Combined uniaxial normal stress σx and shear stress τxy: write the expression for the minimum principal (normal) stress at the point.

Introduction / Context:
When a point in a member is subjected to a normal stress in one direction and an in-plane shear stress, the state of stress must be resolved to principal stresses to assess failure. The minimum (more compressive) principal stress determines safety for brittle materials and affects yielding criteria.

Given Data / Assumptions:

• Plane stress with components σx and τxy; σy = 0 initially.
• Material linear-elastic; small deformations.
• Sign convention: tension positive; shear using standard Mohr’s circle sign.

Concept / Approach:
Principal stresses for a general plane stress state are given by σ1,2 = (σx + σy)/2 ± √( ((σx − σy)/2)^2 + τxy^2 ). Here σy = 0, so σ1,2 = (σx / 2) ± √( (σx/2)^2 + τxy^2 ). The minimum principal stress is the minus-root expression.

Step-by-Step Solution:

Verification / Alternative check:
Construct Mohr’s circle with center at (σx/2, 0) and radius R; the leftmost intersection on the σ-axis equals σ_min.

Why Other Options Are Wrong:
Linear combinations like σx − τxy are not correct transformations.Adding the square-root gives σ_max, not σ_min.The radical expression without averaging does not represent a principal stress.

Common Pitfalls:
Forgetting σy term; sign mistakes in Mohr’s circle; swapping σ_max and σ_min by choosing the wrong sign.

Final Answer:

σ_min = (σx / 2) − √( (σx / 2)^2 + τxy^2 )