Analysis And Design Of Reinforced Concrete Structures.pdf

India’s Mega Online Education Hub for Class 9-12 Students, Engineers, Managers, Lawyers and Doctors. Free Resources for Class 9-12 Students

Free Resources for Engineering Students

Free Resources for MBA/BBA Students

Free Resources for LLB/LLM Students

Free Resources for MBBS/BDS Students

• Lecture Notes

• Lecture Notes

• Lecture Notes

• Lecture Notes

• Lecture Notes

• Project Reports

• Project Reports

• Project Reports

• Project Reports

• Project Reports

• Solved Papers

• Solved Papers

• Solved Papers

• Solved Papers

• Solved Papers

View More »

▼▼

View More »

View More »

View More »

View More »

Scroll Down to View your Downloaded File!

▼▼

Disclaimer Please note none of the content or study material in this document or content in this file is prepared or owned by Studynama.com. This content is shared by our student partners and we do not hold any copyright on this content. Please let us know if the content in this file infringes any of your copyright by writing to us at: [email protected] and we will take appropriate action.

ANALYSIS AND DESIGN OF REINFORCED CONCRETE STRUCTURES-A G+5 BUILDING MODEL

d

yn

S

tu

a

m

a

o c .

m

d

yn

S

tu

a

m

a

o c .

m

MINI PROJECT REPORT ON “ANALYSIS AND DESIGN OF REINFORCED CONCRETE STRUCTURES-A G+5 BUILDING MODEL”

d

yn

S

tu

a

m

a

o c .

m

Abstract Structural design is the primary aspect of civil engineering. The very basis of construction of any building, residential house or dams, bridges, culverts, canals etc. is designing. Structural engineering has existed since humans first started to construct their own structures. The foremost basic in structural engineering is the design of simple basic components and members of a building viz., Slabs, Beams, Columns and Footings. In order to design them, it is important to first obtain the plan of the particular building that is, positioning of the particular rooms (Drawing room, bed room, kitchen toilet etc.) such that they serve their respective purpose and also suiting to the requirement and comfort of the inhabitants. Thereby depending on the suitability; plan layout of beams and the position of columns are fixed. Thereafter, the loads are calculated namely the dead loads, which depend on the unit weight of the materials used (concrete, brick) and the live loads, which according to the code IS:875-1987 is around 2kN/m2. Once the loads are obtained, the component takes the load first i.e the slabs can be designed. Designing of slabs depends upon whether it is a one-way or a two-way slab, the end conditions and the loading. From the slabs, the loads are transferred to the beam. The loads coming from the slabs onto the beam may be trapezoidal or triangular. Depending on this, the beam may be designed. Thereafter, the loads (mainly shear) from the beams are taken by the columns. For designing columns, it is necessary to know the moments they are subjected to. For this purpose, frame analysis is done by Moment Distribution Method. After this, the designing of columns is taken up depending on end conditions, moments, eccentricity and if it is a short or slender column. Most of the columns designed in this mini project were considered to be axially loaded with uniaxial bending. Finally, the footings are designed based on the loading from the column and also the soil bearing capacity value for that particular area. Most importantly, the sections must be checked for all the four components with regard to strength and serviceability.

o c .

m

a

m a Overall, the concepts and procedures of designing the basic components of a multistory building are n described. Apart from that, the planning of the building with regard to appropriate directions for the yof beams respective rooms, choosing position and columns are also properly explained. The future of d structure engineering mainly depends on better and more effective methods of designing the u better tserve structures so that they and are also economical. The advancement of innovative and S environmentally friendly building materials are also coming up. They can give a new direction to the structural engineering field as the availability of concrete and steel is not only decreasing but also they are harmful to the environment. Hence, eco friendly materials which are economical and more effective methods of designing will decide the future of structure engineering.

INTRODUCTION Engineering is a professional art of applying science to the efficient conversion of natural resources for the benefit of man. Engineering therefore requires above all creative imagination to innovative useful application for natural phenomenon. THE DESIGN PROCESS The design process of structural planning and design requires not only imagination and conceptual thinking but also sound knowledge of science of structural engineering besides the knowledge of practical aspects, such as recent design codes, bye laws, backed up by ample experience, intuition and judgement. The purpose of standards is to ensure and enhance the safety, keeping careful balance between economy and safety. The process of design commences with planning of the structure , primarily to meet its functional requirements. Initially, the requirements proposed by the client are taken into consideration. They may be vague, ambiguous or even unacceptable from engineering point of view because he is not aware of the various implications involved in the process of planning and design , and about the limitationa and intricacies of structural science. It is emphasized that any structure to be constructed must satisfy the need efficiently for which it is intended and shall be durable for its desired life span.

o c .

m

Thus, the design of any structure is categorized into the following two main types :1) functional design 2) structural design. FUNCTIONAL DESIGN

yn

a

m

a

The structure to be constructed should be primarily serve the basic purpose for which it is to be used and must have a pleasing look.

d u The building should provide happy environment inside as well as outside. Therefore, the functional planning t of a building must take into account the proper arrangements of rooms / halls to satisfy the need of the S client, good ventilation, lighting, acoustics, unobstructed view in the case of community halls, cinema halls, etc.. sufficient head room, proper water supply and drainage arrangements, planting of trees etc. bearing all these aspects in mind the architect/engineer has to decide whether it should be a load bearing structure or R.C.C framed structure or a steel structure etc..

STRUCTURAL DESIGN Structural design is an art and science of understanding the behaviour of structural members subjected to loads and designing them with economy and elegance to give a safe, serviceable and durable structure.

STRUCTURAL DETAILS OF A FRAMED STRUCTURE

The principle elements of a R.C building frame consists of : 1) 2) 3) 4)

m

slabs to cover large area beams to support slabs and walls columns to support beams footings to distribute concentrated column loads over a large of the supporting soil such that the bearing capacity of soil is not exceeded.

a

m

a

o c .

In a framed structure the load is transferred from slab to beam, from beam to column and then to the foundation and soil below it.

yn

d

tu

STAGES IN STRUCTURAL DESIGN

S

The process of structural design involves the following stages : 1) structural planning 2) action of forces and computation of loads 3) methods of analysis 4) member design 5) detailing, drawing and preparation of schedules.

STRUCTURAL PLANNING After getting an architectural plan of the buildings, the structural planning of the building frame is done. This involves determination of the following :

a) b) c) d) e)

positioning and orientation of column of columns position of beams spanning of slabs layout of stairs selecting proper type of footing

the basic principle in deciding the layout of compent members is that the loads should be transferred to the foundation along the shortest path. POSITION OF COLUMNS 1) Columns should be preferably located at or near the corners of a building and at the intersections of beams/walls. Since the basic function of the columns is to support beams which are normally placed under the walls to support them, their position automatically gets fixed as shown in the figure below.

yn

a

m

a

o c .

m

d

tu

Column position for rectangular pattern building.

S

2) Select the position of columns so as to reduce bending moments in beams. When the locations of two columns are very near, then one column should be provided instead of two at such a position so as to reduce the beam moment. 3) Avoid larger spans of beams. When the centre to centre distance between the intersection of walls is large or when there are no cross walls, the spacing between two columns is governed by limitations of spans of supported beams because spacing of columns decides the span of beam. As the span of the beam increases, the required depth of the beam, and hence its self weight, and the total load on beam increases. It is well known that the moment governing the beam design varies with the square of the span and directly with the load. Hence with the increase in the span, there is considerable increase in the size of the beam. On the other hand, in the case of column, the increase in total load due to increase in length is negligible as long as the column is short. Therefore the cost of the beam per unit length increases rapidly with the

span as compared to beams on the basis of unit cost. Therefore the larger span of the beams should be preferably avoided for economy reasons.

In general, the maximum spans of beams carrying live loads upto 4 kN/m^2 may be limited to the following values. Beam type rectangular flanged

Cantilevers 3meters 5meters

simply supported 6meters 10meters

Fixed/continuous 8meters 12meters

o c .

m

4) Avoid larger centre to centre distance between columns. Larger spacing of columns not only increases the load on the column at each floor posing problem of stocky columns in lower storeys of a multi storeyed building. Heavy sections of column lead to offsets from walls and obstruct the floor area.

a 5) The columns on property line need special treatment. Since column footing requires certain area beyond m footing for such columns. In such cases , the the column, difficulties are encountered in providing a column may be shifted inside along a cross wall to make room for accommodating the footing within the property line. yn d u t S ORIENTATION OF COLUMNS

1) Avoid projection of column outside wall. According requirements of aesthetics and utility, projections of columns outside the wall in the room should be avoided as they not only give bad also obstruct the use of floor space and create problems in furniture flush with the wall. Provide depth of the column in the plane of the wall to avoid such offsets.

2) Orient the column so that the depth of the column is contained in the major plane of bending or is perpendicular to the major axis of bending. When the column is rigidly connected to right angles, it is subjected to moments of addition to the axial load. In such cases, the column should be so oriented that the depth of the column is perpendicular to major axis of bending so as to get larger moment of inertia and hence greater moment resisting capacity. It will also reduce Leff/D ratio resulting in increase in the load carrying capacity of the column.

3) It should be borne in mind that increasing the depth in the plane of bending not only increases the moment carrying capacity but also increases its stiffness, there by more moment is transferred to the column at the beam column junction. 4) However, if the difference in bending moment in two mutually perpendicular directions is not large the depth of the column may be taken along the wall provided column has sufficient strength in the plane of large moment. This will avoid offsets in the rooms. POSITION OF BEAMS 1)

Beams shall normally be provided under the walls or below a heavy concentrated load to avoid these loads directly coming on slabs. Since beams are primarily provided to support slabs, its spacing shall be decided by the maximum spans of slabs.

2)

Slab requires the maximum volume of concrete to carry a given load. Therefore the thickness of slab is required to be kept minimum. The maximum practical thickness for residential/office/public buildings is 200mm while the minimum is 100mm.

3)

The maximum and minimum spans of slabs which decide the spacing of beams are governed by loading and limiting thickness given above. In the case of buildings, with live load less than 5kN/m^2, the maximum spacing of beams may be limited to the values of maximum spans of slabs given below.

o c .

m

Support condition cantilevers Simply supported Fixed/continous One-way Two-way One-way Two-way One-way Two-way One-way Two-way Maximum 1.5m 2.0m 3.5m 4.5m 4.5m 6.0m Recommended span of slabs

m

a

a

4)

n y Avoid larger spacing of beams from deflection and cracking criteria. Larger spans of beams shall also d of controlling the deflection and cracking. This is because it is well be avoided from the considerations u known that deflection varies directly with the cube of span and inversely with the cube of depth i.e., t L /D . Consequently, increase in D is less than increase in span L which results in greater deflection S for larger span. 3

5)

3

However, for large span, normally higher L/D ratio is taken to restrict the depth from considerations of head room, aesthetics and psychological effect. Therefore spans of beams which require the depth of beam greater than one meter should be avoided.

SPANNING OF SLABS This is decided by supporting arrangements. When the supports are only on opposite edges or only in one direction,the slab acts as a one way supported slab. When rectangular slab is supported along its four edges, it acts as one way slab when Ly / Lx > 2 and as two way slab for Ly/Lx < 2. how ever two way action of the slab not only depends on the aspect ratio Ly / Lx and but also on the ratio of reinforcement in the two directions. Therefore, designer is free to decide as to whether the slab should be designed as one way or two way.

