QCC Notes
CLASS 11 · PHYSICSJEE MAIN × NEETहिंदी
§ 8.4NCERT Class 11 · Physics · Chapter 8

Elastic Potential Energy and Engineering Applications

QCC Notes · Class 11 Physics NCERT-aligned Updated 26 Sep 2026 12 min read

When external mechanical work is performed to deform an elastic solid, work is done against internal interatomic restoring forces. Within the elastic limit, this work is stored reversibly as Elastic Potential Energy (Strain Energy). In this module, we derive the exact equations for total strain energy and energy density, and explore four profound practical applications in civil, mechanical, and geological engineering.

1. Mathematical Derivation of Elastic Potential Energy

Consider a wire of original length L and uniform cross-sectional area A clamped rigidly at one end. Let an applied variable force F stretch the wire by an intermediate elongation x.

From Young's modulus definition, the restoring force at elongation x is:

F(x) = (Y × A / L) × x

The infinitesimal work dW done in extending the wire by an additional distance dx is:

dW = F(x) × dx = (Y × A / L) × x × dx

Integrating from x = 0 to the total elongation x = ΔL:

W = ∫ (Y × A / L) × x × dx = (Y × A / L) × [ΔL² / 2] = ½ × [(Y × A × ΔL) / L] × ΔL

Since the final stretching force is F = (Y A ΔL) / L, the stored strain energy U is:

U = W = ½ × Force × Elongation = ½ × F × ΔL

2. Elastic Energy Density (Energy Per Unit Volume)

Elastic energy density (u) is defined as the strain energy stored per unit volume of the material (Volume V = A × L):

u = U / V = (½ × F × ΔL) / (A × L) = ½ × (F / A) × (ΔL / L)
Energy Density (u) = ½ × Stress × Strain

Alternative Equivalent Expressions using Hooke's Law (σ = Y ε):

  • In terms of Young's Modulus and Strain: u = ½ × Y × ε² = ½ × Y × (ΔL / L)²
  • In terms of Young's Modulus and Stress: u = σ² / (2 Y)
  • In terms of Shear Modulus and Shear Strain: u = ½ × G × θ² = τ² / (2 G)
  • In terms of Bulk Modulus and Volume Strain: u = ½ × B × (ΔV / V)² = (ΔP)² / (2 B)
+-----------------------------------------------------------------------------------+ | GRAPHICAL REPRESENTATION OF ELASTIC ENERGY DENSITY | +-----------------------------------------------------------------------------------+ Stress (σ) ^ | / | / σ_final |...................* | / : | / : | / : | / : Area under Stress-Strain graph | / : = ½ × Base × Height | / : = ½ × Strain × Stress | / : = Energy Stored per Unit Volume (u) | / Area = : | / u : O +--------+----------+--------------------------------------> Strain (ε) ε_final +-----------------------------------------------------------------------------------+
Figure 8.6: The area under the linear stress-strain curve directly equals the elastic potential energy density (J/m³).

3. Bending of Beams and Design of I-Shaped Girders

A structural beam supported horizontally at its two ends and loaded centrally with a load W (or mass M, W = Mg) undergoes transverse deflection or sag (δ).

+-----------------------------------------------------------------------------------+ | BEAM DEFLECTION AND I-SHAPED GIRDER ARCHITECTURE | +-----------------------------------------------------------------------------------+ (a) Loaded Rectangular Beam (Length L, Breadth b, Depth d): Support A Load W = Mg Support B /=====\ | /===== +-------+.....................v......................+-------+ | | \ / | | | | \--+-------+--/ | | +-------+ | δ (sag)| +-------+ <---------------------------- L -----------------------------> Sag Equation: δ = (W × L³) / (4 × Y × b × d³) (b) Cross-Section of I-Shaped Girder: +=======================+ <--- Top Flange (Resists Compressive Bending) | | Width = b +===========+===========+ | | | | <-------------- Vertical Web (Resists Shear, Prevents Buckling) | | Depth = d +===========+===========+ | | <--- Bottom Flange (Resists Tensile Bending) +=======================+ +-----------------------------------------------------------------------------------+
Figure 8.7: Mechanical deflection of a center-loaded beam and cross-sectional geometry of an I-shaped girder.

Analytical Breakdown of the Sag Equation:

Sag (δ) = (W × L³) / (4 × Y × b × d³)
  • Effect of Material: Higher Young's modulus Y decreases sag (δ ∝ 1/Y). Steel is vastly superior to aluminum or wood.
  • Effect of Depth vs Breadth: Sag is inversely proportional to breadth b (δ ∝ 1/b), but inversely proportional to the cube of depth d (δ ∝ 1/d³). Doubling the depth decreases sag by a factor of 8! Hence, beams are designed with large vertical depth.
  • The Buckling Challenge: Making a rectangular beam very deep and thin causes it to bend sideways under load, an instability known as buckling.
  • The I-Girder Solution: The I-section provides wide horizontal flanges where bending tensile and compressive stresses are maximized, connected by a slender central web to resist shear. This eliminates buckling, drastically cuts weight, and saves material costs.

