diff --git a/mechanical/mmme1028_statics.md b/mechanical/mmme1028_statics.md
index 8598afc..bb0a0d9 100755
--- a/mechanical/mmme1028_statics.md
+++ b/mechanical/mmme1028_statics.md
@@ -8,7 +8,7 @@ tags:
- maths
- statics
- dynamics
-title: MMME1028 // Statics and Dynamics
+title: MMME1028 // Statics
---
# Lecture L1.1, L1.2
@@ -38,10 +38,14 @@ Can be found here [here](./lecture_exercises/mmme1028_l1.2_exercises_2021-10-04.
- Body is in equilibrium if sum of all forces and moments acting on
body are 0
+
+
### Example
Determine force $F$ and $x$ so that the body is in equilibrium.
+
+
1. Check horizontal equilibrium
@@ -62,6 +66,8 @@ Determine force $F$ and $x$ so that the body is in equilibrium.
$x = 6$
+
+
## Free Body Diagrams
A free body diagram is a diagram of a single (free) body which shows all
@@ -138,3 +144,304 @@ This principle can be useful in determining moments.
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:
+
+ 
+
+- 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
+
+
+
+
+
+### 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:
+
+
+
+And draw a free body diagram for the forces at B:
+
+
+
+$$
+\sum F_y(B) = -\frac{-P} 2 \sin{\frac \pi 4} - T_{BC} = 0 \rightarrow
+T_{BC} = \frac P {\sqrt 2} \times \frac {\sqrt 2} 2 = \frac P 2
+$$
+
+\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
+
+
+
+
+### Example 1
+
+
+
+Draw a virtual cut through the structure, making sure to cut through all the bars whose forces
+you are trying to find:
+
+
+
+Draw the free body diagram, substituting cut bars by forces:
+
+
+
+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:
+
+
+
+\begin{align*}
+\sum M(F) &= \frac P 2 \times 3L -T_{EG}L = 0 \\
+T_{EG} &= \frac {3P} 2
+\end{align*}
+
+
+
+## Zero-Force Members
+
+
+
+Consider the free body diagram for the joint at G:
+
+
+
+$$\sum F_y(G) = T_{FG} = 0$$
+$$\sum F_x(G) = -T_{EG} + T_{GJ} = 0 \rightarrow T_{EG} = T_{GJ}$$
+
+Meaning that the structure is effecively the same as this one:
+
+
+
+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.
+
+
+
+
+
+
+### Example 1
+
+
+
+
+
+Is equivalent to:
+
+
+
+
+
+## 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
+
+
+
+# Lecture L1.6
+
+
+
+
+
+### Example 1
+
+The figure shows a light roof truss oaded by a force $F = 90$ kN at 45\textdegree to the horizontal
+at point $B$.
+
+
+
+
+
+a. Find the reaction forces at A and D using equilibrium applied to the whole structure.
+
+ > 1. Add unknown quantities to the diagram
+ >
+ > 
+ >
+ > 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:
+ >
+ > 
+ >
+ > 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:
+ >
+ > 
+ >
+ > \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:
+ >
+ > 
+ >
+ > 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.
+
+
+