1) A slab normally acts as a one way slab when the aspect ratio Ly/Lx >2 ,since in this case one way action is predominant. In one way slab, main steel is provided along the short span only and the load is transferred to two opposite supports only. The steel along the long span just acts as distribution steel and is not designed for transferring the load but to distribute the load and to resist shrinkage and temperature stresses. 2) A two way slab having aspect ratio Ly / Lx < 2 is generally economical compared to one way slab because steel along the spans acts as main steel and transfers the load to all its four supports. The two way action is advantageous essentially for large spans and for live loads greater than 3kN/m^2. for short spans and light loads, steel required for two way slab does not differ appreciably as compared to steel for one way slab because of the requirement of minimum steel. 3) Spanning of the slab is also decided by the continuity of the slab. 4) Decide the type of the slab. While deciding the type of the slab whether a cantilever or a simply supported slab or a continuous slab loaded by UDL it should be borne in mind that the maximum bending moment in cantilever (M = wL2 / 2) is four times that of a simply supported slab (M=wL2/8) , while it is five to six times that of a continuous slab or a fixed slab (M=wL2/10 or wL2/12) for the same span length. 5) Similarly deflection of a cantilever loaded by a uniformly distributed load is given by : δ = wL4 /8EI = 48/5 *(5wL4 / 38EI) which is 9.6 times that of a simply supported slab = (5wL4 / 384 EI).

m

o c While designing any slab as a cantilever slab, it is . utmost importance to see whether adequate anchorage to the same is available or not. a m CHOICE OF FOOTING TYPE a n y 1) The type of footing depends upon the load carried y the column and bearing capacity of the supporting soil. It may be noted that thed earth under the foundation is susceptible to large variations. Even under one u small building the soil t may vary from a soft clay to hard murum. 2) It is necessaryS to conduct the survey in the area where the proposed structure is to be constructed to determine the soil properties. Drill holes and trail pits should be taken and in situ plate load test may be performed and samples of soil tested in the laboratory to determine the bearing capacity of soil and other properties. 3) For framed structure under study, isolated column footings are normally preferred except in case of soils with very low bearing capacities. If such soil or black cotton soil exists for great depths, pile foundations can be appropriate choice. 4) If columns are very closely spaced and bearing capacity of the soil is low, raft foundation can be an alternative solution. For column on the boundary line, a combined footing or a strap footing may be provided.

ACTIONS OF FORCES AND COMPUTATION OF LOADS BASIC STRUCTURAL ACTIONS The various structural actions which a structural engineer is required to know are as follows :TYPES OF STRUCTURAL ACTIONS

Axial force action :This occurs in the case of one dimensional (discrete) members like columns, arches, cables and members of trusses, and it is caused by forces passing through the centroidal axis and inducing axial (tensile or compressive) stresses only. Membrane action :This occurs in the case of two dimensional (continuum) structures like plates and shells. This induces forces along the axial surface only.

Bending action :-

o c .

m

The force either parallel or transverse, to the membrane axis and contained in the plane of bending induces bending (tensile and compressive) stresses. The bending may be about one or both axes which are perpendicular to the member axis.

m

a

a

The bending action is essentially by transverse forces or by moments about axes lying in the plane of the slab.

yn

d u The shear action is tcaused by in-plane parallel forces inducing shear stresses. S

Shear action :-

Twisting action :This action is caused by out of plane parallel forces i.e., forces not contained in the plane of axis of the member but in a plane perpendicular to axis of the member inducing torsional moment and hence shear stresses in the member Combined action :It is a combination of one or more of above actions. It produces a complex complex stress condition in the member.

ANALYSIS OF A STRUCTURE The different approaches to structural analysis are :-

1) Elastic analysis 2) Limit analysis Elastic analysis is used in working stress method of design. Limit analysis is further bifurcated as plastic theory applied to steel structures and ultimate load method of design, and its modified version namely Limit State Method for R.C.Structures, which includes design for ultimate limit state at which ultimate load theory applies and in service state elastic theory applies and in service elastic theory applies and in services state elastic theory is used. MEMBER DESIGN :- The member design consists of design of slab, beam, column, and footing.

DETAILING, DRAWING, AND PREPARATION OF SCHEDULE Detailing is a process of evolution based on an understanding of structural behavior and material properties. The good detailing ensures that the structure will behave as designed and should not mar the appearance of the exposed surface due to excessive cracking. The skillful detailing will assure satisfactory behaviour and adequate strength of structural members.

o c .

m

a

MARKING OF FRAME COMPONENTS

m

Before starting the structural design of R.C. frame components, it is always necessary to mark or designate them first to facilitate identification, listing and scheduling. The different schemes adopted for marking or identification are given below.

yn

a. b.

a

d u Column reference t scheme Scheme as recommended by IS : 5525 : “ Recommendations for detailing of S in reinforced concrete work ”. This scheme of marking is called as reinforcement 1.5

a grid reference scheme. c. Scheme followed by the private sector. Column reference scheme :In this scheme, columns are first of all numbered serially startin from the column at top left corner proceeding rightwards and then downwards as shown in the figure below. Beams are designated as Bij in which suffix i refers to column number from which beam starts and suffix j refers to the direction in which it runs. (j=1 for beams going northwards in (y) direction, j=2, for beams going eastwards in (x) direction, j=3 is used for cantilever beam going southwards with no column beyond, while j=4 is used for cantilever beam going westward with no column beyond. 1) This scheme is followed by public work department of some states and by steel structures fabricators and erectors. It is not very common with R.C. designers in the

private sector. The government departments which adopt this marking scheme, designate slabs as wSij in which prefixing letter w indicates category number of the slab, while suffix j indicates the type of slab whether one-way or two-way (j=1 for one way slab and j= 2 for two way slab). 2) This practice is useful and advantageous for maintaining a proper record especially when different slab panels are designed for different loads. This record is helpful to avoid wrong usage or over loading of the room in future due to change of user which is very common in government departments or public sector.

Grid reference scheme :1)

In this scheme of marking, starting from the column at the bottom left corner, series of imaginary horizontal grid lines passing through each column are marked as A-A ,B-B, C-C etc, and vertical grid lines passing through each column are marked as 11, 2-2, 3-3 as shown in the below figure.

2)

The columns are designated as Cij in which suffix i and j refer to the horizontal (ith ) and vertical (jth) grid lines intersecting at the column.

3)

Slabs are designated serially as Sb1 , Sb2 , starting from panel in the top left corner, proceeding vertically downwards bay wise and then rightwards.

m

o c 4) This scheme is partially followed in practice. Scheme of marking columns in this . way is very common, but that for beams and a slabs is not very much favored ( especially writing suffixes m and b to mark beam and slab respectively, is m considered to be superfluous). a n y Scheme used in the private sector d u t 1) In this scheme, the columns are marked serially as C1, C2, C3 etc.. or by encircled numbers S such as 1, 2, 3 etc.. by the side of the column starting either from top left corner and moving rightwards and down wards. 2)

Beams are marked serially as B1, B2, etc.. starting from first column and moving rightwards first and then downwards thus numbering first all the beams in horizontal or x- direction and then numbering first all the beams in horizontal or x- direction and then numbering upwards in y – direction starting from left most beams as shown in the figure below.

3)

However, the slabs are not marked serially but are marked according to their categories based on design specifications (namely the thickness, diameter, and spacing of reinforcement along two perpendicular spans). This facilitates scheduling of slabs.

4)

Nevertheless, it requires grouping of slab panel first having nearly equal spans, end conditions and the load so that categories of slabs required to be designed are

reduced to a minimum. The spanning of slabs is shown, by arrows on the plan and specifying separately in the schedules under remarks column.

At present, the loads for which the slabs are designed are many times not shown on the drawings. However, since these drawings form a permanent record with the user or with the licencing bodies like municipal corporations, it is advisable to record the design live load along with the specification of grades of concrete and steel in the notes on the drawings.

DESIGN PHILOSOPHIES

Reinforced concrete structures can be designed by using one of the following design philosophies. 1) Working Stress Method (WSM) 2) Ultimate Load Method (ULM)

o c .

3) Limit State Method (LSM)

m

a

m

Working stress method used over decades is now practically out dated. It is not used at all in many advanced countries of the world because of its inherent drawbacks. The latest I.S. Code gives emphasis on Limit State method which is the modified version of Ultimate load method.

yn

a

d

tu

It is a judicious amalgamation of WSM and ULM removing all drawbacks of both methods but maintaining their good points. It is also based on sound scientific principles backed up by 25 years of research. The limit state method has proved to have an edge over the working stress design from the view point of economy.

S

LOADS AND MATERIALS Loads and properties of materials constitute the basic parameters affecting the design of a R.C. structure. Both of them are basically of varying nature. The correct assessment of loads/forces on a structure is a very important step and serviceable design of structure.

TYPES OF LOADS The loads are broadly classified as vertical loads, horizontal loads, and longitudinal loads. The vertical loads consists of dead load, live load, impact load. The

horizontal loads comprises of wind load and earth quake load. The longitudinal loads (viz, tractive and braking forces are considered in special cases of design of bridges, design of gantry girders etc.) Dead load :Dead loads are permanent or stationary loads which are transferred to the structure throughout their life span. Dead load is primarily due to self weight of structural members, permanent partition walls, fixed permanent equipment and weighs of different materials. Imposed loads or Live loads :Live loads or movable loads with out any acceleration or impact. These are assumed to be produced by the intended use or occupancy of the building including weights of movable partition or furniture etc. The imposed loads to be assumed in buildings Impact load :Impact load is caused by vibration or impact or acceleration. A person walking produces a live load but soldiers marching or frames supporting lifts and hoists produce impact loads. Thus impact load is equal to imposed incremented by some percentage depending on the intensity of impact. Ithe dimension imposed load :-

a

m

a

o c .

m

Impact load is caused by vibration or impact or acceleration. A person walking produces a live load but soldiers marching or frames supporting lifts and hoists produce impact loads. Thus impact load is equal to imposed load incremented by some percentage depending on the intensity of impact.

yn

d

tu

S Wind load :-

Wind load is primary horizontal load caused by movement of air relative to earth. The details of design wind load are given is IS : 875 (part - 3)2.2 Wind load is required to be considered in design especially when the height of the building exceeds two times dimensions transverse to the exposed wind surface. For low rise building say up to 4 to 5 storeys the wind load is not critical because the moment of resistance provided by the continuity of floor system to column connection and walls provided between column connection and walls provided between columns are sufficient to accommodate the effect of these forces. Further in limit state method the factor for design load is reduced to 1.2(DL + LL + WL) when the wind is considered as against the factor of 1.5 (DL + LL) when wind is not considered.

Earth quake load :Earth quake loads are horizontal loads caused by earth quake and shall be computed in accordance with IS : 18932.2 . For monolithic reinforced concrete structures located in seismic zone II and III with out more than 5 storey high, and importance factor less than 1, the seismic forces are not critical.

CHARACTERISTIC LOAD Since the loads are variable in nature they are determined based on statistical approach. But it is impossible to give a guarantee that the loads can not exceed during the life span of the structure. Thus, the characteristic value of the load is obtained based on statistical probabilistic principles from mean value and standard deviation. The characteristic load is defined as that value of load which has 95% probability of not being exceeded during the service span of the structure. However, this requires large amount of statistical data. Code recommends to take the working loads or service loads based on past experience and judgement and are taken as per IS : 875 2.1 and IS : 18932.3 codes.

