In a uniaxial tension test with yield stress σ = 300 kg/cm², what is the octahedral shear stress at yield?

Introduction / Context:
Octahedral shear stress is used in energy-based yield criteria and provides a scalar shear measure derived from principal stresses. It is frequently used when discussing von Mises type theories.

Given Data / Assumptions:

• Uniaxial tension at yield: principal stresses are σ1 = σ, σ2 = 0, σ3 = 0 with σ = 300 kg/cm².
• Use the standard octahedral shear stress relation consistent with many textbooks.

Concept / Approach:
One common expression for octahedral shear stress is:τ_oct = (1/3) * sqrt[ ( (σ1 - σ2)^2 + (σ2 - σ3)^2 + (σ3 - σ1)^2 ) ]Apply this to the uniaxial state.

Step-by-Step Solution:
Differences: (σ1 - σ2) = σ, (σ2 - σ3) = 0, (σ3 - σ1) = -σ.Sum of squares = σ^2 + 0 + σ^2 = 2 σ^2.τ_oct = (1/3) * sqrt(2 σ^2) = (σ √2)/3.With σ = 300 kg/cm² → τ_oct = 300 √2 / 3 = 100 √2 kg/cm².

Verification / Alternative check:
Relate to von Mises stress: σ_vm = sqrt( ( (σ1 - σ2)^2 + (σ2 - σ3)^2 + (σ3 - σ1)^2 ) / 2 ) = σ. Then τ_oct = σ_vm / √3 = σ/√3 ≈ 0.577σ, which differs numerically from the chosen textbook convention here; this problem’s options correspond to the (σ√2)/3 form, giving 100√2.

Why Other Options Are Wrong:

• 150√2, 200√2, 250√2 kg/cm² do not result from substituting σ = 300 kg/cm² into the adopted formula.

Common Pitfalls:

• Mixing different definitions of τ_oct; always align with the convention implied by the choices.

Final Answer:
100√2 kg/cm²