unit 5 ( design of flanged beam: t-beam )

REINFORCED CONCRETE STRUCTURAL DESIGN C4301/UNIT5/

UNIT 5

DESIGN OF FLANGED BEAM: T-BEAM

GENERAL OBJECTIVE

To be able to design reinforced concrete flanged beam in accordance with BS 8110.

SPECIFIC OBJECTIVES:

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

1. calculate the effective width of the flanged beam.

2. calculate the moment of resistance of the flanged beam.

3. calculate the area of reinforcement required for the flanged beam.

4. provide the size and total number of reinforcement bars required for the

flanged beam.

1

OBJECTIVES


5.1 Introduction .

In the construction of reinforced concrete structural element, fresh concrete is

normally poured into the slab and the beam’s form work simultaneously. When

concrete hardens, a homogeneous concrete is formed. This will enable part of the

slab to act as part of the beam. In practice, the flange is often the floor slab and this

is shown in Figure 5.1 below: -

2

T-Beam L

Figure 5.1: Floor Slab

INPUT 1

Actual width

Slab

L-Beam L cc


5.2 Effective width, b

The question arises of what width of the slab is to be taken as the effective width?

The answer is, the width b in Figure 5.2

Clause 3.4.1.5 BS 8110 gives the following recommendation for a T-beam, the

effective width b should be taken as: -

bw + 0.2 z or

The actual flange width or whichever is lesser.

z is the distance between points of zero moment along the span of the beam. For a

continuous beam (we shall discuss this type of beam in Unit 6), z may be

determined from the bending moment diagram, but BS 8110 states that z be taken

as 0.7 times the effective span.

NOW ANSWER THE FOLLOWING QUESTIONS: -

3

b

bw

Figure 5.2: T-Beam


Given the floor plan showing the arrangement of T-beams, calculate the effective

width of the beam.

4

T-Beam T-Beam T-Beam

2.0 m 2.0 m

9.0 m

bw = 250mm

SectionPlan

Figure 5.3: Arrangement of T-Beam

ACTIVITY 5a


You should get the answer as shown below.

= 2050 mm

Actual flange width = 2000mm

The effective width, b is the lesser of these values i.e.

b = 2000 mm

5

FEEDBACK 5a


5.3 Analysis of T-section

There are three cases to be considered. They are as follows: -

5.3.1 Neutral axis in the flange

5.3.2 Neutral axis under the flange

5.3.3 Neutral axis outside the flange

5.4 Neutral axis in the flanged.

This happens when the depth of the stress block is less than the flange thickness(h f).

This is shown in Figure 5.4 below: -

Moment of resistance the section is given by,

6

Fcc=0.405fcubx

As

b

d

bw

0.9x

Z=d – 0.45x

Fst=0.87fyAs

Figure 5.4: Stress Block

INPUT 2


M = Fcc . Z

= 0.405 fcub (0.9x) (d-0.45x)

In this case the maximum depth of the stress block which is equal to 0.9x is

equivalent to hf. Therefore,

M = Mf =

(Where Mf is known as the ultimate moment of resistance of the flange)

It can be shown that when the applied moment, M is lesser or equal to M f, the

neutral axis is located in the flange. In this case the design of the section is similar

to that of the rectangular section which has been discussed in Unit 4.

5.4.1 Design Example (N.A in the flange)

A T-Beam is to carry a design moment (M = 165 kNm). The beam section is shown

in Figure 5.5. If concrete of grade 30 and steel of grade 460 are used, calculate the

area and the total number of bars required.

7

b =1450

d =320

bw=250Figure 5.5: T-Beam


Solution:

Resistance moment of flange:

= 0.45 x 30 x 1450 x 100 (320 – )

=528.5 kNm

Since M< Mf, the neutral axis is located in the flange.

= 0.036 < K’ (=0.156)

8

hf=100 mm


= 0.957d > 0.95d

= 1356 mm 2

Provided 3T25 (As=1470 mm2)

9

ACTIVITY 5b


For the T-Beam shown in Figure 5.6, calculate the area and number of bars required

to carry a design moment of 450 kNm. Use fcu = 30N/mm2and fy=460 N / mm2

Check your answers now!

