Week 12 In-Depth — Why Shear Behaves Like Axial Stress, Just Sideways

Read the cheatsheet first. This note explains why each formula works, walks through one substantial worked example per topic, and shows how the whole week is really one idea — axial mechanics with the force component rotated 90°.

1 — Why the stress-strain curve has the shape it does

The stress-strain curve is a fingerprint of how a material gives way under load. Up to the yield stress ${\sigma }_y$, atomic bonds stretch elastically — every unit of strain costs $E$ units of stress, and unloading returns to the origin. Past ${\sigma }_y$, dislocations begin to move. Now strain accumulates without unloading. The material is plastically deformed.

Three landmark stresses sit on this curve:

• Yield stress ${\sigma }_y$ — onset of plasticity.
• Ultimate tensile stress ${\sigma }_{UTS}$ — the peak. Past this point necking begins and the geometry of the specimen, not its material, dominates.
• Fracture stress ${\sigma }_{failure}$ — the specimen breaks.

The proportional limit (where $\sigma \propto ϵ$ stops holding) and the elastic limit (where unloading stops being reversible) live just below ${\sigma }_y$. For most engineering metals all three are close enough to be treated as one point — this is what slide 3 means by “essentially coincide at the yield point.”

Why toughness and resilience are areas

Work done per unit volume on a uniaxial specimen is $\sigma dϵ$. So the total work per unit volume from undeformed to fracture is the integral:

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

This is exactly the area under the curve. Likewise, the elastic work — the energy you’d get back if you unloaded before plastic flow — is the area only out to ${ϵ}_y$:

$$u_r={\int }_0^{{ϵ}_y}\sigma dϵ$$

In the linear-elastic region $\sigma =Eϵ$, so the integral is the triangle from $(0,0)$ to $({ϵ}_y,{\sigma }_y)$:

$$u_r=\frac{1}{2}{\sigma }_y{ϵ}_y=\frac{{\sigma }_y^2}{2E}.$$

Units: $\sigma (\text{Pa})\times ϵ(\text{dimensionless})={\text{J/m}}^3$. Energy density, not energy.

That explains why ceramics are tough only in a microscopic sense (almost no plastic region, small area) and metals are tough macroscopically (huge plastic region under the curve) — the total area is just a lot bigger.

Worked example — strain hardening of an aluminium alloy (slide 7)

A sample yields at ${\sigma }_{y,1}=450$ MPa with ${ϵ}_y=0.006$. After being loaded into the plastic region and unloaded, the new yield stress is ${\sigma }_{y,2}=600$ MPa. Find $E$, and the modulus of resilience before and after.

Step 1 — $E$ from the elastic slope.

$$E=\frac{{\sigma }_{y,1}}{{ϵ}_y}=\frac{450\text{MPa}}{0.006}=75,000\text{MPa}=75\text{GPa}.$$

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

Step 2 — Resilience before strain hardening.

$$u_{r,1}=\frac{1}{2}{\sigma }_{y,1}{ϵ}_y=\frac{1}{2}(450\times 10^6)(0.006)=1.35{\text{MJ/m}}^3.$$

Step 3 — Resilience after strain hardening. $E$ is unchanged, so the new yield strain is

$${ϵ}_y^{\prime }=\frac{{\sigma }_{y,2}}{E}=\frac{600}{75,000}=\mathrm{0.008.}$$ $$u_{r,2}=\frac{1}{2}(600\times 10^6)(0.008)=2.40{\text{MJ/m}}^3.$$

The crucial conceptual point: strain hardening raises both strength and resilience, even though it doesn’t change the elastic modulus. The triangle gets taller and wider, so its area roughly doubles. The unload line, however, stays parallel to the original elastic line — that’s the geometric statement that $E$ is unchanged.

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2 — Why $\tau =V/A$ is the shear stress, not $N/A$

Cut a body with a plane and resolve the internal force into two components:

• Normal component $N$ — perpendicular to the cut plane → produces normal (axial) stress $\sigma =N/A$.
• Tangential component $V$ — lying in the cut plane → produces shear stress $\tau =V/A$.

