unit 4 ( design of rectangular beam sections )

REINFORCED CONCRETE STRUCTURAL DESIGN C4301/UNIT4/

UNIT 4

DESIGN OF RECTANGULAR BEAM SECTIONS

GENERAL OBJECTIVE

To be able to provide steel reinforcement required in reinforced concrete beams.

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

1. calculate the area and total number of reinforcement required in given

beam sections with known bending moments.

2. calculate the bending moment that can be carried by the beam section

when the grade of steel is already provided.

1

OBJECTIVES

SPECIFIC OBJECTIVES


4.1 Simply Reinforced Rectangular Section

What does it mean?

Simply reinforced section is the steel reinforcement that is provided only at the

tension zone. The form of section, stress distribution and forces on a section are as

shown in Figure 4.1:

2

Z=d-0.45x

Fcc=0.405fcubx

INPUT 1

As

Section

0.87 fy

Stress

Fst=0.87fyAs

Forces

0.45x

0.9x

0.45fcu

x

b

d

Figure 4.1: The form of section, stress distribution and forces on a section


The meaning of symbols used are given in clause 3.4.4.3

For the section to be in balance, the sum of the forces must be equal to zero, i.e.

Fst = Fcc

0.87fyAs = 0.405fcubx

x =

Taking moments about Fcc or Fst, the moment of resistance of the section can be

calculated as follows,

M = Fst. Z

= (0.87fyAs) (d - 0.45x)

or,

M = Fcc. Z

= (0.405fcubx) (d – 0.45 x)

= (0.405 ) ( ) (fcubd2)

= K. fcu bd2

A section whereby the reinforcement is already known, the above equation can be

used to calculate the moment of resistance or the bending moment. Equation 4.1

shows that, x will increase if the area of reinforcement, As increases. The moment

of resistance of the section will also increase. It is clear that a section provided

with more reinforcement will be able to with stand reinforcement load i.e. greater

bending moment.

3


In order to make sure that concrete will not crush or fail when in compression or

when yielding of the steel occurs, BS 8110 limits the value of x to be not greater

than 0.5d.

Therefore, the maximum resistance moment of resistance of the section can be

obtained by equating x = 0.5d. This is shown as follows: -

M = Mu = ) (1- ) (fcubd2)

= 0.156fcubd2

= K’fcubd2

Where Mu is known as the ultimate moment of resistance for singly

reinforced section.

The equation above may be used to calculate the area of reinforcement

required if the bending moment is known.

One important conclusion is that if the bending moment, M applied is less

than Mu, the section requires tension reinforcement in the tension zone.

This is called singly reinforced section.

4.2 Example Simply Reinforced Rectangular Section

A rectangular beam section is required to carry a design moment of 300kNm. The

dimension of the area, b = 250 mm and d = 700 mm. Determine the area and total

number of reinforcement needed if the characteristic strength of reinforcement

used, fy=460N/mm2 and that of concrete, fcu = 40 N/mm2.

4


Solution:

Calculate the ultimate moment of resistance for singly reinforced section as

follows: -

Mu = 0.156fcubd2

= 0.156 (40) (250) (700)2

= 764.4kNm

This shows that Mu > M (=300kNm). Therefore this is a singly reinforced section

Calculate the depth to neutral axis as follows: -

M = 0.405fcubx (d - 0.45x)

300 x 106 = 0.405 (4) (250) (x) (700 - 0.45x)

x2 – 1560x + 16500 = 0

x = 114 mm @ 1441 mm

x is taken as 114mm (the reasonable value).

Calculate Lever arm, Z = d – 0.45x

= 700 – 0.45(114)

= 649 mm

Calculate the area of reinforcement required as follows: -

As =

=

= 1155mm 2

Use 4T20 (As = 1260mm2)

5


4.2.1Example

Calculate the moment of resistance for a beam section shown in Figure 4.1 below:

-

The section is already provided with 4T20 reinforcement whereby fcu and fy are

given as 40N/mm2 and 460 N/mm2 respectively.

