**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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