Unit 1 ( ULTIMATE LIMITS STATE DESIGN CONCEPT ) -REINFORCED CONCRETE STRUCTURAL DESIGN

SPECIFIC OBJECTIVES

REINFORCED CONCRETE STRUCTURAL DESIGN C4301/UNIT1/

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

1

OBJECTIVES


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.

2

INPUT 1


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

Figure 1.1 Design Process

3

Architect’s Drawings

Layout PlanLoad calculations / Structural Analysis

Structural DesignDetailed Drawings


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.

4


2. 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.

5


c) 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.

6


TEST YOUR UNDERSTANDING BEFORE YOU CONTINUE TO THE NEXT

INPUT!

Fill in the blanks with the correct answers:

1.1 The purpose of design may, perhaps be started as the provision of a structure

complying with the __________ and the ________requirements.

1.2 The _________________ is preferred to collapse.

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

1.4 Reinforced concrete is a composite material consisting of _____________ and

_________________.

1.5 Deflection is categorized and checked in accordance to the ________ limit

state.

7

ACTIVITY 1a


1.6 The essential basis of the design method is to consider each ___________ and

to provide a suitable margin of _______________________________.

1.7 The purpose of design consists of finding and dealing with the most

economical structures associated with ________________ and

______________________ requirements.

1.8 BS 8110 defines the limit state as ______________________ and

__________________________.

1.9 BS 6399 is the code of practice for ___________________ and

______________ load.

1.10 Beams, slabs, columns and foundation are reinforced concrete structural

___________________ that have to be designed by the engineer.

8


The answers are given below:

1.1 client’s , user’s

1.2 Ultimate Limit State

1.3 limit state

1.4 concrete , steel reinforcement

1.5 serviceability

1.6 limit state , safety

1.7 safety , serviceability

1.8 Ultimate Limit State (ULS) , Serviceability Limit State ( SLS )

1.9 Dead , Imposed

1.10 elements

9

FEEDBACK 1a


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 ( g k )

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 ( q k )

This load is not fixed but may vary such as the weight of occupants,

equipment, furniture, etc. Characteristic imposed load, gk may also be

obtained from BS 6399: Part 1 Code of Practice for Dead and Imposed

Loads.

10

INPUT 2


3) Wind load ( w k )

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.

11



INPUT!

Answer the following questions

1.11 What is characteristic load?

………………………………………………………………………………

………………………………………………………………………………

………………………………………………………………………………

………………………………………………………………………………

1.12 State the 3 types of load.

a)……………………………………………………………………………

…………………………………………………………………………….

12

ACTIVITY 1b


b)

………………………………………............................................................

....................................................................................................................

c)

………………………………………………………………………………

……………………………………………………...…………………….

1.13 State the code of practice that gives estimates of dead and imposed loads?

………………………………………………………………………………

………………………………………………………………………………

………………………………………………………………………………

………………………………………………………………………………

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

………………………………………………………………………………

………………………………………………………………………………

………………………………………………………………………………

………………………………………………………………………………

13


Please check your answers below:

Answers:

1.11 Minimum load that will not exceed during design life.

1.12 a) Dead load

b) Imposed load

c) Wind load

1.13 BS 6399 : Part 1

1.14 CP3 Chapter 5 : Part 2

14

FEEDBACK 1b


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:

15

Probability distribution Characteristic

strengthMean strength

INPUT 3


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/mm 2

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.

16

Strength

Area = 0.05

Figure 1.2: Characteristic Strength


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.

Design strength is defined as;

Design strength =

=

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;

1.15 Since ultimate limit state is more severe, so the safety factors are much bigger

than those for serviceability limit state.

1.16 Safety factor for steel reinforcement is less than for concrete because steel

produced is of a stringent quality and it is controlled.

1.17 Safety factor for imposed load is more than for dead load because dead load

could be estimated more accurately that imposed load

17



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?

………………………………………………………………………………

………………………………………………………………………………

………………………………………………………………………………

18

ACTIVITY 1c


1.17 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?

………………………………………………………………………………

………………………………………………………………………………

19


………………………………………………………………………………

………………………………………………………………………………

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

………………………………………………………………………………

………………………………………………………………………………

………………………………………………………………………………

………………………………………………………………………………

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 =

1.19 γm = 1.50 (From Table 2.2, BS 8110)

20

FEEDBACK 1c


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.

21

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


4. The design process includes the preparation of the layout plan, structural

analysis, load calculations, structural design and the preparation of

working drawings.

5. 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.

22

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.


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

A. Safety

B. Cost

C. Maintenance

D. Aesthetic

2. Structural concrete should have the following EXCEPT….

A. durable

23

SELF-ASSESSMENT


B. commercial value

C. resistance to misuse

D. fire resistant

3. In the appropriate overall structural scheme, the engineer is guided by the

following codes of practice EXCEPT

A. ISO 9000

B. BS 8110

C. BS 6399

D. CP 3

4. Which of the following failures occur when Ultimate Limit State in

exceeded?

A. Deflection

B. Vibration

C. Collapse

D. Cracking

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

answers.

A. Plastic State Design

B. Conservative Design

24


C. Limit State Design

D. Simple Design

6. Given that fm = 35 N/mm2

σ = 3.5 N/mm2

What is the characteristic strength of concrete?

A. 0.29 N/mm2

B. 2.9 N/mm2

C. 29.0 N/mm2

D. 290.0 N/mm2

7. Given that γf = 1.6 and characteristics load = 58 kN

What is the design load?

A. 928 kN

25


B. 0.928 kN

C. 9.28 kN

D. 92.8 kN

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

equal to...

A. 1.50

B. 1.25

C. 1.40

D. 1.30

9. 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

10. 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

26


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

27

FEEDBACK ON SELF-ASSESSMENT


10. B

28

END OF UNIT 1

Documents

Unit 1 ( ULTIMATE LIMITS STATE DESIGN CONCEPT ) -REINFORCED CONCRETE STRUCTURAL DESIGN