That’s the whole definition. Shear stress is what happens when the internal force has a component sliding along the surface rather than pushing through it.

Why “single” vs “double” shear changes the formula

Draw the FBD of a pin holding two plates together. If both plates pull on the pin from opposite sides, there is one internal cut plane and the whole force $F$ is shear there: $V=F$, $\tau =F/A$. This is single shear.

If a central plate is sandwiched between two outer plates, the pin now has two internal cut planes — one in each gap. By symmetry, each plane carries half the load: $V=F/2$, $\tau =F/(2A)$. This is double shear.

Forgetting the factor of 2 doubles your calculated stress, which can make a safe design look unsafe — or vice versa.

Worked example — sizing a punched disk (slide 15)

A rod with a circular disk on its end passes through a $d=40$ mm hole. The rod is pulled with $P=20$ kN. The allowable shear stress is ${\tau }_{allow}=35$ MPa. Find the minimum disk thickness $t$.

Step 1 — Identify the shear surface.

The disk is being pulled through the hole, so the material being sheared off forms a cylindrical band at the rim of the disk: circumference $\pi d$, height $t$.

$$A_{shear}=\pi d\cdot t.$$

Not the disk face $\pi d^2/4$ — that’s the face you’d pull on, not the surface that’s being sheared.

Step 2 — Apply the sizing equation. With $V=P$:

$${\tau }_{allow}=\frac{V}{A_{shear}}=\frac{P}{\pi dt}⟺t=\frac{P}{{\tau }_{allow}\pi d}.$$

Step 3 — Plug in numbers.

$$t=\frac{20,000\text{N}}{(35\times 10^6\text{Pa})(\pi )(0.040\text{m})}=\frac{20,000}{4,398,230}\approx 4.55\times 10^{-3}\text{m}.$$ $$\boxed{t_{\min }\approx 4.55\text{mm}.}$$

That’s the minimum — choose a larger thickness in practice to leave further margin.

A common parallel — rod embedded in concrete (slide 14)

Glue or bond a rod of diameter $d$ over a length $\ell$. The shear surface is the cylindrical sidewall of the rod inside the concrete: area $\pi d\ell$. Setting ${\tau }_{allow}=P/(\pi d\ell )$ gives

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

Exactly the same formula as the punched disk. Both are sheared along a cylindrical surface — geometry doesn’t know whether you’re punching or bonding.

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3 — Why shear strain is an angle, not a length ratio

Pick two lines through a point, originally perpendicular. After deformation, the angle between them becomes ${\theta }^{\prime }$. The shear strain is just the change:

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

If ${\theta }^{\prime }<\pi /2$ the angle closed up, $\gamma >0$. If ${\theta }^{\prime }>\pi /2$ the angle opened, $\gamma <0$. Units of $\gamma$: radians (it’s an angle).

The intuition: axial strain $ϵ$ measures how a length stretches. Shear strain $\gamma$ measures how perpendicular lines tilt relative to each other. Both are dimensionless, both vanish when the body is undeformed.

Why small-strain analysis works

In most engineering materials, plastic deformation begins at strains of order $10^{-3}$. So angles are tiny — far less than 1 radian. Taylor expansion gives

$$\sin \gamma \approx \gamma -\frac{{\gamma }^3}{6}\approx \gamma ,\tan \gamma \approx \gamma +\frac{{\gamma }^3}{3}\approx \gamma ,\cos \gamma \approx 1-\frac{{\gamma }^2}{2}\approx 1.$$

Errors are $O({\gamma }^3)$ for $\sin$ and $\tan$. At $\gamma =0.01$ rad that’s $10^{-6}$ — utterly negligible.

