Week 11 In-Depth — Why Stress, Strain, and the Tensile Test Tell You Everything

Read the cheatsheet first. This note explains why each idea works and walks through one substantial worked example per topic, with every step on the page.

1 — Why stress and strain exist as separate ideas

A force on a member tells you something about the load, not the material. A 10 kN load tearing through a paperclip and a 10 kN load barely tickling a steel beam are the same force — but obviously a very different experience for the material.

Stress fixes that. By dividing force by cross-sectional area, $\sigma =P/A$, you get an intensity of internal force. Now 10 kN through a 1 mm² paperclip ($10000\text{MPa}$) and 10 kN through a 500 mm² beam ($20\text{MPa}$) read as different severities, which is what the material actually responds to.

Strain does the equivalent for deformation. A 1 mm stretch is dramatic on a 10 mm bar (10%) and invisible on a 10 m bar (0.01%). Dividing by $L_0$ gives a dimensionless intensity that is comparable across specimen sizes:

$\epsilon =\frac{\Delta L}{L_0}.$

Both $\sigma$ and $\epsilon$ are per-unit quantities. That’s why only $\sigma$ and $\epsilon$ — not $P$ and $\Delta L$ — describe a material’s intrinsic behaviour. The tensile-test curve plotted as $\sigma$ vs. $\epsilon$ is geometry-independent.

The unit trap

Combination	Result	Pitfall
$P$ in N, $A$ in m²	Pa	Fine
$P$ in N, $A$ in mm²	MPa	Fine — most engineering use
$P$ in kN, $A$ in mm²	GPa	Off by 1000 — biggest classroom mistake
$P$ in kN, $A$ in m²	kPa	Off by 1000 — same mistake the other way

Lecture slide 5 reads: “Force is ALWAYS in newtons.” Internalise that.

Worked example — two-rod lamp (slide 7)

An 80 kg lamp hangs from joint $B$. Rod $AB$ runs up-left at $30°$ above horizontal ($d=10\text{mm}$); rod $BC$ runs up-right along a 3-4-5 triangle ($d=8\text{mm}$). Find the average normal stress in each rod.

Step 1 — Weight at $B$. $W=mg=80(9.81)=784.8\text{N}$.

Step 2 — Equilibrium at $B$. Let $F_{AB}$, $F_{BC}$ be tensions directed away from $B$:

$\sum F_x=0:-F_{AB}\cos 30°+F_{BC}\frac{4}{5}=0$ $\sum F_y=0:F_{AB}\sin 30°+F_{BC}\frac{3}{5}-784.8=0$

From $\sum F_x$: $F_{AB}=F_{BC}(4/5)/\cos 30°=0.9238F_{BC}$. Substitute:

$0.9238F_{BC}(0.5)+0.6F_{BC}=784.8⟹F_{BC}\approx 739\text{N},F_{AB}\approx 683\text{N}.$

Step 3 — Areas. $A_{AB}=\frac{\pi }{4}(10{)}^2=78.54{\text{mm}}^2$; $A_{BC}=\frac{\pi }{4}(8{)}^2=50.27{\text{mm}}^2$.

Step 4 — Stresses.

${\sigma }_{AB}=\frac{683}{78.54}\approx 8.69\text{MPa},{\sigma }_{BC}=\frac{739}{50.27}\approx 14.7\text{MPa}.$

The thinner, more loaded rod $BC$ carries more stress — exactly the kind of trade-off the design check is meant to catch.

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2 — Why Young’s modulus is the slope of the elastic line

Plot $\sigma$ vs. $\epsilon$ for a metal in a tensile test. The first portion is straight. Straight means proportional: doubling the stress doubles the strain. The slope is a single number characterising the material:

$E=\frac{\Delta \sigma }{\Delta \epsilon }.$

That’s Hooke’s law: $\sigma =E\epsilon$.

$E$ is stiffness, not strength. A material can have a huge $E$ (very stiff — small strain for large stress) but yield at a modest stress. Steel and rubber are at opposite ends.

The engineering-vs.-true distinction

The cross-section $A_0$ measured before loading is what you have on the data sheet. As the bar stretches, it actually narrows — $A<A_0$. So:

• Engineering stress ${\sigma }_{\text{eng}}=F/A_0$ — uses initial area. Easy to compute; what every standard tensile test reports.
• True stress ${\sigma }_{\text{true}}=F/A$ — uses instantaneous area. Requires you to know how $A$ shrinks.

