Chapter 5

 

Analytical techniques and solutions for linear elastic solids

 

 

 

5.7 Energy methods for solving static linear elasticity problems

 

You may recall that energy methods can often be used to simplify complex problems.  For example, to find the equilibrium configuration of a discrete system, you would begin by identifying a suitable set of generalize coordinates q i MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacaWGXbWaaSbaaSqaaiaadMgaaeqaaa aa@34D5@ , and then express the potential energy in terms of these: V( q i ) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacaWGwbGaaiikaiaadghadaWgaaWcba GaamyAaaqabaGccaGGPaaaaa@3713@ .  The equilibrium values of the generalized coordinates could then be determined from the condition that the potential energy is stationary at equilibrium: this gives a set of equations V/ q i =0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacqGHciITcaWGwbGaai4laiabgkGi2k aadghadaWgaaWcbaGaamyAaaqabaGccqGH9aqpcaaIWaaaaa@3AF9@  that could be solved for q i MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacaWGXbWaaSbaaSqaaiaadMgaaeqaaa aa@34D5@ .

 

In this section, we will develop an analogous procedure for solving boundary value problems in linear elasticity.  Our generalized coordinates will be the displacement field u i (x) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacaWG1bWaaSbaaSqaaiaadMgaaeqaaO GaaiikaiaahIhacaGGPaaaaa@373D@ .  We will find an expression for the potential energy of an elastic solid in terms of u i MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacaWG1bWaaSbaaSqaaiaadMgaaeqaaa aa@34D9@ , and then show that the potential energy is stationary if the solid is in equilibrium.  We will find, further, that the potential energy is not only stationary, but is always a minimum, implying that equilibrium configurations in linear elasticity problems are always stable.  (This is because the approximations made in setting up the equations of linear elasticity preclude any possibility of buckling).  This principle will be referred to as the Principle of Minimum Potential Energy

 

The main application of the principle is to generate approximate solutions to linear elastic boundary value problems.  Indeed, the principle will form the basis of the Finite Element Method in linear elasticity.

 

 

5.7.1 Definition of the potential energy of a linear elastic solid under static loading

 

In the following, we consider a generic static boundary value problem in linear elasticity, as shown in the picture. 

 

As always, we assume that we are given:

1.      The shape of the solid in its unloaded condition R MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8XjY=vi pgYlH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqai=hGuQ8ku c9pgc9q8qqaq=dir=f0=yqaiVgFr0xfr=xfr=xb9adbaqaaeGaciGa biaabeqaaiqabaWaaaGcbaGaamOuaaaa@31A4@

2.      The initial stress field in the solid (we will take this to be zero)

3.      The elastic constants for the solid C ijkl MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rkY=vi pgYlH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqai=hGuQ8ku c9pgc9q8qqaq=dir=f0=yqaiVgFr0xfr=xfr=xb9adbaqaaeGaciGa biaabeqaaiqabaWaaaGcbaGaam4qamaaBaaaleaacaWGPbGaamOAai aadUgacaWGSbaabeaaaaa@3581@  and its mass density ρ 0 MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rkY=vi pgYlH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFH e9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabiGaciaa caqabeaacmqaamaaaOqaaiabeg8aYnaaBaaaleaacaaIWaaabeaaaa a@330E@

4.      The thermal expansion coefficients for the solid, and temperature change from the initial configuration ΔT MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeuiLdqKaam ivaaaa@3825@

5.      A body force distribution b MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8YjY=vi pgYlH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqai=hGuQ8ku c9pgc9q8qqaq=dir=f0=yqaiVgFr0xfr=xfr=xb9adbaqaaeGaciGa biaabeqaaiqabaWaaaGcbaGaaCOyaaaa@31C8@  (per unit mass) acting on the solid

6.      Boundary conditions, specifying displacements u * (x) MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8YjY=vi pgYlH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqai=hGuQ8ku c9pgc9q8qqaq=dir=f0=yqaiVgFr0xfr=xfr=xb9adbaqaaeGaciGa biaabeqaaiqabaWaaaGcbaGaaCyDamaaCaaaleqabaGaaiOkaaaaki aacIcacaWH4bGaaiykaaaa@351A@  on a portion 1 R MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rkY=vi pgYlH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqai=hGuQ8ku c9pgc9q8qqaq=dir=f0=yqaiVgFr0xfr=xfr=xb9adbaqaaeGaciGa biaabeqaaiqabaWaaaGcbaGaeyOaIy7aaSbaaSqaaiaaigdaaeqaaO GaamOuaaaa@33FD@  or tractions on a portion 2 R MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rkY=vi pgYlH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqai=hGuQ8ku c9pgc9q8qqaq=dir=f0=yqaiVgFr0xfr=xfr=xb9adbaqaaeGaciGa biaabeqaaiqabaWaaaGcbaGaeyOaIy7aaSbaaSqaaiaaikdaaeqaaO GaamOuaaaa@33FE@  of the boundary of R

 

 

Kinematically Admissible Displacement Fields

 

A ‘kinematically admissible displacement field’ v i (x) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacaWG2bWaaSbaaSqaaiaadMgaaeqaaO GaaiikaiaahIhacaGGPaaaaa@373E@  is any displacement field with the following properties:

1.      v i MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8sk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacaWG2bWaaSbaaSqaaiaadMgaaeqaaa aa@34EA@  is continuous everywhere within the solid

2.      v i MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8sk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacaWG2bWaaSbaaSqaaiaadMgaaeqaaa aa@34EA@  is differentiable everywhere within the solid, so that a strain field may be computed as

ε ^ ij = 1 2 ( v i x j + v j x i ) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacuaH1oqzgaqcamaaBaaaleaacaWGPb GaamOAaaqabaGccqGH9aqpdaWcaaqaaiaaigdaaeaacaaIYaaaamaa bmaabaWaaSaaaeaacqGHciITcaWG2bWaaSbaaSqaaiaadMgaaeqaaa GcbaGaeyOaIyRaamiEamaaBaaaleaacaWGQbaabeaaaaGccqGHRaWk daWcaaqaaiabgkGi2kaadAhadaWgaaWcbaGaamOAaaqabaaakeaacq GHciITcaWG4bWaaSbaaSqaaiaadMgaaeqaaaaaaOGaayjkaiaawMca aaaa@49C1@

3.      v i MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8sk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacaWG2bWaaSbaaSqaaiaadMgaaeqaaa aa@34EA@  satisfies boundary conditions anywhere that displacements are prescribed, i.e. v(x)= u * (x) MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8XjY=vi pgYlH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqai=hGuQ8ku c9pgc9q8qqaq=dir=f0=yqaiVgFr0xfr=xfr=xb9adbaqaaeGaciGa biaabeqaaiqabaWaaaGcbaGaaCODaiaacIcacaWH4bGaaiykaiabg2 da9iaahwhadaahaaWcbeqaaiaacQcaaaGccaGGOaGaaCiEaiaacMca aaa@3969@  on the portion 1 R MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8XjY=vi pgYlH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqai=hGuQ8ku c9pgc9q8qqaq=dir=f0=yqaiVgFr0xfr=xfr=xb9adbaqaaeGaciGa biaabeqaaiqabaWaaaGcbaGaeyOaIy7aaSbaaSqaaiaaigdaaeqaaO GaamOuaaaa@33FB@  on the boundary.

 

Note the v is not necessarily the actual displacement in the solid MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaGqaaKqzGfaeaa aaaaaaa8qacaWFtacaaa@37E6@  it is just an arbitrary displacement field which satisfies any displacement boundary conditions.  You can think of it as a possible displacement field that the solid could adopt.  Out of all these possible displacement fields, it will actually select the one that minimizes the potential energy.

 

The kinematically admissible displacement field can also be thought of as a system of generalized coordinates in the context of analytical mechanics.  Recall that, to use a set of generalized coordinates in Lagranges equations, you must make sure that the system of coordinates satisfies all the constraints.  Similarly, to be admissible, our displacement field must satisfy constraints on the boundary. 

 

Definition of Potential Energy of an Elastic Solid

 

Next, we will define the potential energy of a solid.  The definition may look a bit strange, because it seems to give different values for potential energy depending on how the solid is loaded.  This is true.  But who cares, as long as the definition is useful?

 

For any kinematically admissible displacement field v, the potential energy is

V(v)= V U(v) dV V ρ 0 b i v i dV 2 R t i v i dA MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacaWGwbGaaiikaiaahAhacaGGPaGaey ypa0Zaa8quaeaacaWGvbGaaiikaiaahAhacaGGPaaaleaacaWGwbaa beqdcqGHRiI8aOGaamizaiaadAfacqGHsisldaWdrbqaaiabeg8aYn aaBaaaleaacaaIWaaabeaakiaadkgadaWgaaWcbaGaamyAaaqabaGc caWG2bWaaSbaaSqaaiaadMgaaeqaaaqaaiaadAfaaeqaniabgUIiYd GccaWGKbGaamOvaiabgkHiTmaapefabaGaamiDamaaBaaaleaacaWG PbaabeaakiaadAhadaWgaaWcbaGaamyAaaqabaGccaWGKbGaamyqaa WcbaGaeyOaIy7aaSbaaWqaaiaaikdaaeqaaSGaamOuaaqab0Gaey4k Iipaaaa@581B@

where

U(v)= 1 2 C ijkl ( ε ^ ij α ij ΔT )( ε ^ kl α kl ΔT ) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacaWGvbGaaiikaiaahAhacaGGPaGaey ypa0ZaaSaaaeaacaaIXaaabaGaaGOmaaaacaWGdbWaaSbaaSqaaiaa dMgacaWGQbGaam4AaiaadYgaaeqaaOWaaeWaaeaacuaH1oqzgaqcam aaBaaaleaacaWGPbGaamOAaaqabaGccqGHsislcqaHXoqydaWgaaWc baGaamyAaiaadQgaaeqaaOGaeuiLdqKaamivaaGaayjkaiaawMcaam aabmaabaGafqyTduMbaKaadaWgaaWcbaGaam4AaiaadYgaaeqaaOGa eyOeI0IaeqySde2aaSbaaSqaaiaadUgacaWGSbaabeaakiabfs5aej aadsfaaiaawIcacaGLPaaaaaa@55AA@

is the strain energy density associated with the kinematically admissible displacement field. You can interpret the three terms in the formula for V as the strain energy stored inside the solid; the work done by body forces; and the work done by surface tractions. For the particular case of an isotropic material, with ΔT=0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8sk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacqqHuoarcaWGubGaeyypa0JaaGimaa aa@36D4@ , we see that

U(v)= E 2( 1+ν ) ( ε ^ ij ε ^ ij + ν 12ν ε ^ kk ε ^ mm ) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacaWGvbGaaiikaiaahAhacaGGPaGaey ypa0ZaaSaaaeaacaWGfbaabaGaaGOmamaabmaabaGaaGymaiabgUca Riabe27aUbGaayjkaiaawMcaaaaadaqadaqaaiqbew7aLzaajaWaaS baaSqaaiaadMgacaWGQbaabeaakiqbew7aLzaajaWaaSbaaSqaaiaa dMgacaWGQbaabeaakiabgUcaRmaalaaabaGaeqyVd4gabaGaaGymai abgkHiTiaaikdacqaH9oGBaaGafqyTduMbaKaadaWgaaWcbaGaam4A aiaadUgaaeqaaOGafqyTduMbaKaadaWgaaWcbaGaamyBaiaad2gaae qaaaGccaGLOaGaayzkaaaaaa@54F2@

 

 

 

5.7.2 The principle of stationary and minimum potential energy.

 

