Micromechanics of composites/Proof 3
Surface-volume integral relation 1
editLet be a body and let be its surface. Let be the normal to the surface. Let be a vector field on and let be a second-order tensor field on . Show that
![{\displaystyle \int _{\partial {\Omega }}\mathbf {v} \otimes ({\boldsymbol {S}}^{T}\cdot \mathbf {n} )~{\text{dA}}=\int _{\Omega }[{\boldsymbol {\nabla }}\mathbf {v} \cdot {\boldsymbol {S}}+\mathbf {v} \otimes ({\boldsymbol {\nabla }}\bullet {\boldsymbol {S}}^{T})]~{\text{dV}}~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c526aaf5e74863755e2cc1709c003642c21894e7)
Proof:
Recall the relation
![{\displaystyle {\boldsymbol {\nabla }}\bullet [(\mathbf {v} \cdot \mathbf {a} )({\boldsymbol {S}}\cdot \mathbf {b} )]=\mathbf {a} \cdot [\{{\boldsymbol {\nabla }}\mathbf {v} \cdot {\boldsymbol {S}}+\mathbf {v} \otimes ({\boldsymbol {\nabla }}\bullet {\boldsymbol {S}}^{T})\}\cdot \mathbf {b} ]~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7d08ec551dee6717ea19ddc9673fa3e2b12753f1)
Integrating over the volume, we have
![{\displaystyle \int _{\Omega }{\boldsymbol {\nabla }}\bullet [(\mathbf {v} \cdot \mathbf {a} )({\boldsymbol {S}}\cdot \mathbf {b} )]~{\text{dV}}=\int _{\Omega }\mathbf {a} \cdot [\{{\boldsymbol {\nabla }}\mathbf {v} \cdot {\boldsymbol {S}}+\mathbf {v} \otimes ({\boldsymbol {\nabla }}\bullet {\boldsymbol {S}}^{T})\}\cdot \mathbf {b} ~{\text{dV}}~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b15533ad2beee747f71004042c1acac972038f31)
Since and are constant, we have
![{\displaystyle \int _{\Omega }{\boldsymbol {\nabla }}\bullet [(\mathbf {v} \cdot \mathbf {a} )({\boldsymbol {S}}\cdot \mathbf {b} )]~{\text{dV}}=\mathbf {a} \cdot \left[\left\{\int _{\Omega }[{\boldsymbol {\nabla }}\mathbf {v} \cdot {\boldsymbol {S}}+\mathbf {v} \otimes ({\boldsymbol {\nabla }}\bullet {\boldsymbol {S}}^{T})]~{\text{dV}}\right\}\cdot \mathbf {b} \right]~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c058f794907f341c06123d02ff13e172b34b1506)
From the divergence theorem,
we get
![{\displaystyle \int _{\Omega }{\boldsymbol {\nabla }}\bullet [(\mathbf {v} \cdot \mathbf {a} )({\boldsymbol {S}}\cdot \mathbf {b} )]~{\text{dV}}=\int _{\partial {\Omega }}[(\mathbf {v} \cdot \mathbf {a} )({\boldsymbol {S}}\cdot \mathbf {b} )]\cdot \mathbf {n} ~{\text{dA}}~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/39b9282a15699913a1ef5ebbb995e1ffb395117c)
Using the relation
![{\displaystyle [(\mathbf {v} \bullet \mathbf {a} )({\boldsymbol {S}}\bullet \mathbf {b} )]\cdot \mathbf {n} =\mathbf {a} \cdot [\{\mathbf {v} \otimes ({\boldsymbol {S}}^{T}\bullet \mathbf {n} )\}\cdot \mathbf {b} ]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cf700ebf26bd14bfa1b0fbbc27e201aafa639dc6)
we get
![{\displaystyle \int _{\Omega }{\boldsymbol {\nabla }}\bullet [(\mathbf {v} \cdot \mathbf {a} )({\boldsymbol {S}}\cdot \mathbf {b} )]~{\text{dV}}=\int _{\partial {\Omega }}\mathbf {a} \cdot [\{\mathbf {v} \otimes ({\boldsymbol {S}}^{T}\bullet \mathbf {n} )\}\cdot \mathbf {b} ]~{\text{dA}}~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/02fe364731f404d7107c7af9aaa04b17f48a9308)
Since and are constant, we have
![{\displaystyle \int _{\Omega }{\boldsymbol {\nabla }}\bullet [(\mathbf {v} \cdot \mathbf {a} )({\boldsymbol {S}}\cdot \mathbf {b} )]~{\text{dV}}=\mathbf {a} \cdot \left[\left\{\int _{\partial {\Omega }}\mathbf {v} \otimes ({\boldsymbol {S}}^{T}\bullet \mathbf {n} )~{\text{dA}}\right\}\cdot \mathbf {b} \right]~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/782f0cb1baf5ac445c7fdf605efdc4a1c2e0873a)
Therefore,
![{\displaystyle \mathbf {a} \cdot \left[\left\{\int _{\partial {\Omega }}\mathbf {v} \otimes ({\boldsymbol {S}}^{T}\bullet \mathbf {n} )~{\text{dA}}\right\}\cdot \mathbf {b} \right]=\mathbf {a} \cdot \left[\left\{\int _{\Omega }[{\boldsymbol {\nabla }}\mathbf {v} \cdot {\boldsymbol {S}}+\mathbf {v} \otimes ({\boldsymbol {\nabla }}\bullet {\boldsymbol {S}}^{T})]~{\text{dV}}\right\}\cdot \mathbf {b} \right]~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d1f4c6dcdf170ecc073200cf231db1899bf72b32)
Since and are arbitrary, we have
![{\displaystyle {\int _{\partial {\Omega }}\mathbf {v} \otimes ({\boldsymbol {S}}^{T}\bullet \mathbf {n} )~{\text{dA}}=\int _{\Omega }[{\boldsymbol {\nabla }}\mathbf {v} \cdot {\boldsymbol {S}}+\mathbf {v} \otimes ({\boldsymbol {\nabla }}\bullet {\boldsymbol {S}}^{T})]~{\text{dV}}}\qquad \qquad \qquad \square }](https://wikimedia.org/api/rest_v1/media/math/render/svg/fed13fb1b9f06c03a6df733c44eb0f43f5969921)