Concept Card: Elastic Behaviour of Materials & Stress-Strain Relationship
1. Microscopic Origin of Elasticity & Deforming Forces:
In a crystalline solid, constituent atoms or molecules are situated at stable equilibrium interatomic separations ($r_0$) where the net interatomic force is zero ($F(r_0) = 0$) and the mutual potential energy is a minimum ($U(r_0) = U_{\min}$).
- When external deforming forces pull atoms apart ($r > r_0$), attractive interatomic forces come into play to restore equilibrium.
- When external deforming forces compress atoms together ($r < r_0$), powerful repulsive electrostatic forces resist the interpenetration of electron clouds.
- Elasticity: The property by virtue of which a material body regains its original shape and size when external deforming forces are removed.
- Plasticity: The inability of a material to recover its original configuration upon unloading, retaining permanent deformation (e.g., putty, paraffin wax, lead).
2. Stress: Definition, Tensor Nature & Classification
When an external deforming force is applied to a body, an internal restoring force develops across every internal section.
$$\mathbf{\text{Stress } (\sigma) = \frac{F_{\text{restoring}}}{A} = \frac{F_{\text{ext}}}{A} \quad (\text{in static equilibrium})}$$
- SI Unit: $\text{N/m}^2$ or $\text{Pascal (Pa)}$.
- Dimensional Formula: $[\sigma] = \frac{[M L T^{-2}]}{[L^2]} = \mathbf{[M L^{-1} T^{-2}]}$ (identical to Pressure and Energy Density).
- Tensor Nature: Stress has magnitude, direction of area normal, and direction of force component, making it a second-rank tensor.
Primary Classifications of Stress:
- Longitudinal / Normal Stress ($\sigma_l$): Deforming force acts perpendicular to the cross-sectional area:
- Tensile Stress: Force tends to elongate the body ($L > L_0$).
- Compressive Stress: Force tends to shorten the body ($L < L_0$).
- Tangential / Shearing Stress ($\sigma_s$): Deforming force acts parallel (tangential) to the surface area $A$:
$$\sigma_s = \frac{F_\parallel}{A}$$ - Hydraulic / Volumetric Stress ($\sigma_v$): Uniform compressive force acting normally from all directions per unit surface area by a fluid:
$$\sigma_v = \frac{\Delta F}{A} = \Delta P$$
3. Strain: Classification & Measure of Deformation
Strain ($\varepsilon$) is the ratio of change in configuration to the original configuration:
$$\mathbf{\text{Strain } (\varepsilon) = \frac{\Delta (\text{Configuration})}{\text{Original Configuration}}}$$
As a ratio of two identical physical dimensions, strain is dimensionless and unitless ($[M^0 L^0 T^0]$).
- Longitudinal Strain: $\varepsilon_l = \frac{\Delta L}{L_0}$.
- Shearing Strain: Angular deformation $\theta$ (in radians) between lines originally perpendicular:
$$\theta \approx \tan\theta = \frac{\Delta x}{h}$$ - Volumetric Strain: Fractional change in volume:
$$\varepsilon_v = -\frac{\Delta V}{V_0} \quad (\text{negative sign denotes volume reduction under compression})$$
4. Hooke’s Law & The Modulus of Elasticity
Robert Hooke’s Law (1678): Within the elastic limit of a material, stress is directly proportional to strain:
$$\sigma \propto \varepsilon \implies \mathbf{\sigma = E \cdot \varepsilon} \implies \mathbf{E = \frac{\text{Stress}}{\text{Strain}}}$$
where $E$ is the Modulus of Elasticity of the material.
Crucial Insight:
The Modulus of Elasticity ($E$) is an intrinsic material property. It depends solely on the nature of the material, its crystalline microstructure, and temperature. It is completely independent of the dimensions (length $L$, diameter $d$, cross-sectional area $A$) of the specimen or the applied stress!
