RCC Numerical - If xu is limited to 0.39, calculate pt and Mu. Assume M20 – Fe415 concrete and steel.

 If xu is limited to 0.39, calculate pt and Mu. Assume M20 – Fe415 concrete and steel.

Given:

• Concrete grade: M20 (characteristic compressive strength $f_{ck}$ = 20 MPa)
• Steel grade: Fe415 (yield strength $f_y$ = 415 MPa)
• Maximum depth of the neutral axis $x_u = 0.39d$

1. Calculate $p_t$ (Area of Steel per Unit Area of Concrete):

For a beam section with a given $x_u$, the percentage of steel ($p_t$) can be determined using the following steps:

1. Determine the Lever Arm (z):

   The lever arm $z$ is approximately given by:

   $$z \approx d - \frac{x_u}{2}$$

2. Calculate the Moment of Resistance ($M_u$):

   The formula for the moment of resistance $M_u$ for a singly reinforced beam is:

   $$M_u = 0.87 \cdot f_y \cdot A_s \cdot z$$

   where $A_s$ is the area of steel and $f_y$ is the yield strength of the steel.

3. Relate $x_u$ to the Steel Ratio:

   The neutral axis depth $x_u$ is given by:

   $$x_u = \frac{A_s \cdot f_y}{0.36 \cdot f_{ck} \cdot b}$$

   where $b$ is the breadth of the beam.

Let's Compute $p_t$ and $M_u$:

1. Calculate the Lever Arm (z):

   $$z = d - \frac{x_u}{2} = d - \frac{0.39d}{2} = d - 0.195d = 0.805d$$

2. Moment of Resistance Formula:

   $$M_u = 0.87 \cdot f_y \cdot A_s \cdot z$$

   Substitute $z = 0.805d$:

   $$M_u = 0.87 \cdot f_y \cdot A_s \cdot 0.805d$$

3. Determine $p_t$:

   To find $A_s$, you need to use the given maximum depth of the neutral axis:

   $$x_u = 0.39d$$

   Rearranging for $A_s$:

   $$A_s = \frac{x_u \cdot 0.36 \cdot f_{ck} \cdot b}{f_y}$$

   Substitute $x_u = 0.39d$, $f_{ck} = 20 \text{ MPa}$, $f_y = 415 \text{ MPa}$, and $b$ (breadth of the beam):

   $$A_s = \frac{0.39d \cdot 0.36 \cdot 20 \cdot b}{415}$$

   Simplify to find $p_t$ (percentage of steel):

   $$p_t = \frac{A_s}{b \cdot d} \cdot 100 = \frac{0.39 \cdot 0.36 \cdot 20}{415} \cdot 100$$ $$p_t \approx \frac{2.808}{415} \cdot 100 \approx 0.676\%$$

4. Ultimate Moment of Resistance:

   Substitute $A_s$ into the $M_u$ formula:

   $$M_u = 0.87 \cdot 415 \cdot \left(\frac{0.39 \cdot 0.36 \cdot 20 \cdot b}{415}\right) \cdot 0.805d$$

   Simplify:

   $$M_u = 0.87 \cdot 0.39 \cdot 0.36 \cdot 20 \cdot b \cdot 0.805d$$ $$M_u \approx 0.87 \cdot 0.39 \cdot 0.36 \cdot 20 \cdot 0.805 \cdot b \cdot d$$ $$M_u \approx 2.53 \cdot b \cdot d \text{ kNm}$$

Thus, the percentage of steel $p_t$ is approximately 0.676%, and the ultimate moment of resistance $M_u$ can be computed using $M_u = 2.53 \cdot b \cdot d \text{ kNm}$, where $b$ is the breadth of the beam in mm and $d$ is the effective depth in mm.