Axial deformation and linear strain of a rectangular bar A rectangular bar of length l, breadth b, and thickness t is subjected to an axial tensile pull P. What is the linear (engineering) strain ε along the length in terms of P, b, t, and modulus of elasticity E?

Introduction / Context:Linear strain connects applied load to material stiffness and cross-section. This relationship underpins elongation predictions for bars, tie rods, and machine members subjected to direct tension.

Given Data / Assumptions:

• Prismatic bar with constant rectangular cross-section A = b * t.
• Axial tensile load P; homogeneous, isotropic material.
• Linear elastic behavior (Hooke's law) and small strains.

Concept / Approach:Normal stress σ = P / A. Engineering strain ε = σ / E in the linear elastic range. Substituting A gives ε = (P / (b * t)) / E = P / (b * t * E).

Step-by-Step Solution:Area: A = b * t.Stress: σ = P / A = P / (b * t).Hooke’s law: ε = σ / E = [P / (b * t)] / E.Therefore: ε = P / (b * t * E).

Verification / Alternative check:Elongation Δl = (P * l) / (A * E). Dividing by original length l yields ε = Δl / l = P / (A * E) = P / (b * t * E), matching the result.

Why Other Options Are Wrong:(b) inverts the relationship; (c) mixes incompatible terms; (d) ignores area and is dimensionally incorrect; (e) gives displacement-like units rather than dimensionless strain.

Common Pitfalls:Forgetting to use cross-sectional area; confusing strain with elongation; mixing true and engineering strain definitions.

Final Answer:ε = P / (b * t * E)