10

800 mm

hf=150

bw=200

d=420mm

Figure 5.6: T-Beam

FEEDBACK 5b


Mf = 0.45 x 30 x 800 x 150 x (420 - )

= 558.9 kNm

M < Mf

Therefore, this shows that neutral axis is located in the flange.

= 0.106 < K’

Only tension reinforcement is needed.

= 0.86d < 0.95d

= 3113 mm 2

Provided: 4T32 (As = 3218 mm2)

11


Is your answer similar to the one given here? If your answer is ‘Yes’:

Congratulations! Go on to the next input. If your answer is ‘No’, you should go

through the input once again.

5.5 N.A Located Under the Flange.

12

INPUT 3


When the applied moment, M is greater than the moment of resistance, Mf, the

neutral axis will be located under the flange as shown in Figure 5.7.

Referring to Figure 5.7,

The Internal forces are: -

Fcc1 = (0.45fcu)(bw . 0.9x) = 0.405fcubwx

Fcc2 = (0.405fcu)( b- bw )hf

Fst = 0.87fyAs

Lever Arm:

Z1 = d -0.45x

Z2 = d – 0.5hf

Moment of resistance is given as follows: -

M = Fcc1 . Z1 + Fcc2 . Z2

13

2 21

AS

Fst

Z1Z2

Fcc2

Fcc1

Figure 5.7: N.A Located Under the Flange.


= (0.405fcubwx)(d - 0.45x) + (0.45fcu)(b - bw)hf(d - 0.5hf)

The ultimate moment of resistance of the section (when x = 0.5d) is given by,

Muf = 0.405fcubw0.5d [d - 0.45(0.5d)] + (0.45fcu)(b - bw)hf(d - 0.5hf)

= 0.156fcubwd2 + (0.45fcu)(b-bw) hf (d-0.5hf)

When the applied moment, M < Muf, compression reinforcement is not required.

The area of the tension reinforcement is calculated as shown below: -

Taking the moment of Fcc2,

M = Fst . Z2 - Fcc1. (Z2 - Z1)

= 0.87fyAs(d - 0.5hf) - 0.405fcubw x [(d - 0.5hf)-(d - 0.45x)]

As =

Equating x = 0.5d (at ultimate limit)

Therefore,

This is equation 1, BS 8110 and should be used when hf < 0.45d.

14


5.5.1 Design example (N.A under the flange )

(Use concrete grade 30 and steel grade 460). T-Beam as shown in Figure 5.6 above

is subjected to a design moment of 550kNm. Calculate the area of reinforcement

required.

Solution:

Mf = 528.5 kNm. (Calculated before in 5.5)

Mf < M (=550 kNm)

Muf = 0.156fcubwd2 + (0.45fcu)(b - bw)hf (d-0.5hf)

15

hf=100

d=320

b=1450

bw=250

Figure 5.8: T-Beam


=0.156(30)(250)(320)2 + (0.45)(30)(1450 – 250)(100)(320 – 50)

=552.3 kNm > M

This shows that, only tension reinforcement is required and is calculated as

follows:-

= 5188 mm 2

Use: 3T40 + 3T25 (5240 mm2)

The approximate arrangement of these bars is shown below: -`

16

3T25

3T40Figure 5.9: T-Beam

REINFORCED CONCRETE STRUCTURAL DESIGN C4301/UNIT5/17

Now do the following exercises

ACTIVITY 5c


Calculate the area of reinforcement required for the beam section given below if the

bending moment is 2750 kNm. Use fcu = 30 N/mm2 and fy = 460 N/mm2.

18

b = 1600 mm

bw = 300

d = 700

hf = 250 mm

Figure 5.10: T-Beam

Please do your calculation here!!!

REINFORCED CONCRETE STRUCTURAL DESIGN C4301/UNIT5/19

FEEDBACK 5c


Here are the answers. Please check it out!

= 582.2 kNm.

