Solved Example: Design of 3-3x12 Timber Beam with Southern PineArticles > Solved Example: Design of 3-3x12 Timber Beam with Southern Pine
Length of beam: L = 16 ft
Tributary width: s = 8 ft
Top of beam supported by floor joists at 16 in O.C.
Floor live load: WL = 40 psf
Floor dead load: WD = 10 psf
Superimposed dead load including mechanical and electric load, WSD = 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 = [(WD + WSD+ WL]*s = 464 lb/ft
Design moment: M = W*L2/8 = 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 in3, Modulus of inertia, I = 890 in4.
Bending stress: fb =M/S = 1126 psi
Try Southern pine No. 2, Fb = 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., CL = 1
Wet service factor: CM = 1
Temperature factor: Ct = 1
Load duration factor for dead load: CD = 0.9
Load duration factors for live load: CD = 1.0
Other factors not applicable
Allowable stress, F’b = Fb* CD*CL* CM* Ct = 1500 psi O.K.
Elastic modulus: E = 1600000 psi*CM* Ct = 1600000 psi
Deflection: D = 5*W*L4/(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, CH = 1
Allowable shear stress, Fv = 90 psi * CD*CM* Ct *CH = 90 psi O.K.
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