Week 12 — Shear Stress & Strain + Stress-Strain Curve

What this week is about

This is the final lecture (CTP1, Week 12). It closes the materials & mechanics block and then opens a new one — shear. Two big ideas:

1. The stress-strain curve, fully labelled. Yield, UTS, fracture stress, plus the energy quantities living under the curve: toughness (the whole area, up to fracture) and modulus of resilience (only the elastic triangle). Strain hardening sits on top of this.
2. Shear — the tangential cousin of axial stress. Direct shear ${\tau }_{avg}=V/A$, single vs double shear bookkeeping, sizing bolted/pin/glue connections, shear strain $\gamma$ as a change in angle, and the shear modulus $G$ with its hard-wired link $G=E/[2(1+\nu )]$.

Every shear formula is a direct analogue of an axial-stress formula you already know. See the in-depth note for the parallel.

Notes in this week

• Lecture reconstruction — the slide-by-slide factual source for everything below (definitions, formulas, six worked examples, slide citations).
• Cheatsheet — every rule, table, and recipe in one scannable page. Includes the quiz (mixed difficulty, reshuffles every visit).
• In-depth analysis — why each formula works, full worked example per topic, the axial-shear analogy, the exam-style template.
• Study guide — what’s directly supported vs inferred, common mistakes, practice questions, confidence report.
• Workshop prep — 5-minute and 20-minute revision plans for Portfolio 11.

Any notes you add to this folder will appear here automatically.

What I need to know before the workshop

• Axial stress $\sigma =N/A$ and axial strain $ϵ=\delta /L$ (from Week 11)
• How to draw a free-body diagram of a bolt or pin and identify the cut plane
• Triangle and rectangle areas (the modulus-of-resilience triangle is just $\frac{1}{2}{\sigma }_y{ϵ}_y$)
• Small-angle approximations: $\sin \gamma \approx \tan \gamma \approx \gamma$ in radians
• Young’s modulus $E$, Poisson’s ratio $\nu$ — both reused in the $G$ relation

Assessment relevance

• Exam: shear stress / strain and stress-strain curve questions are near-certain on the CTP1 paper.
• Portfolio 11: completed in the workshop using this lecture.
• There is no Tutorial 12 PDF — slide 23 directs students to Mastering Physics (Shear Stress and Strain, Material Properties), the workshop for Portfolio 11, and exam-prep. For tutorial-style practice, revisit Week 11.

Overview

This is the final lecture (CTP1, Week 12). It closes the materials/mechanics block by mapping the full stress-strain curve (yield, UTS, fracture, strain hardening) and the energy quantities under it (toughness, modulus of resilience), then introduces shear: direct shear stress, sizing of bolted/pin/glue connections, shear strain (with small-strain approximations), and the shear modulus $G$ with its link to Young’s modulus $E$ and Poisson’s ratio $\nu$.

Key concepts

Term	Symbol	Units	Meaning
Yield stress	${\sigma }_y$	Pa (MPa)	Stress at transition from elastic to plastic behaviour
Ultimate tensile stress	${\sigma }_{UTS}$	Pa (MPa)	Largest stress on the stress-strain curve
Fracture stress	${\sigma }_{failure}$	Pa (MPa)	Stress at which the material breaks
Proportional limit	—	Pa	Highest stress where $\sigma \propto ϵ$
Elastic limit	—	Pa	Highest stress with no permanent deformation
Strain energy	$U$	J	Internal energy stored after deformation
Toughness	—	$J/m^3$	Energy absorbed per unit volume up to fracture (= total area under $\sigma$-$ϵ$ curve)
Modulus of resilience	$u_r$	$J/m^3$	Elastic energy per unit volume (= area under elastic part of curve)
Strain hardening	—	—	Increase in yield strength after unloading from the plastic region; $E$ unchanged
Permanent set	—	—	Plastic strain remaining after unloading
Average shear stress	${\tau }_{avg}$	Pa	Internal shear force per unit shear area
Internal shear force	$V$	N	Component of internal force tangent to cut plane
Single shear	—	—	One shear plane carries the full load: $V=F$
Double shear	—	—	Two parallel shear planes share the load: $V=F/2$
Allowable shear stress	${\tau }_{allow}$	Pa	Design shear stress (failure stress divided by safety factor)
Failure shear stress	${\tau }_{fail}$	Pa	Shear stress at which the material fails
Safety factor	$SF$	—	Dimensionless design margin
Shear strain	$\gamma$	rad	Change in angle between two originally perpendicular lines
Shear modulus	$G$	Pa (GPa)	Elastic constant relating $\tau$ and $\gamma$
Poisson’s ratio	$\nu$	—	Lateral-to-axial strain ratio
Young’s modulus	$E$	Pa (GPa)	Elastic modulus in tension

Note from slide 3: in most engineering materials, the proportional limit, elastic limit and yield point essentially coincide at the yield point. Toughness depends on material class — ceramics small (elastic only), metals large (elastic + plastic), polymers very small (slide 4).

Core formulas

Stress-strain energies (slides 4-5):

$\text{Toughness}={\int }_0^{{ϵ}_f}\sigma dϵ$

$u_r={\int }_0^{{ϵ}_y}\sigma dϵ$

For a linear-elastic region, the modulus of resilience reduces to $u_r=\frac{1}{2}{\sigma }_y{ϵ}_y$ (triangular area).

Direct shear (slide 9):

${\tau }_{avg}=\frac{V}{A}$

with ${\tau }_{avg}$ acting in the same direction as $V$.

Shear-plane bookkeeping (slides 11-12):

• Single shear: $V=F$
• Double shear: $V=\frac{F}{2}$

Sizing of simple connections (slides 13-14):

$A=\frac{V}{{\tau }_{allow}},{\tau }_{allow}=\frac{{\tau }_{fail}}{SF}$

For a rod embedded in concrete with bonding glue: $\ell =\frac{P}{{\tau }_{allow}\pi d}$.

Shear strain (slide 16):

${\gamma }_{nt}=\frac{\pi }{2}-{\lim }_{\begin{array}{c}B\to A\text{ along }n\\ C\to A\text{ along }t\end{array}}{\theta }^{\prime }$

For small deformations:

${\gamma }_{nt}=\frac{\pi }{2}-{\theta }^{\prime }$

Small-strain approximations (slide 18): for $ϵ\ll 1$ and small angles in radians,

$\sin \gamma \approx \gamma ,\cos \gamma \approx 1,\tan \gamma \approx \gamma$

Shear modulus / Hooke’s law in shear (slide 20):

$G=\frac{\tau }{\gamma }$

$G=\frac{E}{2(1+\nu )}$

Reference values (slide 20): Aluminium $E=70,G=25,\nu =0.33$; Brass 97/37/0.34; Copper 110/46/0.34; Nickel 207/76/0.31; Steel 207/83/0.30; Titanium 107/45/0.34; Tungsten 407/160/0.28 (E and G in GPa).

Worked examples

Example 1 — Aluminium strain hardening (slide 7)

Given: Yield strength rises from ${\sigma }_{y,1}=450$ MPa (point A) to ${\sigma }_{y,2}=600$ MPa (point B) after strain hardening. From the graph, ${ϵ}_Y=0.006$ at A and the unload line returns through C/D with the plastically hardened curve reaching $ϵ=0.024$ at B (permanent set ${ϵ}_C\approx 0.024-{ϵ}_{y,2}/E$).

(a) Elastic modulus — slope of the linear portion through A: $E=\frac{{\sigma }_{y,1}}{{ϵ}_Y}=\frac{450\text{MPa}}{0.006}=75,000\text{MPa}=75\text{GPa}$

This matches handbook aluminium ($\approx 70$ GPa).

(b) Modulus of resilience — triangular area under elastic part:

Before strain hardening (${\sigma }_y=450$ MPa): $u_{r,1}=\frac{1}{2}{\sigma }_{y,1}{ϵ}_Y=\frac{1}{2}(450\times 10^6)(0.006)=1.35\times 10^6{\text{J/m}}^3=1.35{\text{MJ/m}}^3$

After strain hardening (${\sigma }_y=600$ MPa, same $E$, so ${ϵ}_y^{\prime }=600/75000=0.008$): $u_{r,2}=\frac{1}{2}{\sigma }_{y,2}{ϵ}_{y,2}=\frac{1}{2}(600\times 10^6)(0.008)=2.40\times 10^6{\text{J/m}}^3=2.40{\text{MJ/m}}^3$

Strain hardening raises both strength and resilience while $E$ stays the same.

Example 2 — Glued double-shear joint (slide 10)

Given: $F=20\text{kN}$, glued vertical surfaces AB and CD with $A_{AB}=A_{CD}=1000{\text{mm}}^2$ (double shear).

Internal shear in each plane: $V=F/2=10\text{kN}$.

${\tau }_{avg}=\frac{V}{A}=\frac{10000\text{N}}{1000\times 10^{-6}{\text{m}}^2}=10\times 10^6\text{Pa}=10\text{MPa}$

Example 3 — Sizing a disk in single shear (slide 15)

Given: Pulling load $P=20\text{kN}$ on a rod with a circular disk passing through a $d_{hole}=40$ mm hole. ${\tau }_{allow}=35\text{MPa}$. Find the minimum disk thickness $t$.

The shear surface is the cylindrical edge of the disk being punched out (height $t$, circumference $\pi d_{hole}$): $A_{shear}=\pi d_{hole}t$

Using ${\tau }_{allow}=V/A$ with $V=P$: $t=\frac{P}{{\tau }_{allow}\pi d_{hole}}=\frac{20,000\text{N}}{(35\times 10^6)(\pi )(0.040)}=\frac{20,000}{4,398,230}\approx 4.55\times 10^{-3}\text{m}$

$\boxed{t_{min}\approx 4.55\text{mm}}$

Example 4 — Shear strain on a deformed plate (slide 17)

Given: Rectangular plate $300\text{mm}\times 400\text{mm}$ deforms to a parallelogram: top side shifts $3$ mm horizontally and the right side compresses $2$ mm vertically (top-right corner). Find ${\gamma }_{xy}$ at A relative to the x/y axes.

Two corner rotations contribute at A. Treating angles as small:

• Rotation of side AB from the y-axis: ${\theta }_1\approx 3/400=0.0075$ rad
• Rotation of side AC from the x-axis: ${\theta }_2\approx 2/300\approx 0.00667$ rad (but only if the bottom side rotates — in this slide the 2 mm shortening is at C, so this term enters through the displacement of C relative to A)

For the configuration shown (only the top edge displaces by 3 mm and the right edge shortens 2 mm at the top): ${\gamma }_{xy}\approx \frac{3}{400}=7.5\times 10^{-3}\text{rad}$

(The 2 mm vertical change is a normal strain in the y-direction, ${ϵ}_y\approx -2/400$, not a shear contribution at A.)

Example 5 — Shear strain via small-strain analysis (slide 19)

Given: Rectangle $ABCD$ with $AB=400$ mm (x-direction) and $AD=300$ mm (y-direction). After deformation: D shifts $+5$ mm in x, and B shifts $-5$ mm in y (the corners labelled — interpret from the dashed parallelogram). Find ${\gamma }_{xy}$ at A.

Side AD rotates about A: angle ${\theta }_{AD}\approx 5/300=0.01667$ rad.

Side AB rotates about A: angle ${\theta }_{AB}\approx 5/400=0.0125$ rad.

${\gamma }_{xy}={\theta }_{AD}+{\theta }_{AB}=0.01667+0.0125=0.02917\text{rad}\approx 29.2\times 10^{-3}\text{rad}$

Example 6 — Shear modulus of a polymer block (slide 21)

Given: Block $400\text{mm}\times 200\text{mm}\times 100\text{mm}$ (width). Top plate displaces $\delta =2$ mm under $P=2$ kN; height $h=200$ mm; plan-view shear area $A=400\text{mm}\times 100\text{mm}=40,000{\text{mm}}^2$.

Shear stress: $\tau =\frac{P}{A}=\frac{2000\text{N}}{40,000\times 10^{-6}{\text{m}}^2}=50,000\text{Pa}=50\text{kPa}$

Shear strain (small-angle): $\gamma \approx \tan \gamma =\frac{\delta }{h}=\frac{2}{200}=0.01\text{rad}$

Shear modulus: $G=\frac{\tau }{\gamma }=\frac{50,000}{0.01}=5\times 10^6\text{Pa}=5\text{MPa}$

This is in the realistic range for elastomers (rubbers and soft polymers typically have $G\sim 1$-$10$ MPa).

Things to practise

There is no Tutorial 12 PDF for Week 12 — the slide deck (slide 23) directs students to: Mastering Physics (Shear Stress and Strain, Material Properties), attend the workshop to complete Portfolio 11, attend the tutorial class, and prepare notes and formula sheets for the final exam.

