Читать книгу Experimental Mechanics. A Course of Lectures Delivered at the Royal College of Science for Ireland онлайн

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Fig. 18.

41. We have no means of showing the magnitude of the strain along the strut, but we shall prove that it can be computed by means of the parallelogram of force; this will also explain how it is that the tie is strained by a force three times that of the weight which is used. Through c (ssss1) draw c p parallel to the tie a b, and p q parallel to the strut c b then b p is the diagonal of the parallelogram whose sides are each equal to b c and b q. If therefore we consider the force of 20 lbs. to be represented by b p, the two forces into which it is decomposed will be shown by b q and b c; but a b is equal to b q, since each of them is equal to c p; also b p is equal to a c. Hence the weight of 20 lbs. being represented by a c, the strain along the tie will be represented by the length a b, and that along the strut by the length b c. Remembering that a b is 3' long, c b 3' 6", and a c 1', it follows that the strain along the tie is 60 lbs., and along the strut 70 lbs., when the weight of 20 lbs. is suspended from the hook.