Table of Contents
Homework 10 (due June 25)

— David Wagner 2007/06/21 15:08

Problem 1.

Design a joist system for the floor below. Assume f’c = 4 ksi, Gr. 60 steel, wDL 30 psf (does not include weight of joist system), wLL = 40 psf. The total height of the joist is 14.5 in. The slab is 2.5 in thick. The width of the joist is 5 in at the bottom.

Illustration for Problem 1

Include a sketch with the reinforcement.

f'c=4000, fy=60000, t=2.5 in, h=14.5 in

Slab Design

Assume 30”=2.5' forms.

W_{slab}=2.5/12*150=31.25 psf

W_{u,slab}=1.4*(30+31.25)=85.75 psf

W_{u,slab}=1.2*(30+31.25) + 1.6*40 =73.5+64=0.1375 ksf controls

M_{u,slab} = {W_u{u,slab} {l_n}^2}/12={0.1375 {2.5}^2}/12=0.0716 ft-kip/ft

S_{m,slab}={bt^2}/6={12*2.5^2}/6=12.5

phi M_{n,slab}= phi 5 sqrt{f prime_c} S_{m,slab}=0.55*5*sqrt{4000}*12.5=2.174 kip >= 0.0716✔

A_{s,slab}=0.0018*12*2.5=0.054 in²/ft

Choose welded grade 60 wire mesh from manufacture specification sheets to obtain at least 0.054 in²/ft. For example, 4×12-W2.1xW1.4 is 0.063 in²/ft.

Joist Design

Assume only designing interior spans and the beams supporting the joists are as wide as the columns supporting these beams.

l_n=25'-20”=23.33'

h_min=L/21={12*25}/21=14.286 in → try h_{joist}=15 in

The joist is 5 in at the bottom, and 5+{{15-2.5}/12}=6.0 in at the top. Tributary width = 35”.

W_{joist}=[35*2.5+(15-2.5)*{{6+5}/2}]*{150/144}=[87.5+68.75]*150/1440.163 k/ft

w_u= 1.2(w_D + 1.6w_L) = 1.2*0.163 + 1.2*0.030*{35/12} + 1.6*0.040*{35/12} = 0.1956+0.105+0.187=0.4873 k/ft

M_{u+} = {w_u {l_n}^2}/16={0.4873*23.33^2}/16=16.6 ft-kip = 200 in-kip

M_{u-} = {w_u {l_n}^2}/11={0.4873*23.33^2}/11=24.1 ft-kip = 289 in-kip

Try #4 bars top and bottom.

d^+ = d^- = 14.5-.75-.25=13.5 in

M_{u+}=200 approx 0.9*A_s*60*0.95*13.5 right A_s approx0.2888 in²

Try As=1#5=0.31 in the joist. d^+ = 14.5-.75-{5/16}=13.4375 in

b=b_e=min[16t+b_w=16*2.5+{5+6}/2=45.5,L/4={25*12}/4=75,L_{c-c}=35]=35 in

a={0.31*60}/{0.85*4*35}=0.1563

epsilon_t=0.003{beta_1 d_t -a}/a=0.003*{0.85*13.4375-0.1563}/0.1563=0.216 > 0.005✔

A_{s,min} = max[{3*sqrt{4000}*5.5*13.4375}/60000,{200*5.5*13.4375}/60000] = max[0.234,0.246]=0.246 < 0.31✔

0.9*0.31*60*(13.4375-{0.1563/2})=223.6 >= 200✔

Use 1#5 3/4 in from bottom of joist

M_{u-}=289 approx 0.9*A_s*60*0.9*13.5 right A_s approx0.44 in²

Suppose b ≈ 5.1

a={0.44*60}/{0.85*4*5.1}=1.5225

b=5+{1.5225/12}=5.127

a={0.44*60}/{0.85*4*5.127}=1.51

b=5+{1.51/12}=5.126

s={12*A_{bar}}/{A_s}={12*0.2}/0.44=5.45 rigth s=5 in

A_s={12*0.2}/5=0.48 in²

epsilon_t=0.003{0.85*13.5-1.51}/1.51=0.0198 > 0.005✔

A_{s,min} = max[{3*sqrt{4000}*5.5*13.5}/60000,{200*5.5*13.5}/60000] = max[0.235,0.246]=0.2475 < 0.48✔

phi M_n = 0.9*0.48*60*(13.5-{1.51/2})=330 > 289✔

#4@5” 3/4” from top of flange.

