wall, the support nodes for the beam which is shared by the RC Wall will be shown in the Result Diagram. Consider the following model:

3D View

3D Mesh

The Result Diagram will show the support nodes along the beam as follows: Result Diagram shown is 3D Frame Analysis moment envelop (Auto-Sc aled at 100%).

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Fully supported transfer beam design Transfer beam that does not satisfy any of the requirement below (except if tie beam (perpendicular or with angle not more than 45) is connected to the column node) will be marked as Not Design due to Large Column Offset. 1. More than 95% of perpendicular length (beam width) should be supported by c olumn OR 2. Column centroid is within 5% tolerance of beam width from beam centerline

Illustration of requirements above: 1. 95% of perpendicular length (beam width) should be supported by c olumn

2. Column centroid is within 5% tolerance of beam width from beam centerline

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3. Tie beam (perpendicular or with angle not more than 45) is connected to the same node.

Example: Beam 1b5 (assuming that this beam is a transfer beam) shown below will not be designed. No details will be provided. But beam gb5 will be designed because there exist a tie beam in the connecting node (angle of tie beam allowed not more than 45).

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Beam Deflection Auto Pass Design In Esteem 8, User can let the beam deflection check autopass by enable parameter in Project Parameter> Design > Floor > Beam > Design For: 1. Auto Deflection Control = true(default) 2. As',prov / As,prov = 0.5(default) (0.1 – 1) Steps of how to make the beam Deflection Automatically Passed in Design: 1. Check the Beam Deflection. 2. If passed, stop here. Else, if failed continue the following steps. 3. Get the minimum new modification factor, MF to make it passed 4. If new MF > 1.5, new MF = 1.5 5. Back calculate using the new MF to find the As, prov 6. Report the calculation by using normal way of substituting the new As,prov into the eqn. 7 in BS8110:1985 Table 3.11. Modification factor = By controling the As,prov / As,reqd, the program will find a As' provided that will pass for the deflection check. Example: Below is the example of the deflection calculation when auto deflection check is set to true of beam:

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If user set the auto deflection option to false, the sam e span beam deflection will show fail due to now increase the compression reinforcement.

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Example Beam Deflection Auto Control Example: 1. Compressive steel cannot be increased due to As’,prov/As,prov ratio. 2. Compressive steel increased but still failed and cannot be further increased due to As’,prov/As,prov ratio. 3. Compressive steel increased and pass in deflection.

Example 1 : Compressive steel cannot be increased due to As’,prov/As,prov ratio Increment of compressive steel exceeded As',prov / As,prov = 0.5: Tensile Steel Provided: 7238 mm2 Compressive Steel provided: 3217 mm2 Required Steel Provided = 3546 mm2 (at maximum FSR = 0.49) After test, get new Compressive Steel provided: 4021 mm2 (5T32) exceed FSR x 7238 so switch back to 3217 mm2

Example 2: Compressive steel increased but still failed and cannot be further increased due to As’,prov/As,prov ratio. Increment of compressive steel does not exceed As',prov / As,prov = 0.5 but still failed and cannot be further increased: Tensile Steel Provided: 5984 mm2 Compressive Steel provided: 2412 mm2 Required Steel Provided = 2932 mm2 (at maximum FSR = 0.49) After test, get new Compressive Steel provided: 2945 mm2 (6T25) does not exceed FSR x 5984 = 2992 mm2 so not switc h back to 2412 mm2

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Example 3: Compressive steel increased and pass in deflection. Increment of compressive steel does not exceed As',prov / As,prov = 0.5 and passed: Tensile Steel Provided: 1608 mm2 Compressive Steel provided: 226 mm2 Required Steel Provided = 603 mm2 (at FSR = 0.38) After test, get new Compressive Steel provided: 603 mm2 (3T16)

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Design for torsion and shear reinforcement Refer to Shear and Torsion Design Calculations for computation of link dimensions. Area of link reinforcement, Asv (mm2) According to the BS8110: Part 2; 1985, Clause 2.4.2, the area of reinforcement denoted by Asv is equivalent to the 'area of two legs of closed links at a section'. From design, the required area of reinforcement for torsion and shear are obtained. Example: Description

Notation

Spacing of reinforcement

Sv

Area of links

Asv

Reinforcement area for torsion

TAsv

Reinforcement area for shear

SAsv

Design for torsion and shear reinforcement more than one external loop

Required reinforcement area over spacing, Asv/Sv required (mm2/mm)

Provided Total reinforcement reinforceme area over nt area over spacing, spacing TAsv/Sv required provided 2 (mm /mm) (mm2/mm)

Remark

Torsion TAsv/S v

Shear SAsv/Sv

0.40

0.504

0.904

0.566

Not OK

0.40

0.504

0.904

1.131

OK

1. External loop = 1 Link diameter = 6 mm Spacing = 100 mm Asv/Sv provided = (2*6 2*/4)/100 = 0.566 mm2/mm 2. External loop = 2 Link diameter = 6 mm Spacing = 100 mm Asv/Sv provided = (4*6 2*/4)/100 = 1.131 mm2/mm

Reference: 1. Reinforced Concrete Design, 5 0-333-73956-6), Chapter 7

th

edition, W.H.Mosley, J.H.Bungey, R.Hulse (ISBN

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Behaviour of Deep Beam using the ‘Strut and Tie’ model Transfer girder supporting high shear walls

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Notes

: All deep beam will have NO curtailment of main rebar. For definition of "deep beam", please refer to "Definitions" under "Basic Concepts" above. Reference: 1. Design of Concrete Structures, 13

th

edition. Arthur H. Nilson, David Darwin, Charles p311

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W. Dolan (ISBN 007-123260-5), Chapter 10 rd 2. Reinforced Concrete Mechanics and Design, 3 edition. James G. MacGregor (ISBN 0-13-233974-9), Chapter 18

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Number or external loop in shear reinforcement External Loop for Shear Reinforcement This parameter allows the user to control the number of external loop to be used in design for shear reinforcement with default value of 3. User can input this value ranging from 1 to 10 for this parameter. The term external loop means the loop that borders the perimeter of the section.

Figure 1: Number of external loop = 3 For example in Figure 1 above, user had input value of 3 for the External Loop allowed. In the beam detailing below in Figure 2, it shows that 3R10-125 is used. Same as in cut-section detailing A-A, the external link at ‘a’, 3xR10-125-a for link leg at this section is looped three times. The hook is not included in shear design. When loop for external link provided is more than specified by user, a warning message stating "Nos of loop link > 3, please check the maximum diameter setting" will appear. Example 1: In Figure 1 below, the External Loop specified by user is 3. Thus when the external loop provided in beam shear design is less than or equal to 3, no warning message is shown.

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Figure 2: User input value 3 for External Loop allowed, and the design provided is 3. Example 2: However there are cases the shear reinforcement provided exceeds the value speficied by user. In Figure 2 below, the user input value of 2 for External Loop allowed. But in the design, External Loop provided is 3. Thus, a warning message will appear.

Figure 3: User input value 2 for External Loop allowed, but the design provided is 3 loops. Example 3: User input External Loop allowed to 4 but the design provided is 6 loops.

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Figure 3: User input value 4 for External Loop allowed, but the design provided is 6 loops.

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Shear force diagram known issues In FEM analysis, beams are made up of discrete elements connected together (Figure 2a). Every element will contain 2 nodes: begin node and end node. The end node of the first element will connect to begin node of second element. After analysis, all elements will contain analysis result at begin node and end node. Thus, every node will have two result; one from end node of the left side element and the other the begin node of the right side element. To obtain the value at the node, the average value will be taken. We know in the position of the applied load, the shear force to the left side of the load and the shear force to the right side of the load will be different as per shown in Figure 5. The expected result should be as in Figure 4. The plotting of the shear diagram Esteem 8 begins from one side of the beam. The program will average up the shear force of the end value of the left hand side element and the begin value of the right hand side element resulting in the value as shown in the Figure 6.

Figure 2a : Beam element

Figure 3: loading assignment

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Figure 4 : Conventional analysis result for shear diagram

Figure 5: Node value and average value in the node

Figure 6: Average values was used to plot shear diagram

Figure 7: Shear diagram produced

Notes

:

The average shear force value as shown above will not be characterised in the situation where there is a secondary beam being supported. In this case, the program will plot both the end and begin shear force value independently in view of the fact that the primary beam is p317

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deemed to being sectionalised. (The shear force diagram is plotted section by section.)

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Point load application In Esteem 8, when a load is applied on a beam and if the applied load coincided with the location of a beam node, then load will be fully applied to the beam node. If the load applied does not coincide with the location of the beam node then the load will be distributed to adjacent nodes according the ratio of the distance to the nodes of element from the load point. This concept is similar to the loads applied on mesh nodes.

Figure 1: A point load applied on a beam element in between two nodes

Figure 2: Point load will be distributed according to the ratio

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Shear force diagram in beam-frame element vs. couple beam-shell element

A.

B.

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The shear value shown in the 3D analysis result is not the same as the 2D analysis. This is due the the fact that in 3D analysis, the beam is modelled as a couple-beam shell element and there is a calculative limitation in the post-processing of the shear values. However as far as the design is concerned, the program would still detect the higher shear value from the envelope hence design is still safe.

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Study on the qualitative effects of the tension force in the transfer beam due to the surrounding elements Study qualitative effects of the tension force in the transfer beam due to the surrounding elements The main purpose of this study is to 1. Compare the effects of the tension force in the transfer beam with/without the surrounding elements 2. Find out the tension force value of the adjacent element Here, we create two models.  Model A - a simple 3 storey building consists of 2 storey shear wall which supported by a transfer beam (2/B-C) at lowest floor. Please refer to Figure 1 and Figure 2. Then input a big point load, 500kN (Dead load) onto center of the wall at top floor (2.5m from gridline B).

Figure 1 : Ground Floor Plan for Model A

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Figure 2 : Model A Elevation View The elements dimensions are as follows:Thickness Wall

Width

Depth

150mm

Transfer beam

450mm

1000mm

Column

400mm

400mm

Length

Height

5000mm

3000mm each floor

5000mm 3000mm each floor

 Model B – Similar to model A, but transfer beam (2/B-C) at GB floor is joined to other beams with dimension 230x600. Then, also input a big point load, 500kN (Dead load) onto center of the wall (2.5m from gridline B) at top floor. Please refer to Figure 3 and Figure 4.

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Figure 3 : Ground Floor Plan for Model B

Figure 4 : Model B Elevation View After that, run the batch process for beam analysis  3D frame analysis  wall design  beam design. Then, open the beam checking output for beam 2/B-C (Model A) and beam 2/A-D (Model B) to get the tension force in transfer beam (refer to Figure 5 and 6). Despite of 500kN point load case, we also rerun the models by using the point load 1000kN, 1500kN and 2000kN. Note : Please turn off the notional load analysis.

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Figure 5 : Model A - Beam checking output for tension force in transfer beam

Figure 6 : Model B - Beam checking output for tension force in transfer beam Factored tension force for the transfer beam in model A (kN)

Factored tension force for the transfer beam in model B (kN)

500kN point load

261.1

1000kN point load

Tension force in adjacent beam (kN) Left span

Right span

177.811

0.391

0.37

336.7

322.827

0.637

0.597

1500kN point load

486.8

467.844

0.882

0.824

2000kN point load

891.7

612.861

1.128

1.050

Table 1 : Output summary for the tension force in transfer beam and adjacent beam

Total tension force of the Total tension force of the transfer beam and adjacent beams (kN) adjacent beams (kN)

Percentage of tension force for the adjacent beams

500kN point load

178.572

0.761

0.43%

1000kN point load

324.061

1.234

0.38%

1500kN point load

469.55

1.706

0.36%

2000kN point load

615.039

2.178

0.35%

Table 2 : Percentage of tension force for the adjacent beams p325

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Conclusion : Refer to Table 1, we can notice that the tension force for the transfer beam in model B is lesser if compared to the transfer beam in model A although with the same point load magnitude. Since transfer beam in model B is being tied/joined with other beams, this will increase the stiffness itself. In other words, transfer beam in model B will stiffer and less deflection. Subsequently tension force due to strut and tie effect in transfer beam reduced. Besides, refer to Table 1, we can see that the tension force for the adjacent beam in model B is very minor and negligible if compared to the total amount of point load input. Hence, most of the tension force is taken by transfer beam itself. Adjacent beams may share the tension force which is negligible. From the case study (Table 2), the tension force in adjacent beams is approximately 0.4% to the overall tension force caused by the strut and tie effect. Therefore, we do not consider any tension force design for the adjacent beam in EsteemPlus. The reinforcement provided for the flexural design and minimum steel percentage should have sufficient capacity to resist the minor tension force.

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Alternative deflection check (for cantilever beam only) The program will only do an alternative deflection check (BS8110:1997 clause 3.4.6.3) for cantilever beam when the cantilever beam fails in the deflection check by using span/depth ratio (BS8110:1997 clause 3.4.6.3). For instance when the cantilever beam failed in deflection check based on span/depth ratio,

The program will carry out another alternative deflection check for that failed cantilever beam as shown below.

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The long term deflection coefficient, K , is by default 3. User can change the value between 2 and 5 by going to the Parameter Setting > Design > Floor > Beam > Deflection coefficient as highlighted below.

Notes

: In some cases where a beam behaves like a cantilever beam but physically not, this alternative check will also be considered if the said beam fails in span/depth ratio deflection check.

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Node Value in Full Report Not Tally with Summary Report Please note that:1. Not all values in the summary report is available in the full report calculation because the full report (span, left support, right support etc) is in span basis while the summary report is section basis. Only selected nodes' calculation are shown in full report. However user can compare the detail in the beam node value report for more information. 2. Node selection is based on larger AsReq for moment that have difference not more than 1 decimal point and larger moment if the difference of moments tolerant larger than 1 dec imal point, for example two set of nodes below: A. Mu = 49 kNm, AsReq = 500 mm2 B. Mu = 50 kNm, AsReq = 400 mm2 The node set B still be selected due to larger moment although node set A has larger AsReq (the value maybe available in summary and beam node value reports) but user unable to find in full report.

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Column Column Clear Height, lo Column clear height calculation can be based on either method below: 1. Clear Floor Height 2. Clear Distance between End Restraints

1. Clear Floor Height Clear floor height here means the relative top level distance between the respective two floors. However drops in either or both the top or bottom element can increase/decrease the clear floor height. Restraint element depth is not deducted in this method.

Notes

:

le = (floor to floor height)*constraint condition

le =  lo

2. Clear Distance between End Restraints Column clear height, lo is determined by the distance between end restraints (beam or slab).

Cantilever beam will not be considered as restraint For column that is restrained by wall, the clear height will be taken as half of the current floor height. Example: Column (1,A) is restrained by rc wall in the z direction, therefore the column clear height is taken as half of floor height = 5000/2 = 2500 p330

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Shear in Columns For Rectangular and Square Column, When M/N value does not exceed value stated in BS8110:Part 1 Clause 3.8.4.6 in all the load Cases will not check and design for Shear and Torsion. Column link therefore will design based on BS8110:Part 1 Clause 3.12.7.1 i.e. Link or Ties, at least one quarter the size of the largest com pression bar or 6mm, whichever is greater, should be provided at a maximum spacing of 12 times the size of the sm allest compression bar.

Notes

:

BS 8110-1:1997 - No check is required provided M/N does not exceed 0.6h. BS 8110-1:1985 - No check is required provided M/N does not exceed 0.75h.

Reference:

Example :

Notes 1. design 2. shown

: For EC2 Code, No specific clause is mentioned to check M/N value for Column Link i.e. No additional checking for M/N need to be done. For ACI Code, It does mention that minimum Shear Reinforcement can be waived if by test that required Mn and Vn can be developed when Shear Reinforcement is p332

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omitted. Since there is no clear guideline for this clause, Shear and Torsion for Column Link design is remain.

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Computation of column beta value Introduction Basically, the effective height of column is computed by using the formula le = *lo (refer to BS 8110: Part 1 Clause 3.9.3.2.1) which is following the procedure given in Clause 3.8.

How to determine the  value? The  value of the column can be obtained based on the equations 1-4 below (BS 8110 Part 2 Cl 2.5.4, Cl 2.5.5 and Cl 2.5.6). Braced column The β of the column for framed structures may be taken as the lesser of : ………

(1)

.…….. (2) Unbraced column The β of the column for framed structures may be taken as the lesser of : ………

c,1 the c,2 the

(3)

……… (4) = ratio of the sum of the column stiffness to the sum of the beam stiffness at lower end of a column = ratio of the sum of the column stiffness to the sum of the beam stiffness at

upper end of a column c ,min = min lesser of c,1 and c,2 In specific cases of relative stiffness, c the following simplifying assumptions may be used: c) connection between column and base designed to resist only nominal moment: c to be taken as 10; d) connection between column and base designed to resist column moment: c to be taken as 1.0; Relative Stiffness, c Relative stiffness will be governed by Maximum Relative Stiffness when the ratio of the sum of the column stiffness to the sum of the beam stiffness, c more than the value set in Project Parameter. By default, the Maximum Relative Stiffness is set to 10. This is to ensure m inimum pin fixity of beam to the column.

Example 1 : Column connected to the footing

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Figure 1: Concrete building framing for example 1 Determine the  value for the braced column C1 which is shown in Figure 1.

Assume fixed foundation,

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Take the smaller , hence  = 0.81. Example 2 : Column unrestrained in one direction

Figure 2: Floor plan for example 2 Determine the  value (y-direction) for the column C1 and C2 in Figure 2 which not restrained by beam in y-direction for two floors. Assum e all column size is 230x450mm and column height = 3m. Column C1 and C2 Refer to Figure 2, column C1 and C2 are not restrained by any beam in y-direction. Although column C1 is restrained by slab, but the program won’t consider slab as an element to restrain the column in y axis. Hence, column C1 will have same condition as column C2. Besides, the beam in x-direction also would not contribute any restrain in y-direction. Hence, the beam stiffness in y direction will be none or 0.

Take the smaller  value = 1 p336

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Couple column For the column which connected to the wall in any direction (x or y direction), the wall is assumed as a constraint element. For instance, refer to diagram 1, it consists of one panel of wall 1/A-B and two column (1,A) and (1,B). Wall 1/A-B has thickness of 150mm and height 3000mm. The c olumn size is 150mmx300mm.

Diagram 1 For β value in x-direction (in plane of wall) for column 1/A and 1/B Since the wall is stiff enough to restrain the column in x-direction, hence c,2 = 0 Assume fixed foundation,

Take the smaller , hence  = 0.75 For  value in y-direction (out of plane of wall) Column 1/A (with beam): p338

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Note : Ignore beam stiffness if cantilever beam

Assume fixed foundation,

Take the smaller β, hence β = 0.75 For  value in y-direction (out of plane of wall) Column 1/B (without beam) : Since no beam is tied with column 1/B in y-direction, therefore β value will refer to adjacent wall  value.

Example:  value for column (1,A) =  wall 1/A-B  value for column (1,C) =  wall 1/B-C

 value for column (1,B) =

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For different thickness of wall panel,  value for column 1/B

 value for column (1,B) =

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Column End Conditions (Out of plane cases) For the computation of column end condition, we need to have the following support information for each column on a plan:  Is the column is touching any slab.  Is the column on the RC Wall  Calculation of BeamK.

The average is taken if b and/or h varies along a beam. If there are more than one beam between two supports, then we have to use weighted approach by using the following formula

Resolving BeamK into X and Y Direction with Reference to Column Local x-y Axes

The calculation of BeamK has to be resolved into x and y direction, BeamKx and BeamKy with reference to column local x-y coordinates. The above example assumes that column local x-y coordinates coincide with the global x and y coordinates and there is no rotation. Example, BeamK for c2 is contributed by b1, Kb1 and b2, Kb2. Kb1 has to be resolved into Kb1x = Kb1cos 2, and Kb1y =Kb1sin 2 . We then add the resolved value into BeamKx and BeamKy respectively. p341

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For column with rotation as shown below, we reference the contributing beam with the column local axis.

We need to take the absolute contribution value Kb for each beam section before summing them in BeamKx or BeamKy. Eg. For the diagram above, Beam stiffness = BeamKx in the global X-axis, The absolute contribution value Kb at the ‘big’ column rotated at wrt global X-axis: Column local BeamKx’ = BeamKx .Cos(  ) and Column local BeamKy’ = BeamKx .Sin(  ) If the beam is also rotated at angle, then is the absolute of the difference of the two angles. Where sin (  ) and cos (  ) are the absolute of them. Contribution from beam that branch-off with other beams

In the example above, the end of b1 branch-off with b2, b3. In this case we ignore the contribution of b2 and b3, but take into consideration of b1 and treat its length as 2xl 1 in the calculation. Similarly, the contribution beams for c2, c3, c4 are b2, b3 and b4 respectively with their length doubled in the calculation. However, for c5 and c6, we only consider the b5 with length l5 since we can find a straight beam from the column in consideration to another support. In the above example, b1 might consist of multiple beam sections or beams before it branch-off. In this case we would take the weighted calculation but with length of each section or beam doubled.

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Curved beam framed into column For curved beams framing into column, the calculation of the beam stiffness has to be resolved into x and y direction. Refer to example for column (G1,4) as shown below.

Column's local axis as shown on plan:

The program will also detect the length of the curved beam which is connected to element that can provide lateral restraint such as columns or walls. Otherwise, the curved beam will not be taken into consideration. Therefore in this example, only the highlighted portion of the beam length are taken into consideration when calculating for beam stiffness 3FB74, 3FB94 and 3FB80. Refer to topic "Column End Conditions (Out of plane cases)" on how the beam stiffness is resolved into x and y direction.

Slenderness Calculation About Y-Y Axis Column Clear Length, lo = 8050.0 mm Lateral Force, V = 0, Use Stability Index Q = 0 for Gravity Load The Element is Braced Total Column Stiffness at Top, I/L = 646998 mm3 Total Column Stiffness at Bottom, I/L = 646998 mm3 Floor Element Type Mark b, mm 3FB Column (G1,4) 500 2FB Column (G1,4) 500

h, mm 500 500

L, mm 3650 4400

Total stiffness of restraint elements (beams) at top part of column, KbTop = 2751825.9 mm3 p343

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No. 1 2

Element Type Beam Beam

Mark 3FB74 3FB80

b, mm 200 200

h, mm 750 750

L, mm 8100 3496

Slenderness Calculation About Z-Z Axis Column Clear Length, lo = 8050.0 mm Lateral Force, V = 0, Use Stability Index Q = 0 for Gravity Load The Element is Braced Total Column Stiffness at Top, I/L = 646998 mm3 Total Column Stiffness at Bottom, I/L = 646998 mm3 Floor Element Type Mark b, mm 3FB Column (G1,4) 500 2FB Column (G1,4) 500

Angle, ° 195.1 17.9

h, mm 500 500

K, mm^3 837945 1913881

L, mm 3650 4400

Total stiffness of restraint elements (beams) at top part of column, KbTop = 8161732.5 mm3 No. Element Type Mark b, mm h, mm L, mm Angle, ° K, mm^3 1 Beam 3FB74 200 750 8100 195.1 226474 2 Beam 3FB80 200 750 3496 17.9 618661 3 Beam 3FB94 200 750 961 90.0 7316597

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Is Cantilever false false

K, mm^3 1426941 1183712

Is Cantilever false false false

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Column dimension for shear design Dimensions used for shear design is depending on the column shape. All the dimensions listed below is the outer dimension (concrete profile), for effective dimension, just use the outer dimension – cover – bar diameter / 2. Column Type

Effective dimension

Rectangle

Rectangular Column Dimensions are defined by using B and H For Shear in X-X Axis DIMX = H DIMY = B

Circular

L-Shape

For Shear in Y-Y Axis DIMX = B DIMY = H

C ircular Column Dimensions are defined by using diameter First an equivalent dimension is calculated base on formula

For Shear in X-X Axis

For Shear in Y-Y Axis

DIMX =

DIMX =

DIMY =

DIMY =

Use longer leg in the shear direction as shear design dimension.

For Shear in X-X Axis DIMX = X-Leg DIMY = X Dimension

T-Shape

For Shear in Y-Y Axis DIMX = Y-Leg DIMY = Y-Dimension

Use longer leg in the shear direction as shear design dimension.

For Shear in X-X Axis DIMX = Flange Width DIMY = Flange

Plus Shape

For Shear in Y-Y Axis DIMX = Leg Width DIMY = Leg

Use longer leg in the shear direction as shear design dimension.

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Column Type

Effective dimension

For Shear in X-X Axis DIMX = X-Leg DIMY = X Dimension

U Shape

For Shear in Y-Y Axis DIMX = Y-Leg DIMY = Y Dimension

Use longer leg in the shear direction as shear design dimension.

For Shear in X-X Axis DIMX = Mid-Leg DIMY = X Dimension

V-Shape

For Shear in Y-Y Axis DIMX = Leg 1 + Leg 2 (Need double check) DIMY = Y Dimension

For V-shape column, the design is divided into 2 conditions, base on the inclined angle of the inclined leg, if the angle 60° <= θ <= 120° and 240° <= θ <= 300°, the inclined portion is used to resist the Shear in Y-Y Axis (Case 1), otherwise both Shear in X-X and Y-Y Axis are assumed resisted by the B-Leg portion (Case 2).

C ase 1:

For Shear in X-X Axis DIMX = B-Leg Width DIMY = B-Leg Length

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For Shear in Y-Y Axis DIMX = V-Leg Width DIMY = V-Leg Length Cosine(α)

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Column Type

Effective dimension C ase 2:

Polygon

For Shear in X-X Axis DIMX = B-Leg Width DIMY = B-Leg Length

For Shear in Y-Y Axis DIMX = B-Leg Length DIMY = B-Leg Width

For the V-Shape column, the profile polygon is converted to equivalent rectangle with dimension b Equal For Shear in X-X Axis

For Shear in Y-Y Axis

DIMX =

DIMX =

DIMY =

DIMY =

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Column Nominal Moment

How to derive the moment capacity of nominal rebar formula Mu/bd^2 = p.fyd(1-k2.p.fyd/k1)? From the “beam design” technical manual we know that to find the moment of resistance of the Parabolic stress block is as below.

Where: K1

=

mean conc. stress

= K2*x

=

depth to the centroid of the stress block

X

=

depth of neutral axis

K2

=

factor of x

=

For equilibrium, the ultimate design moment, M, must be balanced by the moment of resistance of the section so that M = Cc x z = Txz Where z is the lever arm between the resultant forces Cc and T

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When user set the pin support at the beam end, the moment transfer to column will be zerobut due to the nominal rebar provide by the beam, it will create a nominal moment that will transfer to the beam.

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Sub frame beam design - fix at both end support

Sub frame beam design - pin at both end support The explanation is as follows. When there is pin end at the sub frame beam, there must be 0 moment transfer to the column, but due to the nominal rebars provided, there is some moment transfer to the column. If let us say that the maximum fix end moment transfer to the c olumn is 30kNm and the nominal beam end bar can resist 20kNm only, then the c olumn will take maximum of 20kNm only and the rest 10kNm will transfer back to the beam and the beam end nominal steel will start yield.

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After column design, user can see the following example calculation in the full report if there is nominal moment at the column.

COLUMN NOMINAL MOMENT FROM BEAM END RELEASE FOR COLUMN C1:(2,A) Number of Beam with Beam End Moment Released = 3 No. 1 2 3

Beam Mark gb1(200x500) gb3(200x500) gb6(200x500)

Width 200 200 200

Eff. Depth 460.0 460.0 460.0

Release Yes Yes Yes

Note

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Angle 0.00 270.00 321.34

As Prov 157 157 157

Mu 10.5 13.4 13.9 Total :

Muy -10.5 0.0 -10.8 -21.3

Muz 0.0 -13.4 -8.7 -22.1

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Width, Eff. Depth - Beam Width, Beam Effective Depth for Top Bar, mm Angle - Angle (Orientation) of beam to the local Z Axis of Column, Degree AsProv - Top Steel Area Provided of Beam, mm² Mu - Total Moment distributed from the Beam Top Bar, kNm Muz - Component of Derived Moment distributed to the local Z-Z Axis from the Beam Top Bar, kNm Muy - Component of Derived Moment distributed to the local Y-Y Axis from the Beam Top Bar, kNm

[Muz, Muy: Derived lower column nominal moment based on column stiffness and beam angle] Sample Calculation for Nominal Moment using Beam gb6(200x500) Back Calculate Beam End Moment from the Top Rebar of Beam Mu / bd² = p × fyd (1 - k2 × p × fyd / k1) fyd = 438 N/mm² Steel Ratio Provided, p = As / (b × d) = 157 / (200 × 460.0) = 0.0017 Mu / bd² = p × fyd × (1 - k2 × p × fyd / k1) = 0.0017 × 438 × (1 - 0.4518 × 0.0017 × 438 / 12.12) = 0.7271 Nmm / mm³ [Formula that used for compute the Mu value which can be found at pg 53, Examples of the design of building to CP110] Design Moment due to Beam Top Bar, Mu = 0.7271 × 200 × 460² / 1000000 = 30.8 kNm Moment Muy = -30.8 × Cos(321.34°) = -24.0 kNm Moment Muz = 30.8 × Sin(321.34°) = -19.2 kNm [Derived total nominal moment based on beam angle] Beam Relative Stiffness, kBeam = 0.00033m³ [ (bh3/12)/l ] Angle of Beam to Column Local Axis, θ1 = 321.34° Current Column Stiffness = |kzz × Sin(θ1)| + |kyy × Cos(θ1)| = |0.00180 × Sin(321.34°)| + |0.00045 × Cos(321.34°)| = 0.00148 Angle of Beam to Top Column Local Axis, θ2 = 321.34° Top Column Stiffness = |kzz × Sin(θ2)| + |kyy × Cos(θ2)| = |0.00180 × Sin(321.34°)| + |0.00045 × Cos(321.34°)| = 0.00148 [Column stiffness calculation based on beam angle] Moment Muy Transferred to Column/Columns = -24.0 × (0.00148 + 0.00148) / (0.00148 + 0.00148 + 0.00033) = -21.6 kNm Moment Muz Transferred to Column/Columns = -19.2 × (0.00148 + 0.00148) / (0.00148 + 0.00148 + 0.00033) = -17.3 kNm Moment Muy Transferred to Column C1:(2,A) = -21.6 × 0.00148 / (0.00148 + 0.00148) = -10.8 kNm Moment Muz Transferred to Column C1:(2,A) = -17.3 × 0.00148 / (0.00148 + 0.00148) = -8.7 kNm [Distribution of column nominal moment based on column stiffness ]

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Follow Top Column detailing

The “Follow Top Column” is to notify user to follow the upper column design. This message will appear when a column passes through one/few floors (due to unrestrained condition in the one/few floors). In this case, Esteem 8 will design the column by considering the restrain conditions and calculates the exact total clear height of column. Hence, user needs to refer the upper column detailing and reporting for the column in question.

Column reporting for gb-1b

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Figure: Column reporting for 1b-2b

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Slender Column ESTEEM 8 COLUMN DESIGN FOR SLENDERNESS MOMENT For every column in Esteem 8, slenderness moment is consider in the following approach:Column design forces (P, Mx, My) are obtained as describe in topic "User Manual > Main Menu > Design > Design Column ". BRACED CONDITION (NON-SWAY) In the case where the column is braced, the Initial Design Ultimate Moment (Mi) value is obtained from center node of column for every design cases separately (eg. P + Mpattern, Pmin + Mpattern, WLC1, 2… etc) * Conventionally formula 0.4 x M1 + 0.6 x M2 (-ve for M1 if double curvature) is used Where M1 is smaller Moment and M2 is larger Moment Since the shape of the columns in Esteem 8 is not symmetrical, following simplification as stated in BS8110 might not be applicable

1. 2. 3. M1 +

/2

4.

, emin <= h / 20 or 20mm If the column is found double curvature (opposite sign of M1 and M2) following design moment is used for column design. The column design/analysis process will be run twice due to the double curvature behavior (tension face appears in two sides within length of colum n), the selected design moments are:-

1. AbsMax (Mtop,

, M3top)

2. AbsMax (Mbot,

, M3bot) ; M3 being the mid-node moment.

M3 moment selected from either:(if Mi and Mtop/Mbot +

same sign)

/ 2 (if Mtop/Mbot and

is different sign)

UN-BRACED CONDITION (SWAY) Un-braced condition is not considered because it has been taken care by P-Delta effect in Analysis Process. Notes: The determination of braced or unbrace condition is explained in topic "Sway Effects" below. For calculation of effective height of column and  value, refert topic "Computation of c olumn beta value" above.

Determine the column whether is short or slender Short column if leff/h or b ≤ 15 for a braced column leff/h or b ≤ 10 for an unbraced column 3.8.1.3) p355

(Reference BS 8110 : Part 1 Clause

Esteem Innovation Sdn Bhd, 2011

ADDITIONAL DESIGN ULTIMATE MOMENT INDUCED BY DEFLECTION OF COLUMN,

M

add is calculated from formula:-

Where

value refer to following topic below. is depth of cross section measured in the plane under consideration. For other shape

column other than rectangle, a h value used is obtained from moment of area for the section in plan under consideration.

where is second

- Design Ultimate Capacity of a section when subjected to axial load only (

0.45fcu*b*h+0.95fyAsc) - Design Axial Load Capacity of a balanced section (for rectangular section) The iteration process for K value only applicable to rectangular and circular column section, for other non-symmetrical polygonal shape column, L, V, T etc, K will be taken as 1 (conservative)

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Checking of Column Slenderness Limit (BS8110) For every load case, a column is determined whether braced or unbraced. The column slenderness should not exceed the limit specified in BS8110: 1997 Part 1, Cl 3.8.1.7 and Cl 3.8.1.8. 3.8.1.7 Slenderness limits for columns Generally, the clear distance, lo between end restraints should not exceed 60*m inimum thickness of the column. 3.8.1.8 Slenderness of unbraced columns Furthermore, for unbraced columns the clear height should NOT exceed the lesser of:

o 60b or o 100*b^2/h

Notes

: h and b are the larger and smaller dimension of the column respectively.

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Column deflection under ultimate conditions (BS8110) As stated in BS 8110 : Part 1 : Clause 3.8.3.1, the deflection of a rectangular or circular column under ultimate conditions may be taken to be :

(equation 32) whereas

(equation 34) Notes : b' is generally the smaller dimension of the c olumn. For slender column with bi-axial bending in column design, Clause 3.8.3.6 in BS 8110 : Part 1 is referred.

"

" Esteem 8 considers bi-axial bending in column design, hence the appropriate equations are consider to the axis accordingly.

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Rectangular/square column i) Moment in X-direction

ii) Moment in Y-direction

whereas rxx = radius of gyration Ixx = second moment of area

whereas ryy = radius of

A = area

gyration Iyy = second moment of area A = area

hence

hence

With the example above, we can notice that l e and r are the main parameters in order to calculate the beta value, βa. le is the effective height which can be evaluated base on the floor height with the consideration of beam depth/slab thickness constraint. Meanwhile r is the radius of gyration and can be calculated by using the formulas below.

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and

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K value Equation 33 of BS 8110: Part 1 gives the expression,

Esteem 8 will calculate the K value iteratively only for the square, rectangular and circular section. K will be taken as 1 for all other sections.

Notes

:

Taking the example of an L-Shaped column for K value consideration. Generally, this section will be stocky with slenderness ratio to within the short column category. So if the odd shaped column falls under the slender column regime, the more conservative value of K = 1 is a safe option.

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Example for column slender moment calculation Column size = 200x400mm Column height = 10000mm Concrete grade: 30N/mm2

The selected loading case for design is Load Case no.1: Load Load Analysis No. Source P My Mz Case Comb. Type 1 G.L. Full LC1 Top 15.8 0.1 -0.5

T

Vy

Vz

Myi

Mzi

0.0

-0.1

0.0

0.0

-0.1

1. Selection of Moment for Design and Analysis My No Selecte 2D/3D NeMin Mi Madd 2D/3D d 1 0.1 0.2 0.0 2.2 2.3 -0.5

Mz NeMin

Mi

Madd

-0.3

-0.1

-1.1

Sele cted -1.2

2. Column Slenderness Result COLUMN SLENDERNESS RESULT Clause 3.8.1.6 : Effective Height of a column Clause 3.8.3 : Deflection induced moments in solid slender columns Column Dimension in Plane of Bending (About Y-Y Axis), Hyy = 200.0 mm Column Dimension in Plane of Bending (About Z-Z Axis), Hzz = 400.0 mm Slenderness Calculation About Y-Y Axis Column Clear Length, lo = 10000.0 mm Lateral Force, V = 0, Use Stability Index Q = 0 for Gravity Load The Element is Braced

Check either column is braced or unbraced. Compute stability index (ACI 10.11.4.2):

Relative Stiffness at Top of Element, αTop = 0.020 Relative Stiffness at Bottom of Element, αBot = 1.000 β1 = 0.7 + 0.05 × (αTop + αBot) = 0.75 β2 = 0.85 + 0.05 × Min (αTop, αBot) = 0.85 β = Min (β1, β2) = 0.75 < 1.0 Effective Length = leff = β × lo = 7500.0 mm leff / H = 7500.0 / 200.0 = 37.5 >= 15 The Column Is Slender About Y-Y Axis

Calculate column beta values, 1, 2. From the minimum beta, get column effective length, le.

