Thermal Stress Analysis in Constrained Systems and Composites

Hello! Welcome back. In our previous lesson, we explored how constraints on thermal expansion lead to the development of internal forces. We saw this with rigid walls, springs, and even buoyancy.

Today, we'll formalize the study of these internal forces. This lesson directly addresses the learning outcome: Analyze thermal stress problems in constrained systems and composite rods. We will define the key mechanical properties of stress and strain, and then develop a systematic approach to solve the types of complex problems you'll encounter in the JEE Advanced exam.

1. A Quick Refresher: Stress, Strain, and Young's Modulus

Before we can quantify the forces from thermal expansion, we need the right language. The concepts of stress, strain, and Young's modulus are fundamental to the mechanics of materials. Since you have a background from JEE Mains, this should be a quick review to set the stage.

[PDF] Elasticity, Thermal Expansion, Calorimetry & Heat Transfer

These notes from the MyRankers PDF provide a concise definition of stress, strain, and Young's Modulus. We'll use these definitions throughout today's lesson.

Please review the following two sections: Start at the top and read down to the definition of 'Tangential Stress or Shear Stress'. This covers the definition of stress. Then, skip to the section 'HOOKE'S LAW' and read through the definition and formula for 'YOUNG'S MODULUS OF ELASTICITY 'Y''. Focus on the mathematical definitions.

To summarize the key relationships:

Stress ( $σ$ ): The internal force per unit area. $σ = \frac{F}{A}$ . Its unit is Pascals (Pa) or N/m².
Strain ( $ϵ$ ): The fractional change in length. It is a dimensionless quantity. $ϵ = \frac{Δ L}{L}$ .
Young's Modulus ( $Y$ ): A measure of a material's stiffness. It's the ratio of stress to strain within the elastic limit (Hooke's Law). $Y = \frac{σ}{ϵ}$ .

2. Thermal Stress in a Rigidly Constrained System

In the last lesson, we noted that heating a rod fixed between two immovable walls induces a compressive force. Let's now derive the magnitude of the resulting stress.

The core idea, which is a powerful problem-solving technique, is superposition. We can think of this as a two-step process:

Step 1 (Unconstrained): Imagine one of the walls is removed. The rod is heated by $Δ T$ and expands freely by an amount $Δ L_{t h er ma l} = L_{0} α Δ T$ .
Step 2 (Mechanical Compression): Now, apply a mechanical force $F$ to compress the expanded rod back to its original length. This compression is $Δ L_{e l a s t i c} = Δ L_{t h er ma l}$ .

Constrained Thermal Expansion and Thermal Shock — This diagram illustrates the principle of superposition. The final state of a heated, constrained rod (top) is equivalent to allowing it to expand freely and then mechanically compressing it back to its original length (bottom).

From this logic, we can find the stress:

The required mechanical compression is $Δ L_{e l a s t i c} = L_{0} α Δ T$ .
The strain corresponding to this compression is $ϵ = \frac{Δ L _{e l a s t i c}}{L _{0}} = \frac{L _{0} α Δ T}{L _{0}} = α Δ T$ .
Using Young's Modulus, the stress required to create this strain is:
$σ = Y ϵ$
$σ_{t h er ma l} = Y α Δ T$
This is the fundamental formula for thermal stress in a fully constrained rod. The force exerted on the walls is simply $F = σ A = Y A α Δ T$ .

Worked Example

Let's apply this to a typical problem.

[PDF] Elasticity, Thermal Expansion, Calorimetry & Heat Transfer

The MyRankers PDF contains a solved problem that is a direct application of this formula.

Locate problem 17 in the document. It starts with 'An iron bar...'. Read the problem and its solution to see the direct application of the formula F = YAαΔT.

In that problem, they ask for the compressive force:

Given: $Y = 1 0^{11} N/m^{2}$ , $α = 1 0^{- 6} /°C$ , $A = 1 0^{- 3} m^{2}$ , $Δ T = 100 °C$ .
$F = Y A α Δ T = (1 0^{11}) (1 0^{- 3}) (1 0^{- 6}) (100) = 1 0^{4} N$ .

3. Thermal Stress in Composite Systems

Things get more interesting when the system involves multiple materials. We will look at two common configurations: rods in series (end-to-end) and rods in parallel (side-by-side).

