Table of Contents
Homework 11 (due June 26)

— David Wagner 2007/06/21 15:05

Problem 1.

Illustration for Problem 8.12 Solve problem 8.13 in your textbook

“In Problems 8.11 to 8.19 for the beams and loads given, select stirrup spacings if f'c = 4000 psi and fy = 60,000 psi. The dead loads shown include beam weights. Do not consider movement of live loads unless specifically requested. Assume #3 U stirrups.”

“Repeat Problem 8.12 if live load postions are considered to cause maximum end shear and maximum FIXME shear. (One ans. 1@3 in., 7@6in., 5@12in.)”

Problem 2.

Illustration for Problem 8.14 Solve problem 8.19 in your textbook

“8.19 If the beam of Problem 8.14 has a factored axial compression load of 120 k in addition to other loads, calculate φVc and redesign the stirrups. (One ans. 1 @ 3 in., 4 @ 9 in., 3 @ 12 in.)”

φ=0.75, Nu = 120 kip, As=5#10= 6.35 in², bw = 15 in, d=dt = 25.5 in, d/2 = 12.75, h = 28 in, f'c=4000, fy=60000, #3U stirrups: Av = 0.22

A_g=bh=15*28=420 in²

phi V_c = phi 2(1+{{N_u}/{500 Ag}})sqrt{f prime_c} b_w d = 0.75*(1+{{120000}/{500*420}})*sqrt{4000}*15*25.5=28.510 kip

{phi V_c}/2=14.26

V_c=38.01 kip

R={1.2*(4*18) + 1.6*(20+20)}/2 = {86.4+64}/2 =75.2 kip

V_u=75.2-1.2*4*{25.5/12}=65 kip >= φVc

V_n >= {V_u/phi}=65/0.75=86.67 kip

V_s=V_n-V_c=86.67-38.01=48.66 kip

s_{crit}={0.22*60*25.5}/{48.66}=6.92 in

V_{u@6-}=75.2-1.2*4*6=46.4

s_{6-}={0.22*60*25.5}/{46.4}=7.25 in

V_{u@6+}=75.2-1.2*4*6-(1.6*20)=26.4

s_{6+}={0.22*60*25.5}/{26.4}=12.75 in

1@2”, 10@7”, 3@12 from both ends”

Problem 3.

Illustration for Problem 7.9 Solve problem 7.9 in your textbook

“In Problems 7.6 to 7.9 determine the development lengths required for the tension bar situations described, (a) using ACI Equation 12-1 and assuming Ktr=0. amd (b) using ACI Equation 12-1 and the calculated balue of Ktr*.”

“7.9 Uncoated top bars in normal-weight concrete. As required = 3.68 in.² (Ans. 59 in., 50 in.)”

db = 1.128 in., As=4.00 in², fy=60000, f'c=6000, Stirrups: #3@8”

c_b=min[3,3/2]=1.5

{c_b+K_{tr}}/{d_b}={1.5+0}/{1.128}=1.33 ⇐ 2.5✔

Psi_t*Psi_e=1.3*1.0=1.3 < 1.7✔

l_d={{3 f_y Psi_t Psi_e Psi_s lambda} / {40 sqrt{f prime_c} (c_b + K_{tr})}} {d_b}^2 {A_{s,req} / A_s} ={{3*60000*1.3*1.0*1.0*1.0 }/{40*sqrt{6000}*1.5 }}*{9/8}^2*{3.68/4.00} =58.62 in

59” using Ktr=0

n=4, s=8, Atr=2#3=0.22

K_{tr}={A_{tr} f_{yt}}/{1500 s n}= {0.22*60000}/{1500*8*4}=0.275

={{3*60000*1.3*1.0*1.0*1.0 }/{40*sqrt{6000}*1.775 }}*{9/8}^2*{3.68/4.00} =49.54 in

50” using Ktr=0.275

Problem 4.

Illustartion for Prblem 7.21 Solve problem 7.21 in your textbook

“Are the uncoated #8 bars shown anchored sufficiently with their 90º hooks? f'c = 3000 psi, fy = 60 ksi. The stress in the bars is 50 ksi. Side and top cover is 2½ on bar extensions. Normal-weight concrete is used. Use ACI Equation 12-1 and assume Ktr =0. (Ans. ldh = 13 in.)”

db = 1 in. ldh provided = 13 in.

c_b=min[2.5 or 2.0,na]=2.5 or 2.0

{2.5+0}/{1}=2.5 ⇐ 2.5✔

Psi_t*Psi_e=1.3*1.0=1.3 < 1.7✔

l_d={{3 f_y Psi_t Psi_e Psi_s lambda} / {40 sqrt{f prime_c} (c_b + K_{tr})}} {d_b}^2 ={{3*60000*1.3*1.0*1.0*1.0 }/{40*sqrt{3000}*2.5 }}*{1}^2 =234000/5477 =42.72 in

ACI Equation 12-1 doesn't help here. Try using the correct equation.

l_{dh,req}={0.7*0.02*1.0*1.0*60000}/{sqrt{3000}*1}=15.34 in > 13 in provided✘

No

Problem 5.

Illustration for Problem 7.26 Solve problem 7.26 in your textbook

“In the column shown, the lower column bars are #8 and the upper ones are #7. The bars are enclosed by ties spaced 12 in. on center. If fy=60,000 and f'c = 4000 psi, what is the minimum lap splice length needed? Normal-weight concrete is to be used for the 12 in. by 12 in. columns.”

Compression splice length of smaller bar = 0.0005*60000*{7/8}=26.25” (>12”✔)

Compression development length of larger bar = l_{dc}=max[{0.02 fy d_b} /{sqrt{f prime_c}} , 0.0003 f_y d_b] = max[{0.02*60000*1}/{sqrt{4000}} , 0.0003*60000*1] = max[18.97,18]=19” (>8”✔)

19”

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