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 (dEtermine internal forces P 0c N R B T M T Bending moment: Mr()=PRsin Torque: MN(O)=PR(-cosP)
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0 M N B T R T O 弯矩:M1(q)= PRsin 扭矩:M()=PR(1-cos)
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: N2( M(x) dx dx t L 2EA L 2GIp L2EⅠ TR PR(1-coS() Rdo+ P2R(Sin p)Rdo 0 2GIp 0 2EI 4G×3 3PRT R兀 dEl Work done by external forces is equal to the strain energy P BPRT PRT Let w 2 fa=uz then JA 2GIE QEI
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: N2( M(x) dx MA(x) dx t dx L 2EA L 2GIp L2EⅠ TR PR(1-coS() Rdo 0 2GIp 0 PR(sm o Rdo 2EI 3P2R3丌PRx 4(× dEl ③外力功等于应变能 P ∴W=fA=U s 3PRI PRT 2 2G1 2EI
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) JL 2EI P M(x) (0≤x≤a) By using symmetry we get: U=2 Jo 2EL 2 dx LeL P et w=U, thenf c 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