Before necking, the two are nearly identical. After necking (slide 13), the engineering curve falls because $A_0$ is now an overestimate, while the true stress curve keeps rising because $A$ is shrinking fast. The material is not “weakening” near fracture — it’s just that the engineering plot misrepresents area.

Worked example — find $E$ from a tensile test

From Tutorial 11 Exercise 2: bar with $L_0=125\text{mm}$, $A=437.5{\text{mm}}^2$, axial force $40\text{kN}$, stretch $0.05\text{mm}$.

Stress. $\sigma =40000/437.5=91.43\text{MPa}$.

Strain. $\epsilon =0.05/125=4.0\times 10^{-4}$.

Modulus.

$E=\frac{\sigma }{\epsilon }=\frac{91.43\text{MPa}}{4.0\times 10^{-4}}=2.286\times 10^5\text{MPa}\approx 229\text{GPa}.$

That value sits in the steel range. Sanity check: had we gotten $229\text{MPa}$, we’d suspect a 1000-factor unit slip.

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3 — Why the $\sigma$–$\epsilon$ curve has the shape it does

The curve has three regimes that correspond to three physical behaviours of the metal’s crystal lattice:

Regime	Physical picture	What changes
Linear elastic	Atomic bonds stretch reversibly	Slope $=E$
Yield + plastic	Dislocations begin to slip on slip planes; lattice rearranges permanently	Curve flattens, then rises gently (strain hardening)
Necking + fracture	Local cross-section thins faster than strain hardens	Engineering stress drops; true stress climbs; bar breaks

This is why the yield point matters in design: cross it, and the part returns deformed when the load is released.

The 0.2% offset method (slide 16)

For metals where the yield knee is gradual, “where the curve becomes nonlinear” is too ambiguous. Convention: pick the strain $\epsilon =0.002$ ($0.2\%$). Draw a line from there parallel to the elastic slope. Where that line intersects the curve = offset yield stress ${\sigma }_y$.

The interpretation: this is the stress at which the material has accumulated $0.2\%$ permanent plastic strain, which is industry-agreed as “definitely yielded.”

Ductility — %EL and %AR

After fracture you measure two things:

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

%EL is plastic strain at fracture as a percentage; %AR captures how much necking occurred. Mild steel sits at %EL $\approx 25\%$; cast iron at $<1\%$.

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4 — Why we use a factor of safety

Real materials have:

• Property variation between batches.
• Imperfections, voids, surface flaws.
• Uncertainty in the applied load (live + dead + impact).
• Long-term degradation (fatigue, corrosion).

Engineering practice is to design so that the allowable stress in service is well below the failure stress:

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

You design to ${\sigma }_{\text{allow}}$, not to ${\sigma }_{\text{failure}}$. The bigger the uncertainty, the bigger $SF$.

Worked example — cable car (slide 20)

Cable car mass $2500\text{kg}$; steel cable $d=12\text{mm}$, $L_0=2.4\text{m}$, $E=207\text{GPa}$, ${\sigma }_y=300\text{MPa}$, $SF=1.2$. Each person $65\text{kg}$.

(a) Strain in the cable. $W=2500(9.81)=24525\text{N}$. $A=\frac{\pi }{4}(12{)}^2=113.10{\text{mm}}^2$.

$\sigma =\frac{24525}{113.10}=216.84\text{MPa},\epsilon =\frac{\sigma }{E}=\frac{216.84\times 10^6}{207\times 10^9}\approx 1.05\times 10^{-3}.$

(b) Extension. $\Delta L=\epsilon L_0=0.00105\times 2400\approx 2.51\text{mm}$.

(c) Number of passengers.

${\sigma }_{\text{allow}}=\frac{300}{1.2}=250\text{MPa},F_{\text{allow}}=250\times 113.10=28275\text{N}.$

Spare capacity $=28275-24525=3750\text{N}$. Per person $=65(9.81)=637.65\text{N}$.

$n=\frac{3750}{637.65}\approx 5.88⟹\boxed{n=5\text{ people}}.$

Always round down. Six people would push the cable past its allowable stress, which is the whole point of $SF$.

