big rename

uni/mmme/1026_maths_for_engineering/calculus.md

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+---
+author: Alvie Rahman
+date: \today
+title: MMME1026 // Calculus
+tags: [ uni, nottingham, mechanical, engineering, mmme1026, maths, calculus ]
+---
+
+# Calculus of One Variable Functions
+
+## Key Terms
+
+<details>
+<summary>
+
+### Function
+
+A function is a rule that assigns a **unique** value $f(x)$ to each value $x$ in a given *domain*.
+
+</summary>
+
+The set of value taken by $f(x)$ when $x$ takes all possible value in the domain is the *range* of
+$f(x)$.
+
+</details>
+
+<details>
+<summary>
+
+### Rational Functions
+
+A function of the type
+
+$$ \frac{f(x)}{g(x)} $$
+
+</summary>
+
+where $f$ and $g$ are polynomials, is called a rational function.
+
+Its range has to exclude all those values of $x$ where $g(x) = 0$.
+
+</details>
+
+<details>
+<summary>
+
+### Inverse Functions
+
+Consider the function $f(x) = y$.
+If $f$ is such that for each $y$ in the range there is exactly one $x$ in the domain,
+we can define the inverse $f^{-1}$ as:
+
+$$f^{-1}(y) = f^{-1}(f(x)) = x$$
+
+</summary>
+</details>
+
+<details>
+<summary>
+
+### Limits
+
+Consider the following:
+
+$$f(x) = \frac{\sin x}{x}$$
+
+The value of the function can be easily calculated when $x \neq 0$, but when $x=0$, we get the
+expression $\frac{\sin 0 }{0}$.
+However, when we evaluate $f(x)$ for values that approach 0, those values of $f(x)$ approach 1.
+
+This suggests defining the limit of a function
+
+$$\lim_{x \rightarrow a} f(x)$$
+
+to be the limiting value, if it exists, of $f(x)$ as $x$ gets approaches $a$.
+
+</summary>
+
+#### Limits from Above and Below
+
+Sometimes approaching 0 with small positive values of $x$ gives you a different limit from
+approaching with small negative values of $x$.
+
+The limit you get from approaching 0 with positive values is known as the limit from above:
+
+$$\lim_{x \rightarrow a^+} f(x)$$
+
+and with negative values is known as the limit from below:
+
+$$\lim_{x \rightarrow a^-} f(x)$$
+
+If the two limits are equal, we simply refer to the *limit*.
+
+
+</details>
+
+## Important Functions
+
+<details>
+<summary>
+
+### Exponential Functions
+
+$$f(x) = e^x = \exp x$$
+
+</summary>
+
+It can also be written as an infinite series:
+
+$$\exp x = e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + ...$$
+
+The two important limits to know are:
+
+- as $x \rightarrow + \infty$, $\exp x \rightarrow +\infty$ ($e^x \rightarrow +\infty$)
+- as $x \rightarrow -\infty$, $\exp x \rightarrow 0$ ($e^x \rightarrow 0$)
+
+Note that $e^x > 0$ for all real values of $x$.
+
+</details>
+
+<details>
+<summary>
+
+### Hyperbolic Functions (sinh and cosh)
+
+The hyperbolic sine ($\sinh$) and hyperbolic cosine function ($\cosh$) are defined by:
+
+$$\sinh x = \frac 1 2 (e^x - e^{-x}) \text{ and } \cosh x = \frac 1 2 (e^x + e^{-x})$$
+$$\tanh = \frac{\sinh x}{\cosh x}$$
+
+</summary>
+
+![[Fylwind at English Wikipedia, Public domain, via Wikimedia Commons](https://commons.wikimedia.org/wiki/File:Sinh_cosh_tanh.svg)](./images/Sinh_cosh_tanh.svg)
+
+Some key facts about these functions:
+
+- $\cosh$ has even symmetry and $\sinh$ and $\tanh$ have odd symmetry
+- as $x \rightarrow + \infty$, $\cosh x \rightarrow +\infty$ and $\sinh x \rightarrow +\infty$
+- $\cosh^2x - \sinh^2x = 1$
+- $\tanh$'s limits are -1 and +1
+- Derivatives:
+  - $\frac{\mathrm{d}}{\mathrm{d}x} \sinh x = \cosh x$
+  - $\frac{\mathrm{d}}{\mathrm{d}x} \cosh x = \sinh x$
+  - $\frac{\mathrm{d}}{\mathrm{d}x} \tanh x = \frac{1}{\cosh^2x}$
+
+</details>
+
+<details>
+<summary>
+
+### Natural Logarithm
+
+$$\ln{e^y} = \ln{\exp y} = y$$
+
+</summary>
+
+Since the exponential of any real number is positive, the domain of $\ln$ is $x > 0$.
+
+</details>
+
+<details>
+<summary>
+
+### Implicit Functions
+
+An implicit function takes the form
+
+$$f(x, y) = 0$$
+
+</summary>
+
+To draw the curve of an implicit function you have to rewrite it in the form $y = f(x)$.
+There may be more than one $y$ value for each $x$ value.
+
+</details>

uni/mmme/1026_maths_for_engineering/complex_numbers.md

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+---
+author: Alvie Rahman
+date: \today
+title: MMME1026 // Complex Numbers
+tags: [ uni, nottingham, mechanical, engineering, mmme1026, maths, complex_numbers ]
+---
+
+# Complex Numbers
+
+## What is a Complex Number?
+
+- $i$ is the unit imaginary number, which is defined by:
+
+  $$ i^2 = -1 $$
+
+- An arbritary complex number is written in the form
+
+  $$z = x + iy$$
+
+  Where:
+
+  - $x$ is the real part of $z$ (which you may seen written as $\Re(z) = x$ or Re$(z) = x$)
+  - $y$ is the imaginary part of $z$ (which you may seen written as $\Im(z) = y$ or Im$(z) = y$)
+
+- Two complex numbers are equal if both their real and imaginary parts are equal
+
+  e.g. $$(3 + 4i) + (1 -2i) = (3+1) + i(4-2) = 2 + 2i$$
+
+### The Complex Conjugate
+
+Given complex number $z$:
+
+$$z = z + iy$$
+
+The complex conjugate of z, $\bar z$ is:
+
+$$\bar{z} = z -iy$$
+
+### Division of Complex Numbers
+
+- Multiply numerator and denominator by the conjugate of the denominator
+
+<details>
+<summary>
+
+#### Example
+
+</summary>
+
+> \begin{align*}
+  z_1 &= 5 + i \\
+  z_2 &= 1 -i \\
+  \\
+  \frac{z_1}{z_2} &= \frac{5+i}{1-i} \cdot \frac{\bar{z_2}}{\bar{z_2}} = \frac{(5+i)(1+i)}{(1-i)(1+i)} \\
+    &= \frac{5 + i + 5i -1}{1 + 1} \\
+    &= \frac{4 + 6i}{2} = 2 + 3i
+> \end{align*}
+
+</details>
+
+### Algebra and Conjugation
+
+When taking complex conjugate of an algebraic expresion, we can replace $i$ by $-i$ before or after
+doing the algebraic operations:
+
+\begin{align*}
+\overline{z_1+z_2} &= \bar{z_1}+\bar{z_2} \\
+\overline{z_1z_2} &= \bar{z_1}\bar{z_2} \\
+\overline{z_1/z_2} &= \frac{\bar{z_1}}{\bar{z_2}}
+\end{align*}
+
+The conjugate of a real number is the same as that number.
+
+#### Application
+
+If $z$ is a root of the polynomial equation
+
+$$0 = az^2 + bz + c$$
+
+with **real** coefficients $a$, $b$, and $c$, then $\bar{z}$ is also a root because
+
+\begin{align*}
+0 &= \overline{az^2 + bz + c} \\
+  &= \bar{a}\bar{z}^2 + \bar{b}\bar{z} + \bar{c} \\
+  &= a\bar{z}^2 + b\bar{z} + c
+\end{align*}
+
+### The Argand Diagram
+
+A general complex number $z = x + iy$ has two components so it can can be represented as a point in
+the plane with Cartesion coordinates $(x, y)$.
