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Homework #3 Due 6/11/07

— David Wagner 2007/06/10 18:17

For the beam cross-sections shown, calculate the nominal and design moments.

Problem 1a

Illustration for Problem 1a Assume #4 stirrups, fy =60,000 psi and f’c = 4,000 psi.

b_w=b=i8 in

A_s=5#8=3.95 in²

d_t=32-1.5-{4/8}-{8/{2*8}}=32-1.5+0.5+0.5=d_t=29.5 in

d_2=32-1.5-{4/8}-{8/8}-1-{8/{2*8}}=27.5 in

d={2.37 d_t + 1.58 d_2}/A_s = {(2.37*29.5)+(1.58*27.5)}/3.95 = {66.915+43.45}/3.95=110.365/3.95=d =27.94 in

a={A_s f_y}/{0.85 {f^prime}_c b}={3.95*60}/{0.85*4*18}= 237/61.2 =3.8725 in

beta_1_4000=0.85

epsilon_t = {0.003(beta_1 d_t - a)}/a ={0.003*(0.85*29.5-3.8725)}/3.8725={0.003*21.2025}/3.8725=0.0636075/3.8725=epsilon_t=0.0164 > 0.005 → φ=0.9

M_n=A_s f_y (d-{a/2}) = 3.95*60*(27.94-{3.8725/2})=3.95*60*26.004 =

M_n=6163 kip-ft

phi M_n= 0.9*6163=

phi M_n=5547 kip-ft

Problem 1b

Illustration for Problem 1b Assume #3 stirrups, fy = 60,000 psi and f’c = 5,500 psi.

b_w=b=12 in, h=24 in, A_s=2.64 in²

d_t=24-1.5-{3/8}-{6/{2*8}}=d_t=21.75 in

d_2=24-1.5-{3/8}-{6/8}-{6/{2*8}}=d_2=21in

d={A_t d_t + A_2 d_2}/A_s = 24-1.5-{3/8}-{6/8}-0.5=d=20.875

a= {A_s f_y}/{0.85 {f^prime}_c b} = {2.64*60}/{0.85*5.5*12}=a=2.8235 in

M_n=A_s f_y(d-{a/2})=2.64*60*{20.875-{2.8235/2}} =

M_n=3083 kip-ft

beta_1_5500 = 0.85-0.05({{f^prime}_c-4000}/1000) = 0.85-0.05*(5500-4000)/1000=0.85-0.075 = beta_1_5500 =0.775

epsilon_t= {0.003(beta_1 d_t - a)}/a = {0.003(0.775*21.75 - 2.8235)}/2.8235 = {0.003*1.4033}/2.8235=0.0149 > 0.005 → φ=0.9

phi M_n = 0.09*3083

phi M_n=2775 kip-ft

Problem 2

Using the cross-section below and the cantilever beam shown below, determine the load PL that the beam could carry. Assume f’c= 5000 psi, fy = 60,000 psi and #4 stirrups. WD = 2.1 kips/ft (it includes the weight of the beam) and WL = 2 kips/ft.

Illustration for Problem 2 Illustration for Problem 2

b_w = b =14in, h =26in, A_s =3.00in

d_t=d=26-1.5-{4/8}-{9/16}=23.4375in

beta_1(5000)=0.85-{{0.05({f^prime}_c-4000)}/1000} = 0.85-{{0.05(5000-4000)}/1000} = beta_1=0.80

a={A_s f_y}/{0.85 {f^prime}_c b}={3*60}/{0.85*5*14}= a =3.0252in

epsilon_t={0.003(beta_1 d_t -a)}/a = {0.003(0.80*23.4375 -3.0252)}/3.0252 = {0.003*15.725}/3.0252 = epsilon_t = 0.01559 > 0.005 → φ=0.9

M_n=A_s f_y(d-{a/2}) = 3.00*60*(23.4375-{3.0252/2})=3*60*21.925=M_n=3946.5kip-ft

phi M_n = 0.9*3946.5 = 3552kip-ft

Assume Live Load » Dead Load

w_u=1.2 w_D + 1.6 w_L = 1.2*2.1 + 1.6*2 =2.52+3.2= w_u=5.72kip/ft

P_u=1.6 P_L

phi M_n >= M_u = {(w_u l^2}/2}+P_u l = {5.72*100}/2 + 1.6*P_L*10 right 16 P_L + 286 <= 3552 → P_L <= {3552-286}/16=204.125

P_L <=204 kip

Check if the following beams satisfy ACI 318-05 cracking control. If any of the beams does not satisfy cracking requirements make the necessary changes. Assume f’c= 5000 psi, fy = 60,000 psi

f_s approx {2 f_y}/3 = {2*60000}/3=40000psi

Assume 1.5” clear cover and #3 stirrups.

c_c = 1.5+{3/8}=1.875 in

S_{max}<= lbrace matrix{2}{1}{{15(40000/f_s)-2.5c_c = 15-2.5*1.875=10.31} {12(40000/f_s)=12}} right S_{max} = 10.3in

Problem 3a

Illustration for Problem 3a

S_{exists}={b-2(a+d_v+c)}/{n-1} = {24-2*(1.5+{3/8}+{6/8})}/2={24-2*2.625}/2= 9.375 in < 10.3 in ✔

h < 3' → side-face reinforcement not required.

Problem 3b

Illustration for Problem 3b

S_{exists}={b-2(a+d_v+c)}/{n-1} = {14-2*(1.5+{3/8}+{6/8})}/2={14-2*2.625}/2=4.375 in < 10.3 in ✔

h > 3' → side-face reinforcement required. Use S=8”.

h/2 = 1.5+{3/8}+{9/16}+3s = 24 = 2.4375+3s

Place 6#3 bars inside either side of the stirrup with pairs spaced 8” apart, at 10”, 18” and 26” from the top tension face of the beam. (It is not clear when rounding the spacing value, where to make up the rounding error. Perhaps these locations should be at 8”, 16”, and 24” from the top tension face of the beam.)

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