DESIGN LOADS

a

m

a

o c .

m

The variation in loads due to unforeseen increase in the loads, constructional inaccuracies, type of limit state etc., are taken into account to define the design load. The design load is given by : Design load = γƒ × characteristic load Where γƒ = partial safety of loads.

yn

d

tu

S

Partial safety factor(γƒ ) for loads (according to IS : 456 – 2000) Load combination limit state of collapse limit state of serviceability DL IL WL DL IL WL DL + IL 1.5 1.5 1.0 1.0 DL + WL 1.5 or 0.9* 1.5 1.0 1.0 DL + IL + WL 1.2 1.2 1.2 1.0 0.8 0.8 * this value is considered when stability against over turning or stress reversal is critical

Notes : (1) DL = dead load IL = imposed load WL = windload (2) while considering earth quake effects. Substitute EL for WL. (3) since the serviceability relates to the behavior of structure at working load the partial safety factors for limit state of serviceability are unity. (4) for limit state of serviceability, the values given in this table are applicable for short term effects. While assessing the long term effects due to creep, the dead load and that part of the dead load and live load likely to be permanent may only be considered.

CRITICAL LOAD COMBINATIONS While designing a structure, all load combinations, in general are required to be considered and the structure is designed for the most critical of all. For building upto 4 storeys, wind load is not considered, the elements are required to be designed for critical combination of dead load and live load only. For deciding critical load arrangements, we are required to use maximum and minimum loads. For this code prescribes different load factors as given below : Maximum load = wmax = 1.5(DL + LL) Minimum load = wmin = DL

o c .

m

The maximum positive moments producing tension at the bottom will occur when the deflection is maximum or curvature producing concavity upwards is maximum. This condition will occur when maximum load (i.e. both DL and LL) covers the whole span while minimum load (i.e. only DL) is on adjacent spans.

a

m

a

(a) consideration may be limited to combination of :

yn

1) Design dead load on all spans will full design live loads on two adjacent spans (for obtaining maximum hogging moment.) 2) Design dead load on all spans with full design imposed load on alternate spans ( to get maximum span moment.) 3) When design imposed load does not exceed three-fourths of the design dead load, the load arrangement may be design dead load and design imposed load on all the spans.

d

tu

S

The loading arrangement giving maximum span moment, say span AB is shown in below figure 1.a and figure 1.b gives the loading arrangements for maximum negative moment at support B

PROPERTIES OF CONCRETE Grade of concrete :Concrete is known by its grade which is designated as M15, M20, M25 etc, in which letter M refers to concrete mix and the number 15, 20, 25 etc. denotes the specified compressive strength (ƒck) of 150mm size cube at 28 days, expressed in N/mm2. Thus, concrete is known by its compressive strength. In R.C. work M20, M25 grades of concrete are common, but higher grades of concrete should be used for severe and very severe and extreme environment. Compressive strength :Like load the strength of concrete is also a quantity which varies considerably for the same concrete mix. There fore a single representative value known as characteristic strength , is arrived at using statistical probabilistic principles. Characteristic strength :It is defined as that value of the strength below which not more than 5% of the test results are suspected to fall,(i.e., there is 95% probability of achieving this value, or only 5% probability of not achieving the same).

o c .

Characteristic strength of concrete in flexural member :-

m

a

It may be noted that the strength of concrete cube does not truly represent the strength of concrete in flexural member because factors namely, the shape effect, the prism effect, state of stress in a member and casting and curing conditions for concrete in the member. Taking this into consideration the characteristic strength of concrete in a flexural member is taken as 0.67times2.6 the strength of concrete cube.

yn

m

a

d

tu

Design strength (ƒd) and partial safety factor(γd) for material strength :-

S

The strength to be taken for the purpose of design is known as design strength and is given by Design strength (ƒd) = characteristic strength(ƒck) Partial safety factor for material Strength (γm) The value of γm depends upon the type of material and upon the type of limit state. According to I.S. code, γm = 1.5 for concrete and γm = 1.15 for steel. Design strength of concrete in member = 0.67ƒck / 1.5 = 0.446 ƒck ≈ 0.45 ƒck

Tensile strength (ƒcr) :The estimate of flexural tensile strength or the modulus of rupture or the cracking strength of concrete from cube compressive strength is obtained from the relation :

ƒcr = 0.7√ ƒck N/mm2 The tensile strength of concrete in direct tension is obtained experimentally by split cylinder strength. It varies between 1/8 to 1/12 of cube compressive strength. Creep :Creep is defined as the plastic deformation under sustained load. The ultimate creep strain is estimated from the creep coefficient θ given by : θ = creep strain / elastic strain = εcc/ εi creep strain εcc depends primarily on the duration of sustained loading. According to the code, the value of ultimate creep coefficient is 1.6 at 28 days of loading. Shrinkage :The property of diminishing in volume during the process of drying and hardening is termed as shrinkage.

o c .

m

It depends mainly on the duration of exposure. If this strain is prevented, it produces tensile stress in the concrete, and hence concrete develops cracks. The shrinkage is measured by shrinkage strain, εcc = 0.0003 for design purposes.

a

m a The secant modulus obtained by testing a concrete specimen at 28 days under specified n of elasticity because inelastic deformations rate of loading is known as short term modulus y under this loading are practically negligible. d u According to tthe code short term modulus of elasticity of concrete is given by : S E = 5000 √ƒ N/mm Short term modulus of elasticity (Ec)

c

ck

2

Long term modulus of elasticity (Ece) :The effect of creep and shrinkage is to reduce modulus of elasticity of concrete with time. Therefore, the long term modulus of elasticity of concrete takes into account the effect of creep and shrinkage and is given by : Ece = Ec / (1+ θ) Where, Ece = long term modulus of elasticity Ec = short term modulus of elasticity θ

= creep coefficient.

Effect of the reduction In E ce with time is to increase deflections and cracking with time. it therefore, plays a very important role in limit state of serviceability and in calculations of deflection and cracking. It is further notified that as Ec changes modular ratio Es / Ec Where, Es = modulus of elasticity of steel = 2 × 105 N/mm2. Ec = 5000 √ƒck N/mm2. As the modulus of elasticity of concrete changes with time, age at loading etc, the modular ratio also changes accordingly. IS : Code gives the following expression for the long term modular ratio also changes accordingly. I.S.Code gives the following expression for the long term modular ratio taking into account the effects of creep and shrinkage partially. Long term modular ratio = m = 280 3ζabc Where ζabc = permissible compressive stress due to bending in concrete in N/mm2

m

This modular ratio is useful only in the working stress design. It is also required for calculating the properties of a transformed section of a R.C. member for the serviceability calculations

a

m

a

o c .

Modular ratio for different grades of concrete Grade of concrete

M20

yn modular ratio ‘m’

d

tu

S

M25

Short term 8.9

8.0

long term 13.3

11.0

CONVENTIONAL METHOD :This involves determination of positions of columns, position of beams, spanning of slabs, and type of footing. The structural plan will be drawn showing therein: 1) Position of columns, beams, and spanning of slabs, 2) Centre to centre dimensions between beams, columns to decide the span lengths of slabs and beams, 3) Marking of slabs, beams, and columns using one of the marking schemes.

1) 2) 3)

After the preparation of structural plan, the calculations will be done for unit loads as:unit loads on slabs of roof, floor, balconies, stairs, w.c and bath rooms, lofts etc, (kN/m) unit loads on walls (external, internal) per metre height,(in kN/m). unit loads on parapet walls, grills, weather sheds etc. (in kN/m).

Once these preliminaries are over design the frame components starting from slab, followed by beams, columns and column footings provided sufficient time is provided. PRACTICAL APPROACH If the work is to be started urgently, it may be necessary to give the sizes of footing and ground floor columns first. In such a case, the design will first be done of footings and columns by estimation of approximate equivalent axial load on columns, giving sufficient allowance for effect of continuity of slabs and beams, uniaxial / biaxial bending in columns due to fixity with beam; slenderness of column etc. where ever necessary.

NOTATIONS

m

o c A : - Area of steel at mid span in shorter direction . a A : - Area of steel at continuous edge in shorter direction A : - Area of steel at mid span m in longer direction a A : - Area of steel at continuous edge in longer direction n b : - Breadth ofy beam B.M :d - Bending Moment u b . t : - Breadth of web or rib S D : -Over all depth A

: - Area

st

stc

sty

styc

w

d

: -Effective depth

D.L.

: - Dead load

fck

: -Characteristic compressive strength of concrete

Fy

:- Characteristic strength of steel

J

: - Lever arm factor

K

: - Neutral axis factor

Leff

: - Effective length

Ld

: - Development length

L.L

: -Live load

M

: - Modular ratio

P

: -Axial load

Q

: - Moment of Resistance factor

S.F.

: - Shear Force

Xu

: - Depth of neutral axis

Z

: -Lever arm

Ocbc

: - Permissible compressive stress in steel

Ost

: -Permissible stress in steel

τc

: - Shear stress in concrete

τv

:- Nominal shear stress

Design codes used: IS:456-2000, SP-16 (Design aid to IS:456), IS:875-1987, SP-34 (Handbook on concrete reinforcement and detailing)

o c .

Introduction to Principle of Planning

m

m

a

The basis of planning of building is to range all the units a building on the floors

a

according to their functional requirement making best use of the space available

n y condition she location,d accommodation requirements. Local by laws etc.., Three are few principles tuof planning listed below which are like to be considered in planning.S for the building the planning is governed by several factors such as climatic

(1) ASPECT: - Aspect means the method of arrangements of door and windows in external walls, of a residential building. This enables the occupations to enjoy the natural gifts. Such as sunshine, breeze, scenery etc., Aspects provides comfort and is important from any particular direction. Different rooms in building needs different aspects.

LIVING ROOM: - Most part of the living room should be towards north. It so because north aspect receives natural north lights which is used in most of the daytime.

KITCHEN: - Eastern aspect to admit morning sun to refresh and purity me the air.

STUDY ROOM: - North aspects this make more light to enter and will be diffused which results in uniform distribution of light. BED ROOM: - North aspects or southwest aspect is very good for bedrooms.

(2) PROSPECT: - Prospect in its proper sense is the impression that a person viewing from outside likely to get. Prospect must not only make outer appearance attractive but also maintain qualities, such as con~{ort cheerfulness, security. economy and up to date. One must feel the sense of pride in having a house, which is pleasing in appearance and is reflecting its individuality.

mfunctional

o c requirement of a room Living room. Living room, . drawing room, kitchen, aetc. will have their own classroom, laboratory room, office room m requirements. a yn d Generally foru the plans, position of beds, furniture pieces like sofas, cub t hoards, dining tables etc. Should be shown, space occupied by hanging S contents from placements should also be given due consideration

(3) FURNITURE

REQUIREMENTS:

-

Furniture

is

the

(4) ROOMINESS: - In planning a building an architect deals with length. width and height of rooms. The feeling of space i.e. whether it is sufficient less or more depends upon suitable and adequate proportions of length. width and height. Maximum benefits should be obtained from maximum dimensions required for the furniture's expected to be achieved from the space.

