Plastic analysis: given plastic modulus Zp = 5 × 10^-4 m^3, shape factor = 1.2, and plastic moment capacity Mp = 120 kN·m, determine the material's yield stress.

Introduction / Context:
In plastic design of steel members, the fully plastic moment capacity Mp is linked directly to yield stress and plastic section modulus. This problem checks the basic relation between Mp, Zp, and σy.

Given Data / Assumptions:

• Zp = 5 × 10^-4 m^3 (plastic modulus).
• Shape factor = 1.2 (not strictly needed once Zp and Mp are given).
• Mp = 120 kN·m = 120,000 N·m.
• Fully plastic stress distribution; σy uniform over yielded blocks.

Concept / Approach:
The fully plastic moment for a section is Mp = σy * Zp. Therefore σy = Mp / Zp. Units must be consistent to obtain σy in pascals, then convert to N/mm^2 (MPa).

Step-by-Step Solution:
Compute σy = Mp / Zp = 120,000 / (5 × 10^-4).Calculate numerator/denominator: 120,000 / 0.0005 = 240,000,000 N/m^2.Convert: 240,000,000 N/m^2 = 240 N/mm^2 (since 1 N/mm^2 = 10^6 N/m^2).Thus σy = 240 N/mm^2.

Verification / Alternative check:
If needed, the elastic modulus Ze = Zp / shape factor = (5 × 10^-4) / 1.2 ≈ 4.167 × 10^-4 m^3; not required for σy but consistent.

Why Other Options Are Wrong:

• 100 and 200 N/mm^2 underpredict Mp for the given Zp.
• 288 N/mm^2 overpredicts Mp for the given Zp.

Common Pitfalls:

• Mixing kN·m with N·mm without proper unit conversion.

Final Answer:
240 N/mm^2