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Design of a Prismatic Beam for Bending

Problem 4: Design of a Prismatic Beam for Bending...A 13 ft long steel cantilever beam carries a distributed load which varies from 0.2 kips/ft to 1.5 kips/ft as shown. Determine the most economical U.S. customary S or W shape if the factor of safety is 2.4.

Solution: For the free body diagram, I draw the distributed load as a rectangle and a triangle (one might alternatively draw it as two triangles)...

Taking the usual directions as positive...

We know that 

I obtained this equation by using the point-slope form of the straight line, where the slope is 0.1 kip/ft and the line passes through x = 0 ft & y = 0.2 kips/ft. Integrating, and noting (using a short mental free body diagram about the left end) that the shear at the left end is zero,

A mental free body about A shows that the shear there (on a left free body) is - 11.05 kips. This equation yields the same thing at x = 13 ft. So it checks!

Since dM/dx = V, and the moment at the left end is zero,

Does this check at the right end? A substitution of x = 13 ft yields - 53.52 kip-ft, as it should!

There is no need to draw the shear and bending moment diagrams to solve this problem, but I will because it may help to understand why the derivative of the moment cannot be used to find the maximum moment. (Try equating the shear to zero to find out where the moment reaches a maximum!)

The largest absolute value of the moment is 53.52 kip-ft, and it is reached at x = 13 ft.

From Appendix B, the ultimate stress for structural steel is 58 ksi. Therefore, the allowable stress is 58/2.4 = 24.17 ksi. The stress must be equal to or smaller than 24.17 ksi:

From Appendix C,

designation Sx (in3)
S 10 x 35
29.4
W 14 x 22
29.0
W 12 x 26
33.4
W 10 x 30
32.4
W 8 x 31
27.5

All of these satisfy the required section modulus. But the W 14 x 22 will be the most economical, since its weight is 22 lb per ft

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