Project 2_Design & Analysis of a Hostel building at Guwahati, Assam

PROJECT 2

NAME OF THE PROJECT: Design & Analysis of a Hostel building.

LOCATION OF THE PROJECT: Guwahati, Assam.

Aim:- To frame the structural design and analysis of a hostel building for the given architectural drawings provided.

Introduction:-

• Softwares uesd for the project may be as follows:-

1. Etabs- This is the software provided by CSI(Computers&Structures Inc.) which is basically used for framing of the superstructure,design and analysis of the beam and column members.
2. Safe:- This is the software provided by CSI(Computers&Structures Inc.) which is basically used for the design and analysis of the substructure which may includes slabs and foundation members.
3. Auto Cad:- This is the software which may be used here for the purpose of drafting the detailings of the reinforcement of superstructure or substructure.

• IS codes Shall be used for the project may be as follows,

1. IS 875 part 1
2. IS 875 part 2
3. IS 13920
4. IS 1893
5. IS 2911 part 2
6. IS 456
7. SP 34

The flow of the project may be in the way that are as follows

1. First the framing of the Superstructure model shall be prepared.
2. Analysis of the superstructure model.
3. Design of the superstructure model members.
4. Extracting the datas required for foundation/Substructure design.
5. Framing of the substructure.
6. Analysis and design of the substructure.
7. Extracting the reinforcement details.
8. Drafting the reinfocement details in the Auto Cad software.

The given Architectural drawings

Procedure:-

1. First the framing of the Superstructure model shall be prepared.

Open the E-tabs software and choose to create a new model.

• Creating the grid system with the reference taken from architectural drawings,

• Creating the Elevation of the structural system frame model with the reference taken from architectural drawings,

• Defining the section materials,

• Defining the concrete material

• Defining the rebar material for longitudnal reinforcement

• Defining the rebar material for Shear/Horizontal reinforcement

• Defining the section properties,

• In the preliminary assumptions of the section members the sections are defined as B400X550, C400X600,Slab150, Slab100 is also defined for providing at the mumty level,Dimensions are taken in mm.
• If the preliminary assumed dimensions satisfied the design its ggod,otherwise we will go ahead to modify the dimensions for the failed members and reanalyse the design checks.

• Defining the beam section,

• Defining the Column section,

• Defining the slab section,

• Assigning the section members.

• Assignment of column to the model.

• Assignment of beam to the model.

• Now,The assignment of slabs into the structural model,

• Now define Load patterns,

• Now, Asssigning the loads to the framing model.

• Preliminary Calculations

• Calculations of the Brickwall loads

Assume thickness of the wall=250 mm

Ht of the wall =3.45 m

Density of AAc block=6kn/m^3

Thickness of plastering both sides assumed to be=25mm

Density of plastering=8kn/m^3

Brickwall load = load imparted by the brickwork+load imparted by both sides plastering

Calculation of the Brickwall loads=height of wall X thickness of wall X density of AAC block+Height of plastering on both sides X density of plastering X thickness of plastering=3.45X0.25X6+0.025X2X8X3.45=6.55KN/m

• Extracting the live load values from IS 875-part (2)

For a Residential building the code of practice for live load are as follows,

All rooms And Kitchen=2 Kn/m^2

Toilet and Bathrooms=2 Kn/m^2

Terrace live load=1 Kn/m^2

Partition wall load=3.5 Kn/m

Corridoor,passsages,Stircase,store rooms,Balconies=2 Kn/m^2

• Assigning Frame loads:-

• Assigning Live loads:-

• Define mass source:-

• Define Modal case:-

• Define Response Spectrum Function:-

• Now,Go to load combinations to include the response spectrum model load cases as follows,

• Now,The Moment Resisting Frame is ready for performing analysis.

• Moment Resisting Frame model after analysis,

• Modal period and frequencies:-

• Modal participating mass ratios:-

• Story Drifts:-

• now,Perform the CONCRETE FRAME DESIGN:-

• Check for Beam Longitudnal Reinforcement:-

• Designing the beam longitudnal Reinforcement and shear reinforcement for maximum Ast value obtained.

Ast max for Longitudnal Reinforcement:-

Beam B34 Maximum Lonitudnal Beam Reinforcement requirement,

Maximum Ast requirement for shear reinforcing for beam B29.

