Question

Two simply supported beams A & B have the same length ‘L’ & subjected to equal bending moment M. The stress induced in the beam A & B are σ_A & σ_B respectively. If the cross-section of beam A is (b × b/2) & that of beam B is (b/2 × b), then the correct relation between σ_A & σ_B is:
1. σ_A = 2σ_B
2. σ_A = 4σ_B
3. σA = 0.5σB
4. σ_A = σ_B

Answer 1

Correct Answer - Option 1 : σ_A = 2σ_B

Concept:

Bending stress is given by:

\(σ = \frac{{My}}{I}\)

Where M is bending moment y is the distance of fibre from the neutral axis and I is the moment of inertia of the cross-section.

Calculation:

Given:

Beam A = (b × b/2), Beam B = (b/2 × b)

\(σ = \frac{{My}}{I}=\frac{M}{Z}\;\)

Where, \(Z=\frac{I}{y}\) is known as sectional-modulus.

Thus for the same bending moment, the bending stress is inversely proportional to section modulus.

Section Modulus for Beam A:

\(Z_A=\frac{I}{y}\Rightarrow\frac{\frac{b\;\times\;\left(\frac{b}{2}\right)^3}{12}}{\frac{b}{2\;\times\;2}}=\frac{b^3}{24}\)

Section Modulus for Beam B:

\(Z_B=\frac{I}{y}\Rightarrow\frac{\frac{\left(\frac{b}{2}\right)\;\times\;b^3}{12}}{\frac{b}{2}}=\frac{b^3}{12}\)

\(\frac{σ_A}{σ_B}=\frac{Z_B}{Z_A}=\frac{\frac{b^3}{12}}{\frac{b^3}{24}}=2\)

∴ σ_A = 2σ_B