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