Solids

(a) $\dfrac{1}{2}\dfrac{YA}{L}x^2$

Elastic potential energy per unit volume is $\frac12 \times \text{stress} \times \text{strain}$. Since stress $=Y\frac{x}{L}$, strain $=\frac{x}{L}$, and volume $=AL$, the total energy is $\frac12\left(Y\frac{x}{L}\right)\left(\frac{x}{L}\right)(AL)=\frac12\frac{YA}{L}x^2$.

Stress-strain curve of a ductile material (e.g. mild steel)

Stress
  ^
  |                    U
  |                 .-'''-.
  |            Y1 .'       '.
  |          E,P /  Y2        '.
  |             /                F
  |            /
  |           /
  |          /
  |         /
  |        /
  +-------+-------------------------> Strain
  O

• $O$ to $P$: straight line (stress $\propto$ strain, Hooke's law).
• $E$: elastic limit, just beyond $P$.
• $Y_1, Y_2$: upper and lower yield points; the curve is nearly flat after $Y_2$.
• $U$: highest point of the curve.
• $F$: end of the curve, where the wire breaks. After $U$ the engineering stress falls because of necking.

(i) Proportional limit

The largest stress up to which stress is directly proportional to strain (point $P$). The curve is a straight line and Hooke's law holds; its slope is Young's modulus.

(ii) Elastic limit

The maximum stress up to which the material fully regains its original length when the load is removed. Beyond it, permanent (plastic) deformation remains. It lies at or just beyond the proportional limit.

(iii) Yield point

The point at which the material begins to deform plastically: strain increases considerably with little or no increase in stress. Many ductile materials show an upper and a lower yield point. The material does not return to its original length after this point.

(iv) Ultimate tensile strength

The maximum stress the material can withstand (point $U$). Beyond it, the wire starts to neck (local thinning) and the stress needed to continue stretching decreases.

(v) Fracture (breaking) point

The point $F$ at which the wire finally breaks. The stress at this point is the breaking stress; it is less than the ultimate tensile strength (when computed on the original area).

Young's modulus of elasticity

Young's modulus of elasticity is the ratio of longitudinal (tensile or compressive) stress to the corresponding longitudinal strain within the elastic limit. It measures the stiffness of a material — the greater the value of Young's modulus, the more difficult it is to stretch or compress the material.

Derivation of (Y = \dfrac{F L}{A \Delta L})

Consider a wire of original length (L) and uniform cross-sectional area (A), fixed at one end and stretched by a force (F) applied along its length. Let the extension produced be (\Delta L).

Longitudinal stress: The restoring force set up per unit area of cross-section is

[ \text{Stress} = \frac{F}{A} ]

Longitudinal strain: The change in length per unit original length is

[ \text{Strain} = \frac{\Delta L}{L} ]

By definition, Young's modulus is the ratio of longitudinal stress to longitudinal strain:

[ Y = \frac{\text{Longitudinal stress}}{\text{Longitudinal strain}} ]

Substituting the expressions for stress and strain,

[ Y = \frac{F/A}{\Delta L/L} ]

[ \boxed{Y = \frac{F,L}{A,\Delta L}} ]

where (F) is the applied force (or the restoring force per unit area), (L) is the original length, (A) is the area of cross-section and (\Delta L) is the extension in length.

SI unit and dimensional formula

• SI unit: (\text{N m}^{-2}) (newton per square metre) or pascal (Pa).
• Dimensional formula: ([Y] = \dfrac{[\text{Force}][\text{Length}]}{[\text{Area}][\text{Length}]} = \dfrac{MLT^{-2}\cdot L}{L^{2}\cdot L} = ML^{-1}T^{-2}).

Thus ([Y] = ML^{-1}T^{-2}).

Given: [ L=2.0\ \text{m},\quad A=1.0\times10^{-6}\ \text{m}^2,\quad F=100\ \text{N} ] [ Y=2.0\times10^{11}\ \text{N m}^{-2} ]

Young's modulus is [ Y=\frac{F/A}{\Delta L/L}=\frac{FL}{A\Delta L} ]

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(i) Elongation

[ \Delta L=\frac{FL}{AY} ]

[ \Delta L=\frac{100\times 2.0}{(1.0\times10^{-6})(2.0\times10^{11})} ]

[ \Delta L=\frac{200}{2.0\times10^5} ]

[ \Delta L=1.0\times10^{-3}\ \text{m} ]

[ \boxed{\Delta L=1.0\ \text{mm}} ]

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(ii) Strain produced

[ \text{Strain}=\frac{\Delta L}{L} ]

[ \text{Strain}=\frac{1.0\times10^{-3}}{2.0} ]

[ \boxed{\text{Strain}=5.0\times10^{-4}} ]

It has no unit.

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(iii) Elastic potential energy stored

[ U=\frac12 F\Delta L ]

[ U=\frac12\times100\times1.0\times10^{-3} ]

[ \boxed{U=0.05\ \text{J}} ]

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(iv) Force required for elongation of (2.0\ \text{mm})

Within the elastic limit, force is directly proportional to elongation: [ F\propto \Delta L ]

So, [ \frac{F_2}{F_1}=\frac{\Delta L_2}{\Delta L_1} ]

[ F_2=F_1\frac{\Delta L_2}{\Delta L_1} ]

[ F_2=100\times\frac{2.0\ \text{mm}}{1.0\ \text{mm}} ]

[ \boxed{F_2=200\ \text{N}} ]

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Final answers:

[ \boxed{\Delta L=1.0\ \text{mm}} ]

[ \boxed{\text{Strain}=5.0\times10^{-4}} ]

[ \boxed{U=0.05\ \text{J}} ]

[ \boxed{F=200\ \text{N}} ]