Homework 15 (due July 5)

— David Wagner 2007/07/01 10:59

Problem 1.

Solve problem 11.8 in your textbook

“For Problems 11.8 to 11.12 and the braced tied columns given, select reinforcing bars if the distance from the column edge to the c.g. of the bars is 2.5 in. fy = 60 ksi and f'c = 4 ksi for all problems, Place bars in two faces only.”

Problem No.	Column size b x d (in.)	lu (ft)	k	Pu (k)	Not factored PD (k)	Factored M1b (ft-k)	Factored M2b (ft-k)	Curvature
11.8	14 x 14	12	1.0	300	100	80	90	Single

${k l_u}/{r} <= 34-12{M_1 / M_2} right {1*12*12}/{0.3*14} <= 34-12*{80/90} right 34.3 <= 23.3$ ✘ → slender; < 100 → moment magnification OK

$E_c = 57000 sqrt{f prime_c} = 57000 sqrt{4000}=$ 3605 ksi

$I_g = {bh^3/12} = {14*14^3}/12=$ 3201 in4

$beta_d = {1.2*100}/{300}=$ 0.4000

$EI={0.4 E_c I_g}/{1+beta_d} = {0.4*3605*3201}/{1.4}=$ 3,297,000 kip-in2

$P_c={pi^2 EI}/{k^2 l_u^2} = {pi^2*3297000}/{1^2 *12*12*12*12}=$ 1569 kip

$C_m=0.6+0.4{M_1 / M_2}=0.6+0.4{80/90}=$ 0.9556

$delta_{ns} = {C_m}/{1-{P_u/{0.75P_c}}} = {0.9556}/{1-{300/{0.75*1569 }}} = {0.9556}/{0.7451}=$ 1.283

$M_{2,min} = P_u(0.6+0.03h) = 300*(0.6+0.03*14)=$ 306.0 in-kip = 25.50 ft-kip

$M_c = delta_{ns}*M_2 = 1.283*90=$ 115.5 ft-kip

e={12*115.5}/{300} = 4.62 in

{e/h} = 4.62/14= 0.330

gamma = {14-2*2.5}/14 = 0.6429

P_n = P_u / phi = 300/0.65= 461.5 kip

$K_n = {P_n}/{f prime_c A_g}= {461.5}/{4*14*14}=$ 0.5886

rho_g = 0.571*0.031+0.429*0.028= 0.02971

A_s = rho_g A_g = 0.02971*14*14= 5.82 in2

6#9=6.00 in2

Use 6#9

Problem 2.

Solve problem 11.12 in your textbook.

Problem No.	Column size b x d (in.)	lu (ft)	k	Pu (k)	Not factored PD (k)	Factored M1b (ft-k)	Factored M2b (ft-k)	Curvature
11.12	15 x 18	18	0.90	500	140	110	120	Double

${k l_u}/{r} <= 34-12{M_1 / M_2} right {0.9*18*12}/{0.3*18} <= 34-12{110/120} right 36 <= 23$ ✘ → slender; < 100 → moment magnification OK

$E_c = 57000 sqrt{f prime_c} = 57000 sqrt{4000}=$ 3605 ksi

$I_g = {bh^3/12} = {15*18^3/12}=$ 7290 in4

$beta_d = {1.2*140}/{500}=$ 0.336

$EI={0.4 E_c I_g}/{1+beta_d} = {0.4*3605*7290}/{1.336}$ 7868000 kip-in2

$P_c={pi^2 EI}/{k^2 l_u^2} = {pi^2*7868000}/{0.9*0.9*18*18*12*12} 77650/37.79=$ 2055 kip

$C_m=0.6+0.4{M_1 / M_2}=0.6+0.4*{110/120}=$ 0.9667

$delta_{ns} = {C_m}/{1-{P_u/{0.75P_c}}} ={0.9667}/{1-{500/{0.75*2055}}}= 0.9667/0.6756 =$ 1.431

$M_{2,min} = P_u(0.6+0.03h) =500*(0.6+0.03*18)=$ 570 ft-kip

$M_c = delta_{ns}*M_2 =1.431*120=$ 171.7 ft-kip

$e = {M_c}/{P_u} = {171.7*12}/500=$ 4.121 in

{e/h} =4.121/18= 0.2289

gamma = 13/18= 0.7222

P_n = P_u / phi =500/0.65= 769.2 kip

$K_n = {P_n}/{f prime_c A_g}=769.2/{4*15*18}=$ 0.7123

rho_g =0.778*0.023 +0.222*0.022 = 0.02278

A_s = rho_g A_g =0.02278*15*18= 6.15 in2

8#8=6.32 in2

8#8

Problem 3.

Solve problem 12.4 in your textbook. (Assume a 26-in footing with a d = 22.5 in). Do not forget to check development length and to add temperature and shrinkage steel.

“For Problems 12.1 to 12.30, assume reinforced concrete weighs 150 lb/ft3, plain concrete 145 lb/ft3, and soil 100 lb/ft3.”

“For Problems 12.1 to 12.5, design wall footings for the values given. The walls are to consist of reinforced concrete.”