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=∂Π∂q1δq1+∂Π∂q2δq2+...δΠ=n∑i=1∂Π∂qiδqi=0
Since δqi are virtual displacements, or arbitrary kinematically admissible displacements: ∂Π∂qiδqi=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: ∂Π∂qiδqi=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=(uC−uB)2l=uC2l ε2=(uD−uC)l
- Then strain energy: Ui=12σi⋅εi⋅AiLi=12(Ei⋅εi)⋅εi⋅AiLi=12(ε2i⋅Ei⋅AiLi) U=U1+U2=12(ε21⋅E1⋅A1L1)+12(ε22⋅E2⋅A2L2) =12(uC2l)2EA⋅2l+12(uD−uCl)24EA⋅l
- Therefore: U=(uC)2⋅EA4l+(uD−uC)2⋅2EAl
- Also, the external work: V=−PuD
- The energy: Π=U+V
- Thus: ∂Π∂uC=0 2uC⋅EA4l+2(uD−uC)⋅(−1)⋅2EAl=0 ∂Π∂uD=0 2(uD−uC)⋅2EAl−P=0
- By adding both equations: 2uC⋅EA4l−P=0 uC=P⋅2lEA=P⋅2L3EA 2(uD−uC)⋅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