C4301 Unit 1

1 REINFORCED CONCRETE STRUCTURAL DESIGN

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UNIT 1 ULTIMATE LIMITS STATE DESIGN CONCEPT

GENERAL OBJECTIVE:

GENERAL OBJECTIVE To understand the philosophy and structural design concepts of reinforced structural design in Ultimate Limit States.

At the end of this unit, you should be able to:





state the aim of the structural design.



explain the design philosophy

explain the statistical relationships between the strength of materials and characteristic load. 

calculate the characteristic strength of materials and characteristic load.



identify and use partial safety factors

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INPUT 1

1.1 Definition What is structural design? Design involves the selection of materials and determination of structural element sizes. The aim of design is to build structures which are safe and that can be used as intended, with minimum cost, during constructions and maintenance.

Reinforced concrete is made by reinforcing concrete with steel reinforcements. Concrete is a composite material of cement, aggregate, sand and water. The compressive strength of concrete is higher than its tensile strength but it is very weak in resisting compressive force and lateral stability. Reinforced concrete is a very strong durable and versatile structural material. It is a combination of concrete and steel.

1.2 Design Process Normally a project starts when the client (individual, government, or company) wants to build a structure for his/ her intended purpose. The client will consult an architect or an engineer who will then transform the project into drawings after taking careful considerations into accounts and possibility in choosing the materials, and the method of construction.

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Based on the architect’s drawings, the engineer will determine the structural layout such as the structural frame, structural elements, sizes and dimensions e.g. beams, columns, foundation etc.

The engineer will carry out the structural analysis. This involves the calculation and determination of the force on each structural element. After that, the sizes of the elements, their positions and their numbers will be calculated as shown in the detailed drawings. The design process and how the project is accomplished is shown in Figure 1.1 Load calculations Detailed Layout Structural Architect’s Drawings Plan / Structural Drawings Design Analysis

Figure 1.1 Design Process

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1.3 Reinforced concrete structures

Generally, reinforced concrete structures consist of various elements e.g. beam, slab, column, wall, staircase and foundation. Each of them will be specifically located at a particular place as designed by the engineer and architect.

1.4 Code of Practice

During the design stage, the engineer will continuously refer to a particular code of practice. Code of practice is a document with contains the standard practices i.e. best practices experienced by enquiries and research which have been compiled and documented. The British Standard is generally referred to when the designing reinforced concrete structures.

As far as reinforced concrete design is concerned, the codes of practice that are normally referred to are as follow:-

1.

BS 8110: 1985 : Structural Use of Concrete Part 1:

Code of Practice for Design and Constructions.

Part 2:

Code of Practice for Special Circumstances

Part 3: Design charts for singly beams, doubly beams and rectangular columns.

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2.

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BS 6399 : 1984 : Design Loading for building Part 1: Code of Practice for Dead and Imposed Loads.

3.

CP3 : 1972 : Chapter V : Loading Part 2: Wind Loads

1.5 Design Method

Design which is in accordance with BS 8110 is based on Limit States Method. This method ensure that the structures are safe and fit for use i.e. they will not achieve their limit states during their service life.

The limit states mentioned earlier are as follows: a)

Ultimate Limit State (ULS) ULS in concerned with the maximum load – carrying capacity of the structure within the limits of strength of the materials used.

b)

Serviceability Limit State (SLS) SLS is concerned with the appearance of the structure, the effects of deflection or deformation on other elements and the durability of the structure.

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c)

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Additional Limit State The overall stability of the building is, of course, of paramount importance. It is of no use when an individual satisfies all the design criteria if the connections between the structure elements are incapable of withstanding at least some effect from misuse or accidents. In this context it is difficult to consider structural elements in isolation, as they form an integral part of the structure.

Generally speaking, reinforced concrete structures are designed to fulfill the ultimate limit state and are checked against the serviceability limit state. This process is of paramount importance because the main purpose of structural elements of a building is to withstand any load without jeopardizing the safety of the occupants.

For water retaining structures, cracks are the most important criteria. They have to be watertight. Therefore the structure should be designed for Ultimate Limit State (ULS) and checked against the Ultimate Limit State.

In addition to that, the durability of reinforced concrete structures should not be ignored. They should be given some attention during the selection of materials and when designing details.

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ACTIVITY 1a

TEST YOUR UNDERSTANDING BEFORE YOU CONTINUE TO THE NEXT INPUT! Fill in the blanks with the correct answers:

The purpose of design may, perhaps be started as the provision of a structure complying with the __________ and the ________requirements.

The _________________ is preferred to collapse.

The purpose of design is to ensure that it will not reach a _____________.

Reinforced concrete is a composite material consisting of _____________ and _________________.

Deflection is categorized and checked in accordance to the ________ limit state.

The essential basis of the design method is to consider each ___________ and to provide a suitable margin of _______________________________.

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The purpose of design consists of finding and dealing with the most economical structures

associated

with

________________

and

______________________

the

limit

______________________

requirements.

