q: Solution:qlq.2BM(xxX22qlq2Elw"x+X22wqlq.3+CElw'XX46qlq4Elw+Cx + DxX1224· Boundary conditions: w(x = O) = O, w(x = I) = O= 9/3=C-D=02416
• Solution: M x ql x q ( ) x 2 2 2 2 2 2 x q x ql EIw x C q x ql EIw 2 3 4 6 x Cx D q x ql EIw 3 4 12 24 • Boundary conditions: w x w x l 0 0, 0 3 , 0 24 ql C D 16 x q l w A B
q: Equations of beam deflection and slopeBq(13—6lx2+4x3)0=(24EIqx13-2lx2 + x3)W24EIW. The maximum deflection and slopeq130=0Rmax24EI5q141W=Wx=max2,384EI17
• Equations of beam deflection and slope • The maximum deflection and slope ( 2 ) 24 3 2 3 l lx x EI qx w ( 6 4 ) 24 3 2 3 l lx x EI q - + x q l w A B 3 max 24 A B ql EI 4 max 5 2 384 l ql w w x EI 17 x q l A B
Sample Problem: Given: flexural rigidity (ED) of a cantilever beam under aconcentrated load acting at its free end: Find: eguations of deflections and slopes, and their maximum values0.max,Wmax)BAxw18
w P A B x l • Given: flexural rigidity (EI) of a cantilever beam under a concentrated load acting at its free end • Find: equations of deflections and slopes, and their maximum values (θmax, wmax) Sample Problem 18
. SolutionM(x) = -P(l - x)EIw"=-Px +PlB4xPElw'+ Plx+CXIw2PPl2Elw+Cx+ DXX26w'(x=0)=0? Boundary conditions: w(x = O) = O,=C=D=019
• Solution: M(x) P(l x) EIw Px Pl x Plx C P EIw 2 2 x Cx D Pl x P EIw 3 2 6 2 • Boundary conditions: w x w x 0 0, 0 0 C D 0 19 w P A B x l x
: Equations of beam deflection and slopePxA(21 - x)2EIBxPx?(3 - x)WW6EI. The maximum deflection and slopeP[20LBmax2EIP[3WWBmax3EI20
(2 ) 2 l x EI Px (3 ) 6 2 l x EI Px w • The maximum deflection and slope EI Pl B 2 2 max EI Pl w wB 3 3 max • Equations of beam deflection and slope 20 w P A B x l x