@ -8,8 +8,9 @@
@@ -8,8 +8,9 @@
\renewcommand { \captionfont } { \small }
\usepackage [english] { babel}
\usepackage { tocloft}
\usepackage { titlesec}
\usepackage { enumitem}
\usepackage { marginnote}
\usepackage { graphicx}
@ -39,7 +40,7 @@
@@ -39,7 +40,7 @@
\usepackage { amssymb}
% for mathbb
\usepackage [english] { babel}
\usepackage [top=2cm, bottom=3cm, outer=7cm, inner=1cm, heightrounded,
marginparwidth=5cm, marginparsep=0.3cm]{ geometry}
@ -47,6 +48,8 @@ marginparwidth=5cm, marginparsep=0.3cm]{geometry}
@@ -47,6 +48,8 @@ marginparwidth=5cm, marginparsep=0.3cm]{geometry}
\title { MATH3102 Lecture Notes}
\author { Alistair Michael}
\titleformat { \chapter } { \normalfont \huge \bf } { \thechapter .} { 20pt} { \huge \bf }
\graphicspath { { images/} }
\newcommand { \sforall } { \enspace \forall }
@ -148,8 +151,10 @@ marginparwidth=5cm, marginparsep=0.3cm]{geometry}
@@ -148,8 +151,10 @@ marginparwidth=5cm, marginparsep=0.3cm]{geometry}
\maketitle
This document has its \LaTeX { } source attached. \attachfile [description=Document
text source,icon=Paperclip]{ notes.tex}
This document can be found at
\href { https://alistairmichael.com/notes/applied.pdf} { alistairmichael.com/notes/applied.pdf} with source available at
\href { https://git.topost.net/alistair/applied-notes} { git.topost.net/alistair/applied-notes} .
It contains many typos and errors.
\tableofcontents
\listoflecture
@ -1749,7 +1754,7 @@ $$R_0 U_{tt} = R_0 F + T'(U_A) U_{AA}$$
@@ -1749,7 +1754,7 @@ $$R_0 U_{tt} = R_0 F + T'(U_A) U_{AA}$$
$$ \lambda = \frac { \ell } { \ell _ 0 } = \frac { \ell - \ell _ 0 + \ell _ 0 } { \ell _ 0 } \sim
\epsilon + 1 $$
$ \e implies $ the reference state has $ \lambda = 1 \Leftrightarrow \epsilon = 0 $
$ \implies $ the reference state has $ \lambda = 1 \Leftrightarrow \epsilon = 0 $
For a small enough $ \epsilon $ any material will exhibit a linear stress--strain
relation.
@ -1908,7 +1913,7 @@ For example for rubber the stress strain raltion looks something like
@@ -1908,7 +1913,7 @@ For example for rubber the stress strain raltion looks something like
\begin { figure} [htpb]
\centering
\begin { tikzpicture} { testppop}
\begin { axis} [ytick=none,xtick=none ,xlabel=$ \epsilon $ , ylabel=$ T $ ,]
\begin { axis} [ytick={ } ,xtick={ } ,xlabel=$ \epsilon $ , ylabel=$ T $ ,]
\addplot [domain=-1:2, samples=90,color=blue,] { (x -0)^ 3 + 0} ;
\end { axis}
\end { tikzpicture}
@ -2075,8 +2080,8 @@ $\implies$ the solution is given by D'Alembert's formula where $\rho=0$ and $f$
@@ -2075,8 +2080,8 @@ $\implies$ the solution is given by D'Alembert's formula where $\rho=0$ and $f$
mode of the forcing term.
\begin { center}
\begin { tikzpicture} { hy }
\begin { axis} [xtick={ 3.141} ,xticklabels={ $ \ell $ } , ytick=none ,]
\begin { tikzpicture} { notes-figure9 }
\begin { axis} [xtick={ 3.141} ,xticklabels={ $ \ell $ } , ytick={ } ,]
\addplot [domain=0:3.141 , samples=90, color=green] { cos(deg(x))} ;
\addlegendentry { $ n = 1 $ }
\addplot [domain=0:3.141 , samples=90, color=black] { cos(2 * deg(x))} ;
@ -2432,7 +2437,7 @@ viscoelastic models explicitly very easiliy}
@@ -2432,7 +2437,7 @@ viscoelastic models explicitly very easiliy}
Use laplace transform on $ ( * ) $ , $ \hat U = L ( U ) $
$$ \begin { cases }
R_ 0 s^ 2 \hat U = L(G) s \hat U_ { A A} $ where $ L(G) = \fraC { \hat E} { s +
R_ 0 s^ 2 \hat U = L(G) s \hat U_ { A A} $ where $ L(G) = \frac { \hat E} { s +
\frac 1 { \tau _ 0} } \\
\hat U_ A (A = 0, s) = \hat F(s) \\
\lim _ { A \to \infty } \hat U(A, s) = 0
@ -2482,7 +2487,7 @@ viscoelastic models explicitly very easiliy}
@@ -2482,7 +2487,7 @@ viscoelastic models explicitly very easiliy}
& =\int _ { \sqrt { \frac { R_ 0} { \hat E} } A} ^ t Q(A, \hat t) F(t - \hat t) \: d\hat t \times H\left (t - \sqrt
{ \frac { R_ 0} { \hat E} } A\right ) \\
\implies & U \text { is zero wherever } t < \sqrt { \frac { R_ 0} A} , \\
\implies & A > t\sqrt { \frac { \haa t E} { R_ 0} } = t \sqrt { \frac { E \tau _ 1} { \tau _ 0}
\implies & A > t\sqrt { \frac { \hat E} { R_ 0} } = t \sqrt { \frac { E \tau _ 1} { \tau _ 0}
\frac 1 { R_ 0} } \\
\text { For } F(t) & = \sin (t) \\
\end { align*}
@ -2664,7 +2669,7 @@ $$J = \det F = \det \underbrace{\nabla_A X}_{\text{Jacobi matrix}} \ne 0$$
@@ -2664,7 +2669,7 @@ $$J = \det F = \det \underbrace{\nabla_A X}_{\text{Jacobi matrix}} \ne 0$$
This is essentially the impenetrability assumption from 1D continuum mechanics in
three dimensions.
At $ t = 0 $ , $ X ( A, t = 0 ) = A \implies \ F = \mathbb I \implies J = 1 $
At $ t = 0 $ , $ X ( A, t = 0 ) = A \implies F = \mathbb I \implies J = 1 $
Since $ J ( t = 0 ) = 1 $ , $ J \ne 0 $ always implies $ J ( t ) > 0 $