In the previous section, the method for determining the size of load that a beam will support was illustrated. In some situations, it is necessary to determine the size beam that is necessary to support a given load. The same equations are used, but they are rearranged to solve for the section modulus. Once the section modulus is known, the beam size can be determined from Table 24.1 if one dimension is 2 inches (nominal), and the orientation is known.
Problem: What is the smallest size of 2-inch, No. 1, southern pine simple beam that will support a uniform load of 2,400 lb if the beam is 100 inches long?
Solution: Using Table 24.1 and Appendix XI, and rearranging the equation:
Referring to Table 24.1, the smallest beam size with a section modulus equal to or larger than 30 cubic inches is a beam 2 inches wide and 12 inches deep. Notice that a 2 x 14 inch beam also would carry the load, but is larger than necessary.
The procedures presented in the previous sections are not limited to boards 2 inches wide. If the section modulus and one dimension are known, the width of a beam greater than 2 inches can be determined if the equation is rearranged.
Problem: What depth of board is needed in the previous problem if a board 4 inches wide is used?
Solution: Rearranging the equation:
K = - x a x b2 b = J-= J-2— = V45 = 6.708 ... or 6.7 in
This example illustrates that a beam 4 x 6.7 inches (actual size) will support the same load as a beam 2 x 12 inches (nominal size).
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