Let v be any kinematically admissible displacement field.  Let u be the actual displacement field MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaGqaaKqzGfaeaa aaaaaaa8qacaWFtacaaa@37E6@  i.e. the one that satisfies the equilibrium equations within the solid as well as all the boundary conditions.  We will show the following:

1. V(v) is stationary (i.e. a local minimum, maximum or inflexion point) for v=u.

2. V(v) is a global minimum for v=u.

 

As a preliminary step, recall that the actual displacement field satisfies the following equations

ε ij = 1 2 ( u i x j + u j x i ) σ ij = C ijkl ( ε kl α kl ΔT) σ ij x i + ρ 0 b j =0 u i = u i * on 1 R σ ij n i = t j * on 2 R MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rkY=vi pgYlH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqai=hGuQ8ku c9pgc9q8qqaq=dir=f0=yqaiVgFr0xfr=xfr=xb9adbaqaaeGaciGa biaabeqaaiqabaWaaaGceaqabeaacqaH1oqzdaWgaaWcbaGaamyAai aadQgaaeqaaOGaeyypa0ZaaSaaaeaacaaIXaaabaGaaGOmaaaadaqa daqaamaalaaabaGaeyOaIyRaamyDamaaBaaaleaacaWGPbaabeaaaO qaaiabgkGi2kaadIhadaWgaaWcbaGaamOAaaqabaaaaOGaey4kaSYa aSaaaeaacqGHciITcaWG1bWaaSbaaSqaaiaadQgaaeqaaaGcbaGaey OaIyRaamiEamaaBaaaleaacaWGPbaabeaaaaaakiaawIcacaGLPaaa caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVl aaykW7caaMc8UaaGPaVlaaykW7cqaHdpWCdaWgaaWcbaGaamyAaiaa dQgaaeqaaOGaeyypa0Jaam4qamaaBaaaleaacaWGPbGaamOAaiaadU gacaWGSbaabeaakiaacIcacqaH1oqzdaWgaaWcbaGaam4AaiaadYga aeqaaOGaeyOeI0IaeqySde2aaSbaaSqaaiaadUgacaWGSbaabeaaki abfs5aejaadsfacaGGPaGaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7 caaMc8UaaGPaVpaalaaabaGaeyOaIyRaeq4Wdm3aaSbaaSqaaiaadM gacaWGQbaabeaaaOqaaiabgkGi2kaadIhadaWgaaWcbaGaamyAaaqa baaaaOGaey4kaSIaeqyWdi3aaSbaaSqaaiaaicdaaeqaaOGaamOyam aaBaaaleaacaWGQbaabeaakiabg2da9iaaicdacaaMc8oabaGaaGPa VlaaykW7caaMc8UaamyDamaaBaaaleaacaWGPbaabeaakiabg2da9i aadwhadaqhaaWcbaGaamyAaaqaaiaacQcaaaGccaaMc8UaaGPaVlaa ykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaG PaVlaaykW7caqGVbGaaeOBaiaaykW7caaMc8UaaGPaVlaaykW7caaM c8UaeyOaIy7aaSbaaSqaaiaaigdaaeqaaOGaamOuaiaaykW7caaMc8 UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7 caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlabeo8aZnaaBaaaleaaca WGPbGaamOAaaqabaGccaWGUbWaaSbaaSqaaiaadMgaaeqaaOGaeyyp a0JaamiDamaaDaaaleaacaWGQbaabaGaaiOkaaaakiaaykW7caaMc8 UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7 caaMc8UaaGPaVlaab+gacaqGUbGaaGPaVlaaykW7caaMc8UaaGPaVl aaykW7cqGHciITdaWgaaWcbaGaaGOmaaqabaGccaWGsbaaaaa@F66E@

Next, re-write the kinematically admissible displacement field in terms of u as

v i = u i +δ u i MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacaWG2bWaaSbaaSqaaiaadMgaaeqaaO Gaeyypa0JaamyDamaaBaaaleaacaWGPbaabeaakiabgUcaRiabes7a KjaadwhadaWgaaWcbaGaamyAaaqabaaaaa@3CA3@

where δ u i MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacqaH0oazcaWG1bWaaSbaaSqaaiaadM gaaeqaaaaa@367E@  is the difference between the kinematically admissible field and the correct equilibrium field.  Observe that

v i = u i * u i = u i * on  1 R δ u i =0on  2 R MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakqaabeqaaiaadAhadaWgaaWcbaGaamyAaa qabaGccqGH9aqpcaWG1bWaa0baaSqaaiaadMgaaeaacaGGQaaaaOGa aGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7ca aMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaa ykW7caaMc8UaamyDamaaBaaaleaacaWGPbaabeaakiabg2da9iaadw hadaqhaaWcbaGaamyAaaqaaiaacQcaaaGccaaMc8UaaGPaVlaaykW7 caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVl aab+gacaqGUbGaaeiiaiabgkGi2oaaBaaaleaacaaIXaaabeaakiaa dkfaaeaacqGHshI3cqaH0oazcaWG1bWaaSbaaSqaaiaadMgaaeqaaO Gaeyypa0JaaGimaiaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPa VlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8 UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7 caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVl aaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caqGVbGa aeOBaiaabccacqGHciITdaWgaaWcbaGaaGOmaaqabaGccaWGsbaaaa a@B7CE@

i.e. the difference between the kinematically admissible field and the actual field is zero wherever displacements are prescribed.

 

Now, note that V(v) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacaWGwbGaaiikaiaahAhacaGGPaaaaa@35F8@  can be expressed in terms of u i MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacaWG1bWaaSbaaSqaaiaadMgaaeqaaa aa@34DA@  and δ u i MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacqaH0oazcaWG1bWaaSbaaSqaaiaadM gaaeqaaaaa@367F@  as

V( u+δu )=V(u)+δV+ 1 2 δ 2 V MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacaWGwbWaaeWaaeaacaWH1bGaey4kaS IaeqiTdqMaaCyDaaGaayjkaiaawMcaaiabg2da9iaadAfacaGGOaGa aCyDaiaacMcacqGHRaWkcqaH0oazcaWGwbGaey4kaSYaaSaaaeaaca aIXaaabaGaaGOmaaaacqaH0oazdaahaaWcbeqaaiaaikdaaaGccaWG wbaaaa@4722@

where

δV= V C ijkl ( ε ij α ij ΔT )δ ε kl dV V b i δ u i dV 2 R t i δ u i dA δ 2 V= V C ijkl δ ε ij δ ε kl ε ij = 1 2 ( u i x j + u j x i )δ ε ij = 1 2 ( δ u i x j + δ u j x i ) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rkY=xq qrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8 fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGa aiaabeqaaiWabaWaaaGceaqabeaacqaH0oazcaWGwbGaeyypa0JaaG PaVlaaykW7daWdrbqaaiaadoeadaWgaaWcbaGaamyAaiaadQgacaWG RbGaamiBaaqabaGcdaqadaqaaiabew7aLnaaBaaaleaacaWGPbGaam OAaaqabaGccqGHsislcqaHXoqydaWgaaWcbaGaamyAaiaadQgaaeqa aOGaeuiLdqKaamivaaGaayjkaiaawMcaaiabes7aKjabew7aLnaaBa aaleaacaWGRbGaamiBaaqabaGccaWGKbGaamOvaaWcbaGaamOvaaqa b0Gaey4kIipakiabgkHiTmaapefabaGaamOyamaaBaaaleaacaWGPb aabeaakiabes7aKjaadwhadaWgaaWcbaGaamyAaaqabaaabaGaamOv aaqab0Gaey4kIipakiaadsgacaWGwbGaeyOeI0Yaa8quaeaacaWG0b WaaSbaaSqaaiaadMgaaeqaaOGaeqiTdqMaamyDamaaBaaaleaacaWG PbaabeaakiaadsgacaWGbbaaleaacqGHciITdaWgaaadbaGaaGOmaa qabaWccaWGsbaabeqdcqGHRiI8aOGaaGPaVlaaykW7caaMc8UaaGPa VlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8 UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7cqaH0oazdaahaaWcbeqa aiaaikdaaaGccaWGwbGaeyypa0Zaa8quaeaacaWGdbWaaSbaaSqaai aadMgacaWGQbGaam4AaiaadYgaaeqaaOGaeqiTdqMaeqyTdu2aaSba aSqaaiaadMgacaWGQbaabeaakiabes7aKjabew7aLnaaBaaaleaaca WGRbGaamiBaaqabaaabaGaamOvaaqab0Gaey4kIipaaOqaaiabew7a LnaaBaaaleaacaWGPbGaamOAaaqabaGccqGH9aqpdaWcaaqaaiaaig daaeaacaaIYaaaamaabmaabaWaaSaaaeaacqGHciITcaWG1bWaaSba aSqaaiaadMgaaeqaaaGcbaGaeyOaIyRaamiEamaaBaaaleaacaWGQb aabeaaaaGccqGHRaWkdaWcaaqaaiabgkGi2kaadwhadaWgaaWcbaGa amOAaaqabaaakeaacqGHciITcaWG4bWaaSbaaSqaaiaadMgaaeqaaa aaaOGaayjkaiaawMcaaiaaykW7caaMc8UaaGPaVlaaykW7caaMc8Ua aGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7ca aMc8UaaGPaVlaaykW7caaMc8UaaGPaVlabes7aKjabew7aLnaaBaaa leaacaWGPbGaamOAaaqabaGccqGH9aqpdaWcaaqaaiaaigdaaeaaca aIYaaaamaabmaabaWaaSaaaeaacqGHciITcqaH0oazcaWG1bWaaSba aSqaaiaadMgaaeqaaaGcbaGaeyOaIyRaamiEamaaBaaaleaacaWGQb aabeaaaaGccqGHRaWkdaWcaaqaaiabgkGi2kabes7aKjaadwhadaWg aaWcbaGaamOAaaqabaaakeaacqGHciITcaWG4bWaaSbaaSqaaiaadM gaaeqaaaaaaOGaayjkaiaawMcaaaaaaa@ED46@