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Deflection of column under ultimate condition, au = βa × Kyy × Hyy Iyy = 266666666.7 mm^4 ryy = √ (Iyy / Ac) = √(266666666.7 / 80000.0) = 57.7 mm βa = (leff / ryy)² / 24000 = (7500.0 / 57.7)² / 24000 = 0.70 Clause 3.8.1.7 Slenderness limits for columns lo / Min(b, h) = 10000 / 200 = 50.0 <= 60 Initial Kyy value = 1.0 Eccentricity, eMin = H / 20 = 200.0 / 20 = 10.0 mm ≤ 20 mm

Calculate column deflection under the ultimate condition, au (BS8110: Part 1 Clause 3.8.3.1) Iyy = bh3/12 = 400*200^3/12 = 266666666.7 mm^4 Calculate minimum eccentricity, eMin

Slenderness Calculation About Z-Z Axis Column Clear Length, lo = 10000.0 mm Lateral Force, V = 0, Use Stability Index Q = 0 for Gravity Load The Element is Braced Relative Stiffness at Top of Element, αTop = 0.079 Relative Stiffness at Bottom of Element, αBot = 1.000 β1 = 0.7 + 0.05 × (αTop + αBot) = 0.75 β2 = 0.85 + 0.05 × Min (αTop, αBot) = 0.85 β = Min (β1, β2) = 0.75 < 1.0 Effective Length = leff = β × lo = 7500.0 mm leff / H = 7500.0 / 400.0 = 18.8 >= 15 The Column Is Slender About Z-Z Axis Deflection of column under ultimate condition, au = βa × Kzz × Hzz Izz = 1066666666.7 mm^4 rzz = √ (Izz / Ac) = √(1066666666.7 / 80000.0) = 115.5 mm βa = (leff / rzz)² / 24000 = (7500.0 / 115.5)² / 24000 = 0.18 Clause 3.8.1.7 Slenderness limits for columns lo / Min(b, h) = 10000 / 200 = 50.0 <= 60 Initial Kzz value = 1.0 Eccentricity, eMin = H / 20 = 400.0 / 20 = 20.0 mm ≤ 20 mm

Calculate column deflection under the ultimate condition, au (BS8110: Part 1 Clause 3.8.3.1) Izz = hb3/12 = 200*400^3/12 = 1066666666.7 mm^4

ITERATION TABLE FOR COLUMN SLENDERNESS MOMENT Design Loading Case No. 1 Slenderness Moment About Z-Z Axis K Mi (kNm) Madd (kNm) 1.00 -0.1 -1.1 1.00 -0.1 -1.1

Mt (kNm) -1.2 -1.2

Nuz (kN) 1278.4 1278.4

Nbal (kN) 562.5 562.5

K-new 1.0 1.0

Slenderness Moment About Y-Y Axis K Mi (kNm) Madd (kNm) 1.00 0.0 2.2 1.00 0.0 2.2

Mt (kNm) 2.3 2.3

Nuz (kN) 1278.4 1278.4

Nbal (kN) 525.0 525.0

K-new 1.0 1.0

2. Sample Calculation for Column Slender Moment a. Slenderness Moment About Z-Z axis Initial K = 1 Mi = 0.1 kNm N = 15.8 kN N*eMin = 15.8*0.02 = 0.3 kNm (value may differ due to decimal places) au = βa × Kzz × Hzz = 0.18*1.0*0.4 = 0.072mm Madd = N*au = 15.8*0.072 = 1.14 kNm Mt = Mi + Madd = 0.1 + 1.14 = 1.23 kNm Nuz = 0.67fcu*b*h*/c + fy*Asc/s = 0.67*30*200*400/1.5 + 460*471.2/1.05 = 1278430.5 = 1278.4 kN

Effective depth, d is taken by considering up to concrete cover only and not to the center of the main bar. This is due to unidentified main bar diameter and link diameter during the iteration process on determination of slenderness moment, and uneven column shape (maybe polygonal). Note: Concrete design stress = 0.67fcu/c

Nbal = 0.25*fcu*b*d =0.25*30*200*(400-25) =562500 N = 562.5 kN p363

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Reinforcement design stress = fy/s

= (1278.4 - 15.8)/(1278.4 - 562.5) = 1.76  1 1. Mz = 0.5 kNm 2. Mi + Madd = 1.23 kNm 3. N*eMin = 0.3 kNm

Based on the obtained bending moments, get the maximum bending moment value:

Selected moment = 1.23 kNm

1. Mz 2. Mzi+Madd 3. N*eMin(minimum eccentricity distance)

b. Slenderness Moment About Y-Y axis

Note: Nuz = 0.45*fcu*Ac + 0.95*fy*Asc (having included the allowance of m) as per defined in Eq. 33 of BS8110 if used may yield a different value as to Nuz = 0.67fcu*b*h*/c + fy*Asc/s (due to rounding of the former to two decimal places)

Initial K = 1 Mi = 0.0 kNm N = 15.8 kN N*eMin = 15.8*0.01 = 0.2 kNm au = βa × Kzz × Hzz = 0.70*1.0*0.2 = 0.14mm Madd = N*au = 15.8*0.14 = 2.21 kNm Mt = Mi + Madd = 0.0 + 2.21 = 2.21 kNm (value differ due to decimal places) Nuz = 0.67fcu*b*h*/c + fy*Asc/s = 0.67*30*400*200/1.5 + 460*471.2/1.05 = 1278430.5 = 1278.4 kN Nbal = 0.25*fcu*b*d =0.25*30*400*(200-25) =525000 N = 525 kN

= (1278.4 - 15.8)/(1278.4 - 525) = 1.67  1 4. My = 0.1 kNm 5. Mi + Madd = 2.21 kNm 6. N*eMin = 0.2 kNm

Based on the obtained bending moments, get the maximum bending moment value:

Selected moment = 2.21 kNm (value may differ due to decimal places)

Notes

1. My 2. Myi+Madd 3. N*eMin(minimum eccentricity distance)

: Value may differ with Report due to decimal places.

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Sway Effects Esteem 8 can automatically perform a sway check to determine whether the column is "braced" or "unbraced" through the concept of "column stability index". Feature: The sway effects check is used to determ ine the bracing condition of the column/ wall. Specification: The sway coefficient will be used to determ ine whether the column/ wall is braced (non- sway) or unbraced (sway). Reference: 1. ACI 3148/318R -99: Section 10.11.2 2. “Braced column stability index”. MC Hee According to ACI 318/318R-99: Section 10:11.21.2006

Sway coefficient,

= Sum of ultimate vertical load (kN) = Drift (relative floor displacement (mm) = Sum of ultimate floor shear (kN) = Floor Height

* To check on individual element in any axis rather than on whole floor. The rational behind is that in some building configuration although a particular floor may be braced or unbraced, a particular element (column) may not be so. Value to read from analysis results (separate calculation for different axis) = drift node displacement (factored) = ultimate shear or the column (internal value) support the floor. = floor to floor height.

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(Unbraced) otherwise BRACED.

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No column design when stump height is zero No column design will be done if in the column Object Viewer, the Default Stump is set to False and Stump Height is set to zero as shown below.

For instance;

Therefore, after running the column analysis for the above structure, the below result will be obtained.

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In Full Report :

Notes

:

Since there is no column design when Default Stump = False and Stump Height = 0, there will be no design for foundation as well.

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Wall Effective Length of Wall in Design of RC Wall Two cases where wall is stressed locally; 1. Wall subjected to high out of plane moment 2. Wall subjected to heavy point load



Wall subjected to high out of plane moment How effective is the long wall panel in taking care of high out of plane moment? 1. Create two models below for comparison. a) Model A: Model wall as long panel 1/A-D. b) Model B: Model as single panels 4/A-B, 4/B-C and 4/C-D connected to each other.

2. A heavy point load on the transverse beams gb1 and gb6 will produce high out of plane moment and shear to the wall. 3. Compare the detailings between the two models. Comparison of wall detailings for Model A and Model B:

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Model A:

Model B:

Discussions: o Due to high localized out of plane moment, the wall may have overstressed locally. o Framing beam is introduced in the wall to help resist the moment and shear. o Wall panel in Model A will be analysed and designed as a single piece of wall. Therefore in the design, the bars wil be distributed evenly throughout the whole panel length. o In Model B, the walls are designed as three pieces of wall panels. Therefore every single wall panel will have the bars distributed according to their own sets of data. Thus the reinforcement in panel 4/A-B is heavier compared to panels 4/B-C and 4/C-D.



Wall subjected to heavy point load What is the effective length/width of the wall (relative to height) that is able to resist the localized stresses? 1. Create two models below. a) Model A: Model as wall panel 1/A-D. b) Model B: Model as single wall panels connected together 4/A-B, 4/B-C and 4/C-D.

2. A heavy point load is applied on the wall panel. p371

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3. Compare the detailings between the two models. Comparison of wall detailings for Model A and Model B: Model A:

Model B:

Discussions: o Similarly for wall subjected to heavy point load, the wall may be overstressed locally if the stresses are not distributed efficiently throughout the whole panel. o Since wall panel in Model A is analyzed and designed as a single piece of wall, the bars will be distributed evenly throughout the whole panel length. o In Model B, the walls are designed as three pieces of wall panels. Therefore every single panel will have the bars distributed according to their own sets of data. Thus the reinforcement in panel 4/B-C is heavier compared to panel 4/A-B and 4/C-D. Conclusions: Both models can produce correct design results as both cases are designed to their section capacity. Somehow modeling the wall panel as in Model B in each case will produce more conservative detailing compared to Model A.

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Wall Detailing Wall Detailing with small openings Wall opening will be reinforced with vertical and horizontal bars at top and side, and also diagonal bar to avoid cracking at the corner of the opening. Below show elevation view of the wall detailing.

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Wall Not Design

Figure 1: Upper floor with shear wall

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Figure 2: Lower floor with transfer beam which will support all the walls from upper wall

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Wall 1 Wall is Not design due to Transfer wall with no side support

Wall 2 p376

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Wall is Not design due to Transfer wall with no side support

Wall 3 Wall is Not design due to Design Failed

Wall 4 Wall is Not design due to Intermediate support in transfer wall

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Wall 5 Wall is Not design due to Intermediate support in transfer wall

Wall 6 Wall is Design with detailing shown

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Not Design due to short transfer wall This is the model with two floor height 3000 mm and span length is 8000 mm. The wall for the first floor will be labelled as "Not Design due to short transfer wall". It is because 3000 mm height is not sufficient to form Strut and Tie in correspondence to 8000 mm length of transfer beam. However, user may go to Project parameter -> Design> wall -> Allow short transfer wall -> select TRUE. Then, run the analysis and design. The wall for first floor will be shown with the detailing. It is because the software now will treat the wall for first floor and second floor as one monolithic wall. So, the required height to form strut and tie is sufficient. The condition to form the strut and tie for first and second floor as one monolithic wall: i) continuity of upper wall and lower wall ii) both of the wall is having same length, same thickness, no offset and no opening

Figure 1: The model

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Figure 2: The wall for first floor is Not Design due to short transfer wall

Figure 3: The required height to form strut and tie for this model is 4618.8 mm

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Figure 4: The wall detailing for second floor is not affected and been shown accordingly

Figure 5: True for Allow Short Transfer Wall

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Figure 6: Wall detailing at first floor is shown after turn to TRUE for Allow Short Transfer Wall

Figure 7: Wall detailing at second floor

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Wall bypass intermediate beam within two support The wall should be able to show the detailing even though there are intermediate beams in between.

However, even after turn it TRUE to allow short transfer wall , the wall still fail and it could be the following reason: 1. If lower transfer wall fail, upper wall which related to lower Strut & Tie will be marked as fail as well "Failure due to lower wall fail in Strut and Tie" 2. If One of the upper related wall which the strut of lower transfer wall is occupied fail, the lower transfer wall marked as fail "Failure in Strut and Tie due to upper wall design fail" 3. For multiple walls on top, or cannot fulfill the conditional of multiple floor strut and tie, the lower transfer wall remains as "Not Design due to short transfer wall". For Example:

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Links for Containment of Large Amounts of Compression Reinforcement of Walls In some cases, hooks will be provided with equal bar size and spacing to horizontal reinforcement in walls as to comply to Clause 3.12.7.58 in BS8110:Part 1. 3.12.7.5 Links for containment of large amounts of compression reinforcement in walls When the vertical compression reinforcement exceeds 2 %, links at least 6 mm or one-quarter the size of the largest compression bar should be provided through the thickness of the wall. The spacing of links should not exceed twice the wall thickness in either the horizontal or vertical direction. In the vertical direction it should be not greater than 16 times the bar size. All vertical compression bars should be enclosed by a link. No bar should be further than 200 mm from a restrained bar, at which a link passes round the bar with an included angle of not more than 90°.

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Wall Design and Detailing using Strut and Tie Model Example:

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Wt is the is the width of the tie and lb is the length of the support. Base on the figure above, Tension force T =2025/tan (60) =2025x 0.577 = 1169kN Compression Force in the strut,C = 2025/sin (60) = 2338kN Assume the cover for the beam is 25mm, the width of tie is the distance from the top reinforcement to bottom reinforcement = 500-25x2=450mm wt= 450mm ws= wtc os60+lbsin60 = 450cos60+500sin60 = 658mm The effective concrete strength in a nodal zone is fcu= 0.45nf'c

n=0.8 fcu=0.45x0.8x35 = 12.6 N/mm2 Compression stress of strut = =11.84N/mm2 < 12.6N/ mm2

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Support Width for Strut and Tie Model Support width is determined based on the supporting beam width and the supporting column offset values. In the following example, changes in column offset values give different support widths which are to be used in the wall strut and tie design. The dimension of adjacent beam connected to the wall is 200mm x 500mm. The c olumn size which supports the wall is 700mm x 300mm.

Column offset (From Wall Detailing) 1. Column and beam with no offset value

Column support width (Full report) STRUT AND TIE DESIGN FOR TRANSFER WALL Left Strut Angle (Gravity), θLG = 60.0° Right Strut Angle (Gravity), θRG = 60.0° Left Strut Angle (Lateral Loads), θLWN = 60.0° Right Strut Angle (Lateral Loads), θRWN = 60.0° Left Support Width, wSup(L) = 450.0 mm Right Support Width, wSup(R) = 450.0 mm

Support width = half column width + half beam width = 700/2 + 200/2 = 450mm 2. Beam not offset, column offset to the left at 250mm (-250mm)

STRUT AND TIE DESIGN FOR TRANSFER WALL Left Strut Angle (Gravity), θLG = 60.0° Right Strut Angle (Gravity), θRG = 60.0° Left Strut Angle (Lateral Loads), θLWN = 60.0° Right Strut Angle (Lateral Loads), θRWN = 60.0°

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Left Support Width, wSup(L) = 200.0 mm Right Support Width, wSup(R) = 200.0 mm

Support width = (half column width - offset value) + half beam width = (700/2 - 250) + 200/2 = 200mm Internal column support (From Wall Detailing) 1. Internal column with no offset value

Column support width (Full report) STRUT AND TIE DESIGN FOR TRANSFER WALL Left Strut Angle (Gravity), θLG = 60.0° Right Strut Angle (Gravity), θRG = 60.0° Left Strut Angle (Lateral Loads), θLWN = 60.0° Right Strut Angle (Lateral Loads), θRWN = 60.0° Left Support Width, wSup(L) = 300.0 mm Right Support Width, wSup(R) = 300.0 mm

Support width = half column width = 600/2 = 300mm

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Shear Contour Centroid Below is a simple model to explain the shear contour centroid where there is a transfer wall (A/2-3) sitting on the beam. Length of wall panel = 2000mm (to centre of support).

The default setting in project parameter for shear contour centroid

This is the report in design result> wall> Full report

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After running the analysis, select on 3D Analysis> right-click> select Nxy from the contour selection

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Example: Right support is a beam To check for the Shear Contour Centroid at the right support, the cut should be at the distance of 1900 mm from the begin grid (100 mm cut from the support)for right support is beam.

Then, the nearest centroid to test is the one nearest to the starting of the cutline, 1A (remember that this force graph is the vertical cut line made to lie sideways where grid mark 1A is the bottom of the wall and grid mark 1b is the top of the wall). Shear contour centroid The first centroid's distance from grid mark 1A is 386. So the percentage for the distance over the wall height is 386/3000 x 100 = 12.86 % Shear Different % Check p394

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Percentage of right centroid shear difference to total shear value = |-87.42-(-81.57)| / |-81.57| x 100 = 7.17 % Allowable shear contour centroid difference The different percentage between two shear contour centroid for right and left support. It is applicable when both end are beam support.

Notes

:

The value that shown in the report will be slightly difference when compare to force graph diagram due to the position of start and end point of the cutline internally.

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Wall Region sign for forces When user input a P point load on a single wall panel 20kN, the value in the wall region for bottom edge will be 20kN and -20kN for top edge. Below show is the sign of the wall region.

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Wall Region local axis for Vx,Vy

In the wall, Project data-3D report user can view the result of the wall region and wall panel. The Vx value is display in the local axis direction. In order to sum all the value, user need to do some conversion. Based on the figure above, for the vertically orientate wall local Vx = y-axis, local Vy = -ve x-axis for the horizontally orientate wall local Vx = x-axis , local Vy = y-axis

Example:

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In order to sum all the result, user need to change as shown below: Wall 1/A-B(Vertical) A/1-2 (Horizontal) 2/A-B (Vertical) B/1-2 (Horizontal) Sum:

Global Vx 0.41 0 -0.21 -0.21 0

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Global Vy 0 0.41 -0.21 -0.21 0

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Wall Region local axis for Mx and My

The longer span of wall is always the local X-axis.

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How RC Wall Self Weight in 3D Mesh Differs from Self Weight in 2D Mesh How RC Wall Self Weight in 3D Mesh Differs from Self Weight in 2D Mesh

Referring to the picture above, Wall 1, Wall 2 and Wall 3 are 3000 mm long and 250 mm thick. That would mean the self weights are Calculation : Height x Length x Thickness x Concrete Density Wall 1: 3 x 3 x 0.25 x 24 = 54 kN Wall 2: 3 x 3 x 0.25 x 24 = 54 kN Wall 3: 3 x 3 x 0.25 x 24 = 54 kN Total: 162 kN This is what we will get in 2D mesh.

However in 3D mesh, because Wall 1 and Wall 2 are located on the same grid, we use an average offset to create the plane which contains Wall 1 and Wall 2. This "offset" could be caused by different wall thickness or different "specific wall offset" in the plane of consideration (could p400

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either be in the same floor or between floors). As an example, Wall 1 offset = 0 mm Wall 2 offset = -100 mm Average offset = (0 + -100)/2 = -50 mm This would mean that Wall 3 would be shorten by 50 mm because Wall 1 will be shifted by 50 mm for 3D mesh generation. So the self weights for the RC walls in 3D mesh would be Calculation : Height x Length x Thickness x Concrete Density Wall 1: 3 x 3 x 0.25 x 24 = 54 kN Wall 2: 3 x 3 x 0.25 x 24 = 54 kN Wall 3: 3 x 2.95 x 0.25 x 24 = 53.1 kN Total: 161.1 kN

So for this case, the RC walls self weight in 3D mesh will be 0.9 kN less than the self weight in 2D mesh. If the offset is positive 100 mm, then it will be 0.9 kN more.

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Wall Restrain Condition 1. Presence of beam o The beam can only be considered as a restrained beam if the connection angle between wall and beam is within 60° ≤ x° ≤ 120°.

o The stiffness of slanting beams will be SB*|sin x|. o For a group of beams attached to a wall, the beam stiffness will be ∑ SB*|sin x|. o Curve beam would not be considered as restrained wall element. 2. Presence of wall o The wall can only be considered as a restrained wall if the connection angle between two walls is within 60° ≤ x° ≤ 120°.

o The wall only can be considered as a restrained wall if the transverse wall length > 0.1*wall length

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Wall Effective Height The wall effective height can be determined using the procedure for computation of c olumn effective height in BS8110: Part 1 1997, Clause 3.8. Computation of relative stiffness in determination of effective height factor,  1. Beam and Wall Stiffness Consideration o The contribution from transverse beam or wall stiffness is only considered when both wall ends are restrained by transverse beam or transverse wall. o The beam is defined as wall restraint if they are restraint by another vertical element (Column or Wall) 2. Presence of both transverse wall and transverse beam For a wall which is restrained by transverse beam and transverse wall, the contribution from the transverse wall stiffness is taken as the minimum of the transverse wall stiffness and transverse beam stiffness. Refer to example below:

Top of wall 1/A-B: Slab stiffness = 100mm Wall thickness = 100mm Wall clear height = 3000mm

Slenderness Result Clear Wall Height, lo = 3000 mm Top Element Stiffness, KeTop = ∑KsTop + ∑KbTop + ∑KtwTop = 16200000.0 mm³ Bottom Foundation Stiffness, KeBot = 1333333.3 mm³ Wall Stiffness, Kw = 1333333.3 mm³

Restraint Elements at the top part of wall zone Elememt Width a, Depth b, Length L, No. Mark Angle α, ° K Type mm mm mm 1 Slab FS1:1 1601 100 2000 90.0 0 2 Beam gb2 - 1 150 600 2000 90.0 8100000 67500000 3 TWall B/1-2 100 3000 2000 90.0 0

Total element stiffness at top = slab stiffness + transverse beam stiffness + min(transverse wall stiffness + transverse beam stiffness) = Ks + Kb + min(Ktw, Kb) = 8100000 + min (8100000 + 675000000) = 16200000 mm³ Percentage of slab FS1 touching wall is < 80%, the slab stiffness is not taken into account (Ks = 0).

Note K = 0 when the element is skipped from computation of restrain condition, please refer to the technical documentation K = |Sin α| × [a × b³ / (2 × L)]

2. Presence of non-orthogonal slab o When non-orthogonal slab is attached to the wall, the length of the slab is taken as the far most distance of the slab from the wall, within the length of the wall.

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o Refer to topic Example 3: Wall Connected to Non-Orthogonal Members below.

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Esteem Innovation Sdn Bhd, 2011

Example 1: Wall Connected to Foundation The example below will explain the contribution of restraining element stiffness in determining the wall effective height in Esteem 8. o Example 1 explains the computation of wall 3/A1-A2 at floor GB which is connected to foundation. o Example 2 explains the computation of wall 3/A1-A2 which is continuous from floor 1B to 2B. Procedure: 1. Determine the wall restrain condition at top and bottom part of the wall. 2. Calculate contribution of top and bottom element stiffnesses in determining the wall effective height.

Plan View

Description

Remarks

Floor GB:

Plan view of wall 3/A1-A2 at GB floor.

Floor 1B:

Plan view of wall 3/A1-A2 at 1B floor (wall is continuous from this floor to floor 2B).

Floor 2B:

Plan view of wall 3/A1-A2 at floor 2B (the bottom part of wall is continuous from floor 1B).

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Description

Remarks

3D View

3D view of wall 3/A1-A2.

Determination of contribution from restraining elements: Example 1: Wall 3/A1-A2 at floor GB. Description/Details

Remark Wall properties

Wall Panel Thickness, t = 150 mm

Sub Panel 1 (Reg. 1)

Wall thickness = 150mm Wall length = 2300mm C lear wall height from floor GB to foundation level = 3000mm

Location in Panel: Wall Zone Length of Sub-Panel, b = 2300 mm Gross Concrete Area, Ac = b × t = 345000 mm²

Wall stiffness = 2300*150^3/3000 = 2587500 mm³

[$X$IF_CON1]RC Wall: 3/A1-A2 (Thickness = 150)

[$X$IF_CON1]In Display mode,

PANEL 0 PANEL TOP Left Walls :

select wall 3/A1-A2 and click Wall Restraint Info. The wall is considered as restraint at top and

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Description/Details

Remark

Right Walls: Left Beam Sections : gb7(200x500) - 1; Middle Beam Sections: Right Beam Sections : gb10(200x500) - 1; Left Sub-slabs : Right Sub-slabs: FS2:1 (Thick = 125, PercentTouch = 100; RESTRAIN INFO Left Wall Exist = FALSE; LeftBeam = 7.41; Right Wall Exist = FALSE; Right Beam = 7.41; SlabThick = 125.00; Slab Percentage = 100.00; MidBeam = 0.00; Slab/Wall Ratio = 0.83; Wall Fixity Case = 12; Fixity Condition = 2; Alternative Fixity Condition = 1;

bottom.

Info for Slab FS2: Ratio of thickness of slab FS2 to wall 3/A1-A2 = 125/150 = 0.83 Slab percentage touching wall = 2300/wall sub-panel length = 2300/2300*100 = 100% Thus the slab stiffness is taken into consideration.

PANEL IS RESTRAINED! PANEL BOTTOM Left Walls : Right Walls: Left Beam Sections : Middle Beam Sections: Right Beam Sections : Left Sub-slabs : Right Sub-slabs: RESTRAIN INFO Left Wall Exist = FALSE; LeftBeam = 0.00; Right Wall Exist = FALSE; Right Beam = 0.00; SlabThick = 0.00; Slab Percentage = 0.00; MidBeam = 0.00; Slab/Wall Ratio = 0.00; Wall Fixity Case = 1; Fixity Condition = 1; Alternative Fixity Condition = 1; PANEL IS RESTRAINED!

Slenderness Result Wall Clear Height, lo = 3000 mm T op Element Stiffness, KeTop = ∑KsTop + ∑KbTop + ∑KtwT op = 3455528.8 mm³ Bottom Foundation Stiffness, KeBot = 2587500.0 mm³ Wall Stiffness, Kw = 2587500.0 mm³

C alculate total element stiffness at top and bottom of wall. Slab stiffness, KsTop: KsTop from FS2: = |sin α|*width*thickness^3/(2L) = sin 90*2300*125^3/(2*650) = 3455529 mm³

Restraint Elements at the top part of wall zone Elememt Width a, Depth Length Angle No. Mark K Type mm b, mm L, mm α, ° 1 Slab FS2:1 2300 125 650 90.0 3455529 Beam stiffness, KbTop: KbTop from each beam gb7 and 2 Beam gb7(200x500) - 1 0 0 0 90.0 0 gb10 are not included since these gb10(200x500) beams are not connected to any 3 Beam 0 0 0 90.0 0 restraint element (column, wall). 1 Note K = 0 when the element is skipped from computation of restrain condition, please refer to the technical documentation K = |Sin α| × [a × b³ / (2 × L)]

Refer to topic "Wall Connected to Beam that is Not Attached to Lateral Restraint Element". Total element stiffness on top of wall, KeTop = ∑KsTop + ∑KbTop + ∑KtwTop = 3455529 + 0 + 0 = 3455529 mm³

Notes

:

Bottom part of wall is connected to foundation (Fixed). Therefore, the stiffness for the bottom part will consider only from the wall. Take KeBot = 2587500 mm³ Determine wall effective length for Braced condition.

Effective Length Factor, β for Braced Condition Relative Stiffness at Top of Element, αTop = 0.749 Relative Stiffness at Bottom of Element, αBot = 1.000 β1 = 0.7 + 0.05 × (αTop + αBot) = 0.79 β2 = 0.85 + 0.05 × Min (αTop, αBot) = 0.89

Relative stiffness at top = wall stiffness/top element stiffness

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Description/Details

Remark = Kw/(∑KsTop+∑KbTop+∑KtwTop) = 2587500/3455529 = 0.0749

β = Min (β1, β2) = 0.79 < 1.0

Relative stiffness at bottom = wall stiffness/bottom element = Kw/(∑KsBot+∑KbBot+∑KtwBot) stiffness = 2587500/2587500 = 1.000 (α for fixed foundation = 1.0) C alculate β1 and β2 using formula: β1 = 0.7 + 0.05 × (αTop + αBot) β2 = 0.85 + 0.05 × Min (αTop, αBot) β = Min (β1, β2) Determine wall effective length for Unbraced condition.

Effective Length Factor, β for Unbraced Condition Relative Stiffness at Top of Element, αTop = 0.749 Relative Stiffness at Bottom of Element, αBot = 1.000 β1 = 1.0 + 0.15 × (αTop + αBot) = 1.26 β2 = 2.0 + 0.3 × Min (αTop, αBot) = 2.22 β = Min (β1, β2) = 1.26

Relative stiffness at top = wall stiffness/top element stiffness = 2587500/3455529 = 0.0749 Relative stiffness at bottom = wall stiffness/bottom element stiffness = 2587500/2587500 = 1.000 (α for fixed foundation = 1.0) C alculate β1 and β2 using formula: β1 = 1.0 + 0.15 × (αTop + αBot) β2 = 2.0 + 0.3 × Min (αTop, αBot) β = Min (β1, β2)

C ompute the wall effective height Wall Effective Height for Braced Condition, le = ßBraced × lo = 2370 mm Wall Effective Height for Unbraced Condition, le = ßUnbraced × lo = 3780 mm for Braced and Unbraced condition. le = ß*lo

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Example 2: Continuous Wall Across Multiple Floors Use the same model from Example 1. Determine the wall effective height for wall 3/A1- A2 which is continuous from floor 1B to floor 2B. You can study the example "WallEffectiveHeight.zip" provided in the installation CD under sub-folder "C:\Program Files\Esteem\Esteem7 ULOCK T .Net 2.0\7.5.xx.0\Help\Examples\WallEffectiveHeight.zip" for clarity.

Plan View

Description

Remarks

Floor GB:

Plan view of wall 3/A1-A2 at GB floor.

Floor 1B:

Plan view of wall 3/A1-A2 at 1B floor (wall is continuous from this floor to floor 2B).

Floor 2B:

Plan view of wall 3/A1-A2 at floor 2B (the bottom part of wall is continuous from floor 1B).

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Description

Remarks

3D View

3D view of wall 3/A1-A2.

Determination of contribution from restraining elements: Example 2: Wall 3/A1-A2 from floor 1B to floor 2B. Description/Details

Remark Wall properties

Wall Panel Thickness, t = 150 mm

Sub Panel 1 (Reg. 1)

Total clear wall height from floor 1B to floor 2B = 3000mm

Location in Panel: Wall Zone Length of Sub-Panel, b = 2300 mm Gross Concrete Area, Ac = b × t = 345000 mm² [$X$IF_CON1]RC Wall: 3/A1-A2 (Thickness = 150)

[$X$IF_CON1]In Display mode,

PANEL 0 PANEL TOP Left Walls : Right Walls: Left Beam Sections : Middle Beam Sections: Right Beam Sections : Left Sub-slabs : Right Sub-slabs:

select wall 3/A1-A2 and click Wall Restraint Info. The wall is restraint at top and bottom.

RESTRAIN INFO Left Wall Exist = FALSE; LeftBeam = 0.00; Right Wall Exist = FALSE; Right Beam = 0.00; SlabThick = 0.00; Slab Percentage = 0.00; MidBeam = 0.00; Slab/Wall Ratio = 0.00; Wall Fixity Case = 30; Fixity Condition = 4; Alternative Fixity Condition = 4;

Slab at the top of wall: No slab connected to wall at floor 1B, thus slab percentage = 0

PANEL IS NOT RESTRAINED! PANEL BOTTOM Left Walls : Right Walls: Left Beam Sections : gb7(200x500) - 1; Middle Beam Sections: Right Beam Sections : gb10(200x500) - 1; Left Sub-slabs : Right Sub-slabs: FS2:1 (Thick = 125, PercentTouch = 100;

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Description/Details

Remark

RESTRAIN INFO Left Wall Exist = FALSE; LeftBeam = 7.41; Right Wall Exist = FALSE; Right Beam = 7.41; SlabThick = 125.00; Slab Percentage = 100.00; MidBeam = 0.00; Slab/Wall Ratio = 0.83; Wall Fixity Case = 12; Fixity Condition = 2; Alternative Fixity Condition = 1;

Slab at the bottom of wall: Slab percentage touching wall = 2300/wall sub-panel length = 2300/2300*100 = 100%

PANEL IS RESTRAINED!

Slenderness Result Wall Clear Height, lo = 3000 mm Top Element Stiffness, KeTop = ∑KsTop + ∑KbTop + ∑Ktw Top = 18161411.2 mm³ Bottom Element Stiffness, KeBot = ∑KsBot + ∑KbBot + ∑Ktw Bot = 3455528.8 mm³ Wall Stiffness, Kw = 2587500.0 mm³

C alculate element stiffness at top of wall. Slab stiffness, KsTop: KsTop from FS2: = [|sin |* width*thickness^3/(2L)] = 1*|sin 90|*2300*125^3/(2*650) = 3455529 mm³

Restraint Elements at the top part of wall zone Elemem Width Depth Length No. Mark Angle α, ° K t Type a, mm b, mm L, mm 1 Slab FS2:1 2300 125 650 90.0 3455529 Beam stiffness, KbTop: 2 Beam 2b7(200x500) - 1 0 0 0 90.0 0 KbTop from each beam gb7 and 2b13(200x500) 3 Beam 200 500 1700 90.0 7352941 gb10 are not included since these 1 beams are not connected to any restraint element (column, wall). 2b10(200x500) 4 Beam 0 0 0 90.0 0 Refer to topic "Wall Connected to 1 Beam that is Not Attached to Lateral 2b14(200x500) 5 Beam 200 500 1700 90.0 7352941 Restraint Element". 1 Note K = 0 when the element is skipped from computation of restrain condition, please refer to the technical documentation K = |Sin α| × [a × b³ / (2 × L)]

KbTop from each beam 2b13 and 2b14: |sin |* 200*500^3/(2*1700) = |sin 90|*200*500^3/(2*1700) = 7352941 mm³ Total element stiffness on top of wall, KeTop = ∑KsTop + ∑KbTop + ∑KtwTop = 3455529 + 0 + 7352941*2 = 18161411 mm³

C alculate element stiffness at top of Restraint Elements at the bottom part of wall zone wall. Elememt Width a, Depth Length Angle α, No. Mark K Type mm b, mm L, mm ° Slab stiffness, KsBot: 1 Slab FS2:1 2300 125 650 90.0 3455529 = [|sin |* 2 Beam gb7(200x500) - 1 0 0 0 90.0 0 width*thickness^3/(2L)] gb10(200x500) = 1*|sin 90|*2300*125^3/(2*650) 3 Beam 0 0 0 90.0 0 1 = 3455529 mm³ Note K = 0 when the element is skipped from computation of restrain condition, please refer to the technical documentation K = |Sin α| × [a × b³ / (2 × L)]

Total element stiffness on top of wall, KeBot = ∑KsTop + ∑KbTop + ∑KtwTop = 3455529 + 0 + 0 = 3455529 mm³ Determine wall effective length for Braced condition.

Effective Length Factor, β for Braced Condition Relative Stiffness at Top of Element, αTop = 0.142 Relative Stiffness at Bottom of Element, αBot = 0.749 β1 = 0.7 + 0.05 × (αTop + αBot) = 0.74 β2 = 0.85 + 0.05 × Min (αTop, αBot) = 0.86 β = Min (β1, β2) = 0.74 < 1.0

Relative stiffness at top = wall stiffness/top element stiffness = Kw/(∑KsTop+∑KbTop+∑KtwTop) = 2587500/18161411 = 0.142 Relative stiffness at bottom = wall stiffness/bottom element stiffness = Kw/(∑KsBot+∑KbBot+∑KtwBot) = 2587500/3455529 = 0.749 C alculate β1 and β2 using formula:

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Description/Details

Remark

β1 = 0.7 + 0.05 × (αTop + αBot) β2 = 0.85 + 0.05 × Min (αTop, αBot) β = Min (β1, β2) Determine wall effective length for Unbraced condition.

Effective Length Factor, β for Unbraced Condition Relative Stiffness at Top of Element, αTop = 0.142 Relative Stiffness at Bottom of Element, αBot = 0.749 β1 = 1.0 + 0.15 × (αTop + αBot) = 1.13 β2 = 2.0 + 0.3 × Min (αTop, αBot) = 2.04 β = Min (β1, β2) = 1.13

Relative stiffness at top = wall stiffness/top element stiffness = Kw/(∑KsTop+∑KbTop+∑KtwTop) = 2587500/18161411 =0.142 Relative stiffness at bottom = wall stiffness/bottom element stiffness = Kw/(∑KsBot+∑KbBot+∑KtwBot) = 2587500/3455529 = 0.749 C alculate β1 and β2 using formula: β1 = 1.0 + 0.15 × (αTop + αBot) β2 = 2.0 + 0.3 × Min (αTop, αBot) β = Min (β1, β2)

C ompute the wall effective height Wall Effective Height for Braced Condition, le = ßBraced × lo = 2220 mm Wall Effective Height for Unbraced Condition, le = ßUnbraced × lo = 3390 mm for Braced and Unbraced condition. le = ß*lo

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Example 3: Wall Connected to Non-Orthogonal Members Wall in model below is connected to transverse beams 1b3 and 1b5 at angle 109.3.

Plan View

Description

Remarks

Floor 1B:

Plan view of wall 2/B-B1 at floor 1B. Both ends of wall 2/B-B1 is restraint by transverse beams 1b3 and 1b5. Both beams are at angle,x=109.3 to the wall. Transverse beams are at angle: 60° ≤ x° ≤ 120°. Thus their stiffness contribution are considered. Refer to Reports below.

For non-orthogonal slab, the length is taken as the far most distance from the slab edge to the wall, within the wall length. Example: Slab FS5

Determination of contribution from restraining elements: Example 1: Wall 2/B-B1 at floor 1B. Description/Details

Remark Wall properties

Wall Panel Thickness, t = 150 mm

Sub Panel 1 (Reg. 1)

Wall height = 3000mm

Location in Panel: Wall Zone Length of Sub-Panel, b = 3000 mm Gross Concrete Area, Ac = b × t = 450000 mm² p413

Esteem Innovation Sdn Bhd, 2011

Description/Details

Remark

Slenderness Result Clear Wall Height, lo = 3000 mm Top Element Stiffness, KeTop = ∑KsTop + ∑KbTop + ∑KtwTop = 11677020.3 mm³ Bottom Foundation Stiffness, KeBot = 3375000.0 mm³ Wall Stiffness, Kw = 3375000.0 mm³ Restraint Elements at the top part of wall zone Width a, Depth Length L, Angle No. Elememt Type Mark mm b, mm mm α, ° 1 Slab FS5:1 3000 150 4720 90.0 2 Beam 1b3 - 1 200 500 4009 109.3 3 Beam 1b5 - 1 200 500 5000 109.3 4 TWall B/1A-2 150 3000 2000 90.0 B1/1A5 TWall 150 3000 2000 90.0 2

K 1072647 2942709 2359477 1012500000 1012500000

The contribution from non-orthogonal slab FS5 is taken as the slab's far most length from the wall (distance = 4720mm) at 90. Total contribution from stiffness of beam 1b3 and 1b5 are taken into account. Total contribution from transverse walls are taken as the minimum of: - total contribution from transverse beams 1b3 and 1b5, and - total contribution from transverse walls B/1A-2 and B2/1A-2

Note K = 0 when the element is skipped from computation of restrain condition, please refer to the technical documentation K = |Sin α| × [a × b³ / (2 × L)]

Effective Length Factor, β for Braced Condition Relative Stiffness at Top of Element, αTop = 0.289 Relative Stiffness at Bottom of Element, αBot = 1.000 β1 = 0.7 + 0.05 × (αTop + αBot) = 0.76 β2 = 0.85 + 0.05 × Min (αTop, αBot) = 0.86 β = Min (β1, β2) = 0.76 < 1.0 Effective Length Factor, β for Unbraced Condition Relative Stiffness at Top of Element, αTop = 0.289 Relative Stiffness at Bottom of Element, αBot = 1.000 β1 = 1.0 + 0.15 × (αTop + αBot) = 1.19 β2 = 2.0 + 0.3 × Min (αTop, αBot) = 2.09 β = Min (β1, β2) = 1.19

Determine wall effective length for Braced condition.

