Strain Energy in Linear Elastic Solids

1 Strain Energy in Linear Elastic SolidsCEE 201L. Uncertainty, Design, and OptimizationDepartment of Civil and Environmental EngineeringDuke UniversityHenri P. GavinSpring, 2015 Consider a force,Fi, applied gradually to a structure. LetDibe the resultingdisplacement at the location and in the direction of the forceFi. If thestructure is Elastic , the force-displacement curve follows the same path onloading and unloading. iDDjFiFj w(x)v(x) DD+D FF 0 DiiiiiiFigure and displacements on the surface of an Elastic increased by Fiand the corresponding increase in the displacementis Di, then as Fi 0, the incremental work, W, done by the loadFipassing through a displacement Diis approximatelyFi Di, or, moreprecisely, W= Di+ DiDiFi(Di)dDi.

2 (1)When the structure iselasticandlinear, that isFi(Di) =kiDi, the work ofa force increasing from 0 toFi, moving through corresponding displacementsfrom 0 toDiisW= Di0 FidDi= Di0kiDidDi=12kiD2i=121kiF2i=12 FiDi.(2)2 CEE 201L. Uncertainty, Design, and Optimization Duke University Spring 2015 GavinIf a Linear Elastic structure is subjected to a system of point forcesF1,F2,..,Fn, iiDFDFjDjn1F1 DFnFDDFFDDDFD1111iinFjFjDjjiiFnFnDnFigur e forces and collocated displacements on Linear Elastic Solids and displacements,D1,D2,..,Dn, in the direction of those forces, thenthe totalexternal work,W, is given byW=12{F1D1+F2D2+ +FnDn}=12{F}T{D}.(3)In the absence of any Energy dissipation, this work is stored in the structurein the form ofstrain Energy . In Elastic structures carrying static loads, theexternal workandstrain energyare Strain Energy = Work of External ForcesUint=Wext(4)Note that forces at fixed reaction points,R, do no work because the displace-ments at the reactions are presumed to be :Small element subjected to normal stress GavinStrain Energy in Linear Elastic Solids3 Strain Energy in a general state of stress and strainA three dimensional Linear Elastic solid with loads supplied by external forcesF1.

3 ,Fn, and through support reactionsR, can be considered to be madeup of small cubic elements as shown below. iiDFDFjDjn1F1 DFnV zzyyxx xyxz yzFigure within a Linear Elastic incremental Strain Energy ,dU, for this elemental cube of volumedVcanbe written:dU=12{ xx xx+ yy yy+ zz zz+ xy xy+ xz xz+ yz yz} the incremental Strain Energy ,dU, over an entire volume,V, thetotal Strain Energy ,U, isU=12 V{ xx xx+ yy yy+ zz zz+ xy xy+ xz xz+ yz yz} the stresses and strains are re-written as vectors,{ }T={ xx yy zz xy xz yz}{ }T={ xx yy zz xy xz yz},then the total Strain Energy can be written compactly asU=12 V{ }T{ }dV.(5)This equation is a general expression for the internal Strain Energy of a linearelastic structure of any type. It can be simplified significantly for structuresbuilt from a number of prismatic members, such as trusses and Gavin4 CEE 201L.

4 Uncertainty, Design, and Optimization Duke University Spring 2015 GavinAxial Strain Energy , xx=Nx/A, xx=u (x)A short section of a bar subjected to an axial forceNxstretches xxdxdxdu = (du/dx) dx xx= ( )dA Figure axial forces, deformation, and stresses in a short section of a Strain along this short section of bar is xx=dudx=u (x).. xx=E normal stress on a small part of the cross section, of areadA, is xx= xx= xx/EThe incremental internal Strain Energy ,dU, in an incremental volume element,dV, in terms of axial forcesNxand axial displacementsu(x) isdU=12 xx xxdV=12(NxA)(u (x))dV=12N2xEA2dV=12E(u (x))2dVand the total Strain Energy in a bar in tension or compression isU=12 lN2xEA2 AdA dxorU=12 lE(u (x))2 AdA AdA,U=12 lN2xEAdxorU=12 lEA(u (x))2dx(6)A prismatic bar with a constant axial force,Nx, and a constant Strain xx= x/L, along its length is like a truss element, and the Strain energycan be expressed asU=12N2xLEAorU=12 EAL 2xor simplyU=12Nx x(7)

5 GavinStrain Energy in Linear Elastic Solids5 Bending Strain Energy , xx= Mzy/Iz, xx v byA short section of a beam subjected to a bending momentMzabout thez-axis bends by an angled . d /2 MMx xxdx yzz d = (d /dx) dx = dx dx+ dx = dx y dx d /2 Figure bending moments, deformation, and stresses in a prismatic normal stress on a cross-section element of areadAat a distanceyfromthe neutral axis is xx(y) = xx= xx/EThe Strain along this short section of bar at a distanceyfrom the neutralaxis is xx(y) = y v by, .. xx=E xxThe incremental internal Strain Energy ,dU, in a volume element,dV, in termsof bending momentsMz(x) and transverse displacementv(x) isdU=12 xx xxdV=12( MzyI)( v b(x)y)dV=12M2zy2EI2zdV=12E(v b(x)y)2dVand the total Strain Energy in a beam under pure bending moments isU=12 lM2zEI2z Ay2dA dxorU=12 lE(v b(x))2 Ay2dA the bending moment of inertia,I, is Ay2dA, provided that the originof the coordinate system lies on the neutral axis of the beam ( Ayz dy dz= 0),U=12 lM2zEIzdxorU=12 lEIz(v b(x))2dx.

