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

objectives of the challenge:

To know the design of rc shear wall using ETABS analysis

To know the basics of shear wall reinforcement detailing with respect to primary standards 

To ensure the ductility reinforcemnt required by IS 13920- 2016 been satisfied. 

Detailing guide line for shear wall:

1.minimum reinforcement for shrinkage and temperature

2. ultimate limit state design rebar requirement 

3. ductility rebar requirement 

(which ever is great should be drawn in RC shear wall detailing )

Procedure:

• Initially, open the given ETABS model.
• Now, go to analysis tab and check the model.

• After that run the analysis.
• Once the analysis is completed we need to carry out the design of the shear wall, for this go to shear wall design.
• Shear wall design to view/revise preferences.

• Now we need to start the design.
• Once, the design is completed go to display design option.
• Then we have checked for any failures in the design.
• Now, we need to extract certain values from the design.
• Go to display design info to pier longitudinal reinforcing.
• Go to display design to pier reinforcing ratios.
• Go to display design ito pier edge/boundary zone width.
• Go to display design to pier shear reinforcing .

• 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.

- Shear wall detailing calculations:

Here, the total height of the wall is 21000mm and 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 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 13128`mm^2`.
  • Since we are using 25mm dia bars, the no. of bars required 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.
  • Hence the total area of reinforcement provided is 13720`mm^2`. Hence, it is safe.
• Now, from the code the minimum vertical reinforcement required is 0.0025+0.01375`((tw)/(Lw))` =0.00525, i.e 0.00525x300x1500 = 2362.5`mm^2` 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 = 3600`mm^2`.
  • The reinforcement provided for boundary element in right side = 4410`mm^2`and
  • At left side = 5390`mm^2`Both are greater than the required value i.e 3600`mm^2`. Hence safe.
• In the web area the minimum reinforcement provided is 0.0025x1500x300=1125`mm^2` and the provided amount of reinofrcement in web area is 3920`mm^2`.Hence it is safe.

- Horizontal reinforcement: 

• From Etabs results the shear reinforcement amount is obtained as 750`mm^2`.
• So let us provide 8mm dia bars @ 100 mm c/c, then the provided area of reinforcement will be 1000`mm^2`. 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/`mm^2` and fy = 500N/`mm^2`.
  • Hence it is obtained as   Ash =`0.05*100*520*25/415`= 156`mm^2`this for one link
  • One link has 2 legs hence area of each leg = 156/2=78`mm^2`
  • Hence, we can choose 10mm dia bars for shear reinforcement in the boundary element width whose area is around 78.5`mm^2`. This can be provided at the left side of the wall.
  • Now, in the right side the boundary element width is 450`mm^2` so, h=450-80=370mm. Hence, Ash requirement will be 
  • Ash = `0.05*100*370*25/415=`111`mm^2` for each link. Hence, for one leg = 55`mm^2`
  • Hence we can provide 8mm dia bars for shear reinforcement in the boundary element width whose area is around 50`mm^2`. 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.