Homework 13 (due July 2)

— David Wagner 2007/06/27 15:35

Problem 1.

Solve problem 10.6 in your textbook

“10.3 Using statics equations, determine the values of Pn and Mn for the column shown, assuming it is strained to -0.00300 on the right-hand edge and to +0.00200 on its left-hand edge. f'c = 4000 psi, fy = 60,000 psi.”

“10.6 Repeat Problem 10.3 if the strain on the left edge is -0.001.”

Firstly, all the steel and concrete are under compression, and the section under consideration is overall in compression sufficient to produce a strain of -0.001 throughout.

Assuming a linear strain distribution and measuring x from the left edge, epsilon(x) = -0.001 - 0.002x/24=-0.001-{0.001x/12}

$epsilon_s=-0.001-{0.001x/12}=-0.001-{0.001*3/12}=$ -0.00125

C_s=29000*0.00125= 36.25 kip

$epsilon prime_s=-0.001-{0.001x/12}=-0.001-{0.001*21/12}=$ -0.00275 < 0.00207 ∴ Compression steel yields. This implies the steel on the right is no longer bearing any load and this is now a singly-reinforced beam under axial compression.

c={{0.003}/{0.003-0.001}}*24= 36 in✘

$a={A_s f_y}/{0.85 f prime_c b}={1.58*60}/{0.85*4*14}=$ 1.9916 in

C_c=0.85*1.9916*14*4= 94.80 kip

P_n= C_c + C_s = 94.80+36.25= 131.05

P_n=131 kip

Taking moments about the center: $Sigma M 0 = M_n + 0*P_n - (12-{a/2})*C_c + {d/2}*C_s right M_n = 11*94.8 - 9*36.25 = 1042.8-326.25=$ 716.55

M_n=717 kip-in

Problem 2.

Solve problem 10.8 in your textbook

“In Problems 10.8 to 10.10 use the interaction curves in the Appendix to select reinforcing for the short columns shown, with f'c = 4000 psi and fy = 60,000 psi.”

h=18, γ=12/18=0.67, e/h=8/18=0.44

A_g=18*12 216 in²

$K_n={400}/{4*216}=$ 0.463

$R_n={400*0.44}/{4*216*18}=$ 0.0113

ρ_g ≈ 0.028 → K_n ≈ 0.5 > 0.463✔, R_n ≈ 0.22 > 0.0113✔

$A_{st}=rho_g*b*h=0.027*12*18=$ 5.832 in²

Select 6#9 = A_st 6.00 in²

6#9

Problem 3.

Solve problem 10.10 in your textbook

“In Problems 10.8 to 10.10 use the interaction curves in the Appendix to select reinforcing for the short columns shown, with f'c = 4000 psi and fy = 60,000 psi.”

h=22”, γ=17/22=0.7727, e/h=10/22=0.4545

$A_g={pi*22^2}/4=$ 380.133 in²

$K_n={500}/{4*380.133}=$ 0.3288

$R_n={500*0.4545}/{4*380.133*22}=$ 0.00679 ? Only minimum ρ ≈ 0.01 needed. (Consider using plain concrete) This makes no sense.

Using Kn=0.33 and e/h=0.45 suggests:

For ρ ≈ 0.015, Kn ≈ 0.33, Rn ≈ 0.15, εt≈0.002

$A_{st}=0.015*380.133=$ 5.702

Use 6#9 = 6.00 in²

6#9

Problem 4.

Solve problem 10.11 in your textbook

“For Problems 10.11 to 10.16 use the interaction diagrams in the Appendix to determine Pn values for the short columns shown which have bending about one axis. fy = 60,000 psi and f'c = 4000 psi.” (Ans. 559 k)

h=21, b=15, Ast=10.16, Ag=315, γ=0.7143, e/h=0.57143

$rho_g={10.16/315}=$ 0.03226