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 fckf_{ck} = 20 MPa)
• Steel grade: Fe415 (yield strength fyf_y = 415 MPa)
• Maximum depth of the neutral axis xu=0.39dx_u = 0.39d

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

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

1. Determine the Lever Arm (z):

   The lever arm zz is approximately given by:

   z≈d−xu2z approx d – frac{x_u}{2}

2. Calculate the Moment of Resistance (MuM_u):

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

   Mu=0.87⋅fy⋅As⋅zM_u = 0.87 cdot f_y cdot A_s cdot z

   where AsA_s is the area of steel and fyf_y is the yield strength of the steel.

3. Relate xux_u to the Steel Ratio:

   The neutral axis depth xux_u is given by:

   xu=As⋅fy0.36⋅fck⋅bx_u = frac{A_s cdot f_y}{0.36 cdot f_{ck} cdot b}

   where bb is the breadth of the beam.

Let’s Compute ptp_t and MuM_u:

1. Calculate the Lever Arm (z):

   z=d−xu2=d−0.39d2=d−0.195d=0.805dz = d – frac{x_u}{2} = d – frac{0.39d}{2} = d – 0.195d = 0.805d

2. Moment of Resistance Formula:

   Mu=0.87⋅fy⋅As⋅zM_u = 0.87 cdot f_y cdot A_s cdot z

   Substitute z=0.805dz = 0.805d:

   Mu=0.87⋅fy⋅As⋅0.805dM_u = 0.87 cdot f_y cdot A_s cdot 0.805d

3. Determine ptp_t:

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

   xu=0.39dx_u = 0.39d

   Rearranging for AsA_s:

   As=xu⋅0.36⋅fck⋅bfyA_s = frac{x_u cdot 0.36 cdot f_{ck} cdot b}{f_y}

   Substitute xu=0.39dx_u = 0.39d, fck=20 MPaf_{ck} = 20 text{ MPa}, fy=415 MPaf_y = 415 text{ MPa}, and bb (breadth of the beam):

   As=0.39d⋅0.36⋅20⋅b415A_s = frac{0.39d cdot 0.36 cdot 20 cdot b}{415}

   Simplify to find ptp_t (percentage of steel):

   pt=Asb⋅d⋅100=0.39⋅0.36⋅20415⋅100p_t = frac{A_s}{b cdot d} cdot 100 = frac{0.39 cdot 0.36 cdot 20}{415} cdot 100
   pt≈2.808415⋅100≈0.676%p_t approx frac{2.808}{415} cdot 100 approx 0.676%

4. Ultimate Moment of Resistance:

   Substitute AsA_s into the MuM_u formula:

   Mu=0.87⋅415⋅(0.39⋅0.36⋅20⋅b415)⋅0.805dM_u = 0.87 cdot 415 cdot left(frac{0.39 cdot 0.36 cdot 20 cdot b}{415}right) cdot 0.805d

   Simplify:

   Mu=0.87⋅0.39⋅0.36⋅20⋅b⋅0.805dM_u = 0.87 cdot 0.39 cdot 0.36 cdot 20 cdot b cdot 0.805d
   Mu≈0.87⋅0.39⋅0.36⋅20⋅0.805⋅b⋅dM_u approx 0.87 cdot 0.39 cdot 0.36 cdot 20 cdot 0.805 cdot b cdot d
   Mu≈2.53⋅b⋅d kNmM_u approx 2.53 cdot b cdot d text{ kNm}

Thus, the percentage of steel ptp_t is approximately 0.676%, and the ultimate moment of resistance MuM_u can be computed using Mu=2.53⋅b⋅d kNmM_u = 2.53 cdot b cdot d text{ kNm}, where bb is the breadth of the beam in mm and dd is the effective depth in mm.