Problem 2.

Illustration for Problem 8.14 Solve problem 8.14 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.”

φ=0.75, wD=4 k/ft, wL=0, b=bw=15 in, d=25.5 in = 2.125', Av=2#3=0.22 in²

phi V_c=phi 2 b_w d sqrt{f prime_c} =0.75*2*15*25.5*sqrt{4000}=32.3 kip

V_{u@supports}={1.4*4*18}/2={100.8}/250.4 kip

V_{u@supports}={1.2*4*18+1.6*20+1.6*20}/2={86.4+64}/275.2 kip controls

x Vu V_s={V_u - phi V_c}/phi={V_u - 32.3}/0.75 s={A_v f_y d}/V_s=336.6/{V_s}
to d=2.125' 75.2-1.125*4*1.2=69.8 50.0 6.732
2 75.2-2*4*1.2=65.6 44.4 7.6
32” 75.2-{32/12}*4*1.2=62.4 40.1 8.4
3 75.2-3*4*1.2=60.8 38 8.9
4 75.2-4*4*1.2=56 31.6 10.7
5 75.2-5*4*1.2=51.2 25.2 13.4
6- 75.2-6*4*1.2=46.4 18.8 17.9
6+ 75.2-6*4*1.2-20*1.6=14.4 - -
9 0 - -
1@2”, 5@6”, 5@8” from both ends

Problem 3.

Illustration for Problem 8.16 Solve problem 8.16 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.”

φ=0.75, wD=2 k/ft, wL=4 k/ft, b=bw=14 in, d=21 in = 1.75', d/2=10.5 in, d/4=5.25 in, Av=2#3=0.22 in²

phi V_c=phi 2 b_w d sqrt{f prime_c} = 0.75*2*14*21*sqrt{4000} =27.9 kip

{phi V_c}/2=13.9 kip

4 b_w d sqrt{f prime_c}=4*14*21*sqrt{4000}=74.377 kip

LL»DL → V_{u@supports}=1.2*12*2 + 1.6*12*4= 28.8 + 76.8105.6 kip w_u=1.2*2+1.6*4=2.4+6.4=8.8 kip/ft

x Vu V_s={V_u - phi V_c}/phi={V_u - 27.9}/0.75 s={A_v f_y d}/V_s=277.2/{V_s}
to d=1.75' 105.6-1.75*8.8=90.2 83.1 3.3”
2 105.6-2*8.8=88.0 80.1 3.5”
35.1” 105.6-2.923*8.8=79.875 69.3 4”
4 105.6-4*8.8=70.4 56.7 4.9”
58.7” 105.6-4.892*8.8=62.55 46.2 6”
6 105.6-6*8.8=52.8 33.2 8.3”
8 105.6-8*8.8=35.2 9.7 28.5 10.5”
10 105.6-10*8.8=17.6 - s={A_v f_y}/{50b_w}=18.9 10.5
12 0 - -
12@3”, 14@4”, 5@10”

Problem 4.

Illustration for Problem 8.14 Solve problem 8.20 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.”

“8.19 If the beam in Problem 8.14 has a factored axial compression load of 120 k in addition to other loads, calculate φVc and redesign the stirrups.”

“8.20 Repeat problem 8.19 if the axial load is tensile. use #4 U stirrups.”

Nu=120 kip

phi V_c=2(1+{N_u/{500 A_g}}) sqrt{f prime_c} b_w d = 0.75*2*(1-{120000/{500*28*15}})*sqrt{4000}*15*25.5=15.551 kip

{phi v_c}/2=7.776 kip

x Vu V_s={V_u - phi V_c}/phi={V_u - 15.55}/0.75 s={A_v f_y d}/V_s=336.6/{V_s}
25.5”to d=2.125' 75.2-1.125*4*1.2=69.8 72.3 4.65
24” 2 75.2-2*4*1.2=65.6 66.7 5.04
32” 75.2-{32/12}*4*1.2=62.4 62.5 5.4
36” 3 75.2-3*4*1.2=60.8 60.3 5.6
48” 4 75.2-4*4*1.2=56 53.9 6.24
60” 5 75.2-5*4*1.2=51.2 47.5 7.08
72”- 6- 75.2-6*4*1.2=46.4 41.1 8.2
72”+ 6+ 75.2-6*4*1.2-20*1.6=14.4 - 12.75
108” 9 0 - -
1@3”, 12@4”, 4@6” 3@11” from both ends

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