QCC Notes
CLASS 11 · PHYSICSJEE MAIN × NEETहिंदी
§ 8.2NCERT Class 11 · Physics · Chapter 8

Hooke's Law and the Stress-Strain Curve

QCC Notes · Class 11 Physics NCERT-aligned Updated 26 Sep 2026 10 min read

The mechanical response of a solid material subjected to progressive tension provides indispensable criteria for engineering design, material selection, and structural safety. In this module, we examine Hooke's empirical law, conduct a granular anatomical breakdown of the complete stress-strain diagram for ductile metals, contrast ductile, brittle, and elastomeric behaviors, and explore elastic hysteresis.

1. Hooke's Law: Formulation and Domain of Validity

In 1676, English scientist Robert Hooke discovered the fundamental proportional relationship governing elastic deformations. Experimentally, when a metal wire is subjected to gradually increasing tensile loads, elongation increases in direct proportion to the applied force up to a characteristic threshold.

Stress ∝ Strain  ⇒  Stress = E × Strain  ⇒  E = Stress / Strain

Here, the constant of proportionality E is called the Modulus of Elasticity of the material. Like stress, E has units of N/m² or Pa, and dimensions [M L⁻¹ T⁻²].

Essential Qualification: Hooke's law is an empirical rule, not an immutable fundamental law of nature. It holds true only within the strictly linear elastic regime (up to the proportional limit). Many real materials, including rubber, biological tissues, and concrete, display non-linear stress-strain relationships even under modest loads.

2. Complete Anatomical Breakdown of the Stress-Strain Curve

To determine the mechanical strength of a material, a standard test specimen (e.g., a cylindrical metal rod of known gauge length and diameter) is clamped in a Universal Testing Machine (UTM) and pulled in tension while continuously recording stress and strain.

+-----------------------------------------------------------------------------------+ | STRESS-STRAIN CURVE FOR A DUCTILE METALLIC SPECIMEN | +-----------------------------------------------------------------------------------+ Stress (σ) ^ | Ultimate Tensile Strength (D) | *-----* | / \ Necking Phase | Elastic Limit (B) / * Fracture Point (E) | (Yield Point) / | * / | / \ / Plastic Region (Ductile Flow) | / \ / Yield Str |.........* *---------* (σ_y) | A (Proportional Limit) | / | / Linear Elastic Region (OA) | / Slope = Young's Modulus (Y) | / | / : | / : Unloading from C creates Permanent Set (OO') | / : O +------+----------------------------------------------------> Strain (ε) O' (Strain < 1%) +-----------------------------------------------------------------------------------+
Figure 8.3: Universal stress-strain diagram for a ductile metal showing proportional limit A, yield point B, permanent set OO', ultimate tensile strength D, and fracture point E.

Detailed Region-by-Region Analysis:

  1. Region OA (Linear Elastic / Hookean Regime):
    From origin O to point A, stress is strictly proportional to strain (σ = Y ε). The graph is a straight line. Point A is the Proportional Limit. The slope of OA directly equals Young's Modulus (Y). If the load is removed at any point in OA, the wire completely retraces its path back to O with zero residual strain.
  2. Region AB (Non-linear Elastic Regime):
    Between A and B, stress is no longer proportional to strain (curve bends slightly), yet the deformation remains fully reversible. Point B is the Elastic Limit or Yield Point. The stress corresponding to point B is termed the Yield Strength (σ_y or S_y). If the load is released at B, the specimen regains its original length.
  3. Region BC (Onset of Plastic Deformation & Yielding):
    Beyond B, even a minuscule increase in stress causes a substantial increase in strain. If the load is removed at point C (beyond B), the material does NOT retrace the original curve; instead, it unloads along a dashed straight line parallel to OA, leaving a residual elongation. The specimen exhibits a Permanent Set (OO'), typically less than 1% strain. The deformation is now plastic.
  4. Region CD (Plastic Flow & Ultimate Strength):
    Beyond C, the material undergoes substantial plastic flow. Dislocation movements within the metallic crystal lattice cause macroscopic elongation. Point D represents the highest point on the curve: the Ultimate Tensile Strength (σ_u or S_u). This is the maximum tensile stress the specimen can sustain before localized instability sets in.
  5. Region DE (Necking and Fracture):
    Beyond point D, a localized constriction or "neck" develops in the specimen. The cross-sectional area in this region rapidly decreases. Even if the applied force is reduced, the local stress continues to climb until the specimen snaps at point E, the Fracture Point / Breaking Point.

