Simple Stresses, Strains & Elastic Constants

Strength of Materials is one of GATE ME's heaviest scorers (~8–12 marks), and it all starts here. Elastic-constant relations and thermal stress are near-guaranteed 1–2 mark items.

Core relations

Stress σ = P/A (normal), strain ε = ΔL/L (dimensionless). Hooke's law: σ = Eε within the elastic limit, where E = Young's modulus.

Poisson's ratio ν = −(lateral strain)/(longitudinal strain), typically 0.25–0.35 for metals (0 ≤ ν ≤ 0.5).

Elastic-constant links (memorise):

• E = 2G(1 + ν) — Young's, shear (rigidity) modulus, Poisson.
• E = 3K(1 − 2ν) — Young's, bulk modulus.
• 9/E = 3/G + 1/K.
  As ν → 0.5, K → ∞ (incompressible). Always G < E < 3G roughly (since ν between 0 and 0.5).

Axial elongation δ = PL/AE (springs in series add δ; in parallel share load). Thermal strain ε_th = αΔT; if free to expand, no stress; if fully constrained, thermal stress σ = EαΔT. Volumetric strain for triaxial loading = (σx + σy + σz)(1 − 2ν)/E.

Exam Tricks & Tips

• 🎯 Two elastic constants fix all four: given any two of E, G, K, ν, get the rest from E = 2G(1+ν) and E = 3K(1−2ν).
• 🎯 Free thermal expansion produces ZERO stress; stress arises only when expansion is restrained (σ = EαΔT), independent of length.
• 🎯 δ = PL/AE: stepped bars → sum PL/AE segment by segment; composite (parallel) bars share load in proportion to AE.
• 🎯 ν = 0.5 means incompressible (K → ∞); rubber is near 0.5, cork near 0.
• 🎯 Factor of safety = ultimate (or yield) stress / working stress — a frequent one-liner.
• ❌ Common mistake: using σ = EαΔT for a partially restrained bar — if a gap allows some free expansion, only the restrained part of the strain generates stress.

Expected exam pattern

A 1-mark elastic-constant conversion or FoS calculation, and a 2-mark stepped/composite bar elongation or a thermal-stress problem (with or without a gap). Volumetric strain under triaxial stress appears occasionally.

Quick recap

σ = P/A, ε = ΔL/L, Hooke σ = Eε. E = 2G(1+ν) = 3K(1−2ν). δ = PL/AE (sum segments). Free expansion → no stress; constrained → σ = EαΔT. ν limits 0–0.5; 0.5 = incompressible.

Cover the answer, recall, then check. 12 cards on stress, strain and elastic constants for GATE ME.

Q1. Define stress and strain.
A1. Stress σ = P/A (force per unit area); strain ε = ΔL/L (fractional deformation, dimensionless).

Q2. State Hooke's law and define Young's modulus.
A2. σ = Eε within the elastic limit; E = σ/ε is Young's modulus (stiffness in tension/compression).

Q3. Define Poisson's ratio and its typical range.
A3. ν = −(lateral strain)/(axial strain); 0 ≤ ν ≤ 0.5, about 0.3 for steel.

Q4. Relate E, G, and ν.
A4. E = 2G(1 + ν), where G is the shear (rigidity) modulus.

Q5. Relate E, K, and ν.
A5. E = 3K(1 − 2ν), where K is the bulk modulus. Also 9/E = 3/G + 1/K.

Q6. What does ν = 0.5 imply?
A6. The material is incompressible: K → ∞ and volume does not change under load (e.g. rubber).

Q7. Formula for axial elongation of a bar.
A7. δ = PL/AE. For stepped bars, sum PL/AE over each segment.

Q8. When does a temperature change cause thermal stress?
A8. Only when expansion is restrained. Free expansion → zero stress; fully constrained → σ = EαΔT.

Q9. Does thermal stress in a fully constrained bar depend on length?
A9. No — σ = EαΔT is independent of length (though the free elongation αΔT·L is not).

Q10. Define factor of safety.
A10. FoS = ultimate (or yield) strength / allowable (working) stress. It quantifies the design margin.

Q11. Volumetric strain under triaxial stress σx, σy, σz?
A11. ε_v = (σx + σy + σz)(1 − 2ν)/E.

Q12. Bars in parallel share an axial load how?
A12. In proportion to their axial stiffness AE/L; both undergo the same elongation (compatibility).

