1 Beam of L= 10 ft length, with uniform load w= 280plf (W = 2800 lbs)
2 Cross-section of wood I-beam
For the formula v= VQ/(Ib) we must find the moment of inertia of the entire cross-section. We could use the parallel axis theorem of Appendix A. However, due to symmetry, a simplified formula is possible, finding the moment of inertia for the overall dimensions as rectangular beam minus that for two rectangles on both sides of the web.
Note c= 10/2 = 5 (half the beam depth due to symmetry)
Static moment Q of flange about the neutral axis:
Shear stress at flange/web intersection:
Static moment Q of flange plus upper half of web about the neutral axis
Maximum shear stress at neutral axis:
Note: Maximum shear stress reaches almost the allowable stress limit, but bending stress is well below allowable bending stress because the beam is very short. We can try at what span the beam approaches allowable stress, assuming L= 30 ft, using the same total load W = 2800 lbs to keep shear stress constant:
At 30 ft span bending stress is just over the allowable stress of 1450 psi. This shows that in short beams shear governs, but in long beams bending or deflection governs.
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