Elastic and Plastic Behaviour of Solids
When a force acts on a solid body, its shape or size may change. This change in configuration of a body is called deformation. Depending on how the material responds to the applied force, solids exhibit two important mechanical behaviours:
- Elastic Behaviour – body returns to original shape after force removal.
- Plastic Behaviour – body undergoes permanent deformation.
Understanding these properties is extremely important in engineering, construction, material science, and mechanical design.
Elasticity
Elasticity is the property of a material by virtue of which it regains its original shape and size after the removal of the deforming force, provided the deformation is within a certain limit known as the elastic limit.
In simple words, a body is elastic if it can recover completely after deformation.
This restoring tendency arises due to interatomic forces that try to bring atoms back to their equilibrium positions.
Examples of Elastic Behaviour
- Stretching a rubber band and releasing it.
- Compression of a spring.
- Slight bending of a steel ruler.
- Vibration of tuning forks and guitar strings.
These materials store the supplied energy as elastic potential energy, which is released when the force is removed.
Plasticity
Plasticity is the property of a material by virtue of which it does not regain its original shape or size after the deforming force is removed.
When the applied stress exceeds the elastic limit, the material undergoes permanent deformation. This behaviour is called plastic behaviour.
Examples of Plastic Behaviour
- Moulding clay into a pot.
- Hammering a metal sheet.
- Permanent bending of a metal wire.
- Forging of steel in industries.
Plasticity is extremely useful in manufacturing processes such as forging, rolling, extrusion, and metal forming.
Microscopic Explanation of Elasticity
In solids, atoms or molecules are arranged in stable equilibrium positions. When an external force is applied:
- The interatomic distance slightly changes.
- Interatomic restoring forces develop.
- These forces try to restore the original configuration.
If the deformation is small, atoms return to their original position and the body behaves elastically. If the deformation becomes large, atomic layers begin to slip permanently, producing plastic deformation.
Elastic deformation disappears after removing force, whereas plastic deformation remains permanently.
Difference Between Elastic and Plastic Behaviour
| Elastic Behaviour | Plastic Behaviour |
|---|---|
| Material regains its original shape after removal of force. | Material does not regain original shape after removal of force. |
| Deformation is temporary and reversible. | Deformation is permanent and irreversible. |
| Occurs within the elastic limit. | Occurs after the elastic limit is exceeded. |
| Obeys Hooke’s Law (stress ∝ strain). | Hooke’s law no longer holds. |
| Energy is stored as elastic potential energy and recovered. | Energy is mostly dissipated as heat. |
| Examples: spring, rubber band, steel wire. | Examples: clay, putty, lead, permanently bent metal. |
Exam Insight
In most competitive exams like CBSE, NEET, and JEE, questions from this topic usually involve:
- Difference between elastic and plastic deformation
- Concept of elastic limit
- Examples of elastic materials
- Relation with Hooke's Law
Stress and Strain
Whenever an external force acts on a solid body, the body tends to change its shape, size, or volume. This change in configuration is called deformation.
In response to deformation, the material develops internal forces that resist the applied force. These internal responses are described using two important physical quantities:
- Stress → Cause of deformation
- Strain → Measure of deformation
Stress
When a deforming force acts on a body, internal restoring forces develop within the material. These forces act in a direction opposite to the applied force and try to restore the original shape.
The restoring force acting per unit area of the material is called stress.
Mathematically,
\[ \text{Stress} = \frac{F}{A} \]
where
- \(F\) = applied force
- \(A\) = cross-sectional area
The SI unit of stress is Pascal (Pa).
\[ 1 \, Pa = 1 \, N m^{-2} \]
Internal restoring forces developed per unit area constitute stress.
Strain
While stress represents the applied cause, strain represents the resulting deformation.
Strain is defined as the ratio of change in dimension to the original dimension of the body.
\[ \text{Strain} = \frac{\text{Change in dimension}}{\text{Original dimension}} \]
Since it is a ratio of two identical physical quantities, strain has no units and is therefore a dimensionless quantity.
Longitudinal Strain
When a rod or wire is stretched or compressed along its length, the resulting strain is called longitudinal strain.
If the original length of the rod is \(L\) and the increase in length is \(\Delta L\), then
\[ \text{Longitudinal strain} = \frac{\Delta L}{L} \]
The corresponding stress produced is
\[ \text{Longitudinal stress} = \frac{F}{A} \]
Shearing Strain
Shearing strain occurs when forces act tangentially to the surface of a body. These forces change the shape of the body without significantly changing its volume.
