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Solved Example: Design of Timber Load Bearing Stud Wall

Articles > Solved Example: Design of Timber Load Bearing Stud Wall

Question:

Design data:

Tributary width of floor supported by wall: B = 20 ft2.

Unsupported height of stud wall, L = 10 ft

Hinge support at top and bottom of stud wall

Design load:

Floor live load: WL = 30 psf

Floor dead load: WD = 10 psf

Superimposed dead load: WSD = 5 psf

Timber: Southern pine, moisture less than 19%, used in normal room temperature.

 

Solution:

Refer to Design of Sawn Timber Columns and Compressive Members article for concepts.

1. Select southern pine, 2"x4" stud grade at 16" O.C. d1 = 3.5 in, d2 = 1.5 in, s = 16 in

Actual cross section: Ac = 5.25 in2.

Allowable compressive stress parallel to grain: Fc = 975 psi

2. Calculate slenderness ratio:

Ke = 1, Lex = Ke´L = 10 ft, Lex/d1=34 < 50

Provide blocking at mid-height in d2 direction

Ley=Ke´(L/2) = 5 ft, Ley/d2=40 < 50 Govern

3. Calculate compressive stress with load duration factor

Load duration factor for dead load: CD = 0.9

Load duration factors for live load: CD = 1.0

Calculate Design load: P = [WD + WSD+ WL]´B´s = 1200 lb

Column compressive stress, fc=P/Ac = 228.5 psi

4. Calculate allowable stress without Cp

CM=1, Ct=1, Cf=1

Fc* = Fc´CD´CM´Ct´CF = 975 psi

5. Calculate elasticity modulus

E’=E´CM´Ct = 1.4´106 psi

6. Calculate FcE

KcE=0.3

FcE= KcE*E’/(Le/d)2= 262.5 psi

7. Calculate Cp

c = 0.8

C_p=frac{1+F_{cB}/F^*_c}{2c}-sqrt{[frac{1+F_{cB}/F^*_c}{2c}]^2-frac{F_{cB}/F^*_c}{c}}

Cp = 0.252

8. Calculate allowable compressive stress

F”c = Fc*´Cp = 246 psi                  >          fc= 228.5 psi        O.K.

 

 


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