Table of Contents
Homework 9 (Due June 21, 2007)

— David Wagner 2007/06/20 00:28

Problem 1.

Illustration for problem 5.12 Solve problem 5.12 in your textbook but instead of 5-#8 bars use 6-#9. Use the T-section equations in the design aids and verify your answer using the general method (centroid of Ac). Assume that As,min and As,max are satisfied.

“In Problems 5.5 to 5.15 determine the design moment strengths of the sections shown. Use fy = 60,000 psi and f'c = 4000 psi[…]. Check each section to see if it is ductile.” be=b=18, bw=12 in, hf=4 in, As=6#9=6.00 in², d=d_t=4+8+17=29 in, beta_{1,4000}=0.85.

T-section Equations

a={A_s f_y}/{0.85f prime_c b_e}={6*60}/{0.85*4*18}=5.8824 > 4

C_2=0.85 f prime_c h_f(b_e - b_w)=0.85*4*4*(18-12)=81.6

a={A_s f_y - C_2}/{0.85 f prime_c b_w}={6*60-81.6}/{0.85*4*12}=6.8235

C_1=0.85 f prime_c a b_w=0.85*4*6.8235*12=278.4

epsilon_t=0.003*(0.85*29-6.8235)/6.8235=0.0078375 > 0.005 → φ=0.9

phi M_n=phi[C_2(d - {{h_f}/2}) + C_1(d - {a/2} )]= 0.9*[81.6*(29 - {{4}/2}) + 278.4*(29 - {6.8235/2} )]= 0.9*[2203+7124]=8394

8394 kip-in

General Method

T=C=A_s f_y = 0.85 f prime_c A_c right A_c={6*60}/{0.85*4}=105.88 in²

A_f=4*18=72 in²

a=h_f+{A_c-A_f}/b_w = 4+{105.88-72}/12=6.8233

overline{y}={72*2 + 2.8233*12*(4+{2.8233/2})}/105.88 = {144+183.34}/105.88=3.0917

Assume φ is still 0.9 from above.

phi M_n = A_s f_y(d-overline{y})=0.9*6*60*(29-3.0917)=8394✔

8394 kip-in

Problem 2.

Illustration for problem 5.13 Solve problem 5.13 in your textbook but instead of 3-#10 use 4-#10. Use the T-section equations in the design aids and verify your answer using the general method (centroid of Ac). Assume that As,min and As,max are satisfied.

“In Problems 5.5 to 5.15 determine the design moment strengths of the sections shown. Use fy = 60,000 psi and f'c = 4000 psi[…]. Check each section to see if it is ductile.”

be=b=14, bw=3 in, hf=6 in, As=4#10=5.08 in², d=d_t=6+18+3=27 in, beta_{1,4000}=0.85.

T-section Equations

a={A_s f_y}/{0.85f prime_c b_e}={5.08*60}/{0.85*4*14}=6.4034 > 6

C_2=0.85 f prime_c h_f(b_e - b_w)=0.85*4*6*(14-3)=224.4

a={A_s f_y - C_2}/{0.85 f prime_c b_w}={5.08*60-224.4}/{0.85*4*3} =7.8824

C_1=0.85 f prime_c a b_w=0.85*4*7.8824*3=80.4

epsilon_t=0.003*(0.85*27-7.8824)/7.8824=0.0057346 > 0.005 → φ=0.9

phi M_n=phi[C_2(d - {{h_f}/2}) + C_1(d - {a/2} )]= 0.9[224.4*(27 - {{6}/2}) + 80.4(27 - {7.8824/2} )]= 0.9*[5386+1854]=6516

6516 kip-in

General Method

T=C=A_s f_y = 0.85 f prime_c A_c right A_c={5.08*60}/{0.85*4}=89.647 in²

A_f=6*14=84 in²

a=h_f+{A_c-A_f}/b_w = 6+{89.647-84}/3=6+1.8824=7.8824

overline{y}={84*3 + 1.8824*3*(6+{1.8824/2})}/89.647 = {252+39.198}/89.647=3.2483

Assume φ is still 0.9 from above.

phi M_n = A_s f_y(d-overline{y})=0.9*5.08*60*(27-3.2483)=6516✔

6516 kip-in

Problem 3.

Illustration for Problem 8.5 Solve problem 8.5 in your textbook

“What is the design shear strength of the beam shown if f'c = 4000 psi and fy=60,000 psi. No shear reinforcing is provided (Ans. 31876 lbs.)”

Av=0, Vs=0, bw=14, d=24, φ=0.75

V_c=2 b_w d sqrt{f prime_c}=2*14*24*sqrt{4000}=42501

phi V_n = phi(V_c + V_s)=0.75*(42501+0)=31876✔

31876 lb

Problem 4.

Illustration for Problem 8.7 Solve problem 8.7 in your textbook

“In problems 8.7 to 8.9, compute φVn for the sections shown if fy of stirrups is 60 ksi and f'c = 4000 psi. (Ans. 79,518 lb)”

Av=2#3=0.22 in², s=8 in, bw=18, d=27, φ=0.75

V_c=2 b_w d sqrt{f prime_c}=2*18*27*sqrt{4000}=61475

8 b_w d sqrt{f prime_c}=245900

V_s={A_v f_y d}/s={0.22*60000*27}/8=44550 < 245900✔

phi V_n = phi(V_c + V_s)=0.75*(61475+44550)=79519✔

79519 lb

Problem 5.

Illustration for Problem 8.9 Solve problem 8.9 in your textbook

“In problems 8.7 to 8.9, compute φVn for the sections shown if fy of stirrups is 60 ksi and f'c = 4000 psi.”

Av=1#4=0.20 in², s=6, bw=5 in, d=5+22+2.5=29.5 in, φ=0.75

V_c=2 b_w d sqrt{f prime_c}=2*5*29.5*sqrt{4000}=18657

8 b_w d sqrt{f prime_c}=74630

V_s={A_v f_y d}/s={0.2*60000*29.5}/6=59000 < 74630✔

phi V_n = phi(V_c + V_s)=0.75*(18657+59000)=58243✔

58243 lb

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