Text-book of mechanics . 0, we may place thesum of the vertical forces acting upon the elementshown as a free body equal to , Qx - wdx - (Qx + dQx) = o, or — wdx — dQx = o, , dQ. whence — w = -^- • dx The negative sign attached to w, when interpretedby means of Fig. 20, where the loading acts downward,shows that a positive rate of loading would act if the sign of the rate of loading, w, is takenpositive when upward, then in an appropriate intervalthis rate of loading is the first derivative of the shearingforce, or dQ*dx 34 MECHANICS OF MATERIALS To recapitulate: AQx , n d
Text-book of mechanics . 0, we may place thesum of the vertical forces acting upon the elementshown as a free body equal to , Qx - wdx - (Qx + dQx) = o, or — wdx — dQx = o, , dQ. whence — w = -^- • dx The negative sign attached to w, when interpretedby means of Fig. 20, where the loading acts downward,shows that a positive rate of loading would act if the sign of the rate of loading, w, is takenpositive when upward, then in an appropriate intervalthis rate of loading is the first derivative of the shearingforce, or dQ*dx 34 MECHANICS OF MATERIALS To recapitulate: AQx , n dMx w = -f- and Qx = —-—jax dx so that w = dQx d2Mx dx dx2 Exercise 35. Sketch a diagram of loading for the beamsin Figs. 15, 16, 17, and 19 and note the geometrical relationsbetween the diagrams of loading, shear, and moments. Theyare derivative curves. 1 A simple beam carrying a total load of P pounds uni-formly increasing, as shown in Fig. 21, will serve toillustrate the application of the above Fig. 21 Here the interval extends from x = o to x = I, as noconcentrated loads are applied between the points andno sudden variation of the distributed load occurs. STRESSES IN BEAMS 35 If w pounds per inch represent the rate of loading 1V0C at x = I, then — is the rate of loading at x; v dQx wx dx I where the minus sign indicates downward loading, thus, Qx = - ^ + d. • 2/ The reactions (as can be found by the principles ofstatics, if we remember that the center of gravity of the whole load is at x 2,\ Wl , Wl = - / are — and — >3 / 6 3 so that Q* wl ,= — when x = 0,0 and therefore Cx _ wl= 6 thus, & wx2 , wl2I 6 And as dMxdx = e., ,- wx2 . wlx , ,-, Mx = - — + — + 6 C2 is now obtained by noting that Mx = o when x = o; ,_, , ,, wx3 , wlx so that, C2 = o, and Mx = — — + — • It is always important to locate the point at whichthe maximum bending moment occurs and then todetermine its value. 36 MECHANICS OF MATERIALS To find the maxi
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