## PHY454H1S Continuum mechanics midterm reflection.

Posted by peeterjoot on March 17, 2012

# Motivation.

I didn’t manage my time well enough on the midterm to complete it (and also missed one easy part of the second question). For later review purposes, here is either what I answered, or what I think I should have answered for these questions.

# Problem 1.

## -waves, -waves, and Love-waves.

\begin{itemize}

\item Show that in -waves the divergence of the displacement vector represents a measure of the relative change in the volume of the body.

**Answer.**

The -wave equation was a result of operating on the displacement equation with the divergence operator

we obtain

We have a wave equation where the “waving” quantity is . Explicitly

Recall that, in a coordinate basis for which the strain is diagonal we have

Expanding in Taylor series to we have for (no sum)

so the displaced volume is

Since

We have

or

The relative change in volume can therefore be expressed as the divergence of , the displacement vector, and it is this relative volume change that is “waving” in the -wave equation as illustrated in the following (\ref{fig:continuumMidtermReflection:continuumMidtermReflectionFig1}) sample 1D compression wave

\begin{figure}[htp]

\centering

\includegraphics[totalheight=0.2\textheight]{continuumMidtermReflectionFig1}

\caption{A 1D compression wave.}

\end{figure}

\item Between a -wave and an -wave which one is longitudinal and which one is transverse?

**Answer.**

-waves are longitudinal.

-waves are transverse.

\item Whose speed is higher?

**Answer.**

From the formula sheet we have

so -waves travel faster than -waves.

\item Is Love wave a body wave or a surface wave?

**Answer.**

Love waves are surface waves, traveling in a medium that can slide on top of another surface. These are characterized by vorticity rotating backwards compared to the direction of propagation as shown in figure (\ref{fig:continuumMidtermReflection:continuumMidtermReflectionFig2})

\begin{figure}[htp]

\centering

\def\svgwidth{0.6\columnwidth}

\caption{Love wave illustrated.}

\end{figure}

\end{itemize}

## (b) constitutive relation, Newtonian fluids, and no-slip conditions.

\begin{itemize}

\item In continuum mechanics what do you mean by \textit{constitutive relation}?

**Answer.** The constitutive relation is the stress-strain relation, generally

for isotropic solids we model this as

and for Newtonian fluids

\item What is the definition of a Non-Newtonian fluid?

**Answer.**

A non-Newtonian fluid would be one with a more general constitutive relationship.

**Grading note.** I lost a mark here. I think the answer that was being looked for (as in [1]) was that a Newtonian fluid is one with a linear stress strain relationship, and a non-Newtonian fluid would be one with a non-linear relationship. According to [2] an example of a non-Newtonian material that we are all familiar with is Silly Putty. This linearity is also how a Newtonian fluid was defined in the notes, but I didn’t remember that (this isn’t really something we use since we assume all fluids and materials are Newtonian in any calculations that we do).

\item What do you mean by \emph{no-slip} boundary condition at a fluid-fluid interface?

**Exam time management note.** Somehow in my misguided attempt to be complete, I missed this question amongst the rest of my verbosity).

**Answer.**

The no slip boundary condition is just one of velocity matching. At a non-moving boundary, the no-slip condition means that we’ll require the fluid to also have no velocity (ie. at that interface the fluid isn’t slipping over the surface). Between two fluids, this is a requirement that the velocities of both fluids match at that point (and all the rest of the points along the region of the interaction.)

\item Write down the continuity equation for an incompressible fluid.

**Answer.**

An incompressible fluid has

but since we also have

A consequence is that for an incompressible fluid. Let’s recall where this statement comes from. Looking at mass conservation, the rate that mass leaves a volume can be expressed as

(the minus sign here signifying that the mass is leaving the volume through the surface, and that we are using an outwards facing normal on the volume.)

If the surface bounding the volume doesn’t change with time (ie. ) we can write

or

so that in differential form we have

Expanding the divergence by chain rule we have

but this is just

So, for an incompressible fluid (one for which ), we must also have .

\end{itemize}

# Problem 2.

## Statement.

Consider steady simple shearing flow as shown in figure (\ref{fig:continuumMidtermReflection:continuumMidtermReflectionFigQ1}) with imposed constant pressure gradient (), being a positive number, of a single layer fluid with viscosity .

\begin{figure}[htp]

\centering

\includegraphics[totalheight=0.2\textheight]{continuumMidtermReflectionFigQ1}

\caption{Shearing flow with pressure gradient and one moving boundary.}

\end{figure}

The boundary conditions are no-slip at the lower plate (). The top plate is moving with a velocity at and fluid is sticking to it, so , being a positive number. Using the Navier-Stokes equation.

\begin{itemize}

\item Derive the velocity profile of the fluid.

