RCC Numerical - Find the ultimate moment of resistance the beam

A singly reinforced beam 230 mm × 480 mm, effective to the centre of 4-20 mm diameter Fe415 bars is cast using M20 concrete. Find the ultimate moment of resistance the beam can offer.

To find the ultimate moment of resistance of the singly reinforced beam, we need to perform the following steps:

Given Data:

• Beam dimensions: Width $𝑏=230\text{ mm}$, Depth $𝐷=480\text{ mm}$
• Effective depth $𝑑=\text{effective depth to the center of reinforcement}$
• Number and diameter of reinforcement bars: 4 bars of 20 mm diameter
• Concrete grade: M20
• Steel grade: Fe415

Steps to Calculate the Ultimate Moment of Resistance:

1. Calculate the Effective Depth (d):

   • Assume the cover is 25 mm.
   • Effective depth $𝑑=𝐷-\text{cover}-\frac{\text{diameter of bars}}{2}$
   • $𝑑=480\text{ mm}-25\text{ mm}-\frac{20\text{ mm}}{2}$
   • $𝑑=480\text{ mm}-25\text{ mm}-10\text{ mm}$
   • $𝑑=445\text{ mm}$

2. Calculate the Area of Steel Reinforcement (Ast):

   • Area of one bar ${𝐴}_{𝑠𝑡}=\frac{𝜋{𝑑}^2}{4}$
   • For 20 mm diameter bars: ${𝐴}_{𝑠𝑡}=\frac{𝜋\times (20{)}^2}{4}$
   • ${𝐴}_{𝑠𝑡}=\frac{3.14\times 400}{4}$
   • ${𝐴}_{𝑠𝑡}=314{\text{ mm}}^2$
   • Total area for 4 bars: ${𝐴}_{𝑠𝑡(𝑡𝑜𝑡𝑎𝑙)}=4\times 314$
   • ${𝐴}_{𝑠𝑡(𝑡𝑜𝑡𝑎𝑙)}=1256{\text{ mm}}^2$

3. Determine the Design Strengths:

   • For M20 concrete, ${𝑓}_{𝑐𝑘}=20\text{ MPa}$ and ${𝑓}_{𝑦}=415\text{ MPa}$ for Fe415 steel.
   • Design strength of concrete ${𝑓}_{𝑐𝑘,𝑑𝑒𝑠𝑖𝑔𝑛}=\frac{{𝑓}_{𝑐𝑘}}{1.5}$
   • ${𝑓}_{𝑐𝑘,𝑑𝑒𝑠𝑖𝑔𝑛}=\frac{20}{1.5}\approx 13.33\text{ MPa}$
   • Design strength of steel ${𝑓}_{𝑦,𝑑𝑒𝑠𝑖𝑔𝑛}=\frac{{𝑓}_{𝑦}}{1.15}$
   • ${𝑓}_{𝑦,𝑑𝑒𝑠𝑖𝑔𝑛}=\frac{415}{1.15}\approx 361\text{ MPa}$

4. Calculate the Moment of Resistance:

   • For singly reinforced sections, the ultimate moment of resistance ${𝑀}_{𝑢}$ is given by:

     $${𝑀}_{𝑢}=0.87\cdot {𝑓}_{𝑦}\cdot {𝐴}_{𝑠𝑡}\cdot (𝑑-\frac{𝑎}{2})$$

     where $𝑎=\frac{{𝐴}_{𝑠𝑡}\cdot {𝑓}_{𝑦}}{0.36\cdot {𝑓}_{𝑐𝑘}\cdot 𝑏}$

   • First, calculate $𝑎$:

     $$𝑎=\frac{1256\cdot 361}{0.36\cdot 13.33\cdot 230}$$$$𝑎=\frac{453816}{11498.8}\approx 39.5\text{ mm}$$

   • Then, calculate ${𝑀}_{𝑢}$:

     $${𝑀}_{𝑢}=0.87\cdot 361\cdot 1256\cdot (445-\frac{39.5}{2})$$$${𝑀}_{𝑢}=0.87\cdot 361\cdot 1256\cdot (445-19.75)$$$${𝑀}_{𝑢}=0.87\cdot 361\cdot 1256\cdot 425.25$$$${𝑀}_{𝑢}\approx 0.87\cdot 361\cdot 1256\cdot 425.25$$$${𝑀}_{𝑢}\approx 62.26\times 425.25\approx 26,536\text{ kN-mm}$$$${𝑀}_{𝑢}\approx 26.54\text{ kNm}$$

Ultimate Moment of Resistance

The ultimate moment of resistance of the beam is approximately 26.54 kNm.