Analysis & Design of RCC shear walls in the model using ETABS

Aim - To calculate the reinforcement from the given etabs file

Procedure - 

Classification of the shearwall according to IS13920 (2016)

Height/length is 14 which is more than 2

Then it is a slender shear wall

-Minimum vertical reinforcement (Table 1)

Is 1440 which is smaller tahn what we got in the etabs

So we should consider the etabs result

-Min horizontal reinforcement (Table 1 iii)

0.0025+0.5x(21000/1500-2)

=250

We got 750 consider 750

-Confinement reinforcement for boundary elements IS13920 - 2016 (10.4.4)

   It may be 1/3rd of minimum member dimension = 100mm

    Or 6 times than main rebar 6x25 = 150

      Or 150 and not more than 200

So consider spacing of 150mm

-Confinement reinforcement at boundaries as per code

0.05 Sv h (Fck/Fy)

We the value of 452mm2

We have greater value as 750 so consider 750mm2 

Longitudinal rebar details-

We have got the value of 13125

Since we have to use only 25mm 

Then 490x28 = 13720

28 bars of 25mm

Shear reinforcement-

Shear reinforcement of 8mm

Spacing of 250mm

Total of 15 no of stirrups

Boundary width reinforcement-

We got the right side value and left side and Naot required for same stories

Let us consider the maximum values

600mm for left side and 450mm on the right side 

Add 12mm bar on each side in the middle