448 lines
12 KiB
Markdown
Executable File
448 lines
12 KiB
Markdown
Executable File
---
|
|
author: Alvie Rahman
|
|
date: \today
|
|
tags:
|
|
- uni
|
|
- nottingham
|
|
- 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>
|
|
|
|
|
|
|
|
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:
|
|
|
|
|
|
|
|
## 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:
|
|
|
|
|
|
|
|
## Principle of Force Transmissibility
|
|
|
|
A force can be move dalong line of action without affecting equilibrium
|
|
of the body which it acts on:
|
|
|
|
|
|
|
|
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:
|
|
|
|
|
|
|
|
> 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:
|
|
|
|
|
|
|
|
> 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:
|
|
|
|
|
|
|
|
- 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
|
|
|
|
<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:
|
|
|
|
|
|
|
|
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*}
|
|
|
|
</details>
|
|
|
|
## 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.
|
|
|
|
|
|
|
|
<details>
|
|
<summary>
|
|
|
|
### Example 1
|
|
|
|
</summary>
|
|
|
|
|
|
|
|
Is equivalent to:
|
|
|
|
|
|
|
|
</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
|
|
|
|
|
|
|
|
# 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>
|
|
|
|
|
|
|
|
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.
|
|
|
|
</details>
|
|
|