**Answer**

Our equations of motion are

\begin{subequations}

\end{subequations}

Here, we’ve used the steady state condition and are neglecting gravity, and kill off our mass compression term with the incompressibility assumption. In component form, what we have left is

with , we must have

which leaves us with just

Having dropped the partials we really just want to integrate our very simple ODE a couple times

Integrate once

and once more to find the velocity

Let’s incorporate an additional constant into

so that we have

(I didn’t do use this way on the exam, nor did I include the factor of in the first integration constant, but both of these should simplify the algebra since we’ll be evaluating the boundary value conditions at .)

Applying the velocity matching conditions we have for the lower and upper plates respectively

Adding these we find

and subtracting find

Our velocity is

or rearranged a bit

\item Draw the velocity profile with the direction of the flow of the fluid when , .

**Answer**

With our velocity has a simple parabolic profile with a max of at

This is plotted in figure (\ref{fig:continuumMidtermReflection:continuumMidtermReflectionFig3})

\begin{figure}[htp]

\centering

\includegraphics[totalheight=0.2\textheight]{continuumMidtermReflectionFig3}

\caption{Parabolic velocity profile.}

\end{figure}

\item Draw the velocity profile with the direction of the flow of the fluid when , .

**Answer**

With , we have a plain old shear flow

This is linear with minimum velocity at , and a maximum of at . This is plotted in figure (\ref{fig:continuumMidtermReflection:continuumMidtermReflectionFig4})

\begin{figure}[htp]

\centering

\includegraphics[totalheight=0.2\textheight]{continuumMidtermReflectionFig4}

\caption{Shear flow.}

\end{figure}

\item Using linear superposition draw the velocity profile of the fluid with the direction of flow qualitatively when , . (i) low , (ii) large .

**Exam time management note.** Somehow I missed this question when I wrote the exam … I figured this out right at the end when I’d run out of time by being too verbose elsewhere. I’m really not very good at writing exams in tight time constraints anymore.

**Answer**

For low we’ll let the parabolic dominate, and can graphically add these two as in figure (\ref{fig:continuumMidtermReflection:continuumMidtermReflectionFig5})

\begin{figure}[htp]

\centering

\includegraphics[totalheight=0.2\textheight]{continuumMidtermReflectionFig5}

\caption{Superposition of shear and parabolic flow (low )}

\end{figure}

For high , we’ll let the shear flow dominate, and have plotted this in figure (\ref{fig:continuumMidtermReflection:continuumMidtermReflectionFig6})

\begin{figure}[htp]

\centering

\includegraphics[totalheight=0.2\textheight]{continuumMidtermReflectionFig6}

\caption{Superposition of shear and parabolic flow (high )}

\end{figure}

\item Calculate the maximum speed when , .

**Answer**

Since our acceleration is

our extreme values occur at

At this point, our velocity is

or just

\item Calculate the flux (the volume flow rate) when , .

**Answer**

An element of our volume flux is

Looking at the volume flux through the width is then

\item Calculate the mean speed when , .

**Answer**

**Exam time management note.** I squandered too much time on other stuff and didn’t get to this part of the problem (which was unfortunately worth a lot). This is how I think it should have been answered.

We’ve done most of the work above, and just have to divide the flux by . That is

\item Calculate the tangential force (per unit width) on the strip of the wall when , .

**Answer**

Our traction vector is

So the directed component of the traction vector is just

We’ve calculated that derivative above in 3.41, so we have

so at we have

To see the contribution of this force on the lower wall over an interval of length we integrate, but this amounts to just multiplying by the length of the segment of the wall

\end{itemize}

# References

[1] Wikipedia. Newtonian fluid — wikipedia, the free encyclopedia [online]. 2011. [Online; accessed 17-March-2012]. http://en.wikipedia.org/w/index.php?title=Newtonian_fluid&oldid=460346447.

[2] Wikipedia. Non-newtonian fluid — wikipedia, the free encyclopedia [online]. 2012. [Online; accessed 17-March-2012]. http://en.wikipedia.org/w/index.php?title=Non-Newtonian_fluid&oldid=479813690.

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