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

Aim: The etabs file for a G+6 building is provided. Run the analysis and design the RCC shear walls.

Procedure:

Step 1:

1. Load the model into Etabs.
2. Before Analysis of Model check parameters for the shear wall.
3. Go to View -> Set display option -> Other assignment -> Lebles.
4. Lables are visible it means all the parameters are defined.

Step 2:

1. Go to Design -> Shear Wall Design -> Start Design/Check.

Step 3:

1. After complition of Design procedure we need to varify the model for failure.
2. Go to Design -> Shear Wall Design -> Display Design Info.
3. Select the Identify P-M failure.
4. Right click on wall to identify.

Step 4: Longitudinal steel ratio

1. Go to Design -> Shear wall Design -> Display Design Info.
2. Select Peir logitudinal reinforcement.

Step 5: Boundary element length

1. Go to Design -> Shear wall Design -> Display Design Info.
2. Select Peir logitudinal reinforcement.

Step 6: Shear reinforcing ratio

1. Go to Design -> Shear wall Design -> Display Design Info.
2. Select Peir/Edge Boundary zones widths.

Calculation:

Base Floor:

`therefore t_e=300mm`

`therefore l_v=1500mm`

Boudary of Element:

`therefore B.E. = 600 mm`

'therefore' Height of Entire wall = 10500mm

{As per IS 13920-2016 (Clause No. 10.1.6)}

Minimum reinforcement in RC Shear wall.

When the total height of the wall divided by the length is more than 2.

If it a Slender wall we have to calculate:

1. Horizontal min reinforcement =`P_hmin`
2. Vertical B.E. reinforcement = `P_v B.E. min`
3. Minimum Web Reinforcement =`P_v Web min`
4. Net Reinforcement = `P_v Net`

`P_h min= 0.025 + 0.5 (h_w/l_w -2)`

`l_v BE min=0.0080`

`l_v web = 0.025`

`l_v Net = 0.0025+0.01375 (t_w/l_w)`

= `0.0025+0.01375(300/1500)`

= 0.2165

For Longitudinal Reinforcement:

`l_w = L_w-2t_w`

= 1500 - 2 (600)

= 300 mm

As per Etabs calculation,

the required rebar area is `A_v`

`therefore` `A_v = 13128 mm^2`

`therefore` Minimum vertical reinforcement `l_v`

`therefore` `l_v`= `13128/ ((C//S) area of steel)`

`l_v`= `13128/(1500xx300)`

`l_v`= 0.029 > 0.008

As per IS 13920-2016

`therefore A_sh`= `0.05xxs_vxxhxx(f_(ck)/f_y)`

`therefore`Minimum dia of Bars = 8

`therefore A_sh`= `(pi)/4xxd^2`

= `(pi/4)xx8^2`

= 50.24 `mm^2`

`therefore 50.24 = 0.05xx100xx h (25/500)`

50.24 = 0.25h

`therefore` h = 200.96mm

Use the Bar of 25mm `phi`

No of Bars required = `13128/((pi)/4xx25^2`

= `13128/490.625`

= 28 Bars.

For Shear Reinforcement:

For 2 legged 16mm `phi` bars

`therefore A_sh`= `4xx (pi/4)xxd^2`

= `4xx (pi/4)xx16^2`

= 803.84 `mm^2`

Provide 6 No. of 12mm `phi` bars in web.

From grounf floor to first floor:

For longitudinal reinforcement `A_v`= 8273 `mm^2`

`therefore A_v = 8273/(1500xx300)`

= `8273/450000`

= 0.018

As per IS 13920-2016,

`therefore A_sh`= `0.05xxs_vxxhxx(f_(ck)/f_y)`

`therefore` h = 200.96 mm.

`therefore` No of 25mm `phi`bars = `8273/((pi/4)xx25^2)`

= `8273/490.87`

= 16.8 `~=` 17 Nos

`therefore` No of 25 mm `phi`bars is 17 Nos.

For Shear reinforcement:

`therefore A_sh=750 mm^2`

By providing 2 legged 8mm `phi`@ 100 mm c/c.

`therefore A_sh = (2xx(pi/4)xxd^2)/0.100`

`therefore A_sh = (2xx(pi/4)xx8^2)/0.100`

= `1005 mm^2`

`therefore` `1005 mm^2 > 750 mm^2`

Similarly for web,

By providing 2 legged 8mm dia @ 125mm c/c

`therefore A_sh = (2xx(pi/4)xx8^2)/0.125`

= `805 mm^2`

Conclusion: