Design data:

Tributary width of floor supported by wall: B = 20 ft^{2}.

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.

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. d_{1} = 3.5 in, d_{2} = 1.5 in, s = 16 in

Actual cross section: A_{c} = 5.25 in^{2}.

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}/d_{1}=34 < 50

Provide blocking at mid-height in d_{2} direction

L_{ey}=K_{e}´(L/2) = 5 ft, L_{ey}/d_{2}=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´10^{6} 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} = 0.252

8. Calculate allowable compressive stress

F”_{c} = F_{c}*´C_{p} = 246 psi > f_{c}= 228.5 psi O.K.

- Design of Sawn Timber Columns and Compressive Members
- Solved Example: Design of Sawn Timber Column
- Design of Sawn Timber Beams or Joists
- Typical Load Combinations For Residential Buildings Design
- Solved Example: Design of a load-bearing brick (unreinforced masonry) wall (BS 5628)