东南大学：《弹性力学》课程教学课件（英文讲稿）14 Energy Method and Variational Principle

Strain Energy for Normal StressIn an element with a nonuniform stress distribution,AUdUU = [U,dV = total strain energyU。= limdvAV→0AVFor values of U.< Uy,i.e., below the proportionallimit,2dV=elasticstrainenergy=E>02EUnder axial loading, x = P/ AdV=AdxI.2dudxEAdx :2AEdx2For a rod of uniform cross-section,2AE6

In an element with a nonuniform stress distribution, 0 0 0 d lim d total strain energy V d U U U U U V   V V        For values of U0 < UY , i.e., below the proportional limit, 2 d elastic strain energy 2 x U V E     Under axial loading, d d  x   P A V A x 2 2 0 0 1 d d d 2 2 d L L P u U x EA x AE x           AE P L U 2 2  For a rod of uniform cross-section, Strain Energy for Normal Stress   E 0 6

Strain Energy for Normal StressFor a beam subjected to a bending load.M:xdVdv2E1?2EKSetting dV = dA dx,I1M1lAdx1My2E1?2E1201M2d"wEIdxdxdx?2EI2=E>0PFor an end-loaded cantilever beam.BM=-PxAp2x?p?L3dx2EI6EI67

I M y x   For a beam subjected to a bending load, 2 2 2 2 d d 2 2 x M y U V V E EI      Setting dV = dA dx, 2 2 2 2 2 2 0 0 2 2 2 2 0 0 d d d d 2 2 1 d d d 2 2 d L L A A L L M y M U A x y A x EI EI M w x EI x EI x                       For an end-loaded cantilever beam, 2 2 2 3 0 d 2 6 L M Px P x P L U x EI EI      Strain Energy for Normal Stress   E 0 7

Strain Energy for Shear StressFor a material subjected to plane shearingstresses,YsIt.dyx10号一xyFor values of trwithin the proportional limit.TMU=IGr,=yYXT2GThe total strain energy is found fromOxj(1+vU=dvU.dyxJE2C→G>0;V>-18

For a material subjected to plane shearing stresses, 0 0 d xy U xy xy      For values of  xy within the proportional limit, 2 1 1 2 0 2 2 2 x y U G xy xy xy G        The total strain energy is found from 2 2 0 (1 ) d d d 2 xy U U V V V xy G E           Strain Energy for Shear Stress     G 0; 1  8

Strain Energy for Shear StressFor a shaft subjected to a torsional load.d2GJ7Setting dV = dA dx,2OdAdx:dx2GJ22G.10TpTxyTdpG.Jdxdx=dx2GJ2In the case of a uniform shaft.T?LU:2GJ9

J T xy    2 2 2 2 d d 2 2 xy T U V V G GJ       For a shaft subjected to a torsional load, Setting dV = dA dx, 2 2 2 2 2 2 0 0 2 2 0 0 d d d d 2 2 1 d d d 2 2 d L L A A L L T T U A x A x GJ GJ T x GJ x GJ x                          In the case of a uniform shaft, GJ T L U 2 2  Strain Energy for Shear Stress 9

Strain Energy for Hydrostatic Stressp3(1- 2v)-3pkk=6 +62 +63=-pE31+2GE31+2G-pK=3△V3(1-2v)p13(1-12v0kkFnm722 K2E→K>0;v<0.510

Strain Energy for Hydrostatic Stress         1 2 3 2 2 0 3 1 2 3 3 2 3 2 3 1 2 3 1 1 1 3 1 2 2 2 2 2 k k kk m kk m m p p E G p E G K V U p K E                                     10    K 0; 0.5 