Consider a rectangular block whose base is fixed. If a tangential force is applied to its top surface, the top layer shifts sideways.
If the height of the block is \(L\) and the lateral displacement is \(\Delta x\), then
\[ \text{Shear strain} = \frac{\Delta x}{L} \]
For small deformation,
\[ \text{Shear strain} = \tan \theta \approx \theta \]
Hydraulic Stress (Volume Stress)
Hydraulic stress is produced when a body is subjected to equal pressure acting normally on all its surfaces.
Since the pressure acts uniformly in all directions, the shape of the body remains unchanged but its volume decreases.
Hydraulic stress is numerically equal to the applied pressure.
\[ \text{Hydraulic stress} = P \]
If the original volume is \(V\) and the decrease in volume is \(\Delta V\), the resulting volume strain is
\[ \text{Volume strain} = \frac{\Delta V}{V} \]
Dimensional Formula
\[ [\text{Stress}] = \frac{[MLT^{-2}]}{[L^2]} = [ML^{-1}T^{-2}] \]
Strain has no dimensions.
Exam Insight
- Stress = Force / Area
- Strain is dimensionless
- Longitudinal strain = ΔL / L
- Shear strain = θ (in radians)
- Hydraulic stress produces volume change only
Hooke’s Law
Hooke’s law is one of the most fundamental principles used to understand the elastic behaviour of solids. It establishes a direct relationship between the internal restoring force in a material and the deformation produced in it.
The law states that:
Within the elastic limit of a material, the strain produced in a body is directly proportional to the stress applied to it.
In simple terms, if the applied stress is increased, the resulting strain also increases in the same proportion, as long as the deformation remains within the elastic range of the material.
Mathematical Expression
Hooke’s law can be written as
\[ \text{Stress} \propto \text{Strain} \]
Introducing a constant of proportionality:
\[ \text{Stress} = k \times \text{Strain} \]
where \(k\) is called the modulus of elasticity. Its value depends on the nature of the material and the type of deformation.
Different types of deformation lead to different elastic constants:
- Young’s Modulus → longitudinal deformation
- Bulk Modulus → volume deformation
- Shear Modulus → shape deformation
Physical Meaning of Hooke’s Law
Hooke’s law implies that the internal restoring force developed in a material is proportional to the deformation produced.
At the microscopic level, this happens because atoms in a solid are held together by interatomic forces. When the body is stretched or compressed, these atomic bonds are slightly disturbed and develop restoring forces that try to return the atoms to their equilibrium positions.
Example: Hooke’s Law in a Spring
A common example of Hooke’s law is a spring. When a force is applied to stretch or compress a spring, the extension produced is proportional to the applied force.
For a spring:
\[ F = kx \]
where
- \(F\) = applied force
- \(x\) = extension produced
- \(k\) = spring constant
Stress–Strain Relation (Hooke’s Law Region)
In the straight-line region of the stress–strain graph, stress is directly proportional to strain. This region obeys Hooke’s law.
Limitations of Hooke’s Law
- Hooke’s law is valid only within the elastic limit.
- Beyond this limit, the stress–strain relation becomes nonlinear.
- Plastic deformation may begin after the elastic limit is exceeded.
- Materials may not return completely to their original shape.
Applications of Hooke’s Law
- Design of springs and shock absorbers
- Mechanical structures such as bridges and buildings
- Measuring forces using spring balances
- Engineering analysis of materials
Exam Insight
- Hooke’s law: Stress ∝ Strain
- Valid only within elastic limit
- Stress–strain graph is linear in Hooke's region
- Elastic modulus = Stress / Strain
- Spring form: \(F = kx\)
Young’s Modulus
Young’s modulus is a measure of the elastic stiffness of a material. It describes how strongly a material resists stretching or compression when a force is applied along its length.
It is defined as the ratio of longitudinal stress to longitudinal strain, provided the deformation is within the elastic limit of the material.
\[ Y = \frac{\text{Longitudinal Stress}}{\text{Longitudinal Strain}} \]
Mathematical Expression
\[ Y = \frac{\sigma}{\epsilon} \]
Substituting the expressions for stress and strain:
\[ Y = \frac{F/A}{\Delta L/L} \]
\[ Y = \frac{F \times L}{A \times \Delta L} \]
where
- \(F\) = applied force
- \(A\) = cross-sectional area
- \(L\) = original length of the rod
- \(\Delta L\) = change in length
Elastic Stretching of a Rod
A tensile force increases the length of a rod by a small amount ΔL.