+
+\begin{align*}
+4-2i &\leftrightarrow (4, -2) \\
+-i &\leftrightarrow (0, -1) \\
+z &\leftrightarrow (x, y) \\
+\bar z &\leftrightarrow (x, -y)
+\end{align*}
+
+### Plotting on a Polar Graph
+
+We can also describe points in the complex plain with polar coordinates $(r, \theta)$:
+
+\begin{align*}
+z &= r(\cos\theta + i\sin\theta) &\text{ polar form of $z$} \\
+r &= \sqrt{x^2+y^2} &\text{(modulus)}\\
+\theta &= \arg z,\text{ where} \tan \theta = \frac y x &\text{(argument)} \\
+x &= r\cos \theta \\
+y &= r\sin \theta
+\end{align*}
+
+Be careful when turning $(x, y)$ into $(r, \theta)$ form as $\tan^{-1} \frac y x = \theta$ does not
+always hold true as there are many solutions.
+
+#### Choosing $\theta$ Correctly
+
+1. Determine which quadrant the point is in (draw a picture).
+2. Find a value of $\theta$ such that $\tan \theta = \frac y x$ and check that it is consistent.
+   If it puts you in the wrong quadrant, add or subtract $\pi$.
+
+## Exponential Functions
+
+- The exponential function $f(x) = \exp x$ may be wirtten as an infinite series:
+
+  $$\exp x = e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots $$
+
+- The function $f(x) = e^{-x}$ is just $\frac 1 {e^x}$
+- Note the important properties:
+
+\begin{align*}
+e^{a+b} &= e^a e^b \\
+(e^a)^b &= e^{ab} 
+\end{align*}
+
+## Euler's Formula
+
+$$e^{i\theta} = \cos\theta + i\sin\theta$$
+
+- Properties of $e^{i\theta}$: For any real angle $\theta$ we have
+
+  $$|e^{i\theta}| = |\cos\theta + i\sin\theta| = \sqrt{\cos^2\theta + \sin^2\theta} = 1$$
+
+  and 
+
+  $$ \arg {e^{i\theta}} = \theta $$
+
+- A complex number in *polar form*, where $r = |z|$, and $\theta = \arg z$, may alternatively be
+  written in its *exponential form*:
+
+  $$z = re^{i\theta}$$
+
+  **Note**: $$\bar z = r\cos\theta - ir\sin\theta = re^{-i\theta}$$
+
+<details>
+<summary>
+
+### Example 1
+
+Write $z = -1 + i$ in exponential form
+
+</summary>
+
+> $\arg z = \frac {3\pi} 4$
+> $|z| = \sqrt 2$
+> 
+> So $z = \sqrt2e^{i\frac{3\pi} 4}$
+
+</details>
+
+<details>
+<summary>
+
+### Example 2
+
+The equations for a mechanical vibration problem are found to have the following mathematical
+solution:
+
+$$z(t) = \frac{e^{i\omega t}}{\omega_0^2-\omega^2 + i\gamma}$$
+
+</summary>
+
+where $t$ represents time and $\omega$, $\omega_0$ and $\gamma$ are all positive real physical
+constants.
+Although $z(t)$
+is complex and cannot directly represent a physical solution, it turns out that the real and
+imaginary parts $x(t)$ and $y(t)$ in $z(t) = x(t) + iy(t)$ can. Polar notation can be used to extract
+this physical information efficiently as follows:
+
+a. Put the denominator in the form
+
+   $$ae^{i\delta}$$
+
+   where you should give explicit expressions for $a$ and $\delta$ in terms of $\gamma$, $\gamma_0$,
+   and $\gamma$.
+
+
+   > \begin{align*}
+     a &= \sqrt{\gamma^2 + (\omega_0^2 - \omega^2)^2} \\
+     \delta &= \tan^{-1}\frac \delta {\omega_0^2 - \omega^2}
+   > \end{align*}
+
+b. Hence find the constants $b$ and $\varphi$ such that
+
+   $$x(t) = b\cos(\omega t + \varphi)$$
+
+   and write a similar expression for $y(t)$.
+
+   > \begin{align*}
+     z &= \frac{e^i\omega t}{ae^{i\delta}} = \frac 1 a e^{i (\omega t - \delta)} \\
+     x + iy &= \frac 1 a \cos(\omega t - \delta) + \frac 1 a \sin(\omega t - \delta) \\
+     \therefore \Re z &= x = \frac 1 a \cos(\omega t - \delta), \\
+     \Im z &= y = \frac 1 a \sin(\omega t - \delta) \\
+     \\
+     b &= \frac 1 a = \frac 1 {\sqrt{\gamma^2 + (\omega_0^2 - \omega^2)^2}} \\
+     \varphi &= -\delta = \tan^{-1}\frac \delta {\omega_0^2 - \omega^2}\\
+     \\
+     y(t) &= \frac 1 a \sin(\omega t - \delta) \\
+   > \end{align*}
+
+</details>
+
+## Products of Complex Numbers
+
+Suppose we have 2 complex numbers:
+
+$$z_1 = x_1 + iy_1 = r_1e^{i\theta_1}$$  
+$$z_2 = x_2 + iy_2 = r_2e^{i\theta_2}$$
+
+Using $e^a e^b = e^{a+b}$, the product is:
+
+\begin{align*}
+z_3 = z_1 z_2 &= (r_1e^{i\theta_1})(r_2e^{i\theta_2}) \\
+&= r_1r_2e^{i\theta_1}e^{i\theta_2} \\
+&= r_1r_2e^{i(\theta_1+\theta_2)} \\
+\\
+|z_1z_2| &= |z_1|\times|z_2| \\
+\arg z_1z_2 &= \arg z_1 \times \arg z_2
+\end{align*}
+
+## de Moivre's Theorem
+
+Let $z = re^{i\theta}$. Consider $z^n$.
+
+Since $z = r(\cos\theta + i\sin\theta)$,
+\begin{align*}
+z^n &= r^n(\cos\theta + i\sin\theta)^n &\text{(1)} \\
+\end{align*}
+
+But also
+
+\begin{align*}
+z^n &= (re^{i\theta})^n \\
+    &= r^n(e^{i\theta})^n \\
+    &= r^ne^{in\theta} \\
+    &= r^n(\cos{n\theta} + i\sin{n\theta}) &\text{(2)} \\
+\end{align*}
+
+By equating (1) and (2), we find de Moivre's theorem:
+
+\begin{align*}
+r^n(\cos\theta +i\sin\theta)^n &= r^n(\cos{n\theta} + i\sin{n\theta}) \\
+(\cos\theta +i\sin\theta)^n &= (\cos{n\theta} + i\sin{n\theta})
+\end{align*}
+
+<details>
+<summary>
+
+### Example 1
+
+Write $1+i$ in polar form and use de Moivre's theorem to calculate $(1+i)^{15}$.
+
+</summary>
+
+> \begin{align*}
+  r &= |1+i| = \sqrt2 \\
+  \theta &= \arg{1+i} = \frac \pi 4 \\
+  \\
+  \text{So } 1 + i &= \sqrt2(\cos{\frac pi 4} + i\sin{\frac \pi 4}) = \sqrt2e^{i\frac \pi 4} \text{ and}\\
+  (i+i)^{15} &= (\sqrt2)^{15}\left(\cos{\frac \pi 4} + i\sin{\frac \pi 4}\right)^{15} \\
+             &= 2^{\frac 15 2} \left(\cos{\frac{15\pi} 4} + i\sin{\frac{15\pi} 4}\right) \\
+             &= 2^{\frac 15 2} \left(\frac 1 {\sqrt2} - \frac i {\sqrt2}\right) \\
+             &= 2^7 (1 - i) \\
+             &= 128 - 128i
+> \end{align*}
+</details>
+
+<details>
+<summary>
+
+### Example 2
+
+Use de Moivre's theorem to show that
+
+\begin{align*}
+\cos{2\theta} &= \cos^2\theta-\sin^2\theta \\
+\text{and} \\
+\sin{2\theta} &= 2\sin\theta\cos\theta
+\end{align*}
+
+</summary>
+
+> Let $n=2$:
+
+> \begin{align*}
+  (\cos\theta+i\sin\theta)^2 &= \cos^2\theta + 2i\sin\theta\cos\theta - \sin^2\theta \\
+  \text{Real part: } \cos^2\theta - \sin^2\theta &= \cos{2\theta}\\
+  \text{Imaginary part: } 2\sin\theta\cos\theta &= \sin{2\theta}
+> \end{align*}
+
+</details>
+
+<details>
+<summary>
+
+### Example 3
+
+Given that $n \in \mathbb{N}$ and $\omega = -1 + i$, show that
+$w^n + \bar{w}^n = 2^{\frac n 2 + 1}\cos{\frac{3n\pi} 4}$ with Euler's formula.