A square room is found to be inconvenient as compared to rectangular room of the same area from utility point of view. Hence the length to

width ratio should be between 1.2 to 1.5 less width with more lengths causes tunnel effect (5) GROUPING: - Grouping means setting different rooms of a building accordance to their inter relationship of the spaces should be such that it is a feeling of invitation and transition rather than feeling of abrupt change. For examples in residential building, dining room should be close to the kitchen. At the same time kitchen should be kept away. from main living room to avoid smoke and smells. Main bedroom should be so located that there is no independent and separate access from each room towards the sanitary units directly or thought other passages. (6) CIRCULATION: - Access or internal through fares between rooms of the same floor (or) between different floors in known as circulation passages, corridors, halls and lobbies serve the purpose of horizontal circulation.

m

Circulation between rooms of the same floor is known as horizontal

o c .

circulation.

a

m a independent. stair case is a very important unit in any building as it forms the n yfloors of the building. They should be well lighted only link among various d and ventilated. Their location should be such that they don't cause any u t disturbance in any part of the building. S Passages, corridors, hall etc., used horizontal circulation's should be

(7) PRIVACY: - In all building some sort of privacy is essential feature. In residential building in particular optimum privacy has to secured in planning. The internal privacy means, screening interior or one room from other room's parts. The extent of privacy of a building from the street lanes and neighboring buildings depends on its function.

Disposition of doors and windows greatly affect internal privacy. lobbies and screens also provide internal privacy toilets lavatories. hath-rooms, require

absolute privacy and as such all these should have in independent access from bedroom without disturbing the others. Doors with angle shutter one preferred to double shutter doors

(8) SANITATION: - Sanitation includes light, cleanliness, ventilation and sanitary conveniences. a. Lighting: - Lighting is required to provide sufficient illumination in the building and to keep hygiene. Lighting may be natural or artificial. Natural lighting is achieved by properly positioning the adequate number of windows to admit the required amount of sum inside the room. Good day lighting means not too much light but sufficient light free from glare. b. Ventilation: - Ventilation is system of supplying (or) removing air by natural or mechanical means to or from any enclosed space to create and maintain comfortable conditions.

m

c. Sanitary Convenience: - Water Closet and bathrooms should be provided

o c include W C. Urinals, Bathrooms and their number . should be sufficient in a relation to the occupant load. m a n y (9) ELEGANCE: - Elegance is related to the effect produced by elevation. d which depends upon the proportion of width height of doors and windows u t choice of materials will also affect the elegance of building. S }1'ith glazed tiles so that they can be deemed regularly sanitary convenience

The other factors are: a. The visualization of elevation should always be kept in mind while preparing plan. b. Architectural design and composition should be studied in detail for achieving success in creating an elegant structure. c. Selection of site for the building greatly affects the elegance.

EX: - In the care of residential buildings, privacy care he secured by carefully planning the entrance path ways and proper grouping of all the rooms in their co-relation building located in the depression will always give

depressed elegance where as that located on the elevated spot gives impressive appearance. d. A slight adjustment or modifications in the elevation through the requirements of the plan are maintained will definitely improve the elegance of building.

(10) ECONOMY: - The economy may not be a principle of planning but definitely a factor affects in it. Economy restricts the liberties which otherwise would have been enjoyed by the planner to .fit the proposed scheme omissions in the original plan have to be affected. But economy should no affect the utility and strength of the structure. Infact, no rules can be framed to achieve economy. It is at ingenuity of an individual. Which he would like to adopt.

m

(11) FLEXIBILITY: - Flexibility means planning the rooms in such a way

o c other purpose also as and when desired. This consideration . is very important a or other building where for designing the houses for middle class families economy is the main consideration. m a n means setting the plan of the proposed (12) ORIETATION: - Good yorientation building in such ad manner and direction that future occupants of the building u t what ever in good and avoid whatever is bad from the natural would enjoy Ssuch as sun wind rain. Orientation also involves proper placement elements which though originally designed for a specific purpose may he used/or

of rooms is relation to the sun wind rain topography and at the same time providing a convenient access both the street and backyard. The placing of the building with respect to the geographical directions. The direction of wind and azimuth of sun is known as orientation building. Orientation is relationship to its environment.

VENTILATION: - The process of the supplying fresh air and removing contaminated air by natural or mechanical means to or .from is termed as Ventilation. Good ventilation is an important factor in providing comfort in building.

NECESSARY OF VENTILATION: Ventilation is necessary for the following reasons: 1. To create air movement. 2. To prevent accumulation of carbon-DI-oxide and moisture in a building. 3. To provide required amount of oxygen in air. 4. To prevent condensation in the building. 5. To prevent the concentration of bacterial carrying particles. 6. To reduce the concentration of body odors fumes. dust and other gases produced during. production in case of industrial building. 7. To prevent suffocating conditions in committee halls. auditorium etc.

FACTORS AFFECTING VENTILATION: From comfort point of view the following factors should be considered they effect ventilation to a greater degree. 1. 2. 3. 4. 5. 6.

Rate of supply of fresh air. Air movements or air change. Temperature of air. Humidity. Purity or quality of air. Use of building.

d

yn

S

tu

a

m

a

o c .

m

ARRANGEMENTS OF ROOMS AND THEIR POSITIONS

Bedroom

North or South West

Verandah

South West or East

Store and Study room

North

Kitchen

East or South East

Rooms mostly used in day time

North or East

Staircase

West

Direction

of

shorter

outer

East-West

direction Direction of Longer outer walls

North South direction

South Side walls

Must

have

verandah

sunshade

o c .

m

West side walls

Only Verandah

Windows

Towards East and South

d

yn

S

tu

a

m

a

and

This procedure involves the design of slab. Primarily to design a slab we have to confirm if it is a one way slab or two way slab a) ONE WAY SLAB :It supports on opposite edges or when Ly/Lx > 2, Predominantly bends in one direction across the span and acts like a wide beam of unit width. If a continuous slab/beam loaded by using UDL has equal spans or if spans do not differ by more than 15% of the longest they are designed using IS:code. For accurate analysis a continuous slab carrying ultimate load is analysed using elastic method with redistribution of moments. b) TWO WAY SLAB :A rectangular slab supported on four edges with ratio of long span to short span less than 2 (Ly/Lx <2) deflects in the form of a dish. It transfers the transverse load to its supporting edges by bending in both directions. DESIGN OF ONE-WAY SLAB:

o c .

STEPS :-

m

a

1)SLAB MARK– write the slab mark or designation such as S1,S2 etc…

mthe end condition No. according to the category of

a

2)END CONDITION – for approximate analysis write the slab.

n y SPAN LENGTH (L)- depending upon end conditions determine the effective span of the slab. In fact, since the depth of slab is not knownd in advance and the width of support is normally greater than the effective u depth of slab, in practice the effective depth of slab is taken equal center to center distance between the t supports to be on safer side. S 3)TRIAL SECTION :Effective depth = effective span L / required(d) basic (L/d) ratio * x where, basic l/d ratio = 7 (for cantilever) = 20 (for simply supported) = 26(for continuous). x = depends upon Pt% and steel stress (fs) Initially assume Pt = 0.5% - 0.9% for steel of steel grade Fe-250 = 0.25% - 0.45% for steel of steel grade Fe-415 = 0.2% - 0.35% for steel

of Fe-500 Obtain the nominal cover from IS:code , and add half the diameter of main steel, to get effective cover. Therefore, Effective cover=d‟=nominal cover + half dia. Total depth of slab = effective depth + effective cover = d + d‟. 4)LOADS :Calculate load in kN/m on one metre wide strip of slab Dead load :Self weight = Ws = 25D Where, D shall be in metre. Floor Finish = FF = 1.5 kN/m Total dead load =DL = Wd = Ws + FF Imposed load = LL Total working load W = DL + LL Total ultimate load Wu = 1.5W 5)DESIGN MOMENTS :Design moment Mu = WL*L/2 (for cantilever) = WL*L/8 (for simply supported) = according to the code (for continuous). 6) CHECK FOR CONCRETE DEPTH :-

yn

Mu,limit = 0.36*fck*Xumax*(d-0.42fck)*b

a

m

a

o c .

m

d

tu

Where,

S

Mu,limit = maximum ultimate moment fck = strength of concrete d = effective depth b = breadth(1meter). If Mu < Mu,limit then we will find area of steel (Ast) from the following formula :Mu = 0.87*Fy*Ast*(d-0.42Xu) If Mu > Mu,limit redesign depth. Minimum area of steel(ast) =0.15%*b*D (for Fy=250) =0.12%*b*D (for Fy=415or500) Assume bar diameter (8mm or 10mm for steel grade Fe415, and 10mm or 12mm for Fe250). Required spacing(S)= 1000*ast/Ast where, ast is area of one bar. Maximum spacing (Smax) < (3d or 300mm) whichever is less.

From practical consideration minimum spacing is 75<s<100. 8) CHECK FOR DEFLECTION:Calculate required Pt% (maximum value at mid-span of continuous slab or simply supported slab). (Pt)assumed < (Pt)required Then the check may be considered to be satisfied else detailed check should be carried out as given in the code as under:-

Calculate steel stress of service load (fs) :fs = 0.58*fy*(Ast)reqd / (Ast)prov. Obtain modification factor (α) corresponding to (Pt)prov and fs. Required depth (d) = L/(basic L/d ratio*α)<effective depth provided. 9) DISTRIBUTION STEEL :Required Ast min= 1.2D for HYSD bars, = 1.5D for Fe250 where D in mm Assume bar diameter(6mm for steel grade Fe 250 and 8mm for Fe 415). Required spacing,s=1000*ast/Ast min,to be rounded off on lower side in multiple of 10mm or 25mm as desired. Maximum spacing,s =<(5d or 450mm) which ever is less. In practice spacing is kept between 150mm to 300mm. 10) CHECK FOR SHEAR :-

yn

a

m

a

o c .

m

d

tu

a) calculate design (maximum) shear.

S

In case of slabs, design shear may be taken equal to maximum shear Vu.max at support and is given by:Vu.max = Wu*L*shear coefficient = Wu*L/2 for simply supported slab. Where, Wu = ultimate UDL on slab/ unit width. In other cases, the maximum shear may be calculated from principles of mechanics. b) calculate shear resistance (Vuc) of slab: This may be obtained from the relation (Vuc) = η.bd*k (b=1000mm in case of slabs). η depends upon Pt = 100Ast /bd. Where Ast = area of tension steel. It is the bottom steel at simply supported end and top steel at Continous end. Ast =Ast /2 if alternate bars from mid span are bent to top at simple support. Check that Vuc > Vu,max. If not increase the depth.

This check for shear is mostly satisfied in all case of slabs subjected to uniformly distributed load and therefore many times omitted in design calculations. It may be noted that when the check of shear is obtained, it is not necessary to provide minimum stirrups as they are required in the case of beams. 11) CHECK FOR DEVELOPMENT LENGTH :Required Ld ≤ 1.3 M/V + Lo For slabs alternate bars are bent at support M = Mu.max / 2 And Lo = b/2-x + 3θ for HYSD bars using 90 degrees bend. = b/2-x + 13θ for mild steel using 180 degrees bend. Where x = end clearance.

DESIGN OF TWO WAY SLABS:

STEPS :-

o c .