To see this, simply substitute into the definition of the potential energy

V(v)= V U(u+δu) dV V b i ( u i +δ u i ) dV 2 R t i ( u i +δ u i )dA = V 1 2 C ijkl ( ε ij α ij ΔT+δ ε ij )( ε kl α kl ΔT+δ ε kl )dV V b i ( u i +δ u i ) dV 2 R t i ( u i +δ u i )dA MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rkY=xq qrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8 fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGa aiaabeqaaiWabaWaaaGceaqabeaacaWGwbGaaiikaiaahAhacaGGPa Gaeyypa0Zaa8quaeaacaWGvbGaaiikaiaahwhacqGHRaWkcqaH0oaz caaMc8UaaCyDaiaacMcaaSqaaiaadAfaaeqaniabgUIiYdGccaWGKb GaamOvaiabgkHiTmaapefabaGaamOyamaaBaaaleaacaWGPbaabeaa kiaacIcacaWG1bWaaSbaaSqaaiaadMgaaeqaaOGaey4kaSIaeqiTdq MaamyDamaaBaaaleaacaWGPbaabeaakiaacMcaaSqaaiaadAfaaeqa niabgUIiYdGccaWGKbGaamOvaiabgkHiTmaapefabaGaamiDamaaBa aaleaacaWGPbaabeaakiaacIcacaWG1bWaaSbaaSqaaiaadMgaaeqa aOGaey4kaSIaeqiTdqMaamyDamaaBaaaleaacaWGPbaabeaakiaacM cacaWGKbGaamyqaaWcbaGaeyOaIy7aaSbaaWqaaiaaikdaaeqaaSGa amOuaaqab0Gaey4kIipaaOqaaiabg2da9maapefabaWaaSaaaeaaca aIXaaabaGaaGOmaaaacaWGdbWaaSbaaSqaaiaadMgacaWGQbGaam4A aiaadYgaaeqaaOWaaeWaaeaacqaH1oqzdaWgaaWcbaGaamyAaiaadQ gaaeqaaOGaeyOeI0IaeqySde2aaSbaaSqaaiaadMgacaWGQbaabeaa kiabfs5aejaadsfacqGHRaWkcqaH0oazcqaH1oqzdaWgaaWcbaGaam yAaiaadQgaaeqaaaGccaGLOaGaayzkaaWaaeWaaeaacqaH1oqzdaWg aaWcbaGaam4AaiaadYgaaeqaaOGaeyOeI0IaeqySde2aaSbaaSqaai aadUgacaWGSbaabeaakiabfs5aejaadsfacqGHRaWkcqaH0oazcqaH 1oqzdaWgaaWcbaGaam4AaiaadYgaaeqaaaGccaGLOaGaayzkaaGaam izaiaadAfaaSqaaiaadAfaaeqaniabgUIiYdGccqGHsisldaWdrbqa aiaadkgadaWgaaWcbaGaamyAaaqabaGccaGGOaGaamyDamaaBaaale aacaWGPbaabeaakiabgUcaRiabes7aKjaadwhadaWgaaWcbaGaamyA aaqabaGccaGGPaaaleaacaWGwbaabeqdcqGHRiI8aOGaamizaiaadA facqGHsisldaWdrbqaaiaadshadaWgaaWcbaGaamyAaaqabaGccaGG OaGaamyDamaaBaaaleaacaWGPbaabeaakiabgUcaRiabes7aKjaadw hadaWgaaWcbaGaamyAaaqabaGccaGGPaGaamizaiaadgeaaSqaaiab gkGi2oaaBaaameaacaaIYaaabeaaliaadkfaaeqaniabgUIiYdaake aaaaaa@B944@

Multiply everything out and use the condition that C ijkl = C klij MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacaWGdbWaaSbaaSqaaiaadMgacaWGQb Gaam4AaiaadYgaaeqaaOGaeyypa0Jaam4qamaaBaaaleaacaWGRbGa amiBaiaadMgacaWGQbaabeaaaaa@3D39@  to get the result stated.

 

Now, to show that V(v) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacaWGwbGaaiikaiaahAhacaGGPaaaaa@35F8@  is stationary at v=u, we need to show that δV=0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacqaH0oazcaWGwbGaeyypa0JaaGimaa aa@3705@ .  This means that, if we add any small change δu MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacqaH0oazcaWH1baaaa@3568@  to the actual displacement field u, the change in potential energy will be zero, to first order in δu MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacqaH0oazcaWH1baaaa@3568@ .

 

To show this, note that

C ijkl ( ε ij α ij ΔT )= σ kl MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacaWGdbWaaSbaaSqaaiaadMgacaWGQb Gaam4AaiaadYgaaeqaaOWaaeWaaeaacqaH1oqzdaWgaaWcbaGaamyA aiaadQgaaeqaaOGaeyOeI0IaeqySde2aaSbaaSqaaiaadMgacaWGQb aabeaakiabfs5aejaadsfaaiaawIcacaGLPaaacqGH9aqpcqaHdpWC daWgaaWcbaGaam4AaiaadYgaaeqaaaaa@4878@

Next, note that

σ kl δ ε kl = σ kl 1 2 ( δ u k x l + δ u l x k )= 1 2 σ lk δ u k x l + 1 2 σ kl δ u l x k = σ kl δ u l x k MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacqaHdpWCdaWgaaWcbaGaam4AaiaadY gaaeqaaOGaeqiTdqMaeqyTdu2aaSbaaSqaaiaadUgacaWGSbaabeaa kiabg2da9iabeo8aZnaaBaaaleaacaWGRbGaamiBaaqabaGcdaWcaa qaaiaaigdaaeaacaaIYaaaamaabmaabaWaaSaaaeaacqGHciITcqaH 0oazcaWG1bWaaSbaaSqaaiaadUgaaeqaaaGcbaGaeyOaIyRaamiEam aaBaaaleaacaWGSbaabeaaaaGccqGHRaWkdaWcaaqaaiabgkGi2kab es7aKjaadwhadaWgaaWcbaGaamiBaaqabaaakeaacqGHciITcaWG4b WaaSbaaSqaaiaadUgaaeqaaaaaaOGaayjkaiaawMcaaiabg2da9maa laaabaGaaGymaaqaaiaaikdaaaGaeq4Wdm3aaSbaaSqaaiaadYgaca WGRbaabeaakmaalaaabaGaeyOaIyRaeqiTdqMaamyDamaaBaaaleaa caWGRbaabeaaaOqaaiabgkGi2kaadIhadaWgaaWcbaGaamiBaaqaba aaaOGaey4kaSYaaSaaaeaacaaIXaaabaGaaGOmaaaacqaHdpWCdaWg aaWcbaGaam4AaiaadYgaaeqaaOWaaSaaaeaacqGHciITcqaH0oazca WG1bWaaSbaaSqaaiaadYgaaeqaaaGcbaGaeyOaIyRaamiEamaaBaaa leaacaWGRbaabeaaaaGccqGH9aqpcqaHdpWCdaWgaaWcbaGaam4Aai aadYgaaeqaaOWaaSaaaeaacqGHciITcqaH0oazcaWG1bWaaSbaaSqa aiaadYgaaeqaaaGcbaGaeyOaIyRaamiEamaaBaaaleaacaWGRbaabe aaaaaaaa@822D@

where we have used the fact that σ kl = σ lk MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacqaHdpWCdaWgaaWcbaGaam4AaiaadY gaaeqaaOGaeyypa0Jaeq4Wdm3aaSbaaSqaaiaadYgacaWGRbaabeaa aaa@3B75@  (angular momentum balance).  Rewrite this as

σ ij δ u j x i = x i ( σ ij δ u j ) σ ij x i δ u j MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacqaHdpWCdaWgaaWcbaGaamyAaiaadQ gaaeqaaOWaaSaaaeaacqGHciITcqaH0oazcaWG1bWaaSbaaSqaaiaa dQgaaeqaaaGcbaGaeyOaIyRaamiEamaaBaaaleaacaWGPbaabeaaaa GccqGH9aqpdaWcaaqaaiabgkGi2cqaaiabgkGi2kaadIhadaWgaaWc baGaamyAaaqabaaaaOWaaeWaaeaacqaHdpWCdaWgaaWcbaGaamyAai aadQgaaeqaaOGaeqiTdqMaaGPaVlaadwhadaWgaaWcbaGaamOAaaqa baaakiaawIcacaGLPaaacqGHsisldaWcaaqaaiabgkGi2kabeo8aZn aaBaaaleaacaWGPbGaamOAaaqabaaakeaacqGHciITcaWG4bWaaSba aSqaaiaadMgaaeqaaaaakiabes7aKjaaykW7caWG1bWaaSbaaSqaai aadQgaaeqaaaaa@5F12@

Substitute back into the expression for δV MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabes7aKjaadA faaaa@38B9@  and rearrange to see that

δV= V x i ( σ ij δ u j )dV V ( σ ij x i + ρ 0 b j )δ u j dV 2 R t i δ u i dA MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rkY=xq qrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8 fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGa aiaabeqaaiWabaWaaaGcbaGaeqiTdqMaamOvaiabg2da9iaaykW7ca aMc8+aa8quaeaadaWcaaqaaiabgkGi2cqaaiabgkGi2kaadIhadaWg aaWcbaGaamyAaaqabaaaaOWaaeWaaeaacqaHdpWCdaWgaaWcbaGaam yAaiaadQgaaeqaaOGaeqiTdqMaamyDamaaBaaaleaacaWGQbaabeaa aOGaayjkaiaawMcaaiaadsgacaWGwbaaleaacaWGwbaabeqdcqGHRi I8aOGaeyOeI0Yaa8quaeaadaqadaqaamaalaaabaGaeyOaIyRaeq4W dm3aaSbaaSqaaiaadMgacaWGQbaabeaaaOqaaiabgkGi2kaadIhada WgaaWcbaGaamyAaaqabaaaaOGaey4kaSIaeqyWdi3aaSbaaSqaaiaa icdaaeqaaOGaamOyamaaBaaaleaacaWGQbaabeaaaOGaayjkaiaawM caaiabes7aKjaadwhadaWgaaWcbaGaamOAaaqabaaabaGaamOvaaqa b0Gaey4kIipakiaadsgacaWGwbGaeyOeI0Yaa8quaeaacaWG0bWaaS baaSqaaiaadMgaaeqaaOGaeqiTdqMaamyDamaaBaaaleaacaWGPbaa beaakiaadsgacaWGbbaaleaacqGHciITdaWgaaadbaGaaGOmaaqaba WccaWGsbaabeqdcqGHRiI8aaaa@7330@

Now, recall the equations of equilibrium

σ ij x i + ρ 0 b j =0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaadaWcaaqaaiabgkGi2kabeo8aZnaaBa aaleaacaWGPbGaamOAaaqabaaakeaacqGHciITcaWG4bWaaSbaaSqa aiaadMgaaeqaaaaakiabgUcaRiabeg8aYnaaBaaaleaacaaIWaaabe aakiaadkgadaWgaaWcbaGaamOAaaqabaGccqGH9aqpcaaIWaaaaa@42F6@

so that the second term vanishes.  Apply the divergence theorem to express the first integral as a surface integral

V x i ( σ ij δ u j )dV= A σ ij δ u j n i dA MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaadaWdrbqaamaalaaabaGaeyOaIylaba GaeyOaIyRaamiEamaaBaaaleaacaWGPbaabeaaaaGcdaqadaqaaiab eo8aZnaaBaaaleaacaWGPbGaamOAaaqabaGccqaH0oazcaWG1bWaaS baaSqaaiaadQgaaeqaaaGccaGLOaGaayzkaaGaamizaiaadAfacqGH 9aqpdaWdrbqaaiabeo8aZnaaBaaaleaacaWGPbGaamOAaaqabaGccq aH0oazcaWG1bWaaSbaaSqaaiaadQgaaeqaaaqaaiaadgeaaeqaniab gUIiYdaaleaacaWGwbaabeqdcqGHRiI8aOGaamOBamaaBaaaleaaca WGPbaabeaakiaadsgacaWGbbaaaa@5539@

Recall that δ u i =0on  1 R MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8sk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacqaH0oazcaWG1bWaaSbaaSqaaiaadM gaaeqaaOGaeyypa0JaaGimaiaaykW7caaMc8Uaae4Baiaab6gacaqG GaGaeyOaIy7aaSbaaSqaaiaaigdaaeqaaOGaamOuaaaa@4122@ , and note that

A dA = 1 R dA + 2 R dA MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8sk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaadaWdrbqaaiaadsgacaWGbbaaleaaca WGbbaabeqdcqGHRiI8aOGaeyypa0Zaa8quaeaacaWGKbGaamyqaaWc baGaeyOaIy7aaSbaaWqaaiaaigdaaeqaaSGaamOuaaqab0Gaey4kIi pakiabgUcaRmaapefabaGaamizaiaadgeaaSqaaiabgkGi2oaaBaaa meaacaaIYaaabeaaliaadkfaaeqaniabgUIiYdaaaa@47E3@

because either tractions or displacements (but not both) must be prescribed on every point on the boundary. 

 

Therefore

V x i ( σ ij δ u j )dV= A σ ij δ u j n i dA= 1 R σ ij n i δ u j dA + 2 R σ ij n i δ u j dA = 2 R σ ij n i δ u j dA MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8sk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaadaWdrbqaamaalaaabaGaeyOaIylaba GaeyOaIyRaamiEamaaBaaaleaacaWGPbaabeaaaaGcdaqadaqaaiab eo8aZnaaBaaaleaacaWGPbGaamOAaaqabaGccqaH0oazcaWG1bWaaS baaSqaaiaadQgaaeqaaaGccaGLOaGaayzkaaGaamizaiaadAfacqGH 9aqpdaWdrbqaaiabeo8aZnaaBaaaleaacaWGPbGaamOAaaqabaGccq aH0oazcaWG1bWaaSbaaSqaaiaadQgaaeqaaaqaaiaadgeaaeqaniab gUIiYdaaleaacaWGwbaabeqdcqGHRiI8aOGaamOBamaaBaaaleaaca WGPbaabeaakiaadsgacaWGbbGaeyypa0Zaa8quaeaacqaHdpWCdaWg aaWcbaGaamyAaiaadQgaaeqaaOGaamOBamaaBaaaleaacaWGPbaabe aakiabes7aKjaadwhadaWgaaWcbaGaamOAaaqabaGccaWGKbGaamyq aaWcbaGaeyOaIy7aaSbaaWqaaiaaigdaaeqaaSGaamOuaaqab0Gaey 4kIipakiabgUcaRmaapefabaGaeq4Wdm3aaSbaaSqaaiaadMgacaWG Qbaabeaakiaad6gadaWgaaWcbaGaamyAaaqabaGccqaH0oazcaWG1b WaaSbaaSqaaiaadQgaaeqaaOGaamizaiaadgeaaSqaaiabgkGi2oaa BaaameaacaaIYaaabeaaliaadkfaaeqaniabgUIiYdGccqGH9aqpda Wdrbqaaiabeo8aZnaaBaaaleaacaWGPbGaamOAaaqabaGccaWGUbWa aSbaaSqaaiaadMgaaeqaaOGaeqiTdqMaamyDamaaBaaaleaacaWGQb aabeaakiaadsgacaWGbbaaleaacqGHciITdaWgaaadbaGaaGOmaaqa baWccaWGsbaabeqdcqGHRiI8aaaa@8ADB@

Finally, recall that

σ ij n i = t j on  2 R MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8sk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacqaHdpWCdaWgaaWcbaGaamyAaiaadQ gaaeqaaOGaamOBamaaBaaaleaacaWGPbaabeaakiabg2da9iaadsha daWgaaWcbaGaamOAaaqabaGccaaMc8UaaGPaVlaaykW7caaMc8UaaG PaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8Uaae4Baiaab6gacaqG GaGaeyOaIy7aaSbaaSqaaiaaikdaaeqaaOGaamOuaaaa@5109@

and substitute back into the expression for δV MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8sk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacqaH0oazcaWGwbaaaa@3555@  to see that

δV= 2 R ( σ ji n j t i δ u i )dA =0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8sk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacqaH0oazcaWGwbGaeyypa0JaaGPaVl aaykW7daWdrbqaamaabmaabaGaeq4Wdm3aaSbaaSqaaiaadQgacaWG Pbaabeaakiaad6gadaWgaaWcbaGaamOAaaqabaGccqGHsislcaWG0b WaaSbaaSqaaiaadMgaaeqaaOGaeqiTdqMaamyDamaaBaaaleaacaWG PbaabeaaaOGaayjkaiaawMcaaiaadsgacaWGbbaaleaacqGHciITda WgaaadbaGaaGOmaaqabaWccaWGsbaabeqdcqGHRiI8aOGaeyypa0Ja aGimaaaa@50A9@

This proves that V(v) is stationary at v=u, as stated.

 

Finally, we wish to show that V(v) is a minimum at v=u.  This is easy.  Note that we have proved that

V( v )=V(u)+ 1 2 δ 2 V δ 2 V= V C ijkl δ ε ij δ ε kl MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakqaabeqaaiaadAfadaqadaqaaiaahAhaai aawIcacaGLPaaacqGH9aqpcaWGwbGaaiikaiaahwhacaGGPaGaey4k aSYaaSaaaeaacaaIXaaabaGaaGOmaaaacqaH0oazdaahaaWcbeqaai aaikdaaaGccaWGwbaabaGaeqiTdq2aaWbaaSqabeaacaaIYaaaaOGa amOvaiabg2da9maapefabaGaam4qamaaBaaaleaacaWGPbGaamOAai aadUgacaWGSbaabeaakiabes7aKjabew7aLnaaBaaaleaacaWGPbGa amOAaaqabaGccqaH0oazcqaH1oqzdaWgaaWcbaGaam4AaiaadYgaae qaaaqaaiaadAfaaeqaniabgUIiYdaaaaa@574A@

Note that

1 2 C ijlk δ ε kl δ ε ij MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8sk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaadaWcaaqaaiaaigdaaeaacaaIYaaaai aadoeadaWgaaWcbaGaamyAaiaadQgacaWGSbGaam4AaaqabaGccqaH 0oazcqaH1oqzdaWgaaWcbaGaam4AaiaadYgaaeqaaOGaeqiTdqMaeq yTdu2aaSbaaSqaaiaadMgacaWGQbaabeaaaaa@43D0@

is the strain energy density associated with a strain δ ε ij MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8sk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacqaH0oazcqaH1oqzdaWgaaWcbaGaam yAaiaadQgaaeqaaaaa@382A@ .  Strain energy density must always be positive or zero, so that

V( v )V(u) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8sk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacaWGwbWaaeWaaeaacaWH2baacaGLOa GaayzkaaGaeyyzImRaamOvaiaacIcacaWH1bGaaiykaaaa@3B30@

 

 

 

5.7.3 Uniaxial compression of a cylinder solved by energy methods

 

Consider a cylindrical bar subjected to a uniform pressure p on one end, and supported on a rigid, frictionless base.  Neglect temperature changes.  Determine the displacement field in the bar.

 

We will solve this problem using energy methods.  We will guess a displacement field of the form

v 1 = λ 1 x 1 , v 2 = λ 2 x 2 v 3 = λ 3 x 3 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacaWG2bWaaSbaaSqaaiaaigdaaeqaaO Gaeyypa0Jaeq4UdW2aaSbaaSqaaiaaigdaaeqaaOGaamiEamaaBaaa leaacaaIXaaabeaakiaacYcacaaMc8UaaGPaVlaaykW7caaMc8UaaG PaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caWG 2bWaaSbaaSqaaiaaikdaaeqaaOGaeyypa0Jaeq4UdW2aaSbaaSqaai aaikdaaeqaaOGaamiEamaaBaaaleaacaaIYaaabeaakiaaykW7caaM c8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaayk W7caaMc8UaaGPaVlaadAhadaWgaaWcbaGaaG4maaqabaGccqGH9aqp cqaH7oaBdaWgaaWcbaGaaG4maaqabaGccaWG4bWaaSbaaSqaaiaaio daaeqaaaaa@6F0B@

This satisfies the boundary conditions on the bottom face of the cylinder, so it is a kinematically admissible displacement field.  The coefficients λ i MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8sk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacqaH7oaBdaWgaaWcbaGaamyAaaqaba aaaa@35A3@  are to be determined, by minimizing the potential energy.  The strains follow as

ε 11 = λ 1 ε 22 = λ 2 ε 33 = λ 3 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacqaH1oqzdaWgaaWcbaGaaGymaiaaig daaeqaaOGaeyypa0Jaeq4UdW2aaSbaaSqaaiaaigdaaeqaaOGaaGPa VlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7cqaH1o qzdaWgaaWcbaGaaGOmaiaaikdaaeqaaOGaeyypa0Jaeq4UdW2aaSba aSqaaiaaikdaaeqaaOGaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7ca aMc8UaaGPaVlaaykW7caaMc8UaaGPaVlabew7aLnaaBaaaleaacaaI ZaGaaG4maaqabaGccqGH9aqpcqaH7oaBdaWgaaWcbaGaaG4maaqaba aaaa@6384@

with all other strain components zero.  The strain energy density is

U= E 2( 1+ν ) { λ 1 2 + λ 2 2 + λ 3 2 + ν 12ν ( λ 1 + λ 2 + λ 3 ) 2 } MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacaWGvbGaaGPaVlaaykW7cqGH9aqpda WcaaqaaiaadweaaeaacaaIYaWaaeWaaeaacaaIXaGaey4kaSIaeqyV d4gacaGLOaGaayzkaaaaamaacmaabaGaeq4UdW2aa0baaSqaaiaaig daaeaacaaIYaaaaOGaey4kaSIaeq4UdW2aa0baaSqaaiaaikdaaeaa caaIYaaaaOGaey4kaSIaeq4UdW2aa0baaSqaaiaaiodaaeaacaaIYa aaaOGaey4kaSYaaSaaaeaacqaH9oGBaeaacaaIXaGaeyOeI0IaaGOm aiabe27aUbaadaqadaqaaiabeU7aSnaaBaaaleaacaaIXaaabeaaki abgUcaRiabeU7aSnaaBaaaleaacaaIYaaabeaakiabgUcaRiabeU7a SnaaBaaaleaacaaIZaaabeaaaOGaayjkaiaawMcaamaaCaaaleqaba GaaGOmaaaaaOGaay5Eaiaaw2haaaaa@5F45@

The boundary conditions are

1.      On the bottom of the cylinder v 2 =0, t 1 = t 3 =0 v i t i =0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacaWG2bWaaSbaaSqaaiaaikdaaeqaaO Gaeyypa0JaaGimaiaacYcacaaMc8UaaGPaVlaaykW7caaMc8UaaGPa VlaaykW7caWG0bWaaSbaaSqaaiaaigdaaeqaaOGaeyypa0JaamiDam aaBaaaleaacaaIZaaabeaakiabg2da9iaaicdacaaMc8UaaGPaVlaa ykW7caaMc8UaaGPaVlaaykW7cqGHshI3caaMc8UaaGPaVlaaykW7ca aMc8UaamODamaaBaaaleaacaWGPbaabeaakiaadshadaWgaaWcbaGa amyAaaqabaGccqGH9aqpcaaIWaaaaa@5EC7@

2.      On the sides of the cylinder, t i =0 v i t i =0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacaWG0bWaaSbaaSqaaiaadMgaaeqaaO Gaeyypa0JaaGimaiaaykW7caaMc8UaaGPaVlaaykW7cqGHshI3caaM c8UaaGPaVlaaykW7caWG2bWaaSbaaSqaaiaadMgaaeqaaOGaamiDam aaBaaaleaacaWGPbaabeaakiabg2da9iaaicdaaaa@49C8@  

3.      On the top of the cylinder v 2 (L)= λ 2 L t 2 =p, t 1 = t 3 =0 v i t i =p λ 2 L MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacaWG2bWaaSbaaSqaaiaaikdaaeqaaO GaaiikaiaadYeacaGGPaGaeyypa0Jaeq4UdW2aaSbaaSqaaiaaikda aeqaaOGaamitaiaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVl aaykW7caaMc8UaaGPaVlaaykW7caWG0bWaaSbaaSqaaiaaikdaaeqa aOGaeyypa0JaeyOeI0IaamiCaiaacYcacaaMc8UaaGPaVlaaykW7ca aMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caWG0bWaaSbaaSqa aiaaigdaaeqaaOGaeyypa0JaamiDamaaBaaaleaacaaIZaaabeaaki abg2da9iaaicdacaaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7 caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlabgkDiElaadAhadaWgaa WcbaGaamyAaaqabaGccaWG0bWaaSbaaSqaaiaadMgaaeqaaOGaeyyp a0JaeyOeI0IaamiCaiabeU7aSnaaBaaaleaacaaIYaaabeaakiaadY eaaaa@82BA@