+
+</summary>
+
+> \begin{align*}
+  r &= \sqrt{2} \\
+  \arg \omega = \theta &= \frac 3 4 \pi \\
+  \\
+  \omega^n &= r^n(cos{n\theta} + i\sin{n\theta}) \\
+  \bar\omega^n &= r^n(cos{n\theta} - i\sin{n\theta}) \\
+  \omega^n + \bar\omega^n &= r^n(2\cos{n\theta}) \\
+                          &= 2^{\frac n 2 + 1}\cos{\frac {3n\pi} 4}
+> \end{align*}
+
+</details>
+
+## Complex Roots of Polynomials
+
+<details>
+<summary>
+
+### Example
+
+Find which complex numbers $z$ satisfy
+
+$$z^3 = 8i$$
+
+</summary>
+
+> 1. Write $8i$ in exponential form,
+> 
+>    $|8i| = 8$ and $\arg{8i} = \frac \pi 2$
+> 
+>    $\therefore 8i = 8e^{i\frac \pi 2}$
+> 
+> 
+> 2. Let the solution be $r = re^{i\theta}$.
+> 
+>    Then $z^3 = r^3e^{3i\theta}$.
+> 
+> 3. $z^3 = r^3e^{3i\theta} = 8e^{i\frac \pi 2}$
+> 
+>    i. Compare modulus:
+> 
+>       $r^3 = 8 \rightarrow r = 2$
+> 
+>    ii. Compare argument:
+> 
+>        $$3\theta = \frac \pi 2$$
+> 
+>        is a solution but there are others since 
+> 
+>        $$e^{i\frac \pi 2} = e^{i \frac \pi 2 + 2n\pi}$$
+> 
+>        so we get a solution whenever
+> 
+>        $$3\theta = \frac \pi 2 + 2n\pi$$
+> 
+>        for any integer `n`
+> 
+>        - $n = 0 \rightarrow z = \sqrt3 + i$
+>        - $n = 1 \rightarrow z = -\sqrt3 + i$
+>        - $n = 2 \rightarrow z = -2i$
+>        - $n = 3 \rightarrow z = \sqrt3 + i$
+>        - $n = 4 \rightarrow z = -\sqrt3 + i$
+>        - The solutions start repeating as you can see
+> 
+>        In general, an $n$-th order polynomial has exactly $n$ complex roots.
+>        Some of these complex roots may be real numbers.
+> 
+> 4. There are three solutions
+
+</details>

uni/mmme/1026_maths_for_engineering/images/Sinh_cosh_tanh.svg

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uni/mmme/1026_maths_for_engineering/systems_and_matrices.md

@@ -0,0 +1,744 @@
+---
+author: Alvie Rahman
+date: \today
+title: MMME1026 // Systems of Equations and Matrices
+tags: [ uni, nottingham, mechanical, engineering, mmme1026, maths, systems_of_equations, matrices ]
+---
+# Systems of Equations (Simultaneous Equations)
+
+## Gaussian Elimination
+
+Gaussian eliminiation can be used when the number of unknown variables you have is equal to the
+number of equations you are given.
+
+I'm pretty sure it's the name for the method you use to solve simultaneous equations in school.
+
+For example if you have 1 equation and 1 unknown:
+
+\begin{align*}
+2x &= 6 \\
+x &= 3
+\end{align*}
+
+### Number of Solutions
+
+Let's generalise the example above to
+
+$$ax = b$$
+
+There are 3 possible cases:
+
+\begin{align*}
+a \ne 0 &\rightarrow x = \frac b a \\
+a = 0, b \ne 0 &\rightarrow \text{no solution for $x$} \\
+a = 0, b = 0 &\rightarrow \text{infinite solutions for $x$}
+\end{align*}
+
+### 2x2 Systems
+
+A 2x2 system is one with 2 equations and 2 unknown variables.
+
+<details>
+<summary>
+
+#### Example 1
+
+\begin{align*}
+3x_1 + 4x_2 &= 2 &\text{(1)} \\
+x_1 + 2x_2 &= 0 &\text{(2)} \\
+\end{align*}
+
+</summary>
+
+\begin{align*}
+3\times\text{(2)} = 3x_1 + 6x_2 &= 0 &\text{(3)} \\
+\text{(3)} - \text{(1)} = 0x_1 + 2x_2 &= -2 \\
+x_2 &= -1
+\end{align*}
+
+We've essentially created a 1x1 system for $x_2$ and now that's solved we can back substitute it
+into equation (1) (or equation (2), it doesn't matter) to work out the value of $x_1$:
+
+\begin{align*}
+3x_1 + 4x_2 &= 2 \\
+3x_1 - 1 &= 2 \\
+3x_1 &= 6 \\
+x_1 &= 2
+\end{align*}
+
+You can check the values for $x_1$ and $x_2$ are correct by substituting them into equation (2).
+
+</details>
+
+### 3x3 Systems
+
+A 3x3 system is one with 3 equations and 3 unknown variables.
+
+<details>
+<summary>
+
+#### Example 1
+
+\begin{align*}
+2x_1 + 3x_2 - x_3 &= 5 &\text{(1)} \\
+4x_1 + 4x_2 - 3x_3 &= 5 &\text{(2)} \\
+2x_1 - 3x_2 + x_3 &= 5 &\text{(3)} \\
+\end{align*}
+
+</summary>
+
+The first step is to eliminate $x_1$ from (2) and (3) using (1):
+
+\begin{align*}
+\text{(2)}-2\times\text{(1)} = -2x_2 -x_3 &= -1 &\text{(4)} \\
+\text{(3)}-\text{(1)} = -6x_2 + 3x_3 &= -6 &\text{(5)} \\
+\end{align*}
+
+This has created a 2x2 system of $x_2$ and $x_3$ which can be solved as any other 2x2 system.
+I'm too lazy to type up the working, but it is solved like any other 2x2 system.
