1 asd design of unreinforced walls loaded laterally in-plane asd design of unreinforced walls loaded...
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3 ASD Design of Unreinforced Walls In-Plane 30 ft ft 20 ft Wind 17 psf 17 psf 1 ft Design Width Uniform Load at Roof diaphragm Uniform Load at Roof diaphragm Produced by wall spanning to roof Produced by wall spanning to roof = Load = 17 lb/ft x 20’ x 20’/2 (1/16.67’) = Load = 17 lb/ft x 20’ x 20’/2 (1/16.67’) = 204 lb/ft Reaction to shear wall = 204 x 30/2 Reaction to shear wall = 204 x 30/2 = 3060 lb = 3060 lbTRANSCRIPT
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ASD Design of Unreinforced Walls loaded laterally
In-Plane
Lateral Load for roof or floor
Applied parallel to walllength
Base Shear
Over turning moment
h
Length of wall = L
2
ASD Design of Unreinforced Walls In-Plane
30 ft
30 ft
16.67 ft20 ft Wind 17 psf
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ASD Design of Unreinforced Walls In-Plane
30 ft
30 ft
16.67 ft20 ft
Wind 17 psf
1 ft Design Width
Uniform Load at Roof diaphragm Produced by wall spanning to roof = Load = 17 lb/ft x 20’ x 20’/2 (1/16.67’)= 204 lb/ft
Reaction to shear wall = 204 x 30/2 = 3060 lb
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ASD Design of Unreinforced Walls In-Plane
16.67 ft
Reaction to shear wall = 3060 lb
Lateral Load for roof or floor
Over turning moment = 3060 x 16.67 ft = 51,010 lb.ft
Length of wall = L= 30 ft
On Board
On Board
Assume 8” Hollow block (CMU) wall with a weight of 37 psf
Check base of wall for flexure - Assume Face shell bedded
Bending (normally check .6D+0.6W & D + 0.75L+ 0.75(0.6W) – first load case governs for tension
Section modulus - you have two strips 1.25” wide going the full length of the wall.
So S = 2 x1.25 (30 x 12)2/6 = 54000 in3
fb(wind) = (0.6) x51,010 x 12/54000 = 0.6 x11.33 = 6.8 psi
fa(dead) = 20 x37x30/(30 x 12x(1.25+1.25)) = 24.67 psi
ft= = -24.67( (0.6) +6.8 = -8.0 psi Still in compression so any mortar would be OK.
Check combined bending and axial compression D+.75L+.75 (06W)
Fb = 500 psi h/r = (16.67 x 12)/2.84 = 70.5 (same as before – for Out-of-plane)
Fa = .25 (1500) [1- 16.67(12)/(140 x2.84)]2= 280 psi
fa = 24.67 psi since all dead load and only wind produces flexure so fb = 0.75 (6.8)
Thus fa/Fa+fb/Fb = 24.67/280+.75(6.8)/500 =
= 0.088 + 0.010 = 0.098 <<<1.0 therefore OK .
No applied load at top of wall so no stability check needed
Check shear
Shear is constant over the height of the shear wall and we want to check the greatest shear load under the lowest axial load (gives lowest capacities)
By inspection 0.6D+0.6W will give the above conditions at the top of the wall
Two rectangular strips go the full length of the wall and are essentially two thin rectangular sections 1.25 in wide. So we can say the maximum shear stress=
fv(max wind) =VQ/Ib = 3/2 (V)/An = 3/2 (3060) / (2.5)(30 x12) = 5.1 psi x 0.6 = 3.06 psi
Allowable Shear = Fv = ? = The smallest of the following
psiAN
psipsif
n
v
m
06.3governs psi37)0(45.03745.0
1201.5815005.1 1205.1 '
mf' ofionDeterminatfor Cement masonry S Type assumedcheck stress
n compressio for the sassumption but thek likely wor ismortar Any bedded shell Face CMU Hollow 8" a Shear Usein OK