Slab base design – required thickness: An axially loaded column bears on a slab base with projection a = 8 cm beyond the column face. Given concrete bearing pressure q = 40 kg/cm² and allowable bending stress in the base plate σ_allow = 1890 kg/cm², what plate thickness t is required (assume cantilever action of the projection)?

Introduction / Context:Slab bases distribute column loads to concrete foundations. The plate projects beyond the column face and behaves like a cantilever strip under uniform bearing pressure. Plate thickness is chosen so bending stress in the projection does not exceed the allowable value.

Given Data / Assumptions:

• Projection a = 8 cm.
• Bearing pressure on concrete q = 40 kg/cm² (uniform).
• Allowable bending stress in steel plate σ_allow = 1890 kg/cm².
• Projection behaves as a cantilever of width 1 cm with uniform load.

Concept / Approach:For a unit-width strip, uniform line load w = q * a (kg/cm). Cantilever moment at the column face: M = w * a / 2 = q * a² / 2. Section modulus for a unit-width plate: Z = b * t² / 6 = t² / 6. Strength check uses M = σ_allow * Z → t = sqrt(6 * M / σ_allow).

Step-by-Step Solution:Compute w = q * a = 40 * 8 = 320 kg/cm.Compute M = q * a² / 2 = 40 * 8² / 2 = 40 * 64 / 2 = 1280 kg·cm per cm width.Compute t from M = σ * t² / 6 → t = sqrt(6 * M / σ_allow).t = sqrt(6 * 1280 / 1890) = sqrt(7680 / 1890) ≈ sqrt(4.0635) ≈ 2.016 cm.Adopt practical size: ≈ 2.0 cm.

Verification / Alternative check:Alternate formula t = a * sqrt(3 * q / σ_allow) gives t = 8 * sqrt(120 / 1890) ≈ 8 * 0.252 ≈ 2.02 cm, confirming the result.

Why Other Options Are Wrong:

• 1.6 cm: underestimates required section modulus; overstresses the plate.
• 2.5–3.5 cm: safe but not the minimum thickness demanded by the given data.

Common Pitfalls:

• Using plate gross width instead of unit-width strip for section modulus.
• Forgetting that projection length is measured from column face to plate edge.

Final Answer:2.0 cm