4. Geological Limit: Maximum Height of Mountains on Earth

Why do no mountains on Earth exceed approximately 10 km in height, while Olympus Mons on Mars towers at 22 km? The answer lies strictly in the elastic yield strength of terrestrial rocks.

Derivation of Mountain Height Ceiling:

Consider a mountain of height h and base area A consisting of rock of average density ρ ≈ 3.0 × 10³ kg/m³.

The downward gravitational compressive pressure at the mountain's base is:

P = ρ × g × h

Because the mountain's rock is unconfined laterally at the base edges, this downward vertical pressure creates a lateral shearing stress of magnitude comparable to ρ g h.

If ρ g h exceeds the elastic shear limit (yield strength σ_y) of rock, the base rock cannot support the load; it fractures and undergoes plastic lateral flow, sinking the mountain.

For typical silicate rock and granite, σ_y ≈ 3.0 × 10⁸ N/m².

h_max ≈ σ_y / (ρ × g) ≈ (3.0 × 10⁸ N/m²) / (3.0 × 10³ kg/m³ × 9.8 m/s²) ≈ 10,204 m ≈ 10 km

Mount Everest stands at 8,848.86 m, remarkably close to this fundamental theoretical mechanical limit! On Mars, where surface gravity is only g_mars ≈ 3.7 m/s², h_max is nearly three times greater, allowing mountains over 20 km to remain mechanically stable.

5. Design of Crane Cables and Factor of Safety

In mobile cranes, tower cranes, and mine shafts, cables must lift heavy payloads M with acceleration a while remaining strictly within the elastic limit.

Total Tension (T) = M × (g + a)

To guard against material imperfections, dynamic shocks, and fatigue, engineers introduce a Factor of Safety (N), typically between 5 and 10:

Permissible Working Stress (σ_perm) = Yield Strength (σ_y) / N
Minimum Cable Radius (r) = √ [ T / (π × σ_perm) ]

6. Solved Examples & Numerical Applications

Example 1: A steel wire of length 2.0 m and cross-sectional diameter 1.0 mm is stretched by 2.0 mm within its elastic limit. Calculate: (a) the elastic potential energy stored in the wire, and (b) the energy density. (Take Young's modulus of steel Y = 2.0 × 10¹¹ N/m²).
Solution:

Step 1: Calculate wire dimensions and strain:

Radius r = d / 2 = 0.5 mm = 5.0 × 10⁻⁴ m.

Area A = π r² = 3.1416 × (5.0 × 10⁻⁴)² = 7.854 × 10⁻⁷ m².

Length L = 2.0 m; Elongation ΔL = 2.0 mm = 2.0 × 10⁻³ m.

Strain ε = ΔL / L = (2.0 × 10⁻³) / 2.0 = 1.0 × 10⁻³.

Step 2: Calculate energy density (u):

u = ½ × Y × ε² = 0.5 × (2.0 × 10¹¹) × (1.0 × 10⁻³)²

u = 1.0 × 10¹¹ × 1.0 × 10⁻⁶ = 1.0 × 10⁵ J/m³ (100 kJ/m³).

Step 3: Calculate total stored energy (U):

Volume V = A × L = 7.854 × 10⁻⁷ × 2.0 = 1.571 × 10⁻⁶ m³.

U = u × V = (1.0 × 10⁵ J/m³) × (1.571 × 10⁻⁶ m³) = 0.157 Joules.

Example 2: A crane is designed to lift a maximum payload of 10 metric tonnes (10,000 kg) with a maximum vertical acceleration of 2.0 m/s². The cable is made of high-tensile steel having a yield strength of 3.0 × 10⁸ N/m². If the engineering design specifies a safety factor of 6, calculate the minimum required radius of the steel cable. (Take g = 9.8 m/s²).
Solution:

Step 1: Calculate maximum dynamic cable tension:

T = M × (g + a) = 10,000 kg × (9.8 + 2.0) m/s² = 10,000 × 11.8 = 118,000 N (118 kN).

Step 2: Calculate permissible safe stress:

σ_perm = σ_y / N = (3.0 × 10⁸ N/m²) / 6 = 5.0 × 10⁷ N/m² (50 MPa).

Step 3: Calculate minimum cable cross-sectional area and radius:

A_min = T / σ_perm = 118,000 / (5.0 × 10⁷) = 2.36 × 10⁻³ m².

r = √ [ A_min / π ] = √ [ (2.36 × 10⁻³) / 3.1416 ] = √ [ 7.512 × 10⁻⁴ ] = 0.0274 m = 2.74 cm (diameter ≈ 5.5 cm).

