Commit a2f7f266 authored by Olivier's avatar Olivier
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Chapter 5: added slides

parent ef6a1b06
../../1/images/shear_force_plate.png
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\documentclass[17pt]{beamer}
\usepackage{fluidmechslides} % from https://git.framasoft.org/u/olivier/sensible-styles
\renewcommand{\documentnumber}{2}
\renewcommand{\titleofthisdocument}{Effects of shear}
\renewcommand{\numberofthischapter}{5}
\renewcommand{\titleofthisdocument}{\namechapterfive}
\renewcommand{\keywordsofthisdocument}{}
\renewcommand{\lastedityear}{2018}
\renewcommand{\lasteditmonth}{04}
\renewcommand{\lasteditday}{19}
\renewcommand{\lastedityear}{2019}
\renewcommand{\lasteditmonth}{05}
\renewcommand{\lasteditday}{09}
% Syntax for single-image slides:
% (the first argument (number) being the maximum fraction of the
% slide width that the image is allowed to have.)
% \figureframe{1}{filename}{Title}{Attribution}
% Do I want the print version? (no \pause, larger peamble)
% Do I want the print version? (no \pause, larger preamble)
\printversion
\usepackage{pdfpc-commands} % experimental: videos in slides
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{document}
\mainslidesbegin
\begin{frame}{Take-home slide for chapter 2}
\begin{frame}{Take-home slide for chapter 5}
\begin{itemize}\pause
\item Shear \emph{has} a direction\pause
\item Net shear force: divergent of tensor: $\divergent{\vec\tau_{ij}}$ \pause
......@@ -29,18 +32,69 @@
\end{itemize}
\end{frame}
\figureframe{0.8}{clutch}{}{\wcfile{Schematic drawing of a simple clutch.svg}{Figure} \cczero \oc}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\setcounter{section}{1}
\section{Shear forces on walls}
%%%%%%%%%
\subsection{Magnitude of the shear force}
\begin{frame}
Simplest case: uniform shear, flat wall
\begin{IEEEeqnarray}{rCcCl}
F_{\text{shear, direction } i} & = & \tau_{\text{uniform, direction } i} \ S_\text{flat wall}\nonumber\\
\end{IEEEeqnarray}
\end{frame}
\begin{frame}
Is $\tau$ not uniform? We need to integrate:\pause
\begin{IEEEeqnarray}{rCl}
F_{\text{shear, direction } i} & = & \int_S \diff F_{\text{shear, direction } i} \\
& = & \int_S \tau_{\text{direction } i} \diff S\nonumber\\
\end{IEEEeqnarray}\pause
What is the function $\tau$?
\end{frame}
\begin{frame}
Typically, in two dimensions $y$ and $z$,
\begin{mdframed}
\begin{IEEEeqnarray}{rCcCl}
F_\text{{shear, direction } i} & = & \iint \tau_{\text{direction } i (x, y)} \diff x \diff y\nonumber\\
\end{IEEEeqnarray}
\end{mdframed}
\includegraphics[width=0.5\textwidth]{shear_force_plate}
\end{frame}
%%%%
\subsection{Direction and position of the shear force}
\begin{frame}
Rigorously:
\begin{IEEEeqnarray}{rCcCl}
\vec F_\text{shear} & = & \int_S \vec \tau_n \diff S \label{eq_shear_force_vector}
\end{IEEEeqnarray}
$\to$ far too tedious for humans
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\setcounter{section}{1}
\section{Concept of shear}
\section{Shear fields in fluids}
\pictureframe{1}{Claudio_Corti_Silverstone_2013}{}{\wcfile{Claudio Corti Silverstone 2013.jpg}{Photo} \ccbysa by \flickruser{smudge9000}}
\pictureframe{1}{Claudio_Corti_Silverstone_2013b}{}{\wcfile{Claudio Corti Silverstone 2013.jpg}{Photo} \ccbysa by \flickruser{smudge9000}}
\begin{frame}{}
Back in chapter 0:
Back in chapter 1:
\begin{IEEEeqnarray}{rCl}
\tau &\equiv& \frac{F_\parallel}{A} \label{eq_first_def_shear_two}
......@@ -204,6 +258,18 @@
&& + & S_1 \vec \tau_{xx\ 1} - S_4 \vec \tau_{xx\ 4}
\end{IEEEeqnarray}
\end{frame}
\begin{frame}
What is the net \emph{force} due to shear on the particle?