This topic is the alphabet of Strength of Materials. Every later result — bending, torsion, columns, thermal stress — is built on the definitions of stress, strain, and the elastic constants that link them. GATE asks direct numericals here and assumes fluency everywhere else.

Core concept: stress is internal force per unit area, strain is fractional deformation, and within the elastic limit they are proportional through the material's stiffness (Young's modulus) — Hooke's law.

Deep explanation

Beginner — stress, strain, Hooke's law

• Normal stress σ = P/A; normal strain ε = δ/L (change in length over original length).
• Hooke's law: σ = E ε, so elongation δ = PL/(AE). E is Young's modulus (stiffness).
• Shear stress τ = V/A; shear strain γ; τ = G γ, where G is the shear (rigidity) modulus.
• Poisson's ratio ν = −(lateral strain)/(longitudinal strain), typically 0.25–0.35 for metals.

Intermediate — the stress-strain curve

Key points on a mild-steel curve: proportional limit, elastic limit, upper/lower yield point, ultimate strength, fracture. Slope in the elastic region = E. Resilience = area under the curve up to yield (energy stored elastically); toughness = total area to fracture. Ductile materials yield and neck; brittle ones fracture with little strain.

Advanced — elastic constant relationships

For an isotropic material the constants are interlinked:
E = 2G(1 + ν) and E = 3K(1 − 2ν),
where K is the bulk modulus. Combining: E = 9KG/(3K + G). These let you find any one constant from two others — a favourite GATE plug-in. Note ν < 0.5 always (else negative K); ν = 0.5 means incompressible.

Thermal strain: a bar heated ΔT wants to expand ε_th = αΔT; if restrained, it develops thermal stress σ = EαΔT (independent of length and area).

Worked example

A steel rod (E = 200 GPa) of 20 mm diameter carries a 60 kN tensile load over 2.5 m. Find the elongation.
Area A = (π/4)(0.020)² = 3.142 × 10⁻⁴ m².
σ = P/A = 60,000/3.142×10⁻⁴ = 190.9 MPa (below yield, elastic OK).
δ = PL/(AE) = (60,000 × 2.5)/(3.142×10⁻⁴ × 200×10⁹) = 150,000/(6.283×10⁷) = 2.39 × 10⁻³ m = 2.39 mm.

GATE relevance

Direct 1–2 mark numericals (δ = PL/AE, thermal stress, E–G–K relations) are near-certain, and the definitions gate every other SOM topic. The elastic-constant relations E = 2G(1+ν) and E = 3K(1−2ν) are among the most reused formulas in the paper.

Exam tricks & shortcuts

• Thermal stress in a fully-restrained bar is σ = EαΔT — length and area cancel out; do not carry them.
• For composite/stepped bars in series, add elongations (same load); in parallel, share load by stiffness (AE/L).
• Mnemonic: "E equals two-G one-plus-nu."

Key formulas

• Stress σ = P/A; strain ε = ΔL/L; Young's modulus E = σ/ε.
• Elongation: δ = PL/AE.
• Elastic constants: E = 2G(1 + ν) = 3K(1 − 2ν); relation E = 9KG/(3K + G).
• Poisson's ratio: ν = lateral strain/longitudinal strain.
• Thermal stress (constrained): σ = EαΔT.

The three elastic constants E, G, K are interrelated through Poisson's ratio ν.

Worked Example

Problem: A steel rod of 20 mm diameter and 2 m length carries an axial tensile load of 40 kN. Given Young's modulus E = 200 GPa, find the axial stress, the strain, and the total elongation.

Solution:
Cross-sectional area:
A = (π/4)·d² = (π/4)(0.020)² = (π/4)(4 × 10⁻⁴) = 3.142 × 10⁻⁴ m².

Axial (normal) stress:
σ = P/A = 40000 / (3.142 × 10⁻⁴) = 1.273 × 10⁸ Pa = 127.3 MPa.

Strain from Hooke's law (σ = E·ε):
ε = σ/E = (127.3 × 10⁶)/(200 × 10⁹) = 6.37 × 10⁻⁴.

Total elongation:
δ = ε·L = (6.37 × 10⁻⁴)(2) = 1.27 × 10⁻³ m = 1.27 mm.
(Equivalently δ = PL/AE.)

Answer: σ = 127.3 MPa, ε = 6.37 × 10⁻⁴, elongation δ = 1.27 mm.