Using the above reinforcement area results we have to design and detail the beam Reinforcement.

• As we know that Reinforcement has to be designed for the maximum Ast.
• The maximum value of Ast

Astmax at top= 1428mm2

Ast bottom R/f=1289 mm2

• Now, we have to adjust the reinforcement area with the available reinforcement diameter of bars in the market.
• For top R/f Ast=1428 mm2 let us provide 4-T20 bars and 2-T12 bars which sums up the total area of 1483 mm2,since it is more than the rerquired, so it satisfies the requirement ,hence ok.
• For bottom R/f Ast=1289 mm2
• let us provide 4T-20 and 1T-12 which sums up the total area of 1369 mm2,since the provided area of reinforcement is higher than the required so it satisfies the requirement hence,ok.
• While providing the reinforcement on the top,it is a good practice to cut the reinforcement in the middle portion but,at the bottom the reinforcement will start from one end and will end at the other end.
• For detailing the rebar in the auto cad development and anchorage length is required,

• Calculation of the anchorage length for the top reinforcements.

Formulae for the anchorage length as per IS 13920

Anchorage length=Ld+10db, Where Ld is development length and db is nothing but the diameter of the bar for which we are calculating the anchorage length.

Ld can be calculated as, Ld = φ σst/4τ bd

In the model,section properties arte provided are concrete of M30 grade while steel rebar for longitudnal reinforcement of HYSD415 are provided.

τ bd value from the IS 456:2000 ,we know=2.72

so,Ld for T-20 rebar,

Ld=((20x500)/(4x2.72))=

Ld=919 mm ,but as per IS 13290 add 10db in the development length so,

anchorage length=919+200=1119 mm.

similarly Ld for T12 rebar = 551mm, anchorage length=551+120=671 mm.

Now,calculations for providing the shear R/f

the depth of column as 500mm,

As per IS 13290,from the end support upto 2d of the beam confined shear reinforcement shall be provided.

=2d, d=500-25-20= 455mm,2d=2x455=910 mm

• Preliminary calculations for Shear reinforcement

The area of shear R/f from the etabs at the support is equal to 1531 mm2.

let's provide 2L-T8 for the shear R/f.

one round of 2L-T8 counts for 100 mm2,so,for 1531 mm2

Total numbers required are=1531/100=15.31, let's take=16 No.s

For spacing we have

Spacing=(910/15)=56.875 mm, let's consider 70 mm c/c

• So,for the Shear reinforcement we are arrived to provide 2L-T8@75mm C/C. upto 2d from the left support.
• As per Is 13290 spacing should be ≤d/4,8db or 100mm,whichever is lesser.
• so,75mm ≤d/4=455/4,8x20,100mm
• Here in our case it satisfies the codal provisions.

• Now,the shear R/f in the middle support,
• Ast=1114 mm2

let's provide 2L-T8 for the shear R/f.

one round of 2L-T8 counts for 100 mm2,so,for 1114 mm2

Total numbers required are=1114/100=12,

For spacing we have

Spacing=(1114/12)=92.83 mm, let's consider 110 mm c/c.

As per Is 13290 spacing should be ≤d/2.

• so,110mm ≤d/2=455/2=227.5 mm
• Here in our case it satisfies the codal provisions.

• Detailing of the beam reinforcement in Auto Cad Software:-

Column Reinforcement:-

1. Longitudinal Reinforcement

As obtained from the ETABS output, the maximum Ast required for column and the value for the same is of 4071mm^2. Hence, if we provide 8 bars of 25mm dia and 2 bars of 20mm dia.

Ast = 8x491 + 2x314 = 4553 > 4071 . Hence OK.

2. Shear Reinforcement

Provide 8mm dia Lateral Ties.

• Anchorage length calculations,

1. For 25 mm dia bars,

Formulae for the anchorage length as per IS 13920

Anchorage length=Ld+10db, Where Ld is development length and db is nothing but the diameter of the bar for which we are calculating the anchorage length.

Ld can be calculated as, Ld = φ σst/4τ bd

In the model,section properties arte provided are concrete of M30 grade while steel rebar for longitudnal reinforcement of HYSD415 are provided.

