Challenge

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

SShivam Patel16 May 2023

AIM:-To Create and Analyse the given building in ETABS 2012, the structural model has to be prepared in accordance with IS 1893 and as per IS 13920. To design the pile cap and pile reinforcement details,Tie beam connecting the pile caps, to sketch the column reinforcements, to produce the framing plan and beam detailing for ground floor.

Given datas:-

The wall are made of AAC blocks Pile capacity -280kN,M25 Concrete Fe 500 steel, Soil type 3.

INTRODUCTIONS:-

Etabs is the software which is being used to create a structural model of given building.
In the first step note down the values of grid spacing from the given structural plans.
Next Create the Structural model and run the Model
Analyse the result and note down the result values from ETABS.
Next is to export the model to SAFE 2016 software in order to design the foundations

PROCEDURE:-

in the first step open the etabs software go to file and click on new model.
Model initialisation window will be appeared.
Click on the use built in setting and select the indian codes and click ok.
Click on the custom grid data and click ok.
Now the values as per the given plans are inserted and our model will be ready.
In the next step material has to be defined. Go to define and click on material properties and add the concrete and rebar material specification.
In the next step section properties has to be defined. Go to define and click on section properties and add the beam and column specification.
450 mm x 600 mm beam size is provided for the all the beams.
450 mm x 600 mm beams are provided for the beams at periphery so as to increase the torsional resistance of the building
Moment of inertia for beams and columns should be reduced by 0.35% and 0.4% respectively.
In the next step define the slab properties by going in to define and slab sections. Add the slab property as While defining the slab for Stair case it should be One way .
Define the stair case slab and provide its thickness as 175 mm( varies from 175-225mm)
Draw all the beams using beam command.
Draw all the columns using column command
In the next step slabs are drawn using slabs command
Now we have to decide the location of shear walls and how much should be the length of the shear wall.
As per Indian standards the length of the wall should be 5-2% in each direction.
The thickness of shear wall is assumed to be 250 mm.
Now go to define and define the loads.
Next select the beams and floors and add the loads.

AAC load of 3.5 kN/m since the aac blocks are used.
The live loads values are got from Is 1875 part 2 for residential buildings.

As per table 1 of IS 875 part 2

For hostel buildings

Living Rooms,Bedrooms and dormitories	2 kN/m^2
Kitchen and Laundries	3 kN/m^2
Billiards and public lounges	3 kN/m^2
Store rooms	5 kN/m^2
Dining rooms,cafeterias and restaurant	4 kN/m^2
Office rooms	2.5 kN/m^2
Bath and toilets	2 kN/m^2
Corridors, passages, staircase	3 kN/m^2
Balconies	4 kN/m^2

As per clause number is 875 part2 reduction factor should be applied on the imposed loads.
Live load at the roof floor needs to be reduced hence 2 kN/m is applied.
Floor finish on all the floors 2 kN/m is applied.
Seismic loads should be define so as per IS 1893 Table no. 8 importance factor should be 1.2
Since the building.
Time period is calculated as per clause number 7.6.2 of IS 1893

T=0.09h/sqrtd

Tx=(0.09*16.85)/(67.7)^0.5=0.184 Seconds. Or Ty=0.09*16.85/(13.3)^0.5=0.415 sec

Now define the load combinations
In the next step we have to define the mass source, for that go to define and click on mass source and add on the new , click on specified load data , in the mass options click on include lateral mass and lump lateral mass at story levels,in the load pattern add multiplier 0.25 for LL and 1 for all the dead and super dead loads.
And select all the floors and apply diaphragm D1
In the next step we have to define for the modal cases for that go to define and click on modal cases click add new model cases select eigen and put the number of modes shape required.
now go to define function and click on response spectrum and add anew function in accordance with IS 1893-2016 check all the values click ok.
Name the function as IS function,Response spectrum IS function IS 1893-2016 tab will appear check all the values Seismic zone factor soil type,importance factor click on linear x linear y click ok.
now go to define Load cases and click on Add new case and Add RSX and Rsy
select the response spectrum click on Add select the function as IS function and U1 for Rx, and U 2 for Rsy and IS function
Go to define click on load combinations and click on default design combo,click on create frame design and click on convert to design combo click ok.
Now check the model if there are any warning correct it and run the model.

Results and discussion

Go to display then click show tables and then structural data click ok.
Click on show tables , select the Analysis results structure output click ok.
Select the base reaction and check the eqx eqy vales , Rx and Ry value see which values are larger if the rX RY values are lesser than the Eqx Eqy unlock the model and then go to define load case and delete the Rx RY cases and again select the load combinations and define that again.
Go to modal mass participation ration and check the values for U1 U2 U3 values sume of these UX Uy UZ for the first 3 modes it should be greater than 65%.
Go to concrete frame design and select the start design or check . and note down the va;ues of reinforcement details.

2.select modal participating ratios we have got that at mode 7 we have got above 90% so its ok.

