Simply supported timber beam with central point load: A 150 cm span wooden beam (rectangular section 16 cm × 24 cm, depth taken as 24 cm) carries a concentrated load at midspan. Permissible bending stress ft = 75 kg/cm² and permissible shear stress fs = 10 kg/cm². What is the safe load based on combined checks?

Introduction / Context:
Design of simply supported beams under a central point load requires checking both bending and shear against permissible stresses. The lower governing value determines the safe load.

Given Data / Assumptions:

• Span L = 150 cm.
• Section b × d = 16 cm × 24 cm (d = vertical depth).
• Permissible bending stress ft = 75 kg/cm².
• Permissible shear stress fs = 10 kg/cm².
• Load W acts at midspan.

Concept / Approach:
For a central point load on a simple span: maximum bending moment M = W * L / 4. Section modulus for a rectangle Z = b * d² / 6. Maximum support shear V = W / 2; for a rectangular section, τ_max = 1.5 * V / (b * d).

Step-by-Step Solution:
Compute section modulus: Z = 16 * 24^2 / 6 = 1536 cm³.Bending check: ft ≥ M / Z = (W * L / 4) / Z → W ≤ 4 * Z * ft / L = 4 * 1536 * 75 / 150 = 3072 kg (≈ 3075 kg).Shear check: τ_max = 1.5 * (W / 2) / (b * d) = 1.5 * W / (2 * 16 * 24) → fs ≥ τ_max gives W ≤ (2 * b * d * fs) / 1.5 = (2 * 16 * 24 * 10) / 1.5 = 5120 kg.Governing value is bending: W_safe ≈ 3072 kg ≈ 3075 kg.

Verification / Alternative check:
Recalculate M = 3075 * 150 / 4 = 115,312.5 kg·cm; σ = M/Z ≈ 75.1 kg/cm², matching allowable within rounding.

Why Other Options Are Wrong:
Values near 3025–3100 kg are distractors; only 3075 kg aligns with the bending limit and rounding of 3072 kg.Higher loads would exceed ft; lower loads are conservative but not the computed safe limit.

Common Pitfalls:
Using average shear instead of τ_max for rectangular beams; treating 16 cm as depth rather than breadth.

Final Answer:
3075 kg.