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

AIM: The aim for this challenge is to run the analysis and to design the RCC shear walls. And to extract the values of the longitudinal steel ratio,boundary element length and the shear reinforcement ratio from the design results.

PROCEDURE:

- So first of all open the ETABS file which is given in the question itself. Open that file into the ETABS software on which we are going to work for designing the shear walls.

- After opening the given file in the ETABS software with the appropriate location just as shown in the below given image:

-Select the shear wall at gris A-11 and show only selected object

- Go to elevation view

- Click on Run Analysis

- After running analysis click on Shear wall Design

- Now that the designing process has been finished in the ETABS software,now we need to collect the all required data which will be helpful in the further detailing calculations.

- So in order to design the shear walls we need three main data:

• Longitudinal Reinforcement.
• Shear Reinforcement.
• Boundary Element Width (BE).

- And all these above data can be extracted from the ETABS software itself.

- An to extract these values from the ETABS software,simply go to the option Shear Wall Design and then there will be a drop-down list and from that list choose option Display Design Info and click on that option just as shown in the below given image:

- After this a new dialouge box opens named as Display Shear Wall Design Results, in that dialouge box change the design Output To Longitudinal Reinforcing just as shown in the below given image:

• Maximum Longitudinal Reinforcement at the base:

- Now in the same dialouge box opens named as Display Shear Wall Design Results, in that dialouge box again change the design Output To Shear Reinforcing just as shown in the below given image:

• Shear Demand throughout the wall height:

- Now in the same dialouge box opens named as Display Shear Wall Design Results, in that dialouge box again change the design Output To Edge/Boundary Zones Widths just as shown in the below given image:

• Boundary Width Data:

DESIGN SUMMARY:

• FLEXURAL INFORMATION:

• SHEAR INFORMATION:

• BOUNDARY WIDTH DATA:

• DETAILED CALCULATIONS FOR THE SHEAR WALLS:

- So the detailed procedure of detailing the reinforcement bars in the shear wall is explained below with brief detailing:

• SECTION CLASSIFICATION:

- So the very first step that we need to check is the geometrical properties of the wall like the thickness of the wall,the length to thickness ratio,the type of shear wall,diameter of the rebar etc. are these properties within the limit or not according to the IS code 13920:2016. So let's check these properties one by one.

• Thickness of the Wall:

- So first property that we are going to check is the Thickness of the wall.

- According to the IS 13920:2016 clause no 10.1.2,this clause state that the minimum allowable thickness of the shear wall should not be less than the 150mm.

- But in our case the thickness of the shear wall given is 300mm which is greater than the 150mm suggested by the IS code 13920:2016. Hence our design is safe.

• Length to Thickness Ratio:

- So the second thing that we need to check is the length to thickness ratio.

- According to the IS 13920:2016,clause 10.1.3, the length to thickness ratio must be or equal to 4.

- So the Length of our shear wall is 1500mm.

- And the thickness of our shear wall is 300mm.

Therefore the Length to Thickness Ratio= LENGTH/WIDTH

Length to Thickness Ratio = 1500/300

Length to Thickness Ratio = 5.

Therefore 5>4 Hence OK .

• Shear wall type:

- According to the IS 13920:2016,clause 10.1.4,suggest that we have three different types of the shear walls based on their geometrical properties and these walls are listed below:

1. Squat Walls: If hw/lw<1
2. Intermediate Walls: If hw/lw<2
3. Slender Walls: If hw/lw>2

- So in our case,we have the height of the wall (hw) = 21000mm.

and the length of the wall (lw) = 1500mm.

∴hw/lw=21000/1500

hw/lw=14

So, hw/lw=14>2

- The fraction here is greater than the 2 hence OK.

• Reinforcement Layer:

- So according to the IS 13920:2016,clause 10.1.7,the reinforcement shall be provided along with the both faces of the wall thickness must be greater than 200mm.

- So in this case we have shear wall thickness of 300mm which is greater than the 200mm,so we have to provide the rebars along with the both faces of the wall.

• Diameter of the Rebar:

- According to the IS 13920:2016,clause 10.1.8,states that the largest diameter of the longitudinal bar shall not exceed 1/10th of its thickness.

- So in our case the thickness of our wall =300mm.

∴1/10th of its thickness = 1/10×300

= 30 mm.

- Which means,we can not use any bar that is exceeding the 30mm diameter value. So we will be using the 25mm diameter bar in our case because we can not use more than the 30mm diameter.

• DETAILED CALCULATION:

- So the steps for the detailed calculations of the reinforcement bars for the shear walls from the ETABS software is discussed below.

- So first of all we have to divide the sections in order to design the walls.

- We know that we can not design the walls for the entire five storey building so we have to divide the walls along its height based on its reinforcement demand.

- I have grouped the walls into two sections i.e Section 1 and Section 2.

• Section 1 = From base upto the First Floor.
• Section 2 = From Second floor upto the Terrace level.

