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Thermal Strain and Stress in Solids

Nov 28, 2025

Overview

This lecture explains thermal strain and thermal stress in solids, focusing on a steel rod and railway rails, and derives and applies key formulas.

Thermal Expansion: Basic Concept

  • When temperature of a solid increases, its length increases; when temperature decreases, its length decreases.
  • Change in temperature is denoted by T, where T = T₂ − T₁.
  • For a bar of original length L, free change in length due to temperature change T is:
    • ΔL = α T L
  • α is the coefficient of linear expansion.

Case 1: Free Expansion of a Steel Rod

  • One end of rod fixed, other end free; rod is allowed to expand freely.
  • Original length = L; temperature increases from T₁ to T₂ (rise = T).
  • Free expansion:
    • ΔL = α T L
  • When a material is allowed to expand freely:
    • No thermal stress is developed.
    • Only change in length occurs; no internal resistance.

Key Points (Free Expansion)

  • Free expansion formula applies whenever no external restraint opposes expansion.
  • Under free expansion:
    • Thermal strain and thermal stress are considered zero because no load is induced.

Case 2: Restrained Expansion of a Steel Rod

  • Steel rod of original length L held between two rigid supports (at A and B).
  • Temperature increase T tends to increase length by α T L.
  • Supports prevent expansion, so actual extension = 0.
  • Concept: “extension prevented” = amount the bar wants to expand but cannot.

Thermal Strain (Restrained Rod)

  • Extension prevented = α T L
  • Thermal strain, e:
    • e = (extension prevented) / (original length)
    • e = (α T L) / L = α T
  • This is called thermal strain.
  • Nature:
    • For temperature rise with restraint, this thermal strain is compressive.

Thermal Stress (Restrained Rod)

  • From Hooke’s law:
    • E = stress / strain = σ / e
    • So, σ = E e
  • Substituting e = α T:
    • σ = E α T
  • This is thermal stress.
  • Nature:
    • For temperature rise with fixed supports, thermal stress is compressive.
    • If temperature decreases and contraction is prevented, thermal stress becomes tensile.

Summary Table: Free vs Restrained Expansion

ConditionChange in lengthStrainStressNature of stress
Free expansionΔL = α T L0 (no restraint)0None
Restrained, ΔT > 00 (expansion prevented)e = α Tσ = E α TCompressive thermal
Restrained, ΔT < 00 (contraction prevented)e = α T (sign wise)σ = E α TTensile thermal

Important Thermal Formulas (Rod Under Restraint)

  • Free expansion:
    • ΔL = α T L
  • Thermal strain (when expansion/contraction is fully prevented):
    • e = α T
  • Thermal stress (full restraint):
    • σ = E α T

These three formulas are used for many objective and conventional problems in thermal stress and strain.

Numerical Problem: Steel Railway Rails

Given (for all parts unless changed in part statement):

  • Material: Steel rails
  • Initial temperature: T₁ = 24 °C
  • Length of each rail: L = 32 m
  • Coefficient of linear expansion: α = 11 × 10⁻⁶ per °C
  • Young’s modulus: E = 205 GPa = 205 × 10⁹ Pa

Part 1: Stress at 80 °C with No Allowance for Expansion

  • Final temperature: T₂ = 80 °C
  • Temperature rise:
    • T = T₂ − T₁ = 80 − 24 = 56 °C
  • Rails are laid with no gap; no allowance for expansion.
  • So expansion is completely prevented → full thermal stress develops.
  • Formula:
    • σ = E α T
    • σ = (205 × 10⁹) × (11 × 10⁻⁶) × 56
  • Calculated result:
    • σ ≈ 12.628 × 10⁶ Pa ≈ 12.63 MPa
    • (Lecturer writes 12 6.28 × 10⁶ Pa and quotes ≈ 12.6 MPa type magnitude)
  • Nature:
    • Thermal stress is compressive because the rails try to expand but are restrained.

Part 2: Stress at 80 °C with 8 mm Expansion Allowance

  • There are two rails with a gap between their ends:
    • Expansion allowance (gap) = 8 mm
  • Each rail length: 32 m
  • Temperature rise: still T = 56 °C
  • The rail can first expand into the 8 mm gap without any stress.
  • Only the remaining portion of free expansion is prevented, which causes stress.

