Challenge

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

ANALYSIS AND DESIGN USING RCC SHEAR WALLS IN THE MODEL USING ETABS

AIM:

To run the analysis and design using the RCC shear walls for the given ETABS MODEL
To extract the values of longitudnal steel ratio,boundary element length and shear reinforcing ratio fron the design results
To provide the cross sectional steel arrangement of shear wall from its case up top terrace in compliance with provision of IS13920:2016
To provide elevation drawings of reinforcement details of the sheat wall from base to terrace

INTRODUCTION

A shear wall consist of a web and boundary elements of eitherr sides.This boundary elements in the shear walls are subjected to alternate compression and tension when it is subjected to lateral force,hence it is necessary to ptovide special reinforcement for the boundary elements

IS 13920-2016 Provides guidelines on several asoects of shear wall detailing
Thickness of wall (clause 10.1.2)
Minimum ratio of wall length to thickness (clause 10.1.3)
Largest diameter of longitudnal steel bar (10.1.8)
Maximum spacing of horizontal and vertical reinforcement(10.1.9)
Minimum reinforcement in RC shear wall(10.1.6)

From clause 10.1.4 walls can be categorised as squat walls ,intermediate walls and slender walls as given below

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

PROCEDURE:

Once the analysis is completed we need to carry out the design of the shear wall
Shear wall design---->view/revise preference

Total height of wall = 21000mm
Total Length of the wall = 1500mm
`21000/1500 = 14`
Ratio is greater than 2 hence it is ''SLENDER'' Wall
Maximum diameter of the bar should not exceed the minimum of the following criteria

a) 1/5 X 1500= 300mm

b) 3 X 300 = 900mm

c)450mm

Hence maximum spacing is limited to 300mm

From ETABS results the maximum longitudnal reinforcement obtained 13128mm^2
We take bar of 25mm diam. no. of bars required will be 13128/490 =28 bar.But we provide additional two elements for proper confinement i.e. longitudnal bars
Now from the coide

`(rho)min = 0.0025+ 0.01375(t /l)`

`0.0025 *300*1500 = 2362.5` mm^2

Reinforcement provided for boundary element in right side =4410mm^2
At left side =5390mm^2 both are greater than required value Henvce safe
In the web area minimum reinforcement provided is 0.0025 X 1500 X 300= 300mm^2 and provided amount of reinforcement in the web area ias 3920mm^2
Then provide 8 mm @ 100mm c/c

From clause 10.4.4. shear reinforcement in the boundary element can be calculated from below formula

Ash is the area of shear reinforcement per link
Sv = 100 mm
Fck =25 N/mm^2 Fy=415N/mm^2
Ash = 0.05 X 100X370 X 25/415 =156MM^2
For two legs 156/2 = 78mm^2
We choose 10 mm dia for shear reinforcement in the boundary element
Boundary element width =450 mm, h = 450-80 =370
Ash = 0.05 X 100 X 370 X25/415 = 111 mm^2 for one linek = 55
Hence we can provide 8mm dia bars for shear reinforcement in the boundary element width whose area is around 59 mm^2.This can be provided at the right side of the wall
For boundary element width special confinement can be provided aas
10 mm dia bars @ 100 mm c/c at left side BE and 8 mm dia bars @100 mm c/c BOUNDARY ELEMENT

The values of longitudnal steel ratio ,boundary element length and shear reinforcing ratio from the design results have been extracted succesfully
Steel arrangement of shear wall from its base up to terrace in compliance with provision of IS 13920 :2016

Ship work like this
in your own course.