Solved Example: Design of Timber Load Bearing Stud WallArticles > Solved Example: Design of Timber Load Bearing Stud Wall
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
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.
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
FcE= KcE*E’/(Le/d)2= 262.5 psi
7. Calculate Cp
c = 0.8
Cp = 0.252
8. Calculate allowable compressive stress
F”c = Fc*´Cp = 246 psi > fc= 228.5 psi O.K.
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