+
+\begin{align*}
+x_2 &= -2
+x_3 &= 5
+\end{align*}
+
+These values can be back-substituted into any of the first 3 equations to find out $x_1$:
+
+\begin{align*}
+-2x_1 + 3x_2 - x_3 = 2x_1 + 6 - 3 = 5 \rightarrow x_1 = 1
+\end{align*}
+
+</details>
+
+<details>
+<summary>
+
+#### Example 2
+
+\begin{align*}
+x_1 + x_2 - 2x_3 &= 1 &R_1 \\
+2x_1 - x_2 - x_3 &= 1 &R_2 \\
+x_1 + 4x_2 + 7x_3 &= 2 &R_3 \\
+\end{align*}
+
+</summary>
+
+1. Eliminate $x_1$ from $R_2$, $R_3$:
+
+   \begin{align*}
+   x_1 + x_2 - 2x_3 &= 1 &R_1' = R_1\\
+   - 3x_2 - 5x_3 &= -1 &R_2' = R_2 - 2R_1 \\
+   3x_2 + 5x_3 &= 1 &R_3' = R_3 - R_1 \\
+   \end{align*}
+
+   We've created another 2x2 system of $R_2'$ and $R_3'$
+
+2. Eliminate $x_2$ from $R_3''$
+
+   \begin{align*}
+   x_1 + x_2 - 2x_3 &= 1 &R_1'' = R_1' = R_1\\
+   - 3x_2 - 5x_3 &= -1 &R_2'' = R_2' = R_2 - 2R_1 \\
+   0x_3 &= 0 &R_3'' = R_3 '+ R_2' \\
+   \end{align*}
+
+   We can see that $x_3$ can be any number, so there are infinite solutions. Let:
+
+   $$x_3 = t$$
+
+   where $t$ can be any number
+
+3. Substitute $x_3$ into $R_2''$:
+
+   $$R_2'' = -3x_2 - 5t = -1 \rightarrow x_2 = \frac 1 3 - \frac{5t} 3$$
+
+4. Substitute $x_2$ and $x_3$ into $R_1''$:
+
+   $$R_1'' = x_1 + \frac 1 3 - \frac{5t} 3 + 2t = 1 \rightarrow x_1 = \frac 2 3 - \frac t 3$$
+
+</details>
+
+## Systems of Equations and Matrices
+
+Many problems in engineering have a very large number of unknowns and equations to solve
+simultaneously.
+We can use matrices to solve these efficiently.
+
+Take the following simultaneous equations::
+
+\begin{align*}
+3x_1 + 4x_2 &= 2 &\text{(1)} \\
+x_1 + 2x_2 &= 0 &\text{(2)}
+\end{align*}
+
+They can be represented by the following matrices:
+
+\begin{align*}
+A &= \begin{pmatrix} 3 & 4 \\ 1 & 2 \end{pmatrix} \\
+\pmb x  &= \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} \\
+\pmb b  &= \begin{pmatrix} 2 \\ 0 \end{pmatrix} \\
+\end{align*}
+
+You can then express the system as:
+
+$$A\pmb x = \pmb b$$
+
+<details>
+<summary>
+
+#### A 3x3 System as a Matrix
+
+</summary>
+
+\begin{align*}
+2x_1 + 3x_2 - x_3 &= 5 \\
+4x_1 + 4x_2 - 3x_3 &= 3 \\
+2x_1 - 3x_2 + x_3 &= -1
+\end{align*}
+
+Could be expressed in the form $A\pmb x = \pmb b$ where:
+
+\begin{align*}
+A &= \begin{pmatrix} 2 & 3 & -1 \\ 4 & 4 & -3 \\ 2 & -3 & -1 \end{pmatrix} \\
+\pmb x  &= \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} \\
+\pmb b  &= \begin{pmatrix} 5 \\ 3 \\ -1 \end{pmatrix} \\
+\end{align*}
+
+</details>
+
+<details>
+<summary>
+
+#### An $m\times n$ System as a Matrix
+
+</summary>
+
+\begin{align*}
+a_{11}x_1 + a_{12}x_2 + \cdots + a_{1n}x_n &= b_1 \\
+a_{21}x_1 + a_{22}x_2 + \cdots + a_{2n}x_n &= b_2 \\
+\cdots \\
+a_{m1}x_1 + a_{m2}x_2 + \cdots + a_{mn}x_n &= b_m \\
+\end{align*}
+
+Could be expressed in the form $A\pmb x = \pmb b$ where:
+
+\begin{align*}
+A = \begin{pmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\
+                a_{21} & a_{22} & \cdots & a_{2n} \\
+                \vdots &        &     & \vdots \\
+                a_{m1} & a_{m2} & \cdots & a_{mn} 
+                \end{pmatrix}, 
+\pmb x  = \begin{pmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{pmatrix}, 
+\pmb b  = \begin{pmatrix} b_1 \\ b_2 \\ \vdots \\ b_m \end{pmatrix}
+\end{align*}
+
+</details>
+
+# Matrices
+
+## Order of a Matrix
+
+The order of a matrix is its size e.g. $3\times2$ or $m\times n$
+
+## Column Vectors
+
+- Column vectors are matrices with only one column:
+
+  $$ \begin{pmatrix} 1 \\ 2 \end{pmatrix} \begin{pmatrix} 4 \\ 45 \\ 12 \end{pmatrix} $$
+
+- Column vector variables typed up or printed are expressed in $\pmb{bold}$ and when it is
+  handwritten it is \underline{underlined}:
+
+  $$ \pmb x = \begin{pmatrix} -3 \\ 2 \end{pmatrix}$$
+
+## Matrix Algebra
+
+### Equality
+
+Two matrices are the same if:
+
+- Their order is the same
+- Their corresponding elements are the same
+
+### Addition and Subtraction
+
+Only possible if their order is the same.
+\begin{align*}
+A + B&= \begin{pmatrix} a_{11} + b_{11} & a_{12} + b_{12} & \cdots & a_{1n} + b_{1n} \\
+                a_{21} + b_{21} & a_{22} + b_{22} & \cdots & a_{2n} + b_{2n} \\
+                \vdots &        &     & \vdots \\
+                a_{m1} + b_{m1} & a_{m2} + b_{m2} & \cdots & a_{mn} + b_{mn}
+                \end{pmatrix} \\
+A - B&= \begin{pmatrix} a_{11} - b_{11} & a_{12} - b_{12} & \cdots & a_{1n} - b_{1n} \\
+                a_{21} - b_{21} & a_{22} - b_{22} & \cdots & a_{2n} - b_{2n} \\
+                \vdots &        &     & \vdots \\
+                a_{m1} - b_{m1} & a_{m2} - b_{m2} & \cdots & a_{mn} - b_{mn}
+                \end{pmatrix}, 
+\end{align*}
+
+### Zero Matrix
+
+This is a matrix whose elements are all zeros.
+For any matrix $A$,
+
+$$A + 0 =A$$
+
+We can only add matrices of the same order, therefore 0 must be of the same order as $A$.
+
+### Multiplication
+
+Let
+
+$$
+\begin{matrix}
+A & m\times n \\
+B & p\times q
+\end{matrix}
+$$
+
+To be able to multiply $A$ by $B$, $n = p$.
+
+If $n \ne p$, then $AB$ does not exist.
+
+$$
+\begin{matrix}
+A & B & = & C \\
+m\times n & p \times q & & m\times q
+\end{matrix}
+$$
+
+When $C = AB$ exists,
+
+$$C_{ij} = \sum_r\! a_{ir}b_{rj}$$
+
+That is, $C_{ij}$ is the 'product' of the $i$th row of $A$ and $j$th column of $B$.
