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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 KN/m³ and 25 KN/m³ 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].

Calculation of Maximum Bending Moment

To find the maximum bending moment in the simply supported beam, we first determine the total uniformly distributed load (UDL) per meter length.

1. Identify Given Parameters and Convert to Meters

  • Span of the beam ($L$): $4\text{ m}$
  • Beam width ($B$): $250\text{ mm} = 0.25\text{ m}$
  • Beam depth ($D$): $600\text{ mm} = 0.6\text{ m}$
  • Wall height ($H_w$): $2\text{ m}$
  • Wall thickness ($t_w$): $250\text{ mm} = 0.25\text{ m}$
  • Density of RCC ($\rho_{\text{rcc}}$): $25\text{ kN/m}^3$
  • Density of brick masonry ($\rho_{\text{brick}}$): $18\text{ kN/m}^3$

2. Calculate the Load per Unit Length ($w$)

The total UDL ($w$) is the sum of the self-weight of the RCC beam and the weight of the brick wall.

A. Self-weight of the RCC beam ($w_{\text{beam}}$):

$$w_{\text{beam}} = (\text{Width} \times \text{Depth}) \times \text{Density} = (0.25\text{ m} \times 0.6\text{ m}) \times 25\text{ kN/m}^3$$ $$w_{\text{beam}} = 0.15\text{ m}^2 \times 25\text{ kN/m}^3 = 3.75\text{ kN/m}$$

B. Weight of the brick wall ($w_{\text{wall}}$):

$$w_{\text{wall}} = (\text{Thickness} \times \text{Height}) \times \text{Density} = (0.25\text{ m} \times 2\text{ m}) \times 18\text{ kN/m}^3$$ $$w_{\text{wall}} = 0.5\text{ m}^2 \times 18\text{ kN/m}^3 = 9.00\text{ kN/m}$$

C. Total UDL ($w$):

$$w = w_{\text{beam}} + w_{\text{wall}} = 3.75 + 9.00 = 12.75\text{ kN/m}$$

3. Calculate Maximum Bending Moment ($M_{\text{max}}$)

For a simply supported beam with a UDL over its entire span, the maximum bending moment occurs at the mid-span:

$$M_{\text{max}} = \frac{wL^2}{8}$$ $$M_{\text{max}} = \frac{12.75\text{ kN/m} \times (4\text{ m})^2}{8}$$ $$M_{\text{max}} = \frac{12.75 \times 16}{8} = 12.75 \times 2$$ $$M_{\text{max}} = 25.50\text{ kN-m}$$

Conclusion

The maximum bending moment in the beam is 25.50 kN-m.

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Important Questions from Design of Structural Elements

  1. A simply supported beam having effective span of 5 meter is carrying a centrally concentrated load of 16 kN. The maximum bending moment in the beam in kN-m is _______________

  2. Area of tensile steel per meter width of a reinforced concrete slab is 335 sq mm. If 8 mm rods are used as reinforcement, then centre to centre spacing of the reinforcement in mm is _______________

  3. The maximum bending moment ($kNm$) in a simply supported beam of 8 $m$ span subjected to a uniformly distributed load of 20 $kN/m$ (inclusive of its self-weight) over the entire span is
  4. Which of the following is NOT a mechanism of bond resistance in reinforced concrete?
  5. A single span simply supported reinforced concrete beam ($250 \text{ mm}$ wide and $480 \text{ mm}$ effective depth) is subjected to a concentrated load of $120 \text{ kN}$ at its mid-span. Neglecting self-weight of the beam, the nominal shear stress (MPa) at the support section is __________

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