Substitute into the expression for strain energy density to see that

V(v)= V E 2( 1+ν ) { λ 1 2 + λ 2 2 + λ 3 2 + ν 12ν ( λ 1 + λ 2 + λ 3 ) 2 }dV A λ 2 L(p) = ALE 2( 1+ν ) { λ 1 2 + λ 2 2 + λ 3 2 + ν 12ν ( λ 1 + λ 2 + λ 3 ) 2 }+A λ 2 Lp MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakqaabeqaaiaadAfacaGGOaGaaCODaiaacM cacqGH9aqpdaWdrbqaamaalaaabaGaamyraaqaaiaaikdadaqadaqa aiaaigdacqGHRaWkcqaH9oGBaiaawIcacaGLPaaaaaWaaiWaaeaacq aH7oaBdaqhaaWcbaGaaGymaaqaaiaaikdaaaGccqGHRaWkcqaH7oaB daqhaaWcbaGaaGOmaaqaaiaaikdaaaGccqGHRaWkcqaH7oaBdaqhaa WcbaGaaG4maaqaaiaaikdaaaGccqGHRaWkdaWcaaqaaiabe27aUbqa aiaaigdacqGHsislcaaIYaGaeqyVd4gaamaabmaabaGaeq4UdW2aaS baaSqaaiaaigdaaeqaaOGaey4kaSIaeq4UdW2aaSbaaSqaaiaaikda aeqaaOGaey4kaSIaeq4UdW2aaSbaaSqaaiaaiodaaeqaaaGccaGLOa GaayzkaaWaaWbaaSqabeaacaaIYaaaaaGccaGL7bGaayzFaaGaamiz aiaadAfacqGHsisldaWdrbqaaiabeU7aSnaaBaaaleaacaaIYaaabe aakiaadYeacaGGOaGaeyOeI0IaamiCaiaacMcaaSqaaiaadgeaaeqa niabgUIiYdaaleaacaWGwbaabeqdcqGHRiI8aaGcbaGaaGPaVlaayk W7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPa VlaaykW7caaMc8UaaGPaVlaaykW7caaMc8Uaeyypa0JaaGPaVlaayk W7daWcaaqaaiaadgeacaWGmbGaamyraaqaaiaaikdadaqadaqaaiaa igdacqGHRaWkcqaH9oGBaiaawIcacaGLPaaaaaWaaiWaaeaacqaH7o aBdaqhaaWcbaGaaGymaaqaaiaaikdaaaGccqGHRaWkcqaH7oaBdaqh aaWcbaGaaGOmaaqaaiaaikdaaaGccqGHRaWkcqaH7oaBdaqhaaWcba GaaG4maaqaaiaaikdaaaGccqGHRaWkdaWcaaqaaiabe27aUbqaaiaa igdacqGHsislcaaIYaGaeqyVd4gaamaabmaabaGaeq4UdW2aaSbaaS qaaiaaigdaaeqaaOGaey4kaSIaeq4UdW2aaSbaaSqaaiaaikdaaeqa aOGaey4kaSIaeq4UdW2aaSbaaSqaaiaaiodaaeqaaaGccaGLOaGaay zkaaWaaWbaaSqabeaacaaIYaaaaaGccaGL7bGaayzFaaGaey4kaSIa amyqaiabeU7aSnaaBaaaleaacaaIYaaabeaakiaadYeacaWGWbaaaa a@B8A7@

Now, the actual displacement field minimizes V.  This requires

V λ 1 = V λ 2 = V λ 3 =0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaadaWcaaqaaiabgkGi2kaadAfaaeaacq GHciITcqaH7oaBdaWgaaWcbaGaaGymaaqabaaaaOGaeyypa0ZaaSaa aeaacqGHciITcaWGwbaabaGaeyOaIyRaeq4UdW2aaSbaaSqaaiaaik daaeqaaaaakiabg2da9maalaaabaGaeyOaIyRaamOvaaqaaiabgkGi 2kabeU7aSnaaBaaaleaacaaIZaaabeaaaaGccqGH9aqpcaaIWaaaaa@49A8@

Evaluate the derivatives to see that

ALE 2( 1+ν ) { 2 λ 1 + 2ν 12ν ( λ 1 + λ 2 + λ 3 ) }=0 ALE 2( 1+ν ) { 2 λ 2 + 2ν 12ν ( λ 1 + λ 2 + λ 3 ) }+ALp=0 ALE 2( 1+ν ) { 2 λ 3 + 2ν 12ν ( λ 1 + λ 2 + λ 3 ) }=0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakqaabeqaamaalaaabaGaamyqaiaadYeaca WGfbaabaGaaGOmamaabmaabaGaaGymaiabgUcaRiabe27aUbGaayjk aiaawMcaaaaadaGadaqaaiaaikdacqaH7oaBdaWgaaWcbaGaaGymaa qabaGccqGHRaWkdaWcaaqaaiaaikdacqaH9oGBaeaacaaIXaGaeyOe I0IaaGOmaiabe27aUbaadaqadaqaaiabeU7aSnaaBaaaleaacaaIXa aabeaakiabgUcaRiabeU7aSnaaBaaaleaacaaIYaaabeaakiabgUca RiabeU7aSnaaBaaaleaacaaIZaaabeaaaOGaayjkaiaawMcaaaGaay 5Eaiaaw2haaiabg2da9iaaicdaaeaadaWcaaqaaiaadgeacaWGmbGa amyraaqaaiaaikdadaqadaqaaiaaigdacqGHRaWkcqaH9oGBaiaawI cacaGLPaaaaaWaaiWaaeaacaaIYaGaeq4UdW2aaSbaaSqaaiaaikda aeqaaOGaey4kaSYaaSaaaeaacaaIYaGaeqyVd4gabaGaaGymaiabgk HiTiaaikdacqaH9oGBaaWaaeWaaeaacqaH7oaBdaWgaaWcbaGaaGym aaqabaGccqGHRaWkcqaH7oaBdaWgaaWcbaGaaGOmaaqabaGccqGHRa WkcqaH7oaBdaWgaaWcbaGaaG4maaqabaaakiaawIcacaGLPaaaaiaa wUhacaGL9baacqGHRaWkcaWGbbGaamitaiaadchacqGH9aqpcaaIWa aabaWaaSaaaeaacaWGbbGaamitaiaadweaaeaacaaIYaWaaeWaaeaa caaIXaGaey4kaSIaeqyVd4gacaGLOaGaayzkaaaaamaacmaabaGaaG OmaiabeU7aSnaaBaaaleaacaaIZaaabeaakiabgUcaRmaalaaabaGa aGOmaiabe27aUbqaaiaaigdacqGHsislcaaIYaGaeqyVd4gaamaabm aabaGaeq4UdW2aaSbaaSqaaiaaigdaaeqaaOGaey4kaSIaeq4UdW2a aSbaaSqaaiaaikdaaeqaaOGaey4kaSIaeq4UdW2aaSbaaSqaaiaaio daaeqaaaGccaGLOaGaayzkaaaacaGL7bGaayzFaaGaeyypa0JaaGim aaaaaa@9C98@

It is easy to solve these equations to see that

λ 1 =p/E λ 2 = λ 3 =νp/E MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacqaH7oaBdaWgaaWcbaGaaGymaaqaba GccqGH9aqpcqGHsislcaWGWbGaai4laiaadweacaaMc8UaaGPaVlaa ykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaG PaVlaaykW7cqaH7oaBdaWgaaWcbaGaaGOmaaqabaGccqGH9aqpcqaH 7oaBdaWgaaWcbaGaaG4maaqabaGccqGH9aqpcqaH9oGBcaWGWbGaai 4laiaadweaaaa@57D6@

 

This is, of course, the exact solution, which is reassuring.  Notice that we never had to calculate stresses or worry about equilibrium MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaGqaaKqzGfaeaa aaaaaaa8qacaWFtacaaa@37E6@  the variational principle takes care of all that for us.

 

Let us solve the same problem, but this time with displacement boundary conditions on the top of the cylinder.

 

The cylinder has unstretched length L and is stretched between frictionless grips to length L+h.  This time, the kinematically admissible displacement field must satisfy boundary conditions on both top and bottom surface of the cylinder.   Therefore, we choose

v 1 = λ 1 x 1 , v 2 =h x 2 /L v 3 = λ 3 x 3 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacaWG2bWaaSbaaSqaaiaaigdaaeqaaO Gaeyypa0Jaeq4UdW2aaSbaaSqaaiaaigdaaeqaaOGaamiEamaaBaaa leaacaaIXaaabeaakiaacYcacaaMc8UaaGPaVlaaykW7caaMc8UaaG PaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caWG 2bWaaSbaaSqaaiaaikdaaeqaaOGaeyypa0JaamiAaiaadIhadaWgaa WcbaGaaGOmaaqabaGccaGGVaGaamitaiaaykW7caaMc8UaaGPaVlaa ykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaG PaVlaadAhadaWgaaWcbaGaaG4maaqabaGccqGH9aqpcqaH7oaBdaWg aaWcbaGaaG4maaqabaGccaWG4bWaaSbaaSqaaiaaiodaaeqaaaaa@6ED6@

Proceeding as before, we now find that the potential energy is

V(v)= ALE 2( 1+ν ) { λ 1 2 + h 2 L 2 + λ 3 2 + ν 12ν ( λ 1 + h L + λ 3 ) 2 } MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacaWGwbGaaiikaiaahAhacaGGPaGaaG PaVlaaykW7caaMc8Uaeyypa0JaaGPaVlaaykW7daWcaaqaaiaadgea caWGmbGaamyraaqaaiaaikdadaqadaqaaiaaigdacqGHRaWkcqaH9o GBaiaawIcacaGLPaaaaaWaaiWaaeaacqaH7oaBdaqhaaWcbaGaaGym aaqaaiaaikdaaaGccqGHRaWkdaWcaaqaaiaadIgadaahaaWcbeqaai aaikdaaaaakeaacaWGmbWaaWbaaSqabeaacaaIYaaaaaaakiabgUca RiabeU7aSnaaDaaaleaacaaIZaaabaGaaGOmaaaakiabgUcaRmaala aabaGaeqyVd4gabaGaaGymaiabgkHiTiaaikdacqaH9oGBaaWaaeWa aeaacqaH7oaBdaWgaaWcbaGaaGymaaqabaGccqGHRaWkdaWcaaqaai aadIgaaeaacaWGmbaaaiabgUcaRiabeU7aSnaaBaaaleaacaaIZaaa beaaaOGaayjkaiaawMcaamaaCaaaleqabaGaaGOmaaaaaOGaay5Eai aaw2haaaaa@674F@

Note that this time there is no contribution to the potential energy from the tractions on the top of the cylinder, because now the displacement is prescribed there, instead of the pressure.  Minimizing the potential energy as before

ALE 2( 1+ν ) { 2 λ 1 + 2ν 12ν ( λ 1 + h L + λ 3 ) }=0 ALE 2( 1+ν ) { 2 λ 3 + 2ν 12ν ( λ 1 + h L + λ 3 ) }=0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakqaabeqaamaalaaabaGaamyqaiaadYeaca WGfbaabaGaaGOmamaabmaabaGaaGymaiabgUcaRiabe27aUbGaayjk aiaawMcaaaaadaGadaqaaiaaikdacqaH7oaBdaWgaaWcbaGaaGymaa qabaGccqGHRaWkdaWcaaqaaiaaikdacqaH9oGBaeaacaaIXaGaeyOe I0IaaGOmaiabe27aUbaadaqadaqaaiabeU7aSnaaBaaaleaacaaIXa aabeaakiabgUcaRmaalaaabaGaamiAaaqaaiaadYeaaaGaey4kaSIa eq4UdW2aaSbaaSqaaiaaiodaaeqaaaGccaGLOaGaayzkaaaacaGL7b GaayzFaaGaeyypa0JaaGimaaqaamaalaaabaGaamyqaiaadYeacaWG fbaabaGaaGOmamaabmaabaGaaGymaiabgUcaRiabe27aUbGaayjkai aawMcaaaaadaGadaqaaiaaikdacqaH7oaBdaWgaaWcbaGaaG4maaqa baGccqGHRaWkdaWcaaqaaiaaikdacqaH9oGBaeaacaaIXaGaeyOeI0 IaaGOmaiabe27aUbaadaqadaqaaiabeU7aSnaaBaaaleaacaaIXaaa beaakiabgUcaRmaalaaabaGaamiAaaqaaiaadYeaaaGaey4kaSIaeq 4UdW2aaSbaaSqaaiaaiodaaeqaaaGccaGLOaGaayzkaaaacaGL7bGa ayzFaaGaeyypa0JaaGimaaaaaa@755A@

Solve these equations to conclude that

λ 1 = λ 3 =ν h L MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacqaH7oaBdaWgaaWcbaGaaGymaaqaba GccqGH9aqpcqaH7oaBdaWgaaWcbaGaaG4maaqabaGccqGH9aqpcqGH sislcqaH9oGBdaWcaaqaaiaadIgaaeaacaWGmbaaaaaa@3E90@

Again, this is the exact solution.

