Installation — business terrible  1 part
September 8th, 2015
Lateral pressure: P = 150 +
P = 150 +
= 679.41
Assume construction joint every 5 ft:
150 x h = 150 x 5 = 750 psf
680 < 3000 and 680 < 150 x h (OK)
The design value for lateral pressure is 680 psf.
Design Criteria
One needs to find the maximum practical span that the design element can withstand.
Stud Spacing
Consider 12in. strip. Load/ft’ = 680 lb/ft2
• Fb = 875 psi
• Flat use factor Cfu = 1.2
• Size factor Cf = 1.2
• Fv = 95 psi (here we have no split)
• Temperature factor = Ct = 1.0
• Load duration factor = CD = 1.25
(Load duration = 7 days for most formwork unless otherwise stated.)
Bending
Allowable stress = F’b = Fb(CfU)(Ct)(Cf)(CD)
Fb = 875 x 1.2 x 1.0 x 0.9 x 1.25 = 1181.25 psf
= 14.824 in.
Shear
Allowable stress = F’v = Fv(CH)(Ct)(CD) Fv = 95 x 2.0 x 1.0 x 1.25 = 237.5 psi
l = 13.3 + 2 x d
From tables we can get:
• d = 0.75 in.
• A = 8.438 in.2
l = 133 x 2375 x 8438 + 2 x 0.75 = 40.7 in



l = 1.69 (■
w
From tables we can get:
• I = 0.396 in.4
•
E = 1,600,000 psi
Hence sheathing will be supported by studs, with a spacing of 12 in. (1 ft).
Wale Spacing:
Load/ft’ = w x (stud spacing) x 1 ft’ of wale span
= 680.0 x 1.0 x 1.0 = 680.0 lb /ft
Try 1 (2 x 4) Douglas Fir.
From tables we can get: [20]
l Shear F’v = Fv(CH)(Ct)(CD) Fv = 95.0 X 2.0 X 1.0 X 1.25 = 237.5 psi F’ A l = 13.3 FvA + 2d w l = 13.3 X 2375 X 525 + 2 x 3.5 = 31.387 in. 680 

















Shear
F’v = Fv(CH)(Ct)(CD)
F’v = 95.0 X 2.0 X 1.0 X 1.25 = 237.5 psi
l = 13.3 (^ w
l = 13.3 (2375 X 525 j + 2 x 3.5 = 31.387 in. 680 /
Deflection










• 3000lb ties will be used.
• Force/tie = (lateral concrete pressure) x (wale spacing) x (tie spacing), which should not be greater than 3000 lb.
• 3000 = 680 x 2 x tie spacing ^ tie spacing < 2.206 ft.
Take tie spacing = 2 ft.