1. SLAB MARK :Write slab designation eg – S1, S2 etc… 2. END CONDITIONS :Write end boundary condition No.

yn

a

m

a

m

d

tu

3. SPANS :-

S

Determine short span Lx, long span Ly, check that Ly / Lx < 2 4. TRIAL DEPTH (D):It will be decided by deflection criteria based on short span Lx and total depth D. The allowable L/D ratio for two way slab with short span up to 3.5m and for loading class up to 3kN/m2

Assuming Pt% between 0.2% to 0.3% and proceeding, 5. LOADS :-

o c .

m

Calculate load for one meter width strip of slab. Wu = 1.5(25D + FF + LL)kN/m 6. DESIGN MOMENTS :-

a

Obtain the bending moments by using the relation Mu = α W Lx*Lx. Using IS CODE.

m a 7. CHECK FOR CONCRETE DEPTH FROM BENDING MOMENT CRITERIA :n ydepths for reinforcement in short span steel and effective depths for In the case of a two way slab, effective d reinforcement in short spanu and long span is placed above short span steel. The effective depth do istfor outer layer of short span steel and effective depth di is for inner layer of long span steel at mid span. SAs far as support section is concerned, the effective depth is do only for both spans. do = D – (nomimal cover + φ/2) where φ = diameter of the bar. di = do – φ for mid span long span steel.

8. MAIN STEEL :Calculate the area of steel required at four different locations. Main steel calculated is provided only in the middle strips of width equal to 3/4 th the slab width. there will be no main steel parallel to the support in edge strip of width equal to 1/8 th of slab width. In this edge strip, only distribution steel will be provided. Distribution steel will be provided for middle strip bars at top of supports.

9. CHECK FOR DEFLECTION :If Lx ≤ 3.5m and L.L≤ 3kN/m2, check that (L/D)prov > (L/D)req then,

For Lx > 3.5m or L.L > 3kN/m2, the deflection check should be similar to that explained in one way slab.

10. TORSION STEEL :At corners where slab is discontinuous over both edges, At =(3/4)Ast. At corners where slab is discontinuous over only one edge , At = (3/8)Ast. At corners where slab is discontinuous over both the edges, At =0.

11. CHECK FOR SHEAR :-

o c .

m

a

m

a) Design maximum shear in two way slab may be obtained using the following relation. At middle of short edge, Vu.max = WLx / 3 per unit width. At middle of long edge, Vumax = WLx [β/(2β+1)] where, β = Lv / Lx.

yn

a

d

Increase above value by 20% for shear at continuous edge and decrease the same by 10% at simply supported discontinuous edge and continuous over the other.

tu

S

b) Shear resistance and hence shear check is obtained in the same way as it is obtained for one way slab. c) load carried by supporting beams of two way slab.

Long edge : Trapezoidal load with ordinate W*Lx /2 Equivalent UD load for bending Weqs = W*Lx[1-(1/3ββ)]/2. Equivalent UD load for shear Weqs = WLx[1-(1/2β)]/2 Short edge : Equivalent UD loading for bending Weqs = WLx/3 Equivalent UD loading for shear Weqs = Wlx/4.

In the case of slab simply supported at one end and continuous at other reduce the loa at simply supported end by 10% (i.e take shear coefficient = 0.45) and increase the same by 20% at the continuous end (i.e take shear coefficient = 0.6) and 25% at continuous end of two span continuous beam.

12. CHECK FOR DEVELOPMENT LENGTH :It will be applied similar to that of one way slab.

EXAMPLE OF SLAB DESIGNING :-

1. Name of slab = S1 2. size of slab = 3.1m *4.5m Lx = 3.1m and Ly = 4.5m. 3. Ly / Lx = 4.5 / 3.1 = 1.45 < 2.

o c .

So two way slab interior panel.

a

Lx < 3.5m and L.L < 3 kN/m2 Fe 415 steel , s.s.b = 28

yn

Let L.L = 2kN/m2 L/γd = 32 3100/γd =40*0.8

d

tu

S

Two way slab Pt = 0.25% fs= 0.58 * 415 =240 N/mm2 γ = 1.7 d = 3100/(1.7*32) = 56.985 mm D = d + φ/2 + cover Mild environment =15mm D = 56.985 + 10/2 + 15 = 76.985mm ≈ 80 mm 4. Loads :-

m

a

m

Dead load of slab of 1m width = Lx * 1* D*2.5 = 3.1*1*80/1000*2.5 = 6.2 kN/m Floor finish

= 1.5*1 =1.5 kN/m

Live load

= 2kN/m

Total load

= 6.2 + 2+1.5 = 9.7 kN/m

Factored load

= 1.5 * 9.7 = 14.55 kN/m.

o c .

Lx =3.1m Ly = 4.5m

a

From IS CODE 456, 1.4 α(x-)

1.5 0.051

α(x+)

0.039

α(y-) α(y+)

0.053

1.5

1.45

yn

d

S

tu

0.051

1.0

x/(1.5-1.45) =(0.053-0.051)/(1.5-1.4) x = 0.001 α(x- ) = 0.052

1.45 m 0.053

a

m

0.041

0.051 0.040 0.0032 0.024

5. Design moments :Mx(-) = α(x-)WLx*Lx = 0.052 *14.55*3.1*3.1 = 7.27 kN-m Mx(+)= α(x+)WLx*Lx =0.040*14.55*3.1*3.1 = 5.59kN-m My(-) = α(y-)WLx*Lx = 0.032*14.55*3.1*3.1 = 4.47kN-m My(+) = α(y+)WLx*Lx = 0.024*14.55*3.1*3.1 = 3.35kN-m Mu.lim = 0.36*(Xumax/d)*[1-(0.42Xumax/d)] *bd*d*Fck = 0.36* 0.48*[1-(0.42*0.48)]*1000*57*57*20 = 8.96 kN-m Xu/d =(0.87fyAst)/(0.36Fckbd) 0.48*0.36*20*1000*57 = 0.87*415*Ast Ast, required = 545.60mm2

o c .

6. Check for deflection :ast = Π*10*10/4 = 78.54 mm2 spacing = 78.54*1000/545.60 = 143.95mm = say 125mm

yn

Max. spacing = 3d =3*57 = 171mm

a

m

a

m

d

tu

Ast provided = 545.60*171/125 = 746.380 mm2

S

Pt% of steel provided = 746.38 *100 /bD = 746.38*100/(1000*80) = 0.93% From table -4 pg-38 of IS CODE, Fs = 145 Fs = 195

1.41

Pt = 1 Pt = 1

1.20

γ=1 γ=1

145

175.5

195

x/(195-175.5)

=

(1.41-1.20)/(195-145)

x

=

γ

=

1.2819

d

=

3100 /(1.2819*32)

=

0.0819

75.57 mm

Provided depth < required depth Assume effective depth = 80mm D = 80 +(10/2)+15 = 100mm. Then, Dead load = 3.1*1*0.1*25 = 7.75 kN/m Floor finish = 1.5kN/m Total dead load = 9.25kN/m Factored load = 1.5 * 11.45 = 17.175 kN/m

m

a

Mx(-) = 17.175*3.1*3.1*0.052 = 8.582691 kN-m

yn

Mx(+) = 17.175*3.1*3.1*0.040 = 6.60207 kN-m

d

tu

My(-) = 17.175*3.1*3.1*0.032 = 5.281 kN-m

S

My(+) = 17.175*3.1*3.1*0.024 = 3.961 kN-m

Mu,lim = 0.36*Xu,max/d* (1-0.42*Xu,max/d)*bd*d*fck = 0.36*0.48*(1-(0.42*0.48))*1000*80*80*20 = 17.62 kN-m 0.48 =0.87*415*Ast/(0.36*20*1000*80) Ast = 765.76*240 = 1837.82mm2 Provide 10mm dia

ast = Π*10*10/4 = 78.5mm2

Spacing = 78.5*1000/765.76 = 102.51mm ≈ 100mm. Maximum spacing = 3d =3*80 = 240mm

a

o c .

m

Ast provided = 765.76* 240 = 1837.82 mm2 % of steel provided = Ast*100/bD = 1837.82*100/1000*100 = 1.80% From table-4 pg 38 ISCODE γ = 1.17 d = 3100/1.17*32 = 78.79. hence safe. 8.58

8.58

6.602

At Support For short Span Mid span Support For short span Longer Span end support Longer Span end support

moment

m bending moment Ast ast o req spacing c . 8.58 309.46 a78.53 253.76 m a 6.602 n 233.48 78.53 336.34 y d

Mx(-)

Mx(+)

tu

240

240

My(-)

S

5.281

184.427

78.53

425.305

240

My(+)

3.961

133.409

78.53

588.640

240

Check for shear :Long edge :

prov spacing

trapezoidal load with ordinate WLx/2.

Equivalent UD load for bending Weqb = WLx(1-(1/3β*β)/2 = 17.175*3.1*(1-(1/(1.451*1.451*3)))/2

= 11.180 Equivalent UD load for shear Weqs = WLx(1-(1/2β))/2 = 17.175*3.1*(1-(1/2*1.451))/2 = 17.46 Short edge : Equivalent UD load for bending = 17.175*3.1/3 = 17.745.

Weqb = WLx/3

Equivalent UD load for shear =17.175*3.1/4 = 13.310

Weqs = WLx/4

Slab No. & Shorter span end Depth(mm) Moment (Mx-) B.M(kN- Ast(mm2) m)

SA; d=100

6.64

SB; d=80

8.58

328.23

4.98

372.05

6.602

Sc; d=80

8.55

SD;d=60

5.664

SE; d=70

12.46

yn

d u t 371.98

S

Shorter span mid Moment (Mx+) B.M(kN- Ast m) (mm2)

261.66

Longer span Mid Moment (My+) B.M(kN- Ast( m) mm2 )

262.06

m

3.62

261.6 2

5.28

184.42

3.961

133.4

co

4.83

.

a

m 233.48

a

Longer span end moment (My-) B.M(kN- Ast(mm2) m)

6.349

319.72

4.416

222.38

3.31

166.6 8

289.4

3.6

176.98

4.57

229.27

3.4

166.8 2

599.55

9.368

404.97

7.724

315.005

5.752

228.1

SF; d=80

12.01

474.09

10.245

394.98

6.535

241.22

6.028

221.2 9

SG;d=80

7.60

284.03

5.77

211.34

5.62

205.522

4.25

153.2 2

SH; d=70

4.266

178.2

3.223

134.05

3.507

146.43

2.654

109.5 63

SI; d=70

-

-

7.66

380

-

-

-

-

d

yn

S

tu

a

m

a

o c .

m

DESIGN OF BEAMS:

A beam is a structural member that is capable of withstanding load by primarily resisting bending. The designing of the beam mainly consists of fixing the breadth and depth of the beam and arriving at the area of steel and the diameter of bars to be used. The breadth of the beam is generally kept equal to the thickness of the wall to avoid offset inside the room. It shall also not exceed the width of the column for effective transfer of load from beam to column. The depth of the beam is taken between L/10 to L/16. The dimensions of the beam that we have chosen are : breadth=230mm and depth=450mm. Procedure to design beams : 1)Analysis : The beam is analyzed first in order to calculate the internal actions such as Bending Moment and Shear Force. A simplified substitute frame analysis can be used for determining the bending moments and shearing forces at any floor or roof level due to gravity loads. The Moment distribution method is used for this purpose.

o c .

