9.3 Thermal stresses and deformations

Thermal Stresses and Deformations

When materials heat up or cool down, they naturally want to expand or contract. If they're free to do so, no stress develops. But if something restrains that movement, internal stresses build up. These thermal stresses are a major concern in axial loading problems because temperature swings can push a member into compression or tension without any external force being applied.

This section covers how to calculate thermal deformations and stresses, how to combine them with mechanical loads, and how to handle both determinate and indeterminate systems.

Thermal Stress in Axial Members

Concept and Effects

Thermal stress is the internal stress that develops when a material tries to expand or contract due to a temperature change but is prevented from doing so by constraints.

Think about a steel rod bolted between two rigid walls. If the temperature rises, the rod wants to get longer, but the walls won't let it. That blocked expansion creates compressive stress inside the rod. If the temperature drops, the rod wants to shorten, the walls resist, and tensile stress develops.

The key principle: if an axially loaded member is free to expand or contract, thermal deformation occurs but no thermal stress develops. If the member is restrained, thermal stress arises because the deformation is partially or fully prevented.

Factors Influencing Thermal Stress

• Coefficient of thermal expansion ($\alpha$): Materials with higher $\alpha$ values expand more per degree of temperature change, leading to greater thermal stresses when restrained.
• Modulus of elasticity ($E$): Stiffer materials resist deformation more strongly, so the same blocked expansion produces higher stress in a stiffer material.
• Temperature change ($\Delta T$): Larger temperature swings produce proportionally larger thermal effects.
• Degree of restraint:
  • Fully restrained members (both ends fixed) experience the maximum thermal stress
  • Partially restrained members experience reduced thermal stress
  • Unrestrained members (free to move) experience zero thermal stress

Real-world examples include steel bridge girders that expand and contract with daily temperature cycles, and concrete pavements that crack under seasonal temperature variations. Expansion joints in bridges and highways exist specifically to manage these effects.

Calculating Thermal Stresses and Deformations

Thermal Deformation Calculation

When a member is free to change length, the thermal deformation is:

$\delta_{thermal} = \alpha \times L \times \Delta T$

• $\alpha$ = coefficient of thermal expansion
• $L$ = original length of the member
• $\Delta T$ = temperature change (positive for heating, negative for cooling)

Example: A 2 m long aluminum rod ($\alpha = 23 \times 10^{-6} /°C$) is heated by 80°C.

$\delta_{thermal} = 23 \times 10^{-6} \times 2 \times 80 = 0.00368 \text{ m} = 3.68 \text{ mm (expansion)}$

Thermal Stress Calculation

When a member is fully restrained (both ends fixed against a rigid support), the thermal deformation is completely blocked. The resulting stress is:

$\sigma_{thermal} = -E \times \alpha \times \Delta T$

The negative sign reflects the sign convention: a positive $\Delta T$ (heating) causes the member to push against its restraints, producing compressive (negative) stress. A negative $\Delta T$ (cooling) produces tensile (positive) stress.

Example: A fully restrained steel rod ($E = 200 \text{ GPa}$, $\alpha = 12 \times 10^{-6} /°C$) undergoes a temperature increase of 50°C.

$\sigma_{thermal} = -(200 \times 10^3) \times (12 \times 10^{-6}) \times 50 = -120 \text{ MPa (compressive)}$

Internal Force from Thermal Restraint

When you need the axial force that develops in a restrained member:

$\Delta F = -E \times A \times \alpha \times \Delta T$

where $A$ is the cross-sectional area.

Example: A fully restrained copper bar ($E = 110 \text{ GPa}$, $\alpha = 17 \times 10^{-6} /°C$, $A = 500 \text{ mm}^2$) cools by 60°C.

$\Delta F = -(110 \times 10^3)(500)(17 \times 10^{-6})(-60) = 56{,}100 \text{ N} = 56.1 \text{ kN (tension)}$

The positive result confirms tension: cooling makes the bar want to shrink, but the restraints pull it back, creating a tensile force.

Combined Mechanical and Thermal Loads

Superposition of Stresses and Deformations

When both mechanical forces and temperature changes act on a member, you find the total response by adding the two effects together (superposition):

$\sigma_{total} = \sigma_{mechanical} + \sigma_{thermal}$

$\delta_{total} = \delta_{mechanical} + \delta_{thermal}$

This works because we're assuming linear elastic behavior, where effects can be added independently.

Calculation Steps

1. Calculate mechanical stress and deformation from the applied force:

   • $\sigma_{mechanical} = \frac{F}{A}$
   • $\delta_{mechanical} = \frac{F \times L}{E \times A}$

2. Calculate thermal stress and deformation from the temperature change:

   • $\sigma_{thermal} = -E \times \alpha \times \Delta T$ (if restrained)
   • $\delta_{thermal} = \alpha \times L \times \Delta T$ (if free to deform)

3. Add them together, keeping careful track of signs (tension = positive, compression = negative).

Worked Example

A 1.5 m long steel rod ($E = 200 \text{ GPa}$, $A = 800 \text{ mm}^2$, $\alpha = 12 \times 10^{-6} /°C$) carries an axial compressive force of 120 kN and also undergoes a temperature increase of 40°C. The rod is free to deform axially (not fixed at both ends), so we're looking at total deformation.

Mechanical effects:

$\sigma_{mechanical} = \frac{120{,}000}{800} = 150 \text{ MPa (compressive)}$

$\delta_{mechanical} = \frac{120{,}000 \times 1500}{200{,}000 \times 800} = 1.125 \text{ mm (shortening)}$

Thermal deformation (free to expand):

$\delta_{thermal} = 12 \times 10^{-6} \times 1500 \times 40 = 0.72 \text{ mm (expansion)}$

Total deformation:

$\delta_{total} = -1.125 + 0.72 = -0.405 \text{ mm}$

The rod still shortens overall, but the thermal expansion partially offsets the mechanical compression. Note that when the rod is free to expand, the thermal deformation doesn't generate thermal stress; it only contributes to the total change in length.

Thermal Stress in Statically Determinate vs. Indeterminate Systems

Statically Determinate Systems

In a statically determinate system, equilibrium equations alone are enough to find all reactions. A classic example is a rod fixed at one end with the other end free.

The critical point: in a determinate system with a uniform temperature change, the member is free to expand or contract. No thermal stress develops. You only get thermal deformation.

Example: A 2 m aluminum rod ($E = 70 \text{ GPa}$, $\alpha = 23 \times 10^{-6} /°C$, $A = 600 \text{ mm}^2$) is fixed at one end and free at the other. It heats up by 60°C.

Since the free end is unconstrained, the rod simply expands:

$\delta_{thermal} = 23 \times 10^{-6} \times 2 \times 60 = 2.76 \text{ mm (expansion at the free end)}$

No reaction force develops at the fixed end beyond what's needed for equilibrium (which is zero, since no external load is applied). The thermal stress is zero.

Statically Indeterminate Systems

A statically indeterminate system has more constraints than equilibrium equations can solve alone. A rod fixed at both ends is the classic case. You need an additional compatibility equation that enforces a geometric condition (e.g., the total change in length must be zero).

Here's the step-by-step approach:

1. Identify the redundant reaction. For a rod fixed at both ends, one of the support reactions is the unknown.
2. Write the compatibility equation. For rigid walls at both ends: $\delta_{total} = \delta_{mechanical} + \delta_{thermal} = 0$
3. Express deformations in terms of the unknown force. The mechanical deformation depends on the unknown reaction force $F$: $\delta_{mechanical} = \frac{F \times L}{E \times A}$
4. Solve for the unknown force.
5. Calculate stresses from the force you found.

Example: A 3 m copper rod ($E = 110 \text{ GPa}$, $\alpha = 17 \times 10^{-6} /°C$, $A = 400 \text{ mm}^2$) is fixed at both ends and cools by 50°C.

Step 1: The rod wants to shrink, but both walls prevent it. A tensile reaction force $F$ develops.

Step 2: Compatibility requires zero net deformation:

$\delta_{mechanical} + \delta_{thermal} = 0$

Step 3: Substitute the deformation expressions:

$\frac{F \times L}{E \times A} + \alpha \times L \times \Delta T = 0$

Step 4: Solve for $F$:

$F = -E \times A \times \alpha \times \Delta T$

$F = -(110 \times 10^3)(400)(17 \times 10^{-6})(-50) = 37{,}400 \text{ N} = 37.4 \text{ kN (tension)}$

Step 5: The stress in the rod:

$\sigma = \frac{F}{A} = \frac{37{,}400}{400} = 93.5 \text{ MPa (tension)}$

You can verify this matches the direct thermal stress formula for a fully restrained member:

$\sigma = E \times \alpha \times |\Delta T| = 110{,}000 \times 17 \times 10^{-6} \times 50 = 93.5 \text{ MPa}$

Notice that in the indeterminate case, the mechanical deformation and thermal deformation cancel each other out (total deformation is zero), but the stress is very real. This is why indeterminate systems under temperature changes require careful analysis: the structure looks like nothing moved, but significant internal stresses can be present.