Volumetric strain for a circular (isotropic) bar: If a circular bar experiences a longitudinal (axial) strain ε_long and an equal lateral strain ε_lat in each of the two transverse directions, what is the volumetric strain?

Introduction / Context:Volumetric strain quantifies the fractional change in volume of a 3D body due to deformation. For bars under uniaxial loading, the axial strain combines with lateral strains (due to Poisson effect) to produce the overall volume change.

Given Data / Assumptions:

• Cylindrical (circular) prismatic bar, isotropic material.
• Small strains; principal (mutually orthogonal) strains are: ε_x = ε_long, ε_y = ε_lat, ε_z = ε_lat.
• Lateral strains are equal in the two transverse directions.

Concept / Approach:For small strains, the volumetric strain ε_v equals the sum of the normal strains in three orthogonal directions: ε_v = ε_x + ε_y + ε_z. Under uniaxial loading with symmetry, this becomes ε_v = ε_long + 2 ε_lat.

Step-by-Step Solution:Write principal strains: ε_x = ε_long; ε_y = ε_lat; ε_z = ε_lat.Sum them: ε_v = ε_x + ε_y + ε_z.Therefore, ε_v = ε_long + 2 ε_lat.

Verification / Alternative check:If Poisson’s ratio ν is known under axial stress σ, then ε_lat = −ν ε_long, giving ε_v = ε_long (1 − 2 ν). This well-known relation confirms the form above.

Why Other Options Are Wrong:Permutations with minus signs or different coefficients do not represent the sum of three orthogonal normal strains for an isotropic bar.

Common Pitfalls:Forgetting there are two equal lateral directions, or confusing engineering strain sign convention (tension positive, lateral contraction negative).

Final Answer:ε_long + 2 ε_lat.