1. 1 ASD Design of Unreinforced Walls loaded laterally In-Plane ASD Design of Unreinforced Walls loaded laterally In-Plane Lateral Load for roof or floor Applied parallel to wall length Base Shear Over turning moment h Length of wall = L

2. 2 ASD Design of Unreinforced Walls In-Plane 30 ft 16.67 ft 20 ft Wind 17 psf 17 psf

3. 3 ASD Design of Unreinforced Walls In-Plane 30 ft 16.67 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 lb

4. 4 ASD Design of Unreinforced Walls In-Plane 16.67 f t Reaction to shear wall = 3060 lb 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

5. On Board

7. 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 in 3 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.

8. 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