+
+#### Multiplication of a Matrix by a Scalar
+
+If $\lambda$ is a scalar, we define
+
+$$
+\lambda a = \begin{pmatrix} \lambda a_{11} & \lambda a_{12} & \cdots & \lambda a_{1n} \\
+                \lambda a_{21} & \lambda a_{22} & \cdots & \lambda a_{2n} \\
+                \vdots &        &     & \vdots \\
+                \lambda a_{m1} & \lambda a_{m2} & \cdots & \lambda a_{mn} 
+                \end{pmatrix}, 
+$$
+
+<details>
+<summary>
+
+#### Example 1
+
+</summary>
+
+$$
+\begin{pmatrix} 1 & -1 \\ 2 & 1 \end{pmatrix}
+\begin{pmatrix} 0 & 1 \\ 3 & 2 \end{pmatrix} = 
+\begin{pmatrix} -3 & -1 \\ 3 & 4 \end{pmatrix}
+$$
+$$
+\begin{pmatrix} 0 & 1 \\ 3 & 2 \end{pmatrix}
+\begin{pmatrix} 1 & -1 \\ 2 & 1 \end{pmatrix} =
+\begin{pmatrix} 2 & 1 \\ 7 & -1 \end{pmatrix}
+$$
+
+</details>
+
+<details>
+<summary>
+
+#### Example 2
+
+</summary>
+
+$$
+A = \begin{pmatrix} 4 & 1 & 6 \\ 3 & 2 & 1 \end{pmatrix},\,
+B = \begin{pmatrix} 1 & 1 \\ 1 & 2 \\ 1 & 0 \end{pmatrix}
+$$
+$$
+AB = \begin{pmatrix} 11 & 6 \\ 6 & 7 \end{pmatrix},\,
+BA = \begin{pmatrix} 7 & 3 & 7 \\ 10 & 5 & 8 \\ 4 & 1 & 6 \end{pmatrix}
+$$
+
+</details>
+
+### Other Properties of Matrix Algebra
+
+- $(\lambda A)B = \lambda(AB) = A(\lambda B)$
+- $A(BC) = (AB)C = ABC$
+- $(A+B)C = AC + BC$
+- $C(A+B) = CA + CB$
+- In general, $AB \ne BA$ even if both exist
+- $AB = 0$ does not always mean $A = 0$ or $B = 0$:
+
+  $$\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} \begin{pmatrix}3 & 0 \\ 0 & 0 \end{pmatrix} = 
+  \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} = 0$$
+
+  <details>
+  <summary>
+
+  It follows that $AB = AC$ does not imply that $B=C$ as
+
+  $$AB = AC \leftrightarrow A(B + C) = 0$$
+
+  and as $A$ and $(B-C)$ are not necessarily 0, $B$ is not necessarily equal to $C$:
+
+  </summary>
+
+  $$AB = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} \begin{pmatrix}0 & 0 \\ 1 & 0 \end{pmatrix} = 
+  \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}$$
+
+  and
+
+  $$AC = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} \begin{pmatrix}1 & 2 \\ 1 & 0 \end{pmatrix} = 
+  \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} = AB$$
+
+  but $B \ne C$
+
+  </details>
+
+## Special Matrices
+
+### Square Matrix
+
+Where $m = n$
+
+<details>
+<summary>
+
+#### Example 1
+
+A $3\times3$ matrix.
+
+</summary>
+
+$$\begin{pmatrix}1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9\end{pmatrix}$$
+
+</details>
+
+<details>
+<summary>
+
+#### Example 2
+
+A $2\times2$ matrix.
+
+</summary>
+
+$$\begin{pmatrix}1 & 2 \\ 4 & 5 \end{pmatrix}$$
+
+</details>
+
+### Identity Matrix
+
+The identity matrix is a square matrix whose eleements are all 0, except the leading diagonal which
+is 1s.
+The leading diagonal is the top left to bottom right corner.
+
+It is usually denoted by $I$ or $I_n$.
+
+The identity matrix has the properties that
+
+$$AI = IA = A$$
+
+for any square matrix $A$ of the same order as I, and
+
+$$Ix = x$$
+
+for any vector $x$.
+
+<details>
+<summary>
+
+#### Example 1
+
+The $3\times3$ identity matrix.
+
+</summary>
+
+$$\begin{pmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{pmatrix}$$
+
+</details>
+
+<details>
+<summary>
+
+#### Example 2
+
+The $2\times2$ identity matrix.
+
+</summary>
+
+$$\begin{pmatrix}1 & 0 \\ 0 & 1 \end{pmatrix}$$
+
+</details>
+
+### Transposed Matrix
+
+The transpose of matrix $A$ of order $m\times n$ is matrix $A^T$ which has the order $n\times m$.
+It is found by reflecting it along the leading diagonal, or interchanging the rows and columns of
+$A$.
+
+![by [Lucas Vieira](https://commons.wikimedia.org/wiki/File:Matrix_transpose.gif)](./images/Matrix_transpose.gif)
+
+Let matrix $D = EF$, then $D^T = (EF)^T = E^TF^T$
+
+#### Example 1
+
+$$
+A = \begin{pmatrix}3 & 2 & 1 \\ 4 & 5 & 6 \end{pmatrix},\,
+A^T = \begin{pmatrix}3 & 4 \\ 2 & 5 \\ 1 & 6\end{pmatrix}
+$$
+
+#### Example 2
+
+$$
+B = \begin{pmatrix}1 \\ 4\end{pmatrix},\,
+B^T = \begin{pmatrix}1 & 4\end{pmatrix}
+$$
+
+#### Example 3
+
+$$
+C = \begin{pmatrix}1 & 2 & 3 \\ 0 & 5 & 1 \\ 2 & 3 & 7\end{pmatrix},\,
+C^T = \begin{pmatrix}1 & 0 & 2 \\ 2 & 5 & 4 \\ 3 & 1 & 7\end{pmatrix}
+$$
+
+### Orthogonal Matrices
+
+A matrix, $A$, such that
+
+$$A^{-1} = A^T$$
+
+is said to be orthogonal.
+
+Another way to say this is
+
+$$AA^T = A^TA = I$$
+
+### Symmetric Matrices
+
+A square matrix which is symmetric about its leading diagonal:
+
+$$A = A^T$$
+
+You can also express this as the matrix $A$, where
+
+$$a_{ij} = a_{ji}$$
+
+is satisfied to all elements.
+
+<details>
+<summary>
+
+#### Example 1
+
+</summary>
+
+$$\begin{pmatrix}
+1 & 0 & -1 & 3 \\
+0 & 3 & 4 & -1 \\
+-2 & 4 & -1 & 6 \\
+3 & -7 & 6 & 2 
+\end{pmatrix}$$
+
+</details>
+
+### Anti-Symmetric
+
+A square matrix is anti-symmetric if
+
+$$A = -A^T$$
+
+This can also be expressed as
+
+$$a_{ij} = -a_{ji}$$
+
+This means that all elements on the leading diagonal must be 0.
+
+<details>
+<summary>
+
+#### Example 1
+
+</summary>
+
+$$\begin{pmatrix}
+0 & -1 & 5 \\
+1 & 0 & 1 \\
+-5 & -1 & 0
+\end{pmatrix}$$
+
+</details>
+
+## The Determinant
+
+### Determinant of a 2x2 System
+
+The determinant of a $2x2$ system is
+
+$$D = a_{11}a_{22} - a_{12}a_{21}$$
+
+It is denoted by
+
+$$
+\begin{vmatrix}
+a_{11} & a_{12} \\
+a_{21} & a_{22}
+\end{vmatrix}
+\text{ or }
+\det
+\begin{pmatrix}
+a_{11} & a_{12} \\
+a_{21} & a_{22}
+\end{pmatrix}
+$$
+
+- A system of equations has a unique solution if $D \ne 0$
+- If $D = 0$, then there are either
+
+  - no solutions (the equations are inconsistent)
+  - intinitely many solutions
+
+### Determinant of a 3x3 System
+
+Let
+
+$$
+A = \begin{pmatrix}
+a_{11} & a_{12} & a_{13} \\
+a_{21} & a_{22} & a_{23} \\
+a_{31} & a_{32} & a_{33}
+\end{pmatrix}
+$$
+
+\begin{align*}
+\det A = &a_{11} \times \det \begin{pmatrix}a_{22} & a_{23} \\ a_{32} & a_{33} \end{pmatrix} \\
+        &-a_{12} \times \det \begin{pmatrix}a_{21} & a_{23} \\ a_{31} & a_{33} \end{pmatrix} \\
+        &+a_{13} \times \det \begin{pmatrix}a_{21} & a_{22} \\ a_{31} & a_{32} \end{pmatrix}
+\end{align*}
+
+The $2x2$ matrices above are created by removing any elements on the same row or column as its corresponding
+coefficient:
+
+![](./images/vimscrot-2021-11-02T16:19:40,013146580+00:00.png)
+
+### Chessboard Determinant
+
+$\det A$ may be obtained by expanding out any row or column.
+To figure out which coefficients should be subtracted and which ones added use the chessboard
+pattern of signs:
+
+$$\begin{pmatrix} + & - & + \\ - & + & - \\ + & - & + \end{pmatrix}$$
+
+### Properties of Determinants
+
+- $$\det A = \det A^T$$
+- If all elements of one row of a matrix are multiplied by a constant $z$, the determinant of the
+  new matrix is $z$ times the determinant of the original matrix:
+
+  \begin{align*}
+  \begin{vmatrix} za & zb \\ c & d \end{vmatrix} &= zad - zbc \\
+       &= z(ad-bc) \\
+       &= z\begin{vmatrix} a & b \\ c & d \end{vmatrix}
+  \end{align*}
+
+  This is also true if a column of a matrix is mutiplied by a constant.
