# Homework #4 Solutions

Feel free to use http://www.continuummechanics.org/interactivecalcs.html when applicable.

1. Show that

$\int_S {\bf n} \cdot \nabla ( {\bf x} \cdot {\bf x} ) dS = 6V$

where $$V$$ is a volume bounded by the surface, $$S$$, and $${\bf n}$$ is the outward unit normal vector.

Transform the term to a volume integral using the divergence theorem.

$\begin{eqnarray} \int_S {\bf n} \cdot \nabla ( {\bf x} \cdot {\bf x} ) dS & = & \int_V \nabla^2 ( {\bf x} \cdot {\bf x} ) dV \\ \\ \\ & = & \int_V ( x_i x_i ),_{jj} dV \\ \\ \\ & = & \int_V ( x_i x_i ),_{jj} dV \\ \\ \\ & = & \int_V ( x_{i,j} x_i + x_i x_{i,j} ),_j dV \\ \\ \\ & = & \int_V ( \delta_{ij} x_i + x_i \delta_{ij} ),_j dV \\ \\ \\ & = & \int_V ( 2 x_j ),_j dV \\ \\ \\ & = & \int_V 2 x_{j,j} dV \\ \\ \\ & = & \int_V 2 \delta_{jj} dV \\ \\ \\ & = & \int_V 6 \, dV \\ \\ \\ & = & 6 \, V \end{eqnarray}$

2. Use Hooke's Law to calculate strain tensors given the following stress tensor. Use E = 50 MPa and calculate two separate strain tensors: (i) one for $$\nu$$ = 0.33 (metals), and (ii) the second for $$\nu$$ = 0.50 (incompressibles).

Thoughts on sensitivity of stress/strain tensors to Poisson's Ratio?

${\bf \sigma} = \left[ \matrix{ 20 & 15 & 5 \\ 15 & 30 & 0 \\ 5 & 0 & 10 } \right] \text{MPa}$

Use the relationship

$\epsilon_{ij} = {1 \over E} \left[ (1+\nu) \sigma_{ij} - \delta_{ij} \, \nu \, \sigma_{kk} \right]$

(i) $$\nu$$ = 0.33

$\begin{eqnarray} \left[ \matrix { \epsilon_{11} & \epsilon_{12} & \epsilon_{13} \\ \epsilon_{21} & \epsilon_{22} & \epsilon_{23} \\ \epsilon_{31} & \epsilon_{32} & \epsilon_{33} } \right] & = & {1 \over \text{50 MPa}} \left\{ (1 + 0.33) \left[ \matrix { 20 & 15 & 5 \\ 15 & 30 & 0 \\ 5 & 0 & 10 } \right] - 0.33 \, \left[ \matrix { 60 & 0 & 0 \\ 0 & 60 & 0 \\ 0 & 0 & 60} \right] \right\} \\ \\ \\ \\ \\ \\ \\ & = & {1 \over \text{50 MPa}} \left[ \matrix { 6.67 & \;\; 20 & \;\;6.67 \\ 20 & \;\; 20 & \;\;0 \\ 6.67 & \;\; 0 & -6.67 } \right] \\ \\ \\ \\ \\ \\ \\ & = & \left[ \matrix { 0.133 & \;\; 0.40 & \;\;\;0.133 \\ 0.400 & \;\; 0.40 & \;\;\;0.000 \\ 0.133 & \;\; 0.00 & -0.133 } \right] \end{eqnarray}$

(ii) $$\nu$$ = 0.50

$\begin{eqnarray} \left[ \matrix { \epsilon_{11} & \epsilon_{12} & \epsilon_{13} \\ \epsilon_{21} & \epsilon_{22} & \epsilon_{23} \\ \epsilon_{31} & \epsilon_{32} & \epsilon_{33} } \right] & = & {1 \over \text{50 MPa}} \left\{ (1 + 0.5) \left[ \matrix { 20 & 15 & 5 \\ 15 & 30 & 0 \\ 5 & 0 & 10 } \right] - 0.5 \, \left[ \matrix { 60 & 0 & 0 \\ 0 & 60 & 0 \\ 0 & 0 & 60} \right] \right\} \\ \\ \\ \\ \\ \\ \\ & = & {1 \over \text{50 MPa}} \left[ \matrix { 0 & 22.5 & \;\;7.5 \\ 22.5 & 15 & \;\;0 \\ 7.5 & 0 & -15 } \right] \\ \\ \\ \\ \\ \\ \\ & = & \left[ \matrix { 0.00 & \; 0.45 & \;\;\;0.15 \\ 0.45 & \; 0.30 & \;\;\;0.00 \\ 0.15 & \; 0.00 & -0.30 } \right] \end{eqnarray}$

The normal strains (on the diagonal) are more sensitive to $$\nu$$ than the shear terms.

3. First, demonstrate that the following coordinate transformation matrix satisfies $${\bf Q} \cdot {\bf Q}^T = {\bf I}$$. Yet, even though it does satisfy $${\bf Q} \cdot {\bf Q}^T = {\bf I}$$, there is something fundamentally wrong with $${\bf Q}$$. Can you identify what it is?

${\bf Q} = \left[ \matrix { 0 & \;\;\;0 & \;\;\;1 \\ 1 & \;\;\;0 & \;\;\;0 \\ 0 & -1 & \;\;\;0 } \right]$

${\bf Q} \cdot {\bf Q}^T = \left[ \matrix { 0 & \;\;\;0 & \;\;\;1 \\ 1 & \;\;\;0 & \;\;\;0 \\ 0 & -1 & \;\;\;0 } \right] \left[ \matrix { 0 & \;\;\;1 & \;\;\;0 \\ 0 & \;\;\;0 & -1 \\ 1 & \;\;\;0 & \;\;\;0 } \right] = \left[ \matrix { 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 } \right]$

Nevertheless, the problem with $${\bf Q}$$ is that it represents a left-handed coordinate system!

4. If $${\bf r} = \theta \, \hat{\bf r}$$ and $$\theta = \omega \, t$$, then determine equations for velocity, $${\bf v}$$, and acceleration, $${\bf a}$$.

$\begin{eqnarray} {\bf r} & = & \omega \, t \, \hat{\bf r} \\ \\ {\bf v} & = & \omega \, \hat{\bf r} + \omega^2 t \, \hat{\boldsymbol{\theta}} \\ \\ {\bf a} & = & 2 \, \omega^2 \, \hat{\boldsymbol{\theta}} - \omega^3 t \, \hat{\bf r} \\ \end{eqnarray}$