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

AIM:

Doing the study and designing the RCC shear walls is the goal of this assignment. additionally to take the values of the boundary element length, shear reinforcement ratio, and longitudinal steel ratio and extract them from the design outcomes.

PROCEDURE:

• Thus, open the ETABS file that is provided in the query itself first. Open the file in the ETABS programme, which is where we will be working on the shear wall design.
• Following the given file's opening in the ETABS programme at the correct location, as indicated in the provided image below:

• The provided model, which we will be using, is as follows:

• The first thing we're going to do is Run Analysis after opening the file.
• Nevertheless, we must first examine the model for any structural flaws before performing the Run analysis.
• To verify the model for any structural defects, navigate to the Analysis tab. There, a drop-down list will appear. Select the option labelled "Check model" by clicking on it, as illustrated in the image below.

• Following this, a new dialogue box called "Check Models" appears. Selecting every option in that dialogue box, clicking OK, then conducting the check will result in the following image being displayed:

• Thus, the image below has been checked, and no warnings are displayed. Additionally, it indicates that we can carry out this G+6 building's analysis going forward.

• Now that the structure has been examined, we will modify the view settings.
• And to change the view settings, we must first choose the shear wall that we are supposed to design and analyse. Once we have chosen the shear wall, right-click anywhere on the screen to bring up a list of options. From that list, select the one labelled "show selected object only," as seen in the image below:

Selected shear wall:

Right click anywhere on the screen:

In 3D view selected shear wall:

• Therefore, after changing the view settings, we must now confirm that the codes we are utilising for design are accurate.
• To accomplish that, simply navigate to the shear wall design symbol and click it. A drop-down list will then appear on the screen; from that list, select the option labelled "View/Revise Reference," as seen in the provided image below.

• Following that selection, as illustrated in the figure below, a new dialogue box labelled Shear Wall Design Preferences for IS 456:2000 opens on the screen:

• Make sure the desired codes are chosen in the dialogue box according to the nation or region where the project is located.
• Once we have confirmed that the specified codes have been fixed, we may proceed with the Run analysis.
• To accomplish this, simply click the Shear Wall Design icon, which is located in the upper-right area of the screen as seen in the image below:

• Now select that icon. The software will then begin to verify the design as indicated in the provided image below.

• After the ETABS software's designing phase is complete, we must gather all the information needed for the subsequent detailed computations.
• Thus, we require three key pieces of information to build the shear walls:

a. longitudinal reinforcement.

b. shear reinforcement.

c. boundary element width (BE).

• Additionally, the ETABS programme itself has all of the aforementioned data.
• Simply select the option Shear Wall Design from the menu, and then click the drop-down list that appears. From there, select Display Design Info, as indicated in the provided image below, to extract these numbers from the ETABS software:

• Following this, the Display Shear Wall Design Results dialogue box opens. In that dialogue box, set the design output to longitudinal reinforcing, as seen in the image below:

Maximum Longitudinal Reinforcement at the base:

• Now, the Display Shear Wall Design Results dialogue box opens. In that dialogue box, alter the design output once more to Shear Reinforcing, as indicated in the image below:

Shear Demand throught the wall height:

• Now that the Display Shear Wall Design Results dialogue box has opened, modify the design output to Edge/Boundary Zones Widths once more, as indicated in the image below:

Boundary Width Data:

Design Summary

• Flexural Information:

• Shear Information:

• Boundary Width Data:

• Detailed calculations for the shear walls:

Thus, the following provides a concise explanation of the thorough process for specifying the shear wall's reinforcement bars:

• Section Classification:

Therefore, the first thing we must do is determine whether or not the wall's geometrical characteristics—such as its thickness, length to thickness ratio, kind of shear wall, rebar diameter, etc.—fall within the parameters specified by IS code 13920:2016. So let's examine each of these attributes separately.

• Thickness of the wall:

a. Thus, the wall's thickness will be the first attribute we examine.

b. The minimum permitted thickness of the shear wall should not be less than 150mm, per IS 13920:2016 clause no. 10.1.2.

c. However, the thickness of the shear wall in our instance is 300 mm, which is more than the IS code 13920:2016's recommended 150 mm. Our design is safe as a result.

• Length to thickness ratio:

a. Thus, the length to thickness ratio is the second item we must examine.

b. Clause 10.1.3 of IS 13920:2016 states that the length to thickness ratio must be 4 or more.

c. Thus, our shear wall is 1500 mm long.

d. Additionally, our shear wall is 300 mm thick.

Therfore the length to thickness ratio= Length/Width

= 1500/300

length to thickness ratio = 5

therefore 5>4

Hence Ok.

• Shear wall type:

Clause 10.1.4 of IS 13920:2016 suggests that there are three distinct types of shear walls, which are outlined below, depending on their geometrical properties:

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.

therefore hw/lw= 21000/1500

= 14

So, hw/lw= 14>2.