5. The Complete Stress-Strain Curve for a Ductile Metal (Mild Steel)
| Region / Characteristic Point | Physical Phenomenon & Material State |
|---|---|
| Region OA (Linear / Proportional Limit) | Stress is strictly proportional to strain ($\sigma \propto \varepsilon$). Hooke’s law holds perfectly. Slope of line $OA$ equals Young’s Modulus ($Y$). Point $A$ is the Proportional Limit. |
| Region AB (Elastic Limit / Yield Point) | Stress and strain are no longer linear, but the deformation remains completely reversible. If the load is removed at or before point $B$, the specimen returns to its exact original length. Point $B$ is the Elastic Limit or Yield Point; the corresponding stress is the Yield Strength ($\sigma_y$). |
| Region BC (Plastic Deformation & Permanent Set) | Beyond point $B$, dislocations glide irreversibly through the crystal lattice. If unloaded at point $C$, the strain does not retrace curve $CBAO$; instead, it follows a parallel line $CO’$ leaving a permanent residual strain $OO’$ called Permanent Set (typically $0.1\% – 0.2\%$). |
| Point D (Ultimate Tensile Strength / UTS) | The maximum nominal stress $\sigma_u$ the material can sustain. Beyond point $D$, local cross-sectional narrowing (**necking**) develops rapidly. |
| Point E (Fracture / Rupture Point) | The specimen snaps into two pieces. Even though applied force decreases past point $D$, true localized stress in the necked region causes structural failure at point $E$. |
Material Categorization via the Stress-Strain Profile:
- Ductile Materials: Possess a large plastic deformation range between yield point $B$ and fracture point $E$ (e.g., copper, mild steel, aluminum). Can be drawn into wires and rolled into sheets.
- Brittle Materials: Fracture point $E$ lies very close to elastic limit $B$, with practically negligible plastic range (e.g., cast iron, glass, ceramic, concrete). Fail catastrophically without warning.
- Elastomers: Can be stretched to enormous strains (several hundred percent) without permanent deformation (e.g., vulcanized rubber, aortic tissue). Do not obey Hooke’s law even at small strains; stress-strain curve exhibits elastic hysteresis (loop area represents energy dissipated as heat per cycle per unit volume).
6. Elastic Strain Energy Density ($u$)
The work done by external forces in deforming an elastic body within its proportional limit is stored internally as elastic potential energy:
$$W = \int F dx = \int_0^{\Delta L} \left(\frac{Y A x}{L}\right) dx = \frac{1}{2} \frac{Y A (\Delta L)^2}{L} = \frac{1}{2} F \Delta L$$
Dividing by the total volume of the wire ($V = A L$):
$$\mathbf{u = \frac{U}{\text{Volume}} = \frac{1}{2} \frac{F \Delta L}{A L} = \frac{1}{2} \times \text{Stress} \times \text{Strain} = \frac{1}{2} Y \varepsilon^2 = \frac{\sigma^2}{2 Y}}$$
The area under the stress-strain curve up to the proportional limit equals the modulus of resilience (strain energy density).
Solved Examples
Example 1 (Direct Young’s Modulus, Strain & Energy Calculation – JEE Main):
A structural steel wire of length $L = 2.0\text{ m}$ and circular cross-sectional area $A = 4.0\text{ mm}^2$ supports a suspended mass $M = 80\text{ kg}$ ($g = 10\text{ m/s}^2$). The Young’s modulus of steel is $Y = 2.0 \times 10^{11}\text{ N/m}^2$.
(a) Determine the tensile stress developed in the wire.
(b) Calculate the fractional strain and total elongation $\Delta L$ of the wire.
(c) Compute the total elastic strain energy stored in the stretched wire.
Solution:
Tensile force: $F = M g = 80 \times 10 = 800\text{ N}$.
Cross-sectional area: $A = 4.0\text{ mm}^2 = 4.0 \times 10^{-6}\text{ m}^2$.
(a) Tensile Stress:
$$\sigma = \frac{F}{A} = \frac{800\text{ N}}{4.0 \times 10^{-6}\text{ m}^2} = \mathbf{2.0 \times 10^8\text{ N/m}^2 = 200\text{ MPa}}$$
(b) Strain and Elongation:
$$\varepsilon = \frac{\sigma}{Y} = \frac{2.0 \times 10^8\text{ Pa}}{2.0 \times 10^{11}\text{ Pa}} = \mathbf{1.0 \times 10^{-3} = 0.10\%}$$
$$\Delta L = \varepsilon L = (1.0 \times 10^{-3})(2.0\text{ m}) = 2.0 \times 10^{-3}\text{ m} = \mathbf{2.0\text{ mm}}$$
(c) Elastic Strain Energy:
$$U = \frac{1}{2} F \Delta L = \frac{1}{2}(800\text{ N})(2.0 \times 10^{-3}\text{ m}) = \mathbf{0.80\text{ J}}$$
Example 2 (Two-Wire Composite Horizontal Bar Suspension – Standard JEE Pattern):
A light, rigid horizontal bar of length $L = 1.0\text{ m}$ is supported at its two ends by two vertical wires: wire 1 (steel) and wire 2 (brass).
