11.7 Minimum potential energy

This is related to the principle of virtual work.


11.7.1 Principle of minimum total potential energy

δV+δU=0=δ(V+U)

  • Total potential energy =U+V=Π δΠ=0Π=Π(q1,q2,...,qi,...,qn) where qi are independent DOFs

  • δΠ=Πq1δq1+Πq2δq2+...δΠ=ni=1Πqiδqi=0


  • Since δqi are virtual displacements, or arbitrary kinematically admissible displacements: Πqiδqi=0

  • Of all the kinematically admissible displacements, those corresponding to a stable equilibrium position make Π a minimum. (Each and every qi must minimize the energy.)

  • Therefore: Πqiδqi=0


11.7.2 Deformable body example

Deformable body virtual work example

  • Given: E1=E2=E

4A1=A2=4A

  • Required: uD

11.7.2.1 Solution:

  • The DOFs are the displacements uD , uC and uB(=0)
  • Write strains first in terms of DOFs: ε1=(uCuB)2l=uC2l ε2=(uDuC)l
  • Then strain energy: Ui=12σiεiAiLi=12(Eiεi)εiAiLi=12(ε2iEiAiLi) U=U1+U2=12(ε21E1A1L1)+12(ε22E2A2L2) =12(uC2l)2EA2l+12(uDuCl)24EAl

  • Therefore: U=(uC)2EA4l+(uDuC)22EAl
  • Also, the external work: V=PuD
  • The energy: Π=U+V
  • Thus: ΠuC=0 2uCEA4l+2(uDuC)(1)2EAl=0 ΠuD=0 2(uDuC)2EAlP=0

  • By adding both equations: 2uCEA4lP=0 uC=P2lEA=P2L3EA 2(uDuC)2EAl=P
  • Therefore: uD=uC+Pl4EA uD=(PlEA)(2+14)=9Pl4EA=3PL4EA=uD
  • Therefore: uD=3PL4EA

Due to overlapping material and to provide continuity, the lectures slides for Lecture 1 are included in the materials for Lecture 0