Analysis & Design of RCC shear walls in the model using ETABS
Aim:
- To run the anlaysis and to design the RCC Shear walls for the given ETABS model.
- To extract the values of longitudinal steel ratio, boundary element length and shear reinforcing ratio from the design results obtained.
- To provide the cross sectional steel arrangement of shear wall from its case up to terrace, in compliance with provisions of IS13920-2016
- To provide elevation drawings of reinforcement details of the shear wall from base to terrace.
Introduction:
IS13920-2016 provides guidelines on several aspects of shear wall detailing, which includes:
- Thickness of wall
- Minimum ratio of wall length to thickness
- Largest diameter of longitudinal steel bar
- Maximum spacing of horizontal and vertical reinforcement
- Conditions for base of shear wall
- Minimum reinforcement in RC shear wall

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Now the minimum shear reinforcement in boundary element of RC shear wall can be determined from the below formula:

Procedure:
- Open the given etabs model.
- Go to analysis tab and check the model.

- Now, run the analysis.

- After the analysis is completed, we need to carry out the design of the shear wall.
- Go to shear wall design.
- Select shear wall design and then select view/revise preferences.

- Now, start the design.
- After the design is completed, go to display design option.
- Then check for failures in the design.
- Now, to extract certain values from the design, go to display design info and select pier longitudinal reinforcing.

- Go to display design info and select pier reinforcing ratios.

- Go to display design info and then select pier edge/boundary zone width.

- Go to display design info and then select pier shear reinforcing .

- Now we need to draft the details of a RC wall in AutoCAD for the wall at A11.
Shear wall detailing calculations:
Here, the total height of the wall is 21000mm and also the width of the wall is 1500mm.
Hence, the ratio is obtained as 21000/1500=14 which is greater than 2
hence, it can be considered as a slender wall.
Maximum diameter of the bar that can be used = 1/10th of thickness of that part = 30mm
Hence, the dia of bar considered is 25mm.
Maximum spacing between the bars should not exceed the minimum of the following:
- 1/5 x 1500 = 300mm
- 3x300 = 900mm
- 450mm
Hence, the maximum spacing is limited to 300mm.
Longitudinal reinforcement :
- From ETABS results, the maximum longitudinal reinforcement obtained for this model is 13128 mm2.
- Since we are using 25mm dia bars, the no. of bars required here will be 13128/490=28 bars. But we are providing two additional bars in the boundary element for proper confinement. i.e total 30 longitudinal bars are used here in this model.
- Hence the total area of reinforcement provided is 13720 mm2. Hence, it is safe.
- Now, from the code the minimum vertical reinforcement required is 0.0025+0.01375(twLw)mm2 =0.00525, i.e 0.00525x300x1500 = 2362.5 mm2 which is less than what is actually provided hence safe.
- Now, minimum vertical reinforcement to be provided in the boundary region = 0.008 x 1500 x 300 = 3600 mm2.
- The reinforcement provided for boundary element in the right side is = 4410 mm2
- At left side = 5390 mm2.As both values are greater than the required value i.e 3600 mm2. Hence it is safe.
- In the web area, the minimum reinforcement provided is 0.0025x1500x300=1125 mm2 and the provided amount of reinforcement in the web area is 3920 mm2.Hence it is safe.
Horizontal reinforcement:
- From the Etabs results obtained, the shear reinforcement is obtained as 750 mm2.
- Let us provide 8mm dia bars @ 100 mm c/c, then the provided area of reinforcement in the model will be 1000 mm2. Hence it is safe.
- This shear reinforcement can be provided for the entire shear wall in the model.
- Now, check for boundary element.
- From clause 10.4.4, the shear reinforcement required in the boundary element can be obtained from the eqn

- Ash is the area of shear reinforcement required per link.
- Sv is the spacing between the shear reinforcement = 100mm
- h is the spacing between the extreme longitudinal reinforcement in the boundary element i.e 600-80=520mm
- fck = 25
">N/mm2 and fy = 500N m m 2 ">N/mm2N m m 2 - Hence it is obtained as Ash =0.05⋅100⋅520⋅25415 mm2= 156 mm2 for one link.
- One link has 2 legs, hence area of each leg = 156/2=78 mm2.
- Hence, we can choose 10mm dia bars for shear reinforcement in the boundary element width whose area is around 78.5 mm2. This can be provided at the left side of the wall.
- Now, the right side boundary element width is 450 so, h=450-80=370mm.
- Hence, Ash requirement will be Ash = 0.05x100x370x25415 = 111 mm2 for each link. Hence, for one leg = 55 mm2
- Hence we can provide 8mm dia bars for shear reinforcement in the boundary element width whose area is around 50 mm2. 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 of the model have been extracted successfully.
- The cross sectional steel arrangement of shear wall from its base up to the terrace, in compliance with the provisions of IS 13920-2016 code have been drafted after the ETABS design calculations obtained.
- The elevation drawings of reinforcement details of the shear wall from the base upto the terrace have also been drafted from the ETABS designs.