Micromechanics of composites/Average stress in a RVE

Average Stress in a RVE

Let the average stress in the RVE be defined as

$⟨\mathit{\sigma }⟩:=\frac{1}{V}\underset{\mathrm{\Omega }}{\int }\mathit{\sigma }(𝐱)\text{dV}$

where $V$ is the volume of $\mathrm{\Omega }$.

We would like to find out the relation between the average stress in a RVE and the applied tractions on the boundary of the RVE. To do that, recall the relation (see Appendix)

$\underset{\partial \mathrm{\Omega }}{\int }𝐯\otimes ({𝑺}^T\bullet 𝐧)\text{dA}=\underset{\mathrm{\Omega }}{\int }[\mathrm{\nabla }𝐯\cdot 𝑺+𝐯\otimes (\mathrm{\nabla }\bullet {𝑺}^T)]\text{dV}.$

If we choose $𝐯$ such that $\mathrm{\nabla }𝐯=1$, we have

$\underset{\partial \mathrm{\Omega }}{\int }𝐯\otimes ({𝑺}^T\bullet 𝐧)\text{dA}=\underset{\mathrm{\Omega }}{\int }[1\cdot 𝑺+𝐯\otimes (\mathrm{\nabla }\bullet {𝑺}^T)]\text{dV}=\underset{\mathrm{\Omega }}{\int }𝑺\text{dV}+\underset{\mathrm{\Omega }}{\int }𝐯\otimes (\mathrm{\nabla }\bullet {𝑺}^T)\text{dV}.$

Therefore,

$\underset{\mathrm{\Omega }}{\int }𝑺\text{dV}=-\underset{\mathrm{\Omega }}{\int }𝐯\otimes (\mathrm{\nabla }\bullet {𝑺}^T)\text{dV}+\underset{\partial \mathrm{\Omega }}{\int }𝐯\otimes ({𝑺}^T\bullet 𝐧)\text{dA}.$

If we choose $𝑺$ to be the stress tensor $\mathit{\sigma }$, and involve the symmetry of the stress tensor, we get

$\underset{\mathrm{\Omega }}{\int }\mathit{\sigma }\text{dV}=-\underset{\mathrm{\Omega }}{\int }𝐯\otimes (\mathrm{\nabla }\bullet \mathit{\sigma })\text{dV}+\underset{\partial \mathrm{\Omega }}{\int }𝐯\otimes (\mathit{\sigma }\bullet 𝐧)\text{dA}.$

Now, the divergence of the stress is zero (from the conservation of linear momentum). Therefore,

$\underset{\mathrm{\Omega }}{\int }\mathit{\sigma }\text{dV}=\underset{\partial \mathrm{\Omega }}{\int }𝐯\otimes (\mathit{\sigma }\bullet 𝐧)\text{dA}.$

Using the traction boundary condition, we have

$\underset{\mathrm{\Omega }}{\int }\mathit{\sigma }\text{dV}=\underset{\partial \mathrm{\Omega }}{\int }𝐯\otimes \overline{𝐭}\text{dA}.$

Now $\mathrm{\nabla }𝐯=1$ if $𝐯=𝐱$. Therefore, we have

$\underset{\mathrm{\Omega }}{\int }\mathit{\sigma }\text{dV}=\underset{\partial \mathrm{\Omega }}{\int }𝐱\otimes \overline{𝐭}\text{dA}.$

Hence the average stress is given by

$⟨\mathit{\sigma }⟩:=\frac{1}{V}\underset{\partial \mathrm{\Omega }}{\int }𝐱\otimes \overline{𝐭}\text{dA}.$

This implies that the average stress is completely determined by the applied tractions!

Symmetry of the average stress and the effect of rigid body translation

Let us now assume that the applied tractions are self equilibrating. Then the resultant forces and moments due to the applied tractions vanish and we have

$\underset{\partial \mathrm{\Omega }}{\int }\overline{𝐭}\text{dA}=\mathrm{𝟎}\text{and}\underset{\partial \mathrm{\Omega }}{\int }𝐱\times \overline{𝐭}\text{dA}=\mathrm{𝟎}.$

From the moment balance equation above we can show that (see Appendix)

$\underset{\partial \mathrm{\Omega }}{\int }𝐱\otimes \overline{𝐭}\text{dA}=\underset{\partial \mathrm{\Omega }}{\int }\overline{𝐭}\otimes 𝐱\text{dA}.$

Therefore the average stress tensor $⟨\mathit{\sigma }⟩$ is symmetric if the applied tractions are self equilibrated.

Now, if we translate the body by a constant amount ${𝐮}_0$ (rigid body translation), we get

$\overline{⟨\mathit{\sigma }⟩}=\frac{1}{V}\underset{\partial \mathrm{\Omega }}{\int }(𝐱+{𝐮}_0)\otimes \overline{𝐭}\text{dA}=\frac{1}{V}\underset{\partial \mathrm{\Omega }}{\int }[𝐱\otimes \overline{𝐭}+{𝐮}_0\otimes \overline{𝐭}]\text{dA}.$

or

$\overline{⟨\mathit{\sigma }⟩}=\frac{1}{V}[\underset{\partial \mathrm{\Omega }}{\int }𝐱\otimes \overline{𝐭}\text{dA}+{𝐮}_0\otimes \underset{\partial \mathrm{\Omega }}{\int }\overline{𝐭}\text{dA}]=⟨\mathit{\sigma }⟩$

Therefore, the average stress is not affected by a rigid body translation only if the applied tractions are self equilibrated.

We can conclude that the average stress $⟨\mathit{\sigma }⟩$ is an acceptable measure of stress in a RVE if the applied tractions are self equilibrated.

Template:Subpage navbar