Elasticity Notes for IOE Entrance — Derivations & Past Questions

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Elasticity & Plasticity

Elastic bodies, plastic bodies & rigid bodies

Elasticity is the property of a body by virtue of which it regains its original configuration (shape and size) after the removal of the deforming force. It arises due to the cohesive force between molecules.

Bodies that show elasticity are elastic bodies. Quartz and phosphor bronze are nearly perfect elastic bodies.
Plasticity is the property by which a body does not regain its original configuration on removal of the deforming force. Putty, mud and paraffin wax are nearly perfect plastic bodies.
A rigid body is one in which the distance between any two particles always remains the same — no external force can deform it. Diamond is the most rigid body.

The elastic after effect is the delay in regaining the original configuration after the deforming force is removed. Glass has the maximum, while quartz and phosphor bronze have the minimum elastic after effect.

Dashed outline = original shape · solid = body after the force is removed An elastic body springs back; a plastic body keeps the new shape

2

Stress

Restoring force per unit area

When a body is deformed by an external force, a restoring force develops inside it. The restoring force per unit area is called stress. As the restoring force is equal and opposite to the deforming force:

$$\text{Stress}=\dfrac{\text{Restoring force}}{\text{Cross-sectional area}} =\dfrac{\text{Deforming force}}{\text{Cross-sectional area}}$$

Its SI unit is $\text{N m}^{-2}$ (pascal) and its dimensional formula is $[ML^{-1}T^{-2}]$. There are three types of stress:

Normal stress $=\dfrac{F_{\text{normal}}}{A}$ Shear stress $=\dfrac{F_{\parallel}}{A}$ Volumetric stress $=\dfrac{F_{\perp}}{A}$

Normal stress — deforming force acts normally (perpendicular) to the surface. It is of two kinds: tensile (stretching) and compressive.
Tangential (shear) stress — deforming force acts parallel to the surface, changing the shape without changing the volume.
Volumetric stress — force acts perpendicular to every point of the surface, changing the volume without changing the shape.

(a) forces normal & outward — stretch (b) forces parallel — change of shape (c) forces inward on all faces — change of volume

3

Strain

The measure of deformation

Strain is the ratio of the change in configuration to the initial configuration. Being a ratio of like quantities, it is dimensionless and has no unit.

$$\text{Strain}=\dfrac{\text{Change in configuration}} {\text{Initial configuration}}$$

There are three types of strain:

Longitudinal strain — change in length per unit original length: $$\text{Longitudinal strain}=\dfrac{\Delta l}{l}$$
Volumetric strain — change in volume per unit original volume: $$\text{Volumetric strain}=\dfrac{\Delta V}{V}$$
Shearing strain — the angular measure $\phi$ of how much a body is distorted by a shear force: $$\text{Shearing strain}=\phi=\dfrac{\Delta l}{l}$$

A tangential force tilts the block through the shearing angle $\phi=\dfrac{\Delta l}{l}$

4

Hooke's Law & Stress–Strain Curve

Proportionality, elastic limit & fracture

Hooke's law: within a certain limit, stress is directly proportional to strain.

$$\text{Stress}\propto\text{Strain}\;\Rightarrow\; \text{Stress}=E\times\text{Strain}\;\Rightarrow\;E=\dfrac{\text{Stress}}{\text{Strain}}$$

where $E$ is the proportionality constant called the modulus of elasticity.

Point a — proportionality limit: the region $oa$ is a straight line and follows Hooke's law.
Point b — elastic limit: up to here the wire regains its original length. Beyond it, plastic behaviour is seen and the wire does not return.
Point d — fracture (breaking) point: the wire breaks here.

Brittle substances (glass, cast-iron) have a short plastic range and break near the elastic limit. Ductile substances (iron, copper, aluminium, silver) have a large plastic range and are used to make wires.

oa obeys Hooke's law b elastic limit d fracture point

5

Moduli of Elasticity

Young's, bulk & rigidity

The modulus of elasticity $E=\dfrac{\text{stress}}{\text{strain}}$ takes three forms depending on the kind of deformation:

Young's modulus — normal stress to longitudinal strain: $$Y=\dfrac{\text{Normal stress}}{\text{Longitudinal strain}} =\dfrac{F/A}{\Delta l/l}=\dfrac{Fl}{A\,\Delta l}$$
Bulk modulus — normal stress to volumetric strain: $$K=\dfrac{\text{Normal stress}}{\text{Volumetric strain}} =\dfrac{F/A}{\Delta V/V}$$
Modulus of rigidity — tangential stress to shearing strain: $$\eta=\dfrac{\text{Tangential stress}}{\text{Shearing strain}} =\dfrac{F/A}{\phi}$$

For a gas the bulk modulus has two values: isothermal $K_{\text{iso}}=P$ and adiabatic $K_{\text{adi}}=\gamma P$, where $\gamma=\dfrac{C_p}{C_v}$. The reciprocal of bulk modulus is the compressibility $C=\dfrac{1}{K}$.

Full derivation — Young's modulus & force constant

A wire of natural length $l$ and uniform cross-sectional area $A$ is fixed at one end. A stretching force $F$ applied at the other end produces an extension $\Delta l$, staying within the elastic limit.

Step 1 — Write the stress and the strain

The force acts normally over the cross-section, and the wire lengthens along its own axis:

$$\text{Normal stress}=\frac{F}{A}, \qquad \text{Longitudinal strain}=\frac{\Delta l}{l}$$

Step 2 — Apply Hooke's law

Within the elastic limit the ratio of stress to strain is the constant $Y$:

$$Y=\frac{\text{Normal stress}}{\text{Longitudinal strain}} =\frac{F/A}{\Delta l/l}$$

$Y=\dfrac{Fl}{A\,\Delta l}$

Step 3 — Force (spring) constant of the wire

Rearranging for the force in terms of the extension:

$$F=\left(\frac{YA}{l}\right)\Delta l = k\,\Delta l$$

so the wire behaves like a spring of force constant $k=\dfrac{YA}{l}$ — the force needed per unit extension.