– Steel wire 1: Length $L_1 = 1.5\text{ m}$, area $A_1 = 1.0\text{ mm}^2$, $Y_1 = 2.0 \times 10^{11}\text{ Pa}$.
– Brass wire 2: Length $L_2 = 1.0\text{ m}$, area $A_2 = 2.0\text{ mm}^2$, $Y_2 = 1.0 \times 10^{11}\text{ Pa}$.
A heavy vertical load $W$ is suspended from the bar at a distance $x$ from the steel wire.
(a) Find $x$ such that both wires develop equal tensile stress ($\sigma_1 = \sigma_2$).
(b) Find $x$ such that both wires undergo equal elongation ($\Delta L_1 = \Delta L_2$), keeping the bar strictly horizontal.
Solution:
Let $T_1$ and $T_2$ be the tensions in wire 1 and wire 2 respectively.
Taking rotational torque equilibrium about the load point:
$$T_1 x = T_2 (1 – x) \implies \frac{T_1}{T_2} = \frac{1 – x}{x}$$
(a) Condition for Equal Stress ($\sigma_1 = \sigma_2$):
$$\frac{T_1}{A_1} = \frac{T_2}{A_2} \implies \frac{T_1}{T_2} = \frac{A_1}{A_2} = \frac{1.0\text{ mm}^2}{2.0\text{ mm}^2} = \frac{1}{2}$$
$$\frac{1 – x}{x} = \frac{1}{2} \implies 2(1 – x) = x \implies 2 – 2x = x \implies 3x = 2 \implies \mathbf{x = \frac{2}{3}\text{ m} \approx 0.67\text{ m}}$$
(b) Condition for Equal Elongation ($\Delta L_1 = \Delta L_2$):
Effective stiffness of a wire: $k = \frac{Y A}{L} \implies \Delta L = \frac{T}{k}$.
$$k_1 = \frac{Y_1 A_1}{L_1} = \frac{(2.0 \times 10^{11})(1.0 \times 10^{-6})}{1.5} = \frac{2.0 \times 10^5}{1.5} = \frac{4}{3} \times 10^5\text{ N/m}$$
$$k_2 = \frac{Y_2 A_2}{L_2} = \frac{(1.0 \times 10^{11})(2.0 \times 10^{-6})}{1.0} = 2.0 \times 10^5\text{ N/m}$$
For equal elongation:
$$\frac{T_1}{k_1} = \frac{T_2}{k_2} \implies \frac{T_1}{T_2} = \frac{k_1}{k_2} = \frac{\frac{4}{3} \times 10^5}{2.0 \times 10^5} = \frac{2}{3}$$
Equating with torque equilibrium:
$$\frac{1 – x}{x} = \frac{2}{3} \implies 3(1 – x) = 2x \implies 3 – 3x = 2x \implies 5x = 3 \implies \mathbf{x = \frac{3}{5}\text{ m} = 0.60\text{ m}}$$
Example 3 (Thermal Stress in Rigidly Clamped Rods – JEE Advanced):
A steel rod of cross-sectional area $A = 2.0\text{ cm}^2$ and initial length $L = 1.0\text{ m}$ is rigidly clamped at both ends between two unyielding walls at $T_1 = 20^\circ\text{C}$. The temperature is raised to $T_2 = 70^\circ\text{C}$ ($\Delta T = 50^\circ\text{C}$). For steel: $Y = 2.0 \times 10^{11}\text{ N/m}^2$ and coefficient of linear expansion $\alpha = 1.2 \times 10^{-5}\text{ /}^\circ\text{C}$.
(a) Derive the thermal stress expression and explain why it is independent of length $L$.
(b) Calculate the compressive thermal stress and the force exerted by the rod on the clamps.
(c) Determine the elastic strain energy stored in the compressed rod.
