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

AIM:

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

PROCEDURE AND DETAILS:

Maximum longitudinal reinforcement at the base

Consistent Shear Demand Throughout the Wall Height

Boundary Width Data

Flexural Information

Shear Info

Boundary Width Info

• Conduct the Design Calculations in Accordance with the IS-13920 and Execute the Ductile Detailing of Shear Wall Using AutoCAD

1. The next step is to go ahead and place the reinforcement bars in the shear wall with proper calculations as per the clauses mentioned in the IS-13920 codebook.
2. So, the exact procedure for ductile detailing of reinforcement bars in the shear wall is explained below and thus the wall design is completed by preparing the good for construction drawings that can be sent to the site.

1. Section Classification

The first thing we need to check before proceeding with the detailing is if the geometrical properties of our wall are within the limits of IS-13920 or not.

THICKNESS

As per clause 10.1.2, the minimum allowable shear wall thickness is 150mm or more. And in our case, we have a 300mm thick shear wall. Our section is safe.

LENGTH TO THICKNESS RATIO

As per clause 10.1.3, the minimum allowable length to thickness ratio must be greater than or equal to 4.

So, in our case, we have 1500mm as the length and 300mm thickness. Therefore 1500/3000 = 5 > 4 Hence OK.

SHEAR WALL TYPE

As per clause 10.1.4, there are primarily three types of shear walls based on their geometrical properties. And the minimum reinforcement requirement also differs for these three types.

1. Squat Walls: If hw/lw < 1
2. Intermediate Walls: If 1 < hw/lw < 2
3. Slender Walls: If hw/lw > 2

So, for our case, hw/lw = 21000/1500 = 14 > 2

Hence OK.

REINFORCEMENT LAYERS

As per clause 10.1.7, reinforcement shall be provided along with both faces of the wall if the wall thickness exceeds 200mm.

So, as we have a 300mm thick shear wall, we have to lay down the rebars along with both the faces of the wall.

LARGEST DIAMETER OF THE REBAR

As per clause 10.1.8, the largest diameter of the longitudinal bar of the shear wall shall not exceed 1/10th of its thickness.

So, 1/10 x 300 = 30mm.

Therefore, we can't use any bar exceeding the 30mm dia. Thus, we have to use the 25mm dia as the largest bar for our project.

2. Ductile Detailing Calculations

Following are the steps to be carried out while calculating the ductile detailing of reinforcement bars for shear walls from the ETABS output.

SECTION DIVISION

As we all know that the entire five-floor shear wall can't be detailed in one go. We have to dive the wall along its height based on the reinforcement demand on each floor.

So, following the ETABS output, we have grouped the wall into two sections. The first section runs from the base uptill the first floor and the second one follows up till the terrace level.

BOUNDARY ELEMENT ALLOCATION

Based on the software output, the boundary width requirement at the base is 600mm.

So, it is only viable to design the entire length for section 1 as the boundary element. Because we only have 300mm as the web width which is very less. Hence for section 1 from the base to the first floor, we have to design the entire wall as the boundary element confining the rebars as per the IS-13920.

But as we go to the upper floors, the boundary element demand reduces to 300mm. So, we can adapt the boundary element of 300mm on either side of the wall for section 2 running from the first floor to the terrace level.

A) DETAILING FROM BASE TO THE FIRST FLOOR (Section 1)

Shear Demand:

As already obtained, we have a consistent shear demand of 750sq.mm per 1000mm height.

So, provide 8mmdia. shear links.

No. of bars = Ast/Ash = 750/100 = 7.5 = 8 bars.

Spacing = h/no. of bars = 1000/8 = 125mm

Therefore, provide 8mm dia shear links @ 125mm c/c as the shear reinforcement for the entire wall.

However, the vertical spacing for shear links within the boundary element shall be governed by clause 10.4.4 which states that it should be 100mm.

Longitudinal Rebar Demand: (Entire length is to be designed as the boundary element)

Required Logitudinal reinforcement ratio = 0.0292

Ag = 1500 x 300 = 450000 sq.mm

Therefore,

Asc = 0.0292 x 450000 = 13140 sq.mm

Also,

Ash = 0.05 Sv h fck/fy

As we have used 8mm dia stirrups@125mm c/c, Ash = 50 sq.mm

Sv = 100mm for boundary element As per Clause 10.4.4

Fck = 25

Fy = 500

Therefore, h = 200mm. Meaning we cannot provide more than 200mm spacing of longitudinal bars in the boundary element which is the entire length of the wall till the first floor.

So, Minimum no. of bars = 1500-40/200 = 8 bars.

Now, to cover the demand of 13140 sq. mm,

Provide 28 numbers of T25 bars.

Therefire, Ast provided = 490 x 28 = 13720 > 13140. Hence OK.

Also the spacing = 120mm < 200mm required. Hence OK.

So, provide 13 bars on one face making 26 on both sides. Also, provide 2 bars along the shorter span reducing the c/c spacing to 150mm. Hence OK.

The detailed GFC drawing for the above calculations is inserted below:

Reinforcement Detailing from the Base upto the First Floor

B) DETAILING FROM THE FIRST FLOOR TILL THE TERRACE LEVEL (Section 2)

1) Design of the Web Steel:

Web width = 900mm

Minimum reinforcement ratio for the web as per IS-13920 = 0.0025

Adopting 0.0030 for the safer calculations.

Ag = 900 x 300 = 270000

Asc = 0.0030 x 270000 = 810 sq.mm

Provide 6 numbers of T16 bars = 6 x 201 = 1206 > 210 Hence OK.

2) Design of Boundary Element

Minimum Reinforcement Ratio as per the code = 0.008

Boundary Width = 300 mm

Ash = 0.05 Sv h fck/fy

As we have used 8mm dia stirrups@125mm c/c, Ash = 50 sq.mm

Sv = 100mm for boundary element As per Clause 10.4.4

Fck = 25

Fy = 500

Therefore, h = 200mm.

Reinforcement required = total Ast - Ast taken by the Web = 4896 - 810 = 4086 sq.mm

Now, there are two boundary elements. So, Ast required for one BE = 4086/2 = 2043 sq.mm

Thus, provde 7 numbers of T25 bars

Ast provided = 7 x 490 = 2940 > 2043 Hence OK.

Reinforcement ratio check:

Ratio = Ast/bwxtw = 2940/300x300 = 0.032 > 0.008 Hence OK.

NOTE: We have carried 14 T25 bars from the floor below satisfying the 50% lapping condition at one go.

The detailed drawing of the above calculations is as follows:

Reinforcement Detailing from the Third Floor Upto the Terrace Level

ELEVATION DRAWING OF THE SHEAR WALL:

• From Base Till First Floor

• From First Floor Till Third Floor

• From Third Floor Till Terrace Level

FINAL OUTPUT

RESULT:

• Analysis & Design of RCC shear walls in the model using ETABS challenge is completed.