τ bd value from the IS 456:2000 ,we know=1.875 (for the deformed bars in compression)

so,Ld for T-32 rebar,

Ld=((25x500)/(4x1.875))

Ld=1667 mm.

Anchorage length=Ld+10db=1667+25*10=1917mm

Similarly for 20mm bars=1534 mm

• Special confining reinforcement from IS 13290-2016, clause 8.1

From there suggessions we have,

1. L0 should be not less than,

L0 is nothing but a particular distance from the top of the beam or bottom of the beam into the column upto which confined reinforcing is suggested by the codal provisions.

• larger column dimension which is 650 mm
• 1/6 of the clear span of the column. (whichever is greater)
• Shorter column dimension which is 450 mm.

So, L0 is considered to be 650mm

2.Confinement spacing nbot more than,

• 1/4 of smaller column dimension.=(1/4)x450=112.5mm
• 6 times dia of column of main bar=6x32=192 mm (whichever is lesser)
• 100mm

So, 100 mm is considered.

Note:-As we are getting the same area of shear reinforcement requirement all along the height of the building. It suggests that the shear reinforcement is not required but, IS code has suggested to provide the minimal shear reinfocements in that perspective so,

• Provide 8 mm lateral ties @200 mm C/C
• In the confined zone Provide 8 mm lateral ties @100mm C/C

Cover in the column is considered to be =40mm

and for the anchorage length in the foundation cover considered is=50 mm

Thus, using the above values the following detailed drawing was prepared in AutoCAD for the column height from building base upto the first slab.

Part B:-

Design and analysis of substructure:-

Required Data for Design and Analysis:-

Let's see for A1 Grid position,

Preliminary calculations

Unfactoreed load=287 Kn.

Dead load Breakup=287/2=143.5 Kn.

Live load Breakup=287/2=143.5 Kn.

Point Spring of Pile= A X E /L=

area of one pile=(πxd^2)/4=(πx500^2)/4=196349.5 mm^2

E=5000 X Fck^0.5=5000x30^0.5=27387N/mm2

L=8000

Point Spring of Pile= A X E /L=196349.5X27387/8000=672176 N/mm=0.672KN/m

• Rebar yield strength(Fy)=500 N/mm^2
• Rebar ultimate strength=575 N/mm^2
• Elastic modulus of M25 concrete= 25000 N/mm^2

Number of piles required= unfactored load with self weight addition/individual pile capacity=287X1.1/280=1.12 Nos.

To distribute evenly, Let's consider 2 piles for symmetric placement of piles under pile cap.

• Considering C/C distance between piles=3 X pile diameter=3x500=1500mm
• Edge distance from pile center= 1 X pile diameter=1X500=500mm.
• Assume pile cap depth=500mm.

Similarly,the no of piles required for other grid locations are also calculated and a reference model for the same is erected in the Auto-Cad software.

• Start the modelling in Safe software,

• Click to start with the new model and the model initialization is as follows,

• Defining the material properties-

• Now, Define the slab properties
• Here, we are required to define three kinds of slab properties, which are depicted in the snaps below

• Now, the geometric reference joints are drawn all along the grids and the Pile_Cap,Column,Pile_500 are assigned to there geometric locations as shown below

• Now,define the load combinations for ULS for 1.5 factor while SLS for without factored definition.

• Now, Assign the loading condition on the column.

• Select the central joint of column then go to assign,joint ,force, give the value of gravity load in negatve direction.
• Do it one by one for all the column joints.
• Apply, ok
• To view the assignment change the view from plan to 3D and can be viewd as follows,

• Assign the spring constant to all the pile joints.
• Defining the spring constant.

• Viewing the spring constant assignment in 3D view.

• Assign the strips along X and Y.

• Now, the model is ready to perform the analysis

Use the play button available above the model space to perform the analysis of the model,

After the analysis the model reflected in the model space as,

Check for settlement-

• The maximum displacement(mm),we have arrived here is far lesser than the maximum allowed from the code of practice that is 25 mm.

Punching Shear check,

As per the provision this value should not cross 1,if it crosses depicts that it has failed in punching shear check.

• Reinforcement Details of the model are as follows,

• Along Strip direction A

• Along Strip direction B