\includegraphics[width=5cm]{particle_shear_tensor_0}
\begin{IEEEeqnarray*}{rCll}
\vec F_{\text{shear}\ x} &=& & S_3 \vec \tau_{zx\ 3} - S_3 \vec \tau_{zx\ 6}\nonumber\\
&& + & S_2 \vec \tau_{yx\ 2} - S_2 \vec \tau_{yx\ 5}\nonumber\\
&& + & S_1 \vec \tau_{xx\ 1} - S_1 \vec \tau_{xx\ 4}
\end{IEEEeqnarray*}
\end{frame}
\begin{frame}
What is the net \emph{force} due to shear on the particle?
......@@ -238,9 +304,9 @@
\includegraphics[width=5cm]{particle_shear_tensor_0}
\begin{IEEEeqnarray*}{rCll}
\vec F_{\text{shear}\ x} &=& &\diff x \diff y (\diff \vec \tau_{zx})\nonumber\\
&& + &\diff x \diff z (\diff \vec \tau_{yx})\nonumber\\
&& + &\diff z \diff y (\diff \vec \tau_{xx})
\vec F_{\text{shear}\ x} &=& &\diff x \diff y \ (\tdiff \vec \tau_{zx})\nonumber\\
&& + &\diff x \diff z \ (\tdiff \vec \tau_{yx})\nonumber\\
&& + &\diff z \diff y \ (\tdiff \vec \tau_{xx})
\end{IEEEeqnarray*}
\end{frame}
......@@ -249,9 +315,9 @@
\includegraphics[width=5cm]{particle_shear_tensor_0}
\begin{IEEEeqnarray*}{rCll}
\vec F_{\text{shear}\ x} &=& &\diff x \diff y \left(\diff z \partialderivative{\vec \tau_{zx}}{z}\right) \nonumber\\
&& + &\diff x \diff z \left(\diff y \partialderivative{\vec \tau_{yx}}{y}\right) \nonumber\\
&& + &\diff z \diff y \left(\diff x \partialderivative{\vec \tau_{xx}}{x}\right)
\vec F_{\text{shear}\ x} &=& &\diff x \diff y \left(\tdiff z \partialderivative{\vec \tau_{zx}}{z}\right) \nonumber\\
&& + &\diff x \diff z \left(\tdiff y \partialderivative{\vec \tau_{yx}}{y}\right) \nonumber\\
&& + &\diff z \diff y \left(\tdiff x \partialderivative{\vec \tau_{xx}}{x}\right)
\end{IEEEeqnarray*}
\end{frame}
......@@ -276,11 +342,11 @@
\begin{frame}
Cool new toy:\\
Mathematical operator \vocab{divergent}
Mathematical operator \vocab{divergent}\pause
\begin{IEEEeqnarray}{rCl}
\divergent{} &\equiv& \partialderivative{}{x} \vec i \cdot \ + \ \partialderivative{}{y} \vec j \cdot \ + \ \partialderivative{}{z} \vec k \cdot \nonumber\\\label{eq_def_divergent}\\\pause
\divergent{\vec A} &\equiv& \partialderivative{}{x} \vec i \cdot \vec A \ + \ \partialderivative{}{y} \vec j \cdot \vec A \ + \ \partialderivative{}{z} \vec k \cdot \vec A \nonumber\\\nonumber\pause\\
\divergent{\vec A} &\equiv& \partialderivative{\vec A_x}{x} \ + \ \partialderivative{\vec A_y}{y} \ + \ \partialderivative{\vec A_z}{z}\nonumber\\
\divergent{\vec A} &\equiv& \partialderivative{A_x}{x} \ + \ \partialderivative{A_y}{y} \ + \ \partialderivative{A_z}{z}\nonumber\\
\end{IEEEeqnarray}
\end{frame}
......@@ -288,11 +354,11 @@
What is the net \emph{force} due to shear on the particle?
