Table of Contents
CE 3253 Optional Homework (Due during Final Exam)

— David Wagner 2007/11/25 15:54

Problem 1.

Solve Problem 7.3 in your Wood Structures textbook.

Since the truss and the loading is symmetrical, the vertical reactions at the supports are the same.

  • P_y=wLs/2= {4*20w}/2=40w

A free-body analysis of joint C shows the tension in the lower chord is equal to the horizontal component of the force in the outer diagonal member, and 12/5 times the vertical component of the force in the diagonal members.

  • sum{}{}{F_x} = 0 right T={12/5}P_y={12/5}40w=96w

A=5.25 in², CF=1.5, FT=575.

a. ASD: CD=1.15 (snow)

  • F prime_t = F_t C_D C_M C_t C_F C_i=575*1.15*1*1*1.5*1=992 psi
  • T=96w=96*(20+55)=7200 lb
  • f_t = T/A = 7200/5.25=1371 > 992 psi✘
Not good by ASD.

b. LRFD: KF=2.16/φ, φ=0.80, λ=0.8

  • F prime_t = F_t C_M C_t C_F C_i K_F phi_t lambda = 575*1*1*1.5*1*{2.16/0.8}*0.8*0.8 =1490 psi
  • w=1.2D+1.6S=1.2*20+1.6*55=24+88=112 lb/ft
  • T=96w=96*112=10752 lb
  • f_t = T/A = 10752/5.25=2048 > 1490 psi✘
Not good by LRFD.

Problem 2.

Solve Problem 7.17 in your Wood Structures textbook.

  • A=131.3, Ix=2461, Sx=328.1, rx=4.330, Iy=837.4, Sy=191.4, ry=2.526.
  • E=1.6e6, Emin=0.83e6, Fc⊥=560, Ft=1250, Fc=1950, Fby=1800, Fvy=230, Fbx=1500, Fvx=265
  • CM=1, Ct=1
  • c=0.9

Buckling about x axis: le=22'=264”, d=15”

  • {l_e}/d = 264/15=17.6

ASD: P=D+Lr=20000+40000=60000 lb, CD=1.25

  • E prime_{min}=E_{min}C_M C_t = 830000*1*1=830000 psi
  • F_{cE}={0.822E prime_{min}}/{17.6^2}={0.822*830000}/{17.6^2}=2203 psi
  • F^{circ}_c = F_c C_D C_M C_t =1950*1.25*1*1=2438 psi
  • {F_{cE}}/{F^{circ}_c} = 2203/2438=0.90
  • {1+0.90}/{2*0.9} =1.056
  • C_P = 1.056-sqrt{1.056^2 - {0.9/0.9}}= 1.056-sqrt{0.115}= 1.056-0.339=0.717

Buckling about y axis: le=12'=144”, d=8.75”

  • {l_e}/d = 144/8.75=16.46
  • F_{cE}={0.822E prime_{min}}/{16.46^2}={0.822*830000}/{16.46^2}=2518psi

ASD:

  • {F_{cE}}/{F^{circ}_c} = 2518/2438=1.03
  • {1+1.03}/{2*0.9} =1.128
  • C_P = 1.128-sqrt{1.128^2 - {1.03/0.9}} = 1.128-sqrt{1.272-1.144} = 1.128-sqrt{0.128} =0.770 > 0.717 → x-axis critical
  • F prime_c = F_c C_D C_M C_t C_P = 1950*1.25*1*1*0.717=1748 psi
  • F prime_c A = 1748*131.3=229500 lb > 60000✔
OK by ASD

LRFD: P = 1.2D+1.6Lr = 1.2*20000+1.6*40000 = 24000+64000= 88000 lb, φ=0.90, λ=0.8

  • E prime_{min}=E_{min}C_M C_t K_F phi = 830000*1*1*{2.16/0.85}*0.85=1792800 psi
  • F^{circ}_c = F_c C_M C_t K_F phi lambda =1950*1*1*{2.16/0.9}*0.9*0.8=3370 psi