Recommended self-practice:

1. Revisit the Week 11 tutorial (axial stress/strain, Hooke’s law in tension) — the shear formulas in Week 12 are direct analogues.
2. Re-do Examples 1-6 from this deck without looking at the solutions to verify you can identify single vs double shear, pick the correct shear area, and apply small-angle approximations.
3. From any handbook value of $E$ and $\nu$ in the slide-20 table, recompute $G=E/[2(1+\nu )]$ and check against the listed $G$ (good consistency check).
4. Sketch a stress-strain curve and mark the proportional limit, elastic limit, ${\sigma }_y$, ${\sigma }_{UTS}$, ${\sigma }_{failure}$, the elastic-region triangle (modulus of resilience), and the total area (toughness).
5. Practise sizing problems by varying $SF$: e.g. recompute Example 3 disk thickness with $SF=2$ on a given ${\tau }_{fail}$.

Common pitfalls

• Single vs double shear: always draw the FBD of the fastener and count the number of cut planes. Forgetting the $V=F/2$ for double shear halves the stress.
• Wrong shear area: for a disk being punched, the shear area is the cylindrical edge $\pi dt$, not the disk face $\pi d^2/4$.
• Toughness vs modulus of resilience: toughness is the total area under the curve (to fracture); resilience is only the elastic triangle (to ${ϵ}_y$).
• Strain hardening misconception: $E$ does not change after strain hardening — only the new effective yield strength does. The unloading path is parallel to the original elastic line.
• Angle units: $\gamma$ must be in radians, never degrees, when used in $G=\tau /\gamma$ or in small-angle approximations.
• Sign of $\gamma$: shear strain is negative when the deformed angle ${\theta }^{\prime }>\pi /2$ (the originally right angle opens up).
• Units in resilience/toughness: $\sigma$ in Pa × $ϵ$ (dimensionless) gives J/m^3, not J. Convert MPa carefully.
• Confusing ${\tau }_{allow}$ and ${\tau }_{fail}$: design with ${\tau }_{allow}={\tau }_{fail}/SF$; never plug ${\tau }_{fail}$ directly into the sizing formula.

Source citations

• Slide 1 — Lecture title: “Lecture 12: Shear stress and strain + Mechanical Properties Shear”.
• Slide 3 — Stress-strain curve definitions (${\sigma }_y$, ${\sigma }_{UTS}$, fracture stress; proportional/elastic limits coincide at yield).
• Slide 4 — Strain energy, toughness, integral formula, material-class behaviour (ceramics/metals/polymers).
• Slide 5 — Modulus of resilience definition and integral.
• Slide 6 — Strain hardening: unload parallel to elastic line; permanent set + elastic recovery; $E$ unchanged.
• Slide 7 — Example 1 (aluminium alloy 450 → 600 MPa, graph with ${ϵ}_Y=0.006$, point at 0.024).
• Slide 9 — Direct shear definition; ${\tau }_{avg}=V/A$; double shear note.
• Slide 10 — Example 2 (glued joint, $F=20$ kN, $A=1000$ mm²).
• Slide 11 — Double shear examples and relation $V=F/2$.
• Slide 12 — Single shear ($V=F$), ${\tau }_{avg}=V/A$.
• Slide 13 — Sizing of simple connections: $A=V/{\tau }_{allow}$, ${\tau }_{allow}={\tau }_{fail}/SF$.
• Slide 14 — Lap joint and embedded-rod sizing $\ell =P/({\tau }_{allow}\pi d)$.
• Slide 15 — Example 3 (rod through 40 mm hole, $P=20$ kN, ${\tau }_{allow}=35$ MPa).
• Slide 16 — Shear strain definition, limit form, small-deformation form ${\gamma }_{nt}=\pi /2-{\theta }^{\prime }$, sign convention.
• Slide 17 — Example 4 (300×400 mm plate, 3 mm + 2 mm displacements).
• Slide 18 — Small-strain analysis approximations.
• Slide 19 — Example 5 (400×300 mm plate, 5 mm + 5 mm displacements).
• Slide 20 — Shear modulus $G=\tau /\gamma$, $G=E/[2(1+\nu )]$, and material table.
• Slide 21 — Example 6 (polymer block 400×200×100 mm, $\delta =2$ mm, $P=2$ kN).
• Slide 23 — Week 12 activities: Mastering Physics, Portfolio 11, tutorial, exam prep.
• Slide 24 — Wolfson, Essential University Physics (4th ed., SI Units), Pearson Prentice Hall, 2020.