Analysis & Design of RCC shear walls in the model using ETABS
Q :- The E-Tabs 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.
Ans :-
AIM :-
- The objective of this assignment is to finish the research and design of the RCC shear walls.
- additionally to extrapolate values from the design outcomes for the boundary element lengths, shear reinforcement ratios, and longitudinal steel ratios.
PROCEDURE :-
Therefore, open the E-TABS file that is provided in the question itself first. Open up that file in the E-TABS program, which is what we'll be using to design the shear walls.
- After opening the provided file in the E-TABS program using the proper location, as shown in the image below,
- Below is the given model in which we are going to work
After opening the file, the analysis will be executed as the first step.
However, we must first check the model for any structural flaws before conducting the Run analysis.
There will be a drop-down menu on the Analysis tab with an option labeled "Examine model" in it. Simply select that choice by clicking as shown in the image below. - We need to do this in order to verify the model's structural integrity.
- Following this, a new dialogue box titled "Check Models" appears; pick all of the options that are present in that dialogue box, click "OK," and then execute the check as indicated in the image below.
The check has been done, and as you can see in the image below, there are no warnings.
It also means that we can carry on with this G+6 building's analysis.
We will now modify the view options after inspecting the structure.
We must first select the shear wall that we will build and analyze in order to change the display settings. Once a drop-down list has appeared, click the option that reads "show selected object alone," as shown in the figure below, by performing a right-click anywhere on the screen.
Selected Shear Wall :-
Right-click anywhere on the screen :-
In 3D view selected Shear Wall :-
We must now verify that the coding we are using for the design is accurate after changing the display settings.
Simply select the shear wall design icon and click it to get started. Following that, a drop-down list will appear; from this list, choose the choice labeled "View/Revise Reference," as shown in the example below.
A new dialogue box with the title "Shear Wall Design Preferences for IS 456:2000" appears on the screen after choosing that option. similar to what is shown in the image provided below
Make sure that the nation or area where the project is located is taken into account when selecting the desired codes in that dialogue box.
After confirming that the chosen codes are fixed, we can now run the analysis.
Simply choose the Shear Wall Design icon from the screen's top-right corner to accomplish this, as shown in the illustration below.
Click that icon now. The software will then begin to check the design as shown in the below-provided image.
Now that the E-TABS program's designing process is complete, we must collect all the information required to be used in the ensuing calculations for the finer details.
Therefore, in order to design the shear walls, we need three primary pieces of information.
the gradual reinforcement of.
Shear's reinforcement.
a boundary element's (BE) width.
- All of the aforementioned information can also be extracted using the ETABS program itself.
- Select Shear Wall Design from the drop-down list that appears after making your selection, as shown in the example below, and then click the Display Design Info option to extract these numbers from the ETABS software:
The Display Shear Wall Design Results dialog box then appears; in that dialog box, change the design output to longitudinal reinforcing as shown in the image below.
- Maximum Longitudinal Reinforcement at the base :-
- The Display Shear Wall Design Results dialogue box will now open. In that dialogue box, set the design output to shear reinforcing as illustrated in the image below.
- Shear Demand throughout the wall height :-
Now that the Display Shear Wall Design Results dialogue box has appeared, change the design output to the edge/boundary zone widths as shown in the illustration below:
- Boundary Width Data :-
DESIGN SUMMARY :-
- FLEXURAL INFORMATION :-
- SHEAR INFORMATION :-
- BOUNDARY WIDTH DATA :-
- DETAILED CALCULATIONS FOR THE SHEAR WALLS :-
So, a brief explanation of the technical procedure for specifying the reinforcing bars in the shear wall is given in the following section:
SECTION CLASSIFICATION :-
To determine whether the geometrical parameters of the wall, such as its thickness, length to thickness ratio, type of shear wall, and rebar's diameter, are within the acceptable range or not, the first thing that needs to be checked is whether or not they are. Let's look at each of these characteristics separately.
- Thickness of the Wall :-.
The thickness of the wall will therefore be the first property that we examine.
- Clause 10.1.2 of the IS 13920:2016 specifies that the shear wall's minimum allowable thickness shall not be less than 150 mm.
However, the thickness of the shear wall in our case is 300 mm, which is greater than the 150 mm recommended by IS code 13920:2016. Our design is therefore secure.
- 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">5 > 4 Hence OK .
- Shear wall type :-
- The IS 13920:2016, clause 10.1.4, suggests that there are three different types of shear walls, which are listed below, based on their geometrical characteristics:
- Squat Walls: If hwlw<1">hw/lw<1.
- Intermediate Walls: If hwlw<2">hw/lw<2.
- Slender Walls: If hwlw>2">hw/lw>2.
- So in our case,we have the height of the wall ( hw ) = 21000mm.
and the length of the wall ( lw ) = 1500mm.
∴hwlw=210001500"> ∴ hw/lw=2100/1500
hwlw=14"> hw/lw=14
So, hwlw=14>2">hw/lw=14>2
- The fraction here is greater than the 2 hence OK.
- Reinforcement Layer :-
In accordance with IS 13920:2016, clause 10.1.7, reinforcement must be provided and both faces of the wall must have a thickness of at least 200 mm.
- Since the shear wall thickness in this instance is 300 mm rather than 200 mm, we must supply rebars for both of the wall's faces.
- 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 110th">1/10thof its thickness.
- So in our case the thickness of our wall = 300mm.
∴110th">∴ 1/10th of its thickness = 110×300">1/10×30 = 30 mm.`
Thus, any bar with a diameter greater than 30 mm cannot be used.
Because we are limited to using no more than a 30mm diameter, we will be using a 25mm diameter bar.
- DETAILED CALCULATION :-
Therefore, the steps for the precise E-TABS software calculations of the reinforcement bars for shear walls are discussed below.
In order to design the walls, we must first divide the sections.
We must divide the walls along their height based on the need for reinforcement because we are aware that we cannot design the walls for the entire five-story building.
- I divided the walls into Sections 1 and 2, or groups of two.
- Section 1 = From base upto the First Floor.
- Section 2 = From Second floor upto the Terrace level.
Additionally, the E-TABS software indicates that the required boundary width is 600 mm.
Therefore, we will design Section 1's entire length, or 600 mm, as the boundary element length.
We have a web width of 300 mm, but since this width is so small, we will use the boundary width in this instance.
Therefore, we will use a boundary width of 600 mm for Section 1, which includes the ground floor up to the first floor.
However, the boundary element demand decreases by up to 300mm on the upper floors.
Therefore, we will use the 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 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">Ast/Ash = 750/100 = 7.5 = 8bars.
Spacing = HeightNo⋅ofBars=10008=125mm">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 = 1500 x 300 = 450000 mm^2
And Asc = Required Longitudinal Reinforcement Ratio X Ag.
Asc = 0.0292 X 450000
Asc = 13140mm2">13140 mm^2
Therefore, Ash=0.05×Sv×h×fckfy">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">∴ Ash = 0.05 × 100 × 200 × 25 / 500
Ash=0.05×40×25"> Ash = 0.05 × 40 × 25
Ash=50mm2"> Ash = 50mm^2
Therefore, Minimum no. of bars = 1500-40200">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">mm^2
Hence provide 28 numbers of T25 bars.
Therefore, Ast provided = 490 X 28
Ast provided = 13720 mm2">mm^2
So 13720>13140mm2">13720>13140mm^2
Hence OK.
Also the spacing = 120<200mm">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 mm2">mm^2
Asc = Ag X Minimum reinforcement ratio.
Asc = 270000 X 0.0030
Asc = 810 mm^2
Hence provide 6 numbers of T 16 bars = 6 X 201.
= 1206 mm
∴1206>210">∴ 1206 > 210
Hence OK.
- Design of Boundary Element :-
Minimum reinforcement ratio according to the IS code = 0.008
Boundary width = 300mm.
Ash=0.05×Sv×h×fckfy">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">∴ Ash = 0.05 × 100 × 200 × 25 / 500
Ash=0.05×40×25"> Ash = 0.05 × 40 × 25
Ash=50mm2"> Ash = 50mm^2
Reinforcement Required = Total Ast - Ast taken by the Web.
= 4896 - 810.
Reinforcement Required = 4086 mm2">mm^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 = 40862">4086 / 2
Ast required for the one Boundary Element = 2043 mm2">mm^2
Thus provide 7 numbers of T25 bars = 7 X 490.
Ast Provided = 3430 mm2">mm^2
So 3430>2043">3430 > 2043
Hence OK.
- Reinforcement Ratio Check :-
Ratio=Astbw×tw">Ratio = Ast / bw × tw
Ratio=3430300×300">Ratio = 3430 / 300×300
Ratio=3430300×300">Ratio=0.038>0.008">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.
Elevation Drawing :-