+
+  **Application** if the fator $z$ appears in each elements of a row or column of a determinant it
+  can be factored out
+
+  $$\begin{vmatrix}2 & 12 \\ 1 & 3 \end{vmatrix} = 2\begin{vmatrix}1 & 6 \\ 1 & 3 \end{vmatrix} = 2 \times 3
+  \begin{vmatrix} 1 & 2 \\ 1 & 1 \end{vmatrix}$$
+
+  **Application** if all elements in one row or column of a matrix are zero, the value of the 
+  determinant is 0.
+
+  $$\begin{vmatrix} 0 & 0 \\ c & d \end{vmatrix} = 0\times d - 0\times c = 0$$
+
+
+  **Application** if $A$ is an $n\times n$ matrix,
+
+  $$\det(zA) = z^n\det A$$
+
+- Swapping any two rows or columns of a matrix changes the sign of the determinant
+
+  \begin{align*}
+  \begin{vmatrix} c & d \\ a & b \end{vmatrix} &= cb - ad \\
+        &= -(ad - bc) \\
+        &= -\begin{vmatrix} a & b \\ c & d \end{vmatrix}
+  \end{align*}
+
+  **Application** If any two rows or two columns are identical, the determinant is zero.
+
+  **Application** If any row is a mutiple of another, or a column a multiple of another column, the
+  determinant is zero.
+
+- The value of a determinant is unchanged by adding to any row a constant multiple of another row,
+  or adding to any column a constant multiple of another column
+
+- If $A$ and $B$ are square matrices of the same order then
+
+  $$\det(AB) = \det A \times \det B $$
+
+## Inverse of a Matrix
+
+If $A$ is a square matrix, then its inverse matrix is $A^{-1}$ and is defined by the property that:
+
+$$A^{-1}A = AA^{-1} = I$$
+
+- Not every matrix has an inverse
+- If the inverse exists, then it is very useful for solving systems of equations:
+
+  \begin{align*}
+  A\pmb{x} = \pmb b \rightarrow A^{-1}A\pmb x &= A^{-1}\pmb b \\
+  I\pmb x &= A^{-1}\pmb b \\
+  \pmb x &= A^{-1}\pmb b
+  \end{align*}
+
+  Therefore there must be a unique solution to $A\pmb x = \pmb b$: $\pmb x = A^{-1}\pmb b$.
+
+- If $D = EF$ then 
+
+  $$D^-1 = (EF)^{-1} = F^{-1}E^{-1}$$
+
+### Inverse of a 2x2 Matrix
+
+If $A$ is the $2x2$ matrix
+
+$$
+A = \begin{pmatrix}
+a_{11} & a_{12} \\
+a_{21} & a_{22}
+\end{pmatrix}
+$$
+
+and its determinant, $D$, satisfies $D \ne 0$, $A$ has the inverse $A^{-1}$ given by
+
+$$
+A^{-1} = \frac 1 D \begin{pmatrix}
+a_{22} & -a_{12} \\
+-a_{21} & a_{11}
+\end{pmatrix}
+$$
+
+If $D = 0$, then matrix $A$ has no inverse.
+
+<details>
+<summary>
+
+#### Example 1
+
+Find the inverse of matrix $A = \begin{pmatrix} -1 & 5 \\ 2 & 3 \end{pmatrix}$.
+
+</summary>
+
+1. Calculate the determinant
+
+   $$\det A = -1 \times 3 - 5 \times 2 = -13$$
+
+   Since $\det A \ne 0$, the inverse exists.
+
+2. Calculate $A^{-1}$
+
+   $$ A^{-1} = \frac 1 {-13} \begin{pmatrix} 3 & -5 \\ -2 & -1\end{pmatrix}$$

uni/mmme/1028_statics_and_dynamics/statics.md

@@ -0,0 +1,578 @@
+---
+author: Alvie Rahman
+date: \today
+tags:
+- uni
+- nottingham
+- mechanical
+- engineering
+- mmme1028
+- maths
+- statics
+- dynamics
+title: MMME1028 // Statics
+---
+
+# Lecture L1.1, L1.2
+
+### Lecture L1.1 Exercises
+
+Can be found [here](./lecture_exercises/mmme1028_l1.1_exercises_2021-09-30.pdf).
+
+### Lecture L1.2 Exercises
+
+Can be found here [here](./lecture_exercises/mmme1028_l1.2_exercises_2021-10-04.pdf)
+
+## Newton's Laws
+
+1.  Remains at constant velocity unless acted on by external force
+
+2.  Sum of forces on body is equal to mass of body multiplied by
+    acceleration
+
+    > 1st Law is a special case of 2nd
+
+3.  When one body exerts a force on another, 2nd body exerts force
+    simultaneously of equal magnitude and opposite direction
+
+## Equilibrium
+
+-   Body is in equilibrium if sum of all forces and moments acting on
+    body are 0
+
+<details>
+<summary>
+
+### Example
+
+Determine force $F$ and $x$ so that the body is in equilibrium.
+
+</summary>
+
+![](./images/vimscrot-2021-10-04T09:14:41,378027532+01:00.png)
+
+1.  Check horizontal equilibrium
+
+    $\sum{F_x} = 0$
+
+2.  Check vertical equilibrium
+
+    $\sum{F_y} = 8 - 8 + F = 0$
+
+    $F = 2$
+
+3.  Take moments about any point
+
+    $\sum{M(A)} = 8\times{}2 - F(2+x) = 0$
+
+    $F(2+x) = 16)$
+
+    $x = 6$
+
+</details>
+
+## Free Body Diagrams
+
+A free body diagram is a diagram of a single (free) body which shows all
+the external forces acting on the body.
+
+Where there are several bodies or subcomponents interacting as a complex
+system, each body is drawn separately:
+
+![](./images/vimscrot-2021-10-04T09:23:03,892292648+01:00.png)
+
+## Friction
+
+-   Arises between rough surfaces and always acts at right angles to the
+    normal reaction force ($R$) in the direction to resist motion.
+-   The maximum value of friction $F$ is $F_{max} = \mu{}R$, where
+    $\mu{}$ is the friction coefficient
+-   $F_{max}$ is also known as the point of slip
+
+## Reactions at Supports
+
+There are three kinds of supports frequently encountered in engineering
+problems:
+
+![](./images/vimscrot-2021-10-04T09:41:56,080077960+01:00.png)
+
+## Principle of Force Transmissibility
+
+A force can be move dalong line of action without affecting equilibrium
+of the body which it acts on:
+
+![](./images/vimscrot-2021-10-04T09:43:04,689667620+01:00.png)
+
+This principle can be useful in determining moments.
+
+## Two-Force Bodies
+
+-   If a body has only 2 forces, then the forces must be collinear,
+    equal, and opposite:
+
+    ![](./images/vimscrot-2021-10-04T09:44:05,581697277+01:00.png)
+
+    > The forces must be collinear so a moment is not created
+
+## Three-Force Bodies
+
+-   If a body in equilibrium has only three forces acting on it, then
+    the lines of actions must go through one point:
+
+    ![](./images/vimscrot-2021-10-04T09:55:59,773394306+01:00.png)
+
+    > This is also to not create a moment
+
+-   The forces must form a closed triangle ($\sum{F} = 0$)
+
+## Naming Conventions
+
+| Term                 | Meaning                                                  |
+|----------------------|----------------------------------------------------------|
+| light                | no mass                                                  |
+| heavy                | body has mass                                            |
+| smooth               | there is no friction                                     |
+| rough                | contact has friction                                     |
+| at the point of slip | one tangential reaction is $F_{max}$                     |
+| roller               | a support only creating normal reaction                  |
+| rigid pin            | a support only providing normal and tangential reactions |
+| built-in             | a support proviting two reaction components and a moment |
+
+## Tips to Solve (Difficult) Problems
+
+1. Make good quality clear and big sketches
+2. Label all forces, dimensions, relevant points
+3. Explain and show your thought process---write complete equations
+4. Follow standard conventions in equations and sketches
+5. Solve everything symbolically (algebraicly) until the end
+6. Check your answers make sense
+7. Don't forget the units
+
+# Lecture L1.4
+
+## Tension and Compression
+
+- The convention in standard mechanical engineering problems is that positive values are for
+  tension and negative values for compression
+- Members in tension can be replaces by cables, which can support tension but not compression
+- Resisting compression is harder as members in compression can buckle
+
+## What is a Pin Joint?