Solution:
(a) Derivation of Thermal Stress:
If the rod were free to expand, its thermal expansion would be:
$$\Delta L_{\text{thermal}} = L \alpha \Delta T$$
Because the rigid clamps permit zero net expansion ($\Delta L_{\text{net}} = 0$), the clamps must exert an equal compressive deformation:
$$\Delta L_{\text{compressive}} = \Delta L_{\text{thermal}} = L \alpha \Delta T$$
Thermal strain:
$$\varepsilon = \frac{\Delta L_{\text{compressive}}}{L} = \frac{L \alpha \Delta T}{L} = \alpha \Delta T$$
Notice that $L$ cancels out completely! Hence:
$$\mathbf{\sigma_{\text{thermal}} = Y \varepsilon = Y \alpha \Delta T}$$
(b) Thermal Stress and Force on Clamps:
$$\sigma_{\text{thermal}} = (2.0 \times 10^{11})(1.2 \times 10^{-5})(50) = \mathbf{1.2 \times 10^8\text{ N/m}^2 = 120\text{ MPa}}$$
Compressive force on clamps ($A = 2.0\text{ cm}^2 = 2.0 \times 10^{-4}\text{ m}^2$):
$$F = \sigma_{\text{thermal}} A = (1.2 \times 10^8\text{ Pa})(2.0 \times 10^{-4}\text{ m}^2) = \mathbf{24,000\text{ N} = 24.0\text{ kN}}$$
(c) Elastic Strain Energy:
$$u = \frac{\sigma^2}{2 Y} = \frac{(1.2 \times 10^8)^2}{2(2.0 \times 10^{11})} = \frac{1.44 \times 10^{16}}{4.0 \times 10^{11}} = 36,000\text{ J/m}^3$$
$$\text{Volume } V = A L = (2.0 \times 10^{-4}\text{ m}^2)(1.0\text{ m}) = 2.0 \times 10^{-4}\text{ m}^3$$
$$U = u \cdot V = (36,000\text{ J/m}^3)(2.0 \times 10^{-4}\text{ m}^3) = \mathbf{7.20\text{ J}}$$
Example 4 (Elongation & Strain Energy Under Self-Weight – Calculus Integration):
A uniform heavy rope of mass $M$, length $L$, cross-sectional area $A$, and Young’s modulus $Y$ hangs vertically from a fixed ceiling under its own weight.
(a) Determine the tension $T(y)$ at a distance $y$ measured from the free bottom tip.
(b) Derive the total elongation $\Delta L$ of the rope due to its own weight.
(c) Derive the total elastic strain energy stored in the hanging rope.
Solution:
Linear mass density: $\lambda = \frac{M}{L}$.
(a) Tension Distribution:
At distance $y$ from the bottom, the tension supports the weight of the lower portion of length $y$:
$$\mathbf{T(y) = (\lambda y) g = \frac{M g y}{L}}$$
At the bottom tip ($y = 0$): $T(0) = 0$.
At the ceiling support ($y = L$): $T(L) = M g$.
(b) Total Elongation:
An infinitesimal element of length $dy$ at distance $y$ experiences elongation $d(\Delta L)$:
$$d(\Delta L) = \frac{T(y) dy}{A Y} = \frac{M g y dy}{A Y L}$$
Integrating from $y = 0$ to $y = L$:
$$\mathbf{\Delta L = \int_0^L \frac{M g y dy}{A Y L} = \frac{M g}{A Y L} \left[\frac{y^2}{2}\right]_0^L = \frac{M g L^2}{2 A Y L} = \frac{M g L}{2 A Y}}$$
Comparison: A weightless wire supporting an end load $M$ elongates by $\Delta L_{\text{load}} = \frac{M g L}{A Y}$. Hence, self-weight elongation is precisely **half** of the end-load elongation!
