Question

A simply supported RCC beam of span 4 m is supporting a brick wall over its entire span. The brick wall is 250 mm thick and 2 m high. The RCC beam has a depth of 600 mm and width of 250 mm. The density of brick masonry and RCC can be assumed as \(18 \, \text{kN/m}^3\) and \(25 \, \text{kN/m}^3\) respectively. Considering the load of the wall and self-weight of the RCC beam, the maximum bending moment in the beam (in kN-m) will be ......... [rounded off to two decimal places]. \begin{center} \includegraphics[width=0.5\textwidth]{08.jpeg} \end{center}

Hint:

For maximum bending moment of a simply supported beam with UDL, always use \(M = \frac{wL^2}{8}\). Carefully convert thickness and depth into meters, and use densities in \(\text{kN/m}^3\) for uniform load calculations.

Answer 1

Step 1: Dimensions of the brick wall.
- Thickness of wall = \(250 \, \text{mm} = 0.25 \, \text{m}\)
- Height of wall = \(2 \, \text{m}\)
- Span (length) of beam = \(4 \, \text{m}\) Thus, volume of brick wall: \[ V_\text{wall} = \text{length} \times \text{thickness} \times \text{height} = 4 \times 0.25 \times 2 = 2.00 \, \text{m}^3 \]

Step 2: Weight of brick wall.
Density of brick masonry = \(18 \, \text{kN/m}^3\). \[ W_\text{wall} = V_\text{wall} \times \gamma = 2.00 \times 18 = 36.00 \, \text{kN} \] Since the beam span is \(4 \, \text{m}\), the wall load is distributed uniformly. \[ w_\text{wall} = \frac{W_\text{wall}}{\text{span}} = \frac{36}{4} = 9.00 \, \text{kN/m} \]

Step 3: Dimensions of RCC beam.
- Width = \(250 \, \text{mm} = 0.25 \, \text{m}\)
- Depth = \(600 \, \text{mm} = 0.60 \, \text{m}\)
- Length = \(4 \, \text{m}\) Volume of beam: \[ V_\text{beam} = 4 \times 0.25 \times 0.60 = 0.60 \, \text{m}^3 \]

Step 4: Self-weight of RCC beam.
Density of RCC = \(25 \, \text{kN/m}^3\). \[ W_\text{beam} = V_\text{beam} \times \gamma = 0.60 \times 25 = 15.00 \, \text{kN} \] Uniformly distributed load from beam: \[ w_\text{beam} = \frac{W_\text{beam}}{\text{span}} = \frac{15}{4} = 3.75 \, \text{kN/m} \]

Step 5: Total UDL on beam.
\[ w_\text{total} = w_\text{wall} + w_\text{beam} = 9.00 + 3.75 = 12.75 \, \text{kN/m} \]

Step 6: Maximum bending moment for a simply supported beam with UDL.
For a simply supported beam with uniform load: \[ M_\text{max} = \frac{w L^2}{8} \] Substitute values: \[ M_\text{max} = \frac{12.75 \times (4^2)}{8} \] \[ M_\text{max} = \frac{12.75 \times 16}{8} = \frac{204}{8} = 25.5 \, \text{kN-m} \] Wait – this seems less than expected. Let us carefully re-check calculation.

Step 7: Double-check loads and units.
- Wall load = \(36 \, \text{kN}\) over span → \(9 \, \text{kN/m}\) ✔
- Beam load = \(15 \, \text{kN}\) over span → \(3.75 \, \text{kN/m}\) ✔
- Total load per unit length = \(12.75 \, \text{kN/m}\). So: \[ M_\text{max} = \frac{12.75 \times (4^2)}{8} = \frac{12.75 \times 16}{8} = 25.5 \, \text{kN-m} \] Thus the correct maximum bending moment is: \[ \boxed{25.50 \, \text{kN-m}} \]