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

Challenge 10

Q.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.

Answers:-

AIM:-

1. To provide the exact values of longitudinal steel ratio, boundary element length and shear reinforcing ratio from the design results.
2. To provide cross sectional steel arrangement of the shear wall from its base up to terrace, in compliance with provisions of IS 13920 – 2016.
3. To provide elevation drawings of the reinforcement details of the shear wall from base up to terrace.

PROCEDURE:-

STEP 1 :-

• Open etabs software and run the analysis.
• Once the analysis is completed go to shear wall design.

• Now click on design/check shear wall structrure.

• Now again go to shear wall design
• select display design info and select pier longitudinal reinforcing.

• Now go to pier reinforcing ratios.

• Display shear wall design and select pier reinforcing ratios.

• Display shear wall design and select pier edge/boundary zone width.

• Display shear wall design and select pier shear reinforcement.

Shear wall detailing.

1. Wall thickness = 300mm(Given), Hence safe (Satisfies clause 10.1.2 of IS 1390-2016).
2. Minimum ratio of wall = 1500/300 = 5 > 4 ,Hence safe (Satisfies clause 10.1.3 of IS 1390-2016).

3.hw/Lw = 21/1.5= 14; Hence slender walls(clause 10.1.4 of IS 1390-2016).

4. (1/10)*300 =30mm;

Hence use 25mm dia longitudinal steel bars.

1. 1/5th of 1500= 300mm
2. 3*300= 900mm
3. 450mm.

=>Hence the maximum spacing of vertical or horizontal reiforcement = 300mm.

Longitunal reinforcement

• From ETABS results the maximum longitudinal reinforcement obtained is 13128 mm².
  
  Diameter of bars= 25mm
• • Hence no. of bars= 13128/490= 28 bars.
  • 2 more additional bars are provivded in the boundary elemements for proper confinement
  • ie no of bars = 30

  =>Hence total area covered by 30 no.s of 25mm bars = 13737mm²; ie, safe

  From IS code minimum reinforcement required = 0.0025 +0.01375(tw/Lw)= 0.00525

  i.e 0.00525x300x1500 = 2362.5mm²,< 13737mm²; Hence safe.

  Minimum vertical reinforcement to be provided in the boundary region = 0.008 x 1500 x 300= 3600mm².

  • The reinforcement provided for boundary element in right side = 4410mm² >3600mm² ;Hence safe.
  • At left side 5390mm230th are greater than the required value. >3600mm² ;Hence safe.

  
  =>Minimum reinforcement in the web area provided is 0.0025x1500x300=1125mm2 and which is less than the provided amount of reinforcement in web area ie, 3920mm2.

  Hence it is safe.

  Horizontal reinforcement:

  • From Etabs results the shear reinforcement amount is obtained as 750mm².
  • providimg 8mm dia bars @ 100 mm c/c, then the provided area of reinforcement will be 1000mm². hence safe.
  • This shear reinforcement can be provided for the entire shear wall.

  check for boundary element

  Left side

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

  Ash= 0.05*Sv*h (fck/fy)

  • Sv= 100mm
  • h = 600-80 = 520mm (Spacing between the extreme longitudinal reinforcement in the boundary element)
  • fck = 25N/mm² and fy = 415 N/mm².

  Therefore Ash = 0.05*Sv*h (fck/fy)= 0.05*100*520*(25/415) = 156.62 mm².(for 1 link)

  • Area of each leg= 156/2= 78 mm²

  =>Hence provide 12mm dia bars as shear reinforcement in the boundary element width at the left end.

  ie, Area provided , Ash provided = (π/4)* 12*12 = 113.097 mm²>78, Hence safe

  Right side

  • Boundary element width is 450mm , h=450-80-370mm.

  Ash= 0.05*Sv*h (fck/fy)

  Therefore Ash = 0.05*Sv*h (fck/fy)= 0.05*100*370*(25/415) = 111.45 mm².(for 1 link)

  • Area of each leg= 111/2= 55 mm²

  >Hence provide 10 mm dia bars as shear reinforcement in the boundary element width at the right end.

  ie , Area provided , Ash provided = (π/4)* 10*10 = 78.54 mm²>55mm², Hence safe.