Question

reinforced concrete design

Problem 2: Design of T-beam Calculate the reinforcing required for the T-beam system below 22 30 15 15 10'0 10'0 10 0 1 of 2 live load is 400 lbf/ft2 and use the dimensions shown in the figure. (a) Calculate the maximum factored moment (note: remember to include the self-weight of slab and beams) (b) Determine the effective flange width (c) Check if a

Answer 1

Given:

Span=10 ft

h_f=8 in

d=22 in

L.L=400 lb/ft²

Self weight of slab=(8 in/12x150) lb/ft³

D.L of slab=100 psf

effective flange width=span/4=60 in(assume it is least)

Area of beam=8*60+22*15=810 in²

D.L of beam=810x150/144=843.75 psf

(a)

W_u=1.2(D.L)+1.6(L.L)

W_u=1.2(843.75)+1.6(400)

W_u=1652.5 psf

M=WL²/8

M=1.6525x(20)²/8

M=82.625 ft-kips

(b)

Effective flange width

Span/4=20x12/4=60 in

16h_f+b_w=16x8+15=143 in

spacing=120 in

The effective flange width is 60 in

(c)

M_n=M/($\phi$)

M_n=82.625/0.9=91.8 ft-kips

Assume a lever arm of z=0.9d or z=d-hf/2

z=0.90x22=19.8 or z=22-8/2=18 in

AsFyZ=M_n

A_s=91.8x12/(60x19.8)

A_s=0.92 in^2, which is less than the minimum area of steel

Provide 2 no of #10 bars of area 2.53 in²

equate the tension to compression

0.85f_cA_c=A_sFy

0.85x4xA_c=2.53x60

A_c=44.64<240 in²

a=44.64/60

a=0.74<h_f

The minimum steel is 1.1 sq in providing 2.53 sq in of #10 no of 2 bars will be enough and a<h_f

(e)

c=0.74/0.85

c=0.87 in

$\epsilon$_t=0.072>0.005

The reduction factor is 0.9

(f)

Assume the cover of concrete to the T beam as 1.5 in. select a #4 diameter stirrup

Check the spacing

S=[b_w-2(clear cover)-0.5(diameter of stirrup)-0.5(diameter of bar)]/2

S=[15-2(1.5)-0.5(0.5)-0.5(1.25)]/2

S=5.5 in

Sketch:

Sit 1、5 いSin

(g)

Moment capacity=0.90TZ

=0.90AsFy(d-a/2)

=0.90x2.53x60x(22-0.74/2)/12

=246.25 ft-kips>91.1 ft kips, this is safe