Homework 14 (due July 3)

— David Wagner 2007/07/01 10:56

Problem 1.

Solve problem 11.5 in your textbook

“The tied column shown is to be used in a braced frame. It is bent about its y axis with the factored moments shown, and lu is 16 ft. If k = 1.0, fy = 60,000 psi and f'c = 4000 psi, select the reinforcing required. Assume PD = 50 k. (One ans. 6 #7)”

Short column max ${k l_u}/r= 34-12{M_1/M_2}=34-12*{75/80}=$ 22.75

${k l_u}/r={1*16*12}/0.3*14=$ 45.71 > 22.75 ∴ slender

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

$I_g={1/12}*14*14^3=$ 3201 in4

$beta_d = {50*1.2}/{100}=$ 0.6

$EI={0.4 E_c I_g}/{1+beta_d}={0.4*3605*3201}/{1+0.6}=$ 2,885,000 k-in2

$P_c = {pi^2 EI}/{k^2{l_u}^2}={pi^2*2885000}/{1^2*16^2*12^2}=$ 772.4 k

$C_m = 0.6+0.4{M_1/M_2}=0.6+0.4*{75/80}=$ 0.975

$delta_{ns}={C_m}/{1-{P_u/{0.75P_c}}} = {0.975}/{1-{100/{0.75*772.4}}}= {0.975}/{0.8274}=$ 1.117

$M_{2,min}=P_u(0.6+0.03h)=100*(0.6+0.03*14)=$ 102.0 in-k = 8.5 ft-k

$M_c = delta_{ns}M_2=1.117*80=$ 89.36 ft-k

e = {12*89.36}/{100}= 10.72 in

gamma={9/14}= 0.643

$P_n = {100}/{0.65}=$ 153.8 k

$K_n = {153.8}/{4*14*14}=$ 0.1962

$R_n = 0.1962*{10.72/14}=$ 0.1503

rho_g approx 0.43*0.017 + 0.57*0.014 = 0.0153

A_s = 0.0153*14*14= 2.997 in2

5#7

Problem 2.

Solve problem 12.8 in your textbook (Assume a 20-in footing with a d = 15.5 in). Do not forget to check development length.

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

“In Problems 12.7 to 12.12, design square single-column footings for the values given.”

Problem	Column size (in.)	D (k)	L (k)	f'c (ksi)	fy (ksi)	qa (ksf)	Distance form bottom of footing to final grade (ft)	Column location
12.8	12 x 12	100	120	3	60	5	5	interior

B_u=1.2*100+1.6*120= 192 k

phi(0.85 f prime_c A_1)=0.65*0.85*3*12*12= 288.7 > 192✔

$phi(0.85 f prime_c A_1)sqrt{A_2 / A_1}=288.7*sqrt{{A_f} / 144} = 192 right sqrt{A_f} = 192*12/288.7=$ 7.981

h=b=8 ft

A_f=8^2= 64 ft2

q_e=

5000-20/12*150-(5-{20/12})*100=5000-250-333= 4417 psf

$A_{req}={100+120}/{4.417} =$ 49.8 ft2 ≤ 64✔

$q_u={B_u}/{A_f}= 192/64 =$ 3 ksf

b_0=4(a+d)=4*(12+15.5)= 110

$min[4 sqrt{f prime_c},2+{4/{beta_c}},{{alpha_s d}/{b_0} + 2}] = min[4*sqrt{3000},2+{4/{1}},{{40*15.5}/{110} + 2}] =min[219,6,7.636]=$ 6

$V_{u2}=(64-{{(12+15.5)^2}/144})*3 =$ 176.2 k

6 phi b_0 d = 6*0.75*110*15.5= 7673 k > 176.2✔

$V_{u1}=8*({l/2}-{a/2}-d)=8*({8/2}-{1/2}-{15.5/12})=$ 17.67 k

$phi 2*sqrt{f prime_c}b_w d=0.75*2*sqrt{3000}*8*12*15.5=$ 122.3 > 17.67✔

M_u=7*8*3*3.5= 588 ft-k

${M_u}/{phi b d^2} = {12*588000}/{0.9*96*15.5^2}=$ 340 psi, rho=0.0061

$rho_{min}= max[200/sqrt{3000}, 3sqrt{3000}/60000]=$ 0.0033✔

A_s=0.0061*96*20=11.71

12#9

psi_t=psi_e=psi_s=lambda=1.0

${c+K_{tr}}/{d_b} =$ 3.11 >2.5

$l_d={9/8}*{3*60000*1*1*1*1}/{40*sqrt{3000}*3.11}=$ 29.72 in < 48 in✔

Dowels A_s > 0.005*12*12= 0.72 in2

As=4#4=0.80 in2, db=0.5 in

$l_d >= {0.02*0.5*60000}/{sqrt{3000}}=$ 10.95 in controls

l_d >= 0.0003*0.5*60000= 9 in

l_d >= 8 in

Dowels: 4#4 extending 11 in up and 11 in down

Problem 3.

Solve problem 12.13 in your textbook. Do not forget to check splice length and development length.

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

“12.13 Design for load transfer from an 18” x 18” column with 6 #8 bars (D = 200 k, L = 350 k) to an 8-ft-0-in. x 8-ft-0-in. footing. f'c = 4 ksi for footing and 5 ksi for column. fy = 60 ksi. (Ans. 4 #6, 13 in. into footing, 12.5 in. into column)”

h=b=8 ft, A_f=8^2= 64 ft2

B_u=1.2*200+1.6*350= 800 k