Design TBeam

7/30/2019 Design TBeam

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Analysis and design of T-beam

Effective Flange Width be

Portions near the webs are more highly stressed than areas away fromthe web.


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ACI Code Provisions for Estimating beff

According to ACI code, effective flange width of a T-beam, beis notto exceed the smallest of: One-fourth the span length of the beam, L/4. Width of web plus 16 times slab thickness, bw+16 hf .

Center-to-center spacing of beams, bactual.


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ACI Code Provisions for Estimating beff

According to ACI code, effective flange width of a L-beam, beis notto exceed the smallest of: bw+ L/12. bw+ 6 hf .

bw+ 0.5x(clear distance to next web).

For isolated T-beams, the flange thickness is not to exceed half the webwidth, bw/2 , and the effective flange width be not more than fourtimes the web width, 4 bw.


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T- versus Rectangular Sections

Compression zoneTension

zone

Sec A-A Sec B-B

When T-shaped sections are subjected to negative moments, the flange

is located in the tension zone. Since concrete strength in tension is

usually neglected in ultimate strength design, the sections are treated asrectangular sections of width bw and when sections are subjected to

positive moments, the flange is located in the compression zone and

the section is treated as a T-section.


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Strength of T-Beam

effc

ys

yseffc

b'f0.85

fA

aTCmequilibriuFrom

fAT&ba'f0.85C

Case 1:when a hf [Same as rectangular section]Assume steel yields

1] Equilibrium


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2

adfAM ysn

004.0003.0c

cd

ac

t

1

Case 1:when a hf

2] Confirm

3] Calculate Mn

Strength of T-Beam


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wc

fwcys

wf

ys

wcw

fwcf

bf0.85

hbbf0.85-fA

CC

fATabf0.85C

hbbf0.85C

a

T

Case 2:when a > hf Assume steel yields

1] Equilibrium

From equilibrium of forces

Strength of T-Beam


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004.0003.0c

cd

ac

t

1

2

hdC

2

adCM ffwn

Case 1:when a > hf

2] Confirm

3] Calculate Mn

Strength of T-Beam


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Minimum Reinforcement, As,min

dbf

14.06

dbf

f8.0

oflargerA

wy

w

y

c

s(min)

bw

dhf

As

beff

bw

d

hfAs

beff

+veMoment

-veMoment

dbf

14.06db

f

f0.8

dbf

f1.6

ofsmallerA

eff

y

eff

y

c

wy

c

s(min)

Flange in compression

Flange in tension


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Flange Reinforcement

Main Reinforcement

min (beff & l/10)Additional

ReinforcementAdditional

Reinforcement

-vemoment

when flanges of T-beams are in tension, part of the flexuralreinforcement shall be distributed over effective flange width, or a

width equal to one-tenth of the span, whichever is smaller

If beff > l/10, some longitudinal reinforcement shall be provided inouter portions of flange.


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Analysis of T-Beam

'f0.85

fAATCmequilibriuFrom

c

ys

c

bw

dhf

beff

Af

1- Check As,used >As,min

2- Compute T = Asfy

3- Determine the area of the concrete in compression (Ac) stressed to

0.85fc

If Ac Af = beff x hf [a < hf ]

If Ac > Af = beff x hf [a > hf ]

4- Calculate a, c, and checkt (t 0.004; < max)

5- Calculate Mn.


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Example 1

Example Problems

2

useds,

2

mins,

mins,

w

y

w

y

c

mins,

cm42.41(7.07)6Acm5.08A

60.725420014.0660.7525

42002800.8A

dbf

14.06db

f

'f0.8A

cm60.752

2.531.0470d

25

630

10

150

70

10

Determine the ACI design moment strength Md (Mn )of the T-beamshown in figure if fc =280 kg/cm

2 and fy= 4200 kg/cm2.

Solution:-

1- Checking min. Steel Percentage


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Example 1

Example Problems

25

630

10

150

7

0

10

ton178.1210

420042.41fAT

3ys

cm10cm0.535.7

178.12a

TCforcesofmequilibriufrom

tona7.5310

a1500.85(280)ba'0.85fC3c

2- Compute T and a

assuming that a < hf= 10cm

i.e. assumption is rightSame as rectangular section


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Example 1

Example Problems

25

630

St.10

150

7

0

10

t.m93.382

5.060.75

10

178.120.9

2

a

dTM

0.90.0050.028

0.0035.88

5.8860.750.003

c

cd

cm5.880.85

5.0

ac

2

d

t

t

1

3- Calculate Mn


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Example 2

Example Problems

2

useds,

2

mins,

mins,

w

y

w

y

c

mins,

cm34.46(8.04)8Acm6.6A

55.6503420014.0655.6503

42002500.8A

dbf

14.06db

f

'f0.8A

cm55.652

2.53.21.0475d

30

832

1075

10

90

Determine the ACI design moment strength Md (Mn )of the T-beamshown in figure if fc =250 kg/cm

2 and fy= 4200 kg/cm2.