\includegraphics[width=5cm]{particle_shear_tensor_0}
{\small The uncool way to write it:}
\begin{IEEEeqnarray}{rCl}
\vec F_{\text{shear}\ x} &=& \diff \vol \left(\partialderivative{\vec \tau_{zx}}{z} + \partialderivative{\vec \tau_{yx}}{y} + \partialderivative{\vec \tau_{xx}}{x}\right)\label{eq_shear_force_x_tmp}\nonumber\\
\end{IEEEeqnarray}
\small\emph{there must be a better way to write this!}
\end{frame}
\begin{frame}
......@@ -322,11 +388,23 @@
|\divergent{\vec \tau_{iz}}| %
\end{array}\right)\nonumber\\\pause
\vec F_{\text{shear}} &=& \diff \vol \ \divergent{\vec \tau_{ij}} \label{eq_shear_force_divergent_shear}
\end{IEEEeqnarray}\pause
\end{IEEEeqnarray}
\end{frame}
\begin{frame}
Net force due to shear, on a volume $\tdiff \vol$:
\begin{mdframed}
\begin{IEEEeqnarray}{rCl}
\frac{1}{\diff \vol} \vec F_\text{net, shear} & = & \divergent{\vec \tau_{ij}}\label{eq_shear_force_divergent_shear}
\end{IEEEeqnarray}
\end{mdframed}\pause
\small net force is volume times \vocab{divergent of shear tensor}.\pause\\
\small $\to$ chapter 4
\small net force per volume is \vocab{divergent of shear tensor}.\pause\\
\small $\to$ chapter 6
\end{frame}
\begin{frame}{Recap 1/2}
......@@ -360,28 +438,15 @@
\pictureframe{1}{Oberaargletscher_from_Oberaar,_2010_07}{}{\wcfile{Oberaargletscher_from_Oberaar,_2010_07.JPG}{Photo} \ccbysa by Simo Räsänen}
%\pictureframe{1}{Clouds_off_the_Aleutian_Islands}{}{\wcfile{File:Clouds off the Aleutian Islands.jpg}{Photo} by Jeff Schmaltz, MODIS, NASA (\pd)}
\begin{comment}
\begin{frame}{And now for a little quizz…}\pause
If you rub your hands on the table,\pause
what is the magnitude of the shear you generate?
\end{frame}
\begin{frame}{If this works, it is thanks to:}
\small
\begin{itemize}
\item OvGU: Claudia Wendt, Anja Hawlitschek, Veit~Köppen, Philipp Berg, Farooq Hussain
\item The server administrators at \textit{Universität Halle}
\item The network administrators (\textsc{urz}) here at \textit{Otto-von-Guericke-Universität Magdeburg}
\item The developers of ARSnova at \textit{Technische Hochschule Mittelhessen}
\end{itemize}
\end{frame}
\end{comment}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Slip and viscosity}
%%%%
\subsection{The no-slip condition}
......@@ -420,22 +485,17 @@
\begin{frame}{}
~
~
\vocab{Viscosity} $\mu$ is the coefficient linking the shear constraint in the flow direction to the spatial rate of change of velocity in a direction perpendicular to the flow\pause
~
\small \vocab{dynamic viscosity} = \vocab{viscosity}
\end{frame}
\begin{frame}{}
Shear stress vs. rate of strain:
\begin{IEEEeqnarray*}{rCcCl}
||\vec \tau_{xy}|| &=& \pause \mu \partialderivative{u_y}{x} \pause &=& \mu \partialderivative{v}{x}
||\vec \tau_{xy}|| &=& \pause \mu \partialderivative{V_y}{x} \pause &=& \mu \partialderivative{v}{x}
\end{IEEEeqnarray*}
\end{frame}
......@@ -445,10 +505,12 @@
\begin{frame}{}
General definition
\begin{IEEEeqnarray}{rCl}
\mu &\equiv& \frac{||\vec \tau_{ij}||}{\left(\partialderivative{u_j}{i}\right)} \\
\mu &\equiv& \frac{||\vec \tau_{ij}||}{\left(\partialderivative{V_j}{i}\right)} \\
\nonumber\\
||\vec \tau_{ij}|| &=& \mu \partialderivative{u_j}{i} \label{eq_shear_velocity_gradient}
\end{IEEEeqnarray}
||\vec \tau_{ij}|| &=& \mu \partialderivative{V_j}{i} \label{eq_shear_velocity_gradient}
\end{IEEEeqnarray}\pause
(\vocab{viscosity} $\equiv$ \vocab{dynamic viscosity})
\end{frame}
......@@ -457,7 +519,9 @@
which is \si{\kilogram\per\metre\per\second}\pause
which is \si{\newton\second\per\metre\squared}
which is \si{\newton\second\per\metre\squared}\pause
\skipinprint{\hspace*{\fill}\includegraphics[width=0.5\textwidth]{friend}}
\end{frame}
\begin{frame}{}
......@@ -468,15 +532,21 @@
\end{IEEEeqnarray*}\pause
\small“my orange juice has a viscosity of one point one centipoise!”
