Challenge

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

Analysis & Design of RCC Shear walls in the model using ETABS.

Aim:

  • To Run the analysis and design the RCC shear walls in the Etabs model.

Introduction:

Column reinforcement according to IS: 13920 code book clause 7

Longitudinal Reinforcement:

  • Minimum dimension of a column shall not be less than 20db largest diameter of longitudinal reinforcement bar in the beam which is passing through column and 300mm.
  • Aspect ratio i.e., B / D should not be less than 0.45
  • In lap splices
  1. Spacing of link shall not exceed 100mm.
  2. Lap splices should not be less than development length of largest reinforcement bar.
  • Lap splices shall be provided at center of column height, not within joint or not within a distance of 2d from joint.
  1. Not more than 50% of area of steel bars shall be spliced at any one section.
  2. Lap splices shall not be used for bars of diameter more than 32mm.

Procedure:

Step: -1

  • Open the Etabs file.
  • After the analysis click concrete frame design

Step: -2

  • Need to check there is any failures occurs
  • Click shear wall design tool.
  • Select display info and choose PM failure.

Step: -3

  • Click any wall and check the details.
  • Here there is no failure occurs all the values of reinforcement ratio is within 4%.

Step: -4

Shear Reinforcing.

  • Go to concrete frame design.
  • Click display design info.
  • Choose shear reinforcing.

Step: -5

Boundary element width.

  • Go to concrete frame design.
  • Click display design info.
  • Choose Pier edge or boundary zone width.

Step: -6

Longitudinal Reinforcing.

  • Go to concrete frame design.
  • Click display design info.
  • Choose pier longitudinal reinforcing.

Storey Assignment tab to refer elevation of Shear wall.

Calculations.

Base floor.

Tw = 300 mm

Lw = 1500 mm

Boundary element = 600 mm for left.

Boundary element = 450 mm for right.

web = 450 mm centre

Height of entire wall, hw = 21000 mm.

Clause 10.1.6 (table wall)

Minimum reinforcement in RC shear wall.

When the total height of the wall divided by the length is more than 2.

From IS: 13920-2016, clause 10.1.4

* Slender walls: hw/Lw >2

: 21000/1500 = 14

Hence, It is a slender walls, we need to calculate.

Horizontal Min reinforcement = `rho`h min

Vertical B.E reinforcement = `rho`v BE min.

Minimum web reinforcement = `rho`v web min.

Net reinforcement = `rho`v net min.

From IS: 13920-2016, clause 10.1.6

Table-1: Minimum Reinforcement in RC Shear Walls.

Therefore,

* `rho`h min = 0.0025+0.5 ((hw/Lw)-2) (`rho`h - 0.0025)

`rho`h = 750/(300x1000) = 0.0025, hw=21000mm, Lw=1500mm

`rho`h min = 0.0025+0.5x((21000/15000)-2) (0.0025-0.0025))

`rho`h min = 0.0025 which is same as 750 mm2

* Minimum vertical reinforcement within boundary element Table-1

For left boundary element;

`rho`v BE min = 0.0080 x (600 x 300) = 1440

For Right boundary element;

`rho`v BE min = 0.0080 x (450 x 300) = 1080

*`rho`v web min = 0.0025

*`rho`v net min = 0.0025 + 0.01375 (tw/Lw)

=0.0025 + 0.01375 (300/1500)

=0.00525 x (1500x300)

= 2362.5 mm2

For Longitudinal Reinforcement:

`rho`w = Lw -2tw`

= 1500 -2(600)

= 300

From Etabs,

The required rebar area, Av = 13128 mm2

Minimum Vertical Reinforcement, `rho`v

= 13128 / (c.s area of steel)

= 13128 / (1500x300)

= 0.029

`rho`v = 0.029 > 0.0080

From IS 13920-2016, clause 10.4.4

Ash = 0.05 Sv h (fck/fy)

Minimum no:of bars =8

Ash =`pi/4 d2`

= `pi/4 (8)2`

Ash = 50.27 mm2

Therefore, 50.27 = 0.05x100xh(25/500)

50.27 = 0.25 h

h = 201.09

Use `phi` = 25 mm

No:of bars required = (13128 ) / `pi/4`(25)2`

= 13128 / 490.625

= 28 bars.

So, Providing 25mm dia 13 no:of bars on each facing & 2 numbers at each side each.

Detailing guideline for Shear reinforcement:

Confining reinforcement within boundary element

From IS 13920-2016, clause 10.4.4

Ash = 0.05 Sv h (fck/fy)

By providing 2 legged 8mm dia @ 100mm c/c

Ash = 2 x `pi/4 (8)2`/ 0.10

Ash = 1005 > 750 mm2

Similarly, for Web.

By providing 2 legged 8mm dia @ 125mm c/c

Ash = 2 x `pi/4 (8)2`/ 0.25

Ash = 805 > 750 mm2

Elevation & Cross section of Shear wall using AutoCAD:

RESULT:

  • The Shear wall is designed successfully in detail from the ETABS software.
  • Provided the details of longitudinal & Shear reinforcement for the shear wall (A-11) successfully.
  • And also provided the Shear wall cross sectional & Elevation details using the AutoCAD software.
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