Buckling about x axis: le=22'=264”, d=15”

  • {l_e}/d = 264/15=17.6
  • F_{cE}={0.822E prime_{min}}/{17.6^2}={0.822*1792800}/{17.6^2}=4757 psi
  • {F_{cE}}/{F^{circ}_c} = 4757/3370=1.41
  • {1+1.41}/{2*0.9} =1.340
  • C_P = 1.340-sqrt{1.340^2 - {1.41/0.9}}= 1.340-sqrt{1.7956-1.567}= 1.340-sqrt{1.7956-1.567}= 1.340-0.478=0.862

Buckling about y axis: le=12'=144”, d=8.75”

  • {l_e}/d = 144/8.75=16.46
  • F_{cE}={0.822E prime_{min}}/{16.46^2}={0.822*1792800}/{16.46^2}=5439 psi

LRFD:

  • {F_{cE}}/{F^{circ}_c} = 5439/3370 =1.61
  • {1+1.61}/{2*0.9} =1.45
  • C_P = 1.45-sqrt{1.45^2 - {1.61/0.9}} = 1.45-sqrt{2.1025-1.789} = 1.45-sqrt{0.3136} =0.890 > 0.862 → x-axis still critical
  • F prime_c = F_c C_M C_t C_P K_F phi lambda= 1950*1*1*0.862*{2.16/0.90}*0.90*0.8 =2905 psi
  • F prime_c A = 2905*131.3=381400 lb > 60000✔
OK by LRFD

Problem 3.

Illustration for Problem 3 For the shear wall shown Service loads are: Hwind = 75 kips, wD = 2.5 kips/ft and wL = 5.5 kips/ft

Assume wind controls shear. Assume Type N mortar. Assume running bond.

a)Determine the required thickness if it is going to be constructed with solid clay bricks (f’m = 6,000 psi). Use both ASD and LRFD.

Assume shear controls thickness.

ASD:

  • w=2500+5500=8000 lb/ft = 667 lb/in
  • f prime_m = 2000-500*{{6200-6000}/{6200-4150}}=2000-500*{200/2050}=1950 psi
  • f_v le 1.5 sqrt{f prime_m}= 1.5*sqrt{1950}=66.2 psi < 120
  • f_v le 60+0.45 {{N_v}/{A_n}}=60+0.45*{667/b}=60+{300/b}

For b<48”, 66.2 psi controls

  • V=H_{wind}=75000 lb → 781 lb/in
  • f_v = {VQ}/{I_n b} = {3V}/{2bd} ={3*75000}/{2b} =781/b le 66.2
  • → b ge 781/66.2=11.8 in
12” (ASD)

LRFD: 1.2D + 1.6W + L, φ=0.80

  • Nu=1.2*2500+5500=8500 lb/ft = 708 lb/in
  • f prime_m = 2000-500*{{6200-6000}/{6200-4150}}=2000-500*{200/2050}=1950 psi
  • V_n le 3.8 A_n sqrt{f prime_m}= 3.8*b*sqrt{1950}=168 b
  • V_n le 300 A_n = 300 b
  • V_n le 90 A_n+0.45 N_u = 90 b +319
  • V_u=1.6 H_{wind}=1.6*75000120000 lb → 1250 lb/in

Check them

  • V_u le phi V_n right 1250 le 0.8*168 b right b ge9 in
  • V_u le phi V_n right 1250 le 0.8*300b right b ge5 in
  • V_u le phi V_n right 1250 le 72b +255 right b ge13.8 in
14” (LRFD)

b)Determine the required amount of vertical and horizontal reinforcement if it is constructed with 6 in. hollow clay masonry (f’m = 3,000 psi). Use either ASD or LRFD. Assume fy = 60,000 psi.

Assume s=8”

  • A_v = {Vs}/{F_s d}={75000*8}/{60000*4*12}=0.21 in²
Vertical0.21 in² every 2'
Horizontal0.21 in² every 8”

Discussion


Personal Tools