 

 

 

 

5.7.4 Variational derivation of the beam equations

 

Variational methods can be used to solve boundary value problems exactly, as described in the preceding section.   The real power of variational methods, however, is to provide a systematic way to find approximate solutions to boundary value problems. 

 

We will illustrate this by re-deriving the equations governing beam bending theory using the principle of minimum potential energy.

 

Consider a slender rod with rectangular cross section, subjected to uniform pressure q(x) on its top surface.  Assume that the rod is an isotropic, linear elastic solid with Young’s modulus E and Poisson’s ratio ν MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8sk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacqaH9oGBaaa@348D@ . The boundary conditions at the ends of the bar will be left unspecified for the time being.

 

We proceed by approximating the displacement field within the bar.  We will suppose that the strains at any given cross section are completely characterized by the local curvature of the beam, so that at a given cross section x

ε 11 = ( x 2 y 0 ) R( x 1 ) ε 22 = ε 33 =ν ε 11 ε 12 = ε 13 = ε 23 =0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacqaH1oqzdaWgaaWcbaGaaGymaiaaig daaeqaaOGaeyypa0ZaaSaaaeaacqGHsislcaGGOaGaamiEamaaBaaa leaacaaIYaaabeaakiabgkHiTiaadMhadaWgaaWcbaGaaGimaaqaba GccaGGPaaabaGaamOuaiaacIcacaWG4bWaaSbaaSqaaiaaigdaaeqa aOGaaiykaaaacaaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7ca aMc8UaeqyTdu2aaSbaaSqaaiaaikdacaaIYaaabeaakiabg2da9iab ew7aLnaaBaaaleaacaaIZaGaaG4maaqabaGccqGH9aqpcqGHsislcq aH9oGBcqaH1oqzdaWgaaWcbaGaaGymaiaaigdaaeqaaOGaaGPaVlaa ykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlabew7aLnaaBaaale aacaaIXaGaaGOmaaqabaGccqGH9aqpcqaH1oqzdaWgaaWcbaGaaGym aiaaiodaaeqaaOGaeyypa0JaeqyTdu2aaSbaaSqaaiaaikdacaaIZa aabeaakiabg2da9iaaicdaaaa@7471@

Here, y 0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacaWG5bWaaSbaaSqaaiaaicdaaeqaaa aa@34A9@  is the height of a fiber in the beam whose length is unchanged.  y 0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8sk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacaWG5bWaaSbaaSqaaiaaicdaaeqaaa aa@34B9@  must be determined as part of the solution.

 

The displacement and strain fields are therefore completely characterized by y 0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8sk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacaWG5bWaaSbaaSqaaiaaicdaaeqaaa aa@34B9@  and R(x). Rather than solve for R, we will approximate the curvature at x by the second derivative of the vertical deflection w, so that

ε 11 = ( x 2 y 0 ) R( x 1 ) ( x 2 y 0 ) d 2 w( x 1 ) d x 1 2 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacqaH1oqzdaWgaaWcbaGaaGymaiaaig daaeqaaOGaeyypa0ZaaSaaaeaacqGHsislcaGGOaGaamiEamaaBaaa leaacaaIYaaabeaakiabgkHiTiaadMhadaWgaaWcbaGaaGimaaqaba GccaGGPaaabaGaamOuaiaacIcacaWG4bWaaSbaaSqaaiaaigdaaeqa aOGaaiykaaaacqGHijYUcqGHsislcaGGOaGaamiEamaaBaaaleaaca aIYaaabeaakiabgkHiTiaadMhadaWgaaWcbaGaaGimaaqabaGccaGG PaWaaSaaaeaacaWGKbWaaWbaaSqabeaacaaIYaaaaOGaam4DaiaacI cacaWG4bWaaSbaaSqaaiaaigdaaeqaaOGaaiykaaqaaiaadsgacaWG 4bWaaSbaaSqaaiaaigdaaeqaaOWaaWbaaSqabeaacaaIYaaaaaaaaa a@550C@

 

Now, we want to find w(x) and  y 0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8sk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacaWG5bWaaSbaaSqaaiaaicdaaeqaaa aa@34B9@  that will best approximate the actual displacement field within the bar.  We will do this by choosing w and  y 0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8sk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacaWG5bWaaSbaaSqaaiaaicdaaeqaaa aa@34B9@  so as to minimize the potential energy of the solid.

 

Begin by computing the potential energy.  It is straightforward to show that the strain energy density is

ϕ= 1 2 E { ( x 2 y 0 ) d 2 w( x 1 ) d x 1 2 } 2 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacqaHvpGzcqGH9aqpdaWcaaqaaiaaig daaeaacaaIYaaaaiaadweadaGadaqaaiaacIcacaWG4bWaaSbaaSqa aiaaikdaaeqaaOGaeyOeI0IaamyEamaaBaaaleaacaaIWaaabeaaki aacMcadaWcaaqaaiaadsgadaahaaWcbeqaaiaaikdaaaGccaWG3bGa aiikaiaadIhadaWgaaWcbaGaaGymaaqabaGccaGGPaaabaGaamizai aadIhadaWgaaWcbaGaaGymaaqabaGcdaahaaWcbeqaaiaaikdaaaaa aaGccaGL7bGaayzFaaWaaWbaaSqabeaacaaIYaaaaaaa@4B1A@

Hence

V(w, y 0 )= 0 L A 1 2 E { ( x 2 y 0 ) d 2 w( x 1 ) d x 1 2 } 2 dAd x 1 0 L bq( x 1 )w( x 1 )dx MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacaWGwbGaaiikaiaadEhacaGGSaGaam yEamaaBaaaleaacaaIWaaabeaakiaacMcacqGH9aqpdaWdXbqaamaa pefabaaaleaacaWGbbaabeqdcqGHRiI8aaWcbaGaaGimaaqaaiaadY eaa0Gaey4kIipakmaalaaabaGaaGymaaqaaiaaikdaaaGaamyramaa cmaabaGaaiikaiaadIhadaWgaaWcbaGaaGOmaaqabaGccqGHsislca WG5bWaaSbaaSqaaiaaicdaaeqaaOGaaiykamaalaaabaGaamizamaa CaaaleqabaGaaGOmaaaakiaadEhacaGGOaGaamiEamaaBaaaleaaca aIXaaabeaakiaacMcaaeaacaWGKbGaamiEamaaBaaaleaacaaIXaaa beaakmaaCaaaleqabaGaaGOmaaaaaaaakiaawUhacaGL9baadaahaa WcbeqaaiaaikdaaaGccaWGKbGaamyqaiaaykW7caaMc8Uaamizaiaa dIhadaWgaaWcbaGaaGymaaqabaGccqGHsisldaWdXbqaaiaadkgaca WGXbGaaiikaiaadIhadaWgaaWcbaGaaGymaaqabaGccaGGPaGaam4D aiaacIcacaWG4bWaaSbaaSqaaiaaigdaaeqaaOGaaiykaiaadsgaca WG4baaleaacaaIWaaabaGaamitaaqdcqGHRiI8aaaa@6E02@

Here, we have neglected the small additional deflection of the beam surface due to ε 22 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8sk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacqaH1oqzdaWgaaWcbaGaaGOmaiaaik daaeqaaaaa@3620@

 

We now wish to minimize V with respect to w and  y 0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8sk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacaWG5bWaaSbaaSqaaiaaicdaaeqaaa aa@34B9@ .  Do the latter first:

V(w, y 0 ) y 0 = 0 L A E { ( x 2 y 0 ) d 2 w( x 1 ) d x 1 2 } dAd x 1 =0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaadaWcaaqaaiabgkGi2kaadAfacaGGOa Gaam4DaiaacYcacaWG5bWaaSbaaSqaaiaaicdaaeqaaOGaaiykaaqa aiabgkGi2kaadMhadaWgaaWcbaGaaGimaaqabaaaaOGaeyypa0Zaa8 qCaeaadaWdrbqaaaWcbaGaamyqaaqab0Gaey4kIipaaSqaaiaaicda aeaacaWGmbaaniabgUIiYdGccaWGfbWaaiWaaeaacaGGOaGaamiEam aaBaaaleaacaaIYaaabeaakiabgkHiTiaadMhadaWgaaWcbaGaaGim aaqabaGccaGGPaWaaSaaaeaacaWGKbWaaWbaaSqabeaacaaIYaaaaO Gaam4DaiaacIcacaWG4bWaaSbaaSqaaiaaigdaaeqaaOGaaiykaaqa aiaadsgacaWG4bWaaSbaaSqaaiaaigdaaeqaaOWaaWbaaSqabeaaca aIYaaaaaaaaOGaay5Eaiaaw2haamaaCaaaleqabaaaaOGaamizaiaa dgeacaaMc8UaaGPaVlaadsgacaWG4bWaaSbaaSqaaiaaigdaaeqaaO Gaeyypa0JaaGimaaaa@621B@

which is evidently satisfied for any w by choosing

y 0 = 1 A A x 2 dA MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacaWG5bWaaSbaaSqaaiaaicdaaeqaaO Gaeyypa0ZaaSaaaeaacaaIXaaabaGaamyqaaaadaWdrbqaaiaadIha daWgaaWcbaGaaGOmaaqabaaabaGaamyqaaqab0Gaey4kIipakiaayk W7caWGKbGaamyqaaaa@3F78@

This is the usual expression for the position of the neutral axis of a beam. We can now simplify our expression for potential energy by defining

