Failure stress of a simply supported cast iron bar under center load A cast iron bar of square section 2.5 cm × 2.5 cm and span 1.0 m is simply supported. It fails when a 100 kg weight is applied at midspan. Determine the maximum bending stress at failure (use kg/cm^2 units).

Introduction / Context:
This is a classic flexure problem for a simply supported member under a central point load. It tests the ability to compute section modulus and relate bending moment to extreme fiber stress.

Given Data / Assumptions:

• Cross-section: square, b = d = 2.5 cm.
• Span L = 1.0 m = 100 cm.
• Central point load W = 100 kgf (kilogram-force), using kg/cm^2 units.
• Material behaves linearly up to failure for stress calculation.

Concept / Approach:
For a simply supported beam with a central load, maximum bending moment occurs at midspan and equals M_max = W * L / 4. The extreme fiber stress follows flexure formula: σ = M / Z, where Z is the section modulus of a rectangle, Z = b * d^2 / 6 (about the strong axis).

Step-by-Step Solution:
Compute M_max = W * L / 4 = 100 * 100 / 4 = 2500 kgf·cm.Compute section modulus Z = b * d^2 / 6 = 2.5 * (2.5)^2 / 6 = 15.625 / 6 = 2.6042 cm^3 (approx).Compute stress σ = M / Z = 2500 / 2.6042 ≈ 960 kg/cm^2.Therefore, the failure stress is about 960 kg/cm^2.

Verification / Alternative check:
Using more precise arithmetic yields the same rounded value (≈ 959.9 kg/cm^2). The answer aligns with typical cast iron flexural strength ranges.

Why Other Options Are Wrong:
980 and 1000 kg/cm^2 are overestimates from rounding errors; 1200 kg/cm^2 is significantly high for the given geometry and load.

Common Pitfalls:
Mixing N and kgf units; using I instead of Z directly; forgetting to convert L to centimeters when working in kg/cm^2.

Final Answer:
960 kg/cm^2