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

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