Case 1: Rods in Series

Imagine a bar made of several different materials joined end-to-end and placed between two rigid walls. When heated, each segment tries to expand by a different amount, but the total length is fixed. The superposition method we just used is perfect for this.

The following video provides an excellent, detailed walkthrough of this exact scenario. The method is just as important as the result.

Mechanics of Materials: Lesson 21 - Thermal Coefficient of Expansion, Axial Elongation

This video from Jeff Hanson provides a clear, step-by-step demonstration of the superposition method for a multi-segment bar constrained between two walls.

Watch from 03:08 to the end (20:05). The core logic is presented between 04:14 and 09:30. Pay close attention to this sequence: He first calculates the total 'free' expansion by summing the individual expansions of each segment as if the end wall D were not there. He then states that the wall must exert a reaction force to push the bar back by this total amount. He sets up the equation: Total free expansion = Total elastic compression. The compression is the sum of compressions in each segment due to the single reaction force F.

The key takeaway from the video is the governing equation that balances thermal expansion with mechanical compression:
$\sum (Δ L_{t h er ma l})_{i} = \sum (Δ L_{e l a s t i c})_{i}$
$\sum_{i} (L_{i} α_{i} Δ T) = \sum_{i} \frac{F _{i} L _{i}}{A _{i} Y _{i}}$
In the video's case, since the segments are in series and there are no external forces other than the walls, the compressive force $F$ is the same in all segments. This simplifies the equation to:
$Δ T \sum_{i} (L_{i} α_{i}) = F \sum_{i} \frac{L _{i}}{A _{i} Y _{i}}$
From this, you can solve for the unknown force $F$ , and subsequently the stress $σ_{i} = F / A_{i}$ in each segment.

Case 2: Rods in Parallel (Composite Rods)

This is the other classic setup, which we previewed in the last lesson. Two or more rods of different materials are rigidly joined side-by-side (e.g., riveted together or a rod within a tube). When the temperature changes, they are forced to have the same final length.

The material with the higher coefficient of expansion ( $α$ ) will be put into compression, while the material with the lower $α$ will be put into tension.

The solution strategy rests on two principles:

Force Equilibrium: Since the composite bar is not accelerating, the internal forces must balance. The total tensile force in the stretched rod(s) must equal the total compressive force in the compressed rod(s).
$F_{t e n s i o n} = F_{co m p r ess i o n} ⟹ σ_{t e n s i o n} A_{t e n s i o n} = σ_{co m p r ess i o n} A_{co m p r ess i o n}$
Deformation Compatibility: The final change in length, $Δ L_{f ina l}$ , must be the same for all components.
- For the compressed rod (higher $α$ ): $Δ L_{f ina l} = Δ L_{t h er ma l} - Δ L_{e l a s t i c} = L α_{co m p} Δ T - \frac{σ _{co m p} L}{Y _{co m p}}$
- For the stretched rod (lower $α$ ): $Δ L_{f ina l} = Δ L_{t h er ma l} + Δ L_{e l a s t i c} = L α_{t e n} Δ T + \frac{σ _{t e n} L}{Y _{t e n}}$

Setting $Δ L_{f ina l} = Δ L_{f ina l}$ gives our second equation. We now have a system of two equations with two unknowns ( $σ_{co m p}$ and $σ_{t e n}$ ) that you can solve.

This approach is very systematic, almost like an algorithm. The following resource provides an excellent, clearly structured set of solved problems using this exact method.

A composite section made up of copper rod 150 mm diameter en...

This solved problem from askfilo.com provides a perfect template for solving thermal stress problems in composite bars. It explicitly lays out the concepts of compatibility and equilibrium.

Read the 'Explanation' and then the 'Step-By-Step Solution' for 'Question 7'. Notice how the solution is built on the two principles we just discussed: Equilibrium (Step 4) and Compatibility (Step 3). Note that the units used are N and mm, but the physics principles are identical.

Test your understanding!

A steel rod of cross-sectional area $A_{s} = 4 cm^{2}$ and a brass rod of area $A_{b} = 6 cm^{2}$ are joined end to end. The composite rod is then placed between two rigid walls with zero initial stress. The temperature of the system is raised by $Δ T = 50 °C$ . Find the stress in the steel and brass rods.