SI Unit
Since strain is dimensionless, the unit of Young’s modulus is the same as that of stress.
\[ \text{Unit of } Y = N m^{-2} = Pascal (Pa) \]
Dimensional Formula
\[ [Y] = [ML^{-1}T^{-2}] \]
Young’s Modulus of Some Materials
| Material | Young's Modulus (Approx) |
|---|---|
| Steel | ≈ 2 × 10¹¹ Pa |
| Copper | ≈ 1.1 × 10¹¹ Pa |
| Aluminium | ≈ 7 × 10¹⁰ Pa |
| Rubber | ≈ 10⁷ Pa |
Materials with larger Young’s modulus are more rigid.
Practical Applications
- Design of bridges and buildings
- Construction of railway tracks
- Design of mechanical springs
- Selection of materials in engineering structures
Exam Insight
- \(Y = \dfrac{F L}{A \Delta L}\)
- Measures stiffness of a material
- Larger Y → more rigid material
- Unit = Pascal (Pa)
Shear Modulus (Modulus of Rigidity)
When a body is subjected to forces acting parallel to its surface, the body tends to change its shape without significantly changing its volume. This type of deformation is called shear deformation.
The ratio of shearing stress to the corresponding shearing strain is called the shear modulus of the material.
It is represented by the symbol G and is also known as the modulus of rigidity.
Mathematical Expression
\[ G = \frac{\text{Shearing Stress}}{\text{Shearing Strain}} \]
\[ G = \frac{\sigma_s}{\theta} \]
Substituting expressions for stress and strain:
\[ G = \frac{F/A}{\Delta x/L} \]
\[ G = \frac{F \times L}{A \times \Delta x} \]
Since
\[ \frac{\Delta x}{L} = \theta \]
\[ G = \frac{F}{A\theta} \]
Thus the shearing stress can also be written as
\[ \sigma_s = G\theta \]
Shear Deformation of a Block
Tangential forces distort the shape of the block producing shear strain.
Physical Meaning of Shear Modulus
Shear modulus measures the rigidity of a material, i.e., its resistance to changes in shape.
- Large value of G → material strongly resists shear deformation
- Small value of G → material easily changes shape
For example, steel has a very large shear modulus compared to rubber.
SI Unit
\[ \text{Unit of } G = N\,m^{-2} = \text{Pascal (Pa)} \]
Since strain is dimensionless, the unit of shear modulus is the same as that of stress.
Dimensional Formula
\[ [G] = [ML^{-1}T^{-2}] \]
Examples of Shear Deformation
- Twisting of a cylindrical shaft
- Distortion of rubber blocks
- Sliding motion between layers of a deck of cards
- Shearing of metal sheets in machinery
Applications
- Design of shafts in mechanical engineering
- Analysis of torsion in machine parts
- Material selection for structural components
- Manufacturing processes involving shear forces
Exam Insight
- \(G = \frac{F L}{A\Delta x}\)
- Also called modulus of rigidity
- Shear strain ≈ angle \( \theta \) (in radians)
- Unit = Pascal (Pa)
Bulk Modulus
When a solid body is subjected to pressure from all directions, such as when it is immersed in a fluid, the pressure acts perpendicularly on every surface of the body.
This produces a hydraulic stress which causes a uniform reduction in the volume of the body without changing its shape.
The resulting fractional change in volume is called volume strain.
\[ \text{Volume strain}=\frac{\Delta V}{V} \]
The ratio of hydraulic stress to the corresponding volume strain is called the bulk modulus.
It is represented by the symbol B.
Mathematical Expression
\[ B = -\frac{p}{\Delta V/V} \]
where
- \(p\) = hydraulic pressure applied
- \(V\) = original volume
- \(\Delta V\) = change in volume
The negative sign indicates that an increase in pressure causes a decrease in volume.
Thus if \(p\) is positive, \( \Delta V \) becomes negative.
For materials in equilibrium, the value of bulk modulus is always taken as positive.
Volume Compression Under Pressure
Uniform pressure compresses the volume of the material without changing its shape.
Physical Meaning of Bulk Modulus
Bulk modulus measures the resistance of a material to uniform compression.
- Large B → material is difficult to compress
- Small B → material compresses easily
For example:
- Liquids have high bulk modulus compared to gases.
- Steel is far less compressible than rubber.
