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

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: to run the analysis and design shear wall for detail of r/f

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

Procedure:

step 1) open the etabs model provided with challenge

then go to plan view of terrace from plan view option

to see all the frame details

step 2) only allow to show grid A-11 shear wall

set working plane as all stories

then select A-11 shear wall and

turn on show selected object only

step 3) Run the analysis

click on run analysis tab to get anlysis of structure

step 4) design of shear wall

click on shear wall design tab

step 5) drop down the shear wall design option

and click on the reinforcement detailing option

step 6) now get information of boundary zones width

from design result option

step 7) same get longitudinal r/f

from shear wall design result

step 8) storey data

open storey data from edit storey data option

from edit tab

step 9) Shear wall detailing calculations:

Since, we have gathered the necessary information, now we need to draft the details of a RC wall in AutoCAD for the wall at A11.

Here, the total height of the wall is 21000mm and the width of the wall is 1500mm.

Hence, the ratio is obtained as 210001500=14210001500=14which is greater than 2 hence, it can be considered as a slender wall.

Maximum diameter of the bar that can be used from IS code = 1/10th of thickness of that part = 30mm

Hence, we take the dia of bar as 25mm.

Maximum spacing between the bars should not exceed the minimum of the following:

a) 1/5 x 1500 = 300mm

b) 3x300 = 900mm

c) 450mm

Hence, the maximum spacing is limited to 300mm.

- Longitudinal reinforcement :

From ETABS results the maximum longitudinal reinforcement obtained is 13128mm2mm2.

Since we are using 25mm dia bars, the no. of bars required will be 13128/490=2813128490=28bars. But we are providing two additional bars in the boundary element for proper confinement. i.e total 30 longitudinal bars.

Hence the total area of reinforcement provided is 13720mm2mm2. Hence, it is safe.

Now, from the code the minimum vertical reinforcement required is 0.0025+0.01375(twLw)(twLw) =0.00525, i.e 0.00525x300x1500 = 2362.5mm2mm2 which is less than actually provided hence safe.

Now, minimum vertical reinforcement to be provided in the boundary region = 0.008 x 1500 x 300 = 3600mm2mm2.

The reinforcement provided for boundary element in right side = 4410mm2mm2and

At left side = 5390mm2mm2Both are greater than the required value i.e 3600mm2mm2. Hence safe.

In the web area the minimum reinforcement provided is 0.0025x1500x300=1125mm2mm2 and the provided amount of reinofrcement in web area is 3920mm2mm2.Hence it is safe.

- Horizontal reinforcement:

From Etabs results the shear reinforcement amount is obtained as 750mm2mm2.

So let us provide 8mm dia bars @ 100 mm c/c, then the provided area of reinforcement will be 1000mm2mm2. hence it is safe.

This shear reinforcement can be provided for the entire shear wall.

Now, we need to check for boundary element

From clause 10.4.4, the shear reinforcement required in the bundary element can be obtained from the eqn

Ash is the area of shear reinforcement required per link.

Sv - spaing between the shear reinforcement = 100mm

h - spacing between the extreme longitudinal reinforcement in the boundary element i.e 600-80=520mm

fck = 25N/mm2mm2 and fy = 500N/mm2mm2.

Hence it is obtained as Ash =0.05⋅100⋅520⋅254150.05⋅100⋅520⋅25415= 156mm2mm2this for one link

One link has 2 legs hence area of each leg = 156/2=78mm2

Hence, we can choose 10mm dia bars for shear reinforcement in the boundary element width whose area is around 78.5mm2. This can be provided at the left side of the wall.

Now, in the right side the boundary element width is 450mm2 so, h=450-80=370mm. Hence, Ash requirement will be

Ash = 0.05⋅100⋅370⋅25415=0.05⋅100⋅370⋅25415=111mm2 for each link. Hence, for one leg = 55mm2

Hence we can provide 8mm dia bars for shear reinforcement in the boundary element width whose area is around 50mm2. This can be provided at the right side of the wall.

That is, for boundary element width special confining reinforcement to be provided is 10mm dia bars @ 100mm c/c at left side BE and 8mm dia bars @100mm c/c BE.

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.