Solution:-

1- Checking min. Steel Percentage


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Example 2

Example Problems

ton23.70210

420034.46fAT

3ys

cm10cm13.1419.13

270.23a

TCforcesofmequilibriufrom

tona13.1910

a090.85(250)ba'0.85fC3c

30

832

1075

10

90

abf0.85C

hbbf0.85C

wcw

fwecf

2- Compute T and a

assuming that a < hf= 10cm

i.e. assumption is wrong


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Example 2

Example Problems

cmaa

C

a

f

39.2238.65.12723.270

CTforcesofmequilibriufrom

ton38.610

a032500.85ba'f0.85C

ton127.510

1030902500.85h)b(b'f0.85C

w

3wcw

3fwecf

855.0)00447.0(3.83483.0

3.83483.00.0040.00447

0.00326.34

34.6255.650.003c

cd

cm34.620.85

22.39

ac

tt

t

1

3- Calculate and Md


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Example 2

Example Problems

mt

hdC

adC

tonaC

f

fw

w

.34.132

2

1055.655.127

2

39.2255.6574.142

10

855.0

22M

74.14239.2238.638.6

2

d

3- Calculate Mn


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Design of T-Beam

bw

dhf

As

beff

2

ef fc

u

5

y

c

db'f0.85

M2(10)11

f

'f0.85

effc

ys

b'f0.85fAa

The design of a T-section involves the determination of 5 unknowns;be , hf , bw , h , and As.

1- Establish hfbased on flexural requirements of the slab.

2- Determine be according to ACI limits.

3- Choose bwand d4- CalculateAs assuming that a < hfwith beam width = beff & =0.90

As = beffd

5- If a < hfthe assumption is right & if not reviseAs using T-beam equations.

6- Check to ensure that t0.004 and min


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Example Problems

Example 1

Spandrel

beam

10

.0m

3.0 m

3.0 m

3.0 m

hf Slab

bw

A floor system consists of a 14.0cmconcrete slab supported by continuous

T-beams with a 10m span, 3.0m on

centers. Web dimensions are bw=30cm

and d=55cm.

Use fc =280 kg/cm2 and fy = 4200 kg/cm2

Requireda] What tensile steel is required at midspanto resist a service dead load moment 32t.mand a service live load moment 25t.m.

b] What tensile steel is required at supportto resist a factored moment 50t.m


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Example 1

Example Problems

30

5514

As

200

2

ef fc

u

5

y

c

db'f0.85

M2(10)11

f

'f0.85

55

30

As

14

200

78.4

t.m

beff is the smallest of:- l/4 = 800/4 = 200 cm

- bw+16(hf) = 30 +16 (14) = 254 cm

- Center-to-center-spacing of beams = 300 cm

beffis taken as 200 cm, as shown in Figurea] for positive moment

Mu+ve = 1.2(32)+1.6(25)=78.4 t.m

Assuming that a


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Example 1

Example Problems

2

2

5

93.385520000354.0

00354.0

550028020.859.0

4.8710211

4200

2800.85

cmdbA effs

OKcm39.27Acm5.52A

55304200

14.065530

4200

2800.8Adb

f

14.06db

f

'f0.8A

14cmhcm3.472002800.85

420039.27

b'f0.85

fAa

2

useds,

2

mins,

mins,w

y

w

y

c

mins,

f

effc

ys

30

5514

As

200

78.4

t.m

Use 825mm (As,used= 39.27 cm2) arranged in two layers.

The assumption is right


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Example 1

Example Problems

OK9.0005.00374.0

0.0034.08

08.4550.003

c

cd

cm08.40.85

3.47

ac

t

1

t.m78.4Mt.m79.072

3.47

5510

420039.270.9

2

adfAM

u5

ysd

55

308

25

14

200

Check=0.9


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Example 1

Example Problems

2wc

u

5

y

c

db'f0.85

M2(10)11

f

'f0.85

30

55As

50t.m

2

2

5

35.2855300172.0

0172.055308020.859.0

5010211

4200

2800.85

cmdbA ws

b] for negative momentMu-ve= 50 t.m

Rectangular section with b=bw

Mu= 50 t.m, bw= 30cm, d=55cm and assume that =0.9

Use 625mm (As,used= 29.45 cm2) arranged in one layers.

Distribute the bars over a distance [min. (200, l/10=800/10=80)]=80cm


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Example 1

Example Problems

55

30

14

200

80

As

Check As,used > As,minFlange in tension

OK2useds,2

s(min)

2

2

s(min)

cm29.45Acm10.52A

cm36.82552004200

14.06

552004200

2800.8

cm10.5255304200

2801.6

A


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Example 1

Example Problems

t.m50Mt.m6.152

17.3355

10

420045.290.9

2adfAM

]004.0[

0.90.0050.0051

0.00320.38

38.02550.003c

cd

cm38.020.85

17.33

ac

33.173028085.0

420045.29

b'f0.85

fAa

u5

ysd

maxt

t

t

1

wc

ys

used

m

55

30

1

4

200

80

As

Design TBeam

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