Micromechanics of composites/Proof 14

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Question

Let 𝑷 be the first Piola-Kirchhoff stress and let 𝑭˙ be the time rate of the deformation gradient in a body whose reference configuration is Ω0 with boundary Ω0. Let 𝐍 be the normal to the boundary. Let V0 be the volume of the body. Let 𝐗 represent the position of points in the reference configuration. Let 𝐱˙ be the material time derivative of 𝐱. Let 𝑨 represent the unweighted volume average of a quantity 𝑨. Show that

𝑭˙𝑷𝑭˙𝑷=1V0Ω0[𝐱˙𝑭˙𝐗]{[𝑷𝑷]T𝐍}dA=1V0Ω0[𝐱˙𝑭˙𝐗](𝑷T𝐍)dA=1V0Ω0𝐱˙{[𝑷𝑷]T𝐍}dA.

Proof

Recall the identity

𝑨𝑩𝑨𝑩=[𝑨𝑨][𝑩𝑩].

Therefore,

𝑭˙𝑷𝑭˙𝑷=[𝑭˙𝑭˙][𝑷𝑷]=1V0Ω0[𝑭˙𝑭˙][𝑷𝑷]dV=1V0Ω0𝑭˙𝑷dV1V0Ω0𝑭˙𝑷dV1V0Ω0𝑭˙𝑷dV+1V0Ω0𝑭˙𝑷dV=1V0Ω0𝑭˙𝑷dV(1V0Ω0𝑭˙dV)𝑷𝑭˙(1V0Ω0𝑷dV)+𝑭˙(1V0Ω01dV)𝑷.

We want express the volume integrals above in terms of surface integrals. To do that, recall that

Ω0𝑭˙𝑷dV=Ω0𝐱˙(𝑷T𝐍)dAΩ0𝑭˙dV=Ω00𝐱˙dV=Ω0𝐱˙𝐍dAΩ0𝑷dV=Ω0𝐗(𝑷T𝐍)dAΩ01dV=Ω00𝐗dV=Ω0𝐗𝐍dA.

Therefore,

1V0Ω0𝑭˙𝑷dV=1V0Ω0𝐱˙(𝑷T𝐍)dA(1V0Ω0𝑭˙dV)𝑷=1V0Ω0(𝐱˙𝐍)𝑷dA=1V0Ω0𝐱˙[𝑷𝑻𝐍]dA𝑭˙(1V0Ω0𝑷dV)=1V0Ω0𝑭˙[𝐗(𝑷T𝐍)]dA=1V0Ω0[𝑭˙𝐗](𝑷T𝐍)dA𝑭˙(1V0Ω01dV)𝑷=1V0Ω0𝑭˙(𝐗𝐍)𝑷dA=1V0Ω0[𝑭˙𝐗][𝑷T𝐍]dA.

Collecting the terms, we have

𝑭˙𝑷𝑭˙𝑷=1V0Ω0{𝐱˙(𝑷T𝐍)𝐱˙[𝑷𝑻𝐍][𝑭˙𝐗](𝑷T𝐍)+[𝑭˙𝐗][𝑷T𝐍]}dA=1V0Ω0{𝐱˙[𝑷T𝐍𝑷𝑻𝐍][𝑭˙𝐗][𝑷T𝐍𝑷T𝐍]}dA=1V0Ω0[𝐱˙𝑭˙𝐗][𝑷T𝐍𝑷𝑻𝐍]dA.

Therefore,

𝑭˙𝑷𝑭˙𝑷=1V0Ω0[𝐱˙𝑭˙𝐗]{[𝑷𝑷]T𝐍}dA.

From the above, clearly

(1V0Ω0𝑭˙dV)𝑷=𝑭˙(1V0Ω0𝑷dV)=𝑭˙(1V0Ω01dV)𝑷.

Therefore,

1V0Ω0𝐱˙[𝑷𝑻𝐍]dA=1V0Ω0[𝑭˙𝐗](𝑷T𝐍)dA=1V0Ω0[𝑭˙𝐗][𝑷T𝐍]dA.

Thus we can alternatively write the expression for the difference as

𝑭˙𝑷𝑭˙𝑷=1V0Ω0{𝐱˙(𝑷T𝐍)[𝑭˙𝐗](𝑷T𝐍)[𝐱˙[𝑷T𝐍][𝑭˙𝐗][𝑷T𝐍]]}dA=1V0Ω0[𝐱˙𝑭˙𝐗](𝑷T𝐍)dA

or,

𝑭˙𝑷𝑭˙𝑷=1V0Ω0{𝐱˙(𝑷T𝐍)𝐱˙[𝑷T𝐍][𝑭˙𝐗](𝑷T𝐍)[𝑭˙𝐗][𝑷T𝐍]]}dA=1V0Ω0𝐱˙[𝑷T𝐍𝑷T𝐍]dA.

Hence,

𝑭˙𝑷𝑭˙𝑷=1V0Ω0[𝐱˙𝑭˙𝐗]{[𝑷𝑷]T𝐍}dA=1V0Ω0[𝐱˙𝑭˙𝐗](𝑷T𝐍)dA=1V0Ω0𝐱˙{[𝑷𝑷]T𝐍}dA.


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