Plane Stress — Orientation for Maximum Normal Stress under Uniaxial Tension A body is subjected to a direct tensile stress σ in one plane only (σy = 0, τxy = 0). The section on which the maximum normal stress acts is inclined at how many degrees to the normal of that section?

Introduction:
Understanding stress transformation is critical for failure analysis. Under uniaxial tension, principal stresses occur on specific planes. This question asks for the orientation that produces the maximum normal stress when only one normal stress is present.

Given Data / Assumptions:

• Plane stress, σx = σ (tension), σy = 0, τxy = 0.
• Linear elasticity and homogeneous material.

Concept / Approach:
For a plane at angle θ (measured from the x-plane normal), the transformed normal stress is: σn = σx * cos^2θ With no other stresses, σn is maximized when cos^2θ = 1, i.e., θ = 0°, meaning the plane whose normal aligns with the applied stress direction (principal plane).

Step-by-Step Solution:
Start with σn = σ * cos^2θ.Derivative wrt θ shows maximum at θ = 0°; physically, a plane perpendicular to the load carries full σ.Thus, maximum normal stress occurs on the plane whose normal is colinear with the load, i.e., θ = 0°.

Verification / Alternative check:
Mohr’s circle degenerates to a point at σ, confirming the single principal stress appears on planes at θ = 0° and 90° (maximum and minimum, respectively).

Why Other Options Are Wrong:

• 45°: Associated with maximum shear under pure shear or combined states, not here.
• 30°, 60°: No extremum occurs at these angles for σn.
• 90°: Gives σn = 0 in this case (plane parallel to load).

Common Pitfalls:
Confusing the angle for maximum shear (often 45°) with the angle for maximum normal stress (0° here).

Final Answer:
0°