3. Classification of Materials based on Stress-Strain Behavior

1. Ductile Materials

Materials that exhibit a large plastic deformation range between the yield point (B) and the fracture point (E). They can be drawn into thin wires or hammered into sheets without fracturing.

Examples: Copper, aluminum, mild steel, gold, silver.

2. Brittle Materials

Materials in which the fracture point (E) lies in close proximity to the elastic limit (B). There is negligible plastic deformation; fracture occurs abruptly without warning or necking.

Examples: Cast iron, glass, ceramic, high-carbon tool steel, rock.

4. Elastomers and Elastic Hysteresis

Certain substances do not obey Hooke's law for almost any part of their deformation, yet they can be stretched reversibly to multiple times their original length. Such materials are termed Elastomers (e.g., vulcanized rubber, tissue of aorta).

+-----------------------------------------------------------------------------------+ | ELASTIC HYSTERESIS LOOP IN VULCANIZED RUBBER | +-----------------------------------------------------------------------------------+ Stress (σ) ^ | / (Loading Curve) | / | / | *----------* | / / | / / | / / | / (Area) / | / / | / / (Unloading Curve) | *----------* | / O +--+--------------------------------------------------------> Strain (ε) +-----------------------------------------------------------------------------------+
Figure 8.4: Elastic hysteresis curve for vulcanized rubber. The area enclosed represents mechanical energy dissipated as thermal energy per cycle.
Engineering Significance of Hysteresis:
  • For rubber, the stress required for a given strain during unloading is smaller than during loading.
  • The work done during stretching exceeds the work recovered during contraction. The difference—the area within the loop—is converted into internal thermal energy (heat).
  • Automobile tires and vibration shock absorbers are fabricated from rubber formulations exhibiting high damping (large hysteresis loop) to dissipate road shocks effectively.

5. Solved Examples & Numerical Applications

Example 1: The stress-strain curves for materials A and B show that material A has a steeper initial linear slope and a large plastic range between yield point and breaking point. Material B has a gentler initial slope and its breaking point is located immediately beyond its elastic limit. Which material has greater Young's modulus, and which is more suitable for structural suspension cables?
Solution:

Step 1: Young's Modulus comparison:

Young's modulus Y = slope of linear elastic region OA = Δσ / Δε. Since the initial slope of curve A is steeper than that of B, Material A has a higher Young's modulus (Y_A > Y_B).

Step 2: Ductility and structural suitability:

Material A has a large plastic range (large distance between yield point B and fracture point E), meaning it is ductile and gives advance warning (visible stretching) before catastrophic failure.

Material B snaps immediately beyond the elastic limit with no plastic warning, meaning it is brittle.

Therefore, Material A is strictly chosen for structural suspension cables, bridges, and cranes.

Example 2: A copper wire of cross-sectional area 2.0 mm² has a yield strength of 2.5 × 10⁸ N/m². Calculate the maximum mass that can be suspended from this wire without producing permanent deformation. (Take g = 9.8 m/s²).
Solution:

Step 1: Relate yield strength to elastic limit:

To avoid permanent deformation, the tensile stress must not exceed the yield strength σ_y.

σ_max = σ_y = 2.5 × 10⁸ N/m².

Step 2: Calculate maximum allowable tensile force:

Area A = 2.0 mm² = 2.0 × 10⁻⁶ m².

F_max = σ_max × A = (2.5 × 10⁸ N/m²) × (2.0 × 10⁻⁶ m²) = 500 N.

