Table of Contents
Homework #6 Due 6/14/07

— David Wagner 2007/06/12 19:36

Problem 1.

Illustration for problem 5.27

Solve problem 5.27 in your textbook.

“In Problem 5.26 to 5.32 compute the design strengths of the beams shown if fy =60000 psi and f'c = 4000 psi. Check the maximum permissible As in each case to ensure tensile failure.”

”(Ans. 679.1 ft-k)”

b=15 in, d=dt=32 in, d'=2.5 in, β1,4000=0.85, As=5.08 in², A's=2.37 in²

5.08-2.37=2.71 >= {0.85*0.85 *4000*2.5*15} / {60000}*{87000}/{87000-60000} = 9.429*10^9/1.62*10^9=5.82 → Compression steel does not yield.

c^2 + ({87000*2.37-5.08*60000}/{0.85*4000*0.85*15})c - {87000*2.37*2.5}/{0.85*4000*0.85*15}=0

=c^2 + ({-98610}/{43350})c - {515480}/{43350}=0

=c^2 - 2.2747c - 11.891=0

c={-2.2747 pm sqrt{-2.2747+4*11.891}}/2 =-1.1374 - pm 3.3649=2.2275

f prime_s=87000*({2.2275-2.5}/2.2275)=10643

A_{s,min,0.004}={0.85*4000*.85*15*3*32}/{60000*7} - {2.37*10643}/{60000} = 9.9086-0.4204=9.4882 > 5.08✔

a={5.08*60000-2.37*10643} / {0.85*4000*15} = 5.4819

M_n=10643*[({5.08*60000}/{10643}-2.37)*(32-{5.4819/2}) + 2.37*(32-2.5)] = 10643*[26.269*(29.259) + 69.915]

M_n= 10643*838.5=8924400=8924 kip-in = 744 kip-ft

phi M_n = 0.9*744=

669 kip-ft✔

Problem 2.

Illustration for problem 5.29

Solve problem 5.29 in your textbook.

“In Problem 5.26 to 5.32 compute the design strengths of the beams shown if fy =60000 psi and f'c = 4000 psi. Check the maximum permissible As in each case to ensure tensile failure.”

”(Ans. 613.0 ft-k)”

b=12 in, d=dt=25 in, d'=2.5 in, β1,4000=0.85, As=6.24 in², A's=1.58 in²

6.24-1.58=4.66 >= {0.85*0.85*4000*2.5*12*87000}/{60000*(87000-60000)} = {7.5429*10^9}/{1.62*10^9}=4.6561 → Compression steel yields.

A_{s,max}={0.85*4000*0.85*12*3*25}/{60000*7} + 1.58 =7.7729 > 6.24✔

a={4.66*60000}/{0.85*4000*12}=6.8529

M_n=60000*[4.66*(25-{6.8529/2})+1.58*(25-2.5)] = 60000*[100.53+35.55] =8165 kip-in = 680.4 kip-in

phi M_n = 0.9*680.4=

612.4 kip-ft✔

Problem 3.

Illustration for problem 5.35

Solve problem5.35 in your textbook. After finding the areas as required in your textbook check if φMn ≥ Mu.

“In Problems 5.35 to 5.38 determine the steel areas required for the sections shown in the accompanying illustrations. In each case the dimensions are limited to the values shown. If compression steel is required, assume it will be placed 3 in. o.c. from the compression face. f'c = 4000 psi and fy = 60000 psi.”

”(Ans. As=8.87 in2, A's=1.77 in2)”

Mu=950 ft-kip=11400 in-kip b=14, d=dt=28 in, beta_{1,4000}=0.85, d'=d'dt=3 in

Assume εt >= 0.005 → φ=0.9

A_{s,max}={0.85 f prime_c beta_1 b 3 d_t}/{8 f_y} = {0.85*4*0.85*14*3*28}/{8*60}=7.08 in²

a={A_s f_y}/{0.85 f prime_c b} = {7.08*60}/{0.85*4*14} =8.924 in

d-{a/2}=28-{8.924/2}=23.54

M_u=phi M_n = 0.9*7.08*60*(28-{8.924/2}) =8999 kip-in < 11400 ✘Compression steel required.

A_s approx 0.9*11400/{0.9*60*23.54}=8.07

As=8.07</m>in²

Assume compression steel yields.

8.07 <= A_{s,max} = 7.08+A prime_s right A prime_s >= 1.01

A_s - A prime_s >= {.85*.85*4*3*14*87}/{87*60-60*60}=6.519 right A prime_s <= 8.07-6.519 = 1.55

11400=phi M_n = 0.9*60*[(7.08 - A prime_s)*23.54+A prime_s(23.54-3)] = 54*[(166.66+23.54A prime_s+20.54A prime_s) = 9000 + 2380A prime_s

right A prime_s={11400-9000}/2380 = 1.01in²

A's=1.01</m>in²

Problem 4.

Illustration for problem 5.37

Solve problem 5.37 ion[sic] your textbook. After finding the areas as required in your textbook check if φMn ≥ Mu.

“In Problems 5.35 to 5.38 determine the steel areas required for the sections shown in the accompanying illustrations. In each case the dimensions are limited to the values shown. If compression steel is required, assume it will be placed 3 in. o.c. from the compression face. f'c = 4000 psi and fy = 60000 psi.”

”(Ans. As=8.02 in2, A's=2.37 in2.)”

Mu<sub>=800 ft-k = 9600 in-k b=12 in, d=d<sub>t<sub>=26 in, d'=d'<sub>t<sub>=3 in, beta_{1,4000}=0.85 Assume ε<sub>t >= 0.005 → φ=0.9

A_{s,max} = {0.85*4*0.85*12*3*26}/{60*8} =3.6355 in²

a={3.6355*60}/{0.85*4*12}=8.2875

M_u=phi M_n = 0.9*3.6355*60*(26-{8.2875/2})=4291 k-in < 9600. Requires compression steel.

Assume it yields at εt=0.005.

A_s-A prime_s >= {0.85*0.85*4*3*12*87}/{60*87-60*60}=9052/{1620}=5.587

A_{s,max}=3.6355+A prime_s

A_s <= A prime_s + 3.6355 right 5.587+A prime_s <= 3.6355+A prime_s✘ ∴Compression steel does **not* yield and As-A primes < 5.587.

But now both As and A primes are unknown…

c^2+({}/{})…But now, I am out of time…

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