CBSE Class 11 Physics • Chapter 08
Chapter Overview: This chapter deals with the property of elasticity of solids. We will understand Stress, Strain, Hooke's Law, and different Moduli of Elasticity (Young's, Bulk, Shear). We will also analyze the Stress-Strain curve.
Deforming Force: A force that changes the configuration (size or shape) of a body.
Restoring Force: An internal force developed within the body that opposes deforming force and tends to restore the body to its original size and shape.
Elasticity: The property of a body to regain its original size and shape strictly after the removal of the deforming force. E.g., Steel, Quartz.
Plasticity: The property by virtue of which a body does not regain its original shape and gets permanently deformed. E.g., Putty, Mud, Paraffin wax.
Q: Which is easier to stretch: Steel or Rubber? Which is more elastic?
Ans: Rubber is easier to stretch. However, Steel is more elastic. In physics, elasticity is defined by the resistance to deformation (Modulus of Elasticity). For the same strain, steel requires a much larger deforming force (stress) than rubber.
It is the internal restoring force acting per unit area of cross-section of the deformed body.
SI Unit: $\text{Nm}^{-2}$ or Pascal (Pa).
Dimensional Formula: $[ML^{-1}T^{-2}]$.
Types of Stress:
It is the ratio of change in configuration to the original configuration.
Unit: Strain is a unitless, dimensionless quantity.
Types of Strain:
Q: A load of 4.0 kg is suspended from a ceiling through a steel wire of radius 2.0 mm. Find the tensile stress developed. ($g \approx 3.14 \pi$)
Ans: $F = mg = 4 \times 3.14 \pi$. Area $A = \pi r^2 = \pi (2 \times 10^{-3})^2 = 4\pi \times 10^{-6} \text{ m}^2$.
Stress $\sigma = \frac{F}{A} = \frac{4 \times 3.14 \pi}{4\pi \times 10^{-6}}$. (Approximating $3.14 \approx \pi$, so $g \approx \pi^2$).
Let's use exact values: $\sigma = \frac{4 \times 9.8}{\pi (2\times 10^{-3})^2} = \frac{39.2}{12.56 \times 10^{-6}} \approx \mathbf{3.1 \times 10^6 \text{ Pa}}$.
Statement: For small deformations (within the elastic limit), Stress is directly proportional to Strain.
Where $E$ is the Modulus of Elasticity of the material.
Stress-Strain Curve for a Ductile Material:
Figure 8.1: Stress-Strain Curve for a ductile material
Note: If D and E are close, material is Brittle (e.g., Glass). If far apart, it is Ductile (e.g., Copper).
Figure 8.2: Three Moduli of Elasticity
Within the elastic limit, the ratio of longitudinal stress to longitudinal strain is called Young's Modulus.
Note: It is a property of the material, not dimensions. Steel has higher Y than Copper.
The ratio of sharing stress to shearing strain.
Ideally for fluids, $\eta = 0$.
The ratio of hydraulic stress to volumetric strain.
The negative sign is included because pressure increase ($+P$) causes volume decrease ($-\Delta V$), making $B$ positive.
Compressibility ($k$): Reciprocal of Bulk Modulus ($k = 1/B$).
Q: A structural steel rod has radius 10 mm and length 1.0 m. A 100 kN force stretches it along its length. Calculate stress and elongation. ($Y_{st} = 2.0 \times 10^{11} \text{ Pa}$)
Ans: $A = \pi (10^{-2})^2 = 3.14 \times 10^{-4} \text{ m}^2$. $F = 10^5 \text{ N}$.
Stress $\sigma = F/A = 10^5 / 3.14 \times 10^{-4} \approx 3.18 \times 10^8 \text{ Pa}$.
Elongation $\Delta L = \frac{FL}{AY} = \frac{(10^5)(1)}{(3.14\times 10^{-4})(2\times 10^{11})} \approx \mathbf{1.59 \text{ mm}}$.
When a wire is loaded, its length increases ($\Delta L$) but its diameter decreases ($\Delta d$).
Lateral Strain: $\beta = - \Delta d/d$
Longitudinal Strain: $\varepsilon = \Delta L/L$
Poisson's Ratio: $\sigma = \frac{\text{Lateral Strain}}{\text{Longitudinal Strain}}$
Limits: Theoretically $-1 \le \sigma \le 0.5$. Practically $0 \le \sigma \le 0.5$. For rubber $\sigma \approx 0.5$.
Goal: Calculate work done in stretching a wire.
Consider a wire of length $L$ and area $A$. Let it be stretched by a small length $x$.
Restoring force $F$ at extension $x$: From $Y = \frac{FL}{Ax} \Rightarrow F = \frac{YAx}{L}$.
Work done in stretching it further by $dx$ is $dW = F dx$.
Total work done in stretching from $0$ to $l$:
$$ W = \int_{0}^{l} F dx = \int_{0}^{l} \frac{YAx}{L} dx = \frac{YA}{L} \left[ \frac{x^2}{2} \right]_{0}^{l} $$
$$ W = \frac{YA l^2}{2L} = \frac{1}{2} \left( \frac{YAl}{L} \right) l $$
But $F_{\text{final}} = \frac{YAl}{L}$ (Force for extension $l$). So:
$$ W = \frac{1}{2} \times \text{Load} \times \text{Extension} $$
Energy Density ($u$): Energy per unit volume ($V = AL$)
$$ u = \frac{W}{AL} = \frac{1}{2} \frac{F l}{AL} = \frac{1}{2} \left( \frac{F}{A} \right) \left( \frac{l}{L} \right) $$
Q: A copper wire of length 2.2 m and a steel wire of length 1.6 m, both of diameter 3.0 mm, are connected end to end. When stretched by a load, net elongation is 0.70 mm. obtain the load applied. ($Y_c = 1.1 \times 10^{11}$, $Y_s = 2.0 \times 10^{11}$)
Ans: Series connection $\Rightarrow$ Force $F$ is same. Total extension $\Delta L = \Delta L_c + \Delta L_s = 0.7 \text{ mm}$.
$\Delta L = \frac{FL_c}{AY_c} + \frac{FL_s}{AY_s} = \frac{F}{A} \left( \frac{L_c}{Y_c} + \frac{L_s}{Y_s} \right)$.
Solving for $F$: $F = \frac{A \Delta L}{(L_c/Y_c + L_s/Y_s)}$.
Substituting values gives $F \approx \mathbf{1.8 \times 10^2 \text{ N}}$.
Question: Why are the springs made of steel and not of copper?
Answer: Young's modulus of steel is greater than that of copper. For a given stress, steel strains less. It returns to its original shape more effectively. Large restoring force is desirable for springs.
Question: Bridges are designed with pillars having distributed surface area at the bottom. Why?
Answer: To reduce stress ($\sigma = F/A$). By increasing Area $A$, the stress on the foundation ground is reduced, preventing sinking.