The Strain Energy Formula gives the total elastic potential energy stored in a deformed body, expressed as Υ = ½ × σ × ε × V, where σ is stress, ε is strain, and V is the volume of the material. This concept is a core part of the Elasticity chapter in NCERT Class 11 Physics (Chapter 9) and is frequently tested in CBSE board exams. JEE Main and JEE Advanced aspirants encounter strain energy in problems involving elastic deformation, springs, and material mechanics. This article covers the formula derivation, a complete formula sheet, three progressive solved examples, CBSE exam tips, common mistakes, and JEE/NEET applications.
Key Strain Energy Formulas at a Glance
Quick reference for the most important strain energy formulas used in CBSE and JEE/NEET exams.
- Strain Energy (general): \( U = \frac{1}{2} \sigma \varepsilon V \)
- Strain Energy Density: \( u = \frac{1}{2} \sigma \varepsilon = \frac{\sigma^2}{2E} \)
- Strain Energy in a spring: \( U = \frac{1}{2} k x^2 \)
- Strain Energy (in terms of stress): \( U = \frac{\sigma^2 V}{2E} \)
- Strain Energy (in terms of strain): \( U = \frac{1}{2} E \varepsilon^2 V \)
- Modulus of Resilience: \( u_r = \frac{\sigma_y^2}{2E} \)
- Strain Energy in torsion: \( U = \frac{T^2 L}{2GJ} \)
What is the Strain Energy Formula?
The Strain Energy Formula defines the elastic potential energy stored inside a material when it undergoes deformation under an applied load. When an external force stretches, compresses, or twists a body, the body stores energy in its intermolecular bonds. This stored energy is called strain energy. When the load is removed, a perfectly elastic body releases this energy and returns to its original shape.
In NCERT Class 11 Physics, Chapter 9 (Mechanical Properties of Solids), strain energy is introduced alongside stress, strain, and Young's modulus. The concept builds directly on Hooke's Law, which states that stress is proportional to strain within the elastic limit.
Mathematically, strain energy is the area under the stress-strain curve up to the point of interest. For a linearly elastic material obeying Hooke's Law, this area is a triangle. The formula therefore carries the factor of one-half. Strain energy is a scalar quantity. Its SI unit is the Joule (J). Understanding the Strain Energy Formula is essential for solving problems in elasticity, structural mechanics, and vibration analysis.
Strain Energy Formula — Expression and Variables
The general Strain Energy Formula for a uniformly stressed elastic body is:
\[ U = \frac{1}{2} \sigma \varepsilon V \]
Since Young's Modulus \( E = \dfrac{\sigma}{\varepsilon} \), we can also write the formula in two alternative forms:
\[ U = \frac{\sigma^2 V}{2E} \quad \text{(in terms of stress)} \]
\[ U = \frac{1}{2} E \varepsilon^2 V \quad \text{(in terms of strain)} \]
The strain energy per unit volume (strain energy density) is:
\[ u = \frac{U}{V} = \frac{1}{2} \sigma \varepsilon = \frac{\sigma^2}{2E} \]
| Symbol | Quantity | SI Unit |
|---|---|---|
| \( U \) | Strain Energy (total) | Joule (J) |
| \( u \) | Strain Energy Density | J/m³ or Pa |
| \( \sigma \) | Stress (force per unit area) | Pascal (Pa) or N/m² |
| \( \varepsilon \) | Strain (dimensionless ratio) | Dimensionless |
| \( V \) | Volume of the body | m³ |
| \( E \) | Young's Modulus of Elasticity | Pascal (Pa) or N/m² |
| \( F \) | Applied Force | Newton (N) |
| \( x \) | Extension or compression | metre (m) |
| \( k \) | Spring constant (for spring) | N/m |
Derivation of the Strain Energy Formula
Consider a rod of length \( L \), cross-sectional area \( A \), and Young's Modulus \( E \). An external force \( F \) stretches it by \( x \).
Step 1: By Hooke's Law, the restoring force at extension \( x \) is \( F = kx \), where \( k = \dfrac{EA}{L} \).
Step 2: The work done by the external force for a small displacement \( dx \) is \( dW = F \, dx = kx \, dx \).
Step 3: Integrate from 0 to total extension \( x_0 \):
\[ U = \int_0^{x_0} kx \, dx = \frac{1}{2} k x_0^2 = \frac{1}{2} F x_0 \]
Step 4: Express in terms of stress and strain. Since \( \sigma = F/A \), \( \varepsilon = x_0/L \), and \( V = AL \):
\[ U = \frac{1}{2} \cdot \frac{F}{A} \cdot \frac{x_0}{L} \cdot AL = \frac{1}{2} \sigma \varepsilon V \]
This completes the derivation. The factor of one-half arises because the force builds up linearly from zero to its final value.
