Locating the neutral axis from extreme fiber stresses In a rectangular beam of total depth 10 cm, the maximum compressive stress at the top is 1600 kg/cm² and the corresponding tensile stress at the bottom is 400 kg/cm² under the same bending moment. How far is the neutral axis from the top surface?

Introduction / Context:Under pure bending with linear elastic behaviour, normal stress varies linearly with distance from the neutral axis (N.A.). Knowing the extreme fiber stresses allows you to back-compute the N.A. position in a homogeneous rectangular section — an essential diagnostic skill when checking combined or unsymmetrical situations.

Given Data / Assumptions:

• Total depth d = 10 cm.
• Top compressive stress = 1600 kg/cm².
• Bottom tensile stress = 400 kg/cm².
• Homogeneous section, linear stress distribution σ = (M/I) y.

Concept / Approach:

For a rectangle, stress magnitudes at the top and bottom are proportional to their distances from the N.A. Let x be the distance from the top surface to the N.A. Then the top distance is x, and the bottom distance is (10 − x). The ratio of stresses equals the ratio of distances from the N.A.

Step-by-Step Solution:

Verification / Alternative check:

Distances 8 cm (top to N.A.) and 2 cm (bottom to N.A.) have ratio 8:2 = 4, matching the stress ratio 1600:400 = 4, confirming consistency.

Why Other Options Are Wrong:

• 2 cm or 4 cm place the N.A. too close to the bottom or mid-depth, contradicting the 4:1 stress ratio.
• 6 cm yields a 6:4 distance ratio = 1.5, not 4.
• 10 cm would put the N.A. at the bottom, impossible for the given opposite stress signs.

Common Pitfalls:

• Forgetting that stress is directly proportional to distance from the N.A., not the full depth.
• Mixing signs; use magnitudes when forming the ratio.

Final Answer:

8 cm from the top.