Step 3: Determine maximum suspended mass:

M_max = F_max / g = 500 N / 9.8 m/s² = 51.02 kg.

Example 3: A cylindrical mild steel test specimen of original gauge length 50.0 mm and diameter 12.5 mm is subjected to tensile testing. The proportional limit is reached at a tensile load of 38.0 kN, and the maximum tensile load recorded before necking is 68.0 kN. Calculate: (a) Proportional limit stress, and (b) Ultimate tensile strength (UTS).
Solution:

Step 1: Compute cross-sectional area:

Diameter d = 12.5 mm = 1.25 × 10⁻² m.

Radius r = d / 2 = 6.25 × 10⁻³ m.

Area A = π r² = 3.1416 × (6.25 × 10⁻³)² = 1.227 × 10⁻⁴ m².

Step 2: Calculate proportional limit stress (σ_p):

σ_p = F_p / A = (38.0 × 10³ N) / (1.227 × 10⁻⁴ m²) = 3.10 × 10⁸ N/m² (310 MPa).

Step 3: Calculate ultimate tensile strength (σ_u):

σ_u = F_max / A = (68.0 × 10³ N) / (1.227 × 10⁻⁴ m²) = 5.54 × 10⁸ N/m² (554 MPa).

Example 4: A 1.0 kg rubber block is cycled through a tension test. The area of the hysteresis loop in the stress-strain plot is found to be 250 J/m³. If the density of rubber is 1.2 × 10³ kg/m³ and its specific heat capacity is 1600 J/(kg·K), estimate the temperature rise of the rubber block after 100 continuous loading-unloading cycles, assuming zero heat loss to the surroundings.
Solution:

Step 1: Calculate energy dissipated per unit volume:

Energy dissipated per cycle per unit volume = Loop Area = 250 J/m³.

For 100 cycles, total volumetric heat generated: q_v = 100 × 250 J/m³ = 25,000 J/m³.

Step 2: Convert to heat per unit mass:

Volume of block V = mass / density = 1.0 kg / (1.2 × 10³ kg/m³) = (1 / 1200) m³.

Total heat Q = q_v × V = 25,000 × (1 / 1200) = 20.83 Joules.

Step 3: Compute temperature rise ΔT:

Q = m × c × ΔT ⇒ ΔT = Q / (m × c) = 20.83 / (1.0 × 1600) = 0.013 °C.

(In high-speed automotive racing where millions of cycles occur, this internal dissipation elevates tire temperature substantially, altering grip and structural integrity).

Frequently Asked Questions (Class 11 & JEE/NEET)

Q1. Why does necking occur only in ductile materials beyond the ultimate tensile strength?
Up to the ultimate tensile strength (UTS), plastic deformation occurs uniformly along the entire length of the specimen as work-hardening compensates for the reducing cross-section. Beyond UTS, work-hardening rate drops below the rate of geometric thinning, causing deformation to localize at the weakest point, forming a distinct neck where fracture subsequently occurs.
Q2. What is meant by the 'proof stress' or 0.2% offset yield stress?
In metals like aluminum, brass, or high-tensile steel, the stress-strain curve does not have a distinct, sharply defined yield point B. To specify yield strength unambiguously, engineers draw a line parallel to the linear elastic slope starting at a strain of 0.002 (0.2% offset). The intersection of this offset line with the stress-strain curve defines the 0.2% proof stress.
Q3. Does an elastomer have a well-defined plastic region?
No. Elastomers like vulcanized rubber can undergo large elastic extensions (up to 700–800% strain) through the uncoiling of tangled polymer chains. When unloaded, they return nearly to their original configuration without undergoing classic metallic plastic flow or work hardening. Thus, they lack a well-defined yield point or plastic region.
Q4. What is the physical significance of the slope of the stress-strain curve?
In the initial linear region (proportional limit), the slope of the stress-strain curve (dσ/dε) represents Young's modulus of elasticity, measuring the material's inherent resistance to longitudinal deformation (stiffness). A steeper slope indicates greater stiffness (higher Young's modulus).
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