Complete Elasticity Formula Sheet
| Formula Name | Expression | Variables | SI Units | NCERT Chapter |
|---|---|---|---|---|
| Strain Energy (general) | \( U = \frac{1}{2} \sigma \varepsilon V \) | σ = stress, ε = strain, V = volume | J | Class 11, Ch 9 |
| Strain Energy (stress form) | \( U = \frac{\sigma^2 V}{2E} \) | σ = stress, V = volume, E = Young's modulus | J | Class 11, Ch 9 |
| Strain Energy (strain form) | \( U = \frac{1}{2} E \varepsilon^2 V \) | E = Young's modulus, ε = strain, V = volume | J | Class 11, Ch 9 |
| Strain Energy Density | \( u = \frac{\sigma^2}{2E} \) | σ = stress, E = Young's modulus | J/m³ | Class 11, Ch 9 |
| Strain Energy in a Spring | \( U = \frac{1}{2} k x^2 \) | k = spring constant, x = displacement | J | Class 11, Ch 14 |
| Young's Modulus | \( E = \frac{\sigma}{\varepsilon} = \frac{F L}{A \Delta L} \) | F = force, L = length, A = area, ΔL = extension | Pa | Class 11, Ch 9 |
| Longitudinal Stress | \( \sigma = \frac{F}{A} \) | F = force, A = cross-sectional area | Pa | Class 11, Ch 9 |
| Longitudinal Strain | \( \varepsilon = \frac{\Delta L}{L} \) | ΔL = change in length, L = original length | Dimensionless | Class 11, Ch 9 |
| Modulus of Resilience | \( u_r = \frac{\sigma_y^2}{2E} \) | σy = yield stress, E = Young's modulus | J/m³ | Class 11, Ch 9 |
| Strain Energy in Shear | \( U = \frac{\tau^2 V}{2G} \) | τ = shear stress, V = volume, G = shear modulus | J | Class 11, Ch 9 |
| Strain Energy in Torsion | \( U = \frac{T^2 L}{2GJ} \) | T = torque, L = length, G = shear modulus, J = polar moment | J | Class 11, Ch 9 |
| Bulk Modulus | \( B = -\frac{\Delta P}{\Delta V / V} \) | ΔP = pressure change, ΔV = volume change | Pa | Class 11, Ch 9 |
Strain Energy Formula — Solved Examples
Example 1 (Class 9-10 Level) — Strain Energy in a Steel Wire
Problem: A steel wire of length 2 m and cross-sectional area 4 × 10²&sup4; m² is stretched by a force of 200 N. The Young's Modulus of steel is 2 × 10¹¹ Pa. Find the strain energy stored in the wire.
Given:
- Length, \( L = 2 \) m
- Area, \( A = 4 \times 10^{-4} \) m²
- Force, \( F = 200 \) N
- Young's Modulus, \( E = 2 \times 10^{11} \) Pa
Step 1: Calculate stress.
\[ \sigma = \frac{F}{A} = \frac{200}{4 \times 10^{-4}} = 5 \times 10^5 \text{ Pa} \]
Step 2: Calculate volume.
\[ V = A \times L = 4 \times 10^{-4} \times 2 = 8 \times 10^{-4} \text{ m}^3 \]
Step 3: Apply the Strain Energy Formula \( U = \dfrac{\sigma^2 V}{2E} \).
\[ U = \frac{(5 \times 10^5)^2 \times 8 \times 10^{-4}}{2 \times 2 \times 10^{11}} \]
\[ U = \frac{25 \times 10^{10} \times 8 \times 10^{-4}}{4 \times 10^{11}} = \frac{2 \times 10^{8}}{4 \times 10^{11}} = 5 \times 10^{-4} \text{ J} \]
Answer
Strain Energy stored in the wire = \( 5 \times 10^{-4} \) J = 0.5 mJ
Example 2 (Class 11-12 Level) — Strain Energy Density Comparison
Problem: Two rods A and B are made of the same material (Young's Modulus \( E = 1.2 \times 10^{11} \) Pa). Rod A has length 1 m and diameter 10 mm. Rod B has length 2 m and diameter 20 mm. Both rods are subjected to the same axial stress of \( 6 \times 10^7 \) Pa. Compare (a) the strain energy density and (b) the total strain energy stored in each rod.