Wall Effective Height for Braced Condition, le = ßBraced × lo = 2280 mm Wall Effective Height for Unbraced Condition, le = ßUnbraced × lo = 3570 mm

C ompute the wall effective height for Braced and Unbraced condition.

Determine wall effective length for Unbraced condition.

le = ß*lo

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Example 4: Wall with Opening For wall with opening, the wall stiffness in couple beam region is ignored. Description

Remarks When couple beam design is implemented for couple beam region (Region 3), the stiffness is ignored in stiffness calculation.

Result for Region 1 and 2 Description/Details

Remark Result for Region 1 and 2

Wall Panel Thickness, t = 150 mm

Sub Panel 1 (Reg. 1, 2)

Wall stiffness contribution from Region 1 and 2: = 750*150^3/3000 = 843750 mm^3

Location in Panel: Wall Zone Length of Sub-Panel, b = 750 mm Gross Concrete Area, Ac = b × t = 112500 mm² Result for Region 4,5 Description/Details

Remark Result for Region 4 and 5

Sub Panel 3 (Reg. 4, 5) Location in Panel: Wall Zone Length of Sub-Panel, b = 750 mm Gross Concrete Area, Ac = b × t = 112500 mm²

Wall stiffness contribution from Region 4 and 5 = 750*150^3/3000 = 843750 mm^3

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Example 5: Wall Connected to Cantilever Beam The cantilever beam segment will provide no lateral restraint to the wall, therefore the stiffness of cantilever beam is excluded from the calculation of wall effective height.

The cantilever beam segment is not included in calculating the beam stiffness to the wall:

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Example 6: Wall Connected to a Beam of Several Sizes For walls that is connected to a beam with several sizes as for beam gb4 (200x500/300x1000), the stiffness of the beam will be the summation of each segment stiffness by liner proportion to their own length in the span.

The stiffness of beam gb4 is obtained through the following calculation: = [(bh^3/2)*(L1/L) + (bh^3/2)*(L2/L)]/L = [(300*1000^3/2)*(3000/6000) + (200*500^3/2)*(3000/6000)]/6000 = 13,541,667

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Example 7: Wall Connected to Beam that is Not Attached to Lateral Restraint Element For walls that is connected to beams which the beams are not connected to element that can provide lateral restraint such as columns or wall, these beams stiffness is not included in the wall effective height calculation.

The beam segment stiffness B/5-6 and C/5-6 is not included in the wall effective height calculation:

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Example 8: Walls Above and Below Not Considered in Wall Relative Stiffness Calculation In the calculation of wall relative stiffness, the walls above and below are not considered. Refer the example below. Example of wall design report:

Notice that only current floor stiffness is considered in the calculation.

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Example 9: Restraint condition for elements at wall end The nearest distance of transverse wall or beam restraint from the wall end which is lesser or equal to the wall clear height will be considered. For example, wall clear height for wall 2/C-D = 3000 mm.

Therefore only beam gb1 is considered in the stiffness calculation.

Slenderness Result Wall Clear Height, lo = 3000 mm Top Element Stiffness, KeTop = ∑KsTop + ∑KbTop + ∑Ktw Top = 6250000.0 mm³ Bottom Foundation Stiffness, KeBot = 14175000.0 mm³ Wall Stiffness, Kw = 14175000.0 mm³

Restraint Elements at the top part of wall zone Elememt No. Mark Width a, mm Depth b, mm Length L, mm Type 1 Beam gb1 - 1 200 500 3000 2 Beam gb1 - 1 200 500 6000 3 Beam gb2 - 1 200 500 3000 4 Beam gb2 - 1 200 500 6000

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Angle α, °

K

90.0 90.0 90.0 90.0

4166667 2083333 0 0

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Example 10: Intermediate transverse wall or beams The intermediate stiffness of the transverse wall or beam will not be considered if any of the distance of intermediate walls or beams with the next restraint elements is more than the wall clear height. For example, wall clear height for wall 1/C-D = 3000 mm.

Therefore beam gb3 is not considered in the stiffness calculation.

Slenderness Result Wall Clear Height, lo = 3000 mm Top Element Stiffness, KeTop = ∑KsTop + ∑KbTop + ∑Ktw Top = 22916666.7 mm³ Bottom Foundation Stiffness, KeBot = 14287500.0 mm³ Wall Stiffness, Kw = 14287500.0 mm³

Restraint Elements at the top part of wall zone Elememt No. Mark Width a, mm Depth b, mm Length L, mm Type 1 Beam gb2 - 1 200 500 3000 2 Beam gb2 - 2 200 500 3000 3 Beam gb3 - 1 200 500 6000 4 Beam gb3 - 2 200 500 3000 5 Beam gb5 - 1 200 500 3000 6 Beam gb6 - 1 200 500 3000 p423

Angle α, °

K

90.0 90.0 90.0 90.0 90.0 90.0

4166667 4166667 0 0 4166667 4166667

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7 8

Beam Beam

gb7 - 1 gb7 - 1

200 200

500 500

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3000 6000

90.0 90.0

4166667 2083333

Esteem Innovation Sdn Bhd, 2011

Foundation

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Pad Unequal bar distribution For rectangular pad footing with ratio D/B  1.2, the reinforcement in the short direction should be distributed as shown below.

Example: Ratio D/B = 2700/1350 = 2  1.2 Pad footing thic kness = 300mm Asmin = 0.3*bh/100 2

As required in short direction = 2430mm 2 Portion of As required in the middle band = 2*2430/(2+1) = 1620mm 2 Portion of As required on both sides = 2430 - 1620 = 810mm Therefore, Using single bar diameter = TRUE 2 2 As provided in the middle band = 9T16 (1810mm ), Asmin = 0.3*1350*300/100 = 1215mm 2 As provided in the both sides band = 8T16 (1608mm ), Asmin = 0.3*(2700-1350)*300/100 2 = 1215mm

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Using single bar diameter = FALSE 2 2 As provided in the middle band = 9T16 (1810mm ), Asmin = 0.3*1350*300/100 = 1215mm 2 As provided in the both sides band = 12T12 (1357mm ), Asmin = 0.3*(2700-1350)*300/100 2 = 1215mm

References: 1. ACI318-05 2. M.Nadim H., Akthem Al-Manaseer. Structural Concrete:Theory and 4th ed., 2008 Chap. 13

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Pile Auto Pile Selection Pile Selection is a feature to allocate pile type to be used for certain range of the column load.

General Formula

where n = 1, 2, 3, …. y = column load y1 = minimum c olumn load yn = maximum column load y(n-1)n = upper / lower bound of the column load so that one type of pile will be allocated to. x = pile capacity for every pile in the list in ascending order. where x1 = pile 1 capacity x2 = pile 2 capacity : : xn = pile n capacity

x(n-1)n = mid value between 2 pile capacity

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Example: A project after analysis, the reaction obtain from the analysis are from 250 kN to 1000kN. Users want to use pile with following capacity 150kN, 250kN and 300kN. Find the range of the reaction load using these 3 types of pile group.

The range The middle boundary capacity of pile 150 and 250, x 12

The middle boundary capacity of pile 250 and 300, x 23

The range of the column load using pile 150kN, y1 – y12

Therefore for column load within range 250 kN to 500 kN using pile 150 kN. The range of the column load using pile 250kN, y12 – y23

Therefore for column load within range 500 kN to 875 kN using pile 250 kN. The range of the column load using pile 300kN, y23 – y3 are from 875 kN to 1000 kN.

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Shear check for Pile Cap Design (BS8110) There are 3 types of shear checks:1. Punching shear check 2. Ultimate shear check 3. Flexural shear check Referring to the code of practice BS 8110 Cl 3.11.4.3, the critical section for punching shear and flexural shear is assumed to be located at 20% of the diameter of the pile inside the face of the pile. However, shear checking for foundation in Esteem 8 is more complete and thorough. Every section with intervals of 0.05d would be checked to ensure that it is sufficient to pass the punching shear and flexural shear. AND, the program has the capability to determine the most critical section for punching shear and flexural shear. The full force from the pile is considered if the section cut lies at the pile center or pile cut off area >= 50%.

Figure 1: Full pile force is considered for shear as section cut lies at the pile center or pile cut off area >= 50%

Figure 2: Only actual pile force (hatch area*pile stress) is considered for shear as pile cut off area < 50%

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Example 1 Pile cap size = 800x800x500mm Pile size: 125x125mm square pile Pile capacity: 15tonne Concrete grade: 30N/mm2 Column size: 150x150mm

Pile cap detailing

1. Punching Shear Check Check for punching shear Shearing capacity = Computation of concrete shear 0.79*((100*As/(b*d)^1/3)*((fcu/25)^1/3)*((400/d)^1/4)/c capacity, vc With As is the minimum steel ratio in the two directions, x and y 100*As/(b*d) should not be taken as greater than 3 (400/d)^1/4 should not be taken as less than 0.67 fcu should not be taken as greater than 40 or less than 25 fcu = min(40, max (25, fcu)) = min (40, max (25, 30)) = 30 Cut Area X = 800*313=250747mm² Cut Area Y = 800*329=263547mm² Average steel % = 100*(Steel Area X + Steel Area Y)/(Cut Area X+Cut Area Y)=100*(1206+1206)/(250747+263547)=0.47 Effective steel percentage = Min(Average steel %, 3) = 0.47% thickness = (313+329)/2=321 Inverted thickness = max(0.67,(400/thickness^(1/4))) Inverted thickness = max(0.67, (400/(321))^(1/4) )= 1 Shearing capacity, vc = 0.79*0.47^(1/3)*1.06*(30/25.0)^(1/3)/1.25 = 0.56 N/mm^2 Check for the case if the critical section occurs at 0.50*d from column face Report shows every section with interval of 0.5d for punching shear Enhancement Shear multiplier, esm = 1.5 Real shear capacity with consideration Enhanced Shear factor = Max(esm/k, 1)= Max(1.5/0.50,1) = 3.00 of enhance shear factor Real shear capacity, vc= vc*enchanced Shear Factor = 0.56*3.00 = 1.67 N/mm^2 k = 0.50 The dimension of the effective critical section B' = B + 2*k*d B' = 150 + 2*0.50*321 = 471 mm L' = L + 2*k*d L' = 150 + 2*0.50*321 = 471 mm

Pile Number 1

Ultimate Coordinate X Coordinate Y Design Load (mm) (mm) (kN) -188

188

206.0

Ultimate Pile Load Inside the Critical Diagonal Zone Ultimate load = 1.4*15(pile (kN) 161.6

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2

188

188

206.0

161.6

3

-188

-188

206.0

161.6

4

188

-188

206.0

161.6

824.0

646.5

Sum

capacity)*9.81 = 206kN

Punching shear force = 824.0-646.5=177.6kN punching shear stress = 10^3*177.56/((2*471+2*471)*321)= 0.29 <=vc = 1.67N/mm² Punching Shear test pass!

Refer to the diagram beside, actual pile load (area*pile stress) is used to calculate the punching shear force as section area (un-hatch area) < 50%.

The most critical section occurs at 0.4d Check for the case if the critical section occurs at 0.40*d from column face Enhancement Shear multiplier, esm = 1.5 Enhanced Shear factor = Max(esm/k, 1)= Max(1.5/0.40,1) = 3.75 Real shear capacity, vc= vc*enchanced Shear Factor = 0.56*3.75 = 2.08 N/mm^2

The program able to detect the most critical section for punching shear which occurs at 0.4d.

k = 0.40 The dimension of the effective critical section B' = B + 2*k*d B' = 150 + 2*0.40*321 = 407 mm L' = L + 2*k*d L' = 150 + 2*0.40*321 = 407 mm Pile Number

Coordinate X (mm)

Coordinate Y (mm)

Ultimate Design Load (kN)

Ultimate Pile Load Inside the Critical Diagonal Zone (kN)

1

-188

188

206.0

0.0

2

188

188

206.0

0.0

3

-188

-188

206.0

0.0

4

188

-188

206.0

0.0

824.0

0.0

Sum

Punching shear force = 824.0-0.0=824.0kN punching shear stress = 10^3*824.04/((2*407+2*407)*321)= 1.57 <=vc = 2.08N/mm² Punching Shear test pass! p432

Full pile force is applied as the section area (un-hatch area) is ≥ 50%. Punching force = 824kN

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2. Ultimate Shear Check Check for Ultimate Shear shear stress = Ultimate axial load/(ColPerimeter*d)<=Min (0.8√fcu, 5) shear stress = 824.0*10^3/(600*321) = 4.27 N/mm² 4.27 <= 4.38 Ultimate Shear Checking pass 3. Flexural Shear Check Check for Flexural Shear Shearing capacity = 0.79*((100*As/(b*d)^1/3)*((fcu/25)^1/3)*((400/d)^1/4)/γc With As is the minimum steel ratio in the two directions, x and y 100*As/(b*d) should not be taken as greater than 3 (400/d)^1/4 should not be taken as less than 0.67 fcu should not be taken as greater than 40 or less than 25 Un-enhanced Shear Capacity in X direction fcu = min(40, max (25, fcu)) = min (40, max (25, 30)) = 30 Effective steel percentage = Min(steel percentage, 3) = 0.48% Inverted thickness = max(0.67,(400/thickness^(1/4))) Inverted thickness = max(0.67, (400/(313))^(1/4) )= 1 Shearing capacity X, vcX = 0.79*0.48^(1/3)*1*(30/25.0)^(1/3)/1.25 = 0.56 N/mm^2

Computation of concrete shear capacity in x and y direction, vc

Un-enhanced Shear Capacity in Y direction fcu = min(40, max (25, fcu)) = min (40, max (25, 30)) = 30 Effective steel percentage = Min(steel percentage, 3) = 0.46% Inverted thickness = max(0.67,(400/thickness^(1/4))) Inverted thickness = max(0.67, (400/(329))^(1/4) )= 1 Shearing capacity Y, vcY= 0.79*0.46^(1/3)*1*(30/25.0)^(1/3)/1.25 = 0.54 N/mm^2

The flexural shear is checked at section =0.50d at left portion of the cap

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Report shows the section (4 sections left, right, top and bottom portion) which defined by user in pile detailing

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parameter. Pile Number

Coordinate X (mm)

Coordinate Y (mm)

1

-188

188

206.0

30.1

2

188

188

206.0

0.0

3

-188

-188

206.0

30.1

4

188

-188

206.0

0.0

824.0

60.3

Sum

Ultimate Design Ultimate pile load at one Load (kN) side of the slab (kN)

Enhancement factor = Max(1.5/k,1) = Max(1.5/0.50,1)=3.00 Enhanced shear = 0.56*3.00 = 1.68N/mm^2 Flexural shear stress = 1.0e3*60.26/(800*313)= 0.24 N/mm^2 <= vc = 1.68 N/mm^2 Flexure Shear test pass!

The most critical section of the face shear is at 0.35d The program able to detect the The flexural shear is checked at section =0.35d at left portion of the cap

most critical section for flexural shear which occurs at 0.35d.

Pile Number

Coordinate X (mm)

Coordinate Y (mm)

Ultimate Design Load (kN)

Ultimate pile load at one side of the slab (kN)

1

-188

188

206.0

206.0

2

188

188

206.0

0.0

3

-188

-188

206.0

206.0

4

188

-188

206.0

0.0

824.0

412.0

Sum

Enhancement factor = Max(1.5/k,1) = Max(1.5/0.35,1)=4.29 Enhanced shear = 0.56*4.29 = 2.40N/mm^2 Flexural shear stress = 1.0e3*412.02/(800*313)= 1.64 N/mm^2 <= vc = 2.40 N/mm^2 Flexure Shear test pass!

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Pile group analysis calculation The example below shows the calculation for pile group analysis used by the program.

Sample data from project: Pile Number, n = 3 piles Pile no. 1 (0, -.26) Pile no. 2 (0.225, 0.13) Pile no. 3 (-0.225, 0.13) Column Moment X, Mx = -1.42 kNm Column Moment Y, My = -3.56 kNm Column Load = 70.20 kN Pile Cap Self-weight Load = [(0.34x0.9) + ½(0.323+0.9)(0.5)]m² x 0.35m x 24kN/m³ = 5.14 kN Total Load Applied = 70.20 kN + 5.14 kN = 75.34 kN Pile Capacity = 20 ton x 9.81 N/kg = 196.20 kN Ixx = 0.260² + (0.130² x 2) = 0.1014 m² Iyy = 0.225² x 2 = 0.1013 m² Vertical load for pile no.1:

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Vertical load for pile no.2:

Vertical load for pile no.3:

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Pile cap size dependency on pile or column size Pile Cap Height Calculation 2 Pile Group when Column is Smaller, Equal or Bigger than Pile Size. Notation Symbol

distance of

lpe

:

Pile edge to pile cap ( user input)

lcah

:

Pile Cap Height

lcaw

:

Pile Cap Width

lph

:

Pile Height

lpw

:

Pile Width

lcw

:

Column Width

lch

:

Column Height

lcva

:

Column Edge to pile edge cutting line

lpp

:

Pile center to pile center

lecv

:

Pile edge to pile cap (vertical)

lech

:

Pile edge to pile cap (horizontal)

Figure 1.1 Pile Cap Layout Drawing

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Figure 1.2 Pile Cap Dimension layout Plan for condition a and b

Figure 1.3 Pile Cap Dimension layouts Plan for condition c

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Figure 1.4 Distance of Pile Cap Edge to Pile Edge The pile cap size was calculated base on following 3 Conditions: a) Column size smaller than Pile size lcah = lpe + lph + lpe lcaw = lpe + lpw/2 + lpp + lpw/2 + lpe b) Column size same with pile size lcah = lpe + lph + lpe lcaw = lpe + lpw/2 + lpp + lpw/2 + lpe c) Column size greater than pile size lcah = lpe + lch + lpe lcaw = lpe + lpw/2 + lpp + lpw/2 + lpe

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Quantity Takeoff Quantity Take Off - Beam Sample Calculation for Quantity Take Off - Beam 1:

3

3

: 0.2 m x 0.6 m x 4 m = 0.48 m

Concrete m

Side Form work m

2

: 0.6 m x 4 m 2 sides:

Bottom Form work m

2

2

= 2.4 m

2

2 x 2.4 = 4.8 m

: 0.2 m x 4 m

2

= 0.8 m

Bottom Bar Length m : 9.19 m (By measurement from end point to the another end point) Top Bar Length m :9.26 m (By measurement from end point to the another end point) 2

Weight of Bottom Bar kg : 9.19 x 78.6/ 9.81 x 1000 x  x 0.006 = 8.328 kg 2 Weight of Top Bar kg : 9.26 x 78.6/ 9.81 x 1000 x  x 0.006 = 8.391 kg

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A = 600 -2(25) = 550 B = 200 -2(25) = 150 L =2(A+B) + 12 d =2(550+150) + 12(10) =1520 mm ~ 1525 mm 16 Nos stirrups x 1525 mm = 24400 mm = 24.400 m 24.4 m x 78.6/9.81 x 1000 x

 x 0.0052 = 15.354 kg

Sample Calculation for Quantity Take Off - Beam 2:

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Side Bar = 50 + 4130 + 50 = 4320 mm (By measurement from end point to the another end point) 2 Nos of Side Bar x 4320 = 8.46 m 8.46 m x 78.6/9.81 x 1000 x

 x 0.0062 = 7.666 kg

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A = 600 -2(25) = 550 B = 200 -2(25) = 150 L = 2A + 3B + 10d = 2(550) + 3(150) + 10(10) = 1650 mm Closed Link 1650 mm x (8 Nos + 8 Nos) = 26400 mm Normal Link 1525 mm x 9 Nos = 13725 mm Closed Link + Normal Link = 26400 + 13725 = 40125 mm = 40.125 m 40.125 m x 78.6/9.81 x 1000 x

 x 0.0052 = 25.250 kg

Sample Calculation for Quantity Take Off - Beam 3:

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A = 600 - 2(25) = 550 B = 450 - 2(25) = 400 L

= 2(A+B) +12 d = 2(550+400)+12(10) = 2020 ~ 2025 mm

A + 2h = 550 + 2(110) = 770 ~ 775 mm Total length = (10 Nos x 2025 mm + 9 Nos x (2025 mm+775 mm))= 45450 mm = 45.450 m

Sample Calculation for Quantity Take Off - Beam 4:

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A = 500-2(25)= 450 B = 600-2(25)= 550 L = 2(A+B) + 12d = 2(450+550)+12(10) = 2120 mm ~ 2125 mm b = {600-2(25)-2(10)-20}/3 = 170 A = 450 B = 170+20+2(10) = 210 L = 2(A+B) + 12d = 2(450+210) + 12(10) = 1440 mm ~ 1450 15 Nos x (2125 + 1450) = 53625 mm p446

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A+2h A = 500-2(25) = 450 h = 110 A+2h = 450+2(110) = 670 mm ~ 675 mm 16 Nos x (2125 + 675) = 44800 mm Total length = 15 Nos + 16 Nos = 53625 + 44800 = 98.425 m

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Quantity Take Off - Column Sample Calculation for Quantity Take Off - Column 1:

3

3

Concrete, m = 0.3 m x 0.3 m x 3.0 m (Height) = 0.270 m

2

2

Formwork, m = 0.3 m x 3.0 m (Height) x 4 Nos = 3.600 m

Length, m = 4T12 at 3.0 m Height = 4 x 3.0 m = 12.000 m 1 kg = 9.81 N 3 unit weight of steel = 78.6 kN/m

Weight, kg = 78.6/ 9.81 x 1000 x = 10.874 kg

 x 0.0062 x 12

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Length, m A = 300 -25 -25 = 250 B = 300 -25 -25 = 250 L = 2(A+B)+ 12d = 2(250 + 250)+12(10) = 1120 mm ~ 1125 mm 3000/125 = 24 Nos + 1 Nos 25 Nos x 1125 mm = 28125 mm = 28.125 m Weight, kg = 78.6/ 9.81 x 1000 x = 17.698 kg

 x 0.0052 x 28.125

Sample Calculation for Quantity Take Off - Column 2:

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A = 600-25-25 =550 B = 600-25-25 =550 L = 2(A+B)+12d = 2(550+550)+12(10) = 2320 ~ 2325mm A+2h = 550 + 2(110) = 770 ~775mm 3000/175 = 17.1 Nos ~ 18 + 1 Nos 19 Nos x (2325+775+775) = 73625 mm = 73.625 m

Sample Calculation for Quantity Take Off - Column 3:

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A = 500 -2(25) = 450 B = 200-2(25) = 150 L = 2(A+B) +12d = 2(450+150) +12(10) = 1320mm ~ 1325mm A+2h A = 200 -2(25) = 150 A+2h = 150 + 2(110) = 370mm ~ 375mm 3000/225 = 13.3 => 14+1 Nos 15 Nos x (1325+1325+375) = 45375mm

Sample Calculation for Quantity Take Off - Column 4:

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Length 600 – 2(25) = 550 3000/175

= 17.1 ~ 18 +1 = 19 Nos

2 Pi (550/2) + 12(10) = 1848 ~ 1850mm 19 Nos x 1850 = 35150 mm Weight 78.6/ 9.81 x 1000 x

 x 0.005^2 x 35.150 = 22.119 kg

Concrete Pi x 0.32 x 3 = 0.848 m3 Formwork 2 x Pi x 0.3 x 3 = 5.655 m2

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Quantity Take Off - Wall The following examples will show the calculation of Quantity Take Off for the R C Wall.

Sample calculation for concrete and formwork: RC wall 1/A-B at floor gb.

Concrete volume Concrete = wall thickness*[(wall length*wall height) - (opening width*opening height)] 3

= 0.2*[(4*3) - (1*2)] = 2.0 m Formwork Exclude opening = 2*(wall length*wall height) 2

= 2*(4*3) = 24 m

Include opening = 2*[(wall length*wall height) - (opening width*opening height)] 2

= 2*[(4*3) - (1*2)] = 20 m

Notes : RC Wall formwork shown in Project Concrete & Formwork is based on formwork excluding the opening.

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Reinforcement (Normal Bar) Sample calculation for reinforcement: RC wall B/1-2 at floor 1b

Vertical and Horizontal Bar

Vertical bar (14T12 - 275 at each side) No. of rebars = 2 sides*14 = 28 Bar length = No. of rebars*height = 28*3000 = 84000 m = 84 m Weight of vertical bars = bar length*7860*rebar area p455

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2

= 84*7860*(0.012 *3.142/4) = 74.68 kg Horizontal bar (10T10 - 300 at each side) Other distance required for computation of bar length is shown in figure below. This also applies to wall designed using Fabric.

No. of rebars = 2 sides*10 = 20 Horizontal Bar length = no. of rebars*[wall length - 2*(half intersecting wall thickness + 100) + 2*tail length] = 20*[4000 - 2*(200/2 + 100) + 2*(50)] = 20*3700 = 74000 mm = 74 m Weight of horizontal bar = horizontal bar length*7860*rebar area 2

= 74*7860*(0.010 *3.142/4) = 45.7 kg Horizontal End U-Bars The distance required for computation of bar length is shown in figure below. This also applies to wall designed using Fabric.

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Horizontal End U Bar at starting and ending of wall panel length Horizontal End U Bar No. of End U bar = 10 (each at the left and start of wall length) U Bar length = 2*no. of End U bar*[2 sides*(lap length + 100 + (intersection wall thickness - cover half bar diameter) + 50) + (wall thickness - 2*cover - 2*half bar diameter)] = 2*10*[2*(400 + 100 + (200 - 25 - 10/2) + 50) + (200 - 2*25 - 2*10/2)] = 20*1580 = 31600 mm = 31.6 m Weight of horizontal bar = U Bar length*7860*rebar area 2

= 31.6*7860*(0.010 *3.142/4) = 19.5 kg Sample calculation for reinforcement: RC wall B/1-2 at floor gb.

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Vertical and Horizontal Bar

Vertical bar No. of rebars = 2 sides*14 = 28 Bar length = No. of rebars*(height + tension lap) = 28*(3000 + 500) = 28*3500 = 98000 mm = 98 m Weight of vertical bars = bar length*7860*rebar area 2

= 98*7860*(0.012 *3.142/4) p458

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= 87.1 kg Horizontal bar No. of rebars = 2 sides*10 = 20 Horizontal Bar length = no. of rebars*[wall length - 2*(half intersecting wall thickness + 100) + 2*tail length] = 20*[4000 - 2*(200/2 + 100) + 2*(50)] = 20*3700 = 74000 mm = 74 m Weight of horizontal bar = horizontal bar length*7860*rebar area 2

= 74*7860*(0.010 *3.142/4) = 45.7 kg Horizontal End U-Bar

Horizontal End U Bar at starting and ending of wall panel length Horizontal End U Bar No. of End U bar = 10 (each at the left and start of wall length) U Bar length = 2*no. of End U bar*[2 sides*(lap length + 100 + (intersection wall thickness - cover half bar diameter) + 50)) + (wall thickness - 2*cover - 2*half bar diameter)] = 2*10*[2*(400 + 100 + (200 - 25 - 10/2) + 50) + (200 - 2*25 - 2*10/2)] = 20*1580 = 31600 mm = 31.6 m Weight of horizontal bar = U Bar length*7860*rebar area 2

= 31.6*7860*(0.010 *3.142/4) = 19.5 kg

Notes

:

1. When there is an upper wall partially sitting on a lower wall, all the bars from the lower wall will be extended.

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No. of rebars = 2 sides*14 = 28 Bar length = No. of rebars*(height + tension lap) = 28*(3000 + 500) = 28*3500 = 98000 mm = 98 m Weight of vertical bars = bar length*7860*rebar area 2

= 98*7860*(0.012 *3.142/4) = 87.1 kg

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Fabric The Quantity Take Off for wall design using fabric is computed based on the total area covered. Similarly to wall designed using normal bars, the distance shown in figure below are required. Refer to sample calculation for Quantity Take Off > Reinforcement (Normal Bar) for further details.

Sample calculation for reinforcement: RC wall 2/A-B at floor 1b

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Fabric

BRC B8

Bar length = 2 sides*[wall length - (2*half of intersecting wall thickness + 2*100) + 2*50] = 2*[4000 - (2*200/2 + 200) + 2*50] = 2*3700 = 7400 mm = 7.4 m p462

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Bar height = wall height = 3000 mm = 3.0 m Zone area = total bar length*bar height = 7.4*3.0 = 22.2 m

2

Weight = 2*(Total zone area provided*unit weight for bar) = 2*(22.2*5.93) = 263.2 kg Sample calculation for reinforcement: RC wall 2/A-B at floor gb

Fabric

Bar length p463

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= 2 sides*[wall length - (2*half of intersecting wall thickness + 2*100) + 2*50] = 2*[4000 - (2*200/2 + 200) + 2*50] = 2*3700 = 7400 mm = 7.4 m 2

fy = 485 N/mm

2

fcu = 30 N/mm  = 0.65 (BS8110 Part 1: Cl 3.12.8.4) 2

fbu = 0.65*30 = 3.56 N/mm

Bar height = wall height + tension lap = 3000 + [(fy/s)/(4*fbu)]*bar diam = 3000 + [(485/1.05)/(4*3.56)]*8 = 3000 + 32.4*8 = 3000 + 300 mm (after rounded) = 3300 mm = 3.3 m Zone area = total bar length*bar height = 7.4*3.3 = 24.4 m

2

Weight = 2*(Total zone area provided*unit weight for bar) = 2*(24.4*5.93) = 289.4 kg

Notes

:

1. The tail length for BRC is calculated since it is based on bars provided in Wall Detailing. 2. Sample calculation for Unit Weight of rebars above: For BRC B8: Sectional areas provided per metre width: 2

8 mm at 100 c/c (503 mm ) 2

8 mm at 200 c/c (251 mm ) Unit Weight for every metre square area = [7860*total area provided/1000000 = [7860*(503 + 251)]/1000000 = 5.93 kg/m

2

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Framing Beam The following examples will show the calculation of the framing beam provided at the top of rc wall. For more details on the framing beam, refer to Technical Documentation > Finite Element > Wall Frame Element.

Sample calculation for framing beam: FB 1/C-D at wall 1/C-D (Floor 1b) LOCATION : SECTION FB 1/C-D-1 Smaller Section Dimension, Dmin = 200.0 mm Larger Section Dimension, Dmax = 400.0 mm.

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Bottom Bar

No. of rebars = 2 Length of bottom bar = no.of rebars*[span length + bend length + bar length passing support] = 2*[4000 + 2*250 + 2*47.5] = 9190 mm = 9.19 m Weight of bottom bar = length of bottom bar*7860*rebar area 2

= 9.19*7860*(0.012 *3.142/4) = 8.17 kg Top Bar p466

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No.of rebar = 2 Length of top bar = no.of rebars*[span length + bend length + bar length passing support] = 2*[4000 + 2*250 + 2*65] = 9260 mm = 9.26 m Weight of top bar = length of bottom bar*7860*rebar area 2

= 9.26*7860*(0.012 *3.142/4) = 8.23 kg Side Bar

No.of rebar = 2 Length of side bar = no.of rebars*[span length + bend length + bar length passing support] = 2*[4000 + (2*50) + 2*0.065] = 8460 mm = 8.46 m Weight of side bar = length of bottom bar*7860*rebar area 2

= 8.46*7860*(0.012 *3.142/4) = 7.52 kg Stirrup Bar

For closed links (Type 79, refer to Quantity Take Off for Beam ): No. of loop = 1 No. of stirrup = 34 A = 400 -2(25) = 350 mm B = 200 -2(25) = 150 mm Stirrup bar length p467

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= 2*A + 3*B + 12 d = 2*0.35 + 3*0.15 + 10*0.010 = 1.25 m Length of stirrup bar = no.of stirrup*stirrup bar length = 34*1.25 = 42.5 m Weight of stirrup bar = length of stirrup bar*7860*rebar area 2

= 42.5*7860*(0.010 *3.142/4) = 26.2 kg

Notes

:

1. The length of the rebars in framing beam are similar to Quantity Take Off for beams, by measurement from end point to the another end point. 2. For the length of the link, refer to Quantity Take Off for Beam.

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Anti-Crack Bar Sample calculation for Anti-Crack bar: Wall 1/A-B with opening (1500w x 2500 height) Wall height = 3 m Opening width = 1.5 m Opening height = 2.5 m

Vertical Bars (T12 - F/B (V) each side) No. of rebars = 2 sides*2 bars =4

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Vertical bar length = No. of rebars*[opening height + 2*(cover + tension anchorage)] = 4*[2500 + 2*(25 + 500) = 4*3550 = 14200 mm = 14.2 m Weight of vertical bars = bar length*7860*rebar area 2

= 14.2*7860*(0.012 *3.142/4) = 12.6 kg Horizontal Bars (T12 F/B (H)) No. of rebars = 2 sides*1 bar = 2 Horizontal bar length = No. of rebars*[opening width + 2*(cover + tension anchorage)] = 2*[1500 + 2*(25 + 500) = 2*2550 = 5100 mm = 5.1 m Weight of horizontal bars = bar length*7860*rebar area 2

= 5.1*7860*(0.012 *3.142/4) = 4.5 kg

Bottom U-Bar (at bottom of opening) Specification for selection of bar size and spacing for the top and bottom U-Bar. U-Bar size: The size will follow the largest diameter of vertical rebars from the left and right side of the opening. U-Bar spacing: The spacing will follow the smallest spacing of vertical rebars from the left and right side of the opening. In this case, Largest vertical rebar diameter from left/right of opening = T12 Smallest vertical rebar spacing from left/right of opening = 275 mm Opening width = 1500mm Therefore no.of rebars required = 1500/275 = 5.5 = 6 bars Provide 6 U-Bars at 275 mm spacing. No. of rebars =6 U-Bar bar length = 6*[2*tension anchorage + (wall thickness - 2*cover)] = 6*[2*500 + (200 - 2*25)] = 6*1150 = 6900 mm = 6.9 m Weight of diagonal bars = U-bar length*7860*rebar area 2

= 6.9*7860*(0.012 *3.142/4) = 6.1 kg p470

Esteem Innovation Sdn Bhd, 2011

Diagonal (T12-Diag) No. of rebars = 2 sides*4 (at each side) = 8 Diagonal bar length = no. of rebars*(2*tension anchorage) = 8*[2*(40*12)] = 8*(2*480) = 8*(2*500) after rounded to 50mm = 8000 mm = 8 m Weight of diagonal bars = diagonal bar length*7860*rebar area 2

= 8*7860*(0.012 *3.142/4) = 7.1 kg Therefore total Anti-crack bar length = vertical bars + horizontal bars + bottom U-Bars + Diagonal bars = 14.2 + 5.1 + 6.9 + 8 = 34.2 m Therefore total Anti-crack bar weight = vertical bars + horizontal bars + bottom U-Bars + Diagonal bars = 12.6 + 4.5 + 7.1 + 6.1 = 30.3 kg

Notes

:

1. The horizontal Anti-crack U-Bar needed at the left and right side of the opening is already provided by the wall pier zones. 2. If wall opening is located below coupled beam, U-Bar will not be provided at the top of the wall opening. This is because the detailing is already provided by the couple beam link. 3. For wall with opening, the Quantity Take Off including anti-crack bars will NOT be provided if either the couple beam or wall piers failed.

p471

Esteem Innovation Sdn Bhd, 2011

Reinforcement (Wall Group Bar) The following examples will show the calculation of the reinforcement provided at wall joints for Wall Groups: o Vertical bars at the joints of wall panels o Horizontal lapping bars across two panels Sample calculation for reinforcement: Wall Group 7 at floor gb

p472

Esteem Innovation Sdn Bhd, 2011

Vertical bars at wall joints Vertical bars at three joints (3,A), (3,B) and (3,C)

p473

Esteem Innovation Sdn Bhd, 2011

No. of rebars =4 Vertical bar length = No. of rebars*wall height = 4*3000 = 12000 = 12 m Total vertical length for three joints; = 3*12 = 36 m Weight of vertical bars = bar length*7860*rebar area 2

= 36*7860*(0.012 *3.142/4) = 32.01 kg Horizontal lapping bars across wall panels 3/A-B and 3/B-C (10T10 - 300 F&B)

p474

Esteem Innovation Sdn Bhd, 2011

No. of rebars = 10*2 sides = 20 Horizontal lapping bar length = No. of rebars*2*(lap length + half wall thickness + 100 + tail length) = 20*2*[400 + 200/2 + 100 + 50] = 26000 mm = 26 m Weight of horizontal lapping bars = Horizontal lapping bar length*7860*rebar area 2

= 26*7860*(0.010 *3.142/4) = 16.05 kg

Notes

:

1. Rebar at wall joints If one of the connecting panels has an upper wall, all the vertical bars at the joint zone will be extended from the lower wall. Example: Upper wall exist at 3/A-B only. Vertical bars at three joints (3,A), (3,B) and (3,C)

p475

Esteem Innovation Sdn Bhd, 2011

No. of rebars =4 Vertical bar length = No. of rebars*(wall height + lapping length) = 4*(3000 + 500) = 14000 = 14 m Total vertical length for three joints; = 3*14 = 42 m Weight of vertical bars = bar length*7860*rebar area 2

= 42*7860*(0.012 *3.142/4) = 37.3 kg 2. Horizontal lapping bars across wall panels will NOT be provided if one of the panels is marked as 'Not Design'. Therefore the Quantity Take Off for the lapping bars in such condition will also not be available.

p476

Esteem Innovation Sdn Bhd, 2011

Quantity Take Off - Slab Sample calculation for concrete and formwork: Slab FS1 and curve slab FS2 This model will be used as an example. Note that the Outer Middle Offset of curve beam gb5 from midpoint of beam gb3 is 2500mm.

