Solved Example: Design of 2x10 Timber Floor Joist with Southern PineArticles > Solved Example: Design of 2x10 Timber Floor Joist with Southern Pine
Length of floor joist: L = 16 ft
Spacing of floor joist: s = 16 in.
Top of joist supported by plywood sheathing.
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 = 77.3 lb/f
Design moment: M = W*L2/8 = 2475 lb-ft
Try 2x10 joist
Nominal dimension, B = 2 in, D = 10 in
Actual dimension, b = 1.5 in, d = 9.25 in
Section modulus: S = 21.39 in3, Modulus of inertia, I = 98.93 in4.
Bending stress: fb =M/S = 1388 psi
Try Southern pine No. 2, Fb = 1500 psi
Load duration factor for dead load: CD = 0.9
Load duration factors for live load: CD = 1.0(Use 1 per NDS 2001)
The depth to width ratio based on nominal dimension, D/B = 5
Since compressive edge is fully supported by plywood floor, CL = 1
Repetition factor for joist: Cr = 1.15
Wet service factor: CM = 1
Temperature factor: Ct = 1
Other factors not applicable
Allowable stress, F’b = Fb*CD* CL* Cr* CM* Ct = 1725 psi O.K.
Elastic modulus: E = 1600000 psi*CM* Ct = 1600000 psi
Deflection: D = 5*W*L4/(384*E*I) = 0.75 in < L/240 O.K.
Check shear stress
Maximum shear force. V = W*L/2 = 640 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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