3.now select the base reactions check that Eqx Eqy values should be same and Rsx and Rsy values also should be atleast equal to Eqx and Eqy respectively. if not then just take the rsx values and divide it by Eq values whatever the values will get that values needs to be multiplied in the load cases rsx tab.

check for the share of shear walls go to base level and click on reaction Ex and ey values will be available take these values and divide it be Total base shear which should be greater than 90%. If now add the shear walls in appropriate directions

Importing the software Safe 2022

Go to file click on export and select the safe v12.fk.
A tab will appear then click on the story as base and select the loads and loads from above click on the combination click ok,
Goto SAFE 2016 and fo to file click import and select the file and save the file as FDB.
Go to dispay click on the show loads select dead above and love above the values will be appeared.

Given data:-

Column of size:- 450mm *600 mm & 600mm*450m m

pile has a load carrying capacity of 280kN

Pile length-16 m and dia of pile -500 mm

SBC-150kN/m^2

Solution:

Number of piles=Total Axial load/pile capacity=(1.5DL+1.5LL)/280= numbers
Piles in the springs are defined as spring so for that the spring constant

K=EA/L

E=5000(fck)^0.5=5000(25)^0.5=25000 N/mm^2

Area of pile =22/7*d^2=3.142*(500)^2/4=196375m^2

L-10000m

K=25000*0.7855/16=306835.9N/mm==419.

Spacing of the piles as per clause number 6.6.2 of IS 2911 the spacing between the piles =2.5dia of the piles =2.5*500=1250mm

Now in the next step we have to define the material properties for go to define and define the material properties of Concrete and Reinforcement properties Fe500 & Fe 415.
Select the values .
Go to define section properties and define pile caps ,pile Dia , and column
Add new , defined pile cap pile 500 as Mat , and Column as Stiff.
Click on the quick draw around the plots a window will open there insert the columns sizes and these columns to their specified positions. Columns should be stated as stiff material.
In the next step pile caps needs to be drawn for that go to Quick draw slab command select pile caps and add it. Piles and pile cap should be stated as stiff material.
Select the pile caps and changed the rebar material as Fe 500 .
Next go to draw and draw point and a window will open up there draw the points where piles are to be drawn.
Now again select the quick dfaw around the pints there select the property circular slab add piles dia in this case 0.5m and add to the points. Select these piles and again check for the rebar materials.
Now we have to define all the piles as spring for that go to define
select soil spring properties add new property there click on compression only and add the calculation value of K=EA?L.
Select all the piles and go to assign and select support data and select spring properties and assign it.
Go to define and select the define beam and define the tie beams whose thickness should be as of pile caps.
Check weather the rebar material is Fe 500 under design preferences
Go to define and define the load combinations sls and uls
Go to design and design preferences
Save the model and run the Analysis
Deformed shapes showed that deformations are negligible
Assign add the design strips
Run the analysis
Deformed shape all the values are less than 25 mm hence ok.
Now go to display and click on the punching shear the values we have go are less than 1 hence ok.
Now go to display and select slab design.
All the reinforcement details will be available
For X DIRECTION MIDDLE STRIP

maximum Reinforcement was

For Top:- 29834 mm^2/m=29834/3.3=9041 mm^2=28 nos. 20 mm dia are provided

For Bottom-13326mm^2/m=4038 mm^2=13 nos. 20 mm dia are provided

For layer Y direction

maximum Reinforcement was

For Top:- 19926= 3831mm^2/m=12 nos of 20 mm dia are provided

For Bottom-22784mm^2/m=4398 mm^2=14 nos 20 mm dia are provided

conditions in f2k format for footing design using SAFE software.

Go to show joints from display option

Show joint force

First create a service load combination for getting the combined and acceptable reaction value to design number of pile requirements for each support

Then select all the joints and assign pin support to each support for getting support reaction

Run the analysis to get the support reaction value and after the analysis

Finally, I have to unlock the model and remove all the pinned supports to no support condition before modeling further.

As on the basis of those above reactions I can crete number of piles-

For the A4 & A1 grid pile-

Axial load = 547 KN

Length of piles 10m

Diameter of pile = 500mm

Pile capacity = 280KN

No of piles = 547/280 = 1.95 = Assume 2 piles

Grade of concrete is M25

Area of each pile = = π4�4X D^2 = 196349.5mm2

E = 5000fck">√fck= 25000

Spring constant (k) = EA/L = (25000 x 196349.5)/10000 = 490873.75 N/mm = 490873.75 Kn/m

Take center to center distance between piles= 2 X Pile diameter(500mm) = 1000mm, edge spacing = 500mm

Assume pile cap depth = 900mm

Length = 2000m and Width = 1000mm

I have done the same calculation for rest of the joints to calculate the number of piles under the pile cap