\skipinprint{\hspace*{\fill}\inlineMovie[loop&autostart]{highfive.mp4}{highfive}{height=0.3\textheight}}
\end{frame}
\begin{frame}{}
\begin{frame}{Also}
\vocab{Kinematic viscosity} $\nu$ defined as
\begin{equation}
\nu \equiv \frac{\mu}{\rho}
\end{equation}
\end{equation}\pause
Greek letter \emph{nu} (not $v$ !!!!)\\
(measured in \si{\metre\squared\per\second})
(measured in \si{\metre\squared\per\second})\pause
~
\textit{\small Fluid dynamicists all would like to apologize for this really dumb decision}
\end{frame}
......@@ -485,7 +555,7 @@
\subsection{Newtonian Fluid}
\begin{frame}{}
Fluids for which $\mu$ is independent from $\frac{\diff u}{\diff y}$ are called \vocab{Newtonian fluids}.
Fluids for which $\mu$ is independent from $\frac{\diff v}{\diff x}$ are called \vocab{Newtonian fluids}.
\end{frame}
\begin{frame}{}
......@@ -529,98 +599,74 @@
\figureframe{1}{quizz1}{Which is which?}{}
\skipinprint{\figureframe{1}{quizz2}{}{}}
\begin{frame}{If this works, it is thanks to:}
\small
\begin{itemize}
\item OvGU: Claudia Wendt, Anja Hawlitschek, Veit~Köppen, Philipp Berg, Farooq Hussain
\item The server administrators at \textit{Universität Halle}
\item The network administrators (\textsc{urz}) here at \textit{Otto-von-Guericke-Universität Magdeburg}
\item The developers of ARSnova at \textit{Technische Hochschule Mittelhessen}
\end{itemize}
\end{frame}
\begin{frame}{send this slide to the past}
\begin{equation*}
\tau = \mu \derivative{u}{y}
\tau = \mu \derivative{v}{x}
\end{equation*}\pause
$\to$ for a static Newtonian fluid, there is no shear $\tau$.\\\pause
(!!!!!)
(huh!)
\end{frame}
\begin{frame}{This Slide Is Sent to the Past}\pause
Contrary to a solid, \textbf{a static Newtonian fluid cannot generate shear on a wall}.\pause
The only forces applying are pressure and gravity.
