Difficulty: Easy
Correct Answer: length increases, width and thickness decreases
Explanation:
Introduction / Context:
In strength of materials, the dimensional response of a bar under uniaxial tension illustrates axial strain and lateral contraction, governed by Young's modulus and Poisson's ratio. This concept is fundamental for predicting deformations in structural members and machine elements.
Given Data / Assumptions:
Concept / Approach:
Axial normal stress is sigma = P / A, where A = b * t. The axial strain is epsilon_long = sigma / E, producing an increase in length. Due to Poisson effect, lateral strain is epsilon_lat = -nu * epsilon_long in directions transverse to the load, causing width and thickness to decrease when nu > 0 (typical solids).
Step-by-Step Solution:
Compute axial stress: sigma = P / (b * t).Axial strain: epsilon_long = sigma / E > 0 → length increases.Lateral strains: epsilon_b = -nu * epsilon_long, epsilon_t = -nu * epsilon_long → width and thickness decrease.Resultant dimensional changes: Δl = epsilon_long * l > 0; Δb < 0; Δt < 0.
Verification / Alternative check:
Energy or compatibility viewpoints lead to the same conclusion: positive axial work elongates the member; volume change is approximately ΔV/V ≈ (1 - 2 nu) * epsilon_long for small strains, supporting lateral contraction for typical nu between 0 and 0.5.
Why Other Options Are Wrong:
Options (a) and (b) change all three dimensions in the same sense, contradicting Poisson effect. Option (d) suggests compression, not tension. Option (e) ignores lateral contraction observed in virtually all structural metals.
Common Pitfalls:
Assuming width and thickness stay constant; confusing engineering strain sign conventions; neglecting that Poisson's ratio is positive for most materials.
Final Answer:
length increases, width and thickness decreases
Discussion & Comments