Question

1. A typical 2x6 stud wall is used for a bearing wall with 2x6 at 16” o.c. Determine the allowable wall load (in lb/ft) for a wall height of L = 10 ft. Assume No.2 SPF, D+L governs, internal exposure with climate control, pin-pin connections, MC<19%, T<100°F. The wall is braced against weak axis buckling at mid-height.
   
   Assume member length of L=4ft

Answer 1

Given,

Length of wall = 10ft

Moisture <19%

T<100 F

Spacing = 16"

For a 2 x 6 SPF No.2 , Area(A_c) = 8.25 in² and E = 1.4 x 10⁶ psi

Allowable compressive stress parallel to grain(F_c) = 975 psi

Step 1) Calculate the slenderness ratio(s),

s = KL/d1

Since both ends are pinned, K = 1

Therefore, s = 1 x 10 x 12/ 5.5

21.81 < 50

Since wall is braced at mid height,therefore slenderness ratio along weak axis(s'),

s' = KL/2d₂

= 1 x10 x12/(2 x 1.50)

= 40 < 50 (Govern)

Step 2 - Calculation of compressive stress with Load duration factor,

Load duration factor for dead load = 0.90

Load duration factor for live load = 1.0

Design load(P) = (DL + LL) x s

Assume dead load of 10psf and live load of 20psf, and width of wall as 20ft

P = (10+20) x 20 x16/12

= 800 lb/ft

Column compressive stress = P/A_c

=96.96 psi

Step 3 Calculate the allowable stress

F"c = Fc x Cd x Cm x Ct x Cf

where Cd is load duration factor

Cm = Wet service factor

Ct = Temperature factor

Cf = size factor

For given temperature, moisture and loading conditions,

Cd = 1 , Cm= 1, Ct = 1 and Cf =1

F"c = 975 x 1 x1 x1 x1

= 975 psf

F_ce = Kc x E/(Le/d)²

Here Kc = 0.30

Fce = 0.30 x 1.40x 10⁶/(120/87.50)²

= 262.5 psi

Step 4) Calculation of Cp

Let Fce/F"c = m

Cp = 1 + m /(2C) -

On solving, Cp= 0.252

Step 5 Allowable compressive stress,

F'c = F"c x Cp

= 975 x 0.252

= 245.70 psi > 96.96 psi

Hence OK

Therefore load on wall = F'c x area

= 96.96 x 10 x 20 x 12 x 12

= 2792.45 kips