RCC Numerical - Find limiting moment of resistance and steel required for a beam

Find limiting moment of resistance and steel required for a beam 300 × 600 mm (effective), if concrete M 25 and Fe 415 steel are used.

To find the limiting moment of resistance and the amount of steel required for a beam with the given dimensions and materials, follow these steps:

Given:

• Beam dimensions: Width $𝑏=300\text{ mm}$, Effective depth $𝑑=600\text{ mm}$
• Concrete grade: M25
• Steel grade: Fe415

1. Limiting Moment of Resistance:

The limiting moment of resistance occurs when the concrete reaches its maximum strain (0.0035) and the steel reaches its yield strain. This is usually calculated using the concept of limiting balanced condition where the beam is on the verge of failure.

For M25 concrete and Fe415 steel, we use the following standard formulas.

Step 1: Calculate the Effective Depth of the Beam

Effective depth $𝑑$ = 600 mm (already given).

Step 2: Assume a Depth for the Neutral Axis

Assume that the neutral axis (na) is at $𝑥$ mm from the top of the beam. For a balanced section, this neutral axis depth can be approximated using:

$𝑥=0.48\cdot 𝑑$

where $𝑑=600\text{ mm}$, so:

$𝑥\approx 0.48\cdot 600=288\text{ mm}$

Step 3: Calculate the Moment of Resistance

The design parameters for concrete and steel are:

• ${𝑓}_{𝑐𝑘}$ = 25 MPa (Concrete compressive strength)
• ${𝑓}_{𝑦}$ = 415 MPa (Yield strength of steel)
• $𝜙$ = 0.87 (Strength reduction factor for steel)
• $𝜆$ = 0.87 (Strength reduction factor for concrete)

The limiting moment of resistance ${𝑀}_{𝑢}$ is given by:

${𝑀}_{𝑢}=0.36{𝑓}_{𝑐𝑘}𝑏𝑥(𝑑-\frac{𝑥}{2})$

where ${𝑓}_{𝑐𝑘}$ is in MPa, $𝑏$ is in mm, and $𝑥$ is the depth of the neutral axis in mm. Here,

${𝑀}_{𝑢}=0.36\times 25\times 300\times 288(600-\frac{288}{2})$

${𝑀}_{𝑢}=0.36\times 25\times 300\times 288\times (600-144)$

${𝑀}_{𝑢}=0.36\times 25\times 300\times 288\times 456$

${𝑀}_{𝑢}\approx 1,584,092,800\text{ Nmm}$

${𝑀}_{𝑢}\approx 1,584.1\text{ kNm}$

2. Steel Required:

The area of steel required ${𝐴}_{𝑠}$ can be found by using the formula for the design moment and the assumed depth of neutral axis.

Step 1: Determine the Moment Capacity

Use the formula for the moment of resistance considering the effective depth:

${𝑀}_{𝑢}=0.87{𝑓}_{𝑦}{𝐴}_{𝑠}(𝑑-\frac{𝑥}{2})$

Rearrange to solve for ${𝐴}_{𝑠}$:

${𝐴}_{𝑠}=\frac{{𝑀}_{𝑢}}{0.87{𝑓}_{𝑦}(𝑑-\frac{𝑥}{2})}$

Substitute the values:

${𝐴}_{𝑠}=\frac{1,584,092,800}{0.87\times 415\times (600-\frac{288}{2})}$

${𝐴}_{𝑠}=\frac{1,584,092,800}{0.87\times 415\times 456}$

${𝐴}_{𝑠}\approx 10,237{\text{ mm}}^2$

Summary

• Limiting Moment of Resistance: Approximately $1,584.1\text{ kNm}$
• Steel Required: Approximately $10,237{\text{ mm}}^2$

These calculations are approximations and should be verified with detailed structural analysis and design codes as needed.