(c) Total Elastic Strain Energy:
Strain energy of element $dy$:
$$dU = \frac{T(y)^2 dy}{2 A Y} = \frac{\left(\frac{M g y}{L}\right)^2 dy}{2 A Y} = \frac{M^2 g^2 y^2 dy}{2 A Y L^2}$$
Integrating over the entire length:
$$\mathbf{U = \int_0^L \frac{M^2 g^2 y^2 dy}{2 A Y L^2} = \frac{M^2 g^2}{2 A Y L^2} \left[\frac{y^3}{3}\right]_0^L = \frac{M^2 g^2 L^3}{6 A Y L^2} = \frac{M^2 g^2 L}{6 A Y}}$$
Practice Worksheet: Elastic Behaviour & Stress-Strain Dynamics
Section A: Single Correct Questions (Q1 to Q5)
Q1. The dimensional formula of Young’s Modulus of elasticity ($Y$) is:
(A) $[M L^{-1} T^{-2}]$
(B) $[M L^1 T^{-2}]$
(C) $[M L^{-2} T^{-1}]$
(D) $[M L^2 T^{-2}]$
Q2. Two wires $A$ and $B$ made of the same metal are subjected to the same tensile load. The ratio of their lengths is $L_A : L_B = 1 : 2$, and the ratio of their diameters is $d_A : d_B = 2 : 1$. The ratio of their elongations $\Delta L_A : \Delta L_B$ is:
(A) $1 : 8$
(B) $1 : 4$
(C) $1 : 2$
(D) $8 : 1$
Q3. In a stress-strain diagram for two materials $A$ and $B$, the initial linear portion for material $A$ has a steeper slope than that for material $B$, and the plastic deformation region between the yield point and the fracture point is much larger for material $A$ than for material $B$. Which of the following conclusions is correct?
(A) Material $A$ has a higher Young’s modulus and is more ductile than material $B$.
(B) Material $B$ has a higher Young’s modulus and is more brittle than material $A$.
(C) Material $A$ is more brittle and has lower tensile strength than material $B$.
(D) Both materials have identical elasticity, but material $A$ is an elastomer.
Q4. A metal wire of original length $L$, cross-sectional area $A$, and Young’s modulus $Y$ is stretched within the proportional limit by an extension $x$. The work done in stretching the wire is:
(A) $\frac{Y A x^2}{2 L}$
(B) $\frac{Y A x^2}{L}$
(C) $\frac{Y A x}{2 L}$
(D) $\frac{2 Y A x^2}{L}$
Q5. A metal rod of cross-sectional area $A$, initial length $L$, and thermal expansion coefficient $\alpha$ is rigidly clamped between two immovable walls. If the temperature is increased by $\Delta T$, the compressive force exerted on the clamps depends on:
(A) $A, Y, \alpha, \Delta T$ and is completely independent of the length $L$
(B) $L, A, Y, \alpha, \Delta T$ and is directly proportional to $L$
(C) $L$ only
(D) $Y$ and $L$ only
Section B: One or More than One Correct Questions (Q6 to Q8)
Q6. Which of the following statements regarding the stress-strain curve of a typical ductile metal (such as mild steel) is/are correct?
(A) Up to the proportional limit, stress is strictly directly proportional to strain (Hooke’s law holds).
(B) Up to the elastic limit, the material returns completely to its original length when the deforming force is removed.
(C) Beyond the elastic limit, unloading results in a permanent residual deformation known as permanent set.
(D) Between the ultimate tensile strength and the fracture point, the wire undergoes necking where local cross-sectional area reduces significantly.
Q7. Regarding the elastic behaviour of vulcanized rubber (an elastomer), which of the following statements is/are correct?
(A) It does not strictly obey Hooke’s law even at small strains.
(B) It can be stretched to several times its original length without breaking.
(C) The stress-strain curve for loading does not retrace itself during unloading, exhibiting elastic hysteresis.
(D) The area enclosed by the elastic hysteresis loop represents mechanical energy dissipated as heat per cycle per unit volume.
Q8. A uniform heavy rope of mass $M$, length $L$, cross-sectional area $A$, and Young’s modulus $Y$ hangs vertically from a ceiling under its own weight. Which of the following statements is/are correct?
(A) The tension in the rope is maximum at the ceiling support ($T = M g$) and zero at the free bottom tip.
(B) The total elongation of the rope under its self-weight is $\Delta L = \frac{M g L}{2 A Y}$.
(C) The elongation of this rope is exactly half of the elongation produced in an identical weightless rope supporting an end load $M$.
(D) The elastic strain energy stored in the rope due to its own weight is $U = \frac{M^2 g^2 L}{6 A Y}$.
Section C: Numerical Value Type Questions (Q9 to Q10)
Q9. A uniform wire of mass $M$ and length $L$ hangs vertically under its own weight and undergoes an elongation $\Delta L_1$. If an external load of mass $M$ is attached to the bottom of the wire (neglecting self-weight), it undergoes an elongation $\Delta L_2$. The ratio $\frac{\Delta L_2}{\Delta L_1}$ is an integer $n$. Find the value of $n$.
