. The strength of materials; a text-book for engineers and architects. Fig. 74.—Inclined Beam with Lower End freely supported. Case 3. Inclined Beam with Vertical Loads—TopReaction Horizontal.—In this case the resultant load mustfirst be found. Let this resultant act down the line x x(Fig. 75). The reaction R^, at B must be horizontal, so drawB X horizontal, then if this meets the line x x, R,v must also BENDING MOMENTS AND SHEARING FORCES 159 pass through x, so that by joining a x we get the directionof R,. The values of R, and R^, are then found by a triangleof forces a, b, c. Now resolve th
. The strength of materials; a text-book for engineers and architects. Fig. 74.—Inclined Beam with Lower End freely supported. Case 3. Inclined Beam with Vertical Loads—TopReaction Horizontal.—In this case the resultant load mustfirst be found. Let this resultant act down the line x x(Fig. 75). The reaction R^, at B must be horizontal, so drawB X horizontal, then if this meets the line x x, R,v must also BENDING MOMENTS AND SHEARING FORCES 159 pass through x, so that by joining a x we get the directionof R,. The values of R, and R^, are then found by a triangleof forces a, b, c. Now resolve the weights and reactions as before along andperpendicular to a b. The perpendicular components will be. Thrust OiaaramFig. 75.—Inclined Beam with Top End freely supported. the same as before, and so the and shear diagrams willbe the same as in the previous case (Fig. 74). The thrusts will be different, and will be as shown on thefigure, which will be clearly followed. Case 4. Sloping Cantilever.—This is worked in a similarmanner. Consider, for example, a uniform load of intensitym; on a cantilever of length I at an inclination 0 (Fig. 76). The 160 THE STRENGTH OF IMATERIALS curve will be a parabola. Its maximum ordinate will be iv I cos 6 rt J because the total weight will be w I, and it acts at a distance ^ from the abutment. The shear diagram
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