BS

8110

defines

state

as

and

__________________________.

BS 6399 is the code of practice for ___________________ and ______________ load.

Beams,

slabs,

columns

and

foundation

are

reinforced

___________________ that have to be designed by the engineer.

concrete

structural

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FEEDBACK 1a

The answers are given below:

client’s , user’s Ultimate Limit State limit state concrete , steel reinforcement serviceability limit state , safety safety , serviceability Ultimate Limit State (ULS) , Serviceability Limit State ( SLS ) Dead , Imposed elements

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INPUT 2

1.6 Load Generally, load on any structural members cannot be determined accurately. For most structures, it is uneconomical to design using anticipated maximum load. Therefore, in normal design practice, the load to be used is based on the characteristic load. Characteristic load is defined as the minimum load that statistically will not exceed during the design life of the structure. There are 3 types of load: 1)

Characteristic dead load ( gk ) Characteristic dead loads are fixed loads that will not be much different from the estimated load. Some examples of dead load are the weight of the elements, finishes, ceiling and fixed equipment, such as water pipes. BS 6399 Part 1 gives some of characteristic dead load to be used in designs.

2)

Characteristic imposed load ( qk ) This load is not fixed but may vary such as the weight of occupants, equipment, furniture, etc. Characteristic imposed load, gkmay also be obtained from BS 6399: Part 1 Code of Practice for Loads.

3)

Wind load ( wk )

Dead and Imposed

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Wind load depends on the location, form, and dimension of the building and the wind velocity of that particular area. CP3 Chapter 5 Part 2 gives some guidelines on how to estimate wind load.

The load is calculated as follows;

Design Load = characteristic load × partial safety factor = ( gk , qk , wk ) × γf

Design load and partial safety factor for various load combination and limit states are given in 2.1, BS 8110 Part 1.

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ACTIVITY 1b

TEST YOUR UNDERSTANDING BEFORE YOU CONTINUE TO THE NEXT INPUT! Answer the following questions

What is characteristic load? …………………………………………………………………………… …………………………………………………………………………… …………………………………………………………………………… …………………………………………………………………………… …………

State the 3 types of load.

a)…………………………………………………………………………… ……………………………………………………………………………. b)…………………………………… …................................................................................................................... .............................................................

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c)…………………………………………………………………………… ………………………………………………………...……………………. State the code of practice that gives estimates of dead and imposed loads? …………………………………………………………………………… …………………………………………………………………………… …………………………………………………………………………… …………………………………………………………………………… …………

The code of practice that gives guideline for calculating wind load? …………………………………………………………………………… …………………………………………………………………………… …………………………………………………………………………… …………………………………………………………………………… …………

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FEEDBACK 1b Please check your answers below:

Answers:

Minimum load that will not exceed during design life.

a) Dead load b) Imposed load c) Wind load

1.13

BS 6399 : Part 1

1.14

CP3 Chapter 5 : Part 2

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INPUT 3

1.7 Strength of Materials

Characteristic strength is used to represent the strength of material. Mean strength is not suitable to be used because normally 50% of test results will fail. Characteristic strength is represented by the area under the normal distribution curve with a value of 0.05. This is shown in Figure 1.2 below: Probability distribution

Area Strength = 0.05

Characteristic strength

Mean strength

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Figure 1.2: Characteristic Strength

From figure 1.2, characteristic strength is calculated as follows:

Characteristic strength = Mean strength – 1.64 × Standard deviation (σ)

It can be shown that 5 % of the test rests will be less than the characteristic strength. For example, to produce a concrete with a characteristic strength of 30 N/ mm2, and a standard deviation of 5 N/mm2, you need to have a mean strength of 38.2 N/mm2. The calculation is as follows:

i.e.: ∴

Characteristic strength = Mean strength – 1.64σ Mean strength

= Characteristic strength + 1.64σ

= 30 + 1.64 (5) = 30 + 8.2 = 38.2 N/mm2

Characteristic strength of concrete ( fcu ) is the strength of concrete at the age of 28 days . Concrete of grade 25, 30, 35, 40 and 50 N/mm2 is normally used.

Characteristic strength of steel reinforcement ( f y) is the yield stress of steel. Three (3) types of steel reinforcement are high yield steel (T) with a strength of 460 N/mm2, mild steel (250 N/mm2 ) and fabric reinforcement (BRC) with a strength of 485 N/mm2.

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Design strength is defined as; Design strength = characteristic strength partial safety factor = ( f cu , f y )

γm

1.8 Partial Safety Factor

Partial Safety Factor is used in design to take into account any variation that could happen to load and strength during design and construction. BS 8110 has established its values in Table 2.2, Part 1. Various factors which contribute to these values are;

Since ultimate limit state is more severe, so the safety factors are much bigger than those for serviceability limit state. Safety factor for steel reinforcement is less than for concrete because steel produced is of a stringent quality and it is controlled. Safety factor for imposed load is more than for dead load because dead load could be estimated more accurately that imposed load

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ACTIVITY 1c TEST YOUR UNDERSTANDING BEFORE YOU CONTINUE TO THE NEXT INPUT!