m

2) Loads: In order to analyze the frame, it is needed to calculate the loads to which the beams are subjected to. The different loadings are as follows:

a

m a The load transferred from the slab per metre length will be either rectangular n from two-way slab. Depending on the from one way slab or trapezoidal/triangular y position of the slab, the loading may be decided. In the case of two way slabs, trapezoidal load comesd from the longer side while the triangular load comes from u the shorter side. t a)Slab on S the Right side : The load transferred from the slab on the right side is i)Uniformly Distributed Load : (w) in kN/m

denoted as ws2 and the slab from the left side is denoted as ws1. The equivalent U.D.L to evaluate shear force from a slab= wlx (1- 1/(3β2))--------(1) 3 The equivalent U.D.L to evaluate bending moment from a slab= wlx (1-1/2β)----(2) 4 Where β=1 for triangular loads & β=ly/lx for trapezoidal loads. b) Masonry wall : Ww=yx t(w)x H(w) where t(w)=thickness in m, H(w)=height in m and y=unit weight of masonry=19.2 kN/m3 c) Self weight : Ws= 25xbxD d) Total working load=(Ws1+Ws2)+Ww+Ws for calculation of B.M and S.F. Design (ultimate) load : wu= 1.5w kN/m.

ii) Point Loads: Given total No. of point loads= Number of secondary beams supported. iii) Design Moment : While designing it should first be noted if it is a flanged section or a rectangular section. Most of the intermediate beams are designed as rectangular sections. The main beams may be designed as flanged sections. For rectangular beams, the maximum depth of N.A lies at the centre. For flanged sections, check if the N.A lies within the flange or not and then proceed to calculate the moment. The dimensions of flanged section as designed as per the code IS: 4562000 as per Cl 23.1. Either way, for a singly reinforced section, Mu = 0.367fck.bf.xu(d-0.42xu)

---------------(3)

If design moment Md calculated through frame analysis is less than Mu, then N.A is known to lie within the flange. This is the case that usually governs the slab-beam construction. iv) Main steel : Ast= ___Md________--------------------------(4) 0.87fy(d-0.42xu) If it is a flanged section, replace d by Df.

o c .

m

The continuous beams at supports are generally required to be designed as a doubly reinforced section.

1. 2. 3.

a

m a Calculate Mumax= 0.367fckbd(d-0.42xumax) n should be as a doubly reinforced. If M>Mumax, then the design y Ast1= ___Mumax______ d 0.87fy(d-0.42xumax) u t S Ast2= (Mu-Mumax)/(0.87fy(d-dc))

Steps to design a doubly reinforced section :

4. 5. Total area of tension = Ast1+Ast2. 6. Calculate Asc= 0.87fyAst2/fsc

fsc can be obtained as Es x 0.0035 x (xumax-d)/xumax v) Detailing of Reinforcement: Select number and diameter of bars. Required spacing may be calculated as per the code. vi) Check for shear & shear reinforcement: 1. Find the shear force(acting),F from the frame analysis. 2. Find the shear strength of the beam given by F‟=k.η.b.d, where the parameters are as designated in the code.

3. If FF‟, then shear reinforcement need to be provided given for F-F‟, with the spacing s=0.87fyAstd/(F-F‟) 5. Incase bars are bent up for provision of shear reinforcement, then the additional force coming in due to the bent up must also be considered. Vusb=0.87fyAsb sinx < 0.5F‟‟ , where F‟‟=F-F‟ vi) Check for deflection: In the case of beam, deflection criteria is normally satisfied, because L/d <16 and hence computations are skipped. Design of Intermediate(secondary) beam B15 : Load calculations : From slab SD, D=60mm; lx=2.7m; ly=3.2m Dead Load = 25x 0.006 = 2 kN/m2 Live load= 2kN/m2 Total load = 4kN/m2 ; factored load = 6 kN/m2 Ly/lx = 1.185 It‟s trapezoidal load from SD. So, β= 3.2/2.7 = Hence, the equivalent U.D.L for B.M is 8.49 kN/m (using (2) )

o c .

The equivalent load for S.F is 6.43 kN/m (using (1) ) Similarly for slab SG : lx = 3.2 ly = 3.8 It‟s a triangular load, so β=1

a

m a The equivalent U.D.L for B.M is 9.6 kN/m. The equivalent U.D.L for S.F is 7.2 ynkN/m. Total slab load on B15=d Weqb= 18.09 kN/m. u t Weqs= 13.63 kN/m. Wall load S = 19.2 x 0.23 x 2.75 = 12.14 kN/m.

Self weight of beam = 0.23 x 0.45 x 25 = 2.58 kN/m. Total (b)= 32.81 kN/m Total (s)= 28.35 kN/m Designing: Mact= 32.81 x 3.2^2/8 = 41.99 kN/m. 0.0035 = Xumax___ 0.0038 430-Xumax From this, Xumax= 205.33 Mu= 0.36 x fckbXu(d-0.42Xu) Substituting the above values, Mulim= 145.88 kN-m.

m

Mact= 0.87fyAst(d-0.42Xu) Ast= 338.31mm^2. Accordingly, main reinforcement, 3 bars of 12Φ @ a spacing of 150mm. Check for shear : The shear force acting= Wlx/2 = 28.35 x 3.2/2= 45.36 kN. The design shear strength of concrete ηc = 0.403 N/mm^2(from table 19 of IS:4562000) Now, maximum shear stress ηcmax = 2.8 N/mm^2 (From table 20 of IS:456-2000) Hence, η < ηcmax. However, according to the code, we must provide a minimum shear reinforcement at a spacing of s= (0.87x 415x50.24)/(0.4x230) = 197.16 mm. Beam details :

12θ bars, 3 in no. main reinforcement

m

Accordingly provide 3 bars of 12Φ at supports as well to resist end moments apart from that provided in the middle. The reinforcement in the middle is provided at the bottom while at the ends is provided at the top because the middle portion should resist compression while the ends should resist tension.

o c .

a

m a Beam 22(1) : bf= (2.1/6 + 6x0.1) + 0.23 n = 1.18m. y Since Max moment is considered, through analysis we get, d u Rax3-(32.72x3x1.5)-18.097+28.77=0 => Ra=45.55 kN; Rb= 52.61 kN t S Mx= Ra.x-18.097-32.72.x^2/2 Design of a beam as a T-beam:

dMx/dx =0 => x=1.40. Hence, Mact=13.6 kN-m; Mlim= 65.5 kN-m. => Mact< Mlim Ast= Mact/(0.87.fy.(d-0.42xu)) Substituting, Mact=13.6 kN-m ; d=0.43; xu=Df=100mm (considering 100mm thick wall) We get,

Ast= 97.08 mm^2

Beam 22 (2): bf= 1.026m To calculate Mact: Rcx2.2-13.66-21.04x2.2^2/2 + 10.59=0

Rc= 24.53 kN; Rb=21.75 kN Mx= Rc.x-13.66-21.04.x^2/2 ; Using dMx/dx =0, we get => x=1.03m Hence, Mact=0.639 kN-m. Mc=0.367.fck.b.Df.(d-0.42.xu)= 59.59 kN-m => Mact< Mc Ast=4.51 mm^2. However, in most of the practical cases, the beams are analyzed and designed as rectangular beams. Only, sometimes where economy is given due consideration, the beams are designed as T-beams. In this particular project, all the beams are designed as rectangular beams.

Beam No.

B1 B2 B3 B4 B4(a)

Size of No. of Ast at Ast at the beam spans in midspan support the beam 230x450 3 4-12Φ 2-12 Φ& 2-16 Φ 230x450 3 4-12 Φ 2-12 Φ& 2-16 Φ 230x450 3 5-12 Φ 2-12 Φ & 3-16 Φ 230x450 3 4-12 Φ 2-12 Φ & 3-16 Φ 230x450 1 2-12 Φ 2-12 Φ

yn

o c .

a

m

a

d

tu

B5

S 230x450

3

5-12Φ

B6

230x450

3

4-12 Φ

B7

230x450

2

4-12 Φ

B8

230x450

4

B9

230x450

1

2-16Φ& 2-12 Φ 2-12 Φ

B10

230x450

2

4-12 Φ

B11

230x450

1

3-12 Φ

B12

230x450

1

3-12 Φ

Diameter& spacing of stirrups at midpspan 2L-8 Φ @ 150mm 2L-8 Φ @ 150mm 2L-8 Φ @ 150mm 2L-8 Φ @ 150mm 2L-8 Φ @ 150mm 2L-8 Φ @ 150mm 2L-8 Φ @ 150mm 2L-8 Φ @ 150mm 2L-8 Φ @ 150mm 2L-8 Φ @ 150mm 2L-8 Φ @ 150mm 2L-8 Φ @ 150mm 2L-8 Φ @ 150mm

2-12 Φ & 2-16 Φ 2-12 Φ & 3-16 Φ 2-12 Φ & 2-16 Φ 2-12 Φ &3-16 Φ 2-12 Φ 2-12 Φ & 3-16 Φ 2-12 Φ & 3-16 Φ 2-12 Φ & 3-16 Φ

Diameter& spacing of stirrups at support 2L-8 Φ @ 125mm 2L-8 Φ @ 125mm` 2L-8 Φ @ 125mm 2L-8 Φ @ 125mm 2L-8 Φ @ 125mm 2L-8 Φ @ 125mm 2L-8 Φ @ 125mm 2L-8 Φ @ 125mm 2L-8 Φ @ 125mm 2L-8 Φ @ 125mm 2L-8 Φ @ 125mm 2L-8 Φ @ 125mm 2L-8 Φ @ 125mm

m

B13

230x450

2

B14

230x450

3

B15

230x450

2

B16

230x450

2

B17

230x450

2

B18

230x450

1

B19

230x450

2

B-20

230x450

1

B-21

230x450

1

B-22

230x450

5

B-23

230x450

1

B-24

230x450

1

B-25

230x450

3

B-25(a)

230x450

1

B-26

230x575

1

B-27

230x450

B-28 B-29 B-30

3

230x450 230x450

1

230x450

6

2L-8 Φ 150mm 5-12 Φ 2-12 Φ & 2L-8 Φ 3-16 Φ 150mm 3-12 Φ 3-12 Φ 2L-8 Φ 150mm 3-12 Φ 3-12 Φ 2L-8 Φ 150mm 3-12 Φ 3-12 Φ 2L-8 Φ 150mm 3-12 Φ 3-12 Φ 2L-8 Φ 150mm 3-12 Φ 2-12 Φ 2L-8 Φ 150mm 3-12 Φ 3-12 Φ 2L-8 Φ 150mm 4-12 Φ 2-12 Φ 2L-8 Φ 150mm 3-12 Φ 3-12 Φ 2L-8 Φ 150mm 3-12 Φ 3-12 Φ 2L-8 Φ 150mm 3-12 Φ 3-12 Φ 2L-8 Φ 150mm 3-12 Φ 3-12 Φ 2L-8 Φ 150mm 5-12 Φ 2-12 Φ 2L-8 Φ 150mm 3-16 Φ & 2-12 Φ 2L-8 Φ 3-20 Φ 150mm 3-12 Φ 5-12 Φ 2L-8 Φ 150mm 3-12 Φ 3-12 Φ 2L-8 Φ 150mm 3-12 Φ 2-12 Φ 2L-8 Φ 150mm 3-12 Φ 3-12 Φ 2L-8 Φ 150mm

yn

d

tu2

S

3-12 Φ

5-12 Φ

a

m

a

.

co

@ 2L-8 Φ 125mm @ 2L-8 Φ 125mm @ 2L-8 Φ 125mm @ 2L-8 Φ 125mm @ 2L-8 Φ 125mm @ 2L-8 Φ 125mm @ 2L-8 Φ 125mm @ 2L-8 Φ 125mm @ 2L-8 Φ 125mm @ 2L-8 Φ 125mm @ 2L-8 Φ 125mm @ 2L-8 Φ 125mm @ 2L-8 Φ 125mm @ 2L-8 Φ 125mm @ 2L-8 Φ 125mm @ 2L-8 Φ 125mm @ 2L-8 Φ 125mm @ 2L-8 Φ 125mm @ 2L-8 Φ 125mm

m

@ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @

The next step after design of beams is the design of the columns. However, before proceeding for design of columns, it is necessary to analyze the frame of the building in order to know how much load is being taken by the column. It is also sometimes done before design of columns to know the moments to which the beams are subjected to.

