98 Mechanics of Materials $5.2 Mx=E d2y dx2=-Wx EI dy Wx2 =-2+A dx assuming El is constant. E=、 W 6+A+B Now when x=bk y=0.A=) and when x=L,y=0.B= WL3 WL2.WL3 6- -L= 2 3 Wx3 WL2x WL3 E-6+2-3 (5.5) This gives the deflection at all values of x and produces a maximum value at the tip of the cantilever when x =0, WL3 i.e. Maximum deflection ymax=- (5.6) 3EI The negative sign indicates that deflection is in the negative y direction,i.e.downwards. Wx2 WL2 Similarly (5.7) and produces a maximum value again when x=0. dy WL2 Maximum slope (5.8) dx 2EI (positive) max (b)Cantilever with uniformly distributed load (Fig.5.6) w/metre Fig.5.6. Mxx=EI d2y wx2 2 2 Eldy wx3 dx 6+A Ely =wxt 24 +Ax+B98 Mechanics of Materials $5.2 .. d2Y M,, = EIy = - WX dx dy Wx2 dx 2 EI-= --+A assuming EI is constant. wx3 EIy= --+Ax+B 6 Now when x=L, -- dy - 0 :. dx w12 2 A=---- and when .. WL3 WLZ w13 6 2 3 x=L,y=Q .’. B=---L= -- --+--- EI 6 2 (5.5) This gives the deflection at all values of x and produces a maximum value at the tip of the cantilever when x = 0, i.e. w13 Maximum deflection = y,= - - 3e1 The negative sign indicates that deflection is in the negative y direction, i.e. downwards. Similarly dY 1 wx2 WL2 dx EI and produces a maximum value again when x = 0. Maximum slope = (2) , =- w12 (positive) 2EI (b) Cantilever with uniformly distributed load (Fig. 5.6) Fig. 5.6. d2y wx2 dx2 2 M =EI-=---- xx dy wx3 dx 6 wx4 EIy= --+Ax+B 24 EI-= --+A (5.7)