Given:

$L_{s} = 20 cm$ , $L_{b} = 30 cm$
$Y_{s} = 2 \times 1 0^{11} Pa$ , $Y_{b} = 1 \times 1 0^{11} Pa$
$α_{s} = 1.2 \times 1 0^{- 5} K^{- 1}$ , $α_{b} = 1.9 \times 1 0^{- 5} K^{- 1}$

Show answer

This is a problem of rods in series, similar to the video example. Let $F$ be the compressive force developed in the system. The stress in steel is $σ_{s} = F / A_{s}$ and in brass is $σ_{b} = F / A_{b}$ .

The condition is that the total free expansion must be cancelled out by the total elastic compression.
$Δ L_{t h er ma l, t o t a l} = Δ L_{e l a s t i c, t o t a l}$
$L_{s} α_{s} Δ T + L_{b} α_{b} Δ T = \frac{F L _{s}}{A _{s} Y _{s}} + \frac{F L _{b}}{A _{b} Y _{b}}$
Let's plug in the numbers (converting to SI units: cm to m, cm² to m²).

$Δ T = 50$ K
$L_{s} = 0.2$ m, $L_{b} = 0.3$ m
$A_{s} = 4 \times 1 0^{- 4} m^{2}$ , $A_{b} = 6 \times 1 0^{- 4} m^{2}$

Calculate thermal expansion:

$Δ L_{t h, s} = (0.2) (1.2 \times 1 0^{- 5}) (50) = 1.2 \times 1 0^{- 4}$ m
$Δ L_{t h, b} = (0.3) (1.9 \times 1 0^{- 5}) (50) = 2.85 \times 1 0^{- 4}$ m
Total thermal expansion = $1.2 + 2.85 = 4.05 \times 1 0^{- 4}$ m

Calculate the elastic compression terms (in terms of F):

$\frac{L _{s}}{A _{s} Y _{s}} = \frac{0.2}{( 4 \times 1 0 ^{- 4} ) ( 2 \times 1 0 ^{11} )} = 2.5 \times 1 0^{- 9} m/N$
$\frac{L _{b}}{A _{b} Y _{b}} = \frac{0.3}{( 6 \times 1 0 ^{- 4} ) ( 1 \times 1 0 ^{11} )} = 5.0 \times 1 0^{- 9} m/N$
Total elastic compression = $F \times (2.5 \times 1 0^{- 9} + 5.0 \times 1 0^{- 9}) = F \times 7.5 \times 1 0^{- 9}$

Equating them:
$4.05 \times 1 0^{- 4} = F \times 7.5 \times 1 0^{- 9}$
$F = \frac{4.05 \times 1 0 ^{- 4}}{7.5 \times 1 0 ^{- 9}} = 5.4 \times 1 0^{4} N$

Now find the stresses:

Stress in steel: $σ_{s} = \frac{F}{A _{s}} = \frac{5.4 \times 1 0 ^{4}}{4 \times 1 0 ^{- 4}} = 1.35 \times 1 0^{8} Pa$ (compressive)
Stress in brass: $σ_{b} = \frac{F}{A _{b}} = \frac{5.4 \times 1 0 ^{4}}{6 \times 1 0 ^{- 4}} = 0.9 \times 1 0^{8} Pa$ (compressive)

Conclusion

Today, we've developed a robust framework for tackling thermal stress problems, a topic that frequently appears in JEE Advanced. By combining the physics of thermal expansion with the mechanics of elastic deformation, we can solve for the internal forces and stresses in constrained systems.

Key Takeaways:

Thermal stress arises when an object is prevented from expanding or contracting freely. For a simple rod rigidly fixed at both ends, the stress is $σ = Y α Δ T$ .
Problems involving composite systems (in series or parallel) can be solved systematically by applying two fundamental principles:
1. Force Equilibrium: Internal tensile and compressive forces must balance.
2. Deformation Compatibility: All parts of the system must deform in a geometrically consistent way (e.g., have the same final length).
The superposition method (calculating free expansion and then the force required to reverse it) is a powerful conceptual tool for analyzing these problems.

Preview of the Next Lesson:
We've seen how differential expansion in parallel rods leads to tension and compression. What happens if the rods are joined along their length, like a single strip made of two metals? The differential expansion will cause the strip to bend. In our next lesson, we will apply thermal expansion concepts to problems with bimetallic strips, analyzing the curvature that results from a change in temperature.

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