RCC Numerical - calculate moment of resistance of the section.

A singly reinforced concrete beam 200 mm width and having effective depth 350 mm is reinforced with 3 bars of 12 mm diameter. Using M15 concrete of Fe415 steel, calculate moment of resistance of the section.

To calculate the moment of resistance of a singly reinforced concrete beam, we use the following steps:

Given Data:

• Width of the beam (b) = 200 mm
• Effective depth (d) = 350 mm
• Reinforcement = 3 bars of 12 mm diameter
• Concrete grade = M15 (i.e., ${𝑓}_{𝑐𝑘}=15$ N/mm²)
• Steel grade = Fe415 (i.e., ${𝑓}_{𝑦}=415$ N/mm²)

Step 1: Calculate the Area of Steel (A${}_{𝑠𝑡}$)

For 3 bars of 12 mm diameter:

$${𝐴}_{𝑠𝑡}=3\times \frac{𝜋}{4}\times (12{)}^2=3\times \frac{𝜋}{4}\times 144=3\times 113.097=339.292{\text{mm}}^2$$

Step 2: Calculate the Neutral Axis (x${}_{𝑢}$)

Using the formula for the neutral axis for singly reinforced sections:

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

Substitute the values:

$${𝑥}_{𝑢}=\frac{0.87\times 415\times 339.292}{0.36\times 15\times 200}$$$${𝑥}_{𝑢}=\frac{122,804.016}{1080}=113.705\text{mm}$$

Step 3: Calculate Limiting Depth of Neutral Axis (x${}_{𝑢𝑚𝑎𝑥}$)

For Fe415 steel, the limiting depth of the neutral axis is:

$${𝑥}_{𝑢𝑚𝑎𝑥}=0.48\times 𝑑=0.48\times 350=168\text{mm}$$

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

Step 4: Calculate Moment of Resistance (M${}_{𝑅}$)

The moment of resistance is given by:

$${𝑀}_{𝑅}=0.87{𝑓}_{𝑦}{𝐴}_{𝑠𝑡}(𝑑-\frac{0.42{𝑥}_{𝑢}}{1000})$$

Substitute the values:

$${𝑀}_{𝑅}=0.87\times 415\times 339.292\times (350-0.42\times 113.705)\mathrm{/}1000$$$${𝑀}_{𝑅}=122,804.016\times (350-47.757)\mathrm{/}1000$$$${𝑀}_{𝑅}=122,804.016\times 302.243\mathrm{/}1000$$$${𝑀}_{𝑅}=37,100,384.54\mathrm{/}1000=37.1\text{kNm}$$

Final Answer:

The moment of resistance of the section is 37.1 kNm.