Slab FS1

Effective area = (slab width - beam width)*(slab length - beam width) = (5 - 0.2)*(5 - 0.2) = 23.04 mm2 Concrete volume Concrete = slab thickness*effective area = 0.175*23.04 = 4.032 m

3

Formwork Slab soffit = Effective area = 23.04 mm2 Curve slab FS2

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Esteem Innovation Sdn Bhd, 2011

By measuring from the keyplan, Curve slab longest radius length = 2399  2400 mm = 2.4 m Curve slab effective radius = curve slab longest length - halfbeamwidth of gb3 = 2.4 - 0.1 = 2.3 m Effective area = 2.3 2*3.142/2 = 8.31 mm2 Concrete volume Concrete = slab thickness*effective area = 0.175*8.31 = 1.454 m

3

Formwork Slab soffit = Effective area = 8.31 mm2

Notes

: The value differ due to different curve slab area. The method used in Esteem 8 is by using equivalent slab bounding polygon, rather than the exact curve slab area. Quantity take off for slab reinforcement The calculation for quantity take off of reinforcem ent is done based on the slab bounding area. Due to complexity in identifying adjacent slabs and their bar connectivity, a configurable parameter is provided to extend the existing slab polygon. This parameter will cover the additional bar coverage area for QS purpose. The boundary is extended from the middle point of the support boundary. Rectangular slab

Curve slab

p478

Esteem Innovation Sdn Bhd, 2011

p479

Esteem Innovation Sdn Bhd, 2011

Reinforcement (Normal Bar) Sample calculation for Slab FS1 and curve slab FS2

Bottom Bar

Slab FS1:

20T10-250 (BB) Extended slab length = 5000 + 2*200 = 5400 mm Rebar area = 102*3.142/4 = 78.54 m2 Steel area (per m) = 78.54*1000/250 = 314.16 mm2/m Slab surface area = 5400*5400 = 2916E4 mm2 = 29.16 m2 Steel volume = steel area*slab surface area = 314.16*29160000 = 9.16E9 mm4/ m p480

Esteem Innovation Sdn Bhd, 2011

= 9.16E6 mm3 Length of steel = steel volume/cross-sectional area of bar = 9.16E6/78.54 = 116640 mm = 116.64 m Weight of bottom bars = steel volume*steel density/1E9 = 9.16E6*7860/1E9 = 71.9 = 72 kg 20T10-250 (BT) Extended slab length = 5000 + 2*200 = 5400 mm Rebar area = 102*3.142/4 = 78.54 m2 Steel area (per m) = 78.54*1000/250 = 314.16 mm2/m Slab surface area = 5400*5400 = 2916E4 mm2 = 29.16 m2 Steel volume = steel area*slab surface area = 314.16*29160000 = 9.16E9 mm4/ m = 9.16E6 mm3 Length of steel = steel volume/cross-sectional area of bar = 9.16E6/78.54 = 116640 mm = 116.64 m Weight of bottom bars = steel volume*steel density/1E9 = 9.16E6*7860/1E9 = 71.9 = 72 kg Total length for bottom bar = 116.64 + 116.64 = 223.28 m Total weight of bottom bars = 72*2 = 144 kg Curve Slab FS2:

20T10-250 (BB) Extended curve slab radius = 2500 - half beam width + 2*200 = 2500 - 100 + 2*200 = 2.8 m Curve slab surface area = 2.82*3.142/2 = 12.31 m2 Rebar area = 102*3.142/4 = 78.54 m2 Steel area (per m) = 78.54*1000/250 = 314.16 mm2/m Steel volume = steel area*curve slab surface area = 314.16*12.31E6 = 3.87E9 mm4/ m = 3.87E6 mm3 Length of steel = steel volume/cross-sectional area of bar = 3.87E6/78.54 = 49266 mm = 49.27 m Weight of bottom bars = steel volume*steel density/1E9 = 3.87E6*7860/1E9 = 30.4 kg p481

Esteem Innovation Sdn Bhd, 2011

9T10-250 (BT) Extended curve slab radius = 2500 - half beam width + 2*200 = 2500 - 100 + 2*200 = 2.8 m Curve slab surface area = 2.82*3.142/2 = 12.31 m2 Rebar area = 102*3.142/4 = 78.54 m2 Steel area (per m) = 78.54*1000/250 = 314.16 mm2/m Steel volume = steel area*curve slab surface area = 314.16*12.31E6 = 3.87E9 mm4/ m = 3.87E6 mm3 Length of steel = steel volume/cross-sectional area of bar = 3.87E6/78.54 = 49266 mm = 49.27 m Weight of bottom bars = steel volume*steel density/1E9 = 3.87E6*7860/1E9 = 30.4 kg Total length for bottom bar = 2*49.27 = 98.54 m Total weight of bottom bars = 2*30.4 = 60.8 kg

Notes

: The value differ due to different curve slab area. The method used in Esteem 8 is by using equivalent slab bounding polygon, rather than the exact curve slab area. Top Distribution Bar

p482

Esteem Innovation Sdn Bhd, 2011

Slab FS1:

At slab edge 1 (5T10-250 TB) Length of steel = No. of bars*length of steel = 5*(L - 2*0.70) = 5*(5 - 2*0.70) = 5*3.6 = 18 m Weight of top distribution bars = length of steel*steel density*steel bar area 2

= 18*7860*(0.010 *3.142/4) = 11.1 kg At slab edge 2 (5T10-250 TB) Length of steel = No. of bars*length of steel = 5*[L - (0.70 + 1.2)] = 5*(5 - 1.9) = 5*3.1 = 15.5 m Weight of top distribution bars = length of steel*steel density*steel bar area p483

Esteem Innovation Sdn Bhd, 2011

2

= 15.5*7860*(0.010 *3.142/4) = 9.6 kg At slab edge 3 (7T10-250 TB) Length of steel = No. of bars*length of steel = 7*(L - 2*0.70) = 5*(5 - 2*0.70) = 7*3.6 = 25.2 m Weight of top distribution bars = length of steel*steel density*steel bar area 2

= 25.2*7860*(0.010 *3.142/4) = 15.5 kg At slab edge 4 (5T10-250 TB) Length of steel = No. of bars*length of steel = 5*[L - (0.70 + 1.2)] = 5*(5 - 1.9) = 5*3.1 = 15.5 m Weight of top distribution bars = length of steel*steel density*steel bar area 2

= 15.5*7860*(0.010 *3.142/4) = 9.6 kg Total length of top distribution bar = 18 + 15.5 + 25.2 + 15.5 = 74.2 m Total weight top distribution bar = 11.1 + 9.6 + 15.5 + 9.6 = 45.8 kg Curve Slab FS2:

At slab edge 3 (9T10-250 TB) Length of steel = No. of bars*length of steel = 9*[curveslablongestlength - cover] = 9*[2*(2.5 -0.025)] = 9*4.95 = 44.55 m Weight of top distribution bars = length of steel*steel density*steel bar area 2

= 44.55*7860*(0.010 *3.142/4) = 27.5 kg

Notes 1. For are the bar

: curve slab, since the length of the distribution bar is gradually shortened as the bars located away from the support, the length of the distribution bar is estimated by using length of the bar located at the center of the range line and m ultiply the number of unit.

p484

Esteem Innovation Sdn Bhd, 2011

Reinforcement (Support Normal Bar) Sample calculation for slab FS1 and curve slab FS2 Top Main Bar Over the Support

Slab FS1:

Top main bar at slab boundary no.1 (20T10-250 TT) Top bar length = no.of rebars*[curtailment + (beam width - cover) + min (bent length, tension anchorage length)] = 20*[1 + (0.2 - 0.025) + min (0.25, 40*0.01)] = 20*1.425 m = 28.5 m Weight of top main bars = length of steel*steel density*steel bar area] p485

Esteem Innovation Sdn Bhd, 2011

2

= 28.5*7860*(0.010 *3.142/4) = 17.6 kg Top main bar at slab boundary no.2 (20T10-250 TT) Top bar length = no.of rebars*[curtailment + (beam width - cover) + min (bent length, tension anchorage length)] = 20*[1 + (0.2 - 0.025) + min (0.25, 40*0.01)] = 20*1.425 m = 28.5 m Weight of top main bars = length of steel*steel density*steel bar area] 2

= 28.5*7860*(0.010 *3.142/4) = 17.6 kg Top main bar at slab boundary no.3 (20T10-250 TT) Top bar length = no.of rebars*[curtailment + longest curve slab length + 0.5*beam width + tension anchorage length)] = 20*[1.5 + (2.5 - cover) + 0.10 + 0.25] = 20*[1.5 + (2.5 - 0.025) + 0.10 + 0.25] = 20*4.325 = 86.5 m Weight of top main bars = length of steel*steel density*steel bar area 2

= 86.5*7860*(0.010 *3.142/4) =53.4 kg Top main bar at slab boundary no.4 (20T10-250 TT) Top bar length = no.of rebars*[curtailment + (beam width - cover) + min (bent length, tension anchorage length)] = 20*[1 + (0.2 - 0.025) + min (0.25, 40*0.01)] = 20*1.425 m = 28.5 m Weight of top main bars = length of steel*steel density*steel bar area] 2

= 28.5*7860*(0.010 *3.142/4) = 17.6 kg Curve Slab FS2: The top main bar for curve slab FS2 is extended from support no.3.

Top main bar at slab boundary no.3 (20T10-250 TT) Top bar length = no.of rebars*[curtailment + longest curve slab length + 0.5*beam width + tension anchorage length)] = 20*[1.5 + (2.5 - cover) + 0.10 + 0.25] = 20*[1.5 + (2.5 - 0.025) + 0.10 + 0.25] = 20*4.325 = 86.5 m Weight of top main bars = length of steel*steel density*steel bar area 2

= 86.5*7860*(0.010 *3.142/4) =53.4 kg

Notes

:

1. For curve slab, since the length of the top m ain bar is gradually shortened as the bars are located away from the support, the length of the main bar is estimated by using the length of the bar located at the center of the range line and m ultiply the number of bar unit.

p486

Esteem Innovation Sdn Bhd, 2011

p487

Esteem Innovation Sdn Bhd, 2011

Reinforcement (Fabric) Similarly like RC Wall, the Quantity Take Off for fabric is calculated based on the area covered. Sample calculation for Slab FS1 and curve slab FS2 Bottom Bar

Slab FS1:

2

fy = 485 N/mm

2

fcu = 30 N/mm  = 0.65 (BS8110 Part 1: Cl 3.12.8.4) p488

Esteem Innovation Sdn Bhd, 2011

2

fbu = 0.65*30 = 3.56 N/mm

BRC B8 (B503) Extended slab length = 5000 + 2*200 = 5400 mm Rebar area = 8 2*3.142/4 = 50.3 m2 Slab surface area = 5400*5400 = 2916E4 mm2 = 29.16 m2 Bar Area in Main Direction = 50.3*1000/100 = 503 mm2/m Fabric Main Direction Volume, VMain =503*1000 = 503000 mm3/m2 Bar Area in Cross Direction = 50.3*1000/200 = 251.5 mm2/m Fabric Cross Direction Volume, VCross = 251.5*1000 = 251500 mm3/m2 Total Volume (per meter square), Vtol = VMain + VCross = 503000 + 251500 = 754500 mm3/m2 Steel Density = 7860 kg/m3 Fabric Unit Weight (kg/m2) = VTol x Steel Density/1E9 = 754500*7860/1E9 = 5.93 kg/m2 Weight of Fabric = Slab Surface Area*fabric Unit Weight = 29.16 x 5.93 = 172.93 kg Curve Slab FS2:

BRC B8 (B503)-250 (BB) Extended curve slab radius = 2500 - half beam width + 2*200 = 2500 - 100 + 2*200 = 2.8 m Curve slab surface area = 2.82*3.142/2 = 12.31 m2 Rebar area = 8 2*3.142/4 = 50.3 m2 Bar Area in Main Direction = 50.3*1000/100 = 503 mm2/m Fabric Main Direction Volume, VMain =503*1000 = 503000 mm3/m2 Bar Area in Cross Direction = 50.3*1000/200 = 251.5 mm2/m Fabric Cross Direction Volume, VCross = 251.5*1000 = 251500 mm3/m2 Total Volume (per meter square), Vtol = VMain + VCross = 503000 + 251500 = 754500 mm3/m2 Steel Density = 7860 kg/m3 Fabric Unit Weight (kg/m2) = VTol x Steel Density/1E9 = 754500*7860/1E9 = 5.93 kg/m2 Weight of Fabric = Slab Surface Area*fabric Unit Weight = 12.31 x 5.93 = 72.9 kg p489

Esteem Innovation Sdn Bhd, 2011

Notes

: The value differ due to different curve slab area. The method used in Esteem 8 is

by using equivalent slab bounding polygon, rather than the exact curve slab area.

p490

Esteem Innovation Sdn Bhd, 2011

Reinforcement (Support Fabric) Sample calculation for slab FS1 and curve slab FS2 Top Main Bar Over the Support

Slab FS1:

Top main bar at slab boundary no.1, 2, and 4 (BRC B6 T) Top bar length = [curtailment + (beam width - cover) + max (bent length, tension anchorage length)] = 1000 + (200 - 25) + max[250, ((fy/s)/(4*fbu))*bar diam)] = 1175 + max [250, ((485/1.05)/(4*3.56))*6)] = 1175 + max [250, 200 (after rounded)] = 1175 + 250 p491

Esteem Innovation Sdn Bhd, 2011

= 1425 mm = 1.425 m Slab length = 5000 = 5 m Rebar area in main direction = 6 2*3.142/4 = 28.3 m2 Rebar area in cross direction = 72*3.142/4 = 38.5 m2 Slab surface area = 1.425*5 = 7.125 m2 Bar Area in Main Direction = 28.3*1000/100 = 283 mm2/m Fabric Main Direction Volume, VMain = 283*1000 = 283000 mm3/m2 Bar Area in Cross Direction = 38.5*1000/200 = 192.4 mm2/m Fabric Cross Direction Volume, VCross = 192.4*1000 = 192447.5 mm3/m2 Total Volume (per meter square), Vtol = VMain + VCross = 283000 + 192447.5 = 475447.5 mm3/m2 Steel Density = 7860 kg/m3 Fabric Unit Weight (kg/m2) = VTol x Steel Density/1E9 = 475447.5*7860/1E9 = 3.73 kg/m2 Weight of Fabric = Slab Surface Area*fabric Unit Weight = 7.125 x 3.73 = 26.6 kg

The top bar over support no.3 is extended to the curve slab FS2. Refer to calculation below. Curve Slab FS2:

Top main bar at slab boundary no.3 (BRC B8 T) Rebar area in main direction = 8 2*3.142/4 = 50.3 m2 Rebar area in cross direction = 82*3.142/4 = 50.3 m2 Bar Area in Main Direction = 50.3*1000/100 = 503 mm2/m Fabric Main Direction Volume, VMain = 503*1000 = 503000 mm3/m2 Bar Area in Cross Direction = 50.3*1000/200 = 251.5 mm2/m Fabric Cross Direction Volume, VCross = 251.5*1000 = 251500 mm3/m2 Total Volume (per meter square), Vtol = VMain + VCross = 503000 + 251500 = 754500 mm3/m2 p492

Esteem Innovation Sdn Bhd, 2011

Steel Density = 7860 kg/m3 Fabric Unit Weight (kg/m2) = VTol x Steel Density/1E9 = 754500*7860/1E9 = 5.93 kg/m2 Top bar surface area in slab FS1 Top bar length = curtailment + half beam width = 1500 + 100 = 1600 mm = 1.6 m Support support length = 5 m Surface area at slab FS1 = 1.6*5 = 8 m2 Top bar surface area in curve slab FS2 Top bar length = curveslablongestlength - cover = 2500 - 25 = 2475 mm = 2.475 m Curve slab surface area = 2.4752*3.142/2 = 9.62 m2 Total surface area = surface area at slab FS1 + curve slab surface area = 8 + 9.62 = 17.6 m2 Weight of Fabric = Total surface Area*fabric Unit Weight = 17.6 x 5.93 = 104.3 kg

Notes

:

1. The coverage area of the slab length is taken as the support length rather than the exact slab range length in detailing. 2. The value differ due to different curve slab area. The method used in Esteem 8 is by using equivalent slab bounding polygon, rather than the exact curve slab area.

p493

Esteem Innovation Sdn Bhd, 2011

Benchmarking Alternative Model Benchmarking for the Alternative Model in Esteem 7 You may input a single span simply supported beam as a benchmark model. The properties of the beam are as follows. Beam size : 150x600mm Span length = 4000mm Beam uniform load : Dead Load = 1kN/m Live Load = 1kN/m Then, run the 2D + 3D analysis. For first run, switch the Alternative Model to False.

You will obtain the following diagrams and result.

Diagram 1: Numbers of nodes generated by non Alternative Model = 11

Diagram 2: Numbers of elements generated by non Alternative Model = 10

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Esteem Innovation Sdn Bhd, 2011

Diagram 3: Moment and shear result diagram Result obtained : Moment = 12.05 kNm Shear = 12.05 kN For the second run, switch the Alternative Model to True.

Diagram 4: Numbers of nodes generated by non Alternative Model = 4

Diagram 5: Numbers of elements generated by non Alternative Model = 3

Diagram 6: Moment and shear result diagram p495

Esteem Innovation Sdn Bhd, 2011

Result obtained : Moment = 12.05 kNm Shear = 12.05 kN Manual Calculation Beam self-weight = 24*0.15*0.6 = 2.16 kN/m Beam uniform dead load = 1 kN/m Total beam uniform deal load = 3.16 kN/m Beam uniform live load = 1 kN/m Beam ultimate uniform load = 1.4*3.16 + 1.6*1 = 6.02 kN/m 2 Moment = 6.02*4 / 8 = 12.05 kNm Shear = 6.02*4 / 2 = 12.05 kN Difference between non Alternative Model and Alternative Model Beam node will reduce from 9 numbers (depends on mesh size) to 2 numbers.

p496

Esteem Innovation Sdn Bhd, 2011

Equivalent Lateral Force Method for Seismic Analysis The example below shows the application of the vertical distribution of seismic forces based on the equivalent lateral force method.

Assumption: 1. Column dimension, 300 mm by 300 mm. 2. Total height of column = 6 m (3 m per floor). Node 5 to 1, 1 to 2, 6 to 3 and 3 to 4 are all 3 m in length. 3. Seismic Load Factor, Cs = 0.1 4. Seismic Distribution Exponent, k = 2 p497

Esteem Innovation Sdn Bhd, 2011

Loading application: In order to apply an equivalent loads of 500 kN at nodes 1 and 5, 50 kN at nodes 3 and 6 (in the X-direction), taking the column self weight (0.3 x 0.3 x 24 x 6 m = 12.96 kN per column of which 3.24 kN is applied onto nodes 2, 5, 4 and 6; 6.48 kN is applied onto nodes 1 and 3), 500 - 6.48 = 493.52 kN is applied onto node 1, 50 - 6.48 = 43.52 kN is applied onto node 3, 500 - 3.24 = 496.76 kN is applied onto node 5 and 50 - 3.24 = 46.76 kN is applied onto node 6.

The lateral force Fx induced at any level can be determined from the following expressions illustrated by Nawy (1).

Cvx = vertical distribution factor V = total design lateral force or shear at the base of the building. Wi and Wx = the portion of the total gravity load of the building, W, located or assigned to Level i or x hi and hx = the height from the base to level i or x k = seismic distribution exponent Calculations:

Force applied on nodes 2, 1 and 5 portrayed graphically,

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Esteem Innovation Sdn Bhd, 2011

The following model shows the application of the lateral loads at the nodes to benchmark to the seismic loading:-

And the following is the comparison of results for lateral load cases (equivalent lateral load & seismic loading):Loads applied laterally according to calculated distribution Bending moments

p499

Seismic Loading based on equivalent lateral force method

Esteem Innovation Sdn Bhd, 2011

Loads applied laterally according to calculated distribution Shear force

p500

Seismic Loading based on equivalent lateral force method

Esteem Innovation Sdn Bhd, 2011

5x5m Slab Benchmarked against: EsteemPlus Project Data: Create a project for the following data for the Benchmark. 1. 2. 3. 4. 5.

Create 5000mm in global x-direction and 5000mm in global y-direction. Add 4 beam at the 4 corner with the size of 200x500. Create a 125mm thickness slab with the finish load 1.2kN/m2 and live load 1.5kN/m2. Create 4 columns on the 4 corner with the size of 200x200. The project will looks as the figure show above.

Change the parameter of the FEM analysis in Esteem Plus and Esteem 7 1. Change the young modulus, E for both project to 24597 for beam, slab3D, Column wall and raft foundation in Esteem Plus. 2. Change the 2D Poisson ratio,V to 0.2 in Esteem Plus and slab Poisson ratio Esteem 8 to 0.2. 3. Change the mesh size to 2500 for both Esteem 7 and Esteem Plus. 4. Change the pin support in Esteem 7 parameter setting>3D Analysis>Pin support to false. 5. Type of mesh change to mix in Esteem 7 and Esteem Plus. 6. Young modulus E, reduction for beam torsion change to 50% for Esteem plus and Esteem 7 Torsion reduction to 0.5. 7. Untick the checkbox beam stiffness increase by 1 times in Esteem Plus and change the beam stiffness in Esteem 7 to1. 8. Lastly run the Project in Esteem Plus in shell option.

Figure 3 Parameter in Esteem Plus to be change

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Esteem Innovation Sdn Bhd, 2011

Figure 4 Parameter for Esteem Plus to change

Figure 5 Esteem 7 Parameter to change

Result Benchmark for Esteem Plus and Esteem 7 Displacement

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Esteem Innovation Sdn Bhd, 2011

Node Number

Esteem Plus Value

Esteem 7 Value

Different %

1

0.08

0.08

0

2

0.08

0.08

0

3

0.08

0.08

0

4

0.08

0.08

0

5

1.22

1.23

-0.8

6

1.22

1.23

-0.8

7

1.22

1.23

-0.8

8

1.22

1.23

-0.8

9

0.88

0.88

0

10

0.88

0.88

0

11

0.88

0.88

0

12

0.88

0.88

0

13

0.88

0.88

0

14

0.88

0.88

0

15

0.88

0.88

0

16

0.88

0.88

0

17

2.44

2.44

0

18

2.44

2.44

0

19

2.44

2.44

0

20

2.44

2.44

0

21

3.83

3.83

0

Node Number

Esteem Plus Value

Esteem 7 Value

Different (%)

1

0.08

0.08

0

2

0.08

0.08

0

3

0.08

0.08

0

4

0.08

0.08

0

5

4.14

4.14

0

6

2.48

2.48

0

7

4.14

4.14

0

8

2.48

2.48

0

9

1.19

1.19

0

10

1.19

1.19

0

Moment X

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Esteem Innovation Sdn Bhd, 2011

Node Number

Esteem Plus Value

Esteem 7 Value

Different (%)

11

0.34

0.34

0

12

0.34

0.34

0

13

1.19

1.19

0

14

1.19

1.19

0

15

0.34

0.34

0

16

0.34

0.34

0

17

3.68

3.68

0

18

3.68

3.68

0

19

3.68

3.68

0

20

3.68

3.68

0

21

4.24

4.24

0

Node Number

Esteem Plus Value

Esteem 7 Value

Different (%)

1

0.08

0.08

0

2

0.08

0.08

0

3

0.08

0.08

0

4

0.08

0.08

0

5

2.48

2.48

0

6

4.14

4.14

0

7

2.48

2.48

0

8

4.14

4.14

0

9

1.19

1.19

0

10

1.19

1.19

0

11

0.34

0.34

0

12

0.34

0.34

0

13

1.19

1.19

0

14

1.19

1.19

0

15

0.34

0.34

0

16

0.34

0.34

0

17

3.68

3.68

0

18

3.68

3.68

0

19

3.68

3.68

0

20

3.68

3.68

0

Moment Y

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Node Number

Esteem Plus Value

Esteem 7 Value

Different (%)

21

4.24

4.24

0

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Slab FEM result : EsteemPlus and Esteem 8 1 Objective: This benchmark result on FEM result for EsteemPlus(6.5.1.4) and Esteem 7(Release 0.0.10.2), is to get the identical results on FEM (same value) 2D analysis in both EsteemPlus and Esteem 7. Project Data For Example:

Figure 1 Esteem 7 Example key plan

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Figure 2 Esteem Plus Example key plan

2 Project Data: 2.1 Create a project for the following data for the Benchmark. 1. 2. 3. 4. 5.

Create 5000mm in global x-direction and 5000mm in global y-direction. Add 4 beam at the 4 corner with the size of 200x500. Create a 225mm thickness slab with the finish load 1.2kN/m2 and live load 1.5kN/m2. Create 4 columns on the 4 corner with the size of 200x200. The project will looks as the figure show above.

2.2 Change the parameter of the FEM analysis in Esteem Plus and Esteem 7 1. Change the young modulus, E for both project to 24597 for beam, slab3D, Column wall and raft foundation in Esteem Plus. 2. Change the 2D Poisson ratio,V to 0.2 in Esteem Plus and slab Poisson ratio Esteem 7 to 0.2. 3. Change the mesh size to 2500 for both Esteem 7 and Esteem Plus. 4. Change the pin support in Esteem7 parameter setting>3D Analysis>Pin foundation (column) to false. 5. Type of mesh change to mix in Esteem 7 and Esteem Plus. 6. Young modulus E, reduction for beam torsion change to 50% for Esteem plus and Esteem 7 Torsion reduction to 0.5. 7. Untick the checkbox beam stiffness increase by 1 times in Esteem Plus and change the p507

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beam stiffness in Esteem 7 to1. 8. Lastly run the Project in Esteem Plus in shell option. 9. On the design parameter, use the below parameter: a. Concrete grade = 30 N/mm2 b. Concrete cover = 30mm c. FE slab above average peak moment design = 100% d. Checked the option Full FEM Design in EsteemPlus.

Figure 3 Parameter in Esteem Plus to be change

Figure 4 Parameter for Esteem Plus to change

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Figure 5: EsteemPlus Design Parameter

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Figure 6 Esteem 7 Parameter to change

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Figure 7: Esteem 7 Parameter to be changed in 3D analysis

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3 Result Benchmark for Esteem Plus and Esteem 7. 3.1 Slab Design Report

Figure 8: Esteem Plus Design report

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Figure 9: Esteem 7 Slab Design Report -1

Figure 10: Esteem 7 Slab Design Report 2 p513

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Item

Esteem Plus Value Esteem 7 Value

Difference

Imposed Load

11.64

11.64

0

Moment X

15.73

15.73

0

Moment Y

15.73

15.73

0

Area Required X

205

202

3 mm2

Area Required Y

226

221

5mm2

Area Provided X

T16-250 -BB

T16-250 -BB

Area Provided Y

T16-250 -BT

T16-250 -BT

4.0 Conclusion: As conclusion from the table show above, we can see that the slab design report show the similar result for Esteem Plus and Esteem 7 on the imposed load, moment x, moment y, area provided. There is some minor different in the Area required between Esteem Plus and Esteem 7, but the different is still within the acceptance range.

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Beam FEM result : EsteemPlus and Esteem 8 1 Objective: This benchmark result on FEM result for EsteemPlus(6.5.1.4) and Esteem 7(Release 7.0.125.0-C), is to get the identical results on FEM (same value) 2D analysis in both EsteemPlus and Esteem 7. Project Data For Example:

Figure 1 Esteem 7 Example key plan

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Figure 2 Esteem Plus Example key plan

2 Project Data: 2.1 Create a project for the following data for the Benchmark. 1. 2. 3. 4. 5.

Create 5000mm in global x-direction and 5000mm in global y-direction. Add 4 beam at the 4 corner with the size of 200x500. Create a 225mm thickness slab with the finish load 1.2kN/m2 and live load 1.5kN/m2. Create 4 columns on the 4 corner with the size of 200x200. The project will looks as the figure show above.

2.2 Change the parameter of the FEM analysis in Esteem Plus and Esteem 7 1. Change the young modulus, E for both project to 24597 for beam, slab3D, Column wall and raft foundation in Esteem Plus. 2. Change the 2D Poisson ratio,V to 0.2 in Esteem Plus and slab Poisson ratio Esteem 7 to 0.2. 3. Change the mesh size to 500 for both Esteem 7 and Esteem Plus. 4. Change the pin support in Esteem7 parameter setting>3D Analysis>Pin foundation (column) to false. 5. Type of mesh change to mix in Esteem 7 and Esteem Plus. 6. Young modulus E, reduction for beam torsion change to 50% for Esteem plus and Esteem 7 Torsion reduction to 0.5. 7. Untick the checkbox beam stiffness increase by 1 times in Esteem Plus and change the beam stiffness in Esteem 7 to1. 8. Lastly run the Project in Esteem Plus in shell option.

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Figure 3 Parameter in Esteem Plus to be change

Figure 4 Parameter for Esteem Plus to change

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Figure 5 Esteem 7 Parameter to change

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Figure 6: Esteem 7 Parameter to be changed in 3D analysis

3 Result Benchmark for Esteem Plus and Esteem 7. 3.1 Moment

Figure 7: Esteem 7 Beam moment result

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Figure 8: Esteemplus Beam moment result

Distance

Esteem 7

EsteemPlus

Difference (kNm)

0

-6.4

-5.7

-0.7

250

5.03

5.6

0.57

500

15.61

16.1

0.49

750

25.29

25.7

0.41

1000

33.56

33.9

0.34

1250

40.91

41.3

0.39

1500

46.72

47.0

0.28

1750

51.52

51.8

0.28

2000

54.71

55.0

0.29

2250

56.88

57.2

0.32

2500

57.39

57.7

0.31

2750

56.88

57.2

0.32

3000

54.73

55.0

0.27

3250

51.52

51.8

0.28

3500

46.71

47.0

0.29

3750

40.91

41.3

0.39

4000

33.57

33.9

0.33

4250

25.28

25.7

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4500

15.59

16.1

0.51

4750

5.03

5.6

0.57

5000

-6.4

-5.7

0.7

3.2 Shear

Figure 9 :Esteem 7 Beam Shear diagram

Figure 10: Esteemplus Beam Shear Diagram p521

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Distance

Esteem 7

EsteemPlus

Difference (kN)

0

44.6

44.3

-0.3

250

42.15

42.0

-0.15

500

38.51

38.3

-0.21

750

34.16

34.0

-0.16

1000

29.72

29.7

-0.02

1250

25.06

25.0

-0.06

1500

20.25

20.2

-0.05

1750

15.24

15.2

-0.04

2000

10.17

10.2

0.03

2250

5.07

5.2

0.13

2500

0.01

0

-0.01

2750

-5.08

-5.0

0.08

3000

-10.24

-10.2

0.04

3250

-15.28

-15.2

-0.08

3500

-20.22

-20.3

-0.08

3750

-25.04

-25.1

-0.06

4000

-29.73

-29.7

0.03

4250

-34.15

-33.9

0.25

4500

-38.42

-38.3

0.12

4750

-42.07

-42.0

0.07

5000

-44.59

-44.3

0.29

4.0 Conclusion The example project used for the above, show that the beam benchmark result between Esteem plus and Esteem 7. From the result, we can show that the result in Esteem plus and Esteem 7 is almost identical and the difference is within the acceptance the range.

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Single Panel Shear Wall Shear wall benchmark documentation EXAMPLE 1: Objective: To prepare a benchmark and to compare results (displacement values) for a shear wall subjected to lateral loading, by using: 1) ESTEEM 7 2) Manual calculation (theoretical) 1) Using ESTEEM 7 Create a shear wall with the properties below: Height = 7500 mm Width = 2400 mm Thickness = 350 mm fcu = 30 N/mm2 Load acting on top of shear wall = 1000 kN

To create the shear wall above: 1. Make a grid 2400mm on the x-direction using the Grid Input icon. 2. Create a wall using Wall Input icon. Length of the wall is 2400mm on the point of intersection of the grid created. 3. Still in the Wall Input mode, click on the wall created, and change the thickness of the wall to 300mm on the Object Viewer Panel.

4. Change the elevation of the wall using the Elevation Input icon. In the Elevation Input mode, click on the wall to change its floor height to 7500mm. (in the Object Viewer Panel)

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5. Input a lateral loading with magnitude 1000kN at the left side of the wall using Lateral Load Input icon. Click Edit Cases in the Wind Load Display mode and add a new wind load (Eg; WL 1). Set the magnitude of the wind load to 1000, and put an Angle of Attack of 0°.

6. The plan view of the wall:

7.Click on the Elevation Input icon and view the elevation of the wall:

8. Analyze the complete wall model using both 2D and 3D analysis: p524

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9. To view the results of the analysis, click on 3D Analysis Display icon. The 3D Contour of the wall is displayed. Select Load Type > Individual Load and select the wind load created previously to view the contours due to this load:

10. Since the wind load is acting in the plane of the wall, the results of blue colored contours indicates zero (0) value of moments. This indicates that the moments produced are very small.

11. Click on the Nod icon to display node numbers of the meshed wall. Change from 3D Contour Display to 3D Displacement to view the displacement of the wall.

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12. Right click on the wall and check the Nodes to display the nodes on the wall. Zoom in to see clear the location of the nodes.

13. Use Zoom Window to view closer the node no. 9. Click until the node is selected. Click again until the displacement values of node 9 (x,y,z) are shown in the Object Viewer panel.

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Results:  Displacements Since the lateral load 1000kN acts in positive-x direction, the top Displacement of the shear wall in x-direction is also positive. (at node 9) Manual calculation

ESTEEM 7

15.00 mm

15.46 mm



Bending and shear of the wall

To check the bending and shear of the wall, click on the Wall and Column Reactions icon. Due to the lateral load (WL 0), the resulting moment, My is 7499.99kNm, and shear Vx in x-drection is -1000kN.

2) Using manual calculation: Ec = 4500√30 = 24.6E3 N/mm2 GAv = 0.42Ec = 10.33E6 kN/m2

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Aspect ratio = 7.5/2.4 = 3.125 ≈ 3 If aspect ratio is 1
ESTEEM 7

Manual calculation

15.46

15.00

Moment (kNm)

1000.00

1000.00

Shear (kN)

7499.99

7500.00

Displacement at node 9 (x-direction), Δ mm

EXAMPLE 2: In this example, the same wall of the same properties is added with two additional loadings 50kN in y-direction. The objective is to check the resulting moments produced by the additional loading. 1. Repeat Step 5 in Example 1 to input two 50kN loading. Set the magnitude of the load to 50, and Angle of Attack 90. Put the first loading on Grid (1,A) and the second on Grid (2,A).

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Plan view of the wall with additional loadings 50kN

2. Run the 2D and 3D Analysis. View the 3D Contours in the 3D Analysis Display icon. Change the Individual Load considered to the additional 50kN loadings. 3. Display the nodes of the wall. Click on the Mx icon to view the bending moment in x-direction. The red contours at the bottom of the wall indicates larger moments compared to the very small (zero) bending moments with blue contours at the top. These moments are also greater than the moments in x-direction obtained due to loading in Example 1.

4. Check the displacement of node 9. Compared to Example 1, the displacement in y-direction is not zero, which is due to the 50kN loadings acting in y-direction. p529

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5. Check the bending moment and shear of the wall in the Wall and Column Reactions icon. The results should show that there is bending moment reaction in x-direction and produced shear in y-direction due the 50kN loading.

Conclusion: The results of reactions of the shear wall obtained from ESTEEM 7 are acceptable.

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Flat Slab

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Verification Column Drop regions overlap or <500mm to each other Integrity check fails when there are two or more column drop regions overlap or have a distance less than 500mm to each other. For example: 1) When two columns have drop regions overlap with each other.

Run integrity check and the check will fail with the error message shown as below.

2) When 2 column drop regions situated less than 500mm apart.

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Integrity check will fail with the error message shown below.

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Column Supporting Flat Slab Integrity checks for columns in Flat Slab design a. A column is not allowed to support two flat slabs at the same time (see below)

b. A column is not allowed to cut across the slab it supports. For example, column (1000mm height) equals the slab dimension in y-direction. This is not allowed because such column input will divide the slab it supports.

c. At least 50% of the area of the column must support the flat slab. For example: Columns at the corner edges. Not allowed

Allowed

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Range of Column Drop B and Drop H values The reason to control the input values of Drop B and Drop H is to prevent the punching shear check area to be out of the custom ized drop area. User is still allowed to change the Drop B and Drop H values when CustomizedDrop = True but these values will be restricted to a minimum value. The default value of Drop B and Drop H , which is also the minimum value, = Max(column width , column height) + 4*h where h = slab thickness + column drop thickness The maximum value of Drop B and Drop H = Max(column width, column height) + 8*h When user input the Drop B and Drop H values out of the acceptable range, a Warning message will appear during Verification process to notify user about this matter after the integrity check is done. However, user is still allowed to proceed to the analysis and design stage by ignoring the warning message. For instance, a column on flat slab key plan having properties as shown,

Minimum Drop B & Drop H = Max(300,300) + 4*125 = 800 Maximum Drop B & Drop H = Max(300,300) + 8*125 = 1300 Acceptable range is between 800 and 1300 mm. However user input value is 500 mm which is not within the acceptable range. Run integrity check will come out a warning message regarding this matter as shown per screen shot.

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Slab Opening in Flat Slab Any slab opening should be at least 300mm from either the c olumn or the Column Drop Region. Otherwise, the slab opening fails in the Verification. Below are some examples where the location of slab openings are not allowed since the distance from the opening to the c olumn or Column Drop Region is less than 300mm. The openings are too near to the Column Drop Region.

The openings are too near to the c olumns.

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Analysis Meshing of column drop area Take a flat slab model below as example. There are column drops at each corner and the middle.

After generating the 2D mesh on plan, the result is as shown below.

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The mesh region will follow the overlapping drop area of the column drops to the slab.

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Design & Detailing Flat Slab not designed Flat slab will not be designed. In the Failed Element list, user is informed of the flat slab which is not designed.

User may use Punching Shear Calculator to check the punching shear at the column locations.

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Punching Shear Calculator Average Moment Over Support For details regarding computation of average slab moment, refer to User Manual - rel="nofollow"> Main Menu -> Tools -> Punching Shear Calculator -> "Average slab moment". For details regarding band strip cut on the moment contour, refer to Technical Documentation -> Raft -> 3D Analysis -> "Band strip contour cut section." The default Interval of Distance (used for obtaining the moment values) along the band width is specified in Project Parameter -> Analysis Setting -> Misc -> Interval of Distance. The Punching Shear reports are available after Slab Design. Refer to topic "Punching Shear Reports". 1. Average Moment Over Column Support: Edge Column Whenever there are hogging and sagging moment obtained from the band strip contour c ut, the maximum hogging moment value will be c onsidered instead of the m aximum sagging moment value. Refer to example below: Average moment over the support = Maximum hogging moment (a, b) = a = -28.31kNm.