Q10. A uniform rigid bar of length $1.0\text{ m}$ is supported horizontally at its ends by two vertical wires: steel wire $A$ ($L_1 = 1.5\text{ m}, A_1 = 1.0\text{ mm}^2, Y_1 = 2.0 \times 10^{11}\text{ Pa}$) and brass wire $B$ ($L_2 = 1.0\text{ m}, A_2 = 2.0\text{ mm}^2, Y_2 = 1.0 \times 10^{11}\text{ Pa}$). In order for both wires to undergo equal elongation, a load must be suspended at a distance $x$ from the steel wire. If $x = \frac{k}{10}\text{ m}$, determine the integer value of $k$.
Comprehensive Solutions and Answer Key
Answer Key
| Question | Correct Answer | Core Physics Concept |
|---|---|---|
| Q1 | (A) | Young’s Modulus Dimensions: $[M L^{-1} T^{-2}]$ (Stress / Pressure) |
| Q2 | (A) | Wire Elongation Ratio: $\Delta L \propto L / d^2 \implies \Delta L_A / \Delta L_B = 1/8$ |
| Q3 | (A) | Stress-Strain Curve Interpretation: Slope = $Y$, Plastic Range = Ductility |
| Q4 | (A) | Work Done in Stretching: $W = \frac{1}{2} F x = \frac{Y A x^2}{2 L}$ |
| Q5 | (A) | Thermal Stress Force: $F = Y A \alpha \Delta T$ (Length Independent) |
| Q6 | (A, B, C, D) | Ductile Mild Steel Stress-Strain Curve Features |
| Q7 | (A, B, C, D) | Elastomers & Elastic Hysteresis Energy Dissipation |
| Q8 | (A, B, C, D) | Self-Weight Hanging Rope Elongation & Strain Energy |
| Q9 | 2 | End Load vs. Self-Weight Elongation Ratio: $\Delta L_2 / \Delta L_1 = 2 \implies n = 2$ |
| Q10 | 6 | Composite Suspension Bar Equilibrium: $x = 3/5\text{ m} = 6/10\text{ m} \implies k = 6$ |
Detailed Explanations
Solution 1:
Young’s Modulus is defined as:
$$Y = \frac{\text{Longitudinal Stress}}{\text{Longitudinal Strain}} = \frac{F / A}{\Delta L / L}$$
Since strain is dimensionless ($[M^0 L^0 T^0]$):
$$[Y] = [\text{Stress}] = \frac{[F]}{[A]} = \frac{[M L T^{-2}]}{[L^2]} = \mathbf{[M L^{-1} T^{-2}]}$$
Correct Answer: (A)
Solution 2:
The elongation of a wire under tensile force $F$ is:
$$\Delta L = \frac{F L}{A Y} = \frac{F L}{\left(\frac{\pi d^2}{4}\right) Y} = \frac{4 F L}{\pi d^2 Y} \propto \frac{L}{d^2}$$
Taking the ratio for wires $A$ and $B$:
$$\frac{\Delta L_A}{\Delta L_B} = \left(\frac{L_A}{L_B}\right) \left(\frac{d_B}{d_A}\right)^2 = \left(\frac{1}{2}\right) \left(\frac{1}{2}\right)^2 = \left(\frac{1}{2}\right) \left(\frac{1}{4}\right) = \mathbf{\frac{1}{8}}$$
$$\Delta L_A : \Delta L_B = 1 : 8$$
Correct Answer: (A)
Solution 3:
1. The slope of the linear elastic region represents Young’s modulus ($Y = \text{slope}$). Because material $A$ has a steeper slope, $Y_A > Y_B$.
2. Ductility is measured by the length of the plastic deformation zone between the yield point and the fracture point. Because this region is much larger for material $A$, material $A$ is more ductile than material $B$.
Correct Answer: (A)
Solution 4:
When a wire is extended by $x$, the restoring force at any instant is $F(x’) = \frac{Y A x’}{L}$.
The work done in stretching from $0$ to $x$:
$$W = \int_0^x F(x’) dx’ = \int_0^x \left(\frac{Y A x’}{L}\right) dx’ = \frac{Y A}{L} \left[\frac{x’^2}{2}\right]_0^x = \mathbf{\frac{Y A x^2}{2 L}}$$
Correct Answer: (A)
Solution 5:
Free thermal expansion of the rod: $\Delta L = L \alpha \Delta T$.