SI Unit
\[ \text{Unit of } B = N\,m^{-2} = \text{Pascal (Pa)} \]
Since volume strain is dimensionless, the unit of bulk modulus is the same as that of pressure.
Dimensional Formula
\[ [B] = [ML^{-1}T^{-2}] \]
Examples
- Compression of liquids in hydraulic systems
- Pressure effects on deep-sea submarine structures
- Elastic behaviour of materials under high pressure
- Sound propagation in fluids (depends on bulk modulus)
Applications
- Design of hydraulic machines
- Deep-sea engineering structures
- Study of compressibility of fluids
- Material selection for pressure vessels
Exam Insight
- \(B = -\frac{p}{\Delta V/V}\)
- Measures resistance to compression
- Unit = Pascal (Pa)
- Liquids have large bulk modulus compared to gases
Compressibility
Compressibility describes how easily the volume of a substance decreases when pressure is applied. It provides a measure of the ability of a material to be compressed.
Mathematically, compressibility is defined as the reciprocal of bulk modulus.
\[ k = \frac{1}{B} \]
It can also be expressed as the fractional change in volume per unit increase in pressure.
Mathematical Expression
\[ k = -\frac{1}{\Delta p}\left(\frac{\Delta V}{V}\right) \]
where
- \(V\) = original volume of the body
- \(\Delta V\) = change in volume
- \(\Delta p\) = increase in pressure
- \(B\) = bulk modulus
The negative sign indicates that an increase in pressure causes a decrease in volume.
Concept of Compressibility
Greater compressibility means the material's volume decreases more easily under pressure.
Physical Meaning
- Large compressibility → material is easily compressed.
- Small compressibility → material strongly resists compression.
Because compressibility is the inverse of bulk modulus:
- Liquids → very small compressibility.
- Gases → very large compressibility.
SI Unit
\[ \text{Unit of } k = Pa^{-1} \]
Since compressibility is the reciprocal of bulk modulus, its unit is the inverse of Pascal.
Dimensional Formula
\[ [k] = [M^{-1}L T^{2}] \]
Examples
- Air in a syringe compresses easily (high compressibility).
- Water in a closed container shows very little compression.
- Hydraulic systems work efficiently because liquids have low compressibility.
Exam Insight
- \(k = \frac{1}{B}\)
- Unit = \(Pa^{-1}\)
- Gases → high compressibility
- Liquids → low compressibility
Poisson’s Ratio
When a wire or rod is stretched by applying a tensile force, its length increases. At the same time, its diameter decreases.
The decrease in dimension perpendicular to the applied force is called lateral contraction, and the corresponding strain produced is called lateral strain.
The French scientist Siméon Denis Poisson observed that within the elastic limit, the lateral strain is proportional to the longitudinal strain.
The ratio of lateral strain to longitudinal strain is called Poisson’s ratio.
Lateral and Longitudinal Strain
If the original diameter of a wire is \(d\) and the decrease in diameter is \(\Delta d\), then
\[ \text{Lateral strain} = \frac{\Delta d}{d} \]
If the original length of the wire is \(L\) and the increase in length is \(\Delta L\), then
\[ \text{Longitudinal strain} = \frac{\Delta L}{L} \]
Mathematical Expression
\[ \text{Poisson’s ratio } (\nu) = \frac{\text{Lateral strain}}{\text{Longitudinal strain}} \]
\[ \nu = \frac{\Delta d/d}{\Delta L/L} \]
\[ \nu = \frac{\Delta d}{d} \times \frac{L}{\Delta L} \]
Because the diameter decreases when the length increases, the lateral strain is negative. Therefore the expression is often written as
\[ \nu = - \frac{\text{Lateral strain}}{\text{Longitudinal strain}} \]
Lateral Contraction in a Stretched Wire
When a rod is stretched, its length increases while its diameter decreases.
Unit
Since Poisson’s ratio is a ratio of two strains, it has no unit and is therefore a dimensionless quantity.
Typical Values
| Material | Poisson's Ratio |
|---|---|
| Steel | ≈ 0.28 – 0.30 |
| Copper | ≈ 0.34 |
| Rubber | ≈ 0.45 – 0.50 |
| Cork | ≈ 0 |
Physical Meaning
- Large Poisson’s ratio → significant lateral contraction.
- Small Poisson’s ratio → very little change in width.
- Materials like cork have almost zero Poisson’s ratio.
