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Question #43530

Difficulty: 8/10 | 🟢 Free

The Question

A cylindrical rod of length LL and constant cross-sectional area AA is rigidly clamped between two immovable walls at x=0x=0 and x=Lx=L. Initially, the rod is at a uniform temperature. The temperature of the rod is then increased such that the temperature change at any cross-section xx is given by ΔT(x)=T0xL\Delta T(x) = T_0 \frac{x}{L}, where T0T_0 is a positive constant. The coefficient of linear thermal expansion of the material varies with position as α(x)=α0xL\alpha(x) = \alpha_0 \frac{x}{L}. Furthermore, the Young's modulus of the material depends on the local temperature change as Y(ΔT)=Y01+ΔTT0Y(\Delta T) = \frac{Y_0}{1 + \frac{\Delta T}{T_0}}. Assuming the rod does not buckle, what is the magnitude of the compressive force FF developed in the rod?

Options

  • A.

    29AY0α0T0\frac{2}{9} A Y_0 \alpha_0 T_0

  • B.

    13AY0α0T0\frac{1}{3} A Y_0 \alpha_0 T_0

  • C.

    12AY0α0T0\frac{1}{2} A Y_0 \alpha_0 T_0

  • D.

    23AY0α0T0\frac{2}{3} A Y_0 \alpha_0 T_0

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Multi-Concept_Bridge

Master problems that require linking physics and math concepts to solve complex competitive questions.

Concepts Behind This Question

Stress vs. Strain Definition|Thermal Stress in Constrained Rods|Calculus Integration

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to abandon the macroscopic formula F=YAαΔTF = Y A \alpha \Delta T and instead apply the fundamental differential definitions of thermal strain and mechanical strain.

The primary trap lies in treating the rod as a single uniform entity or using average values for temperature, expansion coefficient, and Young's modulus.

Students often incorrectly assume that they can calculate an 'effective' α\alpha or YY by simple averaging, or they fail to recognize that while the internal force FF is uniform due to static equilibrium, the local mechanical strain dϵmech=F/(AY(x))d\epsilon_{mech} = F / (A Y(x)) varies because the stiffness Y(T)Y(T) is coupled to the local temperature gradient.

By forcing the integration of both the non-linear thermal expansion and the variable mechanical compliance, the problem rigorously evaluates the concept of infinite infinitesimal springs in series.

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterMechanical Properties of Solids - Class 11 Part 2
ConceptStress vs. Strain Definition
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree