9.3 Thermal stresses and deformations

Thermal stresses and deformations are crucial in understanding how materials respond to temperature changes. When heated or cooled, materials expand or contract, causing internal forces and shape changes. This concept is vital in axial loading scenarios, where temperature fluctuations can significantly impact a structure's behavior.

In axially loaded members, thermal effects can lead to compression or tension, depending on temperature changes and end constraints. Understanding these effects is essential for designing structures that can withstand temperature variations while maintaining their integrity and performance.

Thermal Stress in Axial Members

Concept and Effects

• Thermal stress is the stress induced in a material or structure due to changes in temperature, resulting from the material's tendency to expand or contract when heated or cooled
• Axially loaded members, such as rods, bars, or columns, experience thermal stresses when subjected to temperature changes due to the restraint of their ends or the presence of temperature gradients
• The magnitude of thermal stress depends on the material's coefficient of thermal expansion, the temperature change, and the degree of restraint
• Thermal stresses can cause axial compression or tension in the member, depending on whether the temperature change causes expansion (heating) or contraction (cooling)
• If the axially loaded member is free to expand or contract, no thermal stresses will develop, but if the member is restrained, thermal stresses will arise

Factors Influencing Thermal Stress

• Material properties
  • Coefficient of thermal expansion (α): Higher values lead to greater thermal stresses for a given temperature change
  • Modulus of elasticity (E): Stiffer materials experience higher thermal stresses
• Temperature change (ΔT)
  • Larger temperature changes result in higher thermal stresses
  • Positive ΔT (heating) causes compressive stress, while negative ΔT (cooling) causes tensile stress
• Degree of restraint
  • Fully restrained members experience the highest thermal stresses
  • Partially restrained members experience lower thermal stresses
  • Unrestrained members experience no thermal stresses
• Examples
  • Steel bridge girders exposed to daily temperature fluctuations
  • Concrete pavements subjected to seasonal temperature variations

Calculating Thermal Stresses and Deformations

Thermal Stress Calculation

• The thermal stress in an axially loaded member can be calculated using the formula: $σ_thermal = -E × α × ΔT$
  • $E$ is the modulus of elasticity
  • $α$ is the coefficient of thermal expansion
  • $ΔT$ is the temperature change
• The negative sign in the thermal stress formula indicates that a positive temperature change (heating) results in compressive stress, while a negative temperature change (cooling) results in tensile stress
• Example: A steel rod ($E = 200 GPa$, $α = 12 × 10^{-6} /°C$) subjected to a temperature increase of 50°C will experience a compressive thermal stress of $σ_thermal = -200 GPa × 12 × 10^{-6} /°C × 50°C = -120 MPa$

Thermal Deformation Calculation

• The thermal deformation (expansion or contraction) of an axially loaded member can be calculated using the formula: $δ_thermal = α × L × ΔT$
  • $L$ is the original length of the member
• Example: A 2-meter-long aluminum rod ($α = 23 × 10^{-6} /°C$) subjected to a temperature increase of 80°C will expand by $δ_thermal = 23 × 10^{-6} /°C × 2 m × 80°C = 3.68 mm$
• When the ends of the member are restrained, the thermal deformation causes a change in the internal axial force, which can be calculated using the formula: $ΔF = -E × A × α × ΔT$
  • $A$ is the cross-sectional area of the member
• Example: A restrained copper bar ($E = 110 GPa$, $α = 17 × 10^{-6} /°C$, $A = 500 mm^2$) subjected to a temperature decrease of 60°C will experience an increase in tensile force of $ΔF = -110 GPa × 500 mm^2 × 17 × 10^{-6} /°C × (-60°C) = 56.1 kN$

Combined Mechanical and Thermal Loads

Superposition of Stresses and Deformations

• When an axially loaded member is subjected to both mechanical and thermal loads, the total stress and deformation can be determined by superposition
• The total axial stress is the sum of the mechanical stress (due to applied loads) and the thermal stress: $σ_total = σ_mechanical + σ_thermal$
• The total axial deformation is the sum of the mechanical deformation (due to applied loads) and the thermal deformation: $δ_total = δ_mechanical + δ_thermal$

