LEVELJEE Main

Animated Solution for Physics - Properties of Solids and Liquids: A metal rod of Young's modulus $Y$ and coefficient of thermal expansion $α$ is held at its two ends such that its length remains invariant. If its temperature is raised by $t^{\circ} C$ , the linear stress developed in it is

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Visualized Solution

$Physical Setup$

$Rod is fixed between rigid supports.$
$Net change in length Δ L_{net} = 0$

$Thermal Expansion$

$If free to expand:$
$Δ L_{thermal} = Lα Δ T$
$Here, Δ T = t$

$Elastic Compression$

$Supports apply compressive force F$
$Δ L_{elastic} = \frac{F L}{A Y}$

$Equating Deformations$

$Δ L_{thermal} = Δ L_{elastic}$
$Lα t = \frac{F L}{A Y}$

$Simplification$

$Cancel L from both sides:$
$α t = \frac{F}{A Y}$

$Thermal Stress$

$Stress = \frac{F}{A}$
$Stress = Y α t$

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The Sigma Insight: Thermal Expansion

Solution Diagram

Analyzing the Setup

Imagine you are observing a metal rod perfectly wedged between two immovable, rigid walls. The rod has a natural tendency to respond to temperature changes, but its environment restricts it.

When we raise the temperature of the rod by

t^{\circ} C

, the heat energy causes the atoms within the metal to vibrate more vigorously, pushing against each other. If the rod were lying freely on a table, it would simply expand.

The Master Equation

The natural thermal expansion of the rod is given by the formula:

Δ L_{thermal} = Lα t

where

L

is the original length,

α

is the coefficient of thermal expansion, and

t

is the change in temperature.

However, the rigid walls refuse to yield. To keep the rod's length invariant, the walls must exert a massive compressive force

F

inward. This force essentially "squishes" the rod back to its original length. According to Hooke's Law, the elastic compression is:

Δ L_{elastic} = \frac{F L}{A Y}

Final Calculation

Because the rod's actual length does not change, the thermal expansion must be perfectly counteracted by the elastic compression. We can set these two effects equal to each other:

Lα t = \frac{F L}{A Y}

Notice how the original length

L

appears on both sides of the equation? We can cancel it out! This is a profound physical insight: the thermal stress developed in a constrained rod is completely independent of its length.

α t = \frac{F}{A Y}

We are looking for the linear stress, which is defined as the internal restoring force per unit area (

F / A

). Rearranging our simplified equation, we get:

Stress = \frac{F}{A} = Y α t

This elegant result tells us that the stress depends only on the material properties (

Y

and

α

) and the temperature change (

t

Common Traps & FAQs

Why do we equate thermal expansion to elastic compression?

Because the rod is held between rigid supports, its net change in length is zero. The amount it 'wants' to expand due to heat is exactly the amount the walls must compress it mechanically.

Why doesn't the stress depend on the length of the rod?

Both the thermal expansion and the elastic compression are directly proportional to the original length

L

. When you equate them,

L

cancels out from both sides of the equation.

What if only one end of the rod was fixed?

If only one end was fixed, the other end would be free to move. The rod would simply expand without any resistance, meaning no internal force would develop, and the thermal stress would be zero.

Is the stress developed tensile or compressive?

It is compressive. As the rod tries to expand due to heating, the rigid walls push inward against it to prevent the expansion.

Why do we use Young's modulus here instead of Bulk modulus?

Young's modulus is used because the rod is constrained and compressed only along one dimension (its length). Bulk modulus would be used if the object was subjected to pressure from all sides (volumetric compression).

Similar Questions

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(A)

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(A)

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(B)

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(C)

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The area of cross-section of a railway track is $0.01 m^{2}$ . The temperature variation is $1 0^{\circ} C$ . Coefficient of linear expansion of material of track is $1 0^{- 5} /^{\circ} C$ . The energy stored per metre in the track is ...... $J/m$ . (Take, Young's modulus of material of track is $1 0^{11} Nm^{- 2}$ )

Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Physical Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Thermal Expansion

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Elastic Compression

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Equating Deformations

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Simplification

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Thermal Stress

The Sigma Insight: Thermal Expansion

Analyzing the Setup

The Master Equation

Final Calculation

Similar Questions

$Physical Setup$

$Thermal Expansion$

$Elastic Compression$

$Equating Deformations$

$Simplification$

$Thermal Stress$