Equal inward force on every face changes the volume by $\Delta V$ → $K=\dfrac{F/A}{\Delta V/V}$

6

Elastic Potential Energy

Work stored in a stretched body

The work done by the deforming force in stretching a body is stored as elastic potential energy:

$U=\dfrac{1}{2}kx^2$ $U=\dfrac{1}{2}\times\text{Force}\times\text{extension}$ $U=\dfrac{1}{2}\times\text{Stress}\times\text{Strain}\times\text{Volume}$

The elastic potential energy stored per unit volume is the energy density:

$$u=\dfrac{\text{Elastic P.E.}}{\text{Volume}} =\dfrac{1}{2}\times\dfrac{(\text{stress})^2}{Y} =\dfrac{1}{2}\times(\text{strain})^2\times Y$$

Full derivation — energy stored in a stretched wire

A wire of length $l$, area $A$ and Young's modulus $Y$ is stretched slowly to a final extension $e$. When the extension is $x$, the restoring force (from $F=\tfrac{YA}{l}x$) balances the applied load.

Step 1 — Work for a further small stretch

To increase the extension by $dx$ against the force $F(x)=\dfrac{YA}{l}x$:

$$dW=F(x)\,dx=\frac{YA}{l}\,x\,dx$$

Step 2 — Integrate up to the final extension $e$

$$W=\int_0^{e}\frac{YA}{l}\,x\,dx=\frac{YA}{l}\cdot\frac{e^2}{2} =\frac{1}{2}\left(\frac{YAe}{l}\right)e$$

But $\dfrac{YAe}{l}=F$, the final stretching force, so:

$W=U=\dfrac{1}{2}\,F\,e =\dfrac{1}{2}\times\text{force}\times\text{extension}$

Step 3 — In terms of stress & strain

Writing $F=\text{stress}\times A$ and $e=\text{strain}\times l$:

$$U=\frac{1}{2}(\text{stress}\times A)(\text{strain}\times l) =\frac{1}{2}\times\text{stress}\times\text{strain}\times\underbrace{(Al)}_{\text{volume}}$$

Step 4 — Energy density

$$u=\frac{U}{V}=\frac{1}{2}\times\text{stress}\times\text{strain} =\frac{1}{2}\frac{(\text{stress})^2}{Y}=\frac{1}{2}Y(\text{strain})^2$$

The shaded area under the force–extension line equals the stored elastic energy $\tfrac{1}{2}Fe$

7

Poisson's Ratio & Elastic Relations

Lateral strain & the links between $Y,K,\eta,\sigma$

Poisson's ratio is the ratio of lateral strain to longitudinal strain:

$$\sigma=-\dfrac{\text{Lateral strain}}{\text{Longitudinal strain}} =-\dfrac{\Delta r/r}{\Delta l/l}$$

If $\sigma$ is positive, the radius decreases as the wire elongates.
Theoretically $-1\le\sigma\le\dfrac{1}{2}$; in practice its value lies between $0$ and $0.5$.

The four elastic constants are connected by the relations:

$$Y=3K(1-2\sigma), \qquad Y=2\eta(1+\sigma)$$ $$Y=\dfrac{9K\eta}{3K+\eta}, \qquad \sigma=\dfrac{3K-2\eta}{6K+2\eta}$$

Full derivation — shearing angle & angle of twist

A solid cylinder (wire) of radius $R$ and length $l$ is clamped at the top. The lower face is twisted through an angle $\theta$ (the angle of twist). A straight line drawn on the surface, originally parallel to the axis, tilts through the shearing angle $\phi$.

Step 1 — Displacement of a rim point

As the lower face turns through $\theta$, a point on the circumference of radius $R$ moves along an arc of length:

$$s = R\theta$$

Step 2 — Same displacement as a shear

That surface line tilts by the small shearing angle $\phi$ over the length $l$, so the bottom end is displaced by:

$$s = l\,\phi$$

Step 3 — Equate the two expressions

$$l\,\phi = R\theta$$

$\phi=\dfrac{R\theta}{l}$

The shearing angle grows with the radius and the twist, and falls off with the length of the wire.

Stretching by $\Delta l$ thins the radius by $\Delta r$ → $\sigma=-\dfrac{\Delta r/r}{\Delta l/l}$

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High-Yield Points

Quick revision

Breaking stress depends only on the nature of the material and is independent of dimensions.

Breaking force $=$ breaking stress $\times$ area $=\pi r^2\times$ breaking stress, so it does depend on dimensions $\left(\propto r^2\right)$.

The greater the Young's modulus $Y$, the more elastic the body. For a highly elastic or perfectly rigid body, $Y=\infty$.

On hammering or cold-working, $Y$ increases; on annealing, $Y$ decreases.

On raising the temperature, the Young's modulus of a body decreases.

Adding more elastic impurities increases $Y$; adding less elastic impurities decreases $Y$.

Compressibility $C=\dfrac{1}{K}$, with $K_{\text{solid}}>K_{\text{liquid}}>K_{\text{gas}}$ and $C_{\text{gas}}>C_{\text{liquid}}>C_{\text{solid}}$.

Isothermal bulk modulus of a gas $K_{\text{iso}}=P$; adiabatic $K_{\text{adi}}=\gamma P$.

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Challenge Questions

Slightly harder problems from the chapter

These higher-order questions go a step beyond the past-paper set — combining Young's modulus, elastic energy, Poisson's ratio and breaking-stress ideas. Try each one before revealing the answer.