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: W_L = 30 psf

Floor dead load: W_D = 10 psf

Superimposed dead load: W_SD = 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: A_c = 5.25 in².

Allowable compressive stress parallel to grain: F_c = 975 psi

2. Calculate slenderness ratio:

K_e = 1, L_ex = K_e´L = 10 ft, L_ex/d₁=34 < 50

Provide blocking at mid-height in d₂ direction

L_ey=K_e´(L/2) = 5 ft, L_ey/d₂=40 < 50 Govern

3. Calculate compressive stress with load duration factor

Load duration factor for dead load: C_D = 0.9

Load duration factors for live load: C_D = 1.0

Calculate Design load: P = [W_D + W_SD+ W_L]´B´s = 1200 lb

Column compressive stress, f_c=P/A_c = 228.5 psi

4. Calculate allowable stress without C_p

C_M=1, C_t=1, C_f=1

F_c* = F_c´C_D´C_M´C_t´C_F = 975 psi

5. Calculate elasticity modulus

E’=E´C_M´C_t = 1.4´10⁶ psi

6. Calculate F_cE

K_cE=0.3

F_cE= K_cE*E’/(L_e/d)²= 262.5 psi

7. Calculate C_p

c = 0.8

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

C_p = 0.252

8. Calculate allowable compressive stress

F”_c = F_c*´C_p = 246 psi > f_c= 228.5 psi O.K.

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