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 𝐴bar=𝜋4×𝜙2.

• For 4 bars:

  𝐴st, 4 bars=4×𝜋4×(16)2=4×201.06 mm2=804.24 mm2

• For 5 bars:

  𝐴st, 5 bars=5×201.06=1005.3 mm2

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:

𝐴st,bal=0.36𝑓ck𝑏𝑑0.87𝑓y

Substituting the given values:

𝐴st,bal=0.36×15×230×4000.87×415=1081.35 mm2

• Since 𝐴st, 4 bars=804.24 mm2 is less than 𝐴st,bal, the section is under-reinforced.
• For 5 bars, 𝐴st, 5 bars=1005.3 mm2 is also less than 𝐴st,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:

𝑀u=0.87𝑓y𝐴st(𝑑−0.42𝑥u𝑑)

where:

• 𝑥u=0.87𝑓y𝐴st0.36𝑓ck𝑏

• For 4 bars:

  𝑥u=0.87×415×804.240.36×15×230=95.76 mm

• For 5 bars:

  𝑥u=0.87×415×1005.30.36×15×230=119.65 mm

Since 𝑥u is less than the limiting value for Fe415 steel (which is 𝑥u,lim=0.48𝑑=0.48×400=192 mm), the section is under-reinforced.

• Mu for 4 bars:

  𝑀u, 4 bars=0.87×415×804.24(400−0.42×95.76)=105.6 kNm

• Mu for 5 bars:

  𝑀u, 5 bars=0.87×415×1005.3(400−0.42×119.65)=130.44 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.