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

AIM :

• To extract values of logitudinal steel ratio,boundary element length and shear reinforcing ratio from the design results
• To provide cross sectional steel arrangement of the shear wall from its base up to terrace,in compliance with provisions of IS 13920:2016.
• To provide elevation drawings of the reinforcement details of the shear wall from base up to terrace.
  
  PROCEDURE :
  
  
• Download and open the ETABS file attached in the given question.
  
  
  
  
  
• Check model and Run analysis is done.
• Select Display Menu bar and select Show tables and select Modal Participating Mass ratio and click ok.
  
  
  
• Now the table appears
  
  
  
• Select Display Menu bar and select Show tables and select Base reaction and click OK.
  
  
  
  
• Select Properties from Select option from the Select Menu bar and select wall section,we selected Wall300 (as given in the question).
  
  
  
• Now right click and select view only selected item
• Now from Design Menu bar,select Shear wall design and select Start design/check
  
  
  
• Now select Display Design info from Design menubar > Shear wall design and select Piers reinforcement ratio in Design output.
  
  
  
  
• Select Display Design info from Design menu bar > shear wall design and select Piers Edge/Boundary Zone widths in Design output
  
  
  
  
• Select Display Design info from Design menu bar > shear wall design and select Piers Reinforcing Ratio in design output.
  
  
  
  
• Now right click on the shear wall and the design will be available like as shown below :
  
  
  SHEAR WALL CALCULATIONS :
  
• BASE FLOOR TO FIRST FLOOR :
  
  Lets assume a slender wall since height/length ratio is greater than 2.
  
  i.e.,we have,height of the wall,H=2100mm
  and length of the wall,L=1500mm
  Thickness=300mm
  
  Ph=0.0297*1500*300=13365
  
  Now,Minimum Horizontal Reinforcement,
  (Ph)min=0.0025+0.5(H/L-2)(Ph-0.0025)=0.76
  Minimum Vertical Reinforcement,(Pv)min=0.0025+ 0.01375(T/L)=0.0052
  
  Longitudinal Reinforcement
  
  Using ETABS,the required Rebar Area=13128mm^2
  Minimum Vertial Reinforcement,Pv=13128/(1500*300)=0.0297 > 0.0080
  
  From IS13920:2016,
  Ash=0.05*Sv*H*fck/fy
  
  Assuming minimum no.of bars as 8,thus Ash=50.24mm^2
  50.24=0.05*100*H*(25/100)
  Therefore,H=200.98
  
  Considering 25mm dia bars,no.of bar=13128/area of 25mm dia.bars = 13128/490.87 = 28 bars
  Therefore.28 no.of 25mm dia. bars are provided.
  
  Shear Reinforcement
  
  Assuming 2 legged 8 mm dia. bars at 100mm c/c
  
  Ash = 2*area of 8mm dia.bar/0.100 = 1005.3 >750mm^2
  Thus satisfying the shear reinfocement
  
  
  
  
• FIRST FLOOR TO SECOND FLOOR :
  
  Longitudinal Reinforcement
  
  Using ETABS, the required Rebar Area = 8273mm^2
  
  Minimum Vertical Reinforcement , Pv = 8273/(1500*300) = 0.018 > 0.0080
  
  From IS 13920:2016,
  Ash = 0.05*Sv*H*fck/fy
  
  Assuming minimum no.of bars as 8 , thus Ash = 50.24mm^2
  50.24 = 0.05*100*H*(25/100)
  Therefore, H = 200.98
  
  Considering 20mm dia.bars,no.of bar = 8273/area of 20mm dia.bars = 26 bars
  
  Therefore,26no. of 20mm dia. bars are provided
  
  Shear Reinforcement
  
  Assuming 2 legged 8mm dia.bars at 100mm c/c
  
  Ash = 2*area of 8mm dia. bar/0.100 = 1005.3 > 750mm^2
  
  Thus satisfying the shear reinforcement
  
  
• SECOND FLOOR TO TERRACE :
  
  Longitudinal Reinforcement
  
  Using ETABS,the required Rebar Area = 4896mm^2
  
  Minimum Vertical Reinforcement,Pv = 4896 / (1500*300) = 0.01 > 0.0080
  
  From IS13920:2016,
  Ash = 0.05*Sv*H*fck/fy
  
  Assuming minimum no.of bars as 8,thus Ash = 50.24
  50.24 = 0.05*100*H*(25/100)
  Therefore,H = 200.98
  
  Considering, 16mm dia.bars,no. of bars = 8273/ area of 16mm dia.bars = 26bars
  
  Therefore,26no.of 16mm dia. bars are provided
  
  Shear Reinforcement
  
  Assuming,2 legged 8mm dia.bars at 100mm c/c
  
  Ash = 2 * area of 8mm dia.bar / 0.100 = 1005.3 > 750mm^2
  
  Thus satisfying the shear reinforcement.