Select a wide-flange beam to support the 15-kip load as shown in Fig. 12.14. The allowable normal stress for the steel used is 24 ksi.

Question

Accepted Answer

1. The allowable normal stress is given: [latex]σ_{all} = 24 ksi[/latex].
2. The shear is constant and equal to 15 kips. The bending moment is maximum at B. We have

[latex]|M|_{max} = (15 kips) (8 ft) = 120 kip · ft = 1440 kip · in.[/latex]

3. The minimum allowable section modulus is

[latex]S_{min} = \frac{|M|_{max}}{σ_{all}}= \frac{1440 kip · in.}{24 ksi}= 60.0 in^{3}[/latex]

4. Referring to the table of Properties of Rolled-Steel Shapes in App. B, we note that the shapes are arranged in groups of the same depth and that in each group they are listed in order of decreasing weight. We choose in each group the lightest beam having a section modulus S = I/c at least as large as [latex]S_{min}[/latex] and record the results in the following table.

[latex]S, in^{3}[/latex]	Shape
81.6	W21 × 44
88.9	W18 × 50
64.7	W16 × 40
62.6	W14 × 43
64.2	W12 × 50
60.0	W10 × 54

The most economical is the W16 × 40 shape since it weighs only 40 lb/ft, even though it has a larger section modulus than two of the other shapes.
We also note that the total weight of the beam will be (8 ft) × (40 lb) = 320 lb. This weight is small compared to the 15,000-1b load and can be neglected in our analysis.

Question 12.4: Select a wide-flange beam to support the 15-kip load as show...

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