Exam Insight
- \(\nu = -\frac{\text{lateral strain}}{\text{longitudinal strain}}\)
- Dimensionless quantity
- Typical range: \(0 \leq \nu \leq 0.5\)
- Cork → Poisson’s ratio ≈ 0
Elastic Potential Energy in a Stretched Wire
When a wire is stretched by an external force, work is done on the wire. If the deformation remains within the elastic limit, this work is not lost; instead it is stored in the wire in the form of elastic potential energy.
This stored energy exists because the atoms inside the material are displaced from their equilibrium positions and develop restoring forces that try to bring the wire back to its original length.
Derivation of Elastic Potential Energy
Consider a wire having:
- Original length = \(L\)
- Cross-sectional area = \(A\)
- Young’s modulus = \(Y\)
Let the wire be stretched by a small extension \(l\) within the elastic limit.
From the definition of Young’s modulus,
\[ Y = \frac{\text{stress}}{\text{strain}} \]
\[ Y = \frac{F/A}{l/L} \]
Rearranging the equation,
\[ F = \frac{YA}{L} \, l \]
Thus, the force required to stretch the wire is directly proportional to the extension.
Work Done in Stretching
While stretching the wire, the force gradually increases from 0 to \(F\). Therefore, the average force acting during the extension is
\[ F_{avg} = \frac{F}{2} \]
Work done in stretching the wire by extension \(l\) is
\[ W = F_{avg} \times l \]
\[ W = \frac{1}{2}Fl \]
Substituting \(F = \frac{YA}{L}l\),
\[ U = \frac{1}{2}\frac{YA}{L}l^2 \]
This work done is stored in the wire as elastic potential energy.
Final Expression
\[ U = \frac{1}{2}\frac{YA}{L}l^2 \]
where
- \(U\) = elastic potential energy stored
- \(Y\) = Young’s modulus
- \(A\) = cross-sectional area
- \(L\) = original length
- \(l\) = extension produced
Force–Extension Graph
The elastic potential energy stored equals the area under the force–extension graph.
Elastic Energy Density
Energy stored per unit volume of the material is called elastic energy density.
\[ u = \frac{U}{AL} \]
Using stress \( \sigma = \frac{F}{A} \) and strain \( \epsilon = \frac{l}{L} \),
\[ u = \frac{1}{2}\sigma\epsilon \]
Physical Significance
- Energy is stored due to elastic deformation.
- This energy is completely recoverable within the elastic limit.
- Doubling extension increases stored energy four times.
- Used in suspension cables, springs, and structural wires.
Exam Insight
- \(U = \frac{1}{2}\frac{YA}{L}l^2\)
- Energy = area under force–extension graph
- Energy density = \( \frac{1}{2}\sigma\epsilon \)
- Valid only within elastic limit
Applications of Elastic Behaviour of Materials
Elastic behaviour of materials plays a crucial role in engineering, construction, and mechanical design. A material is said to exhibit elastic behaviour if it regains its original shape and size after the deforming force is removed, provided the deformation remains within the elastic limit.
The quantitative description of elasticity using Young’s modulus, Bulk modulus, and Shear modulus helps engineers select materials capable of withstanding mechanical loads while maintaining structural stability.
1. Design of Bridges, Buildings and Structural Elements
Structural components such as beams, pillars, and girders experience continuous tensile, compressive, and bending stresses. Materials with a large Young’s modulus are preferred because they undergo very small deformation under load.
For a rod or wire of length \(L\), cross-sectional area \(A\), and Young’s modulus \(Y\), the extension produced by a force \(F\) is
\[ Y=\frac{F/A}{l/L} \] \[ l=\frac{FL}{YA} \]
A smaller extension ensures that structures remain stable and safe even under heavy loads.
Elastic stability is essential in bridges and buildings.
2. Suspension Cables and Lifting Machines
Steel cables used in cranes, elevators, and suspension bridges must support large loads without excessive stretching.
Elastic behaviour ensures that cables stretch slightly under load and return to their original length once the load is removed.
The elastic energy stored in stretched cables also helps absorb sudden shocks and vibrations.
3. Springs and Shock Absorbers
Springs are among the most common applications of elastic behaviour. Within the elastic limit, they obey Hooke’s law.
\[ F = kx \]
where \(k\) is the spring constant and \(x\) is the extension.
