意卖 [例1]求下列各等截面直梁的弹性曲线、最大挠度及最大转角。 解: 0建立坐标系并写出弯矩方程 M(x=P(x-l) ②写出微分方程并积分 应用位移边界条件求积分常数 Elf=M(x)=P(L-x) Ef(0)=6 PL3+C,=0 E=-P(L-x)2+ CL 2 E(0)=E(0)=-P2+C=0 Elf-P(L-x)+Cx+C,.C=PL; C2=-PLS
[例1] 求下列各等截面直梁的弹性曲线、最大挠度及最大转角。 建立坐标系并写出弯矩方程 M (x) = P(x − L) 写出微分方程并积分 应用位移边界条件求积分常数 EIf = −M (x) = P(L − x) 1 2 ( ) 2 1 EIf = − P L − x +C 1 2 3 ( ) 6 1 EIf = P L − x +C x +C 0 6 1 (0) 2 3 EIf = PL +C = 0 2 1 (0) (0) 1 2 EI = EIf = − PL +C = 3 2 2 1 6 1 ; 2 1 C = PL C = − PL 解： P L x f x
DEFORMATIONOF BEAMS DUE TO BENDING L x 4 Write out the equation of the elastic curve and plot its curve f(x)= [-x)3+3x-以 6El 5 The maximum deflection and the maximum angle of rotation Pl PL max=O(D)=2EI fmx =f(l)= BEI
Write out the equation of the elastic curve and plot its curve 3 2 3 ( ) 3 6 ( ) L x L x L EI P f x = − + − EI PL f f L 3 ( ) 3 max = = EI PL L 2 ( ) 2 max = = The maximum deflection and the maximum angle of rotation x f P L
L x a写出弹性曲线方程并画出曲线 f(x)= [-x)3+3x-以 6El ⑤最大挠度及最大转角 0=0(L=r23 PL f∫mx=f(L)= 2EI BEI
写出弹性曲线方程并画出曲线 3 2 3 ( ) 3 6 ( ) L x L x L EI P f x = − + − EI PL f f L 3 ( ) 3 max = = EI PL L 2 ( ) 2 max = = 最大挠度及最大转角 x f P L
DEFORMATIONOF BEAMS DUE TO BENDING Solution: O Set up the coordinates and write out the bending- moment equation P(x-a)(0≤x≤a) P M(x)=0 (a≤x≤L) 2 Write out the differential equation and integrate it ∫P(a-x)(0≤x≤a) E!f"=10 (a≤x≤L EV"=12 P(a-x)+C1 EIf P(a-x+Cix+C Dy IDx+D
Solution：Set up the coordinates and write out the bendingmoment equation − = 0 ( ) ( ) (0 ) ( ) a x L P x a x a M x − − + = 1 1 2 ( ) 2 1 D P a x C EIf + − + + = 1 2 1 2 3 ( ) 6 1 D x D P a x C x C EIf − = 0 ( ) ( ) (0 ) a x L P a x x a EIf x f P L a Write out the differential equation and integrate it
解:◎建立坐标系并写出弯矩方程 M(x)= P(x-a)(0≤x≤a) 0 (a≤x≤L) f 2写出微分方程并积分 P(a-x)(0≤x≤a) E”=0 (a≤x≤L EIf-P(a-x)2+C Elf=6 P(a-x+Cix+c 2 D Dx+
解：建立坐标系并写出弯矩方程 − = 0 ( ) ( ) (0 ) ( ) a x L P x a x a M x 写出微分方程并积分 − − + = 1 1 2 ( ) 2 1 D P a x C EIf + − + + = 1 2 1 2 3 ( ) 6 1 D x D P a x C x C EIf − = 0 ( ) ( ) (0 ) a x L P a x x a EIf x f P L a