\begin{equation*}
\gradient{p} = \rho \vec g
\end{equation*}
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Wall shear forces}
\begin{frame}{}
Just like with pressure:\pause
\section{Special case: shear in simple laminar flows}
we go from $F = \tau_\cst S$ to
\begin{IEEEeqnarray}{rCcCl}
\vec F_\text{shear} & = & \int_S \diff \vec F \nonumber\\\pause
& = & \int_S \vec \tau_\text{surface} \diff S \label{eq_shear_force_vector}
\end{IEEEeqnarray}
\end{frame}
\begin{frame}{}
\begin{frame}
\begin{center}
Flat surface? then
\begin{IEEEeqnarray}{rCcCl}
F_{\text{shear} ji} & = & \int_S \diff F \nonumber\\\pause
& = & \int_S ||\vec \tau_{ji}|| \diff S \label{eq_shear_force_vector_flat}
\end{IEEEeqnarray}
Objective: practice calculating shear fields in (very) simple flows\pause
\includegraphics[width=0.6\textwidth]{couette_flow}
Hard part: find $\tau = f(S)$
\end{center}
\end{frame}
\end{frame}{}
\figureframe{0.8}{couette_flow}{}{\wcfile{Couette flow.svg}{Figure} \ccbysa \wcu{Kulmalukko}}
\begin{frame}{}
Split $\diff S$ as $\diff S = \diff x \diff z$ and churn away:\pause
\begin{IEEEeqnarray}{rCl}
F_{\text{shear}ji} &=& \iint \tau_{ji (i, k)} \diff i \diff k \label{eq_shear_force_twod_integration}\\\pause
&=& \mu \iint \partialderivative{u_i}{j} \diff i \diff k\\\pause
F_{\text{shear}yx} &=& \mu \iint \partialderivative{u_x}{y} \diff x \diff z\nonumber
\end{IEEEeqnarray}\pause
\small shear force due to velocity gradient, \emph{not} velocity!
\includegraphics[width=0.4\textwidth]{couette_flow}
Reasonable guess for velocity field:\pause
\begin{IEEEeqnarray*}{rCl}
\left\{ \begin{array}{rcl}
V_x &=& V_\text{bottom wall} + k y\\
V_y &=& 0
\end{array}\right.
\end{IEEEeqnarray*}
\end{frame}
\begin{comment}
\begin{frame}{Moment due to shear}
Just like for pressure
\begin{IEEEeqnarray}{rCcCl}
\vec M_{\text{X} j} & = & \int_S \diff \vec M_\text{X} \nonumber\\\pause
&=& \int_S \vec r_{\text{X}F} \wedge \diff \vec F\nonumber\\\pause
& = & \int_S \vec r_{\text{X}F} \wedge \vec \tau_{ij} \diff S
\end{IEEEeqnarray}
\begin{frame}{}
\includegraphics[width=0.4\textwidth]{couette_flow}
Apply boundary conditions:
\begin{IEEEeqnarray*}{rCl}
\left\{ \begin{array}{rcl}
V_x &=& 0 + \frac{V_\text{top wall}}{H} y\\
V_y &=& 0
\end{array}\right.
\end{IEEEeqnarray*}
\end{frame}
\begin{frame}{Moment due to shear}
Point $X$ along the surface?\pause
\begin{IEEEeqnarray}{rCl}\pause
M_{\text{X} j} &=& \mu \iint r_{\text{X}F} \partialderivative{u_i}{j} \diff i \diff k\\
M_{\text{X} y} &=& \mu \iint r_{\text{X}F} \partialderivative{u_x}{y} \diff x \diff z
\end{IEEEeqnarray}
\begin{frame}
\includegraphics[width=0.4\textwidth]{couette_flow}
Shear in $x$ is viscosity times gradient in $y$ of velocity in $x$:\pause
\begin{IEEEeqnarray*}{rCl}
\tau_{yx} &=& \mu \derivative{}{y} \left(V_x \right)\\\pause
&=& \mu \derivative{}{y} \left(0 + \frac{V_\text{top wall}}{H} y \right)\\\pause
&=& \frac{\mu \ V_\text{top wall}}{H}
\end{IEEEeqnarray*}
(shear field is uniform)
\end{frame}
\end{comment}
\begin{frame}{Take-home slide for chapter 2}
\figureframe{0.7}{clutch}{More complex shear fields?}{\wcfile{Schematic drawing of a simple clutch.svg}{Figure} \cczero \oc}
\begin{frame}{Take-home slide for chapter 5}
\begin{itemize}\pause
\item Shear \emph{has} a direction\pause
\item Net shear force: divergent of tensor: $\divergent{\vec\tau_{ij}}$ \pause
......
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