Springs store energy in the form of elastic potential energy and release it gradually, which is why they are used in:
- Vehicle suspension systems
- Weighing machines
- Mechanical watches and clocks
4. Hydraulic Systems
Hydraulic machines rely on liquids with a high bulk modulus, meaning they resist compression and transmit pressure efficiently.
\[ K = -\frac{\Delta P}{\Delta V/V} \]
Hydraulic systems are used in:
- Hydraulic presses
- Hydraulic brakes
- Heavy lifting machines
5. Musical Instruments
Strings in instruments such as guitars and violins rely on elastic behaviour. The tension and elastic properties of the string determine the frequency of vibration and therefore the pitch of sound produced.
Stable elastic behaviour ensures consistent vibration and high-quality sound.
6. Safety in Mechanical Design
Engineers always ensure that stresses in machines and structures remain well below the elastic limit of the material.
This ensures that components return to their original shape after use and prevents permanent deformation or mechanical failure.
Physical Significance
- Provides structural stability
- Allows storage of elastic potential energy
- Absorbs mechanical shocks
- Ensures safe and predictable mechanical behaviour
Exam Insight
- High Young’s modulus → small deformation
- Springs obey Hooke’s law \(F=kx\)
- Hydraulic systems depend on large bulk modulus
- Elastic behaviour ensures structural safety
Numericals
Example 1: Stress, Strain and Elongation of a Steel Rod
A structural steel rod has a radius of 10 mm and a length of 1.0 m. A force of 100 kN stretches it along its length. Calculate:
- (a) Stress on the rod
- (b) Elongation produced
- (c) Strain in the rod
Young’s modulus of structural steel: \[ Y = 2.0 \times 10^{11}\,N\,m^{-2} \]
A tensile force stretches the rod producing a small extension.
Given Data
Radius of rod: \[ r = 10\,mm = 10\times10^{-3}\,m \] Length of rod: \[ L = 1.0\,m \] Force applied: \[ F = 100\,kN = 100\times10^3\,N \] Young’s modulus: \[ Y = 2.0\times10^{11}\,N\,m^{-2} \]
(a) Stress on the Rod
Cross-sectional area of rod:
\[ A = \pi r^2 \] \[ A = \pi(10\times10^{-3})^2 \] \[ A = \pi\times10^{-4}\,m^2 \]Stress is given by
\[ \sigma = \frac{F}{A} \] \[ \sigma = \frac{100\times10^{3}}{\pi\times10^{-4}} \] \[ \sigma = \frac{100\times10^{7}}{\pi} \] \[ \sigma \approx 3.2\times10^{8}\,N\,m^{-2} \](b) Elongation of the Rod
Using the relation
\[ Y = \frac{FL}{A\Delta L} \]Rearranging,
\[ \Delta L = \frac{FL}{AY} \] \[ \Delta L = \frac{100\times10^{3}\times1} {\pi\times10^{-4}\times2.0\times10^{11}} \] \[ \Delta L = \frac{100}{2\pi}\times10^{-4} \] \[ \Delta L \approx 1.6\times10^{-4}\,m \] \[ \Delta L = 0.16\,mm \](c) Strain Produced
\[ \varepsilon = \frac{\Delta L}{L} \] \[ \varepsilon = \frac{1.6\times10^{-4}}{1} \] \[ \varepsilon = 1.6\times10^{-4} \]In percentage form:
\[ \varepsilon(\%) = 1.6\times10^{-4}\times100 \] \[ \varepsilon = 0.016\% \]Final Answers
- Stress = \(3.2\times10^8\;N\,m^{-2}\)
- Elongation = \(1.6\times10^{-4}\,m = 0.16\,mm\)
- Strain = \(1.6\times10^{-4}\)
Exam Tip
- Always convert mm → m before substitution.
- Remember formula: \[ \Delta L = \frac{FL}{AY} \]
- Strain is dimensionless.
Example 2: Copper and Steel Wires Connected in Series
A copper wire of length 2.2 m and a steel wire of length 1.6 m, both having diameter 3.0 mm, are connected end-to-end. When stretched by a load, the total elongation is 0.70 mm. Find the load applied.
Two wires connected in series experience the same tension.
