东南大学：《建筑力学 Architectural Mechanics》课程教学课件（英文讲稿）A20 Energy Method

Energy Methodmi@sereo.cn1

1 Energy Method mi@seu.edu.cn

Contents·Work and Strain Energy（功与应变能）·StrainEnergyDensity（应变能密度）·Strain Energy due to Normal Stresses（正应力所致应变能)·StrainEnergydue to Shearing Stresses（切应力所致应变能)? Strain Energy due to Bending and Transverse Shear（弯矩和横力所致应变能对比）· Strain Energy due to a General State of Stress （一般应力状态所致应变能）·Work and Energy under a Single Load（单载下的功能互等原理）2

• Work and Strain Energy（功与应变能） • Strain Energy Density（应变能密度） • Strain Energy due to Normal Stresses（正应力所致应 变能） • Strain Energy due to Shearing Stresses（切应力所致应 变能） • Strain Energy due to Bending and Transverse Shear （弯矩和横力所致应变能对比） • Strain Energy due to a General State of Stress（一般应 力状态所致应变能） • Work and Energy under a Single Load（单载下的功能 互等原理） 2 Contents

Contents·Strain Energy cannot be Superposed（应变能的不可叠加性）·Work and Energy under Several Loads（多载下的功能原理）·Castigliano's Second Theorem（卡氏第二定理）·StaticallyIndeterminateTruss（超静定桁架）·Statically Indeterminate Shafts（超静定扭转轴）·Statically Indeterminate Beams（超静定梁）·Method of DummyLoad（虚力法）·Method of Unit Dummy Load（单位虚力法）3

• Strain Energy cannot be Superposed（应变能的不可 叠加性） • Work and Energy under Several Loads（多载下的功 能原理） • Castigliano’s Second Theorem（卡氏第二定理） • Statically Indeterminate Truss（超静定桁架） • Statically Indeterminate Shafts（超静定扭转轴） • Statically Indeterminate Beams（超静定梁） • Method of Dummy Load（虚力法） • Method of Unit Dummy Load （单位虚力法） Contents 3

Work done by External Loads and Strain Energy: A uniform rod is subjected to a slowly increasing load. The elementary work done by the load P as the rodelongates by a small dx isdW = Pdx =elementary workwhich is equal to the area of width dx under the load-deformation diagram: The total work done by the load for a deformation xi,=AreaW=Pdx =total work = strain energyxxiwhich results in an increase of strain energy in the rodP=kxDP. In the case of a linear elastic deformation.W=[Pdx=kxdx=k=Px=UCxxi4

• A uniform rod is subjected to a slowly increasing load • The total work done by the load for a deformation x1 , which results in an increase of strain energy in the rod. W P dx total work strain energy x     1 0 Work done by External Loads and Strain Energy • The elementary work done by the load P as the rod elongates by a small dx is which is equal to the area of width dx under the load￾deformation diagram. dW  Pdx  elementary work 1 1 1 1 2 2 2 1 1 1 0 0 x x W P dx kx dx kx Px U        • In the case of a linear elastic deformation, 4

Strain Energy DensityToeliminatetheeffectsofsize.evaluatethestrainenergyperunitvolume,oXiP dxUVOAL81[ox dex = strainenergy densityu=0EEpEI. The total strain energy density resulting from thedeformation is equal to the area under the curve to j. As the material is unloaded, the stress returns to zerobut there is a permanent deformation. Only the strainenergy represented by the triangular area is recovered.: Remainder of the energy spent in deforming the materialis dissipated as heat.5

Strain Energy Density • To eliminate the effects of size, evaluate the strain￾energy per unit volume, u d strain energy density L dx A P V U x x x      1 1 0 0    • As the material is unloaded, the stress returns to zero but there is a permanent deformation. Only the strain energy represented by the triangular area is recovered. • Remainder of the energy spent in deforming the material is dissipated as heat. • The total strain energy density resulting from the deformation is equal to the area under the curve to 1 . 5

Strain Energy DensityoThe strain energydensity resulting fromModulusof toughnesssetting , = Ep is the modulus of toughness.Rupture? The energy per unit volume required to causethe material torupture is related to its ductilityas well as its ultimate strengthEER2: If the stress remains within the proportionallimit,oEc?u=IEs.daa22E(The strain energy density resulting fromModulussetting , =oy is the modulus of resilienceofresilience2QyOmodulusofresilienceEEYuy2E6

Strain Energy Density • The strain energy density resulting from setting 1  R is the modulus of toughness. • The energy per unit volume required to cause the material to rupture is related to its ductility as well as its ultimate strength. • The strain energy density resulting from setting 1  Y is the modulus of resilience. modulus of resilience E u Y Y   2 2  • If the stress remains within the proportional limit, E E u E x d x 2 2 2 1 2 1 0 1          6

Strain Energy due to Normal Stresses: In an element with a nonuniform stress distribution,AUdUU = [u dV - total strain energylimu=dvAV-0△VFor values of u< uy,i.e., below the proportionallimit,2xdV=elasticstrainenergy2E. Under axial loading, Ox = P/ AdV=Adxp2ax2AE0For a rod of uniform cross-section,p? LU2AE1

Strain Energy due to Normal Stresses • In an element with a nonuniform stress distribution, lim totalstrain energy 0          U u dV dV dU V U u V • For values of u < uY , i.e., below the proportional limit, dV elastic strainenergy E U x 2 2     • Under axial loading, P A dV A dx  x     L dx AE P U 0 2 2 AE P L U 2 2  • For a rod of uniform cross-section, 7

Strain Energy due to Normal StressesForabeam subjectedtoabendingloadx-dd2F2E12Setting dV = dA dx,MyM2U=dAdx12E/22E/202dx2EI0PForanend-loadedcantileverbeamBM=-PxA/D2.2p2r3Xdx6EI2EI08

Strain Energy due to Normal Stresses I M y  x  • For a beam subjected to a bending load,     dV EI M y dV E U x 2 2 2 2 2 2  • Setting dV = dA dx, dx EI M y dA dx EI M dAdx EI M y U L L A L A                 0 2 0 2 2 2 0 2 2 2 2 2 2 • For an end-loaded cantilever beam, EI P L dx EI P x U M Px L 2 6 2 3 0 2 2      8

Strain Energy due to Shearing StressesForamaterial subjected toplane shearingstresses,Yxyu=0一x2. For values of txy, within the proportional limit.Tu-2Txyu=1GyxyYxyxV2G: The total strain energy is found fromOudyU=xy9

Strain Energy due to Shearing Stresses • For a material subjected to plane shearing stresses,   xy xy xy u d    0 • For values of  xy within the proportional limit, G u G xy xy xy xy 2 2 2 2 1 2 1        • The total strain energy is found from     dV G U u dV xy 2 2  9

Strain Energy due to Shearing StressesForashaftsubjectedtoatorsional loadxdVd2GT?? Setting dV = dA dx,dAd20CI0T2dx2GIInthecaseofauniformshaft.T?LU:2GI10

Strain Energy due to Shearing Stresses 2 2 2 2 2 2 x p T U dV dV G GI        • For a shaft subjected to a torsional load, • Setting dV = dA dx, 2 2 2 2 2 2 0 0 2 0 2 2 2 L L A A p p L p T T U dAdx dA dx GI GI T dx GI                • In the case of a uniform shaft, 2 2 p T L U GI  10