The fraction here is greater than the 2 hence ok.

• Reinforcement layer:

Thus, in accordance with IS 13920:2016, clause 10.1.7, both faces of the wall's thickness must be larger than 200 mm, and reinforcement must be given.

Since the shear wall thickness in this instance is 300 mm rather than 200 mm, we must supply the rebars in addition to the wall's two faces.

• Diameter of the rebar:

Clause 10.1.8 of IS 13920:2016 specifies that the longitudinal bar's greatest diameter cannot be greater than 1 /10 the thickness of it.

Thus, in our instance, the wall's thickness is 300 mm.

Therefore 1/10th of its thickness= 1/10X 300

= 30mm.

This implies that no bar larger than 30 mm in diameter may be used. Since we are unable to utilise a bar with a diameter larger than 30 mm, we shall use one with a diameter of 25 mm in this instance.

• Detailed Calculation:

Thus, the following describes the procedures for the in-depth ETABS software computations of the reinforcing bars for the shear walls.

Therefore, in order to design the walls, we must first split the portions.

We must divide the walls along their height in accordance with the need for reinforcement because we are aware that we cannot design the walls for the full five-story building.

The walls are divided into two portions, designated as Section 1 and Section 2.

Base to the First Floor is Section 1.
Section 2 is the area from the second floor to the terrace. Additionally, the ETABS programme indicates that we meet the 600 mm boundary width requirement.

Therefore, we will designate 600 mm, the whole length of Section 1, as the boundary element length.

Although our 300mm web width is quite little, we will be using the boundary width in this instance.

Therefore, we will be employing a boundary width of 600 mm for Section 1, which is the area from base to the first level.

However, the boundary element requirement drops by up to 300 mm on the higher floors.

For this reason, we will be employing a boundary element length of 300mm in Section 2, which is the area from the second floor up to the terrace level.

A) Calculations From Base Upto The First Floor i.e Section1:

Shear Demand:

The shear need of 750 sq. mm per 1000 mm height has already been met.

Give herar links with a diameter of 8 mm.

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

Spacing= Height/No. of Bars=1000/8=125mm

As a result, the entire wall should be shear reinforced using 8 mm diameter shear links spaced 125 mm from centre to centre.

Additionally, IS 13920:2016's clause 10.4.4 provides the vertical spacing for shear links; so, based on this clause, the vertical spacing for 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=450000mm^2

And Asc=Required Longitudinal Reinforcement RatioX Ag.

Asc=0.0292X450000

Asc= 13140mm^2

Therefore, Ash=0.05 X SvX h X fck/fy

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

Assumed data:

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

fck= 25

fy= 500

and h=200mm

Ash= 0.05X100X200X25/500

= 0.05X40X25

Ash= 50mm^2

Therfore, minimum no. of bars=1500-40/200

= 7.3

minimum no. of bars=8 bars.

So we need to cover the demand for 13140mm^2.

Hence provide 28 numbers of T25 bars.

Therefore, Ast provided =490x28

= 13720mm^2.

So 13720>13140mm^2

Hence Ok.

Also the spacing =120<200mm required.

Hence Ok.

• As a result, make 26 bars on both sides by placing 13 bars on one face and 13 bars on the other.
• Along the shorter span, place two bars, bringing the clear spacing down to 150mm. Thus, alright.
• The following is a detailed drawing:

Details of the reinforcement from the base to the first floor:

B) Calculations From Base Upto The First Floor i.e Section2:

Design of the web steel:

Web Width=900mm.

Boundary element width= 300mm.

Minimum reinforcement ratio for the web is 0.0030 for the safer calculations.

Therefore Ag= Web width x boundary element width.

= 900x300

= 270000mm^2

Asc= Ag x Minimum reinforcement ratio.

= 270000x0.0030

= 810mm^2

Hence provide 6 number of T 16 bars= 6x201

= 1206mm

Therefore 1206>210

Hence Ok.

Design of Boundary Element:

Minimum reinforcement ratio according to the IS code= 0.008

Boundary width = 300mm.

Ash= 0.05 x Sv x h x fck/fy

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

Assumed data:

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

fck= 25

fy= 500

and h= 200mm.

Therefore Ash= 0.05 x 100 x 200 x 25/500

= 0.05 x 40 x 25

= 50mm^2

Reinforcement required= 4086mm^2

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

Ast required for the one boundary element= 4086/2

= 2043 mm^2

Thus provided 7 number of T25 bars= 7x490.

Ast Provided= 3430mm^2

So 3430>2043

Hence Ok.

Reinforcement detailing from the second floor upto the Terrace level:

Here all the dimensions are in mm.

Elevation Drawing: