Twenty minutes or less.

5-minute version

Five layers. One sentence each.

• Stress — $\sigma =P/A$. Force in newtons, area in mm², answer in MPa.
• Strain — $\epsilon =\Delta L/L_0$. Dimensionless or percent (remember $0.5\%=0.005$).
• Hooke’s law — $\sigma =E\epsilon$ in the linear region; $E$ is the slope.
• Factor of safety — ${\sigma }_{\text{allow}}={\sigma }_{\text{failure}}/SF$. Round people down.
• Poisson’s ratio — ${\epsilon }_{\text{lat}}=-\nu {\epsilon }_{\text{long}}$. Stretching makes things narrower.

Open the cheatsheet quiz, do 3 easy questions, close it. You’re prepped.

20-minute prep plan

Time	Action
0–5 min	Skim the cheatsheet tables (definitions, formulas, $SF$ ranges, $\nu$ values).
5–10 min	Work Tutorial 11 Exercise 2 (find $E$) longhand. Sanity-check units.
10–15 min	Take the cheatsheet quiz. Don’t worry about score.
15–20 min	Read the in-depth note’s cable-car worked example + the aluminium rod sample.

What to revise first

Most students slip on three specific things in this week:

1. Unit conversion. $1\text{MPa}=1{\text{N/mm}}^2$. If your steel $E$ comes out at $\sim 200\text{MPa}$, you’re off by 1000 — it should be GPa.
2. 0.5% strain. Convert percentages before multiplying by $L_0$.
3. The minus sign on Poisson’s ratio. The definition has it built in; $\nu$ comes out positive for normal materials.

Key formulas

$\sigma =\frac{P}{A}\epsilon =\frac{\Delta L}{L_0}\sigma =E\epsilon$

$\%EL=\frac{L_f-L_0}{L_0}\times 100\%AR=\frac{A_f-A_0}{A_0}\times 100$

$SF=\frac{{\sigma }_{\text{failure}}}{{\sigma }_{\text{allowable}}}{\sigma }_{\text{allow}}=\frac{{\sigma }_{\text{failure}}}{SF}$

$\nu =-\frac{{\epsilon }_{\text{lat}}}{{\epsilon }_{\text{long}}}\Delta d={\epsilon }_{\text{lat}}{d}_0$

Likely workshop tasks

Task type	What the setup usually looks like
Direct stress	Force $P$ + diameter $d$ → $\sigma$ in MPa
Modulus from data	Pair of $(\sigma ,\epsilon )$ or $(F,\Delta L)$ → $E$
Read the diagram	Plot of $\sigma$–$\epsilon$ → slope = $E$; 0.2% offset → ${\sigma }_y$
Free-body joint	Loads at angles ($30°$, $45°$, $60°$ or 3-4-5) → forces in members → stress in each
Safety-factor sizing	Given ${\sigma }_{\text{failure}}$, $SF$, and $F$ → minimum $d$
Combined cable problem	$\sigma$, $\epsilon$, $\Delta L$, allowable load, passengers, then $\Delta d$ via Poisson

Mistakes to avoid

• Mixing kN with mm² — answer ends up in GPa.
• Reporting $E$ in MPa for steel/aluminium — should be GPa.
• Treating $0.5\%$ as $0.5$ instead of $0.005$.
• Forgetting the minus sign in Poisson’s ratio.
• Confusing yield, UTS, and fracture stress.
• Rounding allowable-load passenger counts up.
• Mis-assigning trig sides in a 3-4-5 joint problem.

Mini self-test

Try these without notes. Five minutes total.

1. Bar with $P=25\text{kN}$, $d=16\text{mm}$. Find $\sigma$ in MPa.
2. A steel rod ($E=200\text{GPa}$) under $\sigma =120\text{MPa}$ stretches by how many per-mille?
3. A wire with ${\sigma }_{\text{failure}}=350\text{MPa}$, $SF=2.5$, must carry $1.5\text{kN}$. Minimum diameter?

Answers:

Question	Working	Answer
1	$A=\pi (16{)}^2/4=201.06{\text{mm}}^2$; $\sigma =25000/201.06$	$\sigma \approx 124.3\text{MPa}$
2	$\epsilon =\sigma /E=(120\times 10^6)/(200\times 10^9)$	$\epsilon =6.0\times 10^{-4}$ (i.e. $0.6‰$)
3	${\sigma }_{\text{allow}}=350/2.5=140\text{MPa}$; $A=1500/140=10.71{\text{mm}}^2$; $d=\sqrt{4A/\pi }$	$d_{\min }\approx 3.69\text{mm}$

Done checklist

• Read the cheatsheet tables.
• One tutorial exercise copied out longhand.
• Cheatsheet quiz attempted.
• Mini self-test attempted.
• Confirmed you know which way to round when sizing for ${\sigma }_{\text{allow}}$.

That’s it. Close the laptop.

Source files used

• EGD102-Physics/Lecture11_CTP1.pdf
• EGD102-Physics/Tutorial 11.pdf