So for a top plate displaced $\delta$ over height $h$ — the classic “lap shear” sketch — the angle subtended is

$$\gamma \approx \tan \gamma =\frac{\delta }{h}.$$

Worked example — shear strain on a deformed rectangle (slide 19)

A 400 mm × 300 mm rectangle $ABCD$ (with $AB$ horizontal, $AD$ vertical) deforms so that $D$ shifts $+5$ mm in $x$ and $B$ shifts $-5$ mm in $y$. Find ${\gamma }_{xy}$ at $A$.

Side $AD$ originally pointed along $+y$. After deformation it points to a new $D$ at $x=+5$ mm relative to $A$, so the side rotates clockwise (towards $+x$) by approximately

$${\theta }_{AD}\approx \frac{5\text{mm}}{300\text{mm}}=0.01667\text{rad}.$$

Side $AB$ originally pointed along $+x$. The new $B$ sits at $y=-5$ mm relative to $A$, so the side rotates clockwise (towards $-y$) by

$${\theta }_{AB}\approx \frac{5}{400}=0.0125\text{rad}.$$

Both rotations close the originally right angle at $A$. They add:

$${\gamma }_{xy}={\theta }_{AD}+{\theta }_{AB}=0.01667+0.0125=0.02917\text{rad}.$$

So ${\gamma }_{xy}\approx 29.2\times 10^{-3}$ rad. Positive — the angle at $A$ has closed.

Worked example — when a displacement is not shear (slide 17)

A 300 mm × 400 mm plate deforms so the top edge shifts 3 mm horizontally and the right side shortens by 2 mm vertically (at the top-right corner). Find ${\gamma }_{xy}$ at the bottom-left corner $A$.

The top-edge horizontal shift makes the vertical side $AB$ rotate. Treating it as small:

$${\gamma }_{xy}\approx \frac{3}{400}=7.5\times 10^{-3}\text{rad}.$$

The 2 mm shortening of the right side, however, is a normal strain in the $y$-direction at the right edge:

$${ϵ}_y\approx -\frac{2}{400}=-5\times 10^{-3}.$$

It doesn’t contribute to the shear strain at $A$ because $A$‘s right-angled corner is defined by the bottom-left edges — and those edges aren’t directly affected by a vertical compression at the far corner. The trap: it’s easy to add the 2 mm into the shear strain. Don’t.

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4 — Why $G=E/[2(1+\nu )]$

The shear modulus $G$ is what relates shear stress to shear strain:

$$\tau =G\gamma .$$

Same Hooke’s-law form as $\sigma =Eϵ$, just rotated 90°. So given $E$ in tension you might hope to derive $G$ — and you can, if the material is isotropic (same in every direction).

Here is the argument in one paragraph. Apply a pure tensile stress $\sigma$ along the $x$-axis. The body strains ${ϵ}_x=\sigma /E$ in $x$ and ${ϵ}_y=-\nu {ϵ}_x$ in $y$ (Poisson contraction). Now rotate your coordinate frame by 45°. The same physical deformation decomposes into a pure shear in the rotated frame: a shear stress $\tau =\sigma /2$ and a shear strain $\gamma ={ϵ}_x-{ϵ}_y={ϵ}_x(1+\nu )$. Then

$$G=\frac{\tau }{\gamma }=\frac{\sigma /2}{{ϵ}_x(1+\nu )}=\frac{\sigma /2}{(\sigma /E)(1+\nu )}=\frac{E}{2(1+\nu )}.$$

That single derivation is why only two of the three elastic constants are independent. Once you know $E$ and $\nu$, you know $G$.

Worked example — sanity-checking the table

For copper, slide 20 gives $E=110$ GPa, $\nu =0.34$. Then

$$G_{\text{predicted}}=\frac{110}{2(1+0.34)}=\frac{110}{2.68}\approx 41.0\text{GPa}.$$

Slide 20 lists $G=46$ GPa. Small mismatch (real materials aren’t perfectly isotropic, and handbook values are rounded), but consistent to ~10%. Always worth doing this check on exam tables.