+
+- Pin jointed structures are structures where joints are pinned (free to rotate)
+- Pin joints are represented by a circle (pin) about which members are free to rotate:
+
+  ![](./images/vimscrot-2021-10-18T09:14:14,289274419+01:00.png)
+
+- A pin joint transmits force but cannot carry a moment
+
+## What is a Truss?
+
+- Trusses are an assembly of many bars, which are pin jointed in design but do not rotate due to
+  the geometry of the design. A pylon is a good example of this
+- Trusses are used in engineering to transfer forces through a structure
+- When pin jointed trusses are loaded at the pins, the bars are subjected to pure tensile or
+  compressive forces.
+
+  These bars are two force members
+
+## Equilibrium at the Joints
+
+![](./images/vimscrot-2021-10-18T09:19:44,218728293+01:00.png)
+
+![](./images/vimscrot-2021-10-18T09:20:55,909323283+01:00.png)
+
+### Forces at A
+
+$$\sum F_y(A) = \frac P 2 + T_{AB}\sin{\frac \pi 4} = 0 \rightarrow T_{AB} = -\frac P 2$$
+
+\begin{align*}
+\sum F_x(A) &= T_{AB}\cos{\frac pi 4} + T_{AC} = 0 \\
+T_AC &= -\frac{-P} 2  \times \frac{\sqrt{2}} 2 = \frac P 2
+\end{align*}
+
+### Forces at B
+
+Add the information we just obtained from calculating forces at A:
+
+![](./images/vimscrot-2021-10-18T09:30:57,113760253+01:00.png)
+
+And draw a free body diagram for the forces at B:
+
+![](./images/vimscrot-2021-10-18T09:31:09,513726429+01:00.png)
+
+$$
+\sum F_y(B) = -\frac{-P} 2 \sin{\frac \pi 4} - T_{BC} = 0 \rightarrow
+T_{BC} = \frac P {\sqrt 2} \times \frac {\sqrt 2} 2 = \frac P 2
+$$
+
+\begin{align*}
+\sum F_x(B) &= -\frac{-P}{\sqrt2}\cos{\frac \pi 4} + T_{BD} = 0 \\
+T_{BD} &= -\frac P 2
+\end{align*}
+
+## Symmetry in Stuctures
+
+Symmetry of bar forces in a pin jointed frame depends on to aspects:
+
+1. Symmetry of the stucture
+2. Symmetry of the loading (forces applied)
+
+Both conditions must be met to exploit symmetry.
+
+# Lecture L1.5, L1.6
+
+## Method of Sections
+
+The method of sections is very useful to find a few forces inside a complex structure.
+
+If an entire section is in equilibrium, so are discrete parts of the same structure.
+This means we an isolate substructures and draws free body diagrams for them.
+
+We must add all the forces acting on the substructure.
+Then we make a virtua cut through some of the members, replacing them with forces.
+
+Then we can write 3 equilibrium equations for the substructure:
+
+1. 1 Horizontal, 1 vertical, and 1 moment equation
+2. Either horizonal or vertical and 2 moment equations
+3. 3 moment equations
+
+<details>
+<summary>
+
+### Example 1
+
+</summary>
+
+Draw a virtual cut through the structure, making sure to cut through all the bars whose forces
+you are trying to find:
+
+![](./images/vimscrot-2021-10-18T09:54:16,538381701+01:00.png)
+
+Draw the free body diagram, substituting cut bars by forces:
+
+![](./images/vimscrot-2021-10-18T09:54:44,339423030+01:00.png)
+
+As there are three unknown forces, we need three equilibrium equations.
+
+#### First Equation: Moments about E
+
+\begin{align*}
+\sum M(E) &= \frac P 2 \times 2L + T_{DF}L = 0 \\
+T_{DF} &= -P
+\end{align*}
+
+#### Second Equation: Vertical Equilibrium
+
+\begin{align*}
+\sum F_y &= \frac P 2 + T_{EF}\sin{\frac \pi 4} = 0 \\
+T_{EF} &= -\frac P {\sqrt2}
+\end{align*}
+
+#### Third Equation: Horizontal Equilibrium
+
+\begin{align*}
+\sum F_x = T_{DF} + T_{EF}\cos{\frac \pi 4} + T_{EG} = 0 \\
+T_{EG} = \frac {3P} 2
+\end{align*}
+
+#### Taking Moments from Outside the Structure
+
+If we only needed EG, we could have taken moments about point F, outside our substructure:
+
+![](./images/vimscrot-2021-10-18T10:03:11,111898958+01:00.png)
+
+\begin{align*}
+\sum M(F) &= \frac P 2 \times 3L -T_{EG}L = 0 \\
+T_{EG} &= \frac {3P} 2
+\end{align*}
+
+</details>
+
+## Zero-Force Members
+
+![](./images/vimscrot-2021-10-18T10:24:40,832835425+01:00.png)
+
+Consider the free body diagram for the joint at G:
+
+![](./images/vimscrot-2021-10-18T10:25:19,289848899+01:00.png)
+
+$$\sum F_y(G) = T_{FG} = 0$$
+$$\sum F_x(G) = -T_{EG} + T_{GJ} = 0 \rightarrow T_{EG} = T_{GJ}$$
+
+Meaning that the structure is effecively the same as this one:
+
+![](./images/vimscrot-2021-10-18T10:28:27,972396634+01:00.png)
+
+Why was it there?
+
+- The structure may be designed for other loading patterns
+- The bar may prevent the struture from becoming a mechanism
+- A zero force member may also be there to prevent buckling
+
+## Externally Applied Moments
+
+Externally applied moments are dealt with in the same way as external forces, but they only
+contribute to moment equations and not force equilibrium equations.
+
+## Distributed Load
+
+A distriuted load is applied uniformly to a bar or section of a bar.
+
+It can be represented by a single force through the midpoint its midpoint.
+
+![](./images/vimscrot-2021-10-18T10:40:12,204385456+01:00.png)
+
+<details>
+<summary>
+
+### Example 1
+
+</summary>
+
+![](./images/vimscrot-2021-10-18T10:41:13,789765944+01:00.png)
+
+Is equivalent to:
+
+![](./images/vimscrot-2021-10-18T10:41:49,859420913+01:00.png)
+
+</details>
+
+## Equivalent Loads
+
+When loads are applied within a bar, as far as support reactions and bar forces in *other* bars
+are concered, we can determine *equivalent node forces* using equilibrium
+
+![](./images/vimscrot-2021-10-18T10:44:51,867483875+01:00.png)
+
+# Lecture L1.6
+
+
+<details>
+<summary>
+
+### Example 1
+
+The figure shows a light roof truss loaded by a force $F = 90$ kN at 45\textdegree to the horizontal
+at point $B$.
+
+</summary>
+
+![](./images/vimscrot-2021-10-18T14:42:45,045976929+01:00.png)
+
+a. Find the reaction forces at A and D using equilibrium applied to the whole structure.