Given:
- \( E = 1.2 \times 10^{11} \) Pa for both rods
- \( \sigma = 6 \times 10^7 \) Pa for both rods
- Rod A: \( L_A = 1 \) m, \( d_A = 10 \) mm = 0.01 m
- Rod B: \( L_B = 2 \) m, \( d_B = 20 \) mm = 0.02 m
Step 1: Calculate strain energy density for both rods.
Since both rods have the same stress and the same Young's Modulus:
\[ u = \frac{\sigma^2}{2E} = \frac{(6 \times 10^7)^2}{2 \times 1.2 \times 10^{11}} = \frac{36 \times 10^{14}}{2.4 \ times 10^{11}} = 1.5 \times 10^4 \text{ J/m}^3 \]
The strain energy density is the same for both rods (15,000 J/m³).
Step 2: Calculate volumes.
\[ V_A = \frac{\pi d_A^2}{4} \times L_A = \frac{\pi (0.01)^2}{4} \times 1 = 7.854 \times 10^{-5} \text{ m}^3 \]
\[ V_B = \frac{\pi d_B^2}{4} \times L_B = \frac{\pi (0.02)^2}{4} \times 2 = 6.283 \times 10^{-4} \text{ m}^3 \]
Step 3: Calculate total strain energy.
\[ U_A = u \times V_A = 1.5 \times 10^4 \times 7.854 \times 10^{-5} = 1.178 \text{ J} \]
\[ U_B = u \times V_B = 1.5 \times 10^4 \times 6.283 \times 10^{-4} = 9.425 \text{ J} \]
Step 4: Compare. The ratio \( U_B / U_A = 9.425 / 1.178 \approx 8 \). Rod B stores 8 times more total strain energy than Rod A, even though the strain energy density is identical for both.
Answer
(a) Strain energy density is equal for both rods: \( 1.5 \times 10^4 \) J/m³. (b) Total strain energy: Rod A = 1.178 J, Rod B = 9.425 J. Rod B stores 8 times more total energy due to its larger volume.
Example 3 (JEE/NEET Level) — Spring System and Strain Energy
Problem: A block of mass 2 kg is dropped from a height of 0.5 m onto a spring of spring constant \( k = 800 \) N/m placed vertically on the ground. Find (a) the maximum compression of the spring and (b) the strain energy stored in the spring at maximum compression. Take \( g = 10 \) m/s². Ignore the mass of the spring.
Given:
- Mass of block, \( m = 2 \) kg
- Height of drop, \( h = 0.5 \) m
- Spring constant, \( k = 800 \) N/m
- \( g = 10 \) m/s²
Step 1: Apply conservation of energy. Let \( x \) be the maximum compression. At maximum compression, the block is momentarily at rest. Taking the top of the uncompressed spring as the reference level:
\[ mg(h + x) = \frac{1}{2} k x^2 \]
Step 2: Substitute values.
\[ 2 \times 10 \times (0.5 + x) = \frac{1}{2} \times 800 \times x^2 \]
\[ 20(0.5 + x) = 400 x^2 \]
\[ 10 + 20x = 400 x^2 \]
Step 3: Rearrange into a quadratic equation.
\[ 400 x^2 – 20x – 10 = 0 \]
\[ 40 x^2 – 2x – 1 = 0 \]
Step 4: Solve using the quadratic formula.
\[ x = \frac{2 \pm \sqrt{4 + 4 \times 40 \times 1}}{2 \times 40} = \frac{2 \pm \sqrt{164}}{80} = \frac{2 \pm 12.806}{80} \]
Taking the positive root: \( x = \dfrac{14.806}{80} \approx 0.185 \) m
Step 5: Calculate strain energy stored in the spring at maximum compression.
\[ U = \frac{1}{2} k x^2 = \frac{1}{2} \times 800 \times (0.185)^2 = 400 \times 0.03422 \approx 13.69 \text{ J} \]
Verification: Energy input = \( mg(h + x) = 20 \times (0.5 + 0.185) = 20 \times 0.685 = 13.7 \) J ✓
Answer
(a) Maximum compression of the spring \( x \approx 0.185 \) m (18.5 cm). (b) Strain energy stored in the spring at maximum compression \( U \approx 13.69 \) J.
CBSE Exam Tips 2025-26
- Memorise all three forms: We recommend learning \( U = \frac{1}{2}\sigma\varepsilon V \), \( U = \frac{\sigma^2 V}{2E} \), and \( U = \frac{1}{2}E\varepsilon^2 V \). CBSE 2025-26 questions often give only stress or only strain, so you need the right form quickly.
