Homework #4 Solutions
Feel free to use
http://www.continuummechanics.org/interactivecalcs.html when applicable.
-
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.
\[
\begin{eqnarray}
\int_S {\bf n} \cdot \nabla ( {\bf x} \cdot {\bf x} ) dS
& = & \int_V \nabla^2 ( {\bf x} \cdot {\bf x} ) dV
\\
\\
\\
& = & \int_V ( x_i x_i ),_{jj} dV
\\
\\
\\
& = & \int_V ( x_i x_i ),_{jj} dV
\\
\\
\\
& = & \int_V ( x_{i,j} x_i + x_i x_{i,j} ),_j dV
\\
\\
\\
& = & \int_V ( \delta_{ij} x_i + x_i \delta_{ij} ),_j dV
\\
\\
\\
& = & \int_V ( 2 x_j ),_j dV
\\
\\
\\
& = & \int_V 2 x_{j,j} dV
\\
\\
\\
& = & \int_V 2 \delta_{jj} dV
\\
\\
\\
& = & \int_V 6 \, dV
\\
\\
\\
& = & 6 \, V
\end{eqnarray}
\]
-
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.
-
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!
-
If \({\bf r} = \theta \, \hat{\bf r}\) and \(\theta = \omega \, t\), then determine equations for velocity, \({\bf v}\), and
acceleration, \({\bf a}\).
\[
\begin{eqnarray}
{\bf r} & = & \omega \, t \, \hat{\bf r} \\
\\
{\bf v} & = & \omega \, \hat{\bf r} + \omega^2 t \, \hat{\boldsymbol{\theta}} \\
\\
{\bf a} & = & 2 \, \omega^2 \, \hat{\boldsymbol{\theta}} - \omega^3 t \, \hat{\bf r} \\
\end{eqnarray}
\]