STRUCTURAL ANALYSIS: A brief introduction: We come across various structures in our day to day life ranging from simple ones like the curtain rods and electric poles to more complex ones like multistoried buildings, shell roofs, bridges, dams, heavy machineries, automobiles, aeroplanes and ships. These structures are subjected to various loads like concentrated loads, uniformly distributed loads, uniformly varying loads, random loads, internal or external pressures and dynamic forces. The structure transfers its load to the supports and ultimately to the ground. Treating an entire structure as a single rigid body and finding the reactions from supports is the first step in analyzing a structure. While transferring the loads acting on the structure, the members of the structure are subjected to internal forces like axial forces, shearing forces, bending and torsional moments. Structural analysis deals with analyzing these internal forces in the members of the structures. It is easier to analyze a multistory building with the help of „frame analysis‟ than the analysis of individual beams. The frame analysis of roof, ground floor and an internal frame is done. The results of the internal frame analysis are applied to other internal frames as well and hence the internal forces (namely shear forces and bending moments) are obtained.

m

o c The procedure of „Moment Distribution Method‟ is.used in this case to analyze the multi storey frame. The following steps may bea taken: m a 1) Assuming all ends are fixed, find the fixed end moments developed. n 2) Calculate distributiony factors for all the members meeting at a joint. d 3) Balance a u joint by distributing balancing moment(negative of unbalanced moment t at the joint) to various members meeting at the joint proportional to S their distribution factors. Do similar excersice for all joints. 4) Carry over half the distributed moment to the far ends of the members. This upsets the balance of the joint. 5) Repeat the steps 3 and 4 till distributed moments are negligible.

6) Sum up all the moments at a particular end of the member to get final moment.

If sway is there in the frame, then the following procedure may be adopted.

(a) Assume the sway is prevented by giving external support at beam level. Carry out analysis as explained above. This is called non-sway analysis. Considering the free body diagrams of column, find horizontal forces developed at supports. Then consider the horizontal equilibrium of the entire system to get force „S‟ developed at additional support assumed at beam level. (b) Actually, there is no support at beam level and hence „S‟ is the sway force moving the beam laterally. For the given sway force, it is difficult to find the end moments developed. Hence, an arbitrary sway is assumed, say Δ . Then, fixed end moments developed in column, AB and CD are :

(c) MF1=

-

6EI1Δ/

yn

d

S

tu

L12

a

mand

a

o c .

m

MF2

=

-6EI2Δ/L22

MF1/MF2 = (I1/L12) / (I2/L22) Now, arbitrary proportionate values may be assumed for M F1 and MF2. Then Moment Distribution is carried out to get final moments. Let MAB, MBA, MCD and MDC be the final values HA =( MAB + MBA)/L1 and HD = (MCD + MDC)/ L2 Hence, sway force „S‟ acting in this case is obtained by considering horizontal equilibrium of the frame as shown, we have to multiply by the sway correction factor k=S/S‟ Final Moments= Non Sway Moments + kx sway moments.

d

yn

S

tu

a

m

a

o c .

m

DESIGN OF COLUMNS:

The design of column necessitates determination of loads transferred from beam at different floor levels. Loads are transferred from slabs to beams and then to columns. Hence, slabs and beams are normally designed prior to the design of columns. This method called as „exact method‟ which enables one to assess the loads on columns more accurately and thereby the design of column becomes realistic and economical. However, in practice, many times situations arise which require the design of columns and footings to be given prior to the design of slabs and beams. In such a case, loads on columns and footings are required to be assessed using judgement based on past experience and using approximate methods. The loads can be determined approximately on the basis of floor area shared by each column. These loads are normally calculated on the higher side so that they are not less than the actual loads transferred from slabs/beams. In such cases, the design of column is likely to be uneconomical. Categorization of columns: This is the first step in designing of the columns because the procedure for design of columns in each of the three categories is different.

o c .

m

(I) Internal columns or Axially loaded columns: Internal columns carrying beams either in all four directions or only in opposite directions are predominantly subjected to axial compression because moments due to loads on beams on opposite sides balance each other. Judgement should be used to place a column under this category because if spans and/or loads on beams on opposite sides vary appreciably the beam moments on opposite sides may not balance each other and the column will be subjected to bending moment and it will be required to place under the second category.

yn

a

m

a

d u (II) Side columns t or columns subjected to axial compression and uniaxial bending: S

Columns along the sides of a building which carry beams either in three orthogonal directions or a single beam in one direction are subjected predominantly to axial load and uniaxial bending due to unbalanced moment transferred from a single beam on one side, while the moments from the other two beams in opposite directions balance each other provided their spans and loads on them are nearly equal. If such columns are to be designed as axially loaded columns using approximate method, the axial load is required to be increased to account for the effect of uniaxial bending in column. The load thus arrived is called equivalent acial load for the purpose of design of column section.

(III)

Corner columns or columns subjected to axial compression and biaxial bending : Corner columns or the columns which carry beams in two perpendicular directions are subjected to biaxial bending due to beams in

orthogonal directions. They require large increase in axial load to account for the effect of biaxial bending for obtaining an equivalent axial load. Computation of loads on columns: There are two methods namely for this purpose. They are: 1) Exact Method: This method is used to compute loads if the beam end shears are known prior to the column design. These have been calculated while analyzing the loads on beams and designing them. For columns with axial compression and uniaxial/biaxial bending, the moments on the columns have been obtained from the frame analysis by moment distribution method. Total load (L.C)= V1+V2+V3+V4+Pa+Pself Where V1,V2,V3,V4 = end shears of beams meeting at the column at the floor under consideration from all the four directions. Pa=axial load coming from above Pself= self weight of the column at the floor under consideration.

2) Approximate Method: This method is used when the design of footing is required to be given prior to design of slab and beam and approximate sizes of column are required to be assumed. This is done by knowing the influence area and the load in the area that is borne by that particular column.

o c .

m

a

Pu‟floor= Pus + Puw + Pa Pus= load transferred from slab to column at each floor level= wus x Acol Puw= wall load transferred to column at each floor level= wuw x Lw Pa= load on column from above However, above procedure of column loads does not work well when there are number of secondary beams. In such cases, approximate loads are required to be calculated on beams first and column load are obtained from beam shears.

yn

m

a

d

tu

S

Calculation of Moments in Columns: The moments in the columns are obtained directly and exactly if the entire structural frame is analysed using Moment Distribution method. However, if the building cannot be divided into a number of frames due to peculiar positions of columns, as in some cases of residential buildings or in building frames in which the connections are assumed to be simple, the moments in columns at any floor level can be obtained by considering substitute column frame which consists of only the relevant column together with connected beams fixed at their far end. The moment in the column can be calculated using the equation Mcol=(kc/∑k) x Me Where kc= stiffness of column under consideration= Ic/Lc ∑k = sum of stiffnesses of members ,eeting at the joint = ∑kc + ∑kb/2 Stiffness of the beams kb shall be reduced to half to account for the effect of members beyond the adjacent spans being ignored. Me= unbalanced fixed end moment at the joint.

= wu.L^2/12, if a single beam is rigidly connected to the column on one side. = wu1L‟^2/12 – wu2L”^2/12, if two beams will unequal loads or unequal spans are rigidly connected on opposite sides of the column. Me= WuL^2/24, if a single beam is simply connected to column. Me= Wu1L‟^2/24 – Wu2L”^2/24 , if two beams with unequal loads or unequal spans are simply connected in opposite sides of the column, in which Wu1 and Wu2 are the loads and L1, L2 are lengths of the beams on two sides. The calculated moment in column shall not be less than Mumin. = Pu x emin When column above and below the floor level are of different sizes with their outer faces flush, the load from upper column becomes eccentric with respect to the lower column. However, it may be noted that the moment due to this eccentricity is opposite to the moment transferred by the beam to the column at that level. This, in fact results in reduction of the effective moment and hence the moment due to this eccentricity need not be considered. It needs consideration only when there is no floor beam in the plane of the offset. Grouping of Columns: Once the load on each column and effective lengths are determined, the columns in the same category which have total loads on them not varying by more than 10 to 20% and having the same effective lengths may be grouped together. In such a case, column carrying maximum load may only be designed in that group and the same section be adopted for all the columns in that group. This saves the computational efforts and labour, considerably during the execution of work. This is of prime importance in practical design.

yn

a

m

a

o c .

m

Design of column section: The design of column section may be done by any of the two methods:

d

tu

S

(A) Approximate Equivalent axial load Method: In this approach, total equivalent axial load is obtained by adding calculated approximate axial loads. Preliminary section is designed for this total equivalent axial load using the procedure for design of axially loaded columns. The section so obtained is later on checked by exact method for actual compression and bending moment.

(B) Exact Method: This method of designing column depends upon the type of column (short or slender) and the type of loading and whether the column is subjected to axial load only or subjected to combined axial load and uniaxial bending or combined axial load and biaxial bending. The columns are easy to design using the design aids given in SP-16.

If Leff/h <12, then the column is said to be short and if Leff/h > 12, the column is slender. (I) Axially loaded short columns The column shall be designed as a short axially loaded compression member if the minimum eccentricity does not exceed 0.05 times the lateral dimension. Pu= 0.4.fck.Ac + 0.67.fy.Asc Where, Pu= axial load on the member. fck= characteristic compressive strength of concrete Ac= Area of concrete fy= characteristic strength of compression reinforcement Asc= area of longitudinal reinforcement. Here Ac= Ag-Asc, where Ag is the total cross sectional area of the column. Assume diameter of lateral ties (θtr not less than 5mm or ¼ th the diameter (θ) of main bar, whichever is greater). Normally, 6mm diameter ties are used for main bar diameter less than 25mm. Decide the picth s of ties such „s‟ is not greater than least of (300mm ,width b)

m

(II) Short columns subjected to axial compression and uniaxial bending:

o c .

Determine the bending moments in columns. Assume arrangement of bars.

a

m

a

If the column is subjected to large bending moment M as compared to axial load P (say e/D = M/(PD) ≥ 0.5), assume bars to be equally placed on opposite faces like a doubly reinforced section. On the contrary, if P is large compared to bending moment M ( e/D = M/(PD) < 0.5), assume bars to be uniformly placed all around the periphery.

yn

d u Thesetcharts can be used without significant error for any number of bars S greater than 8, provided the bars are equally distributed on the four sides.

