Analysis & Design of RCC shear walls in the model using ETABS
AIM:
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.
PROCEDURE:
- Then click on the plan view and select the all stories.
- Then select the entire shear seven shells are selected and right click option select the show selected objects only.
- Go to the elevation view along the grid 11 apply and ok.
- Click on the option run the analysis and complete the analysis.
- Then click on shear wall design option.
- Then click on Pier Shear Reinforcing apply and ok.
SHEAR WALL CALCULATION:
BASE FLOOR TO FIRST FLOOR:
Height of wall (h) = 2100
Length of wall(1) = 1500
The wall here has been considered as slender wall
Therefore : height of wall/length of wall >2
2100/1500-14>2
Hence, it is slender wall
we need to calculate for slender wall( for minimum horizontal reinforcement)
(Ph)min=0.0025 +0.5 (hw/Lw -2) (Ph-0.0025)
where:
hw=height of wall = 2100
Lw=length of wall - 1500
Ph=0.0297 x 1500 x 300 = 13365
So, after substituting all the values in eqs of (Ph)min We get:-(Ph)min=0.76
(PvBE) min=0.0080
(Pv web) min=0.025
(Pv net) min=0.0025+0.01375 (tw/hw)
=0.0025+0.01375(300/1500)
=0.0052
we need to calculate for slender wall (for longitudinal reinforcement)
Minimum Vertical reinforcement, ev
ev =13128/(C.S. Area of steel)
ev =13128/(1500 X300)
ev=0.029 0.0080
From IS 13920,
Ash= 0.05 Sv h (fck/fy)
Minimum no. of bars = 8
Ash= π/4 d^2
So, Ash-= 50.24 mm²
50.24= 0-05 x 100 x h (25/500)
50.24=0.025h
h= 200.96
Use dia = 25mm.
No. of bars req. 13128/π/4 25^2
No. of bar 13128/490.625=28 bar
For Shear reinforcement
Ash= 780 mm²
Assuming 2 lagged 8mm dia. @ 100mm C/C
Ash=(2 π/4 8^2)/0.100
Ash= 1005.30> 780 mm²
The shear reinforcement is required
From ground floor to First Floor:
From ground floor to First Floor:
For longitudinal reinforcement
Av: 8273 mm²
Minimum Vertical reinforcement=8273/(1500 x 300)
= 8273/450 000
=0018
As per IS 13920,
Ash=0.05 Sv h fck/fy
h=200.96
No. of bars=8273/(π/4 25^2)
=8273/490.87 =16.8
No. of bar = 17
For Shear reinforcement
Ash = 750mm
By providing 2 legged: 8 mm dia @ 100 c/2
Ash =[2x (π/4) x d^2]/0.100
=[2 x π/4 x 8^2]/0.100 =1005 mm²
1005 > 750mm²
Similarly, For web:
By providing 2 legged 8mm dia @ 125 mm c/c
Ash=(2 x π/4 x 8)/0.125
Ash=805 mm2>750 mm²
From 1st Floor to 2nd floor:
From 1st Floor to 2nd floor:
For longitudinal reinforcement:
Av=4896 mm²
Asv=4896/[1500x300]=0.0108 > 0.025
As per. IS 13920,
Ash= 0.05 Sv h fck/fy
h= 200.96
No. of bar=4896/ (π/4 25^2)
=9
9 no.of 16mm bar.
For 2nd Floor to Terrace.
Boundary Element, be = 300mm
Tw=300mm
For longitudinal reinforcement,
Asv= 2150
2150/[1500x 300]= 0.005 Ash=0.05 Sv h (fck/fy) By providing 8mm dia @ 125 mm c/c No. of bars= 2150/(π/4 25^2) =4.33 =4 nos. Provide 4 no. of 16 mm For Shear Reinforcement Minimum Ash=750 mm^2 Minimum dia. Bar = 8mm By providing 2 legged 8mm dia @100 mm Spacing Ash = [2 x 50.64]/0.100 Ash = 1005 mm2 RESULT: Provided the extracted values of longitudinal steel ratio, boundary element length and shear reinforcing ratio from the design results. Provided cross sectional steel arrangement of the shear wall from its base up to terrace, in compliance with provisions of IS 13920 – 2016. Designed the RCC shear wall using the design results from ETABS file. Provide elevation drawings of the reinforcement details of the shear wall from base up to terrace.