Band strip cut interval = every 1000mm

2. Sagging moment case Whenever there are sagging moment obtained from the band strip contour cut, the average moment value next to the column region (say plot at 1000mm from left grid) will be considered. Refer to example below:

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Average moment over the support = moment value next to c olumn at interval 1000mm = 17.8kNm [$X$IF_CON1]If sagging moment case, the moment at the surface of the drop region/profile or the next point value next to the column region will be given. If hogging moment case, moment value for the next point (if higher moment value can be obtained out of the drop region/column profile) will be given. Refer Case 28136.

Band strip cut interval = every 1000mm

3. Column local axis is not the same as the slab local axis. The program automatically check for the punc hing shear based on the minimum reinforcement percentage required defined in the design parameter for the slab. Refer to example below. Column local axis

The punching shear will be carried out based on the minimum steel area required defined in Parameter -> Design -> Floor -> Slab -> Design -> Minimum Steel Percentage (default 0.13%). A warning message is shown in the punching shear report to inform user. Refer to sample of report below.

Slab local axis

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PUNCHING SHEAR CHECKING REPORT FOR FLAT SLAB - COLUMN (1B,A1) Material and Design Data Code of Practice BS8110 : 1997

fcu (N/mm²) 30

fy (N/mm²) 410

Cover (mm) 25

Enhanced Shear Multiplier 1.5

Column Shape: Rectangular (800 × 800 mm) Finishes = 1.00 kN/m² Live Load = 0.25kN/m² Bar diameter in direction 1 and 2 = 10 mm Effective depth in direction 1 = 170.0 mm Effective depth in direction 2 = 160.0 mm Required steel area in direction 1 = 600 mm² Required steel area in direction 2 = 600 mm² The column local axis is not the same with the slab direction, therefore minimum steel percentage is used for punching shear checking. Steel percentage in direction 1, As1 = 0.35 % Steel percentage in direction 2, As2 = 0.38 % 4. Punching shear perimeter around column overlaps with slab opening. The program will not carry out the punching shear checking for the column location where the punching shear perimeter overlaps with slab opening(s). Therefore in the Slab Detailing output, these locations are marked as "Punching Shear Not Check".

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Raft

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Input Column Drop Refer to topic "Range of Column Drop B and Drop H values" for the range of column drop region allowed. When 3D Analysis > Use Raft Foundation is TRUE, 1) When CustomizedDrop = False, Column Drop B = Column Drop H = Max (Column width , Column height) + 4*h where h = 2 * raft slab thickness

Warning

:

The dimension of the auto-computed column drop when CustomizedDrop = False, is only for viewing and guiding purpose. The auto-computed values will NOT be considered in the Punching Shear Check and Raft Analysis. 2) When CustomizedDrop = True, Column Drop B & Drop H can be defined by user. By default, the values are the same as when CustomizedDrop = False.

For Example:

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3D view:

Notes

:

1. Only when Raft Foundation is set to True, the column profile on the lowest floor (Raft Floor) plan input is drawn as dotted lines without fill as according to the dimension stated. Otherwise, the column input profile is as normal. 2. Customized Drop property is only available when the following Hybrid Raft Type is chosen: a) Column Drop / Column Rigid b) Footing Rigid c ) Pile Rigid

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Column Drop Value Outside Acceptable Range Project verification will return a warning for column drop with value outside of acceptable range. User should notice that when column drop value is outside of acceptable range, punching shear checking will not be performed. Example:

When run Raft Punching Shear

, it will return No raft punching shear results.

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In order to obtain result for Raft Punching Shear, user have to change column drop to a value within acceptable range.

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Slab Opening on Raft Slab There are 3 types of slab openings in Esteem 8. 1) Rectangular Opening 2) Circular Opening 3) Polygonal Opening

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Rectangular Opening This command is to add a rectangular opening on the raft slabs. First click on the rectangular opening on slab input mode.

icon

Select the raft slab that wants a rectangular opening. A red highlight will appear around the slab when it is selected by user.

Click on any corner of the raft slab as the reference for the rectangular opening, a dimension will be shown in x and y directions when the mouse cursor moves within the slab.

Click on the slab to determine the location of the rectangular opening. User can change the properties of the rectangular opening on the object viewer or data sheet input

.

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Circular Opening This command is to add a circular opening on the raft slabs. First click on the circular opening the slab input mode.

icon on

Select the raft slab that wants a circular opening. A red highlight will appear around the slab when it is selected by user.

Click on any corner of the slab as the reference for the circular opening, a dimension will be shown in x and y directions when the mouse cursor moves within the slab.

Click on the slab to determine the location of the circular opening. User can change the properties of the circular opening on the object viewer or data sheet input

.

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Polygonal Opening This command is to add a polygonal opening on the raft slabs. First click on the polygonal opening on the slab input mode.

icon

Select the raft slab that wants a polygonal opening. A red highlight will appear around the slab when it is selected by user.

Click on any corner of the raft slab as the reference for the polygonal opening. A dimension will be shown in x and y directions when the mouse cursor moves within the slab.

Click on the slab to determine the location of the polygonal opening. User can change the properties of the polygonal opening on the object viewer or data sheet input

.

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Column Point Column Point is a point object that has NO value and NO area when being input on plan. It only acts as a point for the program to detect that there is a node on the raft slab when mesh is generated. No spring constant value can be assigned to Column Point. Column Point is only allowed on raft plan when Raft Foundation = True.

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Range of Subgrade Reaction To call out the table of Subgrade Reaction.

User can select from the range of Soil Subgrade Reaction table in the Object Viewer as 3 highlighted below, OR manually insert the value between the range of 1 - 1000000000kN/m

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Verification Column profile sitting outside of raft slab edge When there are columns partly sitting outside the boundary of the raft slab edge, the integrity check will fail showing error in the raft plan when 3D analysis is run. For instance,

When the (2D + 3D) analysis is run, the project integrity fails and an error message will be displayed to highlight the underlying problems. In this case, the error after integrity chec k is as shown.

Tips: User can always run the "Check Plan Integrity" or "Project Verification" before running the (2D+3D) Mesh Analysis to check whether any error or warning exists in the project.

OR

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Column not fully input within raft slab When there are columns partly being input outside the boundary of the raft slab edge, the Plan Integrity check will fail. For example;

When the integrity check is run, it will fail and an error message will be displayed to highlight the underlying problems. In this case, the error after integrity check is as shown.

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Curve beam and curved slab not allowed in raft foundation Curve beams and curve slabs are not allowed on raft floor. Raft floor which contains curve beams gb6 and gb7, and curve slabs FS5 and FS6 will fail in Verification. Refer to example below for an alternative.

As an alternative, use straight beams instead of curve beams. 1. Select the curve beams, and right-click to go to the command "Convert to Straight Beams". 2. After the curve beams have been converted to straight beams, re-input the slabs again.

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Openings in Raft 1) Refer to topic Technical Documentation -> Flat Slab -> Verification -> Slab Opening in Flat Slab. 2) Beam Openings are NOT allowed on raft plan.

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3D Analysis Columns on Raft 1. For an upper transfer column (column which sits on the raft foundation), the upper column must be within the profile of either: o Slab, or o Beam

2. The Column Drop Region can be shared with more than two slabs if there is a beam in between them.

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3. The upper transfer column will be bridged to the beam nearby which is present.

Figure a: 3D View

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Figure b: 3D Analysis with bridging beams 4. Where the Column Drop region edge to the slab edge is less than 150, the drop region created will be extended to the slab edge.

Plan View 3D Mesh

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Raft Foundation Contour The command is known as Raft Foundation Contour as it displays the contour of raft foundation after running of 3D analysis. By default, the raft foundation contour commands will appear after 3D analysis when raft foundation key plan is viewed. Category Raft foundatio n Contour

Command

Action

View/Hide Mesh

This allows user to view the mesh on the contour.

View/Hide Node Number

This allows user to view node number on the contour. This allows user to view element number.

View/Hide Element Number View/Hide Node Value

This allows user to view node value for each node result.

Critical Value

This allows user to view the critical value for each raft contour result.

Displacement

This allows user to view the displacement contour.

Moment X

This allows user to view moment X contour.

Moment Y

This allows user to view moment Y contour.

Shear Stress in XZ plane

This allows user to view shear stress in XZ contour.

Shear Stress in YZ plane

This allows user to view shear stress in YZ contour.

This allows user to view maximum Maximum moment in X- direction moment in X-direction contour. Minimum moment in X- direction

This allows user to view minimum moment in X-direction contour.

Maximum moment in Y- direction

This allows user to view maximum moment in Y-direction contour.

Minimum moment in Y- direction

This allows user to view minimum moment in Y-direction contour.

Maximum Shear Stress in XZ

This allows user to view maximum shear stress in XZ contour.

Maximum Shear Stress in YZ

This allows user to view maximum shear stress in YZ contour.

X-direction Top Steel Area

This allows user to view X-direction top steel area contour.

Y-direction Top Steel Area

This allows user to view Y-direction top steel area contour.

X-direction Bottom Steel Area

This allows user to view X-direction bottom steel area contour.

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Y-direction Bottom Steel Area

This allows user to view Y-direction bottom steel area contour.

Shear reinforcement required in

This allows user to view shear reinforcement required in XZ contour.

Shear Reinforcement required in

This allows user to view shear reinforcement required in YZ contour.

Maximum Soil Pressure

This allows user to view maximum soil pressure contour.

Minimum Soil Pressure

This allows user to view minimum soil pressure contour.

Spring Reaction

This allows user to view the spring reaction of the raft.

XZ

YZ

User is allowed to choose the type of Load Type contour to be displayed. The Load Types that are available for viewing are Load Combination, Individual Load and Seismic Combination.

Tips: By pointing the mouse cursor around the contour plan, the contour value for each corresponding contour value will be displayed. (eg. the value 14.5 displayed below).

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Cut Section on Raft Contour This is a post-processing feature to display the results along a cut section across the raft slab. No 1

Details To define a cut section

Description/Steps/Input

Expected Result/Outcome/Output

1. Move cursor closer to a gridline as reference.

2. Next, click on the gridline to select it as reference grid. C lick the mouse again after setting the position of the cut section.

Tips: Right Click to turn on gridlines if gridlines are not displayed.

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No 2

Details To display results in a cut section

Description/Steps/Input

Expected Result/Outcome/Output

1. Move cursor to a cut section label, "Click to view section" is shown.

2. C lick on the cut section label to open a view on the cut section.

3

To delete a cut section

1. C lick on the line of the cut section to be deleted.

2. Press "Delete". 3. To delete multiple cut section lines, use Window Select. Use Window Select to select multiple cut section lines and press "Delete".

4

To make an arbitrary cut section

1. On the 3D Contour View, right-click and choose "Arbitrary contour cut".

2. Define the location of the section cut position by simply clicking on two points as the starting point [1] and the ending point [2].

3. For orthogonal arbitrary contour cut, right-click and choose "Ortho" while "Arbitrary contour cut" is selected.

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No

Notes

Details

Description/Steps/Input

Expected Result/Outcome/Output

:

The properties of the cut section line can be found in the object viewer. User must first select the cut line then only the Object Viewer will appear.

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Raft slab cut section display A cut section window in view can display multiple results depending on the type of results wanted to be viewed for that one particular cut section only. Users are allowed to select the Loadtype and Load Combination to be displayed. Other view cut sections will be displayed in different windows each (if there are many). Below is the example of the display for raft slab cut section.

Notes

:

Load Types available include Load Combination, Individual Load and Seismic Combination.

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Reset of 3D Mesh Analysis if input object data is edited on raft floor This applies when the whole project has run (2D+3D) batch process. On raft key plan, when there are changes made to the input object data, the program will automatically reset the 3D Mesh/Analysis. The followings are several types of actions done after (2D+3D) batch process that will cause the analysis results to be unsynchronized ('Unsync' state): 1) Editing of column "raft" properties such as: (a) If Column Drop Region is changed. - Customized Drop = True, any changes on Drop B, Drop H and Thickness will 'Unsync' the 2D floor plan. - When Customized Drop is changed from False to True or vice versa, it will also 'Unsync' the 2D floor plan. (b) Change of Spring Constant (c) Change of Customized Support. 2) Changing of slab properties on raft i.e. Soil Subgrade Reaction, Point Loads, Line Loads, Patch Load, etc.

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Benchmark 3D raft analysis with 2D frame analysis Reason : To benchmark and compare the results of 3D analysis with normal 2D frame analysis so as to justify the reliability of the raft analysis result. The input project data is as follow:

The results of 3D Raft analysis and 2D frame analysis:

The results obtained above are when both T-beam effect = 1.0 There are slight difference in the comparison of the node values and contour pattern. p575

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However, it is still within the acceptable range as it is due to the random mesh generated in Esteem 8.

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External Loads on Raft Slab 3D Mesh

After running (2D+3D) Analysis, go to 3D Analysis Display to view the contour lines of the raft slab. To show external loads on the raft, Right click on the 3D analysis display plan and select "External Load" to turn it on.

The example result will be as follow:

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Band strip contour cut section Band width cut section has the following properties : (i) Width (of the band) (ii) Interval of the cut These can be found in the Object Viewer whenever a cut section is made on the raft contour.

By default, the BandStrip is set to False. When user wants to view the Band Strip contour, it has to be changed to True. User can also change the bandstrip properties while the cut section form is in view as shown below.

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The concept of Band Strip is as shown below:

The input range allowed for Interval is between 300mm to 3000mm while for Band Width is between 500mm to 10,000mm.

Tips:

The operation of band width contour cut is the same as normal raft contour cut.

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Full Band Width By default, Full Band Width = False when Band Strip Contour is set to True. The function of Full Band Width depends on the width of the raft slab and only functions when Band Strip Contour = True. The specifications adopted in Esteem 8 is as follow: 1) If Full Band Width = False AND slab width less than 1000mm, the c ontour strip result will not be corresponding to the bandwidth changes. Otherwise, the contour strip varies with BandStrip  slab width. 2) If Full Band Width = True, contour strip will multiply with Min(Bandwidth , Slab width) in meter to get the exact band strip result. (Normally useful for raft slab with width less than 1000mm) For example a slab with width = 600mm (<1000mm):

A horizontal cut in the middle is created on the contour plan.

1) If FullBandWidth = False AND raft slab width < 1000mm, the Mx c ontour graph for Live Load is as shown per screen shot. The result will not correspond to the Bandwidth value. This is because the contour strip produced by default is based on force per meter width (i.e 620kNm / 1 meter width) even though the width of the slab is only 600m m .

2) If FullBandWidth = True, the Mx contour graph for Cut 1 on Live Load is shown as per screen shot. The result now will be based on bandwidth or slab width whichever is lesser. For p581

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example the graph below shows Mx = 620.68 * 600/1000 = 372.41kNm

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Use Rectangular Mesh for Raft Slab Rectangular mesh will automatically be used for raft slab that fulfills these three conditions:

o 3D Analysis > Raft > Use Raft Foundation is TRUE o ONE floor slab ONLY o Square or rectangular slab ONLY (i.e. with orthogonal regions and no grid rotations) Consider the following models for comparison. Model plan view a. Rectangular slab with no columns

3D Raft Slab Mesh (Isometric View) -Use square mesh

b. Slab with non-orthogonal regions

-Use random mesh

c. Two-floor slab

-Use random mesh

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Column Point as fixed node Column Point on raft will be meshed as a Fixed Node with zero spring effect. Example of raft with column points and its mesh are shown below:

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Analysis Failed due to Raft Foundation Uplift Whenever there is an uplift in the raft foundation, the program will not proceed with the analysis. An error message with exception will pop up to inform user as shown below.

The nodes that cause uplift in raft foundation (caused by negative soil pressure) are highlighted in the 3D Analysis Display > list of nodes with Large Displacem ents.

Refer to User Manual > Main Menu > Analysis Result > 3D Analysis Result, for the descriptions regarding 'Large Displacements'.

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Rigid Zone In Esteem 8, the rigid zone on raft pad is generated automatically on raft. The thickness and size of the rigid zone are determined by the equations below:

Edge Length 1 = 2*Max(b,h) Edge Length 2 = 1000*[k*P/(SBP*100)]  B = H = Max(Edge Length 1, Edge Length 2) where b h P SBP B H k

= the width of the c olumn, mm = the height of the colum n, mm = the total loading that the column supports (including self weight), kN 2 = Soil Bearing Pressure, kN/m = the width of the rigid zone, mm = the height of the rigid zone, mm = "Auto Thickening Area" parameter, %

If Edge Length 1 is used, Thickness = Slab Thickness If Edge Length 2 is used, Thickness = 2 * Slab Thickness where Thickness = the property thickness for the rigid zone Slab Thickness = the slab thickness of the raft that the column sits on To view the rigid zone generated after running the (2D+3D) analysis, go to Analysis Result < 3D Analysis Display. The mesh and rigid zone on the raft plan can be seen clearly. For instance:

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Notes

:

1. The rigid zone created is always SQUARE/RECTANGULAR in shape. 2. If the rigid zone size  slab region size, the rigid zone will only be meshed at the overlapping area.

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3. If two rigid zones (eg. from two columns) overlap, a single rigid zone that bounds the two original rigid zones will be used instead. For instance as shown below:

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Meshing of wall on raft edge For any wall sitting less than 300mm on raft from the raft edge, the wall's bottom points will be bridged to the points along the raft edge.

Figure 1: Plan view of wall sitting at 300mm or less from the raft edge

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Figure 2: Bridging beams are provided at the wall bottom

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View Raft support nodes reaction display

The raft support node reactions can be viewed by going to one of the display options below after the 3D analysis is run. Take the above raft key plan as per screenshot as an example for each display option. 1) Raft support nodes reaction can be viewed by clicking Pad or Pile in Project Explorer < Foundation < Pad or Pile. The reactions for Load Combination Load Type displayed in this option are unfactored. User can choose the type of Loadtype to view on the right side bar. Below are the results based on the example above for LC = DL + LL.

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2) Or, another way to display raft support reactions is by clicking on the icon or go to Analysis Result < Wall and Column Reactions. After that, clic k on "3D Option" on the right window to view the reaction values. The reaction values for Load Combination Load Type displayed in this option are of factored values. For instance displayed below is LC = 1.6DL + 1.4LL:

3) The tabulation for the support reactions also available in Analysis Result < 3D Geometry, Load Reactions < Reactions. For example from the same raft key plan above:

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Design Raft Floor Parameter To call out the Floor Parameter for Raft slab, go to Main Menu > Floor Parameter, or right-click on the Raft floor from Project Explorer. This Floor Parameter would make the Raft slab to have its own sets of parameters. For example, to change the Ec value for Raft slab.

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By-pass Batch Design When Raft Foundation = TRUE in parameter setting, the program will not design for the lowest floor when user runs the (2D+3D) batch design even though pad design and pile design are selected. This is because the program will by-pass the batch design for the lowest floor. The batch process will only run the (2D+3D) analysis for whole project including the lowest floor but skip the design process for all elements at the lowest floor including pile and pad. The below screenshot shows the project status with raft foundation = True that underwent 3D batch design process.

Notes

:

N/A means Not Available.

Therefore in this case, the raft floor design, pad footing design and pile footing design are not available when Raft Foundation = True.

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Parameter Edge Spring Multiplier Constant Edge spring multiplier constant is a function to stimulate raft slab edge stiffening effect by adding soil spring along the edge using a constant times the slab thickness. By default, the value is 0 where no edge stiffening is provided to the raft unless user requests for it, then user has to change the value in the parameter setting. The limit range is from 0 to 10.

Below is the illustration on how edge spring multiplier constant functions.

The raft is virtually enlarged as shown in dotted line to demonstrate that the extra raft edge is imaginary. i.e. It does not contribute any finite element stiffness but only add extra soil spring to the edge. This 'extra' soil spring is collected as extra soil stiffness for the boundary condition of original raft slab boundary edge nodes. p598

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For instance; Refer to a simple symmetrical model below with a slab thic kness of 200mm,

1) When the edge spring multiplier constant = 0, There is no slab stiffening around the slab boundary. Therefore, the contour result is as per screenshot.

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2) When the edge spring multiplier constant = 2, The depth of the virtual edge stiffening around the slab boundary = 2 * 200 = 400mm Therefore, smaller values of displacement contour will be obtained since the raft slab edge has been stiffened by added soil spring constant.

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Design Result Punching Shear Perimeter The punching shear checking for raft foundation is done automatically at every section interval, k of 0.5*d, 1.00*d and 1.50*d (shown in Figure 1). The program will give a warning if the shear stress in the section considered exceeds the allowable shear stress.

Figure 1 The total length of punching shear perimeter for calculating the shear stress will depend on the location of the column; a) internal column, b) edge column, c) corner column. Consider a flat slab supported by a square column below: The slab local axis 1 is in the direction shown in Figure 1 while direction 2 is perpendicular to the slab local axis 1. Where, Critical length in direction 1 (mm) = CL1 Critical length in direction 2 (mm) = CL2 Column width = a (mm) Column height = b (mm) For punching shear checking at section equal to k = 1.50 away from the column face. Column type/location a) Internal column

Punching shear perimeter

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CL1 = column width + 2*k*d = a + 2*1.50*d CL2 = column height + 2*k*d = b + 2*1.50*d Total length for punching shear perimeter: Parallel to direction 1, U1 = 2*CL1 Parallel to direction 2, U2 = 2*CL2 b) Corner column

CL1 = column width + k*d = a + 1.50*d CL2 = column width + k*d = b + 1.50*d Total length for punching shear perimeter: Parallel to direction 1, U1 = CL1 Parallel to direction 2, U2 = CL2 c) Edge column

CL1 = column width + k*d = a + 1.50*d CL2 = column width + 2*k*d = b + 2*1.50*d Total length for punching shear perimeter: Parallel to direction 1, U1 = 2*CL1 Parallel to direction 2, U1 = CL2

Notes

:

Punching shear checking will NOT be carried out for the following cases: 1. Punching shear perimeter for upper column overlaps with raft slab opening. The program will not carry out the punching shear checking for the column location where the punching shear perimeter overlaps with slab opening(s).

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2. Column with no Drop Region: The punching shear perimeter at 1.5*d overlaps with an upper RC Wall. 3. Column with Column Drop Region:The Column Drop Region overlaps with an upper RC Wall.

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Average Moment For details regarding computation of average slab moment, refer to User Manual - > Main Menu -> Tools -> Punching Shear Calculator -> "Average slab moment". For details regarding band strip cut on the moment contour, refer to Technical Documentation -> Raft -> 3D Analysis -> "Band strip contour cut section." The default Interval of Distance (used for obtaining the moment values) along the band width is specified in Project Parameter -> Analysis Setting -> Misc -> Interval of Distance. 1. Average Moment: Edge Column Whenever band strip contour cut across the drop region contains positive and negative bending moment value, the positive bending moment value will be considered instead of the negative bending moment value. Refer to example below: Average moment = Maximum sagging moment = 37.88kNm

Band strip cut interval = every 1000mm

2. Average Moment: Column not shared by more than one raft slab Whenever a transfer column on raft is only sitting on one of the raft slab, the cut section is made within the raft slab only. Refer to example below:

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Average moment for column (3,A) which is sitting within slab FS2 is obtained by creating bandstrip cut within slab FS2 only. = Maximum = 24.66kNm

Band strip cut interval = every 500mm Bandwidth = 500mm (column size)

3. Column local axis is not the same as the slab local axis. The program automatically check for the punc hing shear based on the minimum reinforcement percentage required defined in the design parameter for the slab. Refer to example below. Column local axis

The punching shear for the raft slab will be carried out based on the minimum steel area required defined in Parameter -> Design -> Floor -> Slab -> Design -> Minimum Steel p606

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Percentage (default 0.13%). A warning message is shown in the punching shear report to inform user. Refer to sample of report below.

Notes

:

1. The column results are obtained as normal from the Analysis Results -> Plan Loading Results View. Slab local axis

PUNCHING SHEAR CHECKING REPORT FOR RAFT FOUNDATION COLUMN (1B,A1) Material and Design Data Code of Practice BS8110 : 1985

fcu (N/mm²) 30

Ec (N/mm²) 24597

fy (N/mm²) 460

Cover (mm) 25

Enhanced Shear Multiplier 1.5

Column Shape: Rectangular (300 × 300 mm) Finishes = 1.20 kN/m² Live Load = 1.50kN/m² Bar diameter in direction 1 and 2 = 10 mm Effective depth in direction 1 = 170.0 mm Effective depth in direction 2 = 160.0 mm Required steel area in direction 1 = 260 mm² Required steel area in direction 2 = 260 mm² The column local axis is not the same with the slab direction, therefore minimum steel percentage is used for punching shear checking. Steel percentage in direction 1, As1 = 0.15 % Steel percentage in direction 2, As2 = 0.16 %

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Detailing Limitation of band strip cut section 1) Cut along beam / slab edge There is no band strip detailing will be displayed for cut section along beam and raft slab edge as shown in figure below. Cut 1 is along beam and Cut 2 is along raft slab edge.

A message will be displayed on the detailing window when Cut 1 or Cut 2 is viewed.

2) Arbitrary Cut Section No detailing will be generated for arbitrary cut across the raft plan.

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Transfer Slab

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Verification Column is not supporting free edge of slab When an upper column share the same grid with the edge of a transfer slab and does not sit fully on the transfer slab, this is not allowed. An error message will be shown:

To solve, the upper column profile must sit within the transfer slab.

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Column or wall sitting across two transfer slab Upper column or RC Wall is not allowed to cross between two slabs. Integrity check will fail. 1. An upper column is not allowed to sit across two transfer slabs at the same time (see below).

2. An upper RC wall is not allowed to sit across two transfer slabs.

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Analysis Column sitting on a transfer slab Columns sitting on a transfer slab will be meshed as a fixed node in the 3D Mesh. Refer to example below:

The 3D Mesh for the transfer slab:

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Transfer Slab Mesh The transfer slab will be automatically fine-meshed as shell element type: 1. Column sitting on transfer slab

Figure 1a: Before 3D analysis

Figure 1b: 3D mesh for transfer slab after running 3D Analysis 2. Wall sitting on transfer slab (any wall sitting without supporting beam, located either inside the boundary of the slab or on the edge of the slab).

Figure 2a: Before 3D Analysis

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Figure 2b: 3D mesh for transfer slab after running 3D Analysis

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Design & Detailing Design 1. The 2D Slab Design for all floors in a project is not allowed when there exist a transfer slab in any of the floors. A warning message "Transfer elements exist!" will be shown when user click to Design Slab without running 3D analysis. User is first required to run 3D analysis before proceeding with the slab design. This is to ensure that the slab design will be based on the most critical results from either the 2D or 3D analysis.

2. The design of transfer slab will take the most c ritical (maximum) moment value from either the 2D analysis or 3D analysis contour. Full Report Analysis Results a. 2D Analysis FEM Slab Analysis Result Design Bending Moment from FEM Analysis (X) = 54.0 kNm/m Design Bending Moment from FEM Analysis (Y) = 54.5 kNm/m Unfactored Displacement from FEM Analysis = 3.29 mm

b. 3D Analysis

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Therefore maximum results are obtained from 3D Analysis.

Condition where transfer slabs will not be designed 1. Transfer slabs with RC walls sitting directly on the slabs (without beams) will not be designed. User will have to design manually using the analysis results provided in the slab contours and Full Report.

Slab Detail Design Calculation for Slab gb - FS1 MATERIAL AND DESIGN DATA Code of Practice BS8110 : 1997

fcu (N/mm²) 30

Cover (mm) 25

Ec, (N/mm²) 24597

fy (N/mm²) 460

Conc. Unit Weight (kN/m³) 24 p617

γc 1.5

γs 1.05

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SLAB MARK: gb - FS1 Slab Location : - 1/A - 1/B - 2/B - 2/A Slab Shape : Rectangular Dimension: X = 5000 mm, Y = 5000 mm Sub-Slab Thickness, h = 300 mm, Sub-Slab Drop = 0 mm

FEM Slab Analysis Result Design Bending Moment from FEM Analysis (X) = 63.4 kNm/m Design Bending Moment from FEM Analysis (Y) = 60.6 kNm/m Unfactored Displacement from FEM Analysis = 2.34 mm

Design Calculation (Base on F.E.M. Analysis Result) This subslab cannot be designed Transfer wall sitting on a transfer slab

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Punching Shear Checking for Upper Column Punching Shear Checking for Upper Columns The Punching Shear reports are available after Slab Design. Refer to topic "Punching Shear Reports". The punching shear checking for these following cases of transfer slabs will not be carried out. 1. Transfer slab supporting RC Walls. The bottom/top detailing is labelled as "Punching Shear Not Checked for transfer wall".

2. There is an opening in the transfer slab which is overlapped with the punching shear perimeter of the transfer column.

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Bottom detailing:

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Multi-level Foundation Mutli-level Column Footing Column footing can be input at different floors or levels of a structure as long as the column stump is not on top or below of another column stump.

Column stump can be set in the column Object Viewer as shown below. By default, Stump for column is set to False.

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Stump = True to use column footing and False NOT to use column footing. Default Stump = True to use Floor Height and False to use user-defined Stump Height. Stump Height = height defined by user. After running (2D+3D) analysis, the support nodes at the bottom of the column footing can be shown clearly in 3D Analysis Display as per screen shot below.

When analysis is complete, the multi-level column stumps can be designed using pad or pile footings. 1) Example of multi-level column footing with pad design. p622

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2) Example of multi-level column footing with pile design.

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Multi-level Wall Footing Wall footing can be input at different floors or levels of a structure as long as the wall stump is not on top or below of another wall stump.

Wall stump can be set in the wall Object Viewer as shown below. By default, Stum p for wall is set to False.

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Default Stump = True to use Floor Height and False to use user-defined Stump Height. Stump Height = height defined by user. After running (2D+3D) analysis, the support nodes at the bottom of the wall footing can be shown clearly in 3D Analysis Display as per screen shot below.

When analysis is complete, the multi-level wall stumps can be designed using pad or pile footings. 1) Example of multi-level wall footing with pad design.

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2) Example of multi-level wall footing with pile design.

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View When there is a combination of a partial non-stump wall and a partial stump wall in a wall group, the partial non-stump wall will be displayed using dash line in the foundation form. Take a model example as below.

It can be seen from the model that there is only stump wall at level gb. However, at level 1b, there is a combination of both non-stump wall and stump wall. Go to Project Explorer and expand Foundation, double click on Pad or Pile to open the foundation form. At Floor gb, the foundation form will display as follow. No dash line appearance as there is no non-stump wall at this level.

At Floor 1b, the partial non-stump wall from A-B is displayed using dash line while partial stump wall from B-C is displayed using full line to indicate stump area.

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Design When Stump is set to True in Wall Object Viewer, that particular wall will be designed as a wall with footing below. Take the multiple levels wall footing model below for instance.

Wall Group 1 has footing at foundation level. Therefore, design detailing is as per screen shot.

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For Wall Group 4 and Group 6, the design detailing are also similar to Wall Group 1 above with the word "Stump" in the schedule table to indicate the wall is designed as wall footing. Other wall groups than that will not have the word "Stump" in the schedule table. Group 4 :

Group 6 :

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Beam with Opening

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Analysis 2D Mesh 2D mesh for beam with opening(s) is illustrated as shown in Figure 1. Mesh nodes for 2D beam element can be viewed after 2D analysis is run by going to Analysis Result > 2D Contour View > Subframe View

.

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3D Mesh Meshing of 3D Beam with opening(s) element will obey the following specific ation: a) Multiple regions will be used.

b) Free meshing will be used. c) Mesh size will be the opening's dimension. d) Circular opening will be meshed as a square opening.

e) Opening with the ratio as stated below will we ignored in 3D analysis.

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Design To obtain 3D Analysis Result for Top Chord and Bottom Chord of a Beam Opening Consider the following model:

Data input : Code of Practice selected is BS8110:97 Dead line load = 9.68kN/m Live line load = 11.29kN/m Column size = 300x300mm Procedures : 1) Run the (2D+3D) batch analysis and beam design. 2) Open the 2D Subframe at floor gb (Refer to Figure 1).

3) Based on the 2D Subframe result in Figure 1, 3 nodes which are 18, 21 and 20 are located on top of the beam opening. 4) Open the 3D Analysis Display ,

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5) Activate the Contour Selection Mode at the Right Click menu list. 6) Select the top chord region as in Figure 2 and construct 1 vertical cut line at the middle of the beam opening (location at node number 21) as in Figure 3.

7) Open the cut line 1 to get the Axial Force, Moment, Shear and Torsion values for each load combination case as shown in Figure 4. Activate Nx icon to get Axial Force and Moment, activate Nxy icon to get Shear and activate Mx icon to get Torsion.

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8) Refer to Vertical Contour Cut Force Calculation for the derivation of the forces from the contour cut. 9) All the force values will also be shown in the Beam Opening Report under 3D Opening Analysis Values for All Loading Cases as shown in Figure 5.

10) Bottom Chord Analysis Result can be obtained also by repeating the steps above.

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To obtain 3D beam result diagram for Beam with Opening In the Beam Opening full report, there are 3 tables namely Top Chord, Bottom Chord and Combined Chord under the 3D Opening Analysis Value For All Loading Cases. This topic will discuss on how the values in the three tables are related to each other in order to obtain the 3D beam result diagram for beam with opening. Take Figure 1 below as an example:

Make sure the values are taken from the same node number. The equations involved in computing the results for combined chord are as follow: 1) Axial force for combined chord, Pcombined = P top chord + P bottom chord 2) Moment for combined chord, Mcombined = M top chord + M bottom chord + 0.5 x P x z where P = Top Chord Axial Force - Bottom Chord Axial Force z = Distance between centroid of Top Chord and Bottom Chord 3) Shear Force for combined chord, Vcombined = V top chord + V bottom chord 4) Torsion for combined chord, Tcombined = T top chord + T bottom chord Therefore, all the forces' values shown in the Combined Chord table are used to plot out the Beam Result Diagram, for instance the Beam Moment diagram as shown in Figure 2.

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Elaboration of 2D and 3D Opening Node Value Diagram 1 shows the calculation method adopted in Esteem 8 to produce the result values for moment case table of a beam opening in 2D and 3D design. Shear and torsion tables also adopt the same calculation method as in moment case.

Since axial force does not have 2D value, average k value from moment case will be used to calculate the Top(2D) and Bot(2D) values in Axial Force table as shown in Diagram 2.

However, when there are only 2 nodes found within the beam opening, the calculation of k value will be different. Refer Diagram 3 for moment case table when there are only 2 nodes within the beam opening. As there is no intermediate node within the opening, k value is calculated for both nodes. Shear and Torsion tables also adopt the same calculation as in moment case.

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The Axial Force table for 2 nodes within beam opening calculation is the same as being derived in Diagram 2 for beam opening with intermediate nodes. To prevent large k value, calculated k is limited to 2. For instance,

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Additional 2D Load Case for Reverse Moment Case In cases where the hogging moment is larger than the sagging moment at the location of the beam opening, an additional case will be taken into consideration for the hogging moment case. Therefore in this situation, there will be two extra tables namely Additional Moment Case and Additional Axial Force Case under 2D & 3D Opening Node Value For Gravity Load Case in the Beam Opening Full Report.

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Cracking Bar Calculation Below shows how the table of 2D & 3D Cracking Shear Values For All Loading Cases table in the beam opening full report is obtained from the Beam Result Diagram. Figure 1 and Figure 2 show how the 2D and 3D cracking shear values are obtained from the beam shear diagram respectively.

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The calculation of cracking bar design is referred to the book by M.A. Mansur, Concrete Beams with Opening Analysis and Design, on page 183 as shown below in the beam opening full report.

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Detailing User-defined beam opening rebar lapping length User can manually define the coefficient of the beam rebar lapping length by going to Parameter Setting > Detailing > Beam > Anchorage and Lapping (Coefficient) > Tension. The length of the rebar lapping is based on the product of user-defined coefficient for tension anchorage length (as set in the parameter setting) and the diameter of the main rebar. For instance: The coefficient of tension anchorage is manually set as 40 and the dimension round is set to 50 as highlighted below.

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Thus, the calculation for the top and bottom rebar lapping is as shown below:

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Limitation Slenderness moment check Esteem 8 does not check for slenderness effect for beam openings where the opening's chord is subject to compression. User needs to verify and carry out the check manually by complying to the standard code of practice. However, it is suggested that the dimensions of the compression chord be revised so as to eliminate the effects of slenderness so that slenderness moment can be neglected.

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Code of Practice Exception Currently, beam opening design is only available for Code of Practice BS8110 and CP65. Otherwise, user has to manually design the beam opening using the analysis provided when other code of practice is chosen. Take example when code of practice EC2:2004 is chosen. A message below the beam opening detail will state that beam opening design is not available for code of practice EC2:2004.

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Code of Practice Safety factors The safety factors for "Load Combination 1" stipulated from the various codes are as shown below:Codes BS8110:85 & 97 & CP65:99 ACI:99 AS3600:01

Dead Loads (DL) 1.4 1.4 1.25

Live Loads (LL) 1.6 1.7 1.5

Esteem 8 does not allow values lower than what is stated in the codes.

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BS8110 and CP65 Ec (Elastic Modulus) 5500 (fcu / γc) where fcu is in N/mm 2, (cubic strength) γc = 1.5

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Maximum Shear Link Spacing For BS8110 and CP 65 SvMax = Minimum of (0.75 d, 300) where d is effective depth

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Tension anchorage Reference

Description

Clause 3.12.8.3

Design anchorage bond stress

Clause 3.12.9

Curtailment and anchorage of bars

Table 3.27

Ultimate anchorage bond lengths and lap length as multiple of bar size

Clause 3.12.8.23

Effective anchorage length of hook or bend

Clause 3.12.9.4

Anchorage of bars at a simply-supported end of a member

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Enhanced Shear Multiplier The maximum enhance shear coefficient allowed is 1.5 which is also the default value.

