Homework 8 (Due June 20, 2007)

— David Wagner 2007/06/17 13:59

Problem 1.

The T-beam and one-way-slab system shown carries a service live load of 150 lb/ft2 and a service dead load of 110 lb/ft2. The dead load includes the weight of the slab, partitions and utilities. The T-beams are simply supported at each end and span 30 ft on centers. The beam stem is 12 in wide. The brick walls are also 12 in wide. Determine the moments at A, B, C, D, E, F, G, H, and I. Assume f’c = 3000 psi, and fy = 60,000psi

LL»DL → U=1.2D + 1.6L → w_u = 1.2*110 + 1.6*150 = 370 lb/ft

The masonry supports are simple, bearing at midpoint. The figure is laterally symmetric about the center.

For the end spans, l_n=14-0.5= 13.5'

The end supports are considered simple.

0

$M_B=M_H={w_u {l_n}^2}/11 = {370 {13.5}^2}/11 =$ +6.13 kip-ft

+6.13 kip-ft

For the negative moments at interior supports, $l_n={13.5+15}/2=$ 14.25'

$M_C=M_G={w_u {l_n}^2}/10 = {370 {14.25}^2}/10 =$ -7.51 kip-ft

-7.51 kip-ft

$M_D=M_F={w_u {l_n}^2}/11 = {370 {14.25}^2}/11 =$ -6.83 kip-ft

-6.83 kip-ft

For the moment at the center l_n=16-1= 15'

$M_E={w_u {l_n}^2}/16 = M_E={370 {15}^2}/16 =$ +5.20 kip-ft

+5.20 kip-ft

Problem 2.

For the beam shown, compute the design moment capacity. The span length of the beam is 30 ft and the beams are centered 12 ft apart. Tension reinforcement is 10-#11 bars (double layer with an effective depth of 34.5 in, vertical spacing between bars equals 1 in). Assume f’c = 3000 psi and fy = 60,000 psi.

L=30'=360 in, L_cc=12'=144 in, h=38 in, h_f=3 in, d=34.5 in, b_w=18 in, A_s=15.6 in²

$d_t=34.5+0.5+{11/16}=$ 35.7 in

$beta_{1,3000}=0.85+0.05=$ 0.90

$A_{s,min}=larger[{3 sqrt{f prime_c}b_w d}/f_y ,{200*b_w d}/f_y] = larger[{3 sqrt{f 3000}18 *34.5}/60000 ,{200*18*34.5}/60000]$ = larger[1.70,2.07]= 2.07 < 15.6 ✔

$b_e=smallest[L/4,b_w+16*h_f,L_{cc}] = smallest[360/4,18+16*3,144] = smallest[90,66,144]$ = 66 in

$A_{s,max}={0.85f prime_c beta_1 b_e 3 d_t}/{7 fy} + {0.85 f prime_c(b_e-b_w)h_f}/{fy}=$ {0.85*3*0.9*66*3*35.7}/{7*60} + {0.85*3(66-18)*3}/{60} = 38.625+6.120= 44.745 > 15.6 ✔

$a={A_s f_y}/{0.85f prime_c b_e}={15.6*60}/{0.85*3*66}=$ 5.5615 in > 3

C_2=0.85 f prime_c h_f (b_e-b_w)=0.85*3*3*(66-18)= 367.20

$a={A_s f_y - C_2}/{0.85f prime_c b_w}={15.6*60 - 367.2}/{0.85*3*18}=$ 12.392

C_1=0.85 f prime_c a b_w=0.85*3*12.392*18= 568.80

$epsilon_t={0.003*(0.9*35.7-12.392)}/12.392=$ 0.0047784 < 0.005

phi = 0.48+83*0.0047784= 0.87661

$phi M_n=phi [C_2(d-{h_f/2}) + C_1(d-{a/2})]=0.87661*[367.2*(34.5-{3/2}) + 568.8(34.5-{12.392/2})]=$ =0.87661*[12118+16099]= 24735

φM_n=24735 kip-in

Problem 3.

Solve problem 5.23 in your textbook

l_span=32'=384 in, l_cc=12'=144 in, b_w=15 in, d=d_t=24 in, h_f=4 in, beta_1,4000= 0.85.

Design for ε_t > 0.005 and φ=0.9.

h approx 24+0.25+0.75 approx 25 in

$w_D=100*12+150*{12}*{4/12}+150*{15/12}*{{25-4}/12}=1200+600+328=$ 2.128 kip/ft

w_L=80*12= 0.96 kip/ft

w_u=1.4w_D=1.4*2.128= ~~2.979 kip/ft~~

w_u=1.2w_d+1.6w_L=1.2*2.128 + 1.6*0.96= 4.090 kip/ft controls

Assume simple support.

$M_u = {w_u {l_{span}}^2}/{8} = {4.090* {32}^2}/8=$ 523.5 kip-ft = 6282 kip-in

M_n=M_u/phi=xxx/0.9= 581.6 kip-ft = 6980 kip-in

$b_e=smallest[l{span}/4,b_w+16h_f,l_{cc}] = smallest[384/4,15+16*4,144] = smallest[96,79,144]=$ 79 in

Try assuming a ≤ h_f.

$6980 = 0.85*4*a*79*(24-{a/2}) = 268.6*24a-{268.6/2}a^2 = 6446.4a -134.3a^2$ right a^2 -48a +52=0 $right a={48 pm sqrt{48^2-4*52}}/2 =24 pm 22.892=a=$ 1.1083 < 4 ✔