Design data:

Length of beam: L = 16 ft

Tributary width: s = 8 ft

Top of beam supported by floor joists at 16 in O.C.

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.

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

Calculate Design load: W = [(W_{D} + W_{SD}+ W_{L}]*s = 464 lb/ft

Design moment: M = W*L^{2}/8 = 14850 lb-ft

Use 3-3x12 nailed together with 12d nails at 12 in O.C. from both sides staggered.

Nominal dimension, B = 9 in, D = 12 in

Actual dimension, b = 7.5 in, d = 11.25 in

Section modulus: S = 158.2 in^{3}, Modulus of inertia, I = 890 in^{4}.

Bending stress: f_{b} =M/S = 1126 psi

Try Southern pine No. 2, F_{b} = 1500 psi

The depth to width ratio based on nominal dimension, D/B = 1.33

Since compressive edge is supported by floor joist at 16 in O.C., C_{L} = 1

Wet service factor: C_{M} = 1

Temperature factor: C_{t} = 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_{M}* C_{t} = 1500 psi O.K.

Check deflection:

Elastic modulus: E = 1600000 psi*CM* Ct = 1600000 psi

Deflection: D = 5*W*L^{4}/(384*E*I) = 0.48 in < L/240 O.K.

Check shear stress

Maximum shear force. V = W*L/2 = 3840 lb

Shear stress, fv = V/bd = 46 psi

Conservatively assume shear stress factor, C_{H} = 1

Allowable shear stress, Fv = 90 psi * C_{D}*C_{M}* C_{t }*C_{H} = 90 psi O.K.

- Design of Sawn Timber Beams or Joists
- Solved Example: Design of 2x10 Timber Floor Joist with Southern Pine
- Solved Example: Design of Sawn Timber Column
- Solved Example: Design of Timber Load Bearing Stud Wall
- Solved Example: Design of 2x12 Timber Floor Joist with Douglas Fir-Larch