+
+   > 1. Add unknown quantities to the diagram
+   >
+   >    ![](./images/vimscrot-2021-10-18T14:44:02,679972841+01:00.png)
+   >
+   > 2. Consider the number of unknowns --- there are 3 therefore 3 equations are needed
+   > 3. Decide which equilibrium equation to start with
+   >
+   >    Horizontal equilibrium:
+   >
+   >    \begin{align*}
+           \sum F_x &= R_{Ax} - F\cos45 = 0 \\
+           R_{Ax} &= 63.6\text{ kN}
+   >    \end{align*}
+   >
+   >    Vertical equilibrium:
+   >
+   >    \begin{align*}
+           \sum F_y &= R_{Ay} + R_{Dy} - F\sin = 0 \\
+           R_{Ay} + R_{Dy} &= \frac{\sqrt2 F} 2
+   >    \end{align*}
+   >
+   >    Moment equation:
+   >
+   >    \begin{align*}
+   >       \sum M_{xy}(B) &= 4.5R_{Ay} - 4.5R_{Dy} - L_{BC}R_{Ax} = 0 \\
+   >       \frac{L_{BC}}{4.5} &= \tan30 = \frac 1 {\sqrt3} \rightarrow L_{BC} = 2.6 \\
+   >    \end{align*}
+   >
+   > 4. Solve for $R_{Ay}$
+   >
+   >    \begin{align*}
+   >       4.5R_{Ay} &= 4.5R_{Dy} + 2.6\times\frac{\sqrt2 F}{2} & R_{Dy} = \frac{\sqrt2 F}{2} - R_{Ay} \\
+   >       4.5R_{Ay} &= 4.5\left(\frac{\sqrt2 F}{2} - R_{Ay}\right) + 2.6\times\frac{\sqrt2 F}{2} \\
+   >       R_{Ay} &= 0.56F = 50.2\text{ kN}
+   >    \end{align*}
+   >
+   > 5. Substitute to find $R_{Dy}$
+   >
+   >    \begin{align*}
+   >       R_{Dy} = \frac{\sqrt2 F}{2} - R_{Ay} = (0.71-0.56)F = 13.3\text{ kN}\\
+   >    \end{align*}
+   
+   
+b. Use the graphical/trigonometric method o check your answer.
+
+   > Write the reaction at A as a single force with unknown direction:
+   >
+   > ![](./images/vimscrot-2021-10-18T15:13:15,369363473+01:00.png)
+   >
+   > When three forces act on an object in equilibrium, they must:
+   >
+   > 1. Make a triangle of forces
+   > 2. Go through a single point
+   > 
+   > So we can figure out the angle of $R_{A}$ by drawing it such that all the lines of action of 
+   > all forces go through the same point:
+   > 
+   > ![](./images/vimscrot-2021-10-18T15:19:58,074989197+01:00.png)
+   > 
+   > \begin{align*}
+       L_{DE} &= L_{BC} = 2.6 \\
+       L_{EG} &= \tan45\times L_{BE} = 4.5 \\
+       L_{DE} &= 2.6+4.5 = 7.1 \\
+       \\
+       \tan\theta &= \frac{L_{DG}}{L_{AD}} = 0.79 \\
+       \theta &= 38.27
+   > \end{align*}
+   >
+   > Now draw the force triangle:
+   > 
+   > ![](./images/vimscrot-2021-10-18T15:30:35,135421357+01:00.png)
+   > 
+   > Using the sine rule we find out $R_A$ and $R_{Dy}$, which are $81.1$ kN and $13.4$ kN,
+   > respectively.
+   >
+   > Now we can check our answers in part (a) and (b) are the same:
+   >
+   > - $R_{Dy} = 13.4$
+   > - $R_{Ax} = 81.1\cos38.27 = 63.6$
+   > - $R_{Ay} = 81.1\sin.27 = 50.2$
+   > 
+   > The methods agree.
+
+</details>
+
+# Lecture L2.1
+
+## Hooke's Law and Young's Modulus
+
+Hooke's law states that the extension of an object experiencing a force is proportonal to the force.
+
+We can generalize this to be more useful creating:
+
+- Direct stress:
+
+  $$ \sigma = \frac F {A_0} $$
+
+- Direct strain:
+
+  $$ \epsilon = \frac {\Delta L}{L_0} $$
+
+Using these more generalized variables, Young defined Young's Modulus, $E$, which is a universal
+constant of stiffness of a material.
+
+$$ \sigma = E\epsilon $$
+
+<details>
+<summary>
+
+#### Example 1
+
+Calculating Young's Modulus of a Piece of Silicone
+
+</summary>
+
+\begin{align*}
+L_0 &= 4.64 \\
+w_0 &= 0.10 \\
+t_0 &= 150\times10^{-6} \\
+F &= 1.40\times9.81 \\
+L &= 6.33 \\
+w &= 0.086 \\
+t &= 125\times10^{-6} \\
+\\
+\sigma &= \frac F {A_0} = \frac F {w_0t_0} = \frac{1.4\times9.81}{0.1\times150\times10^{-6}} = 915600 \\
+\epsilon &= \frac{\Delta L}{L_0} = \frac{6.33 - 4.64}{4.64} = 0.36422...\\
+E &= \frac \sigma \epsilon = 2513836.686 = 2.5\times10^6 \text{ Pa}
+\end{align*}
+
+</details>
+
+## Stress Strain Curves
+
+![](./images/vimscrot-2021-11-01T09:50:51,184232288+00:00.png)
+
+## Poisson's Ratio
+
+For most materials, their cross sectionts change when they are stretched or compressed.
+This is to keep their volume constant.
+
+$$ \epsilon_x = \frac {\Delta L}{L_0} $$
+$$ \epsilon_y = \frac {\Delta w}{w_0} $$
+$$ \epsilon_z = \frac {\Delta t}{t_0} $$
+
+Poisson's ratio, $\nu$ (the greek letter _nu_, not v), is the ratio of lateral strain to axial
+strain:
+
+$$ \nu = \frac{\epsilon_y}{\epsilon_x} = \frac{\epsilon_z}{\epsilon_x} $$
+
+<details>
+<summary>
+
+#### Example 1
+
+Measuring Poisson's Ratio
+
+</summary>
+
+\begin{align*}
+L_0 &= 4.64 \\
+w_0 &= 0.10 \\
+t_0 &= 150\times10^{-6} \\
+\\
+L &= 6.33 \\
+w &= 0.086 \\
+t &= 125\times10^{-6} \\
+\\
+\epsilon_x &= \frac {\Delta L}{L_0} = 0.364 \\
+\epsilon_y &= \frac {\Delta w}{w_0} = -0.14 \\
+\epsilon_z &= \frac {\Delta t}{t_0} = -0.167 \\
+\\
+\nu_y &= \frac{\epsilon_y}{\epsilon_x} = \frac{-0.14}{0.364} = -0.38 \\
+\nu_z &= \frac{\epsilon_z}{\epsilon_x} = \frac{-0.167}{0.364} = -0.46 \\
+\end{align*}
+
+It's supposed to be that $\nu_y = \nu_z$ but I guess it's close enough right? lol
+
+</details>
+
+## Typical Values of Young's Modulus and Poisson's Ratio
+
+Material | Young's Modulus / GPa | Poisson's Ratio
+-------- | --------------------- | ---------------
+Steel    | 210                   | 0.29
+Aluminum | 69                    | 0.34
+Concrete | 14                    | 0.1
+Nylon    | 3                     | 0.4
+Rubber   | 0.01                  | 0.495
+
+## Direct Stresses and Shear Stresses
+
+![](./images/vimscrot-2021-11-01T10:35:47,339443980+00:00.png)
+
+- A direct stress acts normal to the surface
+- A shear stress acts tangential to the surface
+
+Shear stress is defined in the same way as direct stress but given the symbol $tau$ (tau):
+
+$$ \tau = \frac F A $$
+
+Shear strain is defined as the shear angle $\gamma$:
+
+$$ \gamma \approx \tan\gamma = {\frac x {L_0}} $$
+
+The shear modulus, $G$, is like Young's Modulus but for shear forces:
+
+$$ \tau = G\gamma $$
+
+## Relationship between Young's Modulus and Shear Modulus
+
+$$ G = \frac E {2(1+\nu)} $$
+
+$G \approx \frac E 3$ is a good approximation in a lot of engineering cases