Solution:

6

xd = 700Fcc = 0.405fcubx

Z = d – 0.45x

Fst = 0.87fyAs

Figure 4.2: Simply Reinforced Rectangular Section

b=250mm

d= 700 mm

250 mm

4T20

Figure 4.1: Beam Section


Calculate the depth to neutral axis as follows: -

Fcc = Fst

0.405fcubx = 0.87fyAs

x =

=

= 124 mm

Check,

Therefore concrete will not be crushed during compression.

Calculate the moment of resistance as shown below: -

M = Fst. Z

= 0.87 fy As (d - 0.45x)

= 0.87 (460) (1260) (700 – 0.45 (124))

= 325 kNm

Therefore, the moment of resistance of the beam section, provided with 4T20, is

325 kNm.

7


4.3 Doubly Reinforced Rectangular Section

For singly reinforced section, steel reinforcement is only used for resisting tensile

force. Compressive force is taken by concrete in the compressive zone. BS 8110

allows the maximum value of x to be not greater than 0.5d for the concrete stress

block taking the compressive force.

Based on this maximum limit x = 0.5d, the maximum moment that can be carried

by concrete is given by,

Mu = 0.156fcubd2

If the bending moment, M to be carried by the section is greater than Mu, steel

reinforcement is to be provided in the compression zone to resist the resultant

moment. The stress-strain distribution and the forces acting in a doubly reinforced

rectangular section is shown in Figure 4.3 below: -

8

INPUT 2

As

Fst

Fsc

Fcc

Z Z1

As’

x

d’

Section Strain Stress & Forces

Figure 4.3: The stress-strain distribution and the forces acting in a doubly reinforced rectangular section

d


The condition for equilibrium which must be obtained at any point in the section

is that horizontal forces must equate to zero; i.e.:

Fst = Fcc + Fsc

0.87fyAs = 0.405fcubx + 0.87fyAs’

x = ……..(4.2)

Taking moment (moment of resistance) about the tension steel, we have;

M = Fcc . Z + Fsc . Z1

= (0.405fcubx) (d - 0.45x) + (0.87 fyAs’). (d - d’)

= 0.156fcubd2 + 0.87fyAs’ (d - d’)

= Mu + 0.87fyAs’ (d - d’)

Therefore, area of compression reinforcement is:

As’ =

=

From equation 4.2, it can be shown that:

0.87fyAs = 0.405fcubx + 0.87fyAs’

0.87fyAs.Z = 0.405fcubx. Z + 0.87fyAs’ Z

When x = 0.5d, therefore z = 0.775d

Then,

0.87fyAs. Z = 0.405fcub (0.5d - 0.775d) + 0.87fyAs’. Z

0.87fyAs.Z = 0.156 fcubd2 + 0.87fyAs’. Z

9


Therefore the area of tension reinforcement is given by,

As =

Or As =

4.4 Stress in the compression reinforcement.

In deriving equation 4.3 above, it is assumed that the compression reinforcement

has yielded, i.e. it has achieved its design strength of 0.87fy.

From the strain design diagram, it can be shown that:

For the design strength to achieve the value 0.87fy, therefore;

And

For fy = 460 N/mm2,

= 0.43

10


This implies that for compression reinforcement to achieve its yield point, the

ratio must not exceed 0.43. If > 0.43, the compressive stress, fsc < 0.87 fy

and the value fsc can be obtained from Figure 2.2 BS 8110.

It can be shown that,

fsc = Es . εsc where εsc =

fsc = 200 x 103 x 0.0035 (

= 700

Now do activity 1

11


Answer the following questions.

4.1 What is the limit for depth to neutral axis, x as specified by BS 8110?

__________________________________________________________________

4.2 Write down the formula that is used to calculate the maximum moment that

can be carried by concrete.

__________________________________________________________________

__________________________________________________________________

4.3 What should be done if the applied moment is greater than the ultimate

moment, Mu?

__________________________________________________________________

__________________________________________________________________

4.4 Write down the formula that is used to calculate the area of compression

reinforcement.