I= 1 A A ( x 2 y 0 ) 2 dA MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacaWGjbGaeyypa0ZaaSaaaeaacaaIXa aabaGaamyqaaaadaWdrbqaaiaacIcacaWG4bWaaSbaaSqaaiaaikda aeqaaOGaeyOeI0IaamyEamaaBaaaleaacaaIWaaabeaakiaacMcada ahaaWcbeqaaiaaikdaaaaabaGaamyqaaqab0Gaey4kIipakiaaykW7 caWGKbGaamyqaaaa@437F@

so that

V(w)= 0 L 1 2 EI { d 2 w( x 1 ) d x 1 2 } 2 d x 1 0 L bq( x 1 )w( x 1 )d x 1 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacaWGwbGaaiikaiaadEhacaGGPaGaey ypa0Zaa8qCaeaaaSqaaiaaicdaaeaacaWGmbaaniabgUIiYdGcdaWc aaqaaiaaigdaaeaacaaIYaaaaiaadweacaWGjbWaaiWaaeaadaWcaa qaaiaadsgadaahaaWcbeqaaiaaikdaaaGccaWG3bGaaiikaiaadIha daWgaaWcbaGaaGymaaqabaGccaGGPaaabaGaamizaiaadIhadaWgaa WcbaGaaGymaaqabaGcdaahaaWcbeqaaiaaikdaaaaaaaGccaGL7bGa ayzFaaWaaWbaaSqabeaacaaIYaaaaOGaaGPaVlaadsgacaWG4bWaaS baaSqaaiaaigdaaeqaaOGaeyOeI0Yaa8qCaeaacaWGIbGaamyCaiaa cIcacaWG4bWaaSbaaSqaaiaaigdaaeqaaOGaaiykaiaadEhacaGGOa GaamiEamaaBaaaleaacaaIXaaabeaakiaacMcacaWGKbGaamiEamaa BaaaleaacaaIXaaabeaaaeaacaaIWaaabaGaamitaaqdcqGHRiI8aa aa@60A1@

Now turn to the more difficult problem of finding w that will minimise V.  To do this, let us calculate the change in V when w MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacaWG3baaaa@33C1@  is changed slightly to w+δw MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacaWG3bGaey4kaSIaeqiTdqMaam4Daa aa@3744@

V(w+δw)V(w)= 0 L 1 2 EI { d 2 w( x 1 ) d x 1 2 + d 2 δw( x 1 ) d x 1 2 } 2 d x 1 0 L bq( x 1 ){ w( x 1 )+δw( x 1 ) }d x 1 0 L 1 2 EI { d 2 w( x 1 ) d x 1 2 } 2 d x 1 0 L bq( x 1 )w( x 1 )d x 1 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakqaabeqaaiaadAfacaGGOaGaam4DaiabgU caRiabes7aKjaadEhacaGGPaGaeyOeI0IaamOvaiaacIcacaWG3bGa aiykaiabg2da9maapehabaaaleaacaaIWaaabaGaamitaaqdcqGHRi I8aOWaaSaaaeaacaaIXaaabaGaaGOmaaaacaWGfbGaamysamaacmaa baWaaSaaaeaacaWGKbWaaWbaaSqabeaacaaIYaaaaOGaam4DaiaacI cacaWG4bWaaSbaaSqaaiaaigdaaeqaaOGaaiykaaqaaiaadsgacaWG 4bWaaSbaaSqaaiaaigdaaeqaaOWaaWbaaSqabeaacaaIYaaaaaaaki abgUcaRmaalaaabaGaamizamaaCaaaleqabaGaaGOmaaaakiabes7a KjaadEhacaGGOaGaamiEamaaBaaaleaacaaIXaaabeaakiaacMcaae aacaWGKbGaamiEamaaBaaaleaacaaIXaaabeaakmaaCaaaleqabaGa aGOmaaaaaaaakiaawUhacaGL9baadaahaaWcbeqaaiaaikdaaaGcca aMc8UaamizaiaadIhadaWgaaWcbaGaaGymaaqabaGccqGHsisldaWd XbqaaiaadkgacaWGXbGaaiikaiaadIhadaWgaaWcbaGaaGymaaqaba GccaGGPaWaaiWaaeaacaWG3bGaaiikaiaadIhadaWgaaWcbaGaaGym aaqabaGccaGGPaGaey4kaSIaeqiTdqMaam4DaiaacIcacaWG4bWaaS baaSqaaiaaigdaaeqaaOGaaiykaaGaay5Eaiaaw2haaiaadsgacaWG 4bWaaSbaaSqaaiaaigdaaeqaaaqaaiaaicdaaeaacaWGmbaaniabgU IiYdaakeaacaaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaM c8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaayk W7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPa VlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8 UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7 caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7cqGHsisldaWdXb qaaaWcbaGaaGimaaqaaiaadYeaa0Gaey4kIipakmaalaaabaGaaGym aaqaaiaaikdaaaGaamyraiaadMeadaGadaqaamaalaaabaGaamizam aaCaaaleqabaGaaGOmaaaakiaadEhacaGGOaGaamiEamaaBaaaleaa caaIXaaabeaakiaacMcaaeaacaWGKbGaamiEamaaBaaaleaacaaIXa aabeaakmaaCaaaleqabaGaaGOmaaaaaaaakiaawUhacaGL9baadaah aaWcbeqaaiaaikdaaaGccaaMc8UaamizaiaadIhadaWgaaWcbaGaaG ymaaqabaGccqGHsisldaWdXbqaaiaadkgacaWGXbGaaiikaiaadIha daWgaaWcbaGaaGymaaqabaGccaGGPaGaam4DaiaacIcacaWG4bWaaS baaSqaaiaaigdaaeqaaOGaaiykaiaadsgacaWG4bWaaSbaaSqaaiaa igdaaeqaaaqaaiaaicdaaeaacaWGmbaaniabgUIiYdaaaaa@EDCF@

Expand this out to see that

V(w+δw)V(w)= 0 L EI d 2 w( x 1 ) d x 1 2 d 2 δw( x 1 ) d x 1 2 d x 1 0 L bq( x 1 )δw( x 1 )d x 1 + 0 L 1 2 EI { d 2 δw( x 1 ) d x 1 2 } 2 d x 1 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakqaabeqaaiaadAfacaGGOaGaam4DaiabgU caRiabes7aKjaadEhacaGGPaGaeyOeI0IaamOvaiaacIcacaWG3bGa aiykaiabg2da9maapehabaaaleaacaaIWaaabaGaamitaaqdcqGHRi I8aOGaamyraiaadMeadaWcaaqaaiaadsgadaahaaWcbeqaaiaaikda aaGccaWG3bGaaiikaiaadIhadaWgaaWcbaGaaGymaaqabaGccaGGPa aabaGaamizaiaadIhadaWgaaWcbaGaaGymaaqabaGcdaahaaWcbeqa aiaaikdaaaaaaOWaaSaaaeaacaWGKbWaaWbaaSqabeaacaaIYaaaaO GaeqiTdqMaam4DaiaacIcacaWG4bWaaSbaaSqaaiaaigdaaeqaaOGa aiykaaqaaiaadsgacaWG4bWaaSbaaSqaaiaaigdaaeqaaOWaaWbaaS qabeaacaaIYaaaaaaakiaaykW7caWGKbGaamiEamaaBaaaleaacaaI XaaabeaakiabgkHiTmaapehabaGaamOyaiaadghacaGGOaGaamiEam aaBaaaleaacaaIXaaabeaakiaacMcacqaH0oazcaWG3bGaaiikaiaa dIhadaWgaaWcbaGaaGymaaqabaGccaGGPaGaamizaiaadIhadaWgaa WcbaGaaGymaaqabaaabaGaaGimaaqaaiaadYeaa0Gaey4kIipaaOqa aiaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8 UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7 caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVl aaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8Ua aGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7ca aMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7cqGHRaWkdaWdXbqa aaWcbaGaaGimaaqaaiaadYeaa0Gaey4kIipakmaalaaabaGaaGymaa qaaiaaikdaaaGaamyraiaadMeadaGadaqaamaalaaabaGaamizamaa CaaaleqabaGaaGOmaaaakiabes7aKjaadEhacaGGOaGaamiEamaaBa aaleaacaaIXaaabeaakiaacMcaaeaacaWGKbGaamiEamaaBaaaleaa caaIXaaabeaakmaaCaaaleqabaGaaGOmaaaaaaaakiaawUhacaGL9b aadaahaaWcbeqaaiaaikdaaaGccaaMc8UaamizaiaadIhadaWgaaWc baGaaGymaaqabaaaaaa@D2FD@

Now, if V(w) is a minimum, then

0 L EI d 2 w( x 1 ) d x 1 2 d 2 δw( x 1 ) d x 1 2 d x 1 0 L bq( x 1 )δw( x 1 )d x 1 =0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaadaWdXbqaaaWcbaGaaGimaaqaaiaadY eaa0Gaey4kIipakiaadweacaWGjbWaaSaaaeaacaWGKbWaaWbaaSqa beaacaaIYaaaaOGaam4DaiaacIcacaWG4bWaaSbaaSqaaiaaigdaae qaaOGaaiykaaqaaiaadsgacaWG4bWaaSbaaSqaaiaaigdaaeqaaOWa aWbaaSqabeaacaaIYaaaaaaakmaalaaabaGaamizamaaCaaaleqaba GaaGOmaaaakiabes7aKjaadEhacaGGOaGaamiEamaaBaaaleaacaaI XaaabeaakiaacMcaaeaacaWGKbGaamiEamaaBaaaleaacaaIXaaabe aakmaaCaaaleqabaGaaGOmaaaaaaGccaaMc8UaamizaiaadIhadaWg aaWcbaGaaGymaaqabaGccqGHsisldaWdXbqaaiaadkgacaWGXbGaai ikaiaadIhadaWgaaWcbaGaaGymaaqabaGccaGGPaGaeqiTdqMaam4D aiaacIcacaWG4bWaaSbaaSqaaiaaigdaaeqaaOGaaiykaiaadsgaca WG4bWaaSbaaSqaaiaaigdaaeqaaaqaaiaaicdaaeaacaWGmbaaniab gUIiYdGccqGH9aqpcaaIWaaaaa@66CD@

We are none the wiser as a result of this exercise, but if we integrate the first integral by parts twice, we find that

[ EI d 2 w d x 1 2 dδw d x 1 ] 0 L [ d d x 1 (EI d 2 w d x 1 2 )δw ] 0 L + 0 L ( EI d 4 w( x 1 ) d x 1 4 bq( x 1 ) )δwd x 1 =0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaadaWadaqaaiaadweacaWGjbWaaSaaae aacaWGKbWaaWbaaSqabeaacaaIYaaaaOGaam4DaaqaaiaadsgacaWG 4bWaaSbaaSqaaiaaigdaaeqaaOWaaWbaaSqabeaacaaIYaaaaaaakm aalaaabaGaamizaiabes7aKjaadEhaaeaacaWGKbGaamiEamaaBaaa leaacaaIXaaabeaaaaaakiaawUfacaGLDbaadaqhaaWcbaGaaGimaa qaaiaadYeaaaGccqGHsisldaWadaqaamaalaaabaGaamizaaqaaiaa dsgacaWG4bWaaSbaaSqaaiaaigdaaeqaaaaakiaacIcacaWGfbGaam ysamaalaaabaGaamizamaaCaaaleqabaGaaGOmaaaakiaadEhaaeaa caWGKbGaamiEamaaBaaaleaacaaIXaaabeaakmaaCaaaleqabaGaaG OmaaaaaaGccaGGPaGaeqiTdqMaam4DaaGaay5waiaaw2faamaaDaaa leaacaaIWaaabaGaamitaaaakiabgUcaRmaapehabaaaleaacaaIWa aabaGaamitaaqdcqGHRiI8aOWaaeWaaeaacaWGfbGaamysamaalaaa baGaamizamaaCaaaleqabaGaaGinaaaakiaadEhacaGGOaGaamiEam aaBaaaleaacaaIXaaabeaakiaacMcaaeaacaWGKbGaamiEamaaBaaa leaacaaIXaaabeaakmaaCaaaleqabaGaaGinaaaaaaGccqGHsislca WGIbGaamyCaiaacIcacaWG4bWaaSbaaSqaaiaaigdaaeqaaOGaaiyk aiaaykW7aiaawIcacaGLPaaacqaH0oazcaWG3bGaamizaiaadIhada WgaaWcbaGaaGymaaqabaGccqGH9aqpcaaIWaaaaa@7ABB@

 

Since this is zero for any δw MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacqaH0oazcaWG3baaaa@3566@  we conclude that

EI d 4 w( x 1 ) d x 1 4 bq( x 1 )=0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacaWGfbGaamysamaalaaabaGaamizam aaCaaaleqabaGaaGinaaaakiaadEhacaGGOaGaamiEamaaBaaaleaa caaIXaaabeaakiaacMcaaeaacaWGKbGaamiEamaaBaaaleaacaaIXa aabeaakmaaCaaaleqabaGaaGinaaaaaaGccqGHsislcaWGIbGaamyC aiaacIcacaWG4bWaaSbaaSqaaiaaigdaaeqaaOGaaiykaiabg2da9i aaicdaaaa@462B@

to ensure that the third term in this expression vanishes.  This gives us the required governing equation for w.  However, we still need to deal with the first two boundary terms.

