Monday, July 7, 2014


Since this is not a rectangular beam, shear stress must be computed by the general shear formula.  The maximum shear stress at the neutral axis as well as shear stress at the intersection between flange and web (shear plane As) will be computed.  The latter gives the shear stress in the glued connection.  To compare shear- and bending stress the latter is also computed.

Beam of L= 10 ft length, with uniform load w= 280plf (W = 2800 lbs)
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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