Simple shear of a square block (mechanics of materials): with side AD fixed and a simple shear applied producing shear stress τ and engineering shear strain γ, what is the linear (normal) strain along a diagonal at 45° to the sides?

Introduction / Context:Under simple shear, a square element distorts into a rhombus. Although the pure shear state has zero normal strain in the original x and y directions (for small strains), there are tensile and compressive normal strains along directions rotated by 45°, which correspond to principal strain directions.

Given Data / Assumptions:

• Small (engineering) shear strain γ relates to shear stress by γ = τ / G (for linear elastic isotropic material).
• We examine linear strain along diagonals oriented at ±45° to the original axes.
• Plane strain/plane stress small-strain kinematics apply.

Concept / Approach:For pure/simple shear, principal normal strains are ±γ / 2 along directions at ±45° to the shear directions. Thus, one diagonal experiences tensile normal strain of magnitude γ/2; the orthogonal diagonal experiences compressive normal strain of magnitude γ/2.

Step-by-Step Solution:Start with shear strain tensor having only γ_xy = γ_yx = γ/2 (engineering shear γ).Rotate axes by 45°: the shear components transform to normal components.Result: ε_45° = +γ/2 (tension) and ε_-45° = −γ/2 (compression).Therefore, the linear strain along one diagonal is γ/2 in tension.

Verification / Alternative check:Mohr’s circle for strain shows normal strain extrema at ±γ/2 located 90° apart on the circle, corresponding to physical directions at ±45° in the material.

Why Other Options Are Wrong:

• Both diagonals cannot be in tension; one is tensile, the other compressive.
• Zero strain along diagonals is false for nonzero γ.
• γ/4 is not consistent with the standard transformation relations.

Common Pitfalls:Mixing engineering shear γ with tensor components; forgetting the 45° rotation for principal directions.

Final Answer:γ / 2 (tension) along one diagonal