 

There are several ways to prescribe boundary conditions on the ends of the beam to ensure that V is stationary.

1.      We may prescribe w and its first derivative.  In this case the variation in w MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacaWG3baaaa@33C2@  must satisfy δw=dδw/dx=0 MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacqaH0oazcaWG3bGaeyypa0Jaamizai abes7aKjaadEhacaGGVaGaamizaiaadIhacqGH9aqpcaaIWaaaaa@3E50@  to ensure that w is a kinematically admissible displacement.  The boundary terms are zero under these conditions

2.      Prescribe only the value of w. In this case we must ensure that δw=0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8sk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacqaH0oazcaWG3bGaeyypa0JaaGimaa aa@3736@  on the end of the beam. The  second boundary term is automatically zero.  To ensure that the first boundary term is zero we must set

d 2 w d x 1 2 =0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaadaWcaaqaaiaadsgadaahaaWcbeqaai aaikdaaaGccaWG3baabaGaamizaiaadIhadaWgaaWcbaGaaGymaaqa baGcdaahaaWcbeqaaiaaikdaaaaaaOGaeyypa0JaaGimaiaaykW7aa a@3CC2@

to ensure that V is stationary.  We know from elementary strength of materials courses that this is equivalent to the condition that the shear force vanishes on the end of the beam.

3.      Prescribe only the value of dw/dx MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacaWGKbGaam4Daiaac+cacaWGKbGaam iEaaaa@3744@ .  In this case, we must ensure that dδw/dx=0 MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacaWGKbGaeqiTdqMaam4Daiaac+caca WGKbGaamiEaiabg2da9iaaicdaaaa@3AA9@  so that w+δw MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacaWG3bGaey4kaSIaeqiTdqMaam4Daa aa@3745@  is a kinematically admissible displacement.  The first boundary term vanishes; while the second boundary term is zero if we choose

d d x 1 ( EI d 2 w d x 1 2 )=0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaadaWcaaqaaiaadsgaaeaacaWGKbGaam iEamaaBaaaleaacaaIXaaabeaaaaGcdaqadaqaaiaadweacaWGjbWa aSaaaeaacaWGKbWaaWbaaSqabeaacaaIYaaaaOGaam4Daaqaaiaads gacaWG4bWaaSbaaSqaaiaaigdaaeqaaOWaaWbaaSqabeaacaaIYaaa aaaaaOGaayjkaiaawMcaaiabg2da9iaaicdacaaMc8oaaa@43B3@

This is equivalent to setting the bending moment to zero at the end of the beam.

 

Clearly, one could extend this procedure to account for tractions acting on the ends of the beam.  The details are left as an exercise.  A nice feature of the variational approach that we followed here is that the appropriate boundary conditions follow naturally from the variational principle (indeed, the boundary conditions are called `natural’ boundary conditions).  This turns out to be particularly helpful in setting up approximate theories of plates and shells, where the boundary conditions can be very difficult to determine consistently using any other method.

 

 

 

5.7.5 Energy methods for calculating stiffness

 

Energy methods can also be used to obtain an upper bound to the stiffness of a structure or a component.

 

Begin by reviewing the meaning of stiffness of an elastic solid.  A spring is an example of an elastic solid.  Recall that if you apply a force P to a spring, it deflects by an amount Δ MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8sk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacqqHuoaraaa@343B@ , in proportion to P.  The stiffness k is defined so that

k=P/Δ MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8sk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacaWGRbGaeyypa0Jaamiuaiaac+cacq qHuoaraaa@37B9@

If you apply a load P to any elastic structure (except one which contains two or more contacting surfaces), the point where you apply the load will deflect by a distance that is proportional to the applied load.  For example, for a cantilever beam, the end deflection is

Δ= P L 3 2E a 3 b MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacqqHuoarcqGH9aqpdaWcaaqaaiaadc facaWGmbWaaWbaaSqabeaacaaIZaaaaaGcbaGaaGOmaiaadweacaWG HbWaaWbaaSqabeaacaaIZaaaaOGaamOyaaaaaaa@3C22@

The stiffness of the beam is therefore k=P/Δ=2E a 3 b/ L 3 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacaWGRbGaeyypa0Jaamiuaiaac+cacq qHuoarcqGH9aqpcaaIYaGaamyraiaadggadaahaaWcbeqaaiaaioda aaGccaWGIbGaai4laiaadYeadaahaaWcbeqaaiaaiodaaaaaaa@3F64@

 

To get an upper bound to the stiffness of a structure, one can merely guess its deformed shape, then apply the principle of minimum potential energy.

 

For example, for the beam problem, we might guess that the beam deforms into a circular shape, with unknown radius R.

 

The deflection at the end of the beam is approximately

Δ=R R 2 L 2 L 2 2R MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacqqHuoarcqGH9aqpcaWGsbGaeyOeI0 YaaOaaaeaacaWGsbWaaWbaaSqabeaacaaIYaaaaOGaeyOeI0Iaamit amaaCaaaleqabaGaaGOmaaaaaeqaaOGaeyisIS7aaSaaaeaacaWGmb WaaWbaaSqabeaacaaIYaaaaaGcbaGaaGOmaiaadkfaaaaaaa@4098@

From the preceding section, we know that the potential energy of a beam is

V(w)= 0 L 1 2 EI { d 2 w( x 1 ) d x 1 2 } 2 d x 1 0 L bq( x 1 )w( x 1 )d x 1 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacaWGwbGaaiikaiaadEhacaGGPaGaey ypa0Zaa8qCaeaaaSqaaiaaicdaaeaacaWGmbaaniabgUIiYdGcdaWc aaqaaiaaigdaaeaacaaIYaaaaiaadweacaWGjbWaaiWaaeaadaWcaa qaaiaadsgadaahaaWcbeqaaiaaikdaaaGccaWG3bGaaiikaiaadIha daWgaaWcbaGaaGymaaqabaGccaGGPaaabaGaamizaiaadIhadaWgaa WcbaGaaGymaaqabaGcdaahaaWcbeqaaiaaikdaaaaaaaGccaGL7bGa ayzFaaWaaWbaaSqabeaacaaIYaaaaOGaaGPaVlaadsgacaWG4bWaaS baaSqaaiaaigdaaeqaaOGaeyOeI0Yaa8qCaeaacaWGIbGaamyCaiaa cIcacaWG4bWaaSbaaSqaaiaaigdaaeqaaOGaaiykaiaadEhacaGGOa GaamiEamaaBaaaleaacaaIXaaabeaakiaacMcacaWGKbGaamiEamaa BaaaleaacaaIXaaabeaaaeaacaaIWaaabaGaamitaaqdcqGHRiI8aa aa@60A1@

Here, q( x 1 )=0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8sk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacaWGXbGaaiikaiaadIhadaWgaaWcba GaaGymaaqabaGccaGGPaGaeyypa0JaaGimaaaa@38D2@ , but we need to account for the potential energy of the load P.  Recall that the potential energy of a constant force is PΔ MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8sk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacqGHsislcaWGqbGaeuiLdqeaaa@35FD@ . Recall also that d 2 w/d x 2 1/R MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacaWGKbWaaWbaaSqabeaacaaIYaaaaO Gaam4Daiaac+cacaWGKbGaamiEamaaCaaaleqabaGaaGOmaaaakiab gIKi7kaaigdacaGGVaGaamOuaaaa@3D1F@ . Thus

V(R)= 0 L 1 2 EI R 2 d x 1 PΔ= 1 2 EI R 2 LP L 2 2R MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacaWGwbGaaiikaiaadkfacaGGPaGaey ypa0Zaa8qCaeaadaWcaaqaaiaaigdaaeaacaaIYaaaamaalaaabaGa amyraiaadMeaaeaacaWGsbWaaWbaaSqabeaacaaIYaaaaaaaaeaaca aIWaaabaGaamitaaqdcqGHRiI8aOGaaGPaVlaadsgacaWG4bWaaSba aSqaaiaaigdaaeqaaOGaeyOeI0Iaamiuaiabfs5aejaaykW7caaMc8 UaaGPaVlabg2da9maalaaabaGaaGymaaqaaiaaikdaaaWaaSaaaeaa caWGfbGaamysaaqaaiaadkfadaahaaWcbeqaaiaaikdaaaaaaOGaam itaiabgkHiTiaadcfadaWcaaqaaiaadYeadaahaaWcbeqaaiaaikda aaaakeaacaaIYaGaamOuaaaaaaa@57DC@

Choose R to minimize the potential energy

V R =0 EI R 3 L+P L 2 2 R 2 =0R= 2EI PL MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaadaWcaaqaaiabgkGi2kaadAfaaeaacq GHciITcaWGsbaaaiabg2da9iaaicdacqGHshI3cqGHsisldaWcaaqa aiaadweacaWGjbaabaGaamOuamaaCaaaleqabaGaaG4maaaaaaGcca WGmbGaey4kaSIaamiuamaalaaabaGaamitamaaCaaaleqabaGaaGOm aaaaaOqaaiaaikdacaWGsbWaaWbaaSqabeaacaaIYaaaaaaakiabg2 da9iaaicdacaaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaM c8UaaGPaVlaaykW7caaMc8UaaGPaVlabgkDiElaadkfacqGH9aqpda WcaaqaaiaaikdacaWGfbGaamysaaqaaiaadcfacaWGmbaaaaaa@61AF@

so that

Δ= L 2 2R = L 3 4EI Pk 4EI L 3 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacqqHuoarcqGH9aqpdaWcaaqaaiaadY eadaahaaWcbeqaaiaaikdaaaaakeaacaaIYaGaamOuaaaacqGH9aqp daWcaaqaaiaadYeadaahaaWcbeqaaiaaiodaaaaakeaacaaI0aGaam yraiaadMeaaaGaamiuaiaaykW7caaMc8UaaGPaVlaaykW7caaMc8Ua aGPaVlaaykW7caaMc8UaeyO0H4Taam4AaiabgsMiJoaalaaabaGaaG inaiaadweacaWGjbaabaGaamitamaaCaaaleqabaGaaG4maaaaaaaa aa@5419@

For comparison, the exact solution is k=3EI/ L 3 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8sk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacaWGRbGaeyypa0JaaG4maiaadweaca WGjbGaai4laiaadYeadaahaaWcbeqaaiaaiodaaaaaaa@398E@