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AIM:- Analysis & Design of RCC shear walls in the model using ETABS INTRODUCTION:- Shear wall is a structural member used to resist lateral forces, that is, parallel to the plane of the wall. For slender walls where the flexural deformation is more, shear wall resists the loads due to cantilever action. In other words, shear…
Vishal Mahalle
updated on 30 Aug 2022
AIM:- Analysis & Design of RCC shear walls in the model using ETABS
INTRODUCTION:-
QUESTION STATEMENT:-
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.
PROCEDURE:-
Run Analysis
Pier Longitudinal Reinforcing
]
Pier Shear Reinforcing
Pier Edge/Boundary Zones Widths
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 21000 mm and the width of the wall is 1500 mm.
Hence,
The ratio is obtained as (210001500)=14which 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 =(110)th 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:
Hence,
The maximum spacing is limited to 300mm.
Longitudinal reinforcement :
From ETABS results the maximum longitudinal reinforcement obtained is 13128 mm2
Since ,
We are using 25mm dia bars, the no. of bars required will be 13128490=28bars. 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 mm2. Hence, it is safe.
Now,
From the code the minimum vertical reinforcement required is
0.0025+0.01375(twLw)=0.00525,
i.e
0.00525â‹…300â‹…1500=2362.5mm2...............which is less than actually provided hence safe.
Now,
minimum vertical reinforcement to be provided in the boundary region =0.008â‹…1500â‹…300=3600mm2
The reinforcement provided for boundary element in right side = 4410mm2and
At left side =5390mm2Both are greater than the required value i.e 3600mm2 Hence safe.
In the web area the minimum reinforcement provided is
Horizontal reinforcement:
From Etabs results the shear reinforcement amount is obtained as 750mm2
So let us provide 8mm dia bars @ 100 mm c/c, then the provided area of reinforcement will be 1000mm2. 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=0.05â‹…Svâ‹…hâ‹…(fckfy)
Ash is the area of shear reinforcement required per link.
`S_v =` spaing between the shear reinforcement = 100mm
h= spacing between the extreme longitudinal reinforcement in the boundary element i.e 600-80 = 520mm
fck =25Nmm2and fy= 500Nmm2
Hence it is obtained as ;
Ash=0.05â‹…100â‹…520â‹…(25415)=156mm2this for one link
One link has 2 legs hence area of each leg=1562=78mm2
Hence,
We can choose 10mm dia bars for shear reinforcement in the boundary element width whose area is around 78.5mm2. This can be provided at the left side of the wall`
Now,
In the right side the boundary element width is`450mm^2`
So,
`h=450-80=370mm`.
Hence,
Ashrequirement will be
`A_(sh) = 0.05â‹…100â‹…370â‹…(25/415)=111mm^2` for each link.
Hence,
For one leg =`55mm^2`
Hence'
We can provide 8mm dia bars for shear reinforcement in the boundary element width whose area is around 50mm2
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:
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