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: Brick Wall Dimensions.
\n- Wall Thickness: \(250 \, \text{mm} = 0.25 \, \text{m}\)
\n- Wall Height: \(2 \, \text{m}\)
\n- Beam Span (Length): \(4 \, \text{m}\) \n\nVolume of brick wall: \n\[\nV_\text{wall} = \text{length} \times \text{thickness} \times \text{height} \n= 4 \times 0.25 \times 2 = 2.00 \, \text{m}^3\n\] \n\n \n

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

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

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

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

Step 6: Maximum Bending Moment (Simply Supported Beam with UDL).
\nFormula for a simply supported beam with uniform load: \n\[\nM_\text{max} = \frac{w L^2}{8}\n\] \n\nSubstitute values: \n\[\nM_\text{max} = \frac{12.75 \times (4^2)}{8}\n\] \n\n\[\nM_\text{max} = \frac{12.75 \times 16}{8} = \frac{204}{8} = 25.5 \, \text{kN-m}\n\] \n\nNote: This value appears lower than anticipated. Re-verification of calculations is recommended. \n\n \n

Step 7: Load and Unit Verification.
\n- Wall Load: \(36 \, \text{kN}\) over span → \(9 \, \text{kN/m}\) ✔
\n- Beam Load: \(15 \, \text{kN}\) over span → \(3.75 \, \text{kN/m}\) ✔
\n- Total Load per Unit Length: \(12.75 \, \text{kN/m}\). \n\nTherefore: \n\[\nM_\text{max} = \frac{12.75 \times (4^2)}{8} = \frac{12.75 \times 16}{8} = 25.5 \, \text{kN-m}\n\] \n\nThe confirmed maximum bending moment is: \n\[\n\boxed{25.50 \, \text{kN-m}}\n\]