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BS8110: 1985 only vc (Design concrete shear stress) Shear Stress Calculation 1. Calculate shear stress, v

2. Calculate concrete shear stress, vc

3. Refer to table 3.9 (BS 8110-1:1985), note 2 specified using equation to obtain concrete shear stress, vc. This equation is limited only for concrete grade with grade 25 and above. In BS 8110, it specified that for concrete with grade 25 are made with normal weight aggregates and lower than this grade are made with lightweight concrete (BS 8110-1:1985, cl 3.1.7.2). Therefore for all concrete with grade less than 25 should use BS 8110-2:1985 section 5 to obtained vc. As per specified in the clause the shear stress is limited to 0.63 fcu or 4 N/mm whereby in part 1, concrete grade 25 and above the limiting shear stress should be 0.8 fcu or 5 N/mm2. 4. Note: for shear design only Refer to Table 3.8 Form and area of shear reinforcement in beams to compare the value of v and vc. For Shear and Axial Compression, base on cl. 3.4.5.12

,

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Shear & Torsion Design Calculations The shear and torsion design is done based on BS8110: Part 2: 1985 in Clause 2.4. According to this clause, when the design relies on the torsional resistance of a member, the recommendations provided in Cl 2.4.3, Cl 2.4.4, Cl 2.4.5, Cl 2.4.6, Cl 2.4.7, Cl 2.4.8, Cl 2.4.9 and Cl 2.4.10 should be followed. To c ompute smaller link dimension, x 1 and larger link dimension, y 1 use the initial assumption for link and bar diameter from the "Design" parameter.

The x1 and y1 is the to centre-to-centre sm aller dimension and larger dimension of the link with diameter .

Figure 1: Horizontal link dimension and vertical link dimension

Referring to Figure 1 above, the value of x1 and y1 will be: x1 = b – 2 x Side cover – (2 x link diameter/2) y1 = h – 2 x Cover – (2 x link diameter/2) Where, b = beam width and h = beam height Effective depth of beam section, d;

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Distribution of Torsion Longitudinal Perimeter Bar The distribution of torsion longitudinal bar, Asl to the top, bottom and side bars should follow Clause 2.4.9 in BS 8110: 1985 Part 2. Where, the maximum pitch = 300mm. In Esteem8, the distribution of torsional longitudinal steel area throughout the perimeter of a particular section is done based on the equivalent area method. The steps are listed as below: 1. The perimeter is divided into three portions; top, middle and bottom reinforcement. 2. Get the smaller and larger dimension of the links, x1 and y1 respectively. x1 = b - 2*cover - 2*0.5*link diameter y1 = h - 2*side cover - 2*0.5*link diameter 3. Calculate the clear distance between the surface of the top and bottom most bars, HClear; = h - 2*(cover + bar diameter + link diameter)

Notes

:

o The main bar diameter used here is the minimum main bar diameter specified in Design > Beam > Main Reinforcement > Minimum Main Bar Diameter.

o The link diameter used here is the initial link diameter assumed as specified in Design > General > Link Diameter. 4. If the maximum pitch (300mm) is more than the c lear height, HClear, then no middle bar required from the Asl. 5. Assume the torsional longitudinal bar area, Asl is distributed throughout the perimeter of the links as per unit length. Therefore, compute the torsional longitudinal bar area per unit length, Asl_Unit;

6. Calculate the length of portion (in terms of distance) where the area should be distributed to top and bottom reinforcement, portion; portion = 0.5*maximum pitch + main bar diameter + 0.5*link diameter

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7. Based on the portion above, the perimeter for area Asl to be distributed to top and bottom reinforcement, k

8. Based on the perimeter, k above, the area of Asl distributed to top and bottom reinforcement, AslTopBot; = k*Asl_Unit 9. Therefore the Asl left for middle bar, Asmid = Asl - AslTopBot 10. Divide the portion area AslTopBot equally to the top and bottom reinforcement, Aslt and Aslc; Aslt = AslTopBot/2 Aslc = AslTopBot/2 Example 1: Calculation

Project Data: Beam width, b = 300mm Beam depth, h = 600mm C over = 34mm Side cover = 25mm Main bar diameter = 12mm (minimum main bar diameter) Link diameter = 10mm Longitudinal steel area required for torsion, Asl = 472mm

2

Sample Calculation Smaller link dimension, x1 = 300 - 2 * 25 - 10 = 240 mm Larger link dimension, y1 = 600 - 2 * 34 - 10 = 522 mm C lear height between the surface of top and bottom bars, HClear = 600 - 2 * (34 + 10 + 12) = 488 mm Asl per unit length, 2 Asl_unit = 472 / [2 * (240 + 522)] = 0.3097mm /mm C alculate the length of portion to be distributed to top and bottom reinforcement, portion = 0.5 * 300 + 12 + 0.5*10 = 167mm C alculate the area of Asl distributed to top and bottom reinforcement, 2 AslTopBot = 0.3097 * 2 *[240 + 2*Min(522/3 , 167)] = 355.55mm

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C alculate the area distributed to the middle bar, 2 Asl for middle bar = 472 - 355.55 = 117 mm Therefore divide the top and bottom portion AslTopBot equally to top and bottom reinforcement, 2 Aslt = 355.55 / 2 = 178 mm 2 Aslc = 355.55 / 2 = 178 mm

Esteem 8 Result

BEAM gb1(300x600) SPAN NO. 1 FLEXURAL DESIGN CALCULATION Area of Steel Required for Middle Bar, AsMid = 116 mm² Portion of Longitudinal Torsion Steel Area Distributed to Tension Reinforcement, Aslt = (Asl - AsMid) / 2 = (472 - 116) / 2 = 178 mm² Portion of Longitudinal Torsion Steel Area Distributed to Compression Reinforcement, Aslc = (Asl - AsMid) / 2 = (472 - 116) / 2 = 178 mm²

SHEAR & TORSION DESIGN CALCULATION LOCATION : SECTION 1 LEFT SUPPORT (B:0 mm E:1250 mm from left grid of span) Maximum Torsion within Zone, T = 24.8 kNm Shear at Location of Maximum Torsion, V = 191.8 kN Horizontal Leg, yH = b - 2 × Side Cover - Dialink = 300.0 - 2 × 25 - 10 = 240.0 mm Vertical Leg, yV = h - 2 × Cover - DiaLink = 600.0 - 2 × 34 - 10 = 522.0 mm Longer Leg, y1 = Max (yH, yV) = 522.0 mm Smaller Section Dimension, Dmin = 300.0 mm Larger Section Dimension, Dmax = 600.0 mm Torsion Stress, νst = 2 × T × 106 / (Dmin² × (Dmax - Dmin / 3)) = 1.10 N/mm² Effective depth, d = 540.0 mm Shear Stress due to Loading, νss = V × 1000 / (b × d) = 191.8 × 1000 / (300.0 × 540.0) = 1.18 N/mm² Part 2 : Clause 2.4.5, 2.4.6 Table 2.3 Maximum Combined Stress Allowed, νtu = Min (0.8 × √fcu, 5) = 4.38 N/mm² Total Stress, νTot = νss + νst = 1.18 + 1.10 = 2.29 N/mm² ≤ νtu (4.38 N/mm²) Checking for Combined Stress Allowed Pass Additional Checking While Small Cross Section (y1 < 550 mm) Larger Link Dimension, y1 = 522.0 mm < 550 mm νtu × y1 / 550 = 4.38 × 522.0 / 550 = 4.16 N/mm² νst = 1.10 N/mm² ≤ 4.16 N/mm² Checking for Torsion Stress Allowed Pass Link Horizontal Dimension, h1 = b - 2 × Side Cover - DiaLink = 300 - 2 × 25 - 10 = 240 mm Link Vertical Dimension, v1 = h - 2 × Cover - DiaLink = 600 - 2 × 34 - 10 = 522 mm Dimension x1 = Min (h1, v1) = 240 mm Dimension y1 = Max (h1, v1) = 522 mm Torsion Stress, νst = 1.10 N/mm² Part 2 : Clause 2.4.6 Table 2.3 Torsion Strength contributed by concrete, νt,min = Min (0.067 × √fcu, 0.4) = 0.37 N/mm² Torsion Stress, νst = 1.10 N/mm² > νt,min = 0.37 N/mm² - BS8110: Part 2: Clause 2.4.7 Torsion Steel Area / Spacing Ratio, TAs_Sv = T × 1000000 / (0.8 × x1 × y1 × fyv / γs ) p660

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= 24.821 × 1000000 / (0.8 × 240 × 522 × 0.87 × 460) = 0.62 mm²/mm - BS8110:1985: Part 2 Clause 2.4.7 Longitudinal Steel Area Required for Torsion, Asl = TAs_Sv × fyv × (x1 + y1) / fy = 0.62 × 460 × (240.0 + 522.0) / 460 = 472 mm²

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BS8110: 1997 only vc (Design concrete shear stress) Shear Stress Calculation 1. Calculate shear stress, v

2. Calculate concrete shear stress, vc

3. Refer to table 3.8 (BS 8110-1:1997), note 2 specified using equation to obtain concrete shear stress, vc. This equation is limited only for concrete grade with grade 25 and above. In BS 8110, it specified that for concrete with grade 25 are made with normal weight aggregates and lower than this grade are made with lightweight concrete (BS 8110-1:1997, cl 3.1.7.2). Therefore for all concrete with grade less than 25 should use BS 8110-2:1985 section 5 to obtained vc. As per specified in the clause the shear stress is limited to 0.63 fcu or 4 N/mm whereby in part 1, concrete grade 25 and above the limiting shear stress should be 0.8 fcu or 5 N/mm2. 4. Note: for shear design only Refer to Table 3.7 Form and Area of shear reinforcement in section to compare the value of v and vc. For Shear and Axial Compression, base on cl. 3.4.5.12

,

Notes

BS 8110-1:1997

:

1) Version 1999 and May 2002, - Stress in any bar should not exceed 0.95fyv, 2) Version November 2005, - Stress in any bar should not exceed 0.87fyv

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Column Eccentricity only checked on single axis According to Clause 3.8.2.4, the design of column for biaxial moments due to axial load eccentricities is carried out with the considerations of one axis at a time. Esteem 8 combines the bending in both axis together. Esteem 8 2D sub-frame analysis obtained the envelop of column moments through pattern loading. This is deemed to be a more vigorous analysis than the conventional continuous beam analysis allowed for by the codes as the design moments for columns are guided (for example Clause 3.2.1.2.3 of BS8110: Pt. 1: 1985).

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Minimum Diameter of Longitudinal Reinforcement in Columns According to Clause 3.12.5.3 of BS8110-1:97, the minimum diameter of longitudinal reinforcement in columns is 12 mm. If any value selected by user is less than the minimum allowed, the program automatically overwrites with the default value or previously entered value which is within the allowable range.

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Enhanced Shear Multiplier The maximum enhance shear multiplier allowed is 1.5 which is also the default value.

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Esteem Innovation Sdn Bhd, 2011

CP65: 1999 only Ec (Elastic Modulus) 5500 (fcu / γc) where fcu is in N/mm 2, (cubic strength) γc = 1.5

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Esteem Innovation Sdn Bhd, 2011

ACI 318: 2008

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Material Ec (Elastic Modulus) used in ACI 318 : 2008 For f'c = 3000 psi to 6000 psi ( 21kN/m

2

Ec = 57,000 * (f'c ) = 57,000 *conv * (f'c * conv)  4730 f'c

2

to 42kN/m ) -----f'c in psi -----f'c in psi 2 -----f'c in kN/m

where, f'c is the cylindrical concrete strength in N/mm 2 conv psi to kN/m 2 = 0.006895

2

2

For f'c = 6000 psi to 12,000psi (43kN/m to 80kN/m ) 6

Ec = (40,000 * f'c + 1.0 * 10 ) * (wc / 145)

1.5

* conv

where, f'c is the cylindrical concrete strength in psi conv = 0.006895 3 3 wc is the density of concrete in pounds per cubic foot ( 1 lb/ft = 16.02 kg/m )

Example of Ec Calculation (Imperial Formula) 1) Given f'c = 5000 psi  34.5kN/m 3 wc = 152.714 lb/ft

2

since f'c is between 3000 and 6000 psi, Use Ec = 57,000 * (5000) * conv 2

= 27713.676 kN/m

2) Given f'c = 10,000 psi  69kN/m 3 wc = 152.714 lb/ft

2

since f'c is between 6000 and 12,000 psi, p668

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6

1.5

Use Ec = (40,000 * f'c + 1.0 * 10 ) * (wc / 145) * conv 6 1.5 = (40,000 * 10,000 + 1.0 * 10 ) * (152.714 / 145) * conv 2 = 37262.377 kN/m

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Loading Load factor used in ACI 318: 2008 Below are the default loading cases and partial safety factors for ACI 318: 2008.

Esteem 7 implements this parameter from ACI 318 - Appendix C.

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Strength Reduction Factor for ACI 318 : 2008 Nominal Strength is the strength of a particular structural unit calculated using the current established procedures. Desgin Strength = Strength Reduction Factor,

 x Nominal Strength



The purposes of the strength reduction factor, , (ACI 318-Appendix C) are: 1) to allow for the probability of understrength members due to variations in material strengths and dimensions. 2) to allow for inaccuracies in the design equations. 3) to reflect the degree of ductility and required reliability of the member under the load effects being considered. 4) to reflect the importance of the member in the structure. Table extracted from Nawy's Book, Table 4.5, that summarizes the resistance factors various structural elements as given in ACI code.



Structural Element

Factor,

Beam or slab: bending or flexure Columns with ties Columns with spirals Columns carrying very small axial loads Beam: Shear and torsion Bearing except for strut -and- tie Bearing areas in strut-and-tie Post-tensioned anchorage zone Flexural shear and bearing in plain structural concrete

0.9 0.65 0.75 0.65 - 0.9 0.75 0.65 0.75 0.85 0.60

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 for

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Design

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Design Concept Design ultimate moment resistance of parabolic stress block for ACI 318 : 2008 Reference: PCA Notes on 318-08 : 10.2.7

εu= maximum concrete strain ( Clause 10.2.3) c = depth to neutral axis β1 = factor relating depth of equivalent rectangular compressive stress block to neutral axis depth ACI 318 : 08 Clause 10.2.7.3: For fc′ between 2500 and 4000 psi, β1 shall be taken as 0.85. For fc′ above 4000 psi, β1 shall be reduced linearly at a rate of 0.05 for each 1000 psi of strength in excess of 4000 psi, but β 1 shall not be taken less than 0.65. 2 1psi = 0.006895N/mm Depth of equivalent rectangular stress block, a = β1c (Clause 10.2.7.1) Concrete stress of compression zone = 0.85fc’ (Clause 10.2.7.1) Average concrete stress above neutral axis, k1 = 0.85fc’ * β1 Concrete lever arm factor, k2 = β1/2 Distance from extreme compression fiber to neutral axis, c = cb * d

Limiting effective depth factor, cb = 0.75 *

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(Nawy, page 109)

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Modified span/depth ratio for cantilever spans A span is deemed to be a "cantilever span" when the % Difference of Moment defined above is less than 10%. "Modified" span / depth ratio = 8 + 10.5 (Diff% / 10)

Pure cantilever Maximum difference

Percentage difference, % 0 10

SDR 8 18.5

% Difference of Moment = [ Mc - M ] / [ Mt - M ] If the Diff% > 10%, BSDR will use 16 or 18.5 based on the span condition (simply supported / one end continuous). 3. When MBSDR is TRUE: o The left and right side moment, Mc and Mt will be taken from the moment envelope. o The sagging moment, M will be taken from the Full combination. 4. When MBSDR is FALSE: o If there is a sagging moment detected for the cantilever span, the basic ratio will either be 8 or 18.5. Refer to example below (deflection check for Span 1): The design for moment and shear is based on support face:

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Moment diagram Moment envelop:

Remark The left and right side moment, Mc and Mt will be taken from moment envelope. The sagging moment,M is taken from Full load combination.

Full load combination moment:

DEFLECTION CHECKING FOR SPAN Left Side Moment Value = 0.0kNm Right Side Moment Value = -151.2kNm Maximum Sagging Moment Value = 1.1kNm Percentage Difference (%) = (1.1 / 152.3) × 100% = 0.75 % Modified Allowable Span/Depth Ratio = 8 + 10.5 × (0.75 / 10) = 8.8

Parameter is set to TRUE, calculate percentage difference of moment and get modified span to depth ratio. Percentage difference of moment calculated for Span 1 based on Full Moment. Modified SDR = 8.8

DEFLECTION CHECKING FOR SPAN Actual Span / Depth Ratio, Ar = 4300.0 / 600.0 = 7.2 Allowable Span / Depth Ratio = 18.5 (one end continuous)

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Parameter is set to FALSE, there is sagging moment detected from 2D moment envelope (8.78kNm). Basic span/depth ratio (BSDR) = 18.5 for one end continuous beam

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Slab Design Flexural reinforcement for slab in ACI 318 : 2008 Slab has the similar procedure with beam for determining flexural reinforcement. The only difference is no doubly reinforcement design for slab.

Notes

:

conv = 0.006895 p678

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Strength reduction factor for tension-controlled sections, = 0.9

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Slab deflection check in ACI 318 : 2008

Notes : conv = 0.006895 Allowable span / depth ratio for irregular slab  ASDR = 20 if all slab edges are supported which designed as simply supported slab  ASDR = 10 if ≤ 50% of slab edges are supported which designed as cantilever slab However, users are allowed to change the allowable span / depth ratio which is under slab properties if they disagree the calculation of allowable span / depth ratio for irregular slab shown above.

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Determination of Modified Span Depth Ratio for Irregular Shape Slab To determine the modified basic span depth ratio for irregular slab based on ACI, the following formula is being adopted: Modified Span/Depth Ratio, msdr = 10 + 10*(2*SE - 1) where, SE = Supported Edge Ratio

For example: 1. If the Supported Edge Ratio is 0.5; Modified Span/Depth Ratio = 10 + 10*[2*(0.5) - 1] = 10 (design as cantilever slab) 2. If the Supported Edge Ratio = 0.3; Modified Span/Depth Ratio = 10 + 10*[2*(0.3) - 1] = 10 (minimum msdr = 10) 3 If the Supported Edge Ratio > 0.8, say 1.0, use the formula in item [3a] to get the Final Modified Span/Depth Ratio, msdr. 3a. From the ratio of Continuous Edge Length to the Supported Edge Ratio, get the Final Modified Span/Depth Ratio. Final Modified Span/Depth Ratio, MSDR

= msdr * (1 + 0.4 * c sr)

Where csr,

Example: The Modified Ratio, msdr = 10 + 10 * [2*(1) - 1] = 20 > 10 Continuous Edge Length , c el = 18400mm Supported Edge Ratio, sel = 18400mm Continuous Edge on Supported Edge Ratio, csr = cel / sel = 18400/18400 = 1.0 > 0.8 Final Modified Span / Depth Ratio, MSDR = msdr * (1 + 0.4 * csr) = 20 * (1 + 0.4 * 1.0) = 28 However, users are allowed to change the basic span-depth ratio which is under slab properties if they disagree on the calculation of modified span/depth ratio for irregular slab shown above. Example: Refer to the following example to determine the Modified Span/Depth Ratio for each slab. Plan View

Remarks

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The length of the supported edges, 1 2 3 4 5

= = = = =

2000mm 3000mm 2828.4mm 5000mm 2828.4mm

Total Supported Edge Length for Slab FS1 = 12828.4mm Slab FS3 = 2828.4mm Slab FS4 = 7828.4mm

Slab Design > Full Report

Remarks C ompute the lengths of supported edges.

Data and Result of Slab Mark : gb - FS1 Location : - 1/A - 4/A - 4/C - 1A/C Slab Shape : Irregular-Shape

SubSlab : FS1:1 Location : - 1/A - 4/A - 4/C - 1A/C SubSlab Shape : Polygon Dimension Length of Edge 1 = 5000.0 mm Length of Edge 2 = 2000.0 mm Length of Edge 3 = 3000.0 mm Length of Edge 4 = 2828.4 mm Sub-Slab Thickness, h = 175 mm Sub-Slab Drop = 0 mm Supported edge ratio, SE = 12828.4 / 12828.4 = 1.0

DEFLECTION CHECKING Shorter span in Y Direction (1788.9 mm) Supported Edge Length, sel = 12828.4 mm Continuous Edge Length on Supported Edge, cel = 7828.4 mm Total Edge Length, tel = 12828.4 mm Supported Edge Ratio, SE = sel / tel = 12828.4 / 12828.4 = 1.00 Allowable Span / Depth Ratio = 10 + 10 × (2 × SE - 1) = 20.0 ≥ 10 Continuous Edge on Supported Edge Ratio, csr = cel / sel = 7828.4 / 12828.4 = 0.61 ≤ 0.8 ACI 318 : 2008 Clause 9.5.2.1 and Table 9.5(a) Actual Span /Depth Ratio, Ar = 1788.9 / 175.0 = 10.2 Since fy > 414 N/mm² (60000 psi), Modification Factor for Steel Yield Strength, p682

From the formula, MSDR = 10 + 10*(2*1-1) = 20

Notes

:

For polygonal slabs, the maximum Modified Span/Depth Ratio, can be obtained using the formula above is 20 although the slab should be one end continuous (MSDR = 24).

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MFs = 0.4 + fy / (100000 × conv) = 0.4 + 460.0 / (100000 × 0.006895) = 1.07 New Allowable Span/Depth ratio = 20.0 × 1.07 = 21.34 Deflection Ratio = 21.3 / 10.2 = 2.09 Ratio >= 1 : Deflection Checked PASSED

Data and Result of Slab Mark : gb - FS3 Location : - 1/C - 1/A - 1A/C Slab Shape : Triangular

SubSlab : FS3:1 Location : - 1/C - 1/A - 1A/C SubSlab Shape : Triangular Dimension Length of Edge 1 = 2000.0 mm Length of Edge 2 = 2828.4 mm Length of Edge 3 = 2000.0 mm Sub-Slab Thickness, h = 175 mm Sub-Slab Drop = 0 mm Supported edge ratio, SE = 2828.4 / 6828.4 = 0.41

DEFLECTION CHECKING Shorter span in X Direction (1414.2 mm) Supported Edge Length, sel = 2828.4 mm Continuous Edge Length on Supported Edge, cel = 2828.4 mm Total Edge Length, tel = 6828.4 mm Supported Edge Ratio, SE = sel / tel = 2828.4 / 6828.4 = 0.41 Allowable Span / Depth Ratio = 10 + 10 × (2 × SE - 1) = 8.3 < 10 Use Allowable Span / Depth Ratio = 10 Continuous Edge on Supported Edge Ratio, csr = cel / sel = 2828.4 / 2828.4 = 1.00 > 0.8 Final Allowable Span / Depth Ratio = 10 × (1 + 0.4 * 1.00) = 14.0 ACI 318 : 2008 Clause 9.5.2.1 and Table 9.5(a) Actual Span /Depth Ratio, Ar = 1414.2 / 175.0 = 8.1 Since fy > 414 N/mm² (60000 psi), Modification Factor for Steel Yield Strength, MFs = 0.4 + fy / (100000 × conv) = 0.4 + 460.0 / (100000 × 0.006895) = 1.07 New Allowable Span/Depth ratio = 14.0 × 1.07 = 14.94 Deflection Ratio = 14.9 / 8.1 = 1.85 Ratio >= 1 : Deflection Checked PASSED

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From the formula, Final MSDR = 10 *(1 + 0.4*1) = 14.0

Notes

:

For polygonal slabs, the minimum MSDR is 10. MSDR cannot be lesser than 10.

Esteem Innovation Sdn Bhd, 2011

Data and Result of Slab Mark : gb - FS4 Location : - 1/A - 1A/B - 4/A Slab Shape : Triangular

SubSlab : FS4:1 Location : - 1/A - 1A/B - 4/A SubSlab Shape : Triangular Dimension Length of Edge 1 = 2828.4 mm Length of Edge 2 = 3605.6 mm Length of Edge 3 = 5000.0 mm Sub-Slab Thickness, h = 175 mm Sub-Slab Drop = 0 mm DEFLECTION CHECKING Shorter span in X Direction (2236.1 mm) Supported Edge Length, sel = 7828.4 mm Continuous Edge Length on Supported Edge, cel = 5000.0 mm Total Edge Length, tel = 11434.0 mm Supported Edge Ratio, SE = sel / tel = 7828.4 / 11434.0 = 0.68 Allowable Span / Depth Ratio = 10 + 10 × (2 × SE - 1) = 13.7 ≥ 10 Continuous Edge on Supported Edge Ratio, csr = cel / sel = 5000.0 / 7828.4 = 0.64 ≤ 0.8

Supported edge ratio, SE = 7828.4 / 11434 = 0.68 From the formula, MSDR = 10 + 10*(2*0.68-1) = 13.7

ACI 318 : 2008 Clause 9.5.2.1 and Table 9.5(a) Actual Span /Depth Ratio, Ar = 2236.1 / 175.0 = 12.8 Since fy > 414 N/mm² (60000 psi), Modification Factor for Steel Yield Strength, MFs = 0.4 + fy / (100000 × conv) = 0.4 + 460.0 / (100000 × 0.006895) = 1.07 New Allowable Span/Depth ratio = 13.7 × 1.07 = 14.61 Deflection Ratio = 14.6 / 12.8 = 1.14 Ratio >= 1 : Deflection Checked PASSED Therefore based on the formula for MSDR above, for a polygonal shape slab the minimum MSDR = 10. User can overwrite this value to between the range of 10 to 28 by setting the Span Ratio manually at the Object Viewer.

Plan View

Remarks The 1= 2= 3= 4= 5= 6=

length of the supported edges; 6401mm 2286mm 1829mm 1067m 4572mm 3353mm

Total edge lengths = (1+2+3+4+5+6) = 19508mm Total supported edge lengths = (1+2+3+4+5+6) = 19508mm

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Slab Design > Full Report

Remarks

Area of SubSlab (i): 4878324.0 mm² (1), 10451592.0 mm² (2), 4181094.0 mm² (3) Effective Depth of SubSlab (i): 120.5 mm (1), 120.5 mm (2), 120.5 mm (3) ∑ (Ai × di) = 2351076705.00 ∑ Ai = 19511010.00 Average Effective Depth, d = ∑ (Ai × di) / ∑ Ai = 120.50 Supported edge ratio, SE = ratio of supported edge/total edge length = 19508/19508 = 1

Shorter span in X Direction (3196.9 mm)

Supported Edge Length, sel = 19508.0 mm Continuous Edge Length on Supported Edge, cel = 16155.0 mm Total Edge Length, tel = 19508.0 mm Use formula above to get Supported Edge Ratio, SE = sel / tel = 19508.0 / 19508.0 = 1.00 Final Modified Span/Depth Allowable Span / Depth Ratio = 10 + 10 × (2 × SE - 1) = 20.0 ≥ 10 Continuous Edge on Supported Edge Ratio, csr = cel / sel = 16155.0 / 19508.0 = 0.83 > Ratio = 20 * (1 + 0.4 * 0.83) 0.8 = 26.6 Final Allowable Span / Depth Ratio = 20 × (1 + 0.4 * 0.83) = 26.6 ACI 318 : 2008 Clause 9.5.2.1 and Table 9.5(a) Actual Span /Depth Ratio, Ar = 3196.9 / 150.0 = 21.3 Since fy > 414 N/mm² (60000 psi), Modification Factor for Steel Yield Strength, MFs = 0.4 + fy / (100000 × conv) = 0.4 + 460.0 / (100000 × 0.006895) = 1.07 New Allowable Span/Depth ratio = 26.6 × 1.07 = 28.41 Deflection Ratio = 28.4 / 21.3 = 1.33 Ratio >= 1 : Deflection Checked PASSED

The slab passed in Deflection checking.

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Beam Design Flexural reinforcement for rectangular beam in ACI 318 : 2008

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Notes : conv = 0.006895 Strength reduction factor for tension-controlled sections, = 0.9 In case εs > εsc for doubly reinforced design, compression steel required, Asc = M2 / [fsc × (d – d’)] whereas fsc = εsc × Es

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Flexural reinforcement for flanged beam in ACI 318 : 2008

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Notes : conv = 0.006895 Strength reduction factor for tension-controlled sections, = 0.9 Flanged beam design only will be executed when the flanged beam design is set to True (under Object Viewer) for the selected beam. Besides, users have to compute the flange width and flange depth by themselves as the program not able to auto-detect the flange width and flange depth.

In case εs > εsc for doubly reinforced design, compression steel required, Asc = M2 / [fsc × (d – d’)] whereas fsc = εsc × Es

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Minimum area of tension reinforcement for ACI 318 : 2008 The minimum area of tension reinforcement is checked based on ACI 318 : 2008 Clause 10.5.

Notes : conv = 0.006895 AsMin (%) = 300 × √(f'c × conv) / fy AsMin (%) = 600 × √(f'c × conv) / fy sixth edition, page 135)

for sagging moment for hogging moment and cantilever case (Nawy,

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Beam shear reinforcement in ACI 318 : 2008

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Esteem Innovation Sdn Bhd, 2011

Notes : conv = 0.006895 Vu*d/Mu  1 Strength reduction factor for shear-controlled sections, = 0.75

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Design concrete shear stress, vc for ACI 318 : 2008 Calculate Vc, nominal shear strength provided by concrete, -lb 1. Shear strength Vc shall be computed by provisions of 11.3.1 and 11.3.2. For member subject to shear and moment only use either EQ 11-3 or

EQ 11-5 but not greater than

2. Equation 11-5 will predict a significantly greater value of Vc than equation 11-3 only in regions where the moment Mu is small. 3. Quantity Vud/Mu not 1.0 then limited to 1.0 in computing Vc by EQ (11-5), where Mu is factored moment occurring simultaneously with Vu at section considered.

Shear Reinforcement with Axial compression force Nu 1. R 11.3.2.2 - for member subject to axial compression, it shall permitted to compute Vc using equation 11 - 5 with Mm substituted for Mu

Modified EQ 11 - 5 EQ 11 - 7

2. Instead of using above equation, ACI Code 11.3.1.2 permits the use of an alternative simplified expression: EQ 11 - 4 3. Vud/Mu may be greater than 1.0, but Vc must not exceed

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EQ 11 - 7 Where Ag is the gross area in.

2

Quantity Nu/Ag shall be expressed in psi. when Mm as computed by Eq (11-6 ) is negative, Vc shall be computed by Eq. (11-7) 4. for member subjected to significant axial tension,

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Maximum shear link spacing for ACI 318 : 2008 The maximum shear link spacing is checked based on ACI 318 : 2008 Clause 11.5.5

SvMax = 0.25 d when vd > vMax / 2 SvMax = 0.5 d when vd <= vMax / 2 SvMax < 600mm where d = Effective depth of section vMax = Maximum Shear Stress Allowed vd = shear strength provided by shear reinforcement

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Beam torsion design in ACI 318 : 2008

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Notes : conv = 0.006895 p701

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Strength reduction factor for torsion and shear-controlled sections, = 0.75 Acp = area enclosed by outside perimeter of concrete cross section Pcp = outside perimeter of concrete cross section x1 = link horizontal dimension y1 = link vertical dimension Aoh = area enclosed by centerline of the outmost closed transverse torsional reinforcement = x1 * y1 Ao = gross area enclosed by shear flow path = 0.85 * Aoh ph = perimeter centerline of outmost closed transverse torsional reinforcement = 2*(x1 + y1) ACI 318 : 2008 Eqn. 11- 18 [vss² + {T × ph / (Ø × 1.7 × Aoh²))²]0.5 ≤ vc + 8 × √(f'c × conv)

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Maximum torsion link spacing for ACI 318 : 2008 The maximum torsion link spacing is checked based on ACI 318 : 2008 Clause 11.6.6.1: The spacing of transverse torsion reinforcement shall not exceed the smaller of ph /8 or 12 in. SvMax = Minimum of (ph/8, 300) where ph = perimeter centerline of outmost closed transverse torsional reinforcement

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Beam deflection check for ACI 318 : 2008

Notes : conv = 0.006895

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Column Design Column classification (short or slender column) for ACI 318 : 2008

Notes

:

ψTop = Relative Stiffness at Top of the Element ψBot = Relative Stiffness at Bottom of the Element ψM = Average of the ψ Values at Two Ends of the Element ψ = Relatives Stiffness at the Restrained End Relatives Stiffness = 10 for connection between column to pin foundation or connection between column to slab Relatives Stiffness = 1 for connection between column to fixed foundation lo = Column clear length M1, M2 are the first order moments at the end of the column with |M2| ≥ |M1| r = radius of gyration about the axis considered, (I/A)

0.5

I = Second moment of area of the section about the axis A = Cross section area of column p705

Esteem Innovation Sdn Bhd, 2011

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Esteem Innovation Sdn Bhd, 2011

Example Calculation of Relative Stiffness, Ψ

Take Column (1A,A1) 300mm x 400mm; y-y direction : column = 400*3003 / 12= 900 x 106 mm4 beam

= 300*5003 / 12= 3125 x 106 mm4

ΨTop

= ( column/lcolumn ) /Σ( beam/lbeam ) = ( 900 x 106 / 3000) / [2 * (3125 x 106 / 4000)] = 0.192

ΨBottom = 1.000 (pin connection)

z-z direction :

column = 300*4003 / 12= 1600 x 106 mm4 p707

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beam

= 300*5003 / 12= 3125 x 106 mm4

ΨTop

= ( column/lcolumn ) /Σ( beam/lbeam ) = ( 1600 x 106 / 3000) / [2 * (3125 x 106 / 4000)] = 0.341

ΨBottom = 1.000 (pin connection)

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Short and slender column design in ACI 318 : 2008

Notes

:

N or Pu = Design value of the applied axial force (tension or compression) e0 = Minimum eccentricity moment, see ACI 316 : 2008 Clause 10.12.3.2 H = depth of the section Mend = 1

st

order bending moment at column end

Cm = Correction factor relating the actual moment diagram to an equivalent uniform moment diagram, see ACI 316: 2008 Clause 10.12.3.1 Pc = Critical buckling load, see ACI 316 : 2008 Clause 10.12.3 Q = Stability index for a column, see ACI : 2008 Clause 10.11.4.2 ΣPu and ΣPc value are based on individual column as stability index, Q is also calculated based on individual column.

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Esteem Innovation Sdn Bhd, 2011

Flow chart of column design with consideration of slenderness

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Pad Footing Design Flexural reinforcement design for pad footing in ACI 318 : 2008 Pad footing has the similar procedure with beam for determining flexural reinforcement. The only difference is no doubly reinforcement design for pad footing.

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Esteem Innovation Sdn Bhd, 2011

Notes

: p712

Esteem Innovation Sdn Bhd, 2011

conv = 0.006895 Strength reduction factor for tension-controlled sections, = 0.9

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Punching shear checking for pad footing in ACI 318 : 2008

Notes

:

conv = 0.006895 p714

Esteem Innovation Sdn Bhd, 2011

Strength reduction factor for shear-controlled sections, = 0.75 esm = enhance shear multiplier alpha, α = 20 (assumed) For rectangular column case β = ratio of long side column to short side column Critical perimeter, bo = 2 * (B’ + L’) Area inside critical perimeter = B’ × L’ B’ = B + 2 × k × d L' = L + 2 × k × d

Although the punching shear checking is done at section 0.2d to 0.5d with interval of 0.05d, but pad footing report only displays punching shear checking at 2 sections (0.5d and the most critical section) unless the checked section is out of pad footing profile.

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Esteem Innovation Sdn Bhd, 2011

Bearing strength of column and footing checking for pad footing in ACI 318 : 2008

Notes

:

Earth pressure, ρ = ultimate safety factor * PMax Ultimate axial load = ρ * pad area Strength reduction factor for torsion and bearing-controlled sections,  = 0.65 A1 = loaded area A2 = area of the lower base of the largest frustum of a pyramid, cone, or tapered wedge p716

Esteem Innovation Sdn Bhd, 2011

contained wholly within the support and having for its upper base the loaded area, and having side slopes of 1 vertical to 2 horizontal

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Esteem Innovation Sdn Bhd, 2011

Flexural shear checking for pad footing in ACI 318 : 2008

Notes

:

conv = 0.006895 p718

Esteem Innovation Sdn Bhd, 2011

Strength reduction factor for shear-controlled sections, = 0.75 esm = enhance shear multiplier

Although the flexural shear checking is done at section 0.2d to 1.0d (depends on enhanced shear multiplier value, default is 1.0) with interval of 0.05d, but pad footing report only displays flexural shear checking at preset section and the most critical section unless the checked section is out of pad footing profile

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Esteem Innovation Sdn Bhd, 2011

Pile Footing Design Flexural reinforcement design for pile footing in ACI 318 : 2008 Pile footing has the similar procedure with beam for determining flexural reinforcement. The only difference is no doubly reinforcement design for pile footing.

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Esteem Innovation Sdn Bhd, 2011

Notes

:

conv = 0.006895 Strength reduction factor for tension-controlled sections, = 0.9

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Esteem Innovation Sdn Bhd, 2011

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Esteem Innovation Sdn Bhd, 2011

Punching shear checking for pile footing in ACI 318 : 2008

Notes

:

conv = 0.006895 Strength reduction factor for shear-controlled sections, = 0.75 esm = enhance shear multiplier alpha, α = 20 (assumed) p723

Esteem Innovation Sdn Bhd, 2011

For rectangular column case β = ratio of long side column to short side column Critical perimeter, bo = 2 * (B’ + L’) Area inside critical perimeter = B’ × L’ B’ = B + 2 × k × d L' = L + 2 × k × d

Although the punching shear checking is done at section 0.2d to 0.5d with interval of 0.05d, but pile footing report only displays punching shear checking at 2 sections (0.5d and the most critical section) unless the checked section is out of pile footing profile.

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Esteem Innovation Sdn Bhd, 2011

Bearing strength of column and footing checking for pile footing in ACI 318 : 2008

Notes

:

Ultimate axial load = Ultimate safety factor * num ber of pile provided * pile capacity Strength reduction factor for torsion and bearing-controlled sections,  = 0.65 A1 = loaded area A2 = area of the lower base of the largest frustum of a pyramid, cone, or tapered wedge contained wholly within the support and having for its upper base the loaded area, and having p725

Esteem Innovation Sdn Bhd, 2011

side slopes of 1 vertical to 2 horizontal

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Esteem Innovation Sdn Bhd, 2011

Flexural shear checking for pile footing in ACI 318 : 2008

Notes

:

conv = 0.006895 Strength reduction factor for shear-controlled sections, = 0.75 esm = enhance shear multiplier

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Esteem Innovation Sdn Bhd, 2011

Although the flexural shear checking is done at section 0.2d to 1.0d (depends on enhanced shear multiplier value, default is 1.0) with interval of 0.05d, but pile footing report only displays flexural shear checking at preset section and the most critical section unless the checked section is out of pile footing profile.

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Esteem Innovation Sdn Bhd, 2011

Detailing

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Beam Detailing Distribution of Torsion Longitudinal Bar used in ACI 318 : 2008 Asl = Longitudinal steel area required for torsion Longitudinal Steel Area Required for Torsion at tension face, Aslt = Asl / 2 Longitudinal Steel Area Required for Torsion at compression face, Aslc = Asl / 2 - M / (0.9 × d × fy)

Besides, Clause below also complied for the distribution of torsion longitudinal bar.

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Esteem Innovation Sdn Bhd, 2011

FAQ Does the program carry out shear check for slab design? Question : Does the program carry out shear check for slab design? Answer : No, the program does not carry out shear check in slab design. Users have to manually do the shear check if find it is necessary.

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Why the longitudinal steel area required for torsion at compression face is zero? Q: Why the longitudinal steel are required for torsion at compression face is zero? A: This happens when there is a possibility that the longitudinal steel area required for torsion at compression face, Aslc being calculated having a negative value. Therefore, in order to prevent negative value for Aslc, the program will display the result to be zero instead of negative value calculated. For instance,

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Esteem Innovation Sdn Bhd, 2011

How hanger bar is designed for ACI? Q: How hanger bar is designed for ACI? A: Hanger bar is designed based on minimum tension steel area required by the c ode or minimum tension steel area required defined by user, whichever is greater. Hanger is only needed when the beam is designed as singly reinforced beam. The minimum percentage of tension steel required (user defined) by default is 0.13%. User is allowed to change this value in the parameter setting as shown below.

For example (cantilever beam): Minimum tension steel percentage specified by code, ACI 318 : 2008 Clause 10.5.1 conv = 0.006895 AsMin (%) = 600 × √(f'c × conv) / fy = 600 × √(30 × 0.006895) / 460 = 0.59% ≥ 20000 × conv / fy (0.30%) Use AsMin (%) = 0.59% ≥ 0.13% (User defined) Hence, minimum tension steel percentage = 0.59% Maximum Effective Depth of Section = 457.0 mm Minimum Tension Steel Area Required = 0.59 × 200.0 × 457.0 = 543 mm² Top Tension Steel Area Required = 543 mm² Top Reinforcement Provided = 3T16 (603 mm²) Bottom Reinforcement Provided = 3T12 (339 mm²) p733

Esteem Innovation Sdn Bhd, 2011

2

User defined minimum steel area for hanger bar = 0.13% x 200 x 457 = 119mm

Minimum steel area required by the code for hanger bar = 300 × √(f'c × conv) / fy x 200.0 x 457.0 2 = 274mm 2

Therefore, the bottom reinforcement provided is based on the Max(119,274) = 274mm

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Esteem Innovation Sdn Bhd, 2011

How does Esteem 8 calculate beam shear and torsion link reinforcement required for the beam section in ACI : 08 ? Question : How does Esteem 8 calculate shear and torsion link reinforcement required for the beam section in ACI ? Answer : First, the program will check whether torsion reinforcement is required or not for the beam section. If no, link reinforcement required will be calculated based on shear design only. For instance,

However, if torsion reinforcement is required for the beam section, link reinforcement required is the sum of the link reinforcement required for shear and torsion. For instance,

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Esteem Innovation Sdn Bhd, 2011

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IS456: 2000 Maximum Shear Link Spacing SvMax =Minimum (0.75 d, 450) where d is effec tive depth

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Esteem Innovation Sdn Bhd, 2011

AS3600: 2001

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Esteem Innovation Sdn Bhd, 2011

vc (Design concrete shear stress) Shear Design 1. To design shear strength of a beam, following values are needed: a. Vu b. Vuc c. Vu.min d. Vus e. Vu.max 2. Vu = V* , ultimate shear stress. 3. Clause 8.2.1, the design shear strength of a beam shall be taken øVu where either a.

, taking account of Clause 8.2.3 to 8.2.6 where Vuc is determined from Clause 8.2.7 and Vus is determined from Clauses 8.2.9 and 8.2.10; or b. Vu is calculated by means of a method based on the truss analogy, in which case Clauses 8.2.3 to 8.2.10 may not applied 4. Clause 8.2.7.1, the ultimate shear strength (Vuc) of a reinforced beam, excluding the contribution of shear reinforcement, shall be caluculated from the following equation:

Where β1

=

β2

=

1; or

= for member subject to significant axial tension; or = for member subject to significant axial compression β3

Ast

=

1; or may be taken as

=

2d0 / av but not greater than 2, provided that the applied loads and the support are orientated so as to create diagonal compression over the length av

=

Cross-sectional area of longitudinal reinforcement provided in the tension zone and fully anchored at the cross-section under consideration.

5. Clause 8.2.9, the ultimate shear strength of a beam provided with minimum shear reinforcement (Asv.min) shall be taken as . 6. With Vu and Vuc,

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7. clause 8.2.6, in no case shall the ultimate shear strength (Vu)be taken as greater than df bv

Where = (bw - 0.5Σdd)

Σdd

=

The sum of the diameter of the grouted ducts, if any, in a horizontal plane across the web

Pv

=

The vertical component of the prestressing force at the section under consideration

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Esteem Innovation Sdn Bhd, 2011

Maximum Shear Link Spacing SvMax = Minimum of (0.75 h, 500) when vd < vcMin SvMax = Minimum of (h / 2, 300) when vd >= vcMin where h = Total depth of section vcMin = Minimum Shear Strength of reinforcement (eg. 0.4 for BS, 0.35 for ACI etc ) vd = Design Shear Stress

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NZ3101: 1995 Maximum Shear Link Spacing SvMax = 0.25 d when vd > vMax / 2 SvMax = 0.5 d when vd <= vMax / 2d where d = Effective depth of section vd = Design Shear Stress vMax = Maximum Shear Stress Allowed

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EC2: 2004

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Material Poisson's ratio used in EC2 : 04 According to EC2: 04 Part 1.0 - General Rules for Building, Clause 3.1.2.5.3 - concrete Poisson's ratio will be 1. 0.2 or 2. 0 (provided cracking is permitted for concrete in tension)

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Esteem Innovation Sdn Bhd, 2011

Stress Strain Diagram for reinforcing steel used in EC2 : 04 It should be noted that EC2 permits two types of design stress-strain diagram for reinforcing steel : Standard and Bi-linear. However, only "Standard" type of design stress-strain diagram for reinforcing steel is used throughout the element design in Esteem 7.

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Esteem Innovation Sdn Bhd, 2011

Mean tensile strength of concrete Mean tensile strength of concrete, fctm, can either be determined by referring Table 3.1 Eurocode 2 : 2004 Part 1, or by means of calculation. The table in the code of practice is as shown below, where fctm, can be obtained directly from the table.

In Esteem 7, fctm is being determined by calculation. For instance,

Mean tensile strength of concrete, fctm = 0.3 × fck^⅔ = 0.3 × 25^⅔ = 2.6 N/mm²

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Esteem Innovation Sdn Bhd, 2011

Ecd (Elastic Modulus) used in EC2 : 04 Ec or Ecd (Design value of modulus of elasticity of concrete) = whereas Ecm = (22*(fck+8)/10)^0.3 (in GPa) and γcE = 1.2 Reference: EC2: 04 Clause 5.8.6 and Table 3.10

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Ecm /γcE

Esteem Innovation Sdn Bhd, 2011

Loading Wind Loads for EC1 User is first required to calculate the maximum and minimum wind speed based on the Eurocode 1: Actions of Structures - Part 1-4: General Actions - Wind Actions, Clause 4.2. The default values used in Esteem 8 are as follows: Parameter Maximum wind speed (m/s) Minimum wind speed (m/s) 3

Air density,  (kg/m ) Fundamental value of basic wind velocity, V b,0 (m/s) Directional factor, Cdir

Default values in Esteem 8 25 20 1.226 25 1.0

Season factor, Cseason Probability factor, Cprop

1.0

Roughness factor (for maximum wind speed at top level), C rTop Roughness factor (for minimum wind speed at bottom level), C rBot

1.0

1.0

0.8

Click on the Wind Speed Calculator to automatically calculates the value of the maximum and minimum wind speeds based on the param eter entered by user.

Where, Maximum wind speed (m/s) = V b,0 * Cdir * Cseason * Cprop * CrTop Minimum wind speed (m/s) = V b,0 * Cdir * Cseason * Cprop * CrBot 3 2 The basic velocity pressure, qp (kN/m ) = 0.5 *  * Vb

Notes

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Esteem Innovation Sdn Bhd, 2011

1. The value of the basic wind velocity, Vb varies for every floor depending on the height from the ground level. 2. Refer to Wind Loading for the application of the wind loads on every floor of the structure. Rectangular Load Distribution The calculation for rectangular load distribution: Basic wind velocity, Vb = Vb,0 * Cdir * Cseason * Cprop * Cr 3 2 The basic velocity pressure, qp (kN/m ) = 0.5 *  * Vb

Notes

:

1. The height limit and roughness factor may be overwritten by user as per requirement, while the program will automatically calculates the basic velocity pressure, q b based on the basic wind velocity Vb for the floor height given. . 2. To delete any of the row, click 3. To reset the values in the table and to return to the default setting, click on the button. 4. Refer to Wind Load Input Comply to Rectangular Load Distribution for the example of application.

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Pattern loading for EC2 : 04 Below are the load cases and combination rules which referred from EC2: 04 Part 1.0 General Rules for Building, Clause 2.5.1.2.

If the load factor default value is used (G = 1.35 and Q = 1.5), then it will carry out the pattern loading analysis as shown below.

(i) Loading arrangements for maximum moments in the spans.

(ii) Loading arrangements for maximum support moment in A.

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Esteem Innovation Sdn Bhd, 2011

(iii) Loading for design moments at the supports according to EC2.

Pattern loading result benchmarking between Esteem 8 and StaadPro by using Mosley's text book example (sixth edition - page 39)

Esteem 8 moment/shear diagram

Moment/shear diagram obtained from StaadPro p751

Esteem Innovation Sdn Bhd, 2011

Comparison result between Esteem 8 and StaadPro Hogging moment (from left to right), kNm

Sagging moment (from left to right), kNm

Esteem 8 result

Staad Pro result

Difference

Esteem 8 result

Staad Pro result

Difference

70.67

70.27

0.40 (0.57%)

116.9

117.31

0.41 (0.35%)

144.56

143.94

0.62 (0.43%)

25.74

25.92

0.18 (0.69%)

111.98

111.72

0.23 (0.02%)

10.9

10.89

0.09 (0%)

Shear force (from left to right), kN Esteem 8 result

Staad Pro result

Difference

135.65

135.65

0 (0%)

158.91

158.85

0.04 (0.06%)

105.27

105.11

0.16 (0.15%)

Moment diagram obtained from Mosley’s example

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Esteem Innovation Sdn Bhd, 2011

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Esteem Innovation Sdn Bhd, 2011

Load factor used in EC2 : 04 Below are the default partial safety factors which used in Esteem 8 (refer to Table 2.5 ; Mosley's text book ; sixth edition ; page 25).

However, user still allowed to add in new loading case or change the loading factor (except case 1 for Gravity & Lateral) for consideration of ψ value. User can refer to EC2 : 04 Part 1 :Clause 2.3.3 for detailed explanation of loading partial safty factor.

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Esteem Innovation Sdn Bhd, 2011

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Live load reduction for column, wall and foundation used in Euro Code In accordance with EN 1 Part 1.1 Clause 6.2.2(2) and Clause 6.2.1.3(11) and provided that the area is classified according to Table 6.1 into categories A to D, for columns and walls, the total imposed loads from several storeys may be multiplied by the reduction factor, n. There are two methods provided to calculate the reduction factor in EN 1. One is based on floor area and another one is based on floor number. In Esteem 7, method based on floor number is used.

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Esteem Innovation Sdn Bhd, 2011

Geometric Imperfections (Notional Horizontal Load) In Esteem 8, BS8110 Clause 3.1.4.2 Notional Horizontal Load is adopted in analysis as an alternative to the requirement by EC2: EN1992-1-1 Clause 5.2 Geometric Imperfections. The definition of notional horizontal load in BS8110 and EC2 is as below: BS8110 stated that all buildings should be capable of resisting a notional design ultimate horizontal load applied at each floor or roof level simultaneously equal to 1.5 % of the characteristic dead weight of the structure between mid-height of the storey below and either mid-height of the storey above or the roof surface [i.e. the design ultim ate wind load should not be taken as less than this value] Meanwhile, in EC2, it is stated that a horizontal force should be applied at each level of a structure resulting from a notional inclination of vertical members, and this should be considered in addition to lateral loads. User should aware of the differences between BS8110 and EC2 when considering horizontal load acting on the structure under design.

Notes : User may manually calculate transverse force, Hi according to EC2 and add this value to wind load as to analyse the structure based on EC2 Clause 5.2.

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Esteem Innovation Sdn Bhd, 2011

Design

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Esteem Innovation Sdn Bhd, 2011

Design Concept Procedure for determining flexural reinforcement in EC2 : 04

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Esteem Innovation Sdn Bhd, 2011

Design ultimate moment resistance of parabolic stress block for EC2 design Expressions for the value and location of the compressive force C in the section are given in Reinforced Concrete Design to Eurocode 2, Sixth edition, by Mosley (page 91-93).

o w x k1 k2

= = = = =

the concrete strain at the end of the parabolic section the distance from the neutral axis to strain, o depth to neutral axis mean concrete stress depth to the centroid of the stress block

C

=

k1 * b * x

...e.q 1

(a) To determine the mean concrete stress, k1 From the strain diagram;

Therefore; ...e.q 2 Substitute c2 = 0.002 (refer to EC2 Figure 3.3 and Table 3.1) ...e.q 3

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Substituting for w from the equation 3 gives; ...e.q 4 (b) To determine the depth of the centroid k2x

k2 is determined for a rectangular section by taking area moments of the stress block about the neutral axis – see figure above.

Substituting for w from the equation 3 gives;

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Esteem Innovation Sdn Bhd, 2011

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Esteem Innovation Sdn Bhd, 2011

Modified Structural System for cantilever spans A span is deemed to be a "cantilever span" when the % Difference of Moment defined above is less than 10%. Modified Structural System, K = 0.4 + 0.9 (Diff% / 10)

Pure cantilever Maximum difference

Percentage difference, % 0 10

Structural System, K 0.4 1.3

% Difference of Moment = [ Mc - M ] / [ Mt - M ] If the Diff% > 10%, K will use 1.0 or 1.3 based on the span condition (simply supported \ one end continuous). 5. When MBSDR is TRUE: o The left and right side moment, Mc and Mt will be taken from the moment envelop. o The sagging moment, M will be taken from the Full combination. 6. When MBSDR is FALSE: o If there is a sagging moment detected for the cantilever span, the K value will either be 1.0 or 1.3. Refer to example below (deflection check for Span 1): The design for moment and shear is based on support face:

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Moment diagram Moment envelop:

Remark The left and right side moment, Mc and Mt will be taken from moment envelop. The sagging moment,M is taken from Full load combination.

Full load combination moment:

DEFLECTION CHECKING FOR SPAN Left Side Moment Value = 0.0kNm Right Side Moment Value = -138.2kNm Maximum Sagging Moment Value = 1.1kNm Percentage Difference (%) = (1.1 / 139.3) × 100% = 0.80 % Modified Structural System, K = 0.4 + 0.9 × (0.80 / 10) = 0.5 Span Length, l = 4300.0 mm

Parameter is set to TRUE, calculate percentage difference of moment and get modified structural system, K. Percentage difference of moment calculated for Span 1 based on Full Moment. K = 0.5

DEFLECTION CHECKING FOR SPAN Parameter is set to FALSE, there is sagging moment detected from 2D moment envelop (1.19kNm). Structural System, K = 1.3 for end span continuous

Structural System, K = 1.3 (end span)

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Esteem Innovation Sdn Bhd, 2011

Slab Design Flexural reinforcement for slab in EC2 : 04 Slab has the similar procedure with beam for determining flexural reinforcement. The only difference is no doubly reinforcement design for slab.

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Esteem Innovation Sdn Bhd, 2011

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Slab deflection check in EC2 : 04

whereas p = AsTenReq / (b × d) Notes : Structural system for irregular slab  K = 1 if all slab edges are supported which designed as simply supported slab  K = 0.4 if ≤ 50% of slab edges are supported which designed as cantilever slab However, users are allowed to change the basic span-depth ratio which is under slab properties if they disagree the calculation of structural system for irregular slab shown above .

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Esteem Innovation Sdn Bhd, 2011

Slab Auto Deflection Control for EC2 : 04 Purpose: To automatically satisfy the slab deflection check by increasing tension reinforcement Reference: EC2 : 04 : Part 1 : Clause 7.4.2(2) : eqn. 7.16(a) and 7.16(b) Esteem 7 follows the following flow of logic:1.

Slab Deflection Check

Deflection ratio, (BR*MFSpan*MFs)/(Span/d)

1

MFSpan = Modification factor for long span MFs = Modification factor for tension reinforcement d = effective depth of section BR = Basic span/effective depth ratio 2.

If condition in step 1 is satisfied, stop here. Else, continue the following steps.

3. If slab auto deflection control is turned to False, stop here. Else, continue the following steps. 4. The program will increase provided tension reinforcement area until deflection ratio = 1.

MFs = 500 / fyk * (AsTenProv/AsTenReq) fyk = tension steel strength AsTenProv = Provided tension reinforcement AsTenReq = Required tension reinforcement

7.

Finally check against the AsTenProv/AsTenReq Maximum Limit

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Esteem Innovation Sdn Bhd, 2011

Determination of Structural System, K, for Irregular Shape Slab To determine the structural system, K, for irregular slab based on EC2, the following formula is being adopted: k = 0.4 + 0.6*(2*SE - 1) where, SE = Supported Edge Ratio

For example: 2. If the Supported Edge Ratio is 0.5; K = 0.4 + 0.6*[2*(0.5) - 1] = 0.4 (design as cantilever slab, K=0.4) 2. If the Supported Edge Ratio = 0.3; K = 0.4 + 0.6*[2*(0.3) - 1] = 0.4 (minimum Structural System, K = 0.4) 3 If the Supported Edge Ratio > 0.8, use the formula in item [3a] to get the Final Structural System, K. 3a. From the ratio of Continuous Edge Length to the Supported Edge Ratio , Final Structural System, K can be obtained. Final Structural System, K

=

k * (1 + 0.5 * csr)

Where csr,

Example: The Structural System, k = 0.4 + 0.6*[2*(1) - 1] = 1.0 > 0.4 Continuous Edge Length, c el = 18400mm Supported Edge Ration, sel = 18400 Continuous Edge on Supported Edge Ratio, csr = cel/sel = 18400/18400 = 1.0 > 0.8 Final Structural System, K = k * (1 + 0.5*csr) = 1.0 * (1 + 0.5 * 1) = 1.5 For polygonal and L-shape slab, refer to Example 1 and 2 below. Example: Refer to the following example to determine the Structural System for each slab.

Plan View

Remarks

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Esteem Innovation Sdn Bhd, 2011

The lenght of the supported edges, 1 2 3 4 5

= = = = =

5000mm 5000mm 7071.1mm 10000mm 7071.1mm

Total Supported Edge Length for Slab FS1 = 27071.1mm Slab FS3 = 7071.1mm Slab FS4 = 17071.1mm

Slab Design < Full Report

Data and Result of Slab Mark : gb - FS1

Remarks C ompute the lengths of supported edges.

Location : - 1/A - 4/A - 4/C - 1A/C Slab Shape : Irregular-Shape

SubSlab : FS1:1 Location : - 1/A - 4/A - 4/C - 1A/C SubSlab Shape : Polygon Dimension Length of Edge 1 = 10000.0 mm Length of Edge 2 = 5000.0 mm Length of Edge 3 = 5000.0 mm Length of Edge 4 = 7071.1 mm Sub-Slab Thickness, h = 125 mm Sub-Slab Drop = 0 mm Supported edge ratio, SE = 27071.1 / 27071.1 = 1.0

DEFLECTION CHECKING Shorter span in Y Direction (4330.1 mm) Supported Edge Length, sel = 27071.1 mm Continuous Edge Length on Supported Edge, cel = 17071.1 mm Total Edge Length, tel = 27071.1 mm Supported Edge Ratio, SE = sel / tel = 27071.1 / 27071.1 = 1.00 K = 0.4 + 0.6 × (2 × SE - 1) = 1.0 ≥ 0.4 Continuous Edge on Supported Edge Ratio, csr = cel / sel = 17071.1 / 27071.1 = 0.63 ≤ 0.8 Span Length, l = 4330.1 mm Effective Depth, d = 95.0 mm Actual Span / Depth Ratio, Ar = 45.6 Design Steel Strength, fyk = 500.0 N/mm² Area of Tension Steel Required, AsReq (Ten.) = 207 mm² Area of Tension Steel Provided, AsProv (Ten.) = 314 mm² - Checking for deflection is based on EC2 : 04 Part 1 Clause 7.4.2(2) - Expression 7.16.a or Expression 7.16.b and Table 7.4N: Basic span / effective depth p770

From the formula, K = 0.4 + 0.6*(2*1-1) = 1.0

Esteem Innovation Sdn Bhd, 2011

ratio for rectangular or flange beams - Expression 7.17: Modification factor for steel area provided Structural System, K = 1.0 Required Tension Reinforcement Ratio, ρ = AsTenReq / (b × d) = 207 / (1000.0 × 95.0) = 0.217 × 10 -2 Reference Reinforcement Ratio, ρo = √fck × 0.001 = √30.0 × 0.001 = 0.548 × 10 -2 ρ ≤ ρo => Use Expression (7.16.a) ρo / ρ = 2.519 Basic Span Depth Ratio, BR = K × [11 + 1.5 × √fck × ρo / ρ + 3.2 × √fck × (ρo / ρ - 1)3/2] = 1.0 × [11 + 1.5 × √30.0 × 2.519 + 3.2 × √30.0 × (2.519 - 1)3/2] = 64.5 Modification Factor for Steel Area Provided, MFs = AsTenProv / AsTenReq = 314 / 207 = 1.52 Deflection Ratio = (Br × MFs) / Ar = (64.5 × 1.52) / 45.6= 2.15 Ratio >= 1.0 : Deflection Checked PASSED

Data and Result of Slab Mark : gb - FS3 Location : - 1/C - 1/A - 1A/C Slab Shape : Triangular

SubSlab : FS3:1 Location : - 1/C - 1/A - 1A/C SubSlab Shape : Triangular Dimension Length of Edge 1 = 5000.0 mm Length of Edge 2 = 7071.1 mm Length of Edge 3 = 5000.0 mm Sub-Slab Thickness, h = 125 mm Sub-Slab Drop = 0 mm Supported edge ratio, SE = 7071.1 / 17071.1 = 0.41

DEFLECTION CHECKING Shorter span in X Direction (3535.5 mm) Supported Edge Length, sel = 7071.1 mm Continuous Edge Length on Supported Edge, cel = 7071.1 mm Total Edge Length, tel = 17071.1 mm Supported Edge Ratio, SE = sel / tel = 7071.1 / 17071.1 = 0.41 K = 0.4 + 0.6 × (2 × SE - 1) = 0.3 < 0.4 Use Structural System, K = 0.4 Continuous Edge on Supported Edge Ratio, csr = cel / sel = 7071.1 / 7071.1 = 1.00 > 0.8 Final Structural System, K = 0.4 × (1 + 0.5 × 1.00) = 0.6 Span Length, l = 3535.5 mm Effective Depth, d = 95.0 mm Actual Span / Depth Ratio, Ar = 37.2 Design Steel Strength, fyk = 500.0 N/mm² Area of Tension Steel Required, AsReq (Ten.) = 144 mm² Area of Tension Steel Provided, AsProv (Ten.) = 314 mm² - Checking for deflection is based on EC2 : 04 Part 1 Clause 7.4.2(2) - Expression 7.16.a or Expression 7.16.b and Table 7.4N: Basic span / effective depth ratio for rectangular or flange beams - Expression 7.17: Modification factor for steel area provided Structural System, K = 0.6 Required Tension Reinforcement Ratio, ρ = AsTenReq / (b × d) = 144 / (1000.0 × 95.0) = 0.151 × 10 -2 Reference Reinforcement Ratio, ρo = √fck × 0.001 = √30.0 × 0.001 = 0.548 × 10 -2 ρ ≤ ρo => Use Expression (7.16.a) p771

From the formula, K = 0.4 + 0.6*(2*0.41-1) = 0.3 < 0.4 Use K=0.4 Use Final Structural System, K = 0.4 * (1+0.5*1.0) = 0.6

Notes

:

For polygonal slabs, the minimum Structural System, K, is 0.4. K cannot be lesser than 0.4.

Esteem Innovation Sdn Bhd, 2011

ρo / ρ = 3.637 Basic Span Depth Ratio, BR = K × [11 + 1.5 × √fck × ρo / ρ + 3.2 × √fck × (ρo / ρ - 1)3/2] = 0.6 × [11 + 1.5 × √30.0 × 3.637 + 3.2 × √30.0 × (3.637 - 1)3/2] = 69.5 Modification Factor for Steel Area Provided, MFs = AsTenProv / AsTenReq = 314 / 143 = 2.20 Deflection Ratio = (Br × MFs) / Ar = (69.5 × 2.20) / 37.2= 4.10 Ratio >= 1.0 : Deflection Checked PASSED

Data and Result of Slab Mark : gb - FS4 Location : - 1/A - 1A/B - 4/A Slab Shape : Triangular

SubSlab : FS4:1 Location : - 1/A - 1A/B - 4/A SubSlab Shape : Triangular Dimension Length of Edge 1 = 7071.1 mm Length of Edge 2 = 7071.1 mm Length of Edge 3 = 10000.0 mm Sub-Slab Thickness, h = 125 mm Sub-Slab Drop = 0 mm DEFLECTION CHECKING Shorter span in X Direction (5000.0 mm) Supported Edge Length, sel = 17071.1 mm Continuous Edge Length on Supported Edge, cel = 10000.0 mm Total Edge Length, tel = 24142.1 mm Supported Edge Ratio, SE = sel / tel = 17071.1 / 24142.1 = 0.71 K = 0.4 + 0.6 × (2 × SE - 1) = 0.6 ≥ 0.4 Continuous Edge on Supported Edge Ratio, csr = cel / sel = 10000.0 / 17071.1 = 0.59 ≤ 0.8 Span Length, l = 5000.0 mm Effective Depth, d = 95.0 mm Actual Span / Depth Ratio, Ar = 52.6 Design Steel Strength, fyk = 500.0 N/mm² Area of Tension Steel Required, AsReq (Ten.) = 156 mm² Area of Tension Steel Provided, AsProv (Ten.) = 314 mm² - Checking for deflection is based on EC2 : 04 Part 1 Clause 7.4.2(2) - Expression 7.16.a or Expression 7.16.b and Table 7.4N: Basic span / effective depth ratio for rectangular or flange beams - Expression 7.17: Modification factor for steel area provided Structural System, K = 0.6 Required Tension Reinforcement Ratio, ρ = AsTenReq / (b × d) = 156 / (1000.0 × 95.0) = 0.164 × 10 -2 Reference Reinforcement Ratio, ρo = √fck × 0.001 = √30.0 × 0.001 = 0.548 × 10 -2 ρ ≤ ρo => Use Expression (7.16.a) ρo / ρ = 3.346 Basic Span Depth Ratio, BR = K × [11 + 1.5 × √fck × ρo / ρ + 3.2 × √fck × (ρo / ρ - 1)3/2] = 0.6 × [11 + 1.5 × √30.0 × 3.346 + 3.2 × √30.0 × (3.346 - 1)3/2] = 65.8

p772

Supported edge ratio, SE = 17071.1 / 24142.1 = 0.71 From the formula, K = 0.4 + 0.6*(2*0.71-1) = 0.6

Esteem Innovation Sdn Bhd, 2011

Modification Factor for Steel Area Provided, MFs = AsTenProv / AsTenReq = 314 / 156 = 2.02 Deflection Ratio = (Br × MFs) / Ar = (65.8 × 2.02) / 52.6= 2.53 Ratio >= 1.0 : Deflection Checked PASSED

Therefore based on the formula for K above, for a polygonal shape slab the m inimum K = 0.4. However, user can also overwrite the span ratio to between 6.0 and 36.0 by setting the Span Ratio manually at the Object Viewer. Example 2: L-Shape Slab Refer to the following example on the determination of the Basic Span/Depth Ratio for sub-slabs FS29-1 , FS29-2 and FS29-3. Plan View

Remarks The 1= 2= 3= 4= 5= 6=

length of the supported edges; 6401mm 2286mm 1829mm 1067m 4572mm 3353mm

Total edge lengths = (1+2+3+4+5+6) = 19508mm Total supported edge lengths = (1+2+3+4+5+6) = 19508mm

Slab Design > Full Report

Area of SubSlab (i): 4878324.0 mm² (1), 10451592.0 mm² (2), 4181094.0 mm² (3) Effective Depth of SubSlab (i): 70.5 mm (1), 70.5 mm (2), 70.5 mm (3) ∑ (Ai × di) = 1375526205.00 ∑ Ai = 19511010.00 p773

Remarks

Esteem Innovation Sdn Bhd, 2011

Average Effective Depth, d = ∑ (Ai × di) / ∑ Ai = 70.50 Shorter span in X Direction (3196.9 mm)

Supported edge ratio, SE = ratio of supported edge/total edge length = 19508/19508 = 1

Supported Edge Length, sel = 19508.0 mm Continuous Edge Length on Supported Edge, cel = 16155.0 mm Total Edge Length, tel = 19508.0 mm Supported Edge Ratio, SE = sel / tel = 19508.0 / 19508.0 = 1.00 Use formula above to get K = 0.4 + 0.6 × (2 × SE - 1) = 1.0 ≥ 0.4 Continuous Edge on Supported Edge Ratio, csr = cel / sel = 16155.0 / 19508.0 = 0.83 > Final Structural System, K = 1.0 * (1 + 0.5 * 0.83) 0.8 = 1.4 Final Structural System, K = 1 × (1 + 0.5 × 0.83) = 1.4 Span Length, l = 3196.9 mm Effective Depth, d = 70.5 mm Actual Span / Depth Ratio, Ar = 45.3 Design Steel Strength, fyk = 500.0 N/mm² Area of Tension Steel Required, AsReq (Ten.) = 107 mm² Area of Tension Steel Provided, AsProv (Ten.) = 318 mm² - Checking for deflection is based on EC2 : 04 Part 1 Clause 7.4.2(2) - Expression 7.16.a or Expression 7.16.b and Table 7.4N: Basic span / effective depth ratio for rectangular or flange beams - Expression 7.17: Modification factor for steel area provided Structural System, K = 1.4 Required Tension Reinforcement Ratio, ρ = AsTenReq / (b × d) = 107 / (1000.0 × 70.5) = 0.151 × 10 -2 Reference Reinforcement Ratio, ρo = √fck × 0.001 = √30.0 × 0.001 = 0.548 × 10 -2 ρ ≤ ρo => Use Expression (7.16.a) ρo / ρ = 3.637 Basic Span Depth Ratio, BR = K × [11 + 1.5 × √fck × ρo / ρ + 3.2 × √fck × (ρo / ρ - 1) 3/2 ] = 1.4 × [11 + 1.5 × √30.0 × 3.637 + 3.2 × √30.0 × (3.637 - 1)3/2] = 163.9 Modification Factor for Steel Area Provided, MFs = AsTenProv / AsTenReq = 318 / 106 = 3.00 Deflection Ratio = (Br × MFs) / Ar = (163.9 × 3.00) / 45.3= 10.83 Ratio >= 1.0 : Deflection Checked PASSED

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The slab passed in Deflection checking.

Esteem Innovation Sdn Bhd, 2011

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Beam Design Flexural reinforcement for rectangular beam in EC2 : 04

Notes

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fyd = fyk/s In case εs > εsc for doubly reinforced design, compression steel required, Asc = M2 / [(fsc / γs) × (d – d’)] whereas fsc = εsc × Es

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Flexural reinforcement for flanged beam in EC2 : 04

Notes : Flanged beam design only will be executed when the flanged beam design is set to True (under Object Viewer) for the selected beam. Besides, users have to compute the flange width p778

Esteem Innovation Sdn Bhd, 2011

and flange depth by themselves as the program not able to auto-detect the flange width and flange depth.

In case εs > εsc for doubly reinforced design, compression steel required, Asc = M2 / [(fsc / γs) × (d – d’)] whereas fsc = εsc × Es fyy = 1/s

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Esteem Innovation Sdn Bhd, 2011

Minimum and maximum area of tension reinforcement for EC2 : 04 The minimum and maximum area of tension reinforc ement is checked based on EC2 : 04 ; Part 1 ; Clause 9.2.1.1.

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Esteem Innovation Sdn Bhd, 2011

Beam shear reinforcement in EC2 : 04

Notes : fcd = fck / γc

θ can be computed by using equation below.

θ is range from 22º ~ 45º.

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Esteem Innovation Sdn Bhd, 2011

Maximum shear link spacing for EC2 : 04 The maximum shear link spacing is checked based on EC2 : 04 Clause 9.2.2(6)

SvMax = 0.75d where d = Effective depth of section  = 0

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Beam torsion design in EC2 : 04

where fcd = fck / γc p783

Esteem Innovation Sdn Bhd, 2011

Torsion : θ can be computed by using equation below

Shear : θ can be computed by using equation below

θ is range from 22° ~ 45°.

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Esteem Innovation Sdn Bhd, 2011

Maximum torsion link spacing for EC2 : 04 The maximum torsion link spacing is checked based on EC2 : 04 Clause 9.2.3 (3): The longitudinal spacing of the torsion links should not exceed u / 8 (see 6.3.2, Figure 6.11, for the notation), or the requirement in 9.2.2 (6) or the lesser dimension of the beam cross section. SvMax = Minimum of (u/8, 0.75d, b, h) where u = 2*(b+h) d = section effective depth b = section width h = section depth

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Esteem Innovation Sdn Bhd, 2011

Beam deflection check for EC2 : 04

where p = AsTenReq / (b × d) p' = AsComReq / (b × d)

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Esteem Innovation Sdn Bhd, 2011

Beam auto deflection control for EC2 : 04 Purpose: To automatically satisfy the beam deflection check by increasing compression reinforcement Reference: EC2 : 04 : Part 1 : Clause 7.4.2(2) : eqn. 7.16(a) and 7.16(b) Esteem 7 follows the following flow of logic:3.

Beam Deflection Check

Deflection ratio, (BR*MMF*MFSpan*MFs)/(Span/d) 1



MMF = Modification factor for flange beam MFSpan = Modification factor for long span MFs = Modification factor for tension reinforcement d = effective depth of section BR = Basic span/effective depth ratio 4.

If condition in step 1 is satisfied, stop here. Else, continue the following steps.

3. If beam auto deflection control is turned to False, stop here. Else, continue the following steps. 4. If p  po, stop here. Else, continue the following steps. 5. The program will use compression reinforcement provided ratio as p' to calculate the new BR. If deflection ratio  1, stop here. Else, continue the following steps.

as p>po, BR = K*[11+1.5fck*po/(p-p')+(fck*p'/po)/12] K = Structural system fck = Concrete cylinder strength p = Required tension reinforcement ratio p' = Provided compression reinforcement ratio po = Reference reinforcement ratio,fck 10^-3

6. The program will increase provided compression reinforcement area until deflec tion ratio = 1, provided p > p' (at most p' = 90% of p) 7.

Finally check against the As',Prov/As,Prov Maximum Limit

Notes : 1) Increasing compression bars will increase ductility of strength failure behavior whereas increasing the tension bars will decrease ductility of strength failure behavior. Therefore, increasing the tension steel has been excluded on this basis.

