§5.2 Slope and Deflection of Beams 101 (e)Cantilever subjected to non-uniform distributed load (Fig.5.9) + w 3w Fig.5.9. The loading at section XX is -票-+6w-引-(+) Integrating, 装-(+)+A (1) Erdzy=wt3 (2+3+Ax+B (2) -（+品 /x3,x4 Ax +Bx+C 3 /x4 E1y=-w24+ Ax3 Bx2 + 60L 6* 2+Cx+D (4) Thus,before the slope or deflection can be evaluated,four constants have to be determined; therefore four conditions are required.They are: At x=0,S.F.is zero .from (1)A=0 At x =0,B.M.is zero from (2) B=0 At x =L,slope dy/dx =0(slope normally assumed zero at a built-in support) from (3) 0=-w +) 6 +C C= 4 At x=L,y=0 /L4 from (4) 0=-w ,L4 WL4 24 60 +4+D D=、 23wL4 120$5.2 Slope and DeJIection of Beams 101 (e) Cantilever subjected to non-uniform distributed load (Fig. 5.9) Fig. 5.9. The loading at section XX is w‘ = El- d4Y = - [ w + (3w - w)’] = - w (1 + %) dx4 1 Integrating, E~-=-w d2y (; -+- ;I) +A x+B dx2 (;: 6.6,) Ax3 Bx2 (4) Ely= -W -+- +-+--++x+D 62 (3) Thus, before the slope or deflection can be evaluated, four constants have to be determined; therefore four conditions are required. They are: At x = 0, S.F. is zero .‘. from (1) A=O At x = 0, B.M. is zero .’. from (2) B=O At x = L, slope dyldx = 0 (slope normally assumed zero at a built-in support) .’. from (3) At x=L, y=O ... from (4) o=-w -+- +C (: ti) O= -w($+$)+F+D .. 23wL4 120 D= -~