Design Load

Подпись: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.

Подпись:Design Criteria

One needs to find the maximum practical span that the design element can withstand.

Stud Spacing

Consider 12-in. 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)

Подпись: l Design Load Подпись: 10.95 Подпись: 1181.25 x 1.055 680 Подпись: 1/2

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

Подпись: Copyright © Marcel Dekker, Inc. All rights reserved.l = 13-3 x 237-5 x 8-438 + 2 x 0.75 = 40.7 in

Deflection

 

1/3

 

Подпись:l = 1.69 (■

w

From tables we can get:

• I = 0.396 in.4

Design Load Design Load Подпись: 1.6 x 106 psi = 16.507 in.

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 237-5 X 5-25 + 2 x 3.5 = 31.387 in. 680

Design Load

Deflection

 

As before, E’ = 1.6 X 106 psi.

 

1/3

 

1.6 X 106 X 5.359
680

 

1.69

 

39.336 in.

 

l

 

Bending governs; span = 26.150 in. (take 2 ft). Final stud spacing is 2 ft.

 

Подпись: Copyright © Marcel Dekker, Inc. All rights reserved.

Double Wales

Try 2 (2 X 4) Douglas Fir (properties are same as above).

Load//’ = 680 X (stud spacing)

X 1 /’ of wales = 1360 lb/ft

 

Design Load

Design Load

Design Load

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 (237-5 X 5-25 j + 2 x 3.5 = 31.387 in. 680 /

Deflection

As before, E’ = 1.6 x 106 psi.

 

Подпись: Copyright © Marcel Dekker, Inc. All rights reserved.

1/3

 

1.6 x 106 x 5.359
680

 

l

 

1.69

 

39.336 in.

 

Design Load

Design Load

• 3000-lb 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.