It may be noted that the second arrangement requires large area of steel than that required by the first arrangement. In case of ambiguity of deciding the arrangement, the second one is definitely safer. Procedure: (a) For bending about x-axis bisecting the depth of column

1) Calculate Pu/(fck.b.D) and Mu/(fck.b.D^2) 2) Calculate d‟/D where d‟= effective cover 3) Select appropriate chart corresponding to d‟/D, grade of steel and distribution of reinforcement. Obtain point of intersection of Pu/(fck.b.D) and Mu/(fck.b.D^2) 4) Interpolate the value of p/fck where, p=As/(bD) 5) Calculate total area of steel required= As= fck x (pbD/100)

(b) For bending about y-axis bisecting the width of the column the chart to be referred to is having value of d‟/b and use Mu/(fck.b.D^2). Rest of the procedure is the same as given below.

(III)

Short columns subjected to Axial Compression and Bi-axial bending

(i)

(ii)

(iii)

Assume steel percentage between 1% and 3% and the numberdiameter combination of bars for the same. Assume bars to be placed uniformly all around the periphery as this is better for bi axial bending. Calculate p/fck where p= 100As/(bD) and Pu/(fckbD) Select appropriate chart corresponding to d‟/D. Draw a horizontal line from Pu/(fck.b.D) and continue it till it reaches a point corresponding to the value of p/fck. Drop a perpendicular on x-axis to give the value of Mux1/(fck.b.D^2). Calculate Mux1. Repeat the process by selecting appropriate chart corresponding to d‟/b and obtain the coefficient by dropping the perpendicular on xaxis which gives Muy1/(fck.b^2.D). Calculate Muy1. Calculate Puz = 0.45fck.Ac + 0.75fy.Asc and calculate Pu/Puz and

o c .

hence the value of αn (as per IS:456-2000) (iv)

Check that (Mux/Mux1)

αn

m

+ (Muy/Muy1)αn ≤ 1 -----------------(p)

m

a

If this equation is not satisfied, then the section is unsafe. Increase the section and/or reinforcement and revise the calculations. If the left hand side of the equation (p) is less than 0.8, the section is uneconomical. Reduce the reinforcement or reduce the section and repeat the procedure if desired. Continue with the trials until the section and economical.

yn

a

d

(III)

tu

Slender columns : (i) Calculate additional moment due to slenderness. S Obtain Puz and Pub as mentioned earlier. (ii) Calculate initial moments and obtain total moment Mut. This is now the design moment for the column accompanied by given Pu. (iii) Check the safety of column for combined effect of Pu and total moment Mut using the procedure for axial loading with uni axial bending.

Note: For safe side, most of the columns, which could be designed as axially loaded were designed considering them as axially loaded columns with uniaxial bending. Example : The load on column from roof on the 5th floor : Total axial load from the adjacent beams (i.e the shear forces of the beams) = 148.06+ 45.32 +126.24 + 46.59 = 366.21kN

Self weight of the column= 7.11kN Moments in the column: At top= 18.04kN-m & -5.18kN-m At Bottom= 17.26kN-m & -1.019 kN-m Ex=2.05cm >2cm & ey=1.31cm, hence 2cm. Total load= 366.21 + 7.11= 373.32 kN M= 18.04kN-m. Hence, according to M/PD we have to provide reinforcement equally on all the four sides. Pu/fck.b.D = 0.3 and Mu/fck.b.D2 = 0.02 ; d‟/D =0.11 From the p/fck value from the chart 44 of SP-16, we get p=0.2% Ast= 1300 mm2 Assuming 12Φ bars, 12 no. of bars are required. Now, load on the 4th floor: Load coming from the previous top column+ self weight of the column + shear force from the adjacent beams = 373.21+ 7.11+ 366.21-33%(total load) {for economy purposes} = 725.16 kN 31.66

o c .

Mu= 31.66 kN

a

Pu/fck.b.D = 0.35 Mu/fck.D.b2 =0.06

m

a

m

n y A = 1000mm i.e 5 16Φ bars. d u Hence, all the columns t are designed in this manner and are finally grouped for convenience so that the design of less number of columns may be required. DependingS on their load conditions and reinforcement requirement, they are From this, p/fck = 0.04 => p=0.8% 2

st

categorized.

COLUMN SCHEDULE: S.No. COLUMN No. of SIZE REINFORCEMENT TIES TYPE COLUMNS 1 C1 12 230X450 12-16Φ 8Φ @ 17mm spacing

2

C2

9

230X450 10-16Φ

3

C3

8

230X450 6-16Φ + 2-12Φ

4

C4

12

230X450 8-12Φ

d

yn

S

tu

a

m

a

o c .

m

8Φ @ 17mm spacing 8Φ @ 17mm spacing 8Φ @ 17mm spacing

DESIGN OF FOOTINGS:

Footings are of two types : 1) isolated footing 2) rectangular sloped footing. We have designed isolated footing and the procedure is given below. DESIGN OF ISOLATED FOOTING :The footing for an axially loading column of size b*D is designed as an inverted cantilever outstanding from column and loaded with uniform upward soil pressure. The various steps involved in the design are given below :Proportion of base size :Initially suitable footing dimensions are required to be selected to ensure that under serviceability conditions the soil bearing pressure is not exceeded. The maximum load transferred to the soil is equal to axial load on column plus self weight of the footing. Since the size of the footing is unknown, its self weight is assumed to be equal to 10% of the axial load on the column. If the axial load(working) on column is P then, Area of footing = Aƒ = 1.1P/fb =Lf * Bf Where Lf = Length of the footing Bf = breadth of the footing. fb = safe bearing capacity of soil.

yn

a

m

a

o c .

m

d u Once the area of footingtis known the size of footing gets fixed. The shape of the footing may be square or rectangular or circular. The size of the rectangular base is selected such that the cantilever projections of the Sof the column are equal. This gives approximately the same depth for bending about footing from the faces x and y axes. The length or breadth of the footing based on equal projection is obtained as under : Cantilever projection of footing for bending about x-axis = Cx = (Lf-D)/2 Cantilever projection of footing for bending about y-axis = Cy = (Bf-b)/2 For equal projections , (Lf-D)/2 = (Bf-b)/2 or Bf = Lf – D+b Substituting the value of bf in the below equation and solving quadratic equation in Lf we get ,

Lf = (D-b)/2 +[ [(D-b)/2]^2 +Af ]^0.5 Select the length of the footing by rounding out the value of Lf, Recalculate Cx = (Lf-D)/2 and Cy = (Bf – b)/2

Where, breadth of footing = Bf = b + 2*Cx and Lf and Bf are the length and breadth of footing provided. For square footing , Lf = Bf = √Af Area of the footing provided =Af = Lf * Bf Upward factored soil reaction = Wu = Pu/Af. Where , Pu = load factor * axial force = 1.5 *P Comments : 1) In calculating the upward factored soil reaction the self weight of the footing is not considered because the dead load of the footing acts in the opposite direction of soil pressure and hence does not induce any moment or shear in the footing 2) The value of Wu will work out to be greater than the bearing capacity of the soil. But this is not unsafe because the comparison can be made with the upward working soil reaction which can be obtained by dividing Wu by the load factor of 1.5. then it will be seen that the value of working soil reaction so obtained (Wu/1.5) will be less than the bearing capacity of the soil. Depth of footing from bending moment considerations

o c .

m

The maximum bending moment is calculated at the face of the column or pedestal by passing through the section a vertical plane which extends completely across the footing and computing the moment of forces acting over the entire area of the footing on one side of the said plane.

a

m

a

Bending moment at the column face parallel to x-axis : Mux =Wu*Bf*Cx*Cx/2 Bending moment at the column face parallel to y-axis : Muy = Wu*Bf*Cy*Cy/2

yn

d

tu

Required effective depth for bending about x-axis :

S

Required effective depth for bending about y- axis :

Dx = √(Mux/Ru.max*b1) Dy = √(Muy/Ru.max*D1)

Where , b1 = b + 2e D1 = D + 2e b = width of column , D = depth of column, e = offset provided at the top of footing for seating column form work.

Example of footing design calculation for bending :Data :-

Common data : Concrete grade (mild environment) Steel grade Design constant =2.76N/mm2 Column section D = 450mm Bearing capacity of the soil Minimum depth of footing Offset of footing level Maximum design ultimate column load

M20 Fe 415 Ru.max b = 230mm fb = 250 kN/m2 Df.min = 150mm e = 50mm Pu = 1698.49 kN

Working load P = Pu/1.5 = 1698.49/ 1.5 = 1132.326 kN Area of the footing =P/s.b.c = 1132.326*1.1/250 = 5m2 Lf = 0.45 + (2*0.95) = 2.30 m Bf = 0.23 + (2*0.95) = 2.13m Upward soil pressure = P/Af =(1132.326/5) = 346.7 kN/m2

o c .

Breadth of footing at the top :

a

bx = b + 2e = 330mm by = D + 2e = 550mm

m

Mux =346.7*2.3*0.95*0.95/2 = 322.95 kN-ma n y Muy = 346.7*2.13*0.95*0.95/2 = 299.08 kN-m d u Mu = 0.367*fck*b*Xu(d-Xu) t S 0.0035/0.0038 =Xu/(d-Xu) 0.0035d -0.0035Xu = 0.0038 Xu = 0.479d 322.5 * 106 = 1136.89 d2 Xu = 253.87 Dx = 532.97 mm Dy =512.90mm Whichever is maximum is D Final D =532.97 ≈ 530mm

m

Mux = 0.87FyAst(d-0.42Xu) Ast(x) = 2112.72mm2 Muy = 0.87FyAst(d-0.42Xu) Ast(y) = 1956.56mm2

Check for two way shear :Perimeter = 2 [ (b+Dy) + (D+Dy)] = 2 [230 + 450 + 2 (500)] =

3360mm

Dy – Dfmin / d‟ = L-e/ (Lf – D)/ 2 (480 – 150)/ d‟ = 0.90 / 0.75

o c .

d' = 291.6 mm ≈ 290 mm area of acting for two way shear = perimeter * d‟ = 3280 * 290 = 0.97m2

d

tu

Shear resisted : Ks = 1

yn

a

S

ηu = Ks*ηv = 0.25√Fck = 1.118 N/mm2 Vu = 1.118*290*3360 = 1089.37 kN Design shear = Wu[Lf*Bf - (D+Dy)(b + Dy)] = 1458.04 kN Not safe. Redesign depth : D = 630 mm Perimeter = 2(230 +450 + 2(630 – 55))

m

a

m

= 3660mm d' = 450mm Area of resist shear = 1647000mm^2 Ks = 1, ηv = 0.25√Fck = 1.18N/mm^2 Shear strength = 1841.34 kN Design shear = 346.7 * [2.3*2.13 – (0.45+0.57)(0.23+0.57)] = 1415.17kN D = 600mm Perimeter = 3540 mm

Check for one way shear :Mux = 0.87FyAst(d-0.42Xu)

o c .

322.95 * 10^6 = 0.87 *415*(630 – (0.42*301.7)) Ast = 1777.26 mm^2 Muy = 299.08 * 10^6 Ast = 1645.9mm^2

S

tu

d

yn

a

m

a

m

d

yn

S

tu

a

m

a

o c .

m