For the A3 & A2 grid pile cap-

Axial load = 692 KN

Pile capacity = 280KN

No of piles = 692/280 = 3.33 = Assume 4 piles

For the B4 & B1 grid pile cap-

Axial load = 771 KN

Pile capacity = 280KN

No of piles = 771/280 = 2.75 = Assume 4 piles

For the B3 & B2 grid pile cap-

Axial load = 995 KN

Pile capacity = 280KN

No of piles = 995/280 = 3.55 = Assume 4 piles

For the C4 & C1 grid pile cap-

Axial load = 961 KN

Pile capacity = 280KN

No of piles = 961/280 = 3.43 = Assume 4 piles

For the C3 & C2 grid pile cap-

Axial load = 1249 KN

Pile capacity = 280KN

No of piles = 1249/280 = 4.46 = Assume 6 piles

For the D1 & D4 grid pile cap-

Axial load = 1180 KN

Pile capacity = 280KN

No of piles = 1180/280 = 4.21 = Assume 6 piles

For the D3 & D2 grid pile cap-

Axial load = 1542 KN

Pile capacity = 280KN

No of piles = 1542/280 = 5.5 = Assume 6 piles

For the E4 grid pile cap-

Axial load = 1091 KN

Pile capacity = 280KN

No of piles = 1091/280 = 3.89 = Assume 4 piles

For the E1 grid pile cap-

Axial load = 994 KN

Pile capacity = 280KN

No of piles = 994/280 = 3.55 = Assume 4 piles

For the E3 grid pile cap-

Axial load = 1381 KN

Pile capacity = 280KN

No of piles = 1381/280 = 4.93 = Assume 6 piles

For the E2 grid pile cap-

Axial load = 1302 KN

Pile capacity = 280KN

No of piles = 1302/280 = 4.65 = Assume 6 piles

For the F4 grid pile cap-

Axial load = 850 KN

Pile capacity = 280KN

No of piles = 850/280 = 3.03 = Assume 4 piles

For the F3 grid pile cap-

Axial load = 1062 KN

Pile capacity = 280KN

No of piles = 1062/280 = 3.79 = Assume 4 piles

For the F2 grid pile cap-

Axial load = 985 KN

Pile capacity = 280KN

No of piles = 985/280 = 3.52 = Assume 4 piles

For the F1 grid pile cap-

Axial load = 758 KN

Pile capacity = 280KN

No of piles = 758/280 = 2.70 = Assume 4 piles

For the G4 grid pile cap-

Axial load = 710 KN

Pile capacity = 280KN

No of piles = 710/280 = 2.53 = Assume 4 piles

For the G3 grid pile cap-

Axial load = 912 KN

Pile capacity = 280KN

No of piles = 912/280 = 3.26 = Assume 4 piles

For the G2 grid pile cap-

Axial load = 908 KN

Pile capacity = 280KN

No of piles = /280 = 3.24 = Assume 4 piles

For the G1 grid pile cap-

Axial load = 706 KN

Pile capacity = 280KN

No of piles = 706/280 = 2.52 = Assume 4 piles

For the H4 & H1 grid pile cap-

Axial load = 496 KN

Pile capacity = 280KN

No of piles = 496/280 = 1.77 = Assume 2 piles

For the H3 & H2 grid pile cap-

Axial load = 623 KN

Pile capacity = 280KN

No of piles = 623/280 = 2.22 = Assume 4 piles

For the I4 grid pile cap-

Axial load = 626 KN

Pile capacity = 280KN

No of piles = 626/280 = 2.23 = Assume 4 piles

For the I3 grid pile cap-

Axial load = 760 KN

Pile capacity = 280KN

No of piles = 760/280 = 2.71 = Assume 4 piles

For the I2 grid pile cap-

Axial load = 681 KN

Pile capacity = 280KN

No of piles = 681/280 = 2.43 = Assume 4 piles

For the I1 grid pile cap-

Axial load = 534 KN

Pile capacity = 280KN

No of piles = 534/280 = 1.90 = Assume 2 piles

For the J4 grid pile cap-

Axial load = 1080 KN

Pile capacity = 280KN

No of piles = 1080/280 = 3.86 = Assume 4 piles

For the J3 grid pile cap-

Axial load = 1367 KN

Pile capacity = 280KN

No of piles = 1367/280 = 4.88 = Assume 6 piles

For the J2 grid pile cap-

Axial load = 1289 KN

Pile capacity = 280KN

No of piles = 1289/280 = 4.61 = Assume 6 piles

For the J1 grid pile cap-

Axial load = 985 KN

Pile capacity = 280KN

No of piles = 985/280 = 3.52 = Assume 4 piles

For the K4 grid pile cap-

Axial load = 944 KN

Pile capacity = 280KN

No of piles = 944/280 = 3.37 = Assume 4 piles

For the K3 & K2 grid pile cap-

Axial load = 1227 KN

Pile capacity = 280KN

No of piles = 1227/280 = 4.38 = Assume 6 piles

For the K1 grid pile cap-

Axial load = 943 KN

Pile capacity = 280KN

No of piles = 943/280 = 3.37 = Assume 4 piles

For the L4 & L1 grid pile cap-

Axial load = 960 KN

Pile capacity = 280KN

No of piles = 960/280 = 3.89 = Assume 4 piles

For the L3 & L2 grid pile cap-

Axial load = 1247 KN

Pile capacity = 280KN

No of piles = 1247/280 = 4.45 = Assume 6 piles

For the M4 & M1 grid pile cap-

Axial load = 1009 KN

Pile capacity = 280KN

No of piles = 1009/280 = 3.60 = Assume 4 piles

For the M3 & M2 grid pile cap-

Axial load = 