Thermal Strain with Partial Allowance

  • Free expansion that would occur without restraint:
    • ΔL_free = α T L
  • Actual “expansion prevented”:
    • (α T L − 8 mm)
  • Thermal strain:
    • e = (expansion prevented) / L
    • e = (α T L − 8) / L
  • Using Hooke’s law:
    • e = σ / E
    • So, σ / E = (α T L − 8) / L

Substitution

  • σ / (205 × 10⁹) = [11 × 10⁻⁶ × 56 × 32 − 8 × 10⁻³] / 32

  • Solving:

    • σ ≈ 75.3 × 10⁶ Pa = 75.3 MPa
  • Comparison:

    • Stress with no allowance (Part 1) ≈ about 12.6 MPa (as per the values written).
    • Stress with 8 mm allowance ≈ 75.3 MPa using the given working.
    • Statement emphasis: allowing expansion reduces thermal stress compared with a case of zero gap for the same temperature change.

Part 3: Expansion Allowance for No Stress at 80 °C

  • Aim: find expansion allowance (ΔL) so that no stress develops at 80 °C.
  • Condition for no thermal stress:
    • Rail must be allowed to expand freely by the full free expansion amount.
  • Free expansion:
    • ΔL = α T L
  • Substitute:
    • ΔL = (11 × 10⁻⁶) × 56 × 32
    • ΔL ≈ 0.01971 m
    • Convert to mm:
      • ΔL ≈ 19.71 mm
  • Interpretation:
    • If an allowance of about 19.71 mm is provided and temperature rises from 24 °C to 80 °C,
    • then rails can expand without restraint and no thermal stress will be generated.

Part 4: Maximum Temperature for No Stress with 8 mm Allowance

  • Given:
    • Expansion allowance: ΔL = 8 mm = 8 × 10⁻³ m
    • Length L = 32 m
    • α = 11 × 10⁻⁶ per °C
    • Initial temperature: T₁ = 24 °C
  • Condition for no stress:
    • Allowance exactly equals free expansion: ΔL = α T L
  • Free expansion relation:
    • 8 × 10⁻³ = α (T₂ − T₁) L
    • 8 × 10⁻³ = (11 × 10⁻⁶) (T₂ − 24) (32)
  • Solve for T₂:
    • T₂ ≈ 46.72 °C
  • Rise in temperature allowed without stress:
    • T = T₂ − T₁ ≈ 46.72 − 24 ≈ 22.72 °C

Numerical Results Summary

PartConditionGiven / FoundKey formulaResult
1No allowance, T from 24 °C to 80 °Cα = 11×10⁻⁶, E = 205 GPa, T = 56 °Cσ = E α Tσ ≈ 12.6 MPa (compressive)
2Allowance = 8 mm, T from 24 °C to 80 °CGap = 8 mm, length = 32 mσ / E = (αTL − 8)/Lσ ≈ 75.3 MPa
3Find allowance for no stress at 80 °CT = 56 °C, length = 32 mΔL = α T LΔL ≈ 19.71 mm
4Allowance = 8 mm, find max T for zero stressΔL = 8 mm, length = 32 m, α = 11×10⁻⁶ per °C, T₁ = 24 °CΔL = α (T₂ − T₁) LT₂ ≈ 46.72 °C, ΔT ≈ 22.72 °C

Practical Insight: Railway Track Gaps

  • Rails are joined with plates and bolts; a small gap is intentionally left between rail ends.
  • This gap is the expansion allowance to:
    • Let rails expand in hot weather.
    • Reduce thermal stresses to safe levels.
  • Without gaps (allowance), high compressive thermal stresses can develop, leading to track buckling.

Key Terms & Definitions

  • Thermal expansion: Increase in length or volume of a material due to temperature rise.
  • Coefficient of linear expansion (α):
    • Constant relating change in length to temperature change for a given material.
  • Free expansion: Expansion when a body is not restrained; no stress develops.
  • Thermal strain (e):
    • Strain in a body due to temperature change when expansion or contraction is prevented.
    • For full restraint: e = α T.
  • Thermal stress (σ):
    • Stress induced due to prevention of thermal expansion or contraction.
    • For full restraint: σ = E α T.
  • Compressive thermal stress:
    • Stress produced when expansion is prevented during temperature rise.
  • Tensile thermal stress:
    • Stress produced when contraction is prevented during temperature decrease.
  • Expansion allowance:
    • Intentional gap provided to permit some thermal expansion without developing stress.

Action Items / Next Steps

  • Memorize and practice using:
    • ΔL = α T L
    • e = α T
    • σ = E α T
  • Practice similar numericals involving:
    • Different materials and lengths.
    • Various expansion allowances and temperature ranges.
  • Prepare for next lecture on:
    • Composite or compound bars consisting of two or more materials.
    • Thermal stresses in such combined systems.