Stress-Strain Relation

According to Hooke’s law, “within the elastic limit(for small deformations) stress is directly proportional to stain”.
Thus,
Stress \(\alpha\) strain
Stress \(={k} \times\) Strain ….(9.6),
Where \({k}\) is the proportionality constant called the modulus of elasticity.

\(
{k}=\frac{\text { Stress } \mathrm{s}}{\text { Strain }}
\)

Note: Modulus of elasticity \({k}\) depends on:
(i) Nature of the material of the body and
(ii) Temperature.

It is independent of the dimensions (i.e., length, volume, etc) of the body. S.I unit of ‘ \({k}\) ‘ is \(\mathrm{Nm}^{-2}\) or Pascal \(\left[\mathrm{P}_{\mathrm{a}}\right]\)

Hook’s Law in terms of Applied Force

We can also express Hook’s law in terms of applied force.

Mathematically, Hooke’s law states that the applied force \(F\) equals \(k\) (Stiffness constant) time the displacement or change in length \(x\),
\(
F=-k x
\)
The value of \(k\) depends on the type of elastic material, its dimensions, and its shape. The negative sign represents that the direction of force is opposite to that of \(x\).

Spring Force

When the spring is in natural length, it will not exert any force. But when it is elongated or compressed from its natural length, it will exert force. According to Hooke’s law, the spring force is proportional to the displacement of the object from the equilibrium position (natural length).
Spring force, \(F=-k x\)
Where,
\(k\) is the spring constant,
\(x\) is the displacement from the natural length of the spring.
Here negative sign signifies that the spring force will be developed in the opposite direction to deformation from the natural length. The equation also tells that the force acting at each instant during the compression and extension of the spring is varying with displacement from natural length.

Example 1: A spring is elongated by \(5 \mathrm{~cm}\) by application of force of \(500 \mathrm{~N}\) and held at that place.
(a) What is the spring constant of the spring?
(b) How much force is required to elongate the spring by \(8 \mathrm{~cm}\)?

Solution:

(a) Given,
Deformation in spring, \(x=5 \mathrm{~cm}=0.05 \mathrm{~m}\)
The magnitude of external force, \(F=500 \mathrm{~N}\)
Now, according to Hooke’s law,
\(
F=k x \text {, }
\)
Here \(k\) is the spring constant. Then,
\(
\begin{aligned}
& k=\frac{F}{x} \\
& \Rightarrow k=\frac{500}{0.05}=10 \times 10^3 \mathrm{Nm}^{-1} .
\end{aligned}
\)

(b) In this case, the elongation is \(8 \mathrm{~cm}\). Thus we have,
Also, The value of the spring constant, \(k=10 \times 10^3 \mathrm{~N} \mathrm{~m}^{-1}\)
Now, according to Hooke’s law,
\(
\begin{aligned}
& F=k x \\
& \Rightarrow F=10 \times 10^3 \mathrm{~N} \mathrm{~m}^{-1} \times 0.08 \mathrm{~m} \\
& \Rightarrow F=800 \mathrm{~N} .
\end{aligned}
\)