Homework 2 (Due Monday, September 17)

— David Wagner 2007/09/17 12:22

Redo Example 3.10 changing all the parameters (wind and seismic) to San Antonio, Texas. Assume standard occupancy, Exposure C, Kzt= 1.0 and I = 1.0. For loads assume roof dead load D = 25 psf, roof live load Lr = 20 psf, floor dead load D = 16 psf, floor live load L = 60 psf. the Soil is site Class D. Use both ASD and LRFD.

Wind Forces

San Antonio: 90 mph design wind speed, flat roof

Roof Angle	Load Case	A	B	C	D	E	F	G	H	EOH	GOH
0 to 5°	1	12.8	-6.7	8.5	-4.0	-15.4	-8.8	-10.7	-6.8	-21.6	-16.9

Building Geometry –remains unchanged

h_mean=19 ft; 0.4 h_mean = 7.6 ft; Least horiz. dim. = b=32 ft; 0.1b=3.2 ft

End zone: a = min(0.4 h_mean, 0.1b) = a=3.2 ft

End zone length = 2a=6.4 ft

Wind Pressures I=1.0 ; Kzt=1.0 ; λ= 1.29- (20-19){{1.29-1.21}/{20-15}}=1.29-0.016= 1.274 ; $p_s=lamda K_{zt} I p_{s30}$

Zone A: $p_{s30}=1.274*12.8=$ 16.3 psf
Zone C: $p_{s30}=1.274*8.5=$ 10.8 psf

Load to Diaphragms

w_end= 16.3*6.5=106 lb/ft
w_int= 10.8*6.5=70.2 lb/ft
W_r= $w_{end}2a + w_{int}(L-2a)=$ 4441 lb

$R_{rA}=w_{end}2a(L-a)(1/L) + w_{int}(L-2a)({L/2}-a)(1/L) =$ 106*2*3.2*(60-3.2)*{1/60} + 70.2*(60-3.2*2)({60/2}-3.2)(1/60)= 642.22+1680.68= 2323 lb

$R_{rB}= W_r - R_{rA} = 4441-2323=$ 2118 lb

$w_{min}=p_s=10*6.5=$ 65 lb/ft
$W_{min}=w_{min}L=65*60=$ 3900 lb
$R_{rA}=R_{rB}=3900/2=$ 1950 lb < 2118 < 2323
∴Minimum wind pressure does not govern

R_r= 2323 lb

Second Floor

$w_{end}=16.3*9.5=$ 154.85 ; $w_{int}=10.8*9.5=$ 102.6 ; $W_2=2a w_{end} + w_{int}(L-2a) = 2*3.2*154.85 + 102.6*(60-6.4)=991.04+5499.36=$ 6490 lb

$R_{rA}=w_{end}2a(L-a)(1/L) + w_{int}(L-2a)({L/2}-a)(1/L) =$ {154.85*6.4*56.8}/60 + {102.6*53.6*26.8}/60 =938.2+2456.4= 3395 lb

$R_{2B}=W_2-R_{2A}=6490-3395=$ 3095 lb

$w_{min}=p_s=10*9.5=$ 95 lb/ft
$W_{min}=w_{min}L=95*60=$ 5700 lb
$R_{rA}=R_{rB}=5700/2=$ 2850 < 3095 < 3395
∴Minimum wind pressure does not govern

R₂= 3395 lb

Seismic Forces

T_a=0.182 sec ; S_s=0.10g ; S₁=0.03g ; F_a= 1.57; F_v= 2.4;

S_MS= 0.157g
S_M1 0.072g
S_DS= $2/3 S_{MS}=$ 0.105g
S_D1 $=2/3 S_{M1}=$ 0.048g
T_S $={S_{D1}}/{S_{DS}} = 0.048/0.105=$ 0.46 sec

T_a < T_S → S_DS controls

Redundancy ρ=1.0

Base Shear $C_s=S_{DS}I/R = 0.105g*1.0/5=$ 0.021g > 0.01g✔

W_1,r=25*32 + 780= 1580 lb/ft ; W'_r=1580*60 = 94.8 k ; W_r=94.8+25.0= 119.8 k

W_1,2=16*32+320= 512+320+1140= 1972 lb/ft ; W'₂=1972*60 = 118.3 k ; W₂=1183.+36.5 = 154.8 k

Story	Height hx	Weight wx	wx hx	Story force Fx=0.0015hx wx	Fx Coef.	Story Shear
R	19	119.8	2276.2	3.41 k	0.0285	3.41 k
2	10	154.8	1548.0	2.32 k	0.0150	5.73 k
1	0
Sum		274.6	3824.2	V=CsW=0.021*274.6 = 5.77 k

$F_x={5.77 h_x}/3824.2=$ 0.0015hx

w_ur=0.0285*1580= 45.03 lb/ft
R_ur= $w_{ur}L/2=45.03*60/2=$ 1.35 k
w_u2=0.0150*1972= 29.58 lb/ft
R_u2= $w_{ur}L/2=29.58*60/2=$ 0.88 k

R_ur=1.35 k
R_u2=0.88 k

Midheight Second-story Wall Shear

V_u2r=½F₂=1.71 k

V_u2r=1.71 k

Shearwall wind load: W=Rr = 2323 lb Shearwall Seismic Force: QE=Vu2r=1.71k

ASD Design-Second Story

1.0 W = 2323 lb

0.7 E = 0.7 rho Q E = 1197 lb

LRFD Second Story

1.6W=1.6*2323=3717 lb
1.0E=1.0*rho QE=1.0*1.0*1.71k = 1710 lb

Medheight First-story Wall Shear

V_u12=half*5.77=2.89 k

V_u2r=1.71 k

Shearwall wind load: W=Rr+R2 = 2323+3395=5718 lb Shearwall Seismic Force: QE=Vu12=1.16k

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