A simply supported beam (L = 4 m) carries a concentrated load (= P) at a distance of 1 m from one end. The beam has a square cross-section of 100 mm side. What will be the maximum value of load (= P) if the maximum permissible bending stress is not to exceed 9 MN/m²?

Concept:

The maximum bending stress in a beam is given by:

\(σ = \frac{M}{Z}\)

Where, σ is the bending stress, M is the maximum bending moment, and Z is the section modulus.

Calculation: 

Given:

Beam length, L = 4 m; Load position, a = 1 m; Cross-section: square of side = 100 mm = 0.1 m

Maximum permissible stress, σ = 9 MN/m² = 9 × 10⁶ N/m²

For a point load P at distance a from the left support, reaction at A is:

\(R_A = P \cdot \frac{(L - a)}{L} = P \cdot \frac{3}{4}\)

Maximum moment occurs at point of load:

\(M_{\text{max}} = R_A \cdot a = \frac{3P}{4} \cdot 1 = \frac{3P}{4}~\text{Nm}\)

Section modulus Z for square cross-section of side b = 0.1 m:

\(Z = \frac{b^3}{6} = \frac{(0.1)^3}{6} = 1.6667 × 10^{-4}~\text{m}^3\)

Now, apply bending stress formula:

\(M = σ \cdot Z = 9 × 10^6 \cdot 1.6667 × 10^{-4} = 1500~\text{Nm}\)

Equating maximum moment:

\(\frac{3P}{4} = 1500 \Rightarrow P = \frac{1500 \cdot 4}{3} = 2000~\text{N} = 2~\text{kN}\)