Calculation of Mechanical Stress and Deformation

• The mechanical stress and deformation can be calculated using the basic equations of axial loading: $σ_mechanical = F / A$ and $δ_mechanical = (F × L) / (E × A)$
  • $F$ is the applied axial force
• Example: A 1.5-meter-long steel rod ($E = 200 GPa$, $A = 800 mm^2$) subjected to an axial compressive force of 120 kN will experience a mechanical stress of $σ_mechanical = 120 kN / 800 mm^2 = 150 MPa$ and a mechanical deformation of $δ_mechanical = (120 kN × 1.5 m) / (200 GPa × 800 mm^2) = 1.125 mm$

Combining Mechanical and Thermal Effects

• The thermal stress and deformation are calculated using the equations mentioned in the previous objective
• Example: If the steel rod from the previous example is also subjected to a temperature increase of 40°C ($α = 12 × 10^{-6} /°C$), the thermal stress will be $σ_thermal = -200 GPa × 12 × 10^{-6} /°C × 40°C = -96 MPa$, and the thermal deformation will be $δ_thermal = 12 × 10^{-6} /°C × 1.5 m × 40°C = 0.72 mm$
• The total stress in the rod will be $σ_total = 150 MPa + (-96 MPa) = 54 MPa$ (compressive), and the total deformation will be $δ_total = 1.125 mm + 0.72 mm = 1.845 mm$ (contraction)

Thermal Stress in Statically Determinate vs Indeterminate Systems

Statically Determinate Systems

• Statically determinate axial loading systems, such as a free-standing rod with one end fixed and the other end free to move, can be solved using the equations for thermal stress and deformation directly
• In statically determinate systems, the reaction forces and stresses can be determined using equilibrium equations and the thermal stress and deformation formulas
• Example: A 2-meter-long aluminum rod ($E = 70 GPa$, $α = 23 × 10^{-6} /°C$, $A = 600 mm^2$) with one end fixed and the other end free is subjected to a temperature increase of 60°C. The thermal stress will be $σ_thermal = -70 GPa × 23 × 10^{-6} /°C × 60°C = -96.6 MPa$ (compressive), and the thermal deformation at the free end will be $δ_thermal = 23 × 10^{-6} /°C × 2 m × 60°C = 2.76 mm$ (expansion)

Statically Indeterminate Systems

• Statically indeterminate axial loading systems, such as a rod with both ends fixed or a rod with multiple supports, require additional compatibility equations to solve for the unknown reaction forces and stresses
• In statically indeterminate systems, the compatibility equations are based on the condition that the total deformation (mechanical + thermal) at the supports must be zero
• The unknown reaction forces in indeterminate systems can be determined by solving a system of equations that includes equilibrium equations and compatibility equations
• Once the reaction forces are known, the axial stresses and deformations can be calculated using the equations for mechanical and thermal loads
• Temperature changes in indeterminate systems can cause significant changes in the reaction forces and stresses, as the restraints prevent free expansion or contraction of the members
• Example: A 3-meter-long copper rod ($E = 110 GPa$, $α = 17 × 10^{-6} /°C$, $A = 400 mm^2$) with both ends fixed is subjected to a temperature decrease of 50°C. The compatibility equation states that the total deformation must be zero: $δ_mechanical + δ_thermal = 0$. Solving for the unknown reaction force $F$ yields $F = 93.5 kN$ (tensile). The thermal stress will be $σ_thermal = 110 GPa × 17 × 10^{-6} /°C × (-50°C) = 93.5 MPa$ (tensile), and the mechanical stress will be $σ_mechanical = 93.5 kN / 400 mm^2 = 233.75 MPa$ (tensile). The total stress in the rod will be $σ_total = 93.5 MPa + 233.75 MPa = 327.25 MPa$ (tensile)