ENERGYMETHOD Example 1 A semicircle rod as shown in the figure is lie in horizontal plane A vertical force P act at its point A. Determine the displacement of point A in vertical direction Solution: In energy method ( work done by external forces is equal to the strain energy .L (dEtermine internal forces 0¨-…--P, N R B T eM T Bending moment: Mr()=PRsin Torque: MN()=PR(I-coS()
Solution：In energy method（work done by external forces is equal to the strain energy） ①Determine internal forces MT () = PRsin M () = PR(1−cos) N A Bending moment: Torque: Example 1 A semicircle rod as shown in the figure is lie in horizontal plane. A vertical force P act at its point A. Determine the displacement of point A in vertical direction. P R O Q MN MT A A P N B T O
能量方法 [例1]图示半圆形等截面曲杆位于水平面内,在A点受铅垂力P 的作用,求A点的垂直位移。 解:用能量法(外力功等于应变能) ①求内力O M N B T R M T O 弯矩:Mn(q)=PRsn 扭矩:M(q)=PR(1-cosp)
MN [例1 ] 图示半圆形等截面曲杆位于水平面内，在A点受铅垂力P 的作用，求A点的垂直位移。 解：用能量法（外力功等于应变能） ①求内力 弯矩: MT () = PRsin : M () = PR(1− cos) 扭矩 N A P R O Q MT A A P N B T O
ENERGYMETHOD ② Strain energy: U=J N2(x) dx+-n dx+ M(x) dx 2EA JL 2GIP L2EI P2R(sin p)Rdg cZ P2R(1-cOS )RdP+ 2EI 0 2GIp 3P2R3xP2R丌 4GI24El Work done by external forces is equal to the strain energy P BPRT PRT Let W=f=U 2 then JA 2GIH 2EI
③Work done by external forces is equal to the strain energy ②Strain energy： = + + L L P L x EI M x x G I M x x EA N x U d 2 ( ) d 2 ( ) d 2 ( ) 2 2 n 2 + − = 0 2 2 2 0 2 2 2 d 2 (sin ) d 2 (1 cos ) R EI P R R GI P R P EI P R GI P R 4 P 4 3 2 3 2 3 = + , 2 f U P Let W = A = EI PR GI PR f P A 2 2 3 3 3 then = +
能念法 ②变形能: U= N2(x) dx+ M(x) dx t dx 2EA JL 2GIP L2EI P2R(sin p)Rdg rr P2R2(1-cos()Rdo o 2EI 0 2GIp 3P2R3xP2R丌 4( P 4EI ③外力功等于应变能 21=U P 3PR'T PR'I 2GI2El
③外力功等于应变能 ②变形能： = + + L L P L x EI M x x G I M x x EA N x U d 2 ( ) d 2 ( ) d 2 ( ) 2 2 n 2 + − = 0 2 2 2 0 2 2 2 d 2 (sin ) d 2 (1 cos ) R EI P R R GI P R P EI P R GI P R 4 P 4 3 2 3 2 3 = + f U P W = A = 2 EI PR GI PR f P A 2 2 3 3 3 = +
ENERGYMETHOD Example 2 Determine the deflection of point C by the energy method, where the beam is of equal section and straight P Solution: Work done by external B forces is equal to the strain energy a o w=PfO U= M2(x) L 2El P M(x)=-x:(0≤x≤a By using symmetry we get:U=2 x)2dx 2E2 LeL Pa et w=U, thenfc 6El Thinking: For the distributed load, can we determine the displacement of p by this method? of point C Im L山
Example 2 Determine the deflection of point C by the energy method,where the beam is of equal section and straight. W PfC 2 1 = Solution： Work done by external forces is equal to the strain energy = L x EI M x U d 2 ( ) 2 ;(0 ) 2 ( ) x x a P M x = By using symmetry we get： EI P a x x P EI U a 12 ) d 2 ( 2 1 2 2 3 0 2 = = EI Pa W U thenfC 6 , 3 = = Thinking：For the distributed load ,can we determine the displacement of point C by this method？ q C a a A P B f Let