__________________________________________________________________

__________________________________________________________________

12

ACTIVITY 4a


4.5 Write down the formula that is used to calculate the area of tension

reinforcement.

__________________________________________________________________

__________________________________________________________________

4.6 Write down the formula used to that is calculate fsc when is greater

than 0.43.

__________________________________________________________________

__________________________________________________________________

13


Check your answers below: -

4.1 ≠ 0.5d

4.2 Mu = 0.156fcubd2

4.3 Compression reinforcement should be provided.

4.4 As’ = or

As’ =

4.5 As = or As =

4.6 fsc = 700

14

FEEDBACK 4a


4.3 Design Example (Rectangular Section )

A rectangular section of reinforced concrete beam is to be designed to carry a

design moment of 880kNm. The width of this beam, b=250 mm and the depth,

d=700 mm. Calculate the area and the total number of bars needed if

fcu=40N/mm2 and fy = 460 N/mm2.

Solution:

Ultimate moment of singly reinforced section,

Mu = 0156fcubd2

= 0.156 x 40 x 250 x (700)2

= 764.4 kNm

Mu < M (= 880kNm). This shows that compression reinforcement is needed and

is calculated as follows: -

As’ =

Assuming that d’ = 60 mm,

As’ =

= 451 mm 2 .

15

INPUT 3


Now, we have to check the ratio. This is done as follows: -

x = 0.5d

= 0.5 x 700

= 350 mm

=

= 0.17 < 0.43

This shows that the assumed compression steel achieved its design strength

0.87fy.

The area of tension steel is then calculated as shown below: -

As =

=

= 3972 mm 2

16


Now we have to check from the table below to determine the total number of bars

and then the size.

Provided: 3T16 for compression reinforcement (As’ = 603 mm2)

5T32 for tension reinforcement (As = 4023 mm2)

17

Table 4.1: The Allowable Size and Numbers of Bars


4.7 Given the following information, calculate the area of reinforcement required

for a bending moment of 650 kNm

fcu = 30 N/mm2

fy = 460 N/mm2

b = 300 mm

d = 618 mm

d’ = 60 mm.

18

ACTIVITY 4b


4.7 Solution: -

Step 1

Mu = 0.156fcubd2

= 0.156 x 30 x 300 x (618)2

= 536.2 kNm < M (= 650 kNm)

Compression reinforcement is required.

Step 2

As’ =

=

=509.6 mm 2

Step 3

Check ratio: -

x = 0.5d = 0.5 (618) = 309 mm

= = 0.19 < 0.43

Compression steel has yielded.

19

FEEDBACK 4b


Step 4

As =

=

= 2797 mm 2

Step 5

Provide: 5T12 (As’ = 565 mm2) for compression reinforced

9T20 (As = 2828 mm2) for tension reinforced

20


4.4 Design example (Determine moment capacity of a section)

Calculate the moment of resistance of a rectangular section beam which has been

provided with 5T32 (tension steel) and 3T16 (compression steel). The

characteristic strengths are and .

The section is as shown in Figure 4.4 below:

21

INPUT 3

700 mm

250 mm

3T16

5T32

Figure 4.4: Rectangular Section Beam


Solution:

The stress and forces in the stress block is given below: -

Step:

1. For equilibrium of forces;

2. The depth to neutral axis is given by,

= 338.5 mm

3. Check the steel stress.

22

Figure 4.5: The Stress and Forces in the Stress Block

Fsc

Fcc

Fst

Z Z1=d-d’

x d


This shows that both the compression and tension steel have yielded at

0.87fy

4. Moment of resistance is calculated as shown below;

= 904 kNm.

Therefore the resistance moment of the section is 904 kNm.

Now, do activity 3

23

ACTIVITY 4c


1.8 Determine the moment of resistance for the given section: -

b = 400 mm2

As’ = 800 mm2

d’ = 40 mm

d = 800 mm

As = 5050mm2

fcu = 30N/mm2

fy = 460 N/mm2

24

FEEDBACK 3


4.2 Refer to the answer as shown below: -

Step 1: Equilibrium of forces

= 350 mm

Step 2: Check whether the steel has yielded.