- And according to the ETABS software we have the Boundary Width Requirement i.e, 600 mm.

- So we will be designing the entire length for Section 1 as the boundary element length i.e 600 mm.

- We have the web width of 300mm but this width is very less that's why here we will be using the boundary width.

- Hence for the Section 1 i.e from base upto the first floor we will be using the boundary width of 600mm.

- But as for the upper floors,the boundary element demand reduces upto 300mm.

- Therefore in the Section 2 i.e from second floor upo the terrace level we will be using the boundary element length of 300mm.

(A) CALCULATIONS FROM BASE UPTO THE FIRST FLOOR i.e Section 1:

• Shear Demand:

As already obtained the shear demand of 750 sq.mm per 1000 mm height.

Provide 8mm diameter shear links.

Number of bars =Ast/Ash=750/100= 7.5 = 8 bars

Spacing = Height/No⋅ofBars=1000/8=125mm

Therefore provide 8 mm diameter shear links @125mm c/c as the shear reinforcement of the entire wall.

And the vertical spacing for the shear links can be obtained from clause 10.4.4 of IS 13920:2016, so according to this clause the vertical spacing for the shear links should be 100mm.

• Longitudinal Rebar Demand:

Here entire length is to be designed as the boundary element.

Requirement Longitudinal Reinforcement Ratio = 0.0292.

Therefore , `Ag=1500x300=450000 mm^2`

And Asc = Required Longitudinal Reinforcement Ratio X Ag.

Asc = 0.0292 X 450000

Asc =13140mm2

Therefore,Ash=0.05×Sv×h×fck/fy

Here we have assumed 8mm diameter stirrups@125 c/c.

Assumed data:

Sv = 100mm According to the IS 13920:2016 clause 10.4.4.

fck = 25

fy = 500

and h = 200mm.

∴Ash=0.05×100×200×25/500

Ash=0.05×40×25

Ash=50mm2

Therefore, Minimum no. of bars = 1500−40 /200

Minimum no. of bars = 7.3

Minimum no. of bars = 8 bars.

So we need to cover the demand for 13140 mm2

Hence provide 28 numbers of T25 bars.

Therefore, Ast provided = 490 X 28

Ast provided = 13720 mm2mm2

So 13720>13140mm2

Hence OK.

Also the spacing = 120<200mm required.

Hence OK.

- Hence provide 13 bars on one face and 13 bars on the other face making 26 bars on both the sides.

- Provide 2 bars along the shorter span reducing the clear spacing to 150mm. Hence OK.

- The detailed drawing is given below:

Reinforcement detailing from the Base upto the First floor:

(B) CALCULATIONS FROM SECOND FLOOR UPTO THE TERRACE LEVEL i.e Section 2:

• Design of the Web Steel:

Web Width = 900 mm.

Boundary element width = 300 mm.

Minimum reinforcement ratio for the web as per the IS 13920:2016 = 0.0025.

So here we are using the minimum reinforcement ratio for the web is 0.0030 for the safer calulations.

Therefore Ag = Web width X Boundary element width.

Ag = 900 X 300.

Ag = 270000 mm2mm2

Asc = Ag X Minimum reinforcement ratio.

Asc = 270000 X 0.0030

Asc = 810mm2mm2

Hence provide 6 numbers of T 16 bars = 6 X 201.

= 1206 mm

∴1206>810∴1206>810

Hence OK.

• Design of Boundary Element:

Minimum reinforcement ratio according to the IS code = 0.008

Boundary width = 300mm.

Ash=0.05×Sv×h×fckfyAsh=0.05×Sv×h×fckfy

As we have used 8mm diameter stirrups @125 mm c/c.

Assumed data:

Sv = 100mm According to the IS 13920:2016 clause 10.4.4.

fck = 25

fy = 500

and h = 200mm.

∴Ash=0.05×100×200×25500∴Ash=0.05×100×200×25500

Ash=0.05×40×25Ash=0.05×40×25

Ash=50mm2Ash=50mm2

Reinforcement Required= Total Ast - Ast taken by the Web.

= 4896 - 810.

Reinforcement Required = 4086 mm2mm2

There are two boundary elements, so the Ast required for the one Boundary element = Reinforcement required/2

Ast required for the one Boundary Element = 4086/24086/2

Ast required for the one Boundary Element = 2043mm2mm2

Thus provide 7 numbers of T25 bars = 7 X 490.

Ast Provided = 3430mm2mm2

So 3430>20433430>2043

Hence OK.

• Reinforcement Ratio Check:

Ratio=Ast/bw×twRatio=Astbw×tw

Ratio=3430/300×300Ratio=3430/300×300

Ratio=0.038>0.008Ratio=0.038>0.008

Hence OK.

- The detailed drawing is given below:

Reinforcement detailing from the Second floor upto the Terrace level:

- Here all the dimensions are in mm.

Elevation Drawing: