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

How this week breaks down

Two big blocks. Skim this once, then revise from the in-depth note.

Block	What you do
Stress-strain curve	Identify yield / UTS / fracture; compute toughness and modulus of resilience; understand strain hardening (unload parallel, $E$ unchanged).
Shear	Cut the body, identify the shear plane, compute $\tau =V/A$. Size connections via ${\tau }_{allow}={\tau }_{fail}/SF$. Use $\gamma =\pi /2-{\theta }^{\prime }$ for strain and $G=\tau /\gamma$ for stiffness.

Every shear formula mirrors an axial one — see the analogy below.

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1 — Stress-strain curve

Key points on the curve

Symbol	Name	What it is
${\sigma }_y$	Yield stress	Elastic → plastic transition
${\sigma }_{UTS}$	Ultimate tensile stress	Peak of the curve
${\sigma }_{failure}$	Fracture stress	Stress at the break
Proportional limit	—	Highest stress where $\sigma \propto ϵ$
Elastic limit	—	Highest stress with no permanent set

For most engineering metals the proportional limit, elastic limit and yield point essentially coincide at ${\sigma }_y$ (slide 3).

Energies under the curve

$\text{Toughness}={\int }_0^{{ϵ}_f}\sigma dϵ{\text{(units: J/m}}^3\text{)}$

$u_r={\int }_0^{{ϵ}_y}\sigma dϵ\text{(modulus of resilience)}$

For a linear elastic region:

$\boxed{u_r=\frac{1}{2}{\sigma }_y{ϵ}_y}$

Material class	Toughness behaviour
Ceramics	Small (elastic only, brittle)
Metals	Large (elastic + extensive plastic)
Polymers	Often very small

Strain hardening

Unload from the plastic region → return along a line parallel to the original elastic slope. After reloading, the new effective yield stress is higher but $E$ is unchanged.

Before / after	${\sigma }_y$	${ϵ}_y$	$u_r$	$E$
Before	$450$ MPa	$0.006$	$1.35$ MJ/m³	$75$ GPa
After	$600$ MPa	$0.008$	$2.40$ MJ/m³	$75$ GPa

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2 — Direct shear

The core formula

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

with $V$ tangent to the cut plane and ${\tau }_{avg}$ acting in the same direction as $V$.

Single vs double shear

Configuration	Cut planes	Internal shear $V$ per plane
Single shear	1	$V=F$
Double shear	2	$V=F/2$

Always draw the FBD of the fastener and count the cut planes.

Sizing simple connections

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

Rod glued/embedded over length $\ell$ (cylindrical shear surface, $\pi d\ell$):

$\ell =\frac{P}{{\tau }_{allow}\pi d}$

Disk being punched through a hole (cylindrical edge sheared, height $t$, circumference $\pi d$):

$t=\frac{P}{{\tau }_{allow}\pi d}$

Same formula structure — both shear surfaces are cylindrical edges of area $\pi d\cdot (\text{height})$.

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3 — Shear strain

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

where ${\theta }^{\prime }$ is the deformed angle between two originally perpendicular lines. Units: radians.

Small-strain approximations (for $\gamma \ll 1$)

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

So for a top plate displaced $\delta$ over height $h$:

$\gamma \approx \frac{\delta }{h}$

Common configurations

Setup	${\gamma }_{xy}$ contribution at the corner
Top edge shifts $\delta$ horizontally (height $h$)	$\delta /h$
Side edge shifts $\delta$ vertically (length $L$)	$\delta /L$
Both edges rotate (parallelogram)	sum of both rotations
Normal compression/extension on one edge	not a shear contribution (that’s $ϵ$)

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4 — Shear modulus and the $E$-$G$-$\nu$ relation

Hooke’s law in shear:

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

For an isotropic linear-elastic material:

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

Handbook values (slide 20)

Material	$E$ (GPa)	$G$ (GPa)	$\nu$
Aluminium	70	25	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

(Useful consistency check: plug $E$ and $\nu$ into $G=E/[2(1+\nu )]$ and verify against the table.)

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5 — Axial ↔ shear analogy

Quantity	Axial	Shear
Stress	$\sigma =N/A$	$\tau =V/A$
Strain	$ϵ=\delta /L$	$\gamma =\delta /h$ (small)
Hooke’s law	$\sigma =Eϵ$	$\tau =G\gamma$
Elastic modulus	$E$	$G$
Energy density	$u=\frac{1}{2}\sigma ϵ$	$u=\frac{1}{2}\tau \gamma$

Same machinery, different direction of the internal force.

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Common mistakes

1. Counting cut planes wrong. Single vs double shear changes $V$ by a factor of 2. Always draw the FBD.
2. Wrong shear area. For a punched disk, the shear surface is the cylindrical edge $\pi dt$, not the disk face $\pi d^2/4$.
3. Confusing toughness and resilience. Toughness = whole area (to fracture). Resilience = elastic triangle only (to ${\sigma }_y$).
4. Strain hardening misconception. $E$ does not change — only the new effective yield strength does.
5. Angle units. $\gamma$ must be in radians, never degrees, for $G=\tau /\gamma$ and small-angle approximations.
6. Plugging ${\tau }_{fail}$ into the sizing formula. Always design with ${\tau }_{allow}={\tau }_{fail}/SF$.
7. Mistaking normal strain for shear. A vertical compression on a vertical edge is ${ϵ}_y$, not ${\gamma }_{xy}$.
8. Resilience/toughness units. $\sigma (\text{Pa})\times ϵ(\text{dimensionless})={\text{J/m}}^3$, not J.

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Key formulas

$${\tau }_{avg}=\frac{V}{A}{\tau }_{allow}=\frac{{\tau }_{fail}}{SF}{A}_{\min }=\frac{V}{{\tau }_{allow}}$$ $$\gamma =\frac{\pi }{2}-{\theta }^{\prime }\text{(small: }\gamma \approx \delta /h\text{)}\tau =G\gamma G=\frac{E}{2(1+\nu )}$$ $$u_r=\frac{1}{2}{\sigma }_y{ϵ}_y\text{Toughness}={\int }_0^{{ϵ}_f}\sigma dϵ$$

For the why and worked examples, see the in-depth note.