RCC Numerical - Calculate ultimate moment of resistance of beam & maximum u.d.l

A  RCC beam of rectangular section 250 mm wide and 600 mm deep (effective) is reinforced on tension side by 4 bars of 20 mm diameter. The beam is subjected to mild exposure conditions. Use M20 and Fe415 grade materials. 

(i) Calculate ultimate moment of resistance of beam. 

(ii) Determine the maximum u.d.l., a simply supported beam can carry over a span of 6 m.

Given Data:

• Width of the beam (b) = 250 mm
• Effective depth of the beam (d) = 600 mm
• Number of tension bars = 4
• Diameter of tension bars (Ø) = 20 mm
• Grade of concrete = M20 (fck = 20 MPa)
• Grade of steel = Fe415 (fy = 415 MPa)
• Span of the beam (L) = 6 m
• Mild exposure conditions

Step 1: Calculate the area of tensile reinforcement (Ast)

The area of one bar is given by:

$${𝐴}_{\text{bar}}=\frac{𝜋}{4}\times {𝜙}^2$$

Substituting $𝜙=20$ mm:

$${𝐴}_{\text{bar}}=\frac{𝜋}{4}\times (20{)}^2=314.16{\text{mm}}^2$$

Thus, the total area of steel (Ast):

$${𝐴}_{\text{st}}=4\times 314.16=1256.64{\text{mm}}^2$$

Step 2: Check if the section is under-reinforced or over-reinforced

For an under-reinforced section, the design moment of resistance is governed by the tension steel reaching its yield strength.

Calculate the neutral axis depth factor ${𝑥}_{𝑢}\mathrm{/}𝑑$. The limiting value for Fe415 steel is:

$$\frac{{𝑥}_{𝑢,\text{lim}}}{𝑑}=0.48\text{(for Fe415)}$$

Hence, the limiting depth of the neutral axis is:

$${𝑥}_{𝑢,\text{lim}}=0.48\times 600=288\text{mm}$$

Now, calculate the actual depth of the neutral axis using the equation:

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

Substituting the values:

$$\frac{1256.64\times 415}{0.36\times 20\times 250}={𝑥}_{𝑢}=241.54\text{mm}$$

Since ${𝑥}_{𝑢}<{𝑥}_{𝑢,\text{lim}}$, the section is under-reinforced.

Step 3: Calculate the ultimate moment of resistance (Mu)

The ultimate moment of resistance for an under-reinforced section is given by:

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

Substitute the values:

$${𝑀}_{𝑢}=0.87\times 415\times 1256.64\times (600-0.42\times 241.54)\times 10^{-6}$$$${𝑀}_{𝑢}=453.38\text{kNm}$$

Step 4: Determine the maximum uniform distributed load (w)

For a simply supported beam with span $𝐿$, the maximum moment due to a UDL is given by:

$${𝑀}_{𝑢}=\frac{𝑤\cdot {𝐿}^2}{8}$$

Rearrange to solve for $𝑤$:

$$𝑤=\frac{8\cdot {𝑀}_{𝑢}}{{𝐿}^2}$$

Substituting the values:

$$𝑤=\frac{8\times 453.38\times 10^6}{(6000{)}^2}=100.75\text{kN/m}$$

Final Answers:

1. Ultimate moment of resistance ${𝑀}_{𝑢}$ = 453.38 kNm
2. Maximum UDL the beam can carry = 100.75 kN/m