Challenge

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

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

Aim:

To run the analysis and design the shear walls for the given ETABS Model
To extract the values of longitudinal steel ratio, boundary element length and shear reinforcing ratio from the design results.
To provide the cross -sectional steel arrangement of shear wall from its case up to terrace, in compliance with provisions of IS13920-2016
To provide elevation drawings of reinforcement details of the shear wall from base to terrace.

Introduction:

A shear wall consists of a web and boundary elements at either side This boundary elements in shear walls are subjected to alternate compression and tension when it is subjected to lateral force (EQ), hence it is necessary to provide special reinforcement for the boundary elements.

IS13920-2016 provides guidelines on several aspects 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 longitudinal steel bar (clause 10.1.8)
Maximum spacing of horizontal and vertical reinforcement (clause 10.1.9)
Conditions for base of shear wall (clause 10.1.10)
Minimum reinforcement in RC shear wall (clause 10.1.6)
From clause 10.1.4, walls can be categorised as squat walls, intermediate walls andf slender walls as given below:
Now the 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, for this go to shear wall design.
Shear wall design go to view/revise preferences.
Now we need to start the design.
Once, the design is completed go to display design option.
Then we have checked for any failures in the design.
Now, we need to extract certain values from the design.
Go to display design info and
select pier longitudinal reinforcing.

Since, we have gathered the necessary information, now we need to draft the details of a RC wall in AutoCAD for the wall at All.

Shear wall detailing calculations:

Here, the total height of the wall is 21000mm and the width of the wall is 1500mm.
Hence, the ratio is obtained as 21000 /1500 = 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 = 1/10th 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:

Longitudinal reinforcement:

Since we are using 25mm dia bars, the no. of bars required will be 13128 /490 =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 13720mm². Hence, it is safe.
Now, from the code the minimum vertical reinforcement required is 0.0025+0.01375 (tw/Lw)
= 0.00525, l.e
0.00525x300x1500 = 2362.5mm2 which is less than actually provided hence safe.

Now, 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 = 4410 mm²and
At left side = 5390mm Both are greater than the required value i.e 3600mm². Hence safe.

In the web area the minimum reinforcement provided is 0.0025x1500x300-1125mm² and the provided amount of reinforcement in web area is 3920mm².Hence it is safe.

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.

From clause 10.4.4, the shear reinforcement required in the boundary element can be obtained from the equation

Ash is the area of shear reinforcement required per link.
Sv-spacing between the shear reinforcement = 100mm
h-spacing between the extreme longitudinal reinforcement in the boundary element i.e 600-80=520mm
fck = 25N/mm² and fy = 500N/mm².
Hence it is obtained as Ash = 0.05.100 520 . 25/415 = 156mm² this for one link
One link has 2 legs hence area of each leg=156/2=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 450mm² so, h=450 80-370mm. Hence, Ash requirement will be
Ash = 0.05. 100 . 370 . 25/415 = 11 mm² for each link. Hence, for one leg = 55mm²
Hence we can provide 8mm dia bars for shear reinforcement in the boundary element width whose area is around 50mm². This can be provided at the right side of the wall.
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:

The values of longitudinal steel ratio, boundary element length and shear reinforcing ratio from the design results have been extracted successfully.
The cross - sectional steel arrangement of shear wall from its base up to terrace, in compliance with provisions of IS13920-2016 has been drafted after appropriate calculations.
The elevation drawings of reinforcement details of the shear wall from base to terrace has also been drafted.

Ship work like this
in your own course.