Since rigid walls prevent expansion, compressive thermal strain is:
$$\varepsilon = \frac{\Delta L}{L} = \frac{L \alpha \Delta T}{L} = \alpha \Delta T$$
Compressive thermal stress:
$$\sigma = Y \varepsilon = Y \alpha \Delta T$$
Compressive force on the clamps:
$$F = \sigma A = Y A \alpha \Delta T$$
The force depends only on $A, Y, \alpha,$ and $\Delta T$, and is independent of the length $L$ of the rod.
Correct Answer: (A)
Solution 6:
– (A) True: From origin $O$ to proportional limit $A$, the stress-strain curve is linear and Hooke’s law is obeyed.
– (B) True: Up to the elastic limit $B$, the material undergoes purely elastic deformation without permanent set.
– (C) True: Beyond $B$, unloading follows a path parallel to $OA$ intersecting the strain axis with a non-zero residual strain (permanent set).
– (D) True: Beyond the UTS point $D$, necking develops where the localized area drops sharply until rupture at $E$.
All statements (A, B, C, D) are correct.
Correct Answer: (A, B, C, D)
Solution 7:
– (A) True: In elastomers like vulcanized rubber, stress and strain are non-linear across virtually all regimes.
– (B) True: Rubber can withstand tensile strains of $500\%-700\%$ without structural rupture.
– (C) True: The unloading curve lies below the loading curve, forming an elastic hysteresis loop.
– (D) True: The area enclosed by the hysteresis loop represents work converted into thermal energy, making rubber an ideal material for vehicle vibration dampers and shock absorbers.
All statements (A, B, C, D) are correct.
Correct Answer: (A, B, C, D)
Solution 8:
– (A) True: $T(y) = \frac{M g y}{L}$; tension is maximum at the ceiling ($y = L$, $T = M g$) and zero at the free bottom tip ($y = 0$).
– (B) True: $\Delta L = \int_0^L \frac{T(y) dy}{A Y} = \frac{M g L}{2 A Y}$.
– (C) True: For an end load $M$, $\Delta L_{\text{load}} = \frac{M g L}{A Y} = 2 \Delta L_{\text{self}}$.
– (D) True: Strain energy $U = \int_0^L \frac{T(y)^2 dy}{2 A Y} = \frac{M^2 g^2 L}{6 A Y}$.
All statements (A, B, C, D) are correct.
Correct Answer: (A, B, C, D)
Solution 9:
Self-weight elongation:
$$\Delta L_1 = \frac{M g L}{2 A Y}$$
Elongation under end load $M$:
$$\Delta L_2 = \frac{M g L}{A Y}$$
Taking the ratio:
$$\frac{\Delta L_2}{\Delta L_1} = \frac{\frac{M g L}{A Y}}{\frac{M g L}{2 A Y}} = 2$$
$$\mathbf{n = 2}$$
Correct Answer: 2
Solution 10:
Effective stiffness of wire 1 (steel):
$$k_1 = \frac{Y_1 A_1}{L_1} = \frac{(2.0 \times 10^{11})(1.0 \times 10^{-6})}{1.5} = \frac{4}{3} \times 10^5\text{ N/m}$$
Effective stiffness of wire 2 (brass):
$$k_2 = \frac{Y_2 A_2}{L_2} = \frac{(1.0 \times 10^{11})(2.0 \times 10^{-6})}{1.0} = 2.0 \times 10^5\text{ N/m}$$
For equal elongation $\Delta L_1 = \Delta L_2$:
$$\frac{T_1}{k_1} = \frac{T_2}{k_2} \implies \frac{T_1}{T_2} = \frac{k_1}{k_2} = \frac{4/3 \times 10^5}{2 \times 10^5} = \frac{2}{3}$$
By torque balance about the load suspended at distance $x$ from wire 1 on the $1.0\text{ m}$ bar:
$$T_1 x = T_2 (1 – x) \implies \frac{1 – x}{x} = \frac{T_1}{T_2} = \frac{2}{3}$$
$$3(1 – x) = 2x \implies 3 – 3x = 2x \implies 5x = 3 \implies x = \frac{3}{5}\text{ m} = \mathbf{0.60\text{ m} = \frac{6}{10}\text{ m}}$$
Comparing with $\frac{k}{10}\text{ m}$:
$$\mathbf{k = 6}$$
Correct Answer: 6