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Esteem Innovation Sdn Bhd, 2011

Column Design Concrete Shear Capacity, VRd,c

For Reinforced Concrete Design value of k1 and σcp can be ignored. Since tensile reinforcement area is not able to be obtained, Esteem adopted minimum concrete shear capacity as concrete shear capacity. Example Calculation: EC2 : 04 Part 1 Expression 6.2.b k = 1 + √(200 / d) = 1 + √(200 / 189.0) = 2.0 > 2.0 k = 2.0 Minimum Shear Capacity of Concrete, VRd,cmin = 0.035 × k3/2 × fck1/2 × b × d × 0.001 = 0.035 × 2.03/2 × 30.01/2 × 450.0 × 189.0 × 0.001 = 46.12 kN Note: VRd,c = VRd,cmin as the tensile reinforcement area is not able to be defined Shear Capacity of Concrete, VRd,c = VRd,cmin = 46.12 kN

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Esteem Innovation Sdn Bhd, 2011

Column classification (short or slender column) for EC2 : 04

Notes

:

k1 = Relative Stiffness at Top of the Element k2 = Relative Stiffness at Bottom of the Element Relatives Stiffness = 10 for connection between column to pin foundation or connection between column to slab Relatives Stiffness = 1 for connection between column to fixed foundation lo = Column clear length A = 1 / (1 + 0.2 × φef) φef = Effective creep ratio User is allowed to customize Øef value which is located at Project Parameter > Design > Floor > Column > Design or Floor Parameter > Design > Column > Design B = (1 + 2w) 0.5 w = (As × fyd) / (Ac × fcd) As = Total area of longitudinal reinforcement p789

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fyd = Design yield strength of the reinforcement, fyk/γs Ac = Total area of concrete fcd = Design compressive strength of concrete, αcc × fck / γc αcc = 0.85 C = 1.7 – rm (for braced column) C = 0.7 (for unbraced column) rm = M 01/M02 M01, M02 are the first order moments at the end of the column with M02  M01 n = N / (Ac × fcd) N = Design ultimate axial load in the column i = radius of gyration about the axis considered, (I/A c)

0.5

I = Second moment of area of the section about the axis Ac = Cross section area of column

Clause to be referred: EC2: 04 Clause 5.8.3.1 : Slenderness criterion for isolated members EC2: 04 Clause 5.8.3.2 : Slenderness and effective length of isolated members

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Esteem Innovation Sdn Bhd, 2011

Example calculation of Relative Stiffness, k

Take Column (1A,A1) 300mm x 400mm; y-y direction : column = 400*3003 / 12= 900 x 106 mm4 beam = 300*5003 / 12= 3125 x 106 mm4 k1

= ( column/lcolumn ) /Σ( beam/lbeam ) = ( 900 x 106 / 3000) / [2 * (3125 x 106 / 4000)] = 0.192

k2

z-z direction :

= 1.000 (pin connection)

column = 300*4003 / 12= 1600 x 106 mm4 beam

= 300*5003 / 12= 3125 x 106 mm4

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Esteem Innovation Sdn Bhd, 2011

k1

= ( column/lcolumn ) /Σ( beam/lbeam ) = ( 1600 x 106 / 3000) / [2 * (3125 x 106 / 4000)] = 0.341

k2

= 1.000 (pin connection)

p792

Esteem Innovation Sdn Bhd, 2011

Method to Determine Braced/Unbraced Frames Esteem 8 performs sway check to determine whether the column is "braced" or "unbraced" through the concept of "column stability index" based on ACI318. Kindly refer to Sway Effects.

p793

Esteem Innovation Sdn Bhd, 2011

Short and slender column design in EC2 : 04

Notes

:

N = Design value of the applied axial force (tension or compression) ei = Additional eccentricity covering the effects of geom etrical imperfections, see EC2: 04 5.2(7) leff = Column effective length e0 = Minimum eccentricity moment, see EC2: 04 6.1(4) H = Depth of the section Mend = 1

st

order bending moment at column end

c = Factor depending on the curvature distribution, see EC2: 04 5.8.8.2(4) p794

Esteem Innovation Sdn Bhd, 2011

1/r = Curvature, see EC2: 04 Clause 5.8.8.3 fck = Concrete cylinder strength λ = Slenderness ratio, see EC2: 04 5.8.3.1 Kφ = Factor for taking account of creep, see EC2: 04 5.8.8.3(4) φef =Effective creep ratio, see EC2: 04 5.8.4 User is allowed to customize this value which is under Project Parameter > Design > Floor > Column > Design or Floor Parameter > Design > Column > Design fyk = Characteristic yield strength of reinforcement γs = Steel partial safety factor Es = Elastic modulus for steel d = Section effective depth Kr = Correction factor depending on axial load, see EC2: 04 5.8.8.3(3) Mi = 1

st

order bending moment at mid column

Kr-new = Revised correction factor obtained from iteration n = Relative axial force Ac = Area of concrete cross section fcd = Concrete design strength in compression, αcc fck / γc, αcc = Coefficient taking account of long term effects on the com pressive strength, and of unfavourable effects resulting from the way the load is applied ; 0.85 is use. γc = concrete partial safety factor As = Total area of reinforcement fyd = Design yield strength of reinforcement, fyk/ γs nbal = Value of n at maximum moment resistance, 0.4 may be used

p795

Esteem Innovation Sdn Bhd, 2011

Pad Footing Design Flexural reinforcement design for pad footing in EC2 : 04 Pad footing has the similar procedure with beam for determining flexural reinforcement. The only difference is no doubly reinforcement design for pad footing.

p796

Esteem Innovation Sdn Bhd, 2011

p797

Esteem Innovation Sdn Bhd, 2011

Punching shear checking for pad footing in EC2 : 04

Notes : CRd,c = 0.18 / c esm = enhance shear multiplier

p798

Esteem Innovation Sdn Bhd, 2011

For rectangular column case Critical perimeter = column perimeter + 2 ×  × k × d Area inside critical perimeter = (B' × L') - (4 - ) × (k × d)² B' = B + 2 × k × d L' = L + 2 × k × d

Although the punching shear checking is done at section 0.2d to 2.0d with interval of 0.05d, but pad footing report only displays punching shear checking at 5 sections (0.5d, 1.0d, 1.5d, 2.0d and the most critical section) unless the checked section is out of pad footing profile.

p799

Esteem Innovation Sdn Bhd, 2011

Ultimate shear checking for pad footing in EC2 : 04

Notes

:

Earth pressure, ρ = ultimate safety factor * PMax Ultimate axial load = ρ * pad area

p800

Esteem Innovation Sdn Bhd, 2011

Flexural shear checking for pad footing in EC2 : 04

Notes

:

CRd,c = 0.18 / c esm = enhance shear multiplier

p801

Esteem Innovation Sdn Bhd, 2011

Although the punching shear checking is done at section 0.2d to 1.5d (depends on enhanced shear multiplier value, default is 1.5) with interval of 0.05d, but pad footing report only displays flexural shear checking at preset section and the most critical section unless the checked section is out of pad footing profile.

p802

Esteem Innovation Sdn Bhd, 2011

Pile Footing Design Flexural reinforcement design for pile footing design in EC2 : 04 Pile footing has the similar procedure with beam for determining flexural reinforcement. The only difference is no doubly reinforcement design for pile footing.

p803

Esteem Innovation Sdn Bhd, 2011

Punching shear checking for pile footing in EC2 : 04

Notes : CRd,c = 0.18 / c esm = enhance shear multiplier

For rectangular column case p804

Esteem Innovation Sdn Bhd, 2011

Critical perimeter = column perimeter + 2 ×  × k × d Area inside critical perimeter = (B' × L') - (4 - ) × (k × d)² B' = B + 2 × k × d L' = L + 2 × k × d

Although the punching shear checking is done at section 0.2d to 2.0d with interval of 0.05d, but pile footing report only displays punching shear checking at 5 sections (0.5d, 1.0d, 1.5d, 2.0d and the most critical section) unless the checked section is out of pile footing profile.

p805

Esteem Innovation Sdn Bhd, 2011

Ultimate shear checking for pile footing in EC2 : 04

Notes

:

Ultimate axial load = Ultimate safety factor * num ber of pile provided * pile capacity

p806

Esteem Innovation Sdn Bhd, 2011

Flexural shear checking for pile footing in EC2 : 04

Notes

:

CRd,c = 0.18 / c p807

Esteem Innovation Sdn Bhd, 2011

esm = enhance shear multiplier

Although the punching shear checking is done at section 0.2d to 1.5d (depends on enhanced shear multiplier value, default is 1.5) with interval of 0.05d, but pile footing report only displays flexural shear checking at preset section and the most critical section unless the checked section is out of pile footing profile.

p808

Esteem Innovation Sdn Bhd, 2011

Detailing

p809

Esteem Innovation Sdn Bhd, 2011

Beam Side Bar Detailing According to EC2: 04 Part 1.0 - General Rules for Building, Clause 7.3.3(3) - in order to control cracking,

Notes

:

1. Default setting for the Maximum Depth when EC2 is selected is 999mm. Any beam depth more than this setting will be provided with side bars. 2. Allowable range for the Maximum Depth is from 500mm - 1500mm. User may define the bar diameter and maximum spacing of the side bars in the parameter setting: User Defined Bar Diameter = FALSE  This setting will let the program automatically use side bar diameter same as the bar diameter used in the tension reinforcement.

User Defined Bar Diameter = TRUE



This setting will let user to specify the bar diameter used for the side bar detailing.

p810

Esteem Innovation Sdn Bhd, 2011

FAQ Does the program carry out shear check for slab design? Question : Does the program carry out shear check for slab design? Answer : No, the program does not carry out shear check in slab design. Users have to manually do the shear check if find it is necessary.

p811

Esteem Innovation Sdn Bhd, 2011

Does Esteem 7 do the shear between web and flanges checking for the flanged beam design in EC2 ? Question : Does Esteem 7 do the shear between web and flanges checking for the flanged beam design in EC2 ? Answer : No. Shear between web and flanges does not checked for flanged beam design in EC2. However, user can manually do the shear between web and flanges checking based on EC2 : 04 Part 1 Clause 6.2.4.

Tips: A warning message will appear on top in the full report reminding user that the program does not run shear check for flanged beam design, thus user has to do shear check manually.

p812

Esteem Innovation Sdn Bhd, 2011

How does Esteem 7 calculate beam shear and torsion link reinforcement required for the beam section in EC2 : 04 ? Question : How does Esteem 7 calculate shear and torsion link reinforcement required for the beam section in EC2 ? Answer : First, the program will check whether torsion reinforcement is required or not for the beam section. If no, link reinforcement required will be calculated based on shear design only. For instance,

However, if torsion reinforcement is required for the beam section, link reinforcement required is the sum of the link reinforcement required for shear and torsion. For instance,

p813

Esteem Innovation Sdn Bhd, 2011

p814

Esteem Innovation Sdn Bhd, 2011

Does the program do the shear check based on EC2 : 04 Part 1 Clause 6.2.1 (8) ? Question : Does the program do the shear check based on EC2 : 04 Part 1 Clause 6.2.1 (8) whereas for members subject to predominantly uniformly distributed loading, the design shear force need not to be checked at a distance less than d from the face of the support? Answer : No, the program does not do the shear check based on EC2 : 04 Part 1 Caluse 6.2.1(8). However, it does the shear check (for support zone) either at support face or support center which can be defined by user.

p815

Esteem Innovation Sdn Bhd, 2011

How hanger bar is designed for EC2? Q: How hanger bar is designed in for EC2? A: Hanger bar is designed based on minimum tension steel area required by the c ode or minimum tension steel area required defined by user, whichever is greater. Hanger bar is only needed when the beam is designed as single reinforced beam. The minimum percentage of tension steel required (user defined) by default is 0.13%. User is allowed to change this value in the parameter setting as shown below.

For example (cantilever beam): Minimum tension steel percentage specified by code, EC2 : 04 Part 1 Clause 9.2.1.1 fctm = 0.3 × fck2/3 = 0.3 × 302/3 = 2.9 N/mm² AsMin (%) = 26 × (fctm/fyk) = 26 × (2.9/460) = 0.16% ≥ 0.13% (User defined) Hence, minimum tension steel percentage = 0.16% Maximum Effective Depth of Section = 459.0 mm Minimum Tension Steel Area Required = 0.16 × 200.0 × 459.0 = 151 mm² Top Tension Steel Area Required = 151 mm² Top Reinforcement Provided = 2T12 (226 mm²) Bottom Reinforcement Provided = 2T12 (226 mm²) p816

Esteem Innovation Sdn Bhd, 2011

2

User defined minimum steel area required for hanger bar = 0.13% x 200 x 459 = 119mm 2

Minimum steel area required by the c ode for hanger bar = 151mm

2

Therefore, the bottom reinforcement provided is based on the Max(119,151) = 151mm

p817

Esteem Innovation Sdn Bhd, 2011

EC8: 2003

p818

Esteem Innovation Sdn Bhd, 2011

Parameters Ductility Classes and Material Requirements The design and detailing provision for seismic is done based on EC8. This parameter is available when Code of Practice used is EC2 and parameter Seismic Analysis is TRUE. Three dissipation classes are introduced:  Low (DCL)  Medium (DCM)  High (DCH) Material Requirements - DCL Class Reference: EC8: 2003 Clause 5.3.1 and 5.3.2 i. When DCL class is selected, the requirement for concrete class will be the same as EC2. Material Requirements - DCM Class Reference: EC8: 2003 Clause 5.4.1.1(1) and (2) i. The program will automatically check that concrete of a class lower than C 16/20 shall not be used. Material Requirements - DCH Class Reference: EC8: 2003 Clause 5.5.1.1(1) and (2) i. The program will automatically check that concrete of a class lower than C 20/25 shall not be used. The allowable range for the Characteristic Yield Strength of Reinforcement (fyk) is from 400 2 2 N/mm to 600 N/mm . Refer to EC8: 2003 Clause 5.3.1 and 5.3.2.

p819

Esteem Innovation Sdn Bhd, 2011

Input Beam Geometrical Constraints Beam Geometrical Constraints - DCH class Reference: EC8: 2003 Clause 5.5.1.2.1 (1) and (2) When EC2 is selected and Seismic Analysis is TRUE: i. The program will check in the Verification that the width of any beam element shall be not less than 200 mm. ii. The program will check in the Verification that any beam element depths shall not be more than 3.5 times the width of the beam. Refer to Seismic Design -> Verification for more information.

p820

Esteem Innovation Sdn Bhd, 2011

Column Geometrical Constraints Column Geometrical Constraints - DCH Class Reference: EC8: 2003 Clause 5.5.1.2.2(1) When EC2 is selected and Seismic Analysis is TRUE: iii. The program will check in the Verification that the minimum dimension of a column (width or height) shall not be less than 250mm. iv. The program will check that each dimension of a polygonal column shall not be less than 250mm. Otherwise, Warning will be shown for the c olumn. Refer to Seismic Design -> Verification for more information.

p821

Esteem Innovation Sdn Bhd, 2011

Wall Geometrical Constraints Wall Geometrical Constraints – DCM Class and DCH Class Reference: EC8: 2003 Clause 5.4.1.2.3(1) Ductile Walls v. The program will check in the Verification that the thickness of any wall should not be less than 150mm. Notes : 1. Note that the program will not check for the requirement where the thickness of the wall should not be less than hs/20, where hs is the clear storey height. User must check this c onfiguration manually.

p822

Esteem Innovation Sdn Bhd, 2011

Verification These Warning and Verification shows a list of integrity checking based on the requirements of Seismic Design to Eurocode 8: 2003. Warning Message Type of Integrity Check

Error Message Remarks

“[]” represents the mark “{}” represents the limitation value Beam 1.

Beam Geometrical C onstraints for DCH Class

2.

Beam section [] has width less than 200 mm, which is not allowed when DCH is selected for seismic design (Refer to EC8: 2003 Clause 5.5.1.2.1(1)). Beam section [] has depth that are more than 3.5 times of the section's width, which is not allowed when DCH class is selected for seismic design (Refer to EC8: 2003 C lause 5.5.1.2.1(2)). Column

1. For Warning no.1, any 1. C olumn [] contains an edge Column Geometry C onstraints for DCH Class

that is less than 250mm long, which is not advised as DCH class is selected for seismic design (Refer to EC8: 2003 C lause 5.5.1.2.2(1)). (Refer to Remark no.1).

1. C olumn []'s width is less than

edge length; a, b, c, etc for the polygonal column will be checked.

250mm, which is not allowed when DC H class is selected for seismic design (Refer to EC 8: 2003 C lause 5.5.1.2.2(1)).

RC Wall

1. RC wall [] has thickness less Wall Geometry C onstraints

than 150mm, which is not allowed for seismic design (Refer to EC8: 2003 Clause 5.4.1.2.3(1)).

p823

For error no.1, the program will not check whether the wall thickness satisfies  hs /20; where hs is the clear storey height in meters. [$X$IF_CON1]Refer to Case 28006.

Esteem Innovation Sdn Bhd, 2011

Design and Detailing

p824

Esteem Innovation Sdn Bhd, 2011

Beam Minimum Reinforcement Ratio of the Tension Zone DCM and DCH Class According to EC8: 2003 Clause Clause 5.4.3.1.2 (5), the reinforcement ratio at the tension zone is specified as following:

p825

Esteem Innovation Sdn Bhd, 2011

Maximum Link Distance from Support for DCM and DCH Class DCM and DCH Class According to Clause 5.4.3.1.2 (6)(c) and 5.5.3.1.3(6), the first link distance from beam end section shall not be more than 50mm. Therefore when Code of Practice EC2 and Seismic Analysis is TRUE, for both DCM and DCH Class the default parameter for link distance is 50 mm from the support. Feature/Input

Expected Outcome/Output

p826

Esteem Innovation Sdn Bhd, 2011

Beam Critical Regions for DCM Class Normal beams According to Clause 5.4.3.1.2 (1), the critical regions for beams are from the beam end section up to a distance, lcr = hw, where hw is the beam depth Beams supporting transfer columns According to Clause 5.4.3.1.2 (2), the critical regions for beams supporting transfer columns is from the beam end section at transfer column surface up to a distance, lcr = 2 X h w, where hw is the beam depth, at both sides of the transfer column Cantilever beams Critical region is applied to the whole span of the cantilever beam. Critical regions in beam detailing for DCM Class Co Beam depth = 850 mm nti Critical region = beam depth = 850 mm nu ou s be a m Be Beam depth = 600 mm a Critical region = 2 X 600 = 1200 mm, at both sides of transfer c olumn m su pp ort in g tra nsf er col u m n Ca Whole span of cantilever beam regarded as critical region nti lev er be a m

Ta Tapered beam begin depth at left support = 800 mm pe Critical region at left support= 800 mm re Tapered beam end depth = 1200 mm d Critical region at right support= 1200 mm be a p827

Esteem Innovation Sdn Bhd, 2011

m

p828

Esteem Innovation Sdn Bhd, 2011

Minimum Link Diameter and Maximum Link Spacing for DCM Class According to Clause 5.4.3.1.2(6) within critical region 1. The minimum link diameter shall not be less than 6 m m 2. The maximum link spacing SvMax should not exceed the minimum of the following (all dimensions in millimeters):

o o o o

1/4 X beam depth 24 X link diameter 225 8 X minimum longitudinal bar diameter

p829

Esteem Innovation Sdn Bhd, 2011

Beam Critical Regions for DCH Class Normal beams According to Clause 5.5.3.1.3 (1), the critical regions for beams is from the beam end section up to a distance, lcr = 1.5 X h w, where hw is the beam depth Beams supporting transfer columns According to Clause 5.5.3.1.3 (2), the critical regions for beams supporting transfer columns is following Clause 5.4.3.1.2(2) which is from the beam end section at transfer c olumn surface up to a distance, lcr = 2 X h w, where hw is the beam depth, at both sides of the transfer column Cantilever beams Critical region is applied to the whole span of the cantilever beam. Critical regions in beam detailing for DCM Class Co Beam depth = 850 mm nti Critical region = 1.5 X beam depth = 1275 mm nu ou s be a m Be Beam depth = 600 mm a Critical region = 2 X 600 = 1200 mm, at both sides of transfer c olumn m su pp ort in g tra nsf er col u m n Ca Whole span of cantilever beam regarded as critical region nti lev er sp an

Ta Tapered beam begin depth at left support = 800 mm pe Critical region at left support= 1.5 x 800 = 1200 mm re Tapered beam end depth = 1200 mm d Critical region at right support= 1.5 x 1200 = 1800 mm be a p830

Esteem Innovation Sdn Bhd, 2011

m

p831

Esteem Innovation Sdn Bhd, 2011

Minimum Link Diameter and Maximum Link Spacing for DCH Class According to Clause 5.5.3.1.3(6) within critical region 3. The minimum link diameter shall not be less than 6 m m 4. The maximum link spacing SvMax should not exceed the minimum of the following (all dimensions in millimeters):

o o o o

1/4 X beam depth 24 X link diameter 175 6 X minimum longitudinal bar diameter

p832

Esteem Innovation Sdn Bhd, 2011

Column Minimum and Maximum Compressive Steel Percentage According to EC8: 2003 Clause 5.4.3.2.2(1), the m inimum and maximum compression steel percentage are 1% and 4% respectively. In Esteem 8, when these conditions are TRUE, the default settings for the m inimum and maximum compressive steel percentage are 1% and 4% respectively.

 

Code of Practice EC2 Seismic Analysis = TRUE

p833

Esteem Innovation Sdn Bhd, 2011

Column Normalised Axial Force Check Column Normalized Axial Force Check – DCM Reference: EC8: 2003 Clause 5.4.3.2.1(3)



The program will check in the Full Report that the value of the normalized axial force, vd shall not exceed 0.65. If this is exceeded, the column will fail. COLUMN NORMALISED AXIAL FORCE CHECK FOR DCM CLASS EC8: 03 Clause 5.4.3.2.1(3) fcd = fck / 1.5 = 30.0 / 1.5 = 20.0 N/mm² Normalised Axial Force, vd = NEd / (Ac × fcd) = 806.8 × 1000 / (196350 × 20.0) = 0.21 ≤ 0.65 Checking for Normalised Axial Force Pass

Column Normalized Axial Force Check – DCH Reference: EC8: 2003 Clause 5.5.3.2.1(3)



The program will check in the Full Report that the value of the normalized axial force, vd shall not exceed 0.55. If this is exceeded, the column will fail. COLUMN NORMALISED AXIAL FORCE CHECK FOR DCH CLASS EC8: 03 Clause 5.5.3.2.1(3) fcd = fck / 1.5 = 30.0 / 1.5 = 20.0 N/mm² Normalised Axial Force, vd = NEd / (Ac × fcd) = 806.8 × 1000 / (196350 × 20.0) = 0.21 ≤ 0.55 Checking for Normalised Axial Force Pass

p834

Esteem Innovation Sdn Bhd, 2011

Column Critical Regions for DCM Class According to Clause 5.4.3.2.2,

i) The regions of a primary seismic beam up to a distance lcr = hw (where hw denotes the depth of the beam) from an end cross-section where the beam frames into a beam column joint, as well as from both sides of any other cross-section liable to yield in the seismic design situation, shall be considered as being critical regions.

ii) In the absence of more precise information, the length of the critical region lcr (in meters) may be computed from the following expression: lcr = max{hc ; lcl / 6; 0.45} where; hc is the largest cross-sectional dimension of the column (in meters); and lcl is the clear length of the column (in meters).

iii) If lcl/hc < 3, the entire height of the primary seismic column shall be considered as being a critical region and shall be reinforced accordingly.

iv) At least one intermediate bar shall be provided between corner bars along each column side, to ensure the integrity of the beam-column joints.

v) Within the critical regions of the primary seismic columns, hoops and cross-ties, of at least 6 mm in diameter, shall be provided at a spac ing such that a minimum ductility is ensured and local buckling of longitudinal bars is prevented. The hoop pattern shall be such that the cross-section benefits from the tri-axial stress conditions produced by the hoops.

p835

Esteem Innovation Sdn Bhd, 2011

Column Maximum Link Spacing for DCM Class According to Clause 5.4.3.1.2(6) within critical region, the minimum conditions of (10)P of this sub-clause are deemed to be satisfied if the following conditions are met. a) The spacing, s, of the hoops (in millimeters) does not exceed: s = min{bo/2; 175; 8dbL} (5.18) where, bo (in millimeters) is the minimum dimension of the concrete core (to the c enterline of the hoops); and dbL is the minimum diameter of the longitudinal bars (in m illimeters). b) The distance between consecutive longitudinal bars engaged by hoops or cross-ties does not exceed 200 mm, taking into account EN 1992-1-1:2004, 9.5.3(6).

p836

Esteem Innovation Sdn Bhd, 2011

Column Critical Regions for DCH Class According to Clause 5.5.3.2.2

i)

In the absence of more precise information, the length of the critical region lcr (in meters) may be computed from the following expression: lcr = max{ 1.5hc ; lcl / 6; 0.6} where; hc is the largest cross-sectional dimension of the column (in meters); and lcl is the clear length of the column (in meters).

ii)

If lcl/hc < 3, the entire height of the primary seismic column shall be considered as being a critical region and shall be reinforced accordingly.

iii) At least one intermediate bar shall be provided between corner bars along each column side, to ensure the integrity of the beam-column joints.

p837

Esteem Innovation Sdn Bhd, 2011

Column Maximum Link Spacing for DCH Class iv) The minimum conditions of (10)P of this sub-clause are deemed to be satisfied if the following conditions are met. a) The diameter dbw of the hoops is at least equal to dbw = 0.4 · dbLmax · vfydL / fydw (5.31) b) The spacing, s, of the hoops (in millimeters) does not exceed: s = min{bo/3; 125; 6dbL} (5.32) where; bo (in millimeters) is the minimum dimension of the concrete c ore (to the c enterline of the hoops) dbL is the minimum diameter of the longitudinal bars (in m illimeters). c) The distance between consecutive longitudinal bars restrained by hoops or cross-ties does not exceed 150 mm.

p838

Esteem Innovation Sdn Bhd, 2011

Wall Wall Normalised Axial Force Check Wall Normalized Axial Force Check – DCM Class Reference: EC8: 2003 Clause 5.4.3.4.1(2)



The program will check in the Full Report that the value of the normalized axial force, vd shall not exceed 0.4. If this is exceeded, the wall will fail in Normalised Axial Force checking. WALL NORMALISED AXIAL FORCE CHECK FOR DCM CLASS Sample calculation from EC8: 03 Clause 5.4.3.4.1(2) wall Full Report fcd = fck / 1.5 = 30.0 / 1.5 = 20.0 N/mm² Normalised Axial Force, vd = NEd / (Ac × fcd) = 3907.8 × 1000 / (750000 × 20.0) = 0.26 ≤ 0.40 Checking for Normalised Axial Force Pass

p839

Esteem Innovation Sdn Bhd, 2011

Wall Critical Regions for DCM Class According to Clause 5.4.3.4.2 (1) the critical region height, hcr above the base of the wall is obtained from the maximum of the following:

o lw o hw/6 where lw = wall length hw = total height of the wall above the foundation But, the critical region height, hcr should be less than the following:

o hcr  2 X lw o hcr  hs, for n  6 storeys o hcr  2 X hs, for n  7 storeys where hs = clear storey height for the current floor n = number of storeys

p840

Esteem Innovation Sdn Bhd, 2011

Minimum Link Diameter and Maximum Link Spacing for DCM Class General detailing rules regarding the diameter, spacing and anchorage of hoops and ties for wall boundary elements design ductility according to EC8 are the same as for columns (Ahmed, 2009, pg.135). The link or horizontal bar diameter should not be less than 6 mm based on Clause 5.4.3.2.2(10). In the critical regions, the maximum link or horizontal bar spacing, SvMax based on Clause 5.4.3.2.2(11) where it should not exceed the minimum of the following (all dimensions in millimeters)

o bo/2 o 175 o 8 X dbL where bo = minimum dimension of the concrete core (to the centreline of the hoops) dbL = minimum diameter of the longitudinal bar Reference: 1. Ahmed Y.Elghazouli Seismic Design of Buildings to Eurocode 8 1st Edition, 2009 Chap. 5

p841

Esteem Innovation Sdn Bhd, 2011

Maximum Distance Between Longitudinal Bars for DCM Class According to Clause 5.4.3.2.2(10) the distance between consecutive longitudinal bars restrained by hoops or cross-ties should not exceed 200 mm.

p842

Esteem Innovation Sdn Bhd, 2011

Wall Critical Regions for DCH Class According to Clause 5.5.3.4.5, the critical region height, hcr above the base of the wall is obtained from Clause 5.4.3.4.2(1), the maximum of the following:

o lw o hw/6 where lw = wall length hw = total height of the wall above the foundation But, the critical region height, hcr should be less than the following:

o hcr  2 X lw o hcr  hs, for n  6 storeys o hcr  2 X hs, for n  7 storeys where hs = clear storey height for the current floor n = number of storeys

p843

Esteem Innovation Sdn Bhd, 2011

Minimum Link Diameter and Maximum Link Spacing for DCH Class According to Clause 5.5.3.4.5(10) the minimum link or horizontal bar diameter as per Clause 5.5.3.2.2(12), dw should not be less than the following (all dim ensions in millimeters)

o dw  0.4 X dbLmax X (fydL/fydw) where dw = link or horizontal bar diameter dbLmax = maximum longitudinal bar diamet er fydL = strength of longitudinal bar used fydw = strength of link or horizontal bar used In the critical regions, the maximum link or horizontal bar spacing, SvMax based on Clause 5.5.3.2.2(12)(b) where it should not exceed the minimum of the following (all dimensions in millimeters)

o bo o 125 o 6 X dbL where bo = minimum dimension of the concrete core (to the centreline of the hoops) dbL = minimum diameter of the longitudinal bar

p844

Esteem Innovation Sdn Bhd, 2011

Maximum Distance Between Longitudinal Bars for DCH Class According to Clause 5.5.3.2.2(12)(c) the distance between consecutive longitudinal bars restrained by hoops or cross-ties should not exceed 150 mm.

p845

Esteem Innovation Sdn Bhd, 2011

EC2: Singapore National Annex Wind Loads to Singapore National Annex User is first required to calculate the maximum and minimum wind speed based on the Singapore National Annex to Eurocode 1: Actions of Structures - Part 1-4: General Actions Wind Actions, Clause 4.2. The default values used in Esteem 8 are as follows: Parameter Maximum wind speed (m/s) Minimum wind speed (m/s) 3

Air density,  (kg/m ) Fundamental value of basic wind velocity, V b,0 (m/s) Directional factor, Cdir

Default values in Esteem 8 20 16 1.194 20 1.0

Season factor, Cseason Probability factor, Cprop

1.0

Roughness factor (for maximum wind speed at top level), C rTop Roughness factor (for minimum wind speed at bottom level), C rBot

1.0

1.0

0.8

Click on the Wind Speed Calculator to automatically calculates the value of the maximum and minimum wind speeds based on the param eter entered by user.

Where, Maximum wind speed (m/s) = V b,0 * Cdir * Cseason * Cprop * CrTop Minimum wind speed (m/s) = V b,0 * Cdir * Cseason * Cprop * CrBot 3 2 The basic velocity pressure, qp (kN/m ) = 0.5 *  * Vb

Notes

: p846

Esteem Innovation Sdn Bhd, 2011

3. The value of the basic wind velocity, Vb varies for every floor depending on the height from the ground level. 4. Refer to Wind Loading for the application of the wind loads on every floor of the structure. Rectangular Load Distribution The calculation for rectangular load distribution: Basic wind velocity, Vb = Vb,0 * Cdir * Cseason * Cprop * Cr 3 2 The basic velocity pressure, qp (kN/m ) = 0.5 *  * Vb

Notes

:

5. The height limit and roughness factor may be overwritten by user as per requirement, while the program will automatically calculates the basic velocity pressure, q b based on the basic wind velocity Vb for the floor height given. . 6. To delete any of the row, click 7. To reset the values in the table and to return to the default setting, click on the button. 8. Refer to Wind Load Input Comply to Rectangular Load Distribution for the example of application.

p847

Esteem Innovation Sdn Bhd, 2011

Beam End Condition Setting When Code of Practice for Singapore National Annex to EC2 is selected, the default setting for the Beam End Releases is Fixed.

p848

Esteem Innovation Sdn Bhd, 2011

Design Compressive and Tensile Strengths In Clause 3.1.6 (1)P of the Singapore National Annex to EC2, NA to SS EN 1992-1-1: 2008, the value of cc may be taken as 0.85 for the Design Compressive and tensile strengths.

p849

Esteem Innovation Sdn Bhd, 2011

Maximum Concrete Strength Class In Clause 3.1.2 (2)P of the Singapore National Annex to EC2, NA to SS EN 1992-1-1: 2008, the maximum concrete strength class is limited to C50/60. If any of the concrete strength entered by user is m ore than the maximum allowed, the program automatically overwrites with the default value or previously entered value which is within the allowable range. 1. For design test specimen using cylinder crushing strength, allowable range for concrete strength:

2. For design test specimen using cube strength, allowable range for concrete strength:

p850

Esteem Innovation Sdn Bhd, 2011

Maximum Characteristic Yield Strength of Reinforcement According to Clause 3.1.2 (2)P of the Singapore National Annex to EC2, NA to SS EN 1992-1-1: 2008. the maximum characteristic yield strength of high yield and m ild yield 2 strength is based on the recommended value which is 600 N/mm . The allowable range for high yield steel:

The allowable range for mild steel:

p851

Esteem Innovation Sdn Bhd, 2011

Minimum Diameter of Longitudinal Reinforcement in Columns According to Clause 9.5.2 (1) of the Singapore National Annex to EC2, NA to SS EN 1992-1-1: 2008, the minimum diameter of longitudinal reinforcement in columns is based on the recommended value which is 12 mm. If any value selected by user is less than the minimum allowed, the program automatically overwrites with the default value or previously entered value which is within the allowable range.

p852

Esteem Innovation Sdn Bhd, 2011

Slenderness Limit Calculation Clause 5.8.3.1 (1) of the Singapore National Annex to EC2, NA to SS EN 1992-1-1: 2008, Slenderness limit is given by the expression 5.13,

p853

Esteem Innovation Sdn Bhd, 2011

Foundation Pad According to Clause 9.8.1 (3) of the Singapore National Annex to EC2, NA to SS EN 1992-1-1: 2008, the minimum diameter of longitudinal reinforcement in Pad is based on the recommended value which is 10 mm.

p854

Esteem Innovation Sdn Bhd, 2011

Pile According to Clause 9.8.1 (3) of the Singapore National Annex to EC2, NA to SS EN 1992-1-1: 2008, the minimum diameter of longitudinal reinforcement in Pile is based on the recommended value which is 10 mm.

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p855

Esteem Innovation Sdn Bhd, 2011

Index - Average Moment Over Support 541

-22D Mesh 633

-33D Mesh 634

-AAdditional 2D Load Case for Reverse Moment Case 642 Alternative deflection check (for cantilever beam only) 327 Analysis Failed due to Raft Foundation Uplift 586 Average Moment 605

-BBand strip contour cut section 579 Beam Critical Regions for DCH Class 830 Beam Critical Regions for DCM Class 827 Beam element with opening(s) 83 Beam End Condition Setting 848 Beam Geometrical Constraints 820 Bottom Bar Detailing 280

-CCantilever Slab with drop 248 Checking of Column Slenderness Limit (BS8110) 357 Code of Practice Exception 650 Column Clear Height, lo 330 Column Critical Regions for DCH Class 837 Column Critical Regions for DCM Class 835 Column Drop Value Outside Acceptable Range 548 Column Geometrical Constraints 821 Column is not supporting free edge of slab 611 Column Maximum Link Spacing for DCH Class 838 Column Maximum Link Spacing for DCM Class 836 Column Normalised Axial Force Check 834 column on raft 561, 559 Column or wall sitting across two transfer slab 612 Column Point 557 Column Point as fixed node 585 Column sitting on a transfer slab 613 p856

Esteem Innovation Sdn Bhd, 2011

Columns on Raft 564 Concrete Shear Capacity, VRd,c 788 Cracking Bar Calculation 643 Curve beam and curved slab not allowed in raft foundation 562 Curved beam framed into column 343 cut section on raft 570

-DDeep beam element 81 Design and Detailing Design Compressive and Tensile Strengths 849 design raft 596 Ductility Classes and Material Requirements 819

-EEdge Spring Multiplier Constant 598 Elaboration of 2D and 3D Opening Node Value 640 Enhanced Shear Multiplier 665 Example 305 Example 10: Intermediate transverse wall or beams 423 Example 5: Wall Connected to Cantilever Beam 416 Example 6: Wall Connected to a Beam of Several Sizes 417 Example 7: Wall Connected to Beam that is Not Attached to Lateral Restraint Element 419 Example 8: Walls Above and Below Not Considered in Wall Relative Stiffness Calculation 420 Example 9: Restraint condition for elements at wall end 421 External Loads on Raft Slab 3D Mesh 577

-FFlat Slab not designed 540 Full Band Width 581

-GGeometric Imperfections (Notional Horizontal Load) 757

-IInput 820

-LLimitation of band strip cut section 608 Links for Containment of Large Amounts of Compression Reinforcement of Walls 387

-MMaterial Requirements Maximum Characteristic Yield Strength of Reinforcement 851 Maximum Concrete Strength Class 850 Maximum Distance Between Longitudinal Bars for DCH Class 845 Maximum Distance Between Longitudinal Bars for DCM Class 842 p857

Esteem Innovation Sdn Bhd, 2011

Maximum Link Distance from Support for DCM and DCH Class 826 Meshing of column drop area 538 Meshing of wall on raft edge 591 Method to Determine Braced/Unbraced Frames 793 Minimum and Maximum Compressive Steel Percentage 833 Minimum Diameter of Longitudinal Reinforcement in Columns 852, 664 Minimum Link Diameter and Maximum Link Spacing for DCH Class 832, 844 Minimum Link Diameter and Maximum Link Spacing for DCH Class 832, 844 Minimum Link Diameter and Maximum Link Spacing for DCM Class 829, 841 Minimum Link Diameter and Maximum Link Spacing for DCM Class 829, 841 Minimum Reinforcement Ratio of the Tension Zone 825 Multi-level Wall Footing 624 Mutli-level Column Footing 621

-NNo column design when stump height is zero 367 Node Value in Full Report Not Tally with Summary Report 329 not checked 619 Not Design due to short transfer wall 380

-OOpenings in Raft 563

-PPad 854 Parameters 819 Pile 855 Punching Shear Checking for Upper Columns 619 Punching Shear Perimeter 602

-QQuantity Take Off - Slab 477

-Rraft 564, 574, 587, 579, 575, 597, 546, 561, 557, 585, 559, 570, 596, 598, 577, 581, 546, 598, 596, 593, 550, 583, 559, 593 raft band width 581 raft contour 567 Raft Floor Parameter 596 raft foundation 546, 546 raft parameter 598 raft rigid zone 587 raft section 573 Range of Subgrade Reaction 558 Reinforcement (Fabric) 488 Reinforcement (Normal Bar) 480 Reinforcement (Support Fabric) 491 Reinforcement (Support Normal Bar) 485

p858

Esteem Innovation Sdn Bhd, 2011

-SShear Contour Centroid 392 Shear in Columns 332 Side Bar Detailing 810 Skip Node Within Column Profile 240 Slab Mesh in 3D Analysis 207 Slab Opening in Flat Slab 537 Slab Opening on Raft Slab 550 Slab Top Bar (Support) Analysis Result 231 Slenderness Limit Calculation 853 Slenderness moment check 649

-TTo obtain 3D Analysis Result for Top Chord and Bottom Chord of a Beam Opening 635 To obtain 3D beam result diagram for Beam with Opening 638 Transfer Slab Mesh 614 Transfer wall seated at an eccentricity on a transfer beam 148 Two-way Slabs : Uniformly loaded rectangular panels 175

-UUnequal bar distribution 426 Unsymmetrical SFD for structural with symmetrical shape and loadings 107 Use Rectangular Mesh for Raft Slab 583 Use tension bar lapping for sagging moment at support 281 User-defined beam opening rebar lapping length 646

-VVerification 823 View 627

-WWall bypass intermediate beam within two support 384 Wall Critical Regions for DCH Class 843 Wall Critical Regions for DCM Class 840 Wall Geometrical Constraints 822 Wall Normalised Axial Force Check 839 Wall Not Design 374 Wall Region local axis for Mx and My 399 Wind Loads for EC1 748 Wind Loads to Singapore National Annex 846

p859

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