1.15

How do you define characteristic strength?

…………………………………………………………………………… …………………………………………………………………………… …………………………………………………………………………… ………

1.16

If a concrete of grade 40 is to be produced, what mean strength is needed if σ = 4N/mm2?

…………………………………………………………………………… …………………………………………………………………………… …………………………………………………………………………… ………

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1.17

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State three (3) types of reinforcement used in reinforced concrete structures.

a)…………………………………………………………………………… ………………………………………………………………………………

b)…………………………………………………………………………… ………………………………………………………………………………

c)…………………………………………………………………………… ………………………………………………………………………………

1.18

Given fy = 250 N/mm2, γm = 1.15. What is the steel design strength? …………………………………………………………………………… …………………………………………………………………………… …………………………………………………………………………… …………………………………………………………………………… …………

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1.19

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What is the partial safety factor for concrete in flexure? …………………………………………………………………………… …………………………………………………………………………… …………………………………………………………………………… …………………………………………………………………………… …………

FEEDBACK 1c

Let’s check the answers together! Answers: 1.15

Characteristic strength = Mean strength - 1.64 × Standard deviation

1.16

Mean strength = 40 – 1.64 (4) = 33.44N/mm2

1.17

a) high yield steel b) mild steel c) BRC

1.18

Steel design strength = 250 = 217.39N / mm 2 1.15

1.19

γm = 1.50 (From Table 2.2, BS 8110)

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Congratulations! You have successfully completed UNIT 1. Please check to the answers that you have fulfilled to all the objectives of this unit. If you haven’t done this, do not hesitate to go through this unit again. Now you should do the self-assessment section. But before doing so, please read the summary of UNIT 1.

SUMMARY

1.

The aim of structural design is to build structures, which are safe and that which can be used for intended purposes during it service life.

2.

Reinforced concrete is an important construction material, as it is a very strong, durable, fire resistant and versatile.

3.

The design process starts immediately after the client proposes his or her intention to the architect or engineer.

4.

The design process includes the preparation of the layout plan, structural analysis, load calculations, structural design and the preparation of working drawings.

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5.

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BS 8110 is the code of practice that should be referred to when designing reinforced concrete structural elements.

6.

BS 8110 uses the Ultimate Limit State method of design.

7.

Loads are categorized as dead load, imposed load, and wind load.

8.

Design load is equal to characteristic load multiplied by partial safety factor.

9.

Characteristic strength is used to represent the strength of materials.

10.

Partial Safety Factors is used to take into account any inaccuracies during the design and construction of reinforced concrete structures.

Are you ready to do the self –assessment? This test is in the form of multiple-choice questions. You have to answer all the questions given in 30 minutes. Award 1 mark for every correct answer. You may refer to BS 8110 if you want to.

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SELF-ASSESSMENT

1.

Which of the following is not included in the aims of structural design? A.

Safety

B.

Cost

C.

Maintenance D.

2.

Aesthetic

Structural concrete should have the following EXCEPT…. A.

durable

B.

commercial value

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3.

C.

resistance to misuse

D.

fire resistant

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In the appropriate overall structural scheme, the engineer is guided by the following codes of practice EXCEPT

4.

5.

A.

ISO 9000

B.

BS 8110

C.

BS 6399

D.

CP 3

Which of the following failures occur when Ultimate Limit State in exceeded?

A.

Deflection

B.

Vibration

C.

Collapse

D.

Cracking

The design method used in BS 8110 is one of the following. Select the answers.

A.

Plastic State Design

B.

Conservative Design

C.

Limit State Design

D.

Simple Design

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6.

Given that fm = 35 N/mm2 σ = 3.5 N/mm2 What is the characteristic strength of concrete?

7.

A.

0.29 N/mm2

B.

2.9 N/mm2

C.

29.0 N/mm2

D.

290.0 N/mm2

Given that γf = 1.6 and characteristics load = 58 kN What is the design load?

A.

928 kN

B.

0.928 kN

C.

9.28 kN

D.

92.8 kN

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8.

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Partial safety factor, γm for concrete in shear without shear reinforcement is equal to...

9.

10.

A.

1.50

B.

1.25

C.

1.40

D.

1.30

Given that γm = 1.15, fy = 460 N/mm2. What is the design strength?

A.

400 N/mm2

B.

450 N/mm2

C.

500 N/mm2

D.

550 N/mm2

Given that Gk = 50 kN, Qk = 100 kN. What is the design load?

A.

23 kN

B.

230 kN

C.

2300 kN

D.

23, 000 kN

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FEEDBACK ON SELF-ASSESSMENT

Award one mark for every correct answer. The total is 100% for 10 marks. 1.

D

2.

B

3.

A

4.

C

5.

C

6.

C

7.

D

8.

B

9.

A

10.

B

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Now calculate the marks you have got. You should get at least 80% to pass this unit. If you have got more than 80%, congratulations! You are a good student. But if you have got less than 80%, you should go through this unit again.

END OF UNIT 1