Example 3: A rectangular wooden beam of span L = 4.0 m, breadth b = 10 cm, and depth d = 20 cm is supported horizontally at both ends. When a central load of 2000 N is placed on it, it exhibits a central sag of 1.25 mm. Calculate: (a) Young's modulus of the wood, and (b) what the sag would be if the beam is laid flat so that its breadth is 20 cm and depth is 10 cm.
Solution:

Step 1: Calculate Young's modulus from sag formula:

L = 4.0 m; b = 0.10 m; d = 0.20 m; W = 2000 N; δ₁ = 1.25 mm = 1.25 × 10⁻³ m.

δ = (W L³) / (4 Y b d³) ⇒ Y = (W L³) / (4 δ b d³).

Numerator = 2000 × (4.0)³ = 2000 × 64 = 128,000 N·m³.

Denominator = 4 × (1.25 × 10⁻³) × 0.10 × (0.20)³ = 5.0 × 10⁻³ × 0.10 × 0.008 = 4.0 × 10⁻⁶.

Y = 128,000 / (4.0 × 10⁻⁶) = 3.2 × 10¹⁰ N/m² (32 GPa).

Step 2: Compute sag when turned flat (b' = 0.20 m, d' = 0.10 m):

Since δ ∝ 1 / (b d³), we take the ratio:

δ₂ / δ₁ = (b₁ d₁³) / (b₂ d₂³) = [ 0.10 × (0.20)³ ] / [ 0.20 × (0.10)³ ] = [ 0.10 × 0.008 ] / [ 0.20 × 0.001 ] = 0.0008 / 0.0002 = 4.

δ₂ = 4 × δ₁ = 4 × 1.25 mm = 5.0 mm.

(Turning the beam flat increases the sag by 400% for the exact same amount of wood!).

Example 4: A 5.0 kg steel ball falls from a height h = 1.0 m onto a vertical steel rod of length L = 1.0 m and cross-sectional area A = 2.0 cm² rigidly clamped at the bottom. Calculate the maximum compressive strain and maximum instantaneous compressive stress developed in the rod. (Young's modulus of steel Y = 2.0 × 10¹¹ N/m², g = 9.8 m/s²).
Solution:

Step 1: Formulate energy conservation during impact:

Gravitational potential energy lost by the falling mass is converted into elastic strain energy of the compressed rod:

M g (h + ΔL) = ½ (Y A / L) ΔL².

Since ΔL ≪ h, we can accurately approximate M g (h + ΔL) ≈ M g h.

Step 2: Solve for maximum dynamic compression ΔL:

½ (Y A / L) ΔL² = M g h ⇒ ΔL² = (2 M g h L) / (Y A).

M g h = 5.0 × 9.8 × 1.0 = 49.0 J.

Y A / L = (2.0 × 10¹¹ × 2.0 × 10⁻⁴) / 1.0 = 4.0 × 10⁷ N/m.

ΔL² = (2 × 49.0) / (4.0 × 10⁷) = 98.0 / (4.0 × 10⁷) = 2.45 × 10⁻⁶ m².

ΔL = √(2.45 × 10⁻⁶) = 1.565 × 10⁻³ m = 1.57 mm.

Step 3: Calculate maximum strain and instantaneous stress:

ε_max = ΔL / L = (1.565 × 10⁻³) / 1.0 = 1.57 × 10⁻³.

σ_max = Y × ε_max = (2.0 × 10¹¹) × (1.565 × 10⁻³) = 3.13 × 10⁸ N/m² (313 MPa).

Frequently Asked Questions (Class 11 & JEE/NEET)

Q1. Where does the missing mechanical energy go when a wire stretches under a hanging weight Mg?
When a load of weight Mg descends by elongation ΔL, the loss in gravitational potential energy is Mg ΔL. However, the stored elastic potential energy in the wire is only ½ Mg ΔL. The remaining ½ Mg ΔL is converted into kinetic energy of oscillation and ultimately dissipated as internal heat through microscopic friction and air resistance as oscillations damp out.
Q2. Why is a hollow cylinder stronger against twisting than a solid cylinder of the same mass and length?
The torsional restoring couple (torque required per unit angle of twist) is C = (π G / 2L) (r_ext⁴ − r_int⁴). For identical mass and length, a hollow shaft places its material at a larger average radial distance from the central axis, dramatically amplifying its polar moment of inertia and torsional stiffness without adding weight.
Q3. Why does increasing beam length cause such a drastic increase in sag?
The sag formula contains length cubed (δ ∝ L³). Doubling the span length of a bridge beam while maintaining identical cross-section increases deflection by a factor of 2³ = 8. Long spans therefore require intermediate pillars, trusses, or cable-stayed suspension systems.
Q4. How does the concept of resilience relate to elastic potential energy?
Resilience is the capacity of a material to absorb energy when deformed elastically and release that energy upon unloading. Proof resilience is the maximum strain energy that can be absorbed per unit volume up to the elastic limit without undergoing permanent deformation. Spring steels are engineered specifically for high proof resilience (high yield strength and moderate Young's modulus).
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