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Stress Restricting Thermal Expansion

Reference data and engineering information about stress restricting thermal expansion for material properties applications.

stressrestrictingthermalexpansionCalculatorData Table

Overview

Engineering reference data for Stress Restricting Thermal Expansion in material science and properties.

Key Formulas

Stress

σ=FA\sigma = \frac{F}{A}

Force per unit area.

Strain

ε=ΔLL0\varepsilon = \frac{\Delta L}{L_0}

Change in length per original length.

Hooke's Law

σ=Eε\sigma = E \varepsilon

Stress proportional to strain in elastic region.

Thermal Expansion

ΔL=αL0ΔT\Delta L = \alpha L_0 \Delta T

Length change due to temperature.

Variables

SymbolDescriptionUnit
σ\sigmaStressPa
ε\varepsilonStrain—
EEYoung's modulusPa
α\alphaThermal expansion coefficient1/°C
ΔT\Delta TTemperature change°C

Worked Examples

Example 1: Heated Steel Pipe

A DN150 Std. (6 in) steel pipe with length 50 m is heated from 20°C to 90°C.

Given Properties:

  • Expansion coefficient: α=12×10−6\alpha = 12 \times 10^{-6} m/mK
  • Young's Modulus: E=200E = 200 GPa
  • Outside diameter: Do=168.275D_o = 168.275 mm
  • Wall thickness: t=7.112t = 7.112 mm

Metric Calculations:

Unrestricted expansion: Δl=α⋅lo⋅Δt=(12×10−6)(50)(90−20)=0.042 m\Delta l = \alpha \cdot l_o \cdot \Delta t = (12 \times 10^{-6})(50)(90 - 20) = 0.042 \text{ m}

Thermal stress (restricted): σdt=E⋅α⋅Δt=(200×109)(12×10−6)(70)=168 MPa\sigma_{dt} = E \cdot \alpha \cdot \Delta t = (200 \times 10^9)(12 \times 10^{-6})(70) = 168 \text{ MPa}

Cross-sectional area of pipe wall: A=π(Do2)2−π(Do−2t2)2=3598 mm2=3.6×10−3 m2A = \pi\left(\frac{D_o}{2}\right)^2 - \pi\left(\frac{D_o - 2t}{2}\right)^2 = 3598 \text{ mm}^2 = 3.6 \times 10^{-3} \text{ m}^2

Axial force at restricted ends: F=σdt⋅A=(168×106)(3.6×10−3)=604.8 kNF = \sigma_{dt} \cdot A = (168 \times 10^6)(3.6 \times 10^{-3}) = 604.8 \text{ kN}

Imperial Calculations:

Δl=(6.7×10−6)(1669)(194−68)=1.4 in\Delta l = (6.7 \times 10^{-6})(1669)(194 - 68) = 1.4 \text{ in}

σdt=(29×106)(6.7×10−6)(126)=24,481 psi\sigma_{dt} = (29 \times 10^6)(6.7 \times 10^{-6})(126) = 24{,}481 \text{ psi}

A=5.3 in2A = 5.3 \text{ in}^2

F=(24,481)(5.3)=129,749 lbF = (24{,}481)(5.3) = 129{,}749 \text{ lb}

Example 2: Reinforced PVC Bar with Steel Rod

A PVC bar of 10 m length reinforced with a steel rod, subjected to Δt=100°C\Delta t = 100°C.

Materialα\alpha (m/mK)EE (Pa)
PVC50.4×10−650.4 \times 10^{-6}2.8×1092.8 \times 10^9
Steel12×10−612 \times 10^{-6}200×109200 \times 10^9

Free expansion: ΔlPVC=(50.4×10−6)(10)(100)=0.054 m\Delta l_{PVC} = (50.4 \times 10^{-6})(10)(100) = 0.054 \text{ m} Δlsteel=(12×10−6)(10)(100)=0.012 m\Delta l_{steel} = (12 \times 10^{-6})(10)(100) = 0.012 \text{ m}

Assuming the steel rod is much stronger than the PVC bar, the thermal tension in the PVC due to differential expansion:

σPVC=EPVC⋅ΔlPVC−Δlsteello=(2.8×109)⋅0.054−0.01210=11.8 MPa\sigma_{PVC} = E_{PVC} \cdot \frac{\Delta l_{PVC} - \Delta l_{steel}}{l_o} = (2.8 \times 10^9) \cdot \frac{0.054 - 0.012}{10} = 11.8 \text{ MPa}

Note: The tensile yield strength of PVC is approximately 55 MPa. This thermal stress represents about 21% of the yield capacity.

Combined Stress Considerations

When a pipe carries internal pressure, the total axial stress must account for multiple contributions. The axial (longitudinal) stress from pressure is:

σaxial=p⋅Di4t\sigma_{axial} = \frac{p \cdot D_i}{4t}

The hoop (circumferential) stress from pressure is:

σhoop=p⋅Di2t\sigma_{hoop} = \frac{p \cdot D_i}{2t}

Important: If there is pressure in the pipe, the axial and circumferential (hoop) stresses must be added to the restricted temperature expansion stress using vector addition, not simple scalar addition.

References

Source: engineeringtoolbox.com