美国麻省理工大学：《结构分析与设计技术》教学资源（讲义）notes10

16.21 Techniques of Structural Analysis and esig Spring 2003 Unit #10- Principle of minimum potential energy and Castigliano's First Theorem Principle of minimum potential energy The principle of virtual displacements applies regardless of the constitutive law. Restrict attention to elastic materials(possibly nonlinear). Start from the Pvd ai dv=ti S+/fii dv, Vi/i=0 on S. Replacing the expression for the stresses for elastic materials: aUo and assuming that the virtual displacement field is a variation of the equili brated displacement field i=du, Ei;= deir i Su ds+/fiuid

¯ ¯ � � � � � � 16.21 Techniques of Structural Analysis and Design Spring 2003 Unit #10 - Principle of minimum potential energy and Castigliano’s First Theorem Principle of minimum potential energy The principle of virtual displacements applies regardless of the constitutive law. Restrict attention to elastic materials (possibly nonlinear). Start from the PVD: σij �ijdV = tiu¯idS + fiu¯idV, ∀u/¯ u¯ = 0 on Su (1) V S V Replacing the expression for the stresses for elastic materials: ∂U0 σij = ∂�ij and assuming that the virtual displacement field is a variation of the equili￾brated displacement field u¯ = δu, �ij = δ�ij . ∂U0 δ�ij dV = tiδuidS + fiδuidV V ∂�ij V � �� � S 1

The expression over the brace is the variation of the strain energy density 6 aUo Using the properties of calculus of variations8/0=50 SUodv=8/ Uodv=8u=8(/ tiu;ds+/fu;dv)=8(-v where v is the potential of the external loads. Therefore 6=6(U+V which is known as the Principle of minimum potential energy(PMPE). In fact this expression only says that II is stationary with respect to variations in the displacement field when the body is in equilibrium We can prove that it is indeed a minimum in the case of a linear elastic material: Uo=2Ci]kIEN. We want to show I(v)≥∏(u),v I(v)=I(u)分v=t Consider i=u+ du t4(u2+6u)ds-/t(2+6)dV I()+2/cnn6 The second, fourth and fifth term disappear after invoking the PVD and we are left with II(u+ Su)=II()+/Ciju dei SEudV

� � � � The expression over the brace is the variation of the strain energy density δU0: ∂U0 δU0 = δ�ij ∂�ij Using the properties of calculus of variations δ () = δ(): � � �� � � δU0dV = δ U0dV = δU = δ tiuidS + fiuidV = δ(−V ) S V where V is the potential of the external loads. Therefore: δΠ = δ(U + V ) = 0 which is known as the Principle of minimum potential energy (PMPE). In fact this expression only says that Π is stationary with respect to variations in the displacement field when the body is in equilibrium. We can prove that it is indeed a minimum in the case of a linear elastic mat 1 erial: U0 = 2Cijkl�kl. We want to show: Π(v) ≥ Π(u), ∀v Π(v) = Π(u) ⇔ v = u Consider u¯ = u + δu: � �1 � Π(u + δu) = Cijkl(�ij + δ�ij )(�kl + δ�kl) dV V 2 � � − ti(ui + δui)dS − ti(ui + δui)dV S � V � 1 1 =Π(u)+ �2 V �2 C � ijkl�ijδ�kldV + V 2 Cijklδ�ijδ�kldV − tiδuidS − fiδuidV S V The second, fourth and fifth term disappear after invoking the PVD and we are left with: 1 Π(u + δu) = Π(u) + Cijklδ�ijδ�kldV V 2 2

The integral is always >0, since Cijkl is positive definite. Therefore I(+6n)=I(u)+a,a≥0,a=0分6u=0 and I()≥II(),vt [I(o)=Il(u) as sought Castigliano's First theorem M Given a body in equilibrium under the action of N concentrated forces FI. The potential energy of the external forces is given by Fur where the ur are the values of the displacement field at the point of applica- tion of the forces F. Imagine that somehow we can express the strain energy as a function of the ul, i.e U=U(a 3

� The integral is always ≥ 0, since Cijkl is positive definite. Therefore: Π(u + δu) = Π(u) + a, a ≥ 0, a = 0 ⇔ δu = 0 and Π(v) ≥ Π(u), ∀v Π(v) = Π(u) ⇔ v = u as sought. Castigliano’s First theorem Given a body in equilibrium under the action of N concentrated forces FI . The potential energy of the external forces is given by: V = − N I=1 FIuI where the uI are the values of the displacement field at the point of applica￾tion of the forces FI . Imagine that somehow we can express the strain energy as a function of the uI , i.e.: U = U(u1, u2, . . . , uN ) = U(uI ) 3

I=I(un)=U(u)+V=U(un)-> I=1 Invoking the pmpe dII=0 a8u11 ∑F Frou OU F)6 V SuI +Fr du Theorem: If the strain energy can be expressed in terms of n displacements corresponding to N applied forces, the first derivative of the strain energy with respect to displacement ur is the applied force Example: EA

� � ���� � � � Then: Π = Π(uI ) = U(uI ) + V = U(uI ) − N I=1 FIuI Invoking the PMPE: δΠ = 0 = N I=1 ∂U δuI − ∂uI ∂uI FI ∂uJ δuJ N I=1 ∂U δuI − ∂uI = FI δIJ δuJ N I=1 ∂U δuI − ∂uI = FI δuI � ∂U − FI ∂uI = δuI ∂U ∂uI ∀ δuI ⇔ FI = Theorem: If the strain energy can be expressed in terms of N displacements corresponding to N applied forces, the first derivative of the strain energy with respect to displacement uI is the applied force. Example: 4

(L+)2+2 ∈rI (L+u)2+(L-)2 L U==AEL L)+AEV2I 2("乙)} Note that we have written U=U(u, v). According to the theorem 0 aU See solution in accompanying mathematica file. 5

� � (L + u)2 + v2 u �I = − 1 ∼ L2 L (L + u)2 + (L − v)2 1 u − v �I I = − 1 ∼ 2L2 2 L 1� � u �2 √ �1�u − v��2� U = AEL + AE 2L 2 L 2 L Note that we have written U = U(u, v). According to the theorem: ∂U 0 = ∂u ∂U F = ∂v See solution in accompanying mathematica file. 5