RCC Numerical - Find the ultimate moment of resistance for a singly reinforced concrete beam

Find the ultimate moment of resistance for a singly reinforced concrete beam having 230 mm wide and 400 mm effective depth. It is reinforced with 4 bars of 16 mm diameter. Use M15 and Fe415 grade materials. What will be the moment of resistance if 5 bars of 16 mm diameter are used? Will the moment of resistance increase further with increase in number of bars.

To find the ultimate moment of resistance (Mu) for a singly reinforced concrete beam, we use the following steps:

1. Given Data:

• Width of beam (b) = 230 mm
• Effective depth (d) = 400 mm
• Number of bars (n):
  • Case 1: 4 bars of 16 mm diameter
  • Case 2: 5 bars of 16 mm diameter
• Diameter of bars (ϕ) = 16 mm
• Grade of concrete (fck) = M15 (15 N/mm²)
• Grade of steel (fy) = Fe415 (415 N/mm²)

2. Determine Area of Steel (Ast):

The formula for the area of a circular bar is ${𝐴}_{\text{bar}}=\frac{𝜋}{4}\times {𝜙}^2$.

• For 4 bars:

  $${𝐴}_{\text{st, 4 bars}}=4\times \frac{𝜋}{4}\times (16{)}^2=4\times 201.06{\text{mm}}^2=804.24{\text{mm}}^2$$

• For 5 bars:

  $${𝐴}_{\text{st, 5 bars}}=5\times 201.06=1005.3{\text{mm}}^2$$

3. Check the type of section (Under or Over-reinforced):

To check whether the beam is under-reinforced or over-reinforced, we compare the steel percentage with the balanced steel percentage.

The balanced steel area (Ast,bal) is given by:

$${𝐴}_{\text{st,bal}}=\frac{0.36{𝑓}_{\text{ck}}𝑏𝑑}{0.87{𝑓}_{\text{y}}}$$

Substituting the given values:

$${𝐴}_{\text{st,bal}}=\frac{0.36\times 15\times 230\times 400}{0.87\times 415}=1081.35{\text{mm}}^2$$

• Since ${𝐴}_{\text{st, 4 bars}}=804.24{\text{mm}}^2$ is less than ${𝐴}_{\text{st,bal}}$, the section is under-reinforced.
• For 5 bars, ${𝐴}_{\text{st, 5 bars}}=1005.3{\text{mm}}^2$ is also less than ${𝐴}_{\text{st,bal}}$, so it is still under-reinforced.

4. Calculate the Ultimate Moment of Resistance (Mu):

For under-reinforced sections, the ultimate moment of resistance (Mu) is calculated using the formula:

$${𝑀}_{\text{u}}=0.87{𝑓}_{\text{y}}{𝐴}_{\text{st}}(𝑑-\frac{0.42{𝑥}_{\text{u}}}{𝑑})$$

where:

• ${𝑥}_{\text{u}}=\frac{0.87{𝑓}_{\text{y}}{𝐴}_{\text{st}}}{0.36{𝑓}_{\text{ck}}𝑏}$

• For 4 bars:

  $${𝑥}_{\text{u}}=\frac{0.87\times 415\times 804.24}{0.36\times 15\times 230}=95.76\text{mm}$$

• For 5 bars:

  $${𝑥}_{\text{u}}=\frac{0.87\times 415\times 1005.3}{0.36\times 15\times 230}=119.65\text{mm}$$

Since ${𝑥}_{\text{u}}$ is less than the limiting value for Fe415 steel (which is ${𝑥}_{\text{u,lim}}=0.48𝑑=0.48\times 400=192\text{mm}$), the section is under-reinforced.

• Mu for 4 bars:

  $${𝑀}_{\text{u, 4 bars}}=0.87\times 415\times 804.24(400-0.42\times 95.76)=105.6\text{kNm}$$

• Mu for 5 bars:

  $${𝑀}_{\text{u, 5 bars}}=0.87\times 415\times 1005.3(400-0.42\times 119.65)=130.44\text{kNm}$$

5. Effect of Increasing Number of Bars:

As we increase the number of bars (area of steel), the moment of resistance increases until the section becomes balanced. Once the section becomes balanced, adding more steel won't significantly increase the moment of resistance, and the section will eventually become over-reinforced, which is undesirable as it leads to brittle failure.

In this case, since both 4 bars and 5 bars make the section under-reinforced, increasing the number of bars from 4 to 5 increases the moment of resistance from 105.6 kNm to 130.44 kNm. However, beyond a certain point, the section may become over-reinforced, and further increases in reinforcement will not lead to a proportional increase in moment capacity.