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5 — Why Poisson’s ratio exists and why it has a minus sign

Stretch a rubber band: it gets longer and thinner. Most solids do this. The longitudinal strain ${\epsilon }_{\text{long}}$ is positive in tension; the lateral strain ${\epsilon }_{\text{lat}}$ is negative (the diameter shrinks).

If we defined $\nu ={\epsilon }_{\text{lat}}/{\epsilon }_{\text{long}}$ raw, we’d always get a negative number for normal materials — annoying. So the convention puts a minus sign in:

$\nu =-\frac{{\epsilon }_{\text{lat}}}{{\epsilon }_{\text{long}}}.$

That makes $\nu >0$ for normal materials. The theoretical bound for an isotropic linear-elastic material is $0\le \nu \le 0.5$. At $\nu =0.5$ the material is incompressible (volume preserved exactly under strain); rubber comes close. Most metals are around $0.3$.

Worked example — cable car continued (slide 23)

Same cable as above with $\nu =0.30$, ${\epsilon }_{\text{long}}=0.00104$.

Lateral strain. ${\epsilon }_{\text{lat}}=-\nu {\epsilon }_{\text{long}}=-0.30(0.00104)=-3.12\times 10^{-4}$.

Diameter change. $\Delta d={\epsilon }_{\text{lat}}{d}_0=-3.12\times 10^{-4}(12)=-3.744\times 10^{-3}\text{mm}$.

New diameter. $d_f=12-0.003744=11.9963\text{mm}$.

New cross-section. $A_f=\frac{\pi }{4}(11.9963{)}^2=113.027{\text{mm}}^2$.

Change. $\Delta A=A_f-A_0\approx -0.0706{\text{mm}}^2$. About $0.07{\text{mm}}^2$ reduction — small but measurable.

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6 — How these ideas connect

Concept	What’s it doing?
Stress + strain	The two per-unit intensities; everything else is computed from them.
Hooke’s law / $E$	The slope of $\sigma$–$\epsilon$ in the linear region; the material’s stiffness.
Yield, UTS, fracture	Successive milestones on the same curve; each tells you a different design limit.
Ductility (%EL, %AR)	A shape descriptor for the curve — how much plastic strain before failure.
Factor of safety	A margin drawn back from the failure stress, by which you should design.
Poisson’s ratio	The lateral-deformation companion to axial strain.

Once you’ve got the diagram in your head, every property is just “which feature of the curve do you want today?“

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7 — Exam-style sample, end-to-end

An aluminium rod, $L_0=500\text{mm}$, $d_0=40\text{mm}$, $E=70\text{GPa}$, $\nu =0.33$, is loaded by an axial force $P=100\text{kN}$. Find (a) the axial stress, (b) the elongation, and (c) the change in diameter.

(a) Axial stress.

$A=\frac{\pi }{4}(40{)}^2=1256.64{\text{mm}}^2.$ $\sigma =\frac{100000}{1256.64}\approx 79.58\text{MPa}.$

(b) Elongation. Longitudinal strain:

${\epsilon }_{\text{long}}=\frac{\sigma }{E}=\frac{79.58\times 10^6}{70\times 10^9}\approx 1.137\times 10^{-3}.$ $\Delta L={\epsilon }_{\text{long}}{L}_0=1.137\times 10^{-3}(500)\approx 0.569\text{mm}.$

(c) Diameter change. Lateral strain:

${\epsilon }_{\text{lat}}=-\nu {\epsilon }_{\text{long}}=-0.33(1.137\times 10^{-3})\approx -3.75\times 10^{-4}.$ $\Delta d={\epsilon }_{\text{lat}}{d}_0=-3.75\times 10^{-4}(40)\approx -0.0150\text{mm}.$

Answer. Stress $\approx 79.6\text{MPa}$; elongation $\approx 0.57\text{mm}$; diameter shrinks by $\approx 0.015\text{mm}$.

That layout — Geometry → $\sigma$ → ${\epsilon }_{\text{long}}\to \Delta L$ → ${\epsilon }_{\text{lat}}\to \Delta d$ — is the template for every multi-part tensile question on the paper.

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Where to look next

• The cheatsheet for fast revision + the quiz.
• The lecture summary for slide-by-slide reconstruction with citations.
• Tutorial 11 worksheet (Exercises 1–6) for further drill.