Compression and tension steel have yielded.

Step 3: Calculate the moment of resistance.

25


= 15.87 + 1092.9 kNm

= 1108.77 kNm

26

SUMMARYSUMMARY


To calculate the area of reinforcement of a beam (having a size of b x d), with a

bending moment of M, using concrete strength of fcu and steel strength fy , you

should use the following steps: -

1. Calculate

2. Calculate

3. If K< K’, compression reinforcement is not required.

i) Calculate but z < 0.95d

ii) Calculate As =

4. If K > K’, compression reinforcement is required.

i) Calculate

ii) Calculate

iii) Check ratio.

iv) Calculate

if

27


or if

v) Calculate

28

SELF-ASSESSMENT


Test your understanding by solving the given problem.

1. Calculate the area, size and total number of bars required by the given

section shown in Figure 4.6. The section is to carry a design moment of

90kNm. Use concrete of grade 30 and high yield steel.

2. Calculate the area, size and total number of bars required for the given

section. The bending moment, fcu and fy to be used are the same as in

question 1. Use d’ = 50 mm.

29

200 mm

400 mm

(4 Marks)

350 mm

150 mm(7 Marks)

Figure 4.6: Rectangular Bar


3. The figure below shows a rectangular beam section which is to be designed

using concrete of grade 25 and high yield steel reinforcement. Calculate the

moment of resistance of the section.

4. Calculate the moment of resistance of the given section using concrete of

grade 25 and high yield steel reinforcement.

30

600 mm

300 mm

4T25

d=450mm

250mm

2T12

3T20 (4 Marks)

(3 Marks)


Check your answers below and then total up your marks. The marks allocated are

shown at every step:

(TOTAL 100 MARKS)

1.

31

FEEDBACK ON SELF-ASSESSMENT

1

1


= 0.094 ……………………. 1

Since, K < K’

Therefore compression reinforcement is not required.

= ……………………..

= 380 mm

…………………………..

Use 6T12 (As = 679 mm2)

2.

32

1

1

1


= 0.163 ………………………..

K > K’

Compression reinforcement is required.

=272 mm ………………………….

0.95d = 0.95(350)

=332.5 mm

Z < 0.95d

45.0

Zdx

45.0

272350

=173.3 mm

33

1

1

1

1


………………………….

……………………………

= 822 mm 2 ………………………………

Use : 2T8 (As’=101 mm2) as compression reinforcement . ………………

8T12 (As=905 mm2) as tension reinforcement ………………..

3. For equilibrium of forces ;

Fcc = Fst

0.405 fcu b x = 0.87 fy As

x =

=

34

1

1

1


= 258.8 mm

= 0.43 < 0.5

Moment of resistance,

M = Fst Z

= 0.87 x 460 x 1964 (600 – 0.45 x 258.8)

= 380.06 kNm

4. Fst = Fcc + Fsc

0.87 fy As = 0.405 fcu b x + 0.87 fy As’

x =

=

= 113.4 mm ……………………………

…………………………

35

1


…………...………………

Moment of resistance is,

M =

= {0.87 x 460 x 226 x (450 - 40)} + {0.405 x 25 x 250 x 113.4 x [(450-

(0.45x113.4)]}

151.6 kNm ……………………………….

Total = 18 marks

NOW TOTAL UP YOUR MARKS. FULL MARKS = 18 MARKS

Calculate your score as shown below:

Score = total marks obtained x 100% 18

36

You should score 80% or more to pass this unit.

If you have scored 80% or more, you may

proceed to unit 5.

If you have scored less than 80%, you should

go through unit 4 again until you pass this unit.See you in UNIT 5.

REINFORCED CONCRETE STRUCTURAL DESIGN C4301/UNIT4/37

END OF UNIT 4

unit 4 ( design of rectangular beam sections )

Documents

reinforced section

x d fcc

section figure

form of section

design moment

fcubx d

fyas d

reinforced concrete