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

The etabs file for a G+6 building is provided. Run the analysis and design the RCC shear walls (show in the etabs model). The following challenge deals specifically with the shear wall of size 300x1500 at grid A-11 in the structural model.

• Extract values of longitudinal steel ratio, boundary element length and shear reinforcing ratio from the design results.
• Provide cross sectional steel arrangement of the shear wall from its base up to terrace, in compliance with provisions of IS 13920 – 2016. AutoCAD application can be used for this exercise. Alternatively, it can be drawn by hand as well.
• Provide elevation drawings of the reinforcement details of the shear wall from base up to terrace.

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.

Introduction: Design a shear wall and extract reinforcement detailing and draft the r/f detail drawing by doing calculation.

PROCEDURE:

•  open the etabs model provided with challenge
• 
•  After opening the file the very first step that we are going to do is Run Analysis.
• But before doing the Run analysis first we have to check the model for any structural defects.
• In order to check the model for any structural defect we need to go to the Analysis tab there will be a drop-down list and in that list there will be an option called Check model .
• 
•  After this a new dialouge box opens named as Check Models in that dialouge box select all the options which is present in that dialouge box and then click OK and run .
• So the check has been done and there are no warnings shown in the below image. And it also means that we can proceed further for the analysis of this G+6 building.
• 
• 
•  In order to adjust the view settings first we need to select the shear wall that we are expected to design and analyse and after selecting the shear walls,right-click anywhere on the screen and then there will be a drop-down list in that list choose the option called show selected object.
• 
• Right-click anywhere on the screen.
• 
• 
• check now after adjusting the view settings,now we will have to make sure that the codes that we are using in desinging is correct or not.
• In order to do that,just go to the shear wall design icon and click on that icon after clicking on that icon a drop-down list appears on the screen from that list click on the option View/Revise Reference option.
• 
• After clicking on that option a new dialouge box appears on the screen named as Shear Wall Design Preferences for IS 456:2000 
• 
• In that dialouge box make sure that the desired codes are selected based on the country/region where the project is located.After making sure that the codes that are selected are corrected,now we can perform the Run analysis.
• In order to do so just click on the Shear Wall Design icon .
• 
• Now click on that icon,after clicking that icon the software starts checking the 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:

1. Longitudinal Reinforcement.
2. Shear Reinforcement.
3. 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 .
• 
• 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.
• 

Shear Demand throughout the wall height:

• 
• 
• 

• DESIGN SUMMARY:

  • FLEXURAL INFORMATION
  • SHEAR INFORMATION
  • BOUNDARY WIDTH DATA

• 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.
• 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.
• 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.

        the thickness of our shear wall is 300mm.

  Therefore the Length to Thickness Ratio=length/ width

                   Length to Thickness Ratio    = 1500300</math>">1500/300

                    Length to Thickness Ratio   = 5.

  Therefore 5>4</math>">5>45>4 OK .

• 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

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

 So in our case,we have the height of the wall (hw) = 21000mm.                                                                                                                                the length of the wall (lw) = 1500mm.

hw/lw=21000/1500

hw/lw=14

hw/lw=14>2

• 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.
• 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.

• According to the IS 13920:2016,clause 10.1.8,states that the largest diameter of the longitudinal bar shall not exceed 110th</math>">1/10thof its thickness. 

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

  Thickness=30mm
• 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.

• 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.
•  according to the ETABS software we have the Boundary Width Requirement i.e, 600 mm.so according to that we will design.
• We have the web width of 300mm but this width is very less that's why here we will be using the boundary width.
• From base upto the first floor we will be using the boundary width of 600mm.
• as for the upper floors,the boundary element demand reduces upto 300mm.
• 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 =AstAsh=750100=7.5=8</math>">Ast/Ash=750/100=8bars.

Spacing = HeightNo⋅ofBars=10008=125mm</math>">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=1500xx300=450000 mm^2

And Asc = Required Longitudinal Reinforcement Ratio X Ag.

      Asc = 0.0292 X 450000

      Asc =13140mm2</math>">13140mm

Therefore,Ash=0.05×Sv×h×fckfy</math>">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=0.05×100×200×25500</math>">∴Ash=0.05×100×200×25/500

Ash=0.05×40×25</math>">Ash=0.05×40×25

Ash=50mm2</math>">Ash=50mm²

Therefore, Minimum no. of bars = (1500-40200</math>">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</math>">mm²

Hence provide 28 numbers of T25 bars.

Therefore, Ast provided = 490 X 28

                Ast provided = 13720 mm2</math>">mm²

So 13720>13140mm2</math>">13720>13140mm²

Also the spacing = 120<200mm</math>">120<200mmrequired.

 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. 

 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 mm2</math>">mm²

Asc = Ag X Minimum reinforcement ratio.

Asc = 270000 X 0.0030

Asc = 810mm2</math>">mm²

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

                                                        = 1206 mm

∴1206>210</math>">∴1206>210∴1206>210

• Design of Boundary Element:

Minimum reinforcement ratio according to the IS code = 0.008

Boundary width = 300mm.

Ash=0.05×Sv×h×fckfy</math>">Ash=0.05×Sv×h×fck/fy

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</math>">∴Ash=0.05×100×200×(25/500)

Ash=0.05×40×25</math>">Ash=0.05×40×25

Ash=50mm2</math>">Ash=50mm²

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

                                        = 4896 - 810.

Reinforcement Required = 4086 mm2</math>">mm²

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

                                                                      Ast required for the one Boundary Element = (40862</math>">4086/2)

Ast required for the one Boundary Element = 2043mm2</math>">mm²

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

                                       Ast Provided = 3430mm2</math>">mm²

mm2</math>">^{So 3430>2043</math>">3430>2043}

• Reinforcement Ratio Check:

Ratio=Astbw×tw</math>">Ratio=Ast/(bw×tw)

Ratio=Astbw×tw</math>">Ratio=3430300×300</math>">Ratio=3430300×300

Ratio=0.038>0.008</math>">Ratio=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.

Result:

• The values of longitudinal steel ratio, boundary element length and shear reinforcing ratio from the design results have been extracted successfully.
• The cross sectional steel arrangement of shear wall from its base up to terrace, in compliance with provisions of IS13920-2016 has been drafted after appropriate calculations.
• The elevation drawings of reinforcement details of the shear wall from base to terrace has also been drafted.

40862</math>">