Project 1

 

 

 

 

 

Development of structural system & framing plan of a Sample Precast building (G+ 6) in Bhopal, India as per Indian Structural codes.

• Collaboration of course content from week 01 - Week 06
• Understanding of architectural layouts of the building
• Creating the load path of the various forces to foundation
• Developing the structural scheme (Precast) for the building as per architectural plans
• Preparing structural framing plans of the building along with the structural sections
• Load calculations for all the forces acting on the building. 
• Calculation for seismic and wind forces
• Identifying and dealing the irregularities of the building
• Preparing the design basis report of the building design

Answer:

 

Aim: To Develop structural system & framing plan of a Sample Precast building (G+ 6) in Bhopal, India as per Indian Structural codes.

 

1.Preparing the design basis report of the building design:

Design Basis Report consists the follwing

 

a.Cover page with Over view

b.Introduction of the Building

Building Description:Sample Precast building (G+ 6) 

location:Bhopal, India

Building Dimensions: 20m*8.21m

Building Height: 24m

Building usage: Residential purpose

Building Construction type: Precast construction

 

c.Standards used

Codes used :

• IS 875 part 1 : Dead loads
• IS 875 part 2 : Live loads
• IS 875 part 4 : Wind loads
• IS 875 part 5 : Special loads and load combinations
• IS 13920       : Ductile design of Buildings
• IS 1893         : Earthquake resistant design of Structures
• IS 456           : Plain and Reinforced concrete structures

 

d.Loads

 

Dead load

Super Imposed load

Live load 

Seismic load 

Wind load

Load combinations

 

e.Structural system and stability 

f.Calculation for element design

g.Calculation for connection design

 

Procedure:

 

2.Understanding of architectural layouts of the building and Strutural plan is drawn accordingly:

 

Select the Indian Design parameters

 

 

 

Go to options > Dispaly units

This helps to change the units into Inches as per the Architectural Drawings

 

 

 

 

 

 

 

Grid system is created as per the Given plan in the Auto CAD drawing

 

 

 

 

 

 

 

 

 

 

 

Define the Material Properties

 

 

 

 

 

 

 

Material M40 Concrete is created

 

 

 

 

 

 

 

Material M35 Concrete is created

 

 

 

 

 

 

Material Fe 500 is created

 

 

 

 

 

 

 

 

Material Fe 415 is created

 

 

 

 

 

 

 

 

 

 

 

View of all materials

 

 

 

 

 

 

 

 

 

 

 

 

StoryData is created as per the AutoCAD Drawing

 

 

 

 

 

 

 

 

 

 

Frame Data Column500*500 mm is assumed with M 40 Concrete Material

 

 

 

 

 

 

 

Frame Data Beam 230*500 mm is assumed with M 35 Concrete Material

 

 

 

 

 

 

 

Frame Data Beam 230*600 mm is assumed with M 35 Concrete Material

 

 

 

 

 

 

 

 

 

Properties of Column and Beams are shown

 

 

 

 

 

 

 

Pinned connections are made for

1.Column-Beam connections

2.Beam slab connections

There should be moment Release for Precast Members

 

 

 

 

 

 

Slab Property Data is shown here

 

 

 

 

 

 

 

 

 

Wall Property data is created

 

 

 

Grid view of Structural Plan from the Architectural plans

 

 

 

 

Replicate the floors for the other storeys

Using Edit > Replicate

 

 

 

 

 

 

 

 

 

 

 

 

 

 

3.Develope the structural scheme (Precast) for the building as per architectural plans

4.Preparing structural framing plans of the building along with the structural sections

View of the Structural Frame in 3D view

 

 

 

 

 

 

Elevation View

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Assignment of joint Restriant

Fixed supports the Foundation

 

 

5.Creating the load path of the various forces to foundation

Load is transferred from

a.Slabs and all floors to Beams 

b.Beams to columns

c.Columns to the foundation

 

 

Define the Load Patterns for Various type of loads as shown

 

 

Seismic load patterns are added as per IS 1893-2016

 

 

 

Wind load Patterns are added as per IS 875-2015

 

 

 

 

 

 

6.Load calculations for all the forces acting on the building. 

7.Calculation for seismic and wind forces

 

As per the code IS 875 Part I Dead load and SuperImposed loads are calculated:

 

Dead loads:

Assume unit weight of Precast concrete having Reinforcement = 25 KN/m3

Concrete doesn't have reinforcement , so unit weight of screed = 24 KN/m3

 

Assume Total Depth of slab D = 250mm

(Taken in our calculation also)

Dead load = 0.25*25 = 6.25KN/m2

 

 

 

Super Imposed Dead loads:

Assume 230 mm thick wall 

Floor Height = 3.0 m     (118.11 inches = 3m)

Beam size = 0.23 m*0.5 m

Assume standard brick having density = 18 KN/m3

Now Height of wall = Floor Height - Beam Depth

                            = 3.0 - 0.5 = 2.5m

Total SuperImposed Load = Height of wall * Thickness of wall * Unit weight of bricks with Plastering

                                     = 2.5 * 0.23 * 18 = 10.35 KN/m

 

 

 

IS 875 Part 2

Live Loads 

Living Room and Bedrooms = 3KN/m2

Kitchen Room = 3 KN/m2

Dining Room = 4 KN/m2

Toilet and Bath rooms = 2 KN/m2

Corridors and Passage  = 3 KN/m2

Store Room = 5 KN/m2

Balconies = 4 KN/m2

 

 

 

 

 

 Consider Live load = KN/m2

Considering only for Gravity loading and no lateral loading for preliminary dimensions

Load combination = 1.5DL + 1.5LL

                          = 1.5*(6.25+10.35) + 1.5*5 = 32.4 KN/m2

Design load = 32.4 KN/m2

 

 

Earthquake loads IS 1893:2016:

Consider Zone II as per the Map in IS code 1893 for Bhopal (from fig 1 IS 1893)   

Seismic Zone factor Z = 0.10 Table 2 of IS 1893

Consider OMRF for the given Building

Based on IS 1893 for Ordinary Moment Resisting Frames we have 

Response Reduction Factor R = 3 From Table 9 IS 1893

Assume site Type = Type 2

Importance Factor of the Building I = 1 (from Table 8 IS 1893)

Moment Resisting Frames with Masonry Infills clause 7.6.2 c     

 

Height of the Building = 944.8819 inches = 24 m

Sa/g = Design Acceleration coefficient for Soil types     (Clause 6.4.2  (a) IS 1893) and Table 2

For use in Equivalent static method

Assume Medium stiff soils Sa/g = 1.36/T 

Clause 7.6.2 (c) Buildings with RC structural walls have Time period

T = 0.09h/d^0.5 

h = height of Building = 24 m 

d = Base Dimension of the Building at the Plinth level along the considered direction of the Earthquake

Considering in both directions Direction X = 323 inches = 8.21 m

                                           Direction Y = 787.25inches = 20 m

Ta1 = 0.09*24/8.21^0.5

Ta1 = 0.753 = 0.76

Ta2 = 0.09*24/20^0.5

Ta2 = 0.483 = 0.49

 

Now Sa/g = 1.36/T 

For direction 1 Sa/g = 1.36/0.76 = 1.79 

For direction 2 Sa/g = 1.36/0.49 = 2.78

 

Now For direction X

Ah = (Z/2)*(I/R)*(Sa/g)

Ah = (0.10/2)*(1/3)*1.79 = 0.023

Now For direction Y

Ah = (0.10/2)*(1/3)*2.78 = 0.047

 

 

Base shear = Vb = Ah*w

Vb = 2.3% of Seismic weight for Direction X

Vb = 4.7% of Seismic weight for Direction Y

 

 

Wind load Calculation

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

8.Identifying and dealing the irregularities of the building:

 

Irregularities in the Plan

In the Structural grid plan it is observed that there are both Verical and Horizontal ,Torsional Irregularities in the Plan.

Vertical Irregularities:

Soft Storey at the Bottom Stilt Floor is seen as per the Architectural plan.

So to strengt the Stilt floor it necessary to provide Shear walls or Strength the Columns by Capacity Design.

This helps in the reducing the Torsion of the whole building during earthquake.

Horizontal Irregularities:

The plan shows so may Open spaces in the Architectural drawing which causes Irregularities in the Building.

Torsional irregularities:

The combination of both the Horzontal and vertical Irrgularties should be greater than Torsional Irregularity to find a structure to be stable.