Problem 1

[Formula derivation]

Problem 2

150 psf → d=0.75” [tbl.13-6]; EI=0.298×10⁶; Fsks=0.878×10³; Fsib=0.517×10³

Bending, plywood >3' span: $l=10.95*sqrt{Fsks/w} = 10.95*sqrt{0.878*10^3/150}=$ 26.5”
Shear, plywood: $l=20*{Fsib/w}+2d=20*{0.517*10^3 / 150}+2*0.75=$ 70.4”
deflection: $l=1.94*root{3}{EI/w}=1.94*root{3}{0.298*10^6 /150}=$ 24.4”

24”

cw=1; cc=1; E=1.6E6*0.97=1.55E6; h=8; wmax=150h=1200psf;

$P=c_w c_c(150+{900R}/T)=1*1*(150+{9000*53/72})=$ 734 psf

$P_{min}=600 c_w = 600*1=$ 600 < 734 → w=734 psf <1200✔

F_b=1400*1.25*.86= 1505 psi

F_c=975*1.25*.7= 853 psi

$F_{c⊥}=405*1.25*0.67=$ 339 psi

$F_{vc}=180*1.25*0.97=$ 218 psi

F_t=675*1.25*0.84= 866 psi

d=0.75; EI=0.298E6; Fbms=878.9; Fhlb/Q=517;

l=12”

p= 1.25wh=1.25*150*18= 3375

3375 psf = 23.4 psi

1” nominal lumber sheathing → t=0.75”; assume 7-day wet conditions.

w=600 lb/ft=50 lb/in; b=12”; d=0.75”; S=3.063 in³; Fb=1150 lb/in²; Fv=150 lb/in²; A=12*0.75 in²; EI=1.4E6*12*0.75^3

Bending: l= $4.46 d sqrt{{F_b b}/w} = 4.46*0.75*sqrt{{1150*12}/{50}}=$ 55.57”
Shear: l= $13.3{F_v A/w} +2d = 13.3*{150*12*0.75/50} +2*0.75=359.1+1.5=$ 360.6”
Deflection: l= $1.94 root{3}{EI/w} = 1.94*root{3}{1.4*10^6*12*0.75^3/50}=$ 101.2”

55.6”

w=1000 lb/ft=83.33 lb/in; b=1.5”; d=3.5”; S=3.063 in³; Fb=1450 lb/in²; Fv=185 lb/in²; A=5.25 in²; EI=1.7E6*5.359

Bending: l= $10.95 sqrt{{F_b S}/w} = 10.95*sqrt{{1450*3.063}/{83.33}}=$ 79.94”
Shear: l= $13.3{F_v A/w} +2d = 13.3*{185*5.25/83.33} +2*3.5=155.1+7=$ 162.1”
Deflection: l= $1.69 root{3}{EI/w} = 1.69*root{3}{1.7*10^6*5.359/83.33}=$ 80.8”