6 (8) Gavin6 CEE 201L. Uncertainty, Design, and Optimization Duke University Spring 2015 GavinShear Strain Energy , xy=VyQ(y)/Izt(y), xy=v s(x)A short section beam subjected to a shear forceVydeflects by an amountdvs. yxyyVVy xyyt(y)ddxsv ssdv = (dv /dx)dx = v dxsFigure shear forces, deformation, and stresses, if a short section of a beam. xy(y) =VyQ(y)Izt(y).. xy= xy/GQ(y) = Moment of Area of Cross Section = d/2yt(y)y dydU=12 xy xydV=12 2xyGdV=12V2yQ(y)2I2zGt(y)2dA dxU=12 lV2yI2zG AQ(y)2t(y)2dA dx=12 lV2yGA AI2z AQ(y)2t(y)2dA dxThis last integral reduces to a constant that depends only upon the shape ofthe cross-section. This constant is given the variable name . =AI2z AQ(y)2t(y)2dAValues of for some common cross-section shapes are given below ( >1).

7 Solid circular sections: rectangular sections: circular tubes: square tubes: in strong-axis shear: A/(td)With this simplification, the internal Strain Energy due to shear forces isU=12 l V2yGAdx=12 lV2yG(A/ )dx .(9) GavinStrain Energy in Linear Elastic Solids7 The term (A/ ) is called theeffective shear a review of shear stresses in beams, consider the shear stress in a rectan-gular section (with sectiond b). xy=VyQ(y)Izt(y)Q(y) = d/2yt(y)y dy=b d/2yy dy=b y22 d/2y=b d28 y22 xy=Vy2Iz d24 y2 .This stress varies parabolically along the direction of the applied shear. It ismaximum at the centroid of the section and zero at the analogy, the corresponding shear Strain Energy equation in terms of dis-placements isU=12 lG(A/ )(v s(x))2dx(10)where the total transverse displacement is a combinastion of bending-relatedvb(x) and shear-relatedvs(x) displacements,v(x) =vb(x) +vs(x).

8 For exam-ple,vb(x) = Mzz(x)EIzz(x)dxandvs(x) = Vz(x)GA(x)/ dx . Gavin8 CEE 201L. Uncertainty, Design, and Optimization Duke University Spring 2015 GavinTorsional Strain Energy , x =Txr/J, x =r A short section of a circular shaft loaded with a torqueTxtwists by anangled .yxTyz Txxx rdx d/dx) = RR(d d Figure torsional moments, deformation, and stresses in a short section of a circumferential shear stress x (r) is x (r) = x = x /Gand the corresponding shear Strain is x (r) = r .. x =G x The incremental internal Strain energydUin terms of torsional momentsTx(x) and torsional rotations (x) isdU=12 x x dV=12(TxrJ)( r)dV=12T2xr2GJ2dV=12G( r)2dVand the total Strain Energy for the shaft isU=12 lT2xJ2G Ar2dA dxorU=12 lG( )2 Ar2dA the term Ar2dAis the same as the polar moment of inertia,J,U=12 lT2xGJdxorU=12 lGJ( (x))2dx.

9 (11)For a prismatic shaft with a constant torque along its lengthTx, and a totaltwist , the Strain Energy can be expressed asU=12T2xLGJorU=12 GJL 2or simplyU=12Tx (12) GavinStrain Energy in Linear Elastic Solids9 Total Strain Energy arising from Combined Axial StressesAs a review of the material above, consider a three-dimensional bendingproblem with a super-imposed normal force, axial force and bending moments in a prismatic beam. xx=NxA MzyIz+ total Strain Energy arising from axial and bending effects isUn=12 V xx xxdV=12 V 2xxEdV=12 l1E A 2xxdA term 2xxin the integral above can be expanded as follows. A 2xxdA= A N2xA2+M2zy2I2z+M2yz2I2y 2 NxMzyAIz+ 2 NxMyzAIy 2 MzMyzyIzIy , since the coordinate axes are assumed to pass through the centroid ofthe cross-sectional area, Ay dA= Az dA= Ayz dA= 0 Therefore, the total potential Energy is simply the sum of the potential ener-gies due to axial and bending moments lN2xEAdx+12 lM2zEIzdx+12 Gavin10 CEE 201L.

10 Uncertainty, Design, and Optimization Duke University Spring 2015 GavinTotal Strain Energy arising from Combined Shear StressesJust as a structural element can be subjected to combined normal and bend-ing stresses, combined shear stresses can also act shear forces and torsional moment in a short section of a beam. xy=VyQy(y)Iztz(y) xz=VzQz(z)Iyty(z) x =TxrJThrough mathematical manipulations similar to those above, it can be shownthatUv=12 lV2yG(A/ y)dx+12 lV2zG(A/ z)dx+12 lT2xGJdx ,where y=AI2z A Qy(y)tz(y) 2dA z=AI2y A Qz(z)ty(z) 2dATotal Strain EnergyThe total Strain Energy for Solids subjected to axial, bending, shear, andtorsional forces is the sum GavinStrain Energy in Linear Elastic Solids11 SummaryStrain Energy is a kind of potential Energy arising from stressanddeformationof Elastic Solids .