Bearing Stress

1. Studs on wales:

Bearing area = bwale x bstal x 2.0

(we multiply by 2 because of the double wale).

Load transmitted by bearing = pressure intensity x stud spacing x wale spacing = 680.0 x 1.0 x 2.0 = 1360.0 lb

From tables we can get:

• Bearing area factor = CB = 1.25

• Temperature factor = Ct = 1.0

• FC1 = 625.0 psi

Подпись:FCі = Fd (CB) (Ct)

F’C1 = 625 x 1.25 x 1.0 = 781.25 psi

The calculated value of bearing stress is:

Fcі,,„„ = д1”/ 2 = 302.22 < 781.25 (safe)

Подпись: 2720.0 781.25 2. Tie plate:

Tie load = (pressure intensity) x (tie spacing)

x (wale spacing) = 680.0 x 2 x 2 = 2720.0 lb

Allowable stress in bearing is 781.25 psi.

Area of bearing shown in the figure =

(width of wale) x (required width of tie plate) x 2 (since we have 2 wales)

B, the required width of the tie plate, can be obtained as follows: 2 x 1.5 x B

B > 1.161 in.; take B

Strut Load

Подпись: 1.5 in.

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

Try 4 x 4 in.

0.3E’ = 0.3 x 1.6 x 106 (l/d )2 (48.48)2

Подпись: FcE in strut204.228 psi

F* = Fc x (all factors except Cp)

F* = Fc(Cd)(Cm)(Q(Cf)

F* = 1300.0 x 1.25 x 1.0 x 1.0 x 1.5 = 2437.5 psi Cp = column stability factor

Bearing Stress2437’5 = 0.0823 0.8

Подпись: d Bearing Stress

F’c = F* (CP) = 2437.5 x 0.0823 = 200.629 psi (unsafe) Try 5 x 5 in. Following the same procedure, we get

Cp = 0.03784 ^ F’c = 922.2754 psi

Allowable P is 922.2754 x (4.5)2 = 18676.07 lb. and spacing be­tween struts = 18676.07/2121.30 = 8.8041 ft. Take spacing = 8 ft.

EXAMPLE 2

Rework Example 1 with sheathing that has the following character­istics:

Plywood is Type: APA B-B PLYFORM CLASS 1 with species group of face ply = 1.

Подпись: Copyright © Marcel Dekker, Inc. All rights reserved.Dry condition. Thickness: 7/8 in.

Подпись: 680 =From Tables 3.11 and 3.12 of geometric properties, we get:

• KS = 0.515 in.

• I = 0.278 in.4

• lb/Q = 8.05 in.2

From Table 3.14 of mechanical properties, we can obtain the fol­lowing design values:

• Fb = 1650.0 psi.

• Fv = 190.0 psi

• E = 1.8 X 106 psi

Concrete pressure is 680.0 psf. [see Example 1]

STUD SPACING:

Bending

KS

Wb = 680 = 120 X Fb X KS-

11

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

120 X 1650.0 X 0.515

Due to bending only

Подпись: A A A. 360 l3 J_

360

wl 3

1743EI

_______ 680 x 14______

1743 x 1.8 x 106 x 0.278 15.273 in.

Bending governs. Stud spacing = 12.246 in. rounded down to 12 in. = 1 ft. The rest of the design is as before.

Bearing Stress

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