M > Mf. (N.A. under the flange)

= 687.96 + 2522.8

= 3210.8 kNm

Muf > M (=2750 kNm). This indicates that only tension reinforcement is required

Check that hf < 0.45d

0.45d = 0.45(700)

=315 mm

hf < 0.45d and we are going to use equation 1, BS 8110 shown below:

To calculate the area of reinforcement required.

20


= 12128.5 mm 2

Use: 7T40 + 7T32 (As = 12238 mm2)

21

INPUT 4


5.6 Neutral Axis Outside The Flange

When the applied moment M > Muf, the compression reinforcement should be

provided so that the depth of neutral axis, does not exceed 0.5d.

The stress distribution for a flanged beam provided with compression reinforcement

is shown in Figure 5.11 below:-

Referring to Figure 5.11, the forces in the stress block are as follows: -

Fcc1 = 0.405fcubwx

Fcc2 = 0.45fcu(b - bw)hf

Fst = 0.87fyAs

Fsc = 0.87fyAs’

The lever arm is: -

Z1 = d - 0.45x

Z2 = d - 0.5hf

Z3 = d – d’

22

Fsc

Fcc2

Fcc1

Fst

Figure 5.11: The stress distribution for a flanged beam


Taking moments about the tension reinforcement, we have:

M = Fsc . Z3 + Fcc1 . Z1 + Fcc2 . Z2

= 0.87fyAs’(d -d’)+(0.405fcubwx)(d - 0.45x) + (0.45fcu)(b - bw)hf (d - 0.5hf)

At ultimate limit state, x = 0.5d,

M = 0.87fyAs’(d - d’) + Muf

Therefore, As’ =

For Equilibrium of the forces: -

Fst = Fcc1 + Fcc2 + Fsc

0.87fyAs = 0.405fcubw( )+ 0.45fcu(b -bw)hf + 0.87fyAs’

Therefore, As = [0.2fcubwd + 0.45fcuhf (b - bw)+ 0.87

“This is a very important formula for you to memorize. Good Luck! ”

Design Example (N.A outside the Flange)

Given the beam section and data, calculate the area of reinforcement required so

that beam can carry a design moment of 650 kNm. Assume that d’ = 50 mm

23

fcu=30 N/mm2

fy = 460 N/mm2

Figure 5.12

d = 320

b = 1450

hf = 100

bw = 250


Solution:

Mf = 528.5 kNm < M (=650 kNm)

Neutral axis located outside the flange.

Muf = 552.3 kNm < M (=650kNm)

Compression reinforcement is required.

= 904 mm 2

= 5247 + 904

= 6151 mm 2

24


Use: 3T20 (As = 943 mm2) for compression reinforcement

5T40 (As = 6286 mm2) for tension reinforcement.

Calculate the reinforcement required for a flanged beam with the following

information:

25

Now you should be able to calculate the tension and compression reinforcement for flanged beam.

Now do the following exercise and then compare your answers with the answers given in the feedback section!.....

ACTIVITY 5d

Please do your calculation here!


Compare your answers below: -

26

M = 1700 kNm

fcu = 30 N/mm2

fy = 460 30 N/mm2

bw = 300 mm

d = 618 mm

d’ = 60 mm (embedment of compression reinforcement)

b = 1200 mm

hf = 150 mm

FEEDBACK 5d


=1319.5 kNm

Mf < M (=1700)

=1525.8 kNm

Muf < M (=1700 kNm)


=780 mm 2

= 3590 mm 2

Use: 7T12 (As = 792 mm2) as compression reinforcement

5T25 + 4T20 (As = 3712 mm2) as the tension reinforcement

27


To design flanged beam, the following procedure can be used: -

Calculate Mf = 0.45fcubhf(d - 0.5hf)

Compare Mf with M (applied moment)

If M<Mf, neutral axis is located in the flange. Calculate: -

28

Go through the summary of Unit 5 now

SUMMARY

Flip through Unit 5 once again and when you are ready, do the following Self-Assessment. Read the instructions before you start! Good Luck!