Worked example — shear modulus from a block test (slide 21)

A polymer block 400 mm long, 100 mm wide, 200 mm tall has its top face displaced $\delta =2$ mm sideways under a force $P=2$ kN. Find $G$.

Shear stress. Plan-view area in the shear plane: $A=400\times 100=40,000$ mm² $=4\times 10^{-2}$ m².

$$\tau =\frac{P}{A}=\frac{2000\text{N}}{4\times 10^{-2}{\text{m}}^2}=50,000\text{Pa}=50\text{kPa}.$$

Shear strain. Small-angle approximation with $h=200$ mm:

$$\gamma \approx \frac{\delta }{h}=\frac{2}{200}=0.01\text{rad}.$$

Shear modulus.

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

A nice sanity check: rubbers and soft polymers typically have $G$ between $1$ and $10$ MPa. The answer is in the right physical ballpark.

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5 — How the whole week connects

The single unifying idea: shear is axial stress rotated 90°.

Concept	Axial	Shear
Stress	$\sigma =N/A$	$\tau =V/A$
Strain	$ϵ=\delta /L$	$\gamma =\delta /h$
Hooke’s law	$\sigma =Eϵ$	$\tau =G\gamma$
Elastic modulus	$E$	$G$
Energy density (elastic)	$\frac{1}{2}\sigma ϵ=\frac{{\sigma }^2}{2E}$	$\frac{1}{2}\tau \gamma =\frac{{\tau }^2}{2G}$
Failure stress	${\sigma }_{fail}$	${\tau }_{fail}$
Allowable	${\sigma }_{allow}={\sigma }_{fail}/SF$	${\tau }_{allow}={\tau }_{fail}/SF$

Then $G$ and $E$ are themselves linked by $\nu$. So everything in Week 12 — including the stress-strain curve story — is the same machinery you’ve already used for axial members, with the force component lying in the cut plane instead of perpendicular to it.

That’s why every shear formula has an axial twin you already know.

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6 — Exam-style sample, end-to-end

A steel pin of diameter $d=12$ mm joins two plates in double shear, carrying a load $F=24$ kN. The steel has ${\tau }_{fail}=240$ MPa. (a) What is the average shear stress in the pin? (b) What is the safety factor against shear failure? (c) If the same pin had been used in single shear, what would the safety factor have been?

Setup. Cross-sectional area of the pin:

$$A=\frac{\pi d^2}{4}=\frac{\pi (0.012{)}^2}{4}\approx 1.131\times 10^{-4}{\text{m}}^2.$$

(a) Average shear stress. Double shear → $V=F/2=12$ kN per cut plane.

$${\tau }_{avg}=\frac{V}{A}=\frac{12,000}{1.131\times 10^{-4}}\approx 1.061\times 10^8\text{Pa}\approx 106\text{MPa}.$$

(b) Safety factor in double shear.

$$SF=\frac{{\tau }_{fail}}{{\tau }_{avg}}=\frac{240}{106}\approx \mathrm{2.27.}$$

(c) Single shear case. Now $V=F=24$ kN per cut plane, so ${\tau }_{avg}$ doubles to $\approx 212$ MPa:

$$S{F}_{single}=\frac{240}{212}\approx \mathrm{1.13.}$$

Discussion. Going from double shear to single shear halves the safety factor from $\sim 2.3$ to $\sim 1.1$ — a marginal design rather than a safe one. This is the most common practical use of single vs double shear: it lets you reuse the same fastener for twice the load with the same safety factor. Always identify the configuration from the FBD before computing.

That layout — identify config → compute $V$ → compute $\tau$ → compare to ${\tau }_{fail}$ or ${\tau }_{allow}$ — is the template for every shear-sizing question on the paper.

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Where to look next

• The cheatsheet for fast revision + the quiz.
• The lecture reconstruction for slide-by-slide citations and the full set of six worked examples.
• Whatever you add to this folder — Lecture Atlas picks up any new Markdown note automatically.