Homework #12 Solutions
Reminder - you may need these webpages:
http://www.continuummechanics.org/cm/techforms/index.html
to do this homework.
-
Start with the following equation from the
Navier-Stokes page
\[
\rho \, \left( {\partial {\bf v} \over \partial t} +
{\bf v} \cdot \nabla {\bf v} \right) =
-\nabla P + 2 \mu \nabla \cdot {\bf D}' + \rho {\bf f}
\]
and use tensor notation to show how to get to the equation below,
which occurs farther down the page.
\[
\rho \, \left( v_{i,t} + v_k v_{i,k} \right) =
-P,_x + 2 \, \mu \, D'_{ij,j} + \rho f_i
\]
Substitute \(D_{ij,j} - {1 \over 3} \delta_{ij} D_{kk,j}\)
for \(D'_{ij,j}\).
\[
\rho \, \left( v_{i,t} + v_k v_{i,k} \right) =
-P,_x + 2 \, \mu \, \left( D_{ij,j} - {1 \over 3} \delta_{ij} D_{kk,j} \right)
+ \rho f_i
\]
Expand out and swap out \(\delta_{ij}\).
\[
\rho \, \left( v_{i,t} + v_k v_{i,k} \right) =
-P,_x + 2 \, \mu \, D_{ij,j} - {2 \over 3} \mu \, D_{kk,i}
+ \rho f_i
\]
Replace \(D_{ij}\) with \({1 \over 2} (v_{i,j} + v_{j,i})\), and
replace \(D_{kk}\) with \(v_{k,k}\).
\[
\rho \, \left( v_{i,t} + v_k v_{i,k} \right) =
-P,_x + \mu \, (v_{i,jj} + v_{j,ij}) - {2 \over 3} \mu \, v_{k,ki}
+ \rho f_i
\]
But \(v_{j,ij}\) and \(v_{k,ki}\) are the same thing, so combine to get
\[
\rho \, \left( v_{i,t} + v_k v_{i,k} \right) =
-P,_x + \mu \, v_{i,jj} + {1 \over 3} \mu \, v_{j,ji}
+ \rho f_i
\]
But \(v_{j,ji}\) is the same thing as \((v_{j,j}),_i\) and for incompressible
materials, \(v_{j,j}\) is zero. So this term drops out to give
\[
\rho \, \left( v_{i,t} + v_k v_{i,k} \right) =
-P,_x + \mu \, v_{i,jj}
+ \rho f_i
\]
which can be written in matrix notation as
\[
\rho \, \left( {\partial {\bf v} \over \partial t} +
{\bf v} \cdot \nabla {\bf v} \right) =
-\nabla P + \mu \nabla^2 {\bf v}
+ \rho \, {\bf f}
\]
-
The (made up) test data below is for tension tests of a
rubber sample at two temperatures. Propose a Helmholtz
function in terms of strain and temperature and show that
it reproduces the measured data. (Note - The Helmholtz
function must be a function of temperature in Kelvin,
not Celsius.)
| 27°C | 77°C |
Strain | Stress (MPa) | Stress (MPa) |
0 | 0 | 0 |
0.05 | 0.3 | 0.2 |
0.10 | 0.7 | 0.4 |
0.15 | 1.0 | 0.7 |
0.20 | 1.5 | 0.8 |
0.25 | 1.8 | 1.1 |
0.30 | 2.5 | 1.6 |
0.35 | 3.0 | 1.9 |
0.40 | 3.5 | 2.4 |
0.45 | 4.2 | 2.9 |
0.50 | 4.9 | 3.3 |
0.55 | 5.7 | 3.8 |
0.60 | 6.7 | 4.3 |
0.65 | 7.5 | 4.9 |
0.70 | 8.5 | 5.5 |
0.75 | 9.4 | 6.2 |
0.80 | 10.4 | 6.9 |
0.85 | 11.5 | 7.6 |
0.90 | 12.7 | 8.4 |
Propose the following function for the Helmholtz free energy.
\[
\Psi = { \text{20E6} \, \epsilon^2 + \text{30E6} \, \epsilon^3 \over T^{2.8} }
\]
Keep in mind that temperature must be in Kelvin, use 300K and 350K.
Take the derivative to get
\[
\sigma = {\partial \Psi \over \partial \epsilon} =
{ \text{40E6} \, \epsilon + \text{90E6} \, \epsilon^2 \over T^{2.8} }
\]
Plot this on the graph.