Elasticity

Elastic limit: the maximum stress up to which a body completely regains its original size and shape when the deforming force is removed.

Work done in stretching a wire. Let a wire of length $L$ and cross-section $A$ have Young's modulus $Y$. When it is stretched by $x$, the tension is

$F=\frac{YA}{L}\,x$

This force grows uniformly from 0, so a small extension $dx$ needs work $dW = F\,dx$. The total work to stretch it by $l$ is

$W=\int_0^{l}\frac{YA}{L}\,x\,dx=\frac{YA\,l^2}{2L}$

Equivalently, since the average force is $\tfrac12 F$ (with $F$ the final force),

$W=\tfrac12 F\,l$

Because $AL$ is the volume, $W=\tfrac12\times\text{stress}\times\text{strain}\times\text{volume}$. This work is stored in the wire as elastic potential energy.

No. Young's modulus $Y=\dfrac{\text{stress}}{\text{strain}}=\dfrac{F/A}{\Delta l/l}$ is a property of the material only. Doubling the load doubles the stress, and within the elastic limit (Hooke's law) the strain also doubles, so their ratio is unchanged. It depends neither on the load nor on the length or cross-section of the wire.

Joined end to end, the two wires carry the same force $F$ and have the same area $A$. From $Y=\dfrac{FL}{A\,\Delta L}$,

$\Delta L=\frac{FL}{AY}\;\Rightarrow\;\frac{\Delta L_c}{\Delta L_s}=\frac{L_c/Y_c}{L_s/Y_s}=\frac{1/1.2}{2/2}=\frac{0.833}{1}=\frac{5}{6}$

The total elongation is $\Delta L_c+\Delta L_s=1\ \text{cm}$. Hence

$\Delta L_c=\frac{5}{11}\times 1\ \text{cm}\approx 0.45\ \text{cm},\qquad \Delta L_s=\frac{6}{11}\times 1\ \text{cm}\approx 0.55\ \text{cm}$

(b) 5 J

Work done in stretching the spring equals the area under the force–extension graph, which is a triangle:

$W=\tfrac12 Fx=\tfrac12\times 100\ \text{N}\times 0.1\ \text{m}=5\ \text{J}$

(The extension must be converted from 100 mm to 0.1 m.)

(a) Under repeated loading and unloading by traffic, wind and temperature changes, the material of a bridge undergoes elastic fatigue. Its elastic strength falls, and it may break under a load it could once bear. So after long use it is declared unsafe.

(b) Strain at the elastic limit:

$\frac{\Delta l}{l}=\frac{\text{stress}}{Y}=\frac{8\times10^{8}}{2\times10^{11}}=4\times10^{-3}$

$\Delta l=4\times10^{-3}\times0.5=2\times10^{-3}\ \text{m}=2\ \text{mm}$

(c) No. A liquid cannot sustain a shearing stress, as it flows, so it has no rigidity. Its modulus of rigidity is zero.

(c) remain the same

Young's modulus is a property of the material, $Y=\dfrac{FL}{A\,\Delta L}$. Doubling the length doubles the extension for a given force, so the ratio stays the same.

(a) Energy stored. Let a wire of length $L$ and area $A$ be stretched by $l$ under force $F$, with $F=\dfrac{YA}{L}x$ at extension $x$. The work done in stretching by a further $dx$ is $F\,dx$, so

$W=\int_0^{l}\frac{YA}{L}x\,dx=\frac{YA\,l^2}{2L}=\tfrac12\,Fl$

This work is stored as elastic potential energy: $U=\tfrac12Fl$.

Energy density $u$ is the elastic energy stored per unit volume of a strained body:

$u=\frac{U}{AL}=\tfrac12\times\text{stress}\times\text{strain}=\frac{1}{2}Y(\text{strain})^2$

(b) Force $F=100\ \text{kgf}=100\times1000\times980\ \text{dyne}=9.8\times10^{7}\ \text{dyne}$.

Radius $=0.125\ \text{cm}$, so $A=\pi(0.125)^2=0.0491\ \text{cm}^2$.

$\text{stress}=\frac{9.8\times10^{7}}{0.0491}=2.0\times10^{9}\ \text{dyne cm}^{-2}$

$\text{strain}=\frac{\text{stress}}{Y}=\frac{2.0\times10^{9}}{12.5\times10^{11}}=1.6\times10^{-3}$

Percentage increase in length $=1.6\times10^{-3}\times100\approx0.16\%$.

(c) Elasticity is the ability to resist deformation and regain shape. For the same stress iron shows a much smaller strain than rubber, i.e. it has a much larger Young's modulus ($Y\approx2\times10^{11}\ \text{Pa}$ against about $10^{6}$ to $10^{7}\ \text{Pa}$ for rubber), so it is the more elastic.

(a) — the graph with the steepest (most nearly vertical) straight line.

Stiffness is measured by Young's modulus, $Y = \dfrac{\text{stress}}{\text{strain}}$, which is the slope of the stress–strain graph. The steeper the slope, the greater the stress needed for the same strain, so the material is stiffer. Among the four graphs, the one with the largest slope represents the stiffest material.

(a) Hooke's law

For small deformations, the stress produced in a body is directly proportional to the strain within the elastic limit, i.e.

$\text{stress} \propto \text{strain} \quad\Rightarrow\quad \frac{\text{stress}}{\text{strain}} = E \ (\text{modulus of elasticity})$

In terms of extension, the extension produced in a wire is directly proportional to the applied load, provided the elastic limit is not exceeded.

(b) Warming of the tyre walls

Fig. 1.2 shows that the stress–strain curve for the rubber compound is not the same while the rubber is being stretched and while it is allowed to contract:

• For the same value of stress, the strain during loading (stretching) is greater than the strain during unloading (contracting); the unloading curve lies below the loading curve.
• The two curves therefore enclose a closed loop, called the hysteresis loop.

The area of this loop represents the mechanical work done per unit volume on the rubber in one complete stretch–contract cycle that is not returned as elastic potential energy — it is dissipated as internal energy (heat) in the rubber. This is known as elastic hysteresis.

As the car moves, each part of the tyre wall is repeatedly bent and straightened, so the rubber undergoes many such loading–unloading cycles every second. The energy lost in each cycle (the loop area) is added to the rubber as heat. Since the cycles occur very rapidly and rubber is a poor conductor of heat, this heat cannot escape quickly, and the tyre walls become warm. (This is why rubber has a large hysteresis loss and tyres heat up during running.)

Hooke's law: Within the elastic limit, the extension (or compression) produced in a body is directly proportional to the applied load (deforming force).

$F \propto x \quad \Rightarrow \quad F = kx$

where $k$ is the force constant (spring constant) of the body.

Young's modulus is $Y=\dfrac{Fl}{A\,\Delta l}$, so for the same load $F$ and the same extension $\Delta l$,

$Y\propto\frac{l}{A}$

Hence

$\frac{Y_s}{Y_c}=\frac{l_s/A_s}{l_c/A_c}=\frac{4.7/(3\times10^{-5})}{3.5/(4\times10^{-5})}=\frac{4.7\times4}{3\times3.5}=\frac{18.8}{10.5}\approx1.79$

The Young's modulus of steel is about $1.79$ times that of copper.