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Question #42294

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The Question

A uniform rod of length $L$ and cross-sectional area $A$ is rigidly fixed between two immovable walls at $x=0$ and $x=L$ . Initially, the rod is unstressed at a uniform temperature $T_0$ . The temperature of the rod is then changed to a steady state distribution $T(x) = T_0(1 + k x^2)$ , where $k$ is a positive constant. The coefficient of linear thermal expansion of the rod is $\alpha$ . The Young's modulus of the rod varies with position as $Y(x) = \frac{Y_0}{1 + \beta x^2}$ , where $Y_0$ and $\beta$ are positive constants. What is the compressive force $F$ developed in the rod?

Options

A.
$\frac{A Y_0 \alpha T_0 k L^2}{3 + \beta L^2}$
B.
$\frac{A Y_0 \alpha T_0 k L^2}{1 + \beta L^2}$
C.
$\frac{3 A Y_0 \alpha T_0 k L^2}{3 + \beta L^2}$
D.
$\frac{A Y_0 \alpha T_0 k L^2}{3(1 + \beta L^2)}$

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Variable_Constraint

Develop pure symbolic logic skills by mastering algebraic constraints without relying on numerical data.

Concepts Behind This Question

Young's Modulus Sensitivity|Thermal Stress Calculation|Integration

Why This Question Is Tricky

Common Mistakes

This question tests the coupling of spatially varying thermal strain and material properties under a global constraint.

Students often incorrectly assume that either the stress or the strain is constant across the rod.

While the internal force $F$ must be uniform across the cross-section to maintain static equilibrium, the local strain $\epsilon(x)$ must vary to compensate for the changing Young's modulus $Y(x)$ and the non-linear temperature gradient $\Delta T(x)$ .

The trap lies in failing to integrate the differential deformation $dL = (\frac{F}{A Y(x)} + \alpha \Delta T(x))dx$ over the length $L$ , or incorrectly averaging $Y(x)$ independently before calculating the force.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Mechanical Properties of Solids - Class 11 Part 2
Concept	Young's Modulus Sensitivity
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free