Solved Example: Design of 2x12 Timber Floor Joist with Douglas Fir-Larch

Articles > Solved Example: Design of 2x12 Timber Floor Joist with Douglas Fir-Larch

Question:

Design data:

Length of floor joist: L = 16 ft

Spacing of floor joist: s = 16 in.

Top of joist supported by plywood sheathing.

Design load:

Floor live load: W_L = 40 psf

Floor dead load: W_D = 10 psf

Superimposed dead load including mechanical and electric load, W_SD = 8 psf

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

Solution:

Refer to Design of Sawn Timber Beams or Joists article for the concepts.

Calculate Design load: W = [W_D + W_SD+ W_L]*s = 77.3 lb/f

Design moment: M = W*L²/8 = 2475 lb-ft

Try 2x12 joist

Nominal dimension, B = 2 in, D = 12 in

Actual dimension, b = 1.5 in, d = 11.25 in

Section modulus: S = 31.64 in³, Modulus of inertia, I = 178 in⁴.

Bending stress: f_b =M/S = 938.5 psi

Try Douglas Fir-Larch No. 1, F_b = 1000 psi

The depth to width ratio based on nominal dimension, D/B = 6

Since compressive edge is fully supported by plywood floor,

Provide solid blocking at both ends, and cross bracing at mid-span,

Maximum spacing = 8 ft, C_L = 1

Repetition factor for joist: C_r = 1.15

Wet service factor: C_M = 1

Temperature factor: C_t = 1

From NDS Table, size factor, C_F = 1

Load duration factor for dead load: C_D = 0.9

Load duration factors for live load: C_D = 1.0

Other factors not applicable

Allowable stress, F’_b = F_b* C_D*C_L* C_r* C_M* C_t*C_F = 1150 psi O.K.

Check deflection:

Elastic modulus: E = 1700000 psi*C_M* C_t = 1700000 psi

Deflection: D = 5*W*L⁴/(384*E*I) = 0.38 in < L/240 O.K.

Check shear stress

Maximum shear force. V = W*L/2 = 640 lb

Shear stress, fv = V/bd = 38 psi

Conservatively assume shear stress factor, C_H = 1

Allowable shear stress, Fv = 95 psi * C_D*C_M* C_t *C_H = 95 psi O.K.

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