. The strength of materials; a text-book for engineers and architects. E> M for Unijorm Load. C. Combined 65a. To jfind the reactions take moments round a (Fig. 65a), Then 18 R„ = 6 X 12 + 2 X 24 = 120 •• ^ 18^ ^ ^^ ^^^^.-. R^ = 8 - 6f = 11 tons. BENDING MOMENTS AND SHEARING FORCES 147 The shear at c will be = 2 tons. It then increases until thepoint B is reached, when its value becomes equal to 35 then suddenly changes sign to a value 3*17 tons and thendecreases uniformly to the end A, where the value comes shear diagram then curves as shown in the figure, thedotte
. The strength of materials; a text-book for engineers and architects. E> M for Unijorm Load. C. Combined 65a. To jfind the reactions take moments round a (Fig. 65a), Then 18 R„ = 6 X 12 + 2 X 24 = 120 •• ^ 18^ ^ ^^ ^^^^.-. R^ = 8 - 6f = 11 tons. BENDING MOMENTS AND SHEARING FORCES 147 The shear at c will be = 2 tons. It then increases until thepoint B is reached, when its value becomes equal to 35 then suddenly changes sign to a value 3*17 tons and thendecreases uniformly to the end A, where the value comes shear diagram then curves as shown in the figure, thedotted lines indicating what occurs in practice owing to theimpossibility of getting the loads and reactions concentratedon a mathematical point. Considering first the for the isolated and uniform loads separately, the curve due to the isolated load will come as shown in the figure, the at b being equal to 6 x 2 = 12 ft. tons. Now, considering the uniform load, the diagram for the portion b c will be a parabola with vertex wP 1 6^at c, the ordinate b^ d at B^ being = = -^ x ^
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