Given
\[ L_c = 2.2\,m \] \[ L_s = 1.6\,m \] Diameter \(=3\,mm\) \[ r=1.5\times10^{-3}\,m \] \[ \Delta L =0.70\times10^{-3}\,m \] \[ Y_s = 200\times10^9\,Nm^{-2} \] \[ Y_c =110\times10^9\,Nm^{-2} \]
Relation between elongations Stress is same because tension is same. \[ Y_s\frac{\Delta L_s}{L_s} = Y_c\frac{\Delta L_c}{L_c} \] \[ \frac{\Delta L_s}{\Delta L_c} = \frac{Y_c}{Y_s}\frac{L_s}{L_c} \] \[ = \frac{110}{200}\times\frac{1.6}{2.2} \] \[ = \frac{2}{5} \] \[ \Delta L_s=\frac{2}{5}\Delta L_c \] Using total elongation \[ \Delta L_c+\Delta L_s=0.70\times10^{-3} \] \[ \Delta L_c+\frac{2}{5}\Delta L_c=0.70\times10^{-3} \] \[ \frac{7}{5}\Delta L_c=0.70\times10^{-3} \] \[ \Delta L_c=0.50\times10^{-3}m \] \[ \Delta L_s=0.20\times10^{-3}m \] Load calculation Cross-sectional area \[ A=\pi r^2 \] \[ =\pi(1.5\times10^{-3})^2 \] \[ =\pi\times2.25\times10^{-6} \] Using steel wire \[ F=Y_sA\frac{\Delta L_s}{L_s} \] \[ F\approx180\,N \] Final Answer- Load applied = 180 N
- Series wires → same force
- Total extension = sum of individual extensions
Example 3: Compression of Human Thighbone
In a human pyramid, the performer at the bottom supports the entire weight of the group using his legs. Find the compression in each thighbone (femur).
Given Total mass of group \[ 280\,kg \] Mass of bottom performer \[ 60\,kg \] Effective mass on legs \[ m=220\,kg \] Weight supported \[ W=220\times9.8 \] \[ W=2156\,N \] Force on each leg \[ F=\frac{2156}{2}=1078\,N \] Bone dimensions Length \[ L=0.50\,m \] Radius \[ r=2\times10^{-2}m \] Young's modulus of bone \[ Y=9.4\times10^9\,Nm^{-2} \] Area \[ A=\pi r^2 \] \[ =\pi(2\times10^{-2})^2 \] \[ =\pi\times4\times10^{-4} \] Compression \[ \Delta L=\frac{FL}{YA} \] \[ \Delta L = \frac{1078\times0.5} {9.4\times10^9\times\pi\times4\times10^{-4}} \] \[ \Delta L \approx4.6\times10^{-5}m \] Final Result \[ \Delta L\approx4.6\times10^{-5}m \] \[ \Delta L\approx0.046\,mm \] Each thighbone compresses by **0.046 mm**. Exam Insight- Load shared by both legs
- Use compression formula \( \Delta L = FL/YA \)
Example-4
A square lead slab of side 50 cm and thickness 10 cm is subject to a shearing force on its narrow face of \(9.0 \times 10^{4}\,N\). The lower edge is riveted to the floor. Calculate the displacement of the upper edge.
Given Data
Side of square slab \[ 50\,cm = 0.50\,m \] Thickness of slab \[ L = 10\,cm = 0.10\,m \] Shearing force \[ F = 9.0 \times 10^{4}\,N \] Shear modulus of lead \[ G = 5.6 \times 10^{9}\,N\,m^{-2} \]Step-1: Area of the Face
Area on which force acts \[ A = 0.50 \times 0.50 \] \[ A = 0.25\,m^{2} \]Step-2: Formula for Shear Modulus
\[ G = \frac{F}{A}\frac{L}{\Delta x} \] Rearranging, \[ \Delta x = \frac{FL}{GA} \]Step-3: Substitution
\[ \Delta x = \frac{9.0 \times 10^{4} \times 0.10} {5.6 \times 10^{9} \times 0.25} \] \[ = \frac{9.0 \times 10^{3}} {1.4 \times 10^{9}} \] \[ \Delta x \approx 6.4 \times 10^{-6}\,m \]Final Answer
Displacement of the upper edge \[ \Delta x \approx 6.4 \times 10^{-6}\,m \] \[ \Delta x \approx 6.4\,\mu m \]Example-5
The average depth of the Indian Ocean is about 3000 m. Calculate the fractional compression \( \Delta V / V \) of water at the bottom of the ocean. Bulk modulus of water \( B = 2.2 \times 10^{9}\,N\,m^{-2} \). (Take \( g = 10\,m\,s^{-2} \))
Given Data
Average depth of ocean \[ h = 3000\,m \] Bulk modulus of water \[ B = 2.2 \times 10^{9}\,N\,m^{-2} \] Acceleration due to gravity \[ g = 10\,m\,s^{-2} \] Density of water \[ \rho = 1000\,kg\,m^{-3} \]Step-1: Pressure at the Bottom of Ocean
Pressure at depth \(h\) \[ P = \rho g h \] Substituting values \[ P = 1000 \times 10 \times 3000 \] \[ P = 3.0 \times 10^{7}\,N\,m^{-2} \]Step-2: Bulk Modulus Relation
\[ B = \frac{P}{\Delta V/V} \] Rearranging \[ \frac{\Delta V}{V} = \frac{P}{B} \]Step-3: Substitution
\[ \frac{\Delta V}{V} = \frac{3.0 \times 10^{7}} {2.2 \times 10^{9}} \] \[ \frac{\Delta V}{V} \approx 1.36 \times 10^{-2} \]Final Answer
Fractional compression of water at the bottom \[ \frac{\Delta V}{V} \approx 1.4 \times 10^{-2} \]Mechanical Properties of Solids – All Formulas in One Place
This section provides a quick revision of all important formulas from the chapter Mechanical Properties of Solids. These formulas are frequently used in numerical problems and competitive examinations.
1. Stress
Stress is defined as the internal restoring force per unit area developed in a body.
\[ \text{Stress} = \frac{F}{A} \]Unit: Pascal (Pa)
2. Strain
Strain is the fractional change in dimension of a body.
\[ \text{Strain} = \frac{\text{Change in dimension}}{\text{Original dimension}} \]Strain is a dimensionless quantity.
3. Hooke’s Law
Within the elastic limit, stress is directly proportional to strain.
\sigma \propto \epsilon4. Young’s Modulus
Ratio of longitudinal stress to longitudinal strain.
\[ Y = \frac{\text{Stress}}{\text{Strain}} \] \[ Y = \frac{F L}{A \Delta L} \]5. Shear Modulus (Modulus of Rigidity)
Ratio of shearing stress to shearing strain.
\[ G = \frac{\text{Shear Stress}}{\text{Shear Strain}} \] \[ G = \frac{F L}{A \Delta x} \]6. Bulk Modulus
Ratio of hydraulic stress to volume strain.
\[ B = - \frac{P}{\Delta V / V} \]7. Compressibility
\[ k = \frac{1}{B} \]Compressibility represents the fractional change in volume per unit increase in pressure.
8. Poisson’s Ratio
Ratio of lateral strain to longitudinal strain.
\[ \nu = - \frac{\text{Lateral strain}}{\text{Longitudinal strain}} \]9. Elastic Potential Energy of a Stretched Wire
Energy stored due to elastic deformation.
\[ U = \frac{1}{2} F l \] \[ U = \frac{1}{2}\frac{YA}{L}l^2 \]10. Elastic Energy Density
Energy stored per unit volume.
\[ u = \frac{1}{2}\sigma \epsilon \]Quick Exam Reminder
- Stress unit = Pascal (Pa)
- Strain has no unit
- Young’s modulus measures stiffness
- Bulk modulus measures resistance to compression
- Shear modulus measures resistance to shape change
- Elastic energy equals the area under the force–extension graph
Top 10 Conceptual Mistakes Students Make in Mechanical Properties of Solids
Students often lose marks in this chapter not because formulas are difficult, but because of conceptual misunderstandings. The following are the most common mistakes observed in exams like CBSE, NEET, and JEE.
Stress is the internal restoring force per unit area developed inside a material, whereas pressure is an external force per unit area applied on a surface.
Strain is the ratio of two lengths. Therefore it has no unit. Many students incorrectly write units for strain.
Hooke’s law is valid only within the elastic limit of a material. Beyond this region, stress is no longer proportional to strain.
\sigma \propto \epsilonWhen wires are connected end-to-end, the same tension acts throughout the system. However, the elongations can be different because materials and lengths differ.
Lengths are often given in mm or cm in numerical problems. These must be converted to meters before substitution in formulas.
The expression for bulk modulus contains a negative sign because an increase in pressure leads to a decrease in volume.
Shear strain represents angular deformation and is approximately equal to the angle of shear in radians.
The elastic potential energy stored in a stretched wire equals the area under the force–extension graph.
When a rod is stretched, its diameter decreases. Many students forget that lateral strain is negative relative to longitudinal strain.
Extensions in metals are usually very small (often in micrometers). If your answer is several centimeters, it is probably incorrect.
Always follow the sequence:
- Identify the correct elastic constant (Y, G, or B)
- Write the appropriate formula
- Convert all quantities to SI units
- Substitute values carefully
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