RCC Numerical - Ultimate moment of resistance

A singly reinforced rectangular beam is 450 mm (wide) and 560 mm (effective depth) with tensile reinforcement of 4 mild steel bars of 20 mm diameter. The effective cover to tensile reinforcement is 40 mm. Establish whether the section is under-reinforced or over-reinforced and calculate ultimate moment of resistance of beam. Use M20 concrete.

Given Data:

• Beam width (b): 450 mm
• Effective depth (d): 560 mm
• Diameter of tensile reinforcement bars: 20 mm
• Number of tensile bars (n): 4 bars
• Effective cover (d’): 40 mm
• Grade of concrete: M20
• Yield strength of mild steel (fy): 250 N/mm² (for mild steel)
• Characteristic strength of concrete (fck): 20 N/mm²

Step 1: Calculate Area of Tensile Reinforcement (Ast)

$\text{Area of steel (Ast)} = n \times \frac{\pi}{4} \times d_{bar}^2$ $\text{Ast} = 4 \times \frac{\pi}{4} \times (20)^2$ $\text{Ast} = 4 \times 314.16 = 1256.64 \, \text{mm}^2$

Step 2: Calculate Limiting Area of Steel (Ast,lim)

The limiting area of steel (Ast,lim) is calculated using the following formula: $\text{Ast,lim} = 0.36 \times fck \times b \times x_{u,\lim} / f_y$ Where:

• $fck$ = 20 N/mm²
• $fy$ = 250 N/mm² (for mild steel)
• $b$ = 450 mm
• $x_{u,\lim}$ is the limiting depth of the neutral axis for under-reinforced sections.

For Fe-250 (mild steel), the limiting neutral axis depth factor for M20 concrete is 0.53d: $x_{u,\lim} = 0.53 \times 560 = 296.8 \, \text{mm}$

Now, calculate the limiting steel area: $\text{Ast,lim} = \frac{0.36 \times 20 \times 450 \times 296.8}{250}$ $\text{Ast,lim} = 3830.8 \, \text{mm}^2$

Step 3: Check Whether the Section is Under-Reinforced or Over-Reinforced

If $Ast < Ast,lim$, the section is under-reinforced. Otherwise, it is over-reinforced.

In this case:

• $Ast = 1256.64 \, \text{mm}^2$
• $Ast,lim = 3830.8 \, \text{mm}^2$

Since $Ast < Ast,lim$, the section is under-reinforced.

Step 4: Calculate Ultimate Moment of Resistance (Mu)

For under-reinforced sections, the moment of resistance (Mu) can be calculated as: $\text{Mu} = 0.87 \times f_y \times \text{Ast} \times \left( d - 0.42 \times x_u \right)$ Where $x_u$ is the actual depth of the neutral axis: $x_u = \frac{0.87 \times f_y \times \text{Ast}}{0.36 \times fck \times b}$ $x_u = \frac{0.87 \times 250 \times 1256.64}{0.36 \times 20 \times 450}$ $x_u = 1058.044 / 3240$ $x_u = 97.46 \, \text{mm}$

Now, calculate the ultimate moment: $\text{Mu} = 0.87 \times 250 \times 1256.64 \times (560 - 0.42 \times 97.46)$ $\text{Mu} = 0.87 \times 250 \times 1256.64 \times (560 - 40.93)$ $\text{Mu} = 273318 \times 519.07$ $\text{Mu} = 141.89 \times 10^6 \, \text{Nmm}$ $\text{Mu} = 141.89 \, \text{kNm}$

Final Answer:

• The beam section is under-reinforced.
• The ultimate moment of resistance of the beam is 141.89 kNm.

Let me know if you need further clarification or additional calculations!