Building construction and superintendence, by F.E. Kidder ... (1896)
Kidder, Frank E. (Frank Eugene), 1859-1905 · New York, W.T. Comstock · leaf 252
d in the same way as for trusses supported atthe ends, but to make the method perfectly plain we will give afew examples. EXAMPLE 11.—To find the respective loads on Px and P2,Fig- 257. Ans.—We can take moments about any point in the beam, butin this particular case it will be simplest to take… … Fig. 264. beam in the opposite direction, and we will call its moment minus. 10 x 22 8 x 13... 7x 7 9 x 3 Total moment Multiplying each weight by its arm, we have: = +220 = +104 = +49 = -27 +373 — 27 = 346. This moment must be resisted by Px acting with an arm of 17 ft. 346 Consequently Px must…
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