big rename

uni/mmme/1028_statics_and_dynamics/statics.md

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+---
+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