If M > Mf, neutral axis is located at outside of flange. Calculate:

f from Table 3.7 or equation 2 BS8110

Mu f = ffcubd2

Compare M with Muf

Check that hf < 0.45d

If M < Muf, compression reinforcement is not required. Calculate:

If M > Muf, compression reinforcement is required. Calculate:

29


There are five questions within this test. Marks are allocated for each question. You

have to answer all the questions. You should use your own writing paper and be

ready with your calculator.

Questions:

30

SELF-ASSESSMENT


1. The figure given below shows section of a beam and floor slab of a building.

Calculate the effective width of beam A, B and C assuming they are simply

supported. The beam span is 8.0 m.

2. For the given beam section below, calculate the area of reinforcement

required to carry a design moment of 600 kNm. Use fcu = 30 N/mm2 and fy =

460 N/mm2.

31

A

250 250 250

3500 3500 500

500

BC

(3 marks)

1450

d = 330

250

125


3. For the given beam section below, calculate the area of reinforcement

required to carry an ultimate design moment of 600 kNm. Use fcu=30N/mm2,

fy= 460 N/mm2 and d’ = 40 mm.

4. Calculate moment of resistance of the given beam section. The characteristic

material strengths are 25 N/mm2 for concrete and 460 N/mm2 for steel

reinforcement.

32

(5 marks)

500

100

300

200

(9 marks)

(2 marks)

2T25

2T32

1000

d = 500

100

250


5. Using the same characteristic material strength in question 4, calculate the

moment of resistance of the beam section given below. Note that this is an

edge beam or known as L- Beam.

1. For beam A , the section is shown below;

33

500

b

250

Beam A

2T16

4T32

d’=40

600

100

d =400

(2 marks)250

FEEDBACK ON SELF-ASSESSMENT


……………………………………………………….

For beam B:

…………………………………………………………

For beam C:

34

250

250

b

b

Beam B

Beam C

1

1

1


…………………………………………………….

2. Resistance moment of flange,

…………………………………………………..

M = 600 kNm < Mf………………………………………………………………………..

The neutral axis is in the flange.

= 0.127……………………………………………………………..

= 0.83d > 0.95d…………………………………………………

35

1

1

1

1


= 4782 mm 2 …………………………………………………….

Use 6T32 (As = 4827 mm2)

3.

= 202.5 kNm……………………………………………….

M = 600 kNm > Mf

Neutral axis is outside the flange.

= 0.0661 + 0.06

= 0.13……………………………………………………………..

Muf = 0.13 (30) (500) (350)2

= 238.9 kNm……………………………………………………..

M = 600 kNm > Muf ………………………………………………..


hf = 100 mm

0.45 d = 0.45 (350)

= 157.5 mm………………………………………………….

hf < 0.45 d…………………………………………………………

36

1

1

1

1

1

1

1


= 2911 mm 2 …………………………………………………….

Use: 6T25 (As’ = 2946 mm2) as the compression reinforcement.

= 1555 + 2911

= 4466 mm 2 ……………………………………………………

Use: 6T32 (As = 4827 mm2) as the tension reinforcement………….

The approximate arrangement of the reinforcement is shown below: -

4. fcu = 25 N/mm2

fy = 460 N/mm2

2T25 As = 982 mm2

2T32 As = 1609 mm2

37

6T25

6T32

1

1

1


Total As = 2591 mm2

= 243.75 + 379.96

= 623.44 kNm………………………………………………….

Therefore, the resistance moment is 623.44 kNm.

5. Muf

= 156 + 137.8

38

4T32

2T16

600

d = 400

100

250

2

1


= 293.8 kNm………………………………………………………..

= 351.72 kNm…………………………………………………..

Therefore, the moment of resistance of the section is 351.7 kNm.

Now calculate your score as shown below: -

Score = Total marks obtained x 100 % 21

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

39

1


If your score is less than 80%, you should go through all or part of

this unit once again.

Now you have completed unit 5.

You may proceed to unit 6.

END OF UNIT 5

GLOSSARY

ENGLISH MALAY

flanged beam rasuk berbibir

web web

effective width lebar berkesan

neutral axis paksi nutral

stress block bungkah tegasan

40


resistance moment momen rintangan

stress distribution taburan tegasan

41

unit 5 ( design of flanged beam: t-beam )

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