- Strain energy density is key: Questions in Section B of CBSE papers frequently ask for energy per unit volume. Always remember \( u = \sigma^2 / 2E \). This is the most commonly tested variant.
- Check units carefully: Stress is in Pa (N/m²), volume in m³, and Young's Modulus in Pa. The unit of strain energy is always Joules. A unit mismatch is the most common cause of lost marks.
- Link to Hooke's Law: Strain energy is valid only within the elastic limit. CBSE frequently asks whether a material obeys Hooke's Law before you can apply the formula. Always state this assumption in your answer.
- Spring problems: For spring-related strain energy questions, use \( U = \frac{1}{2}kx^2 \). This formula appears in both Chapter 9 (Elasticity) and Chapter 14 (Oscillations). Our experts suggest practising both contexts.
- Draw the stress-strain graph: In 3-mark or 5-mark questions, always draw the stress-strain curve and shade the area representing strain energy. This earns you diagram marks even if your calculation has a minor error.
Common Mistakes to Avoid
- Forgetting the factor of ½: Many students write \( U = \sigma \varepsilon V \) without the half. This mistake happens because students confuse strain energy with work done by a constant force. The force here builds from zero, so the average force is half the final value. Always include the ½ factor.
- Using diameter instead of radius for area: In solved examples, the cross-sectional area \( A = \pi r^2 = \pi d^2 / 4 \). Students often substitute \( d \) directly as \( r \). This gives an area four times too large and an energy sixteen times too large.
- Confusing strain energy with potential energy: Gravitational potential energy is \( mgh \). Strain energy is elastic potential energy stored in the material's bonds. These are different forms of energy. In spring-drop problems, both are present and must be handled separately.
- Ignoring the elastic limit condition: The Strain Energy Formula applies only when the material is within its elastic limit. If the applied stress exceeds the yield stress, the formula is no longer valid. Always check whether the problem states the material is elastic or obeys Hooke's Law.
- Wrong substitution in the quadratic (spring problems): In problems where a mass falls onto a spring, students forget to include \( mgx \) (the work done by gravity during compression) alongside \( mgh \). The correct energy equation is \( mg(h + x) = \frac{1}{2}kx^2 \), not \( mgh = \frac{1}{2}kx^2 \).
JEE/NEET Application of Strain Energy Formula
In our experience, JEE aspirants encounter the Strain Energy Formula in at least two to three questions per year across JEE Main and JEE Advanced papers. The formula bridges elasticity, energy methods, and simple harmonic motion. Here are the three most important application patterns.
Pattern 1: Comparing Strain Energy in Rods of Different Dimensions
JEE frequently gives two or more rods made of the same or different materials, subjected to the same force or same stress. You must compare their strain energies using \( U = \sigma^2 V / 2E \). The key insight is: at the same stress, strain energy is proportional to volume. At the same force, strain energy depends on both geometry and material. Practice identifying which quantity is held constant.
Pattern 2: Energy Conservation with Springs and Masses
NEET and JEE Main regularly test energy conservation problems involving springs. A mass dropped onto a spring, two masses connected by a spring, or a spring compressed by a moving block — all require the strain energy formula \( U = \frac{1}{2}kx^2 \). The trick is to set up the energy equation correctly, accounting for gravitational PE, kinetic energy, and spring strain energy at each stage. In our experience, the most common error is forgetting the \( mgx \) term during vertical spring compression.
Pattern 3: Strain Energy Density and Modulus of Resilience
JEE Advanced occasionally tests the concept of modulus of resilience, which is the maximum strain energy density a material can absorb without permanent deformation. It equals \( u_r = \sigma_y^2 / 2E \), where \( \sigma_y \) is the yield stress. Questions may ask you to compare two materials and determine which can absorb more energy per unit volume. A material with high yield stress and low Young's Modulus (like rubber) has a high modulus of resilience. This concept also appears in materials science questions in NEET biology-adjacent physics sections.
We recommend solving at least 20 previous year JEE questions on elasticity and strain energy. Refer to the official NCERT textbook at ncert.nic.in for the standard derivations and solved examples that form the basis of most board and competitive exam questions.
FAQs on Strain Energy Formula
Explore More Physics Formulas
Now that you have mastered the Strain Energy Formula, strengthen your understanding of related topics. Read our detailed guide on Bernoulli's Equation Formula to explore energy conservation in fluid mechanics. Study Buoyancy Formula to understand pressure and force in fluids. For optical properties of materials, visit our article on Brewster's Law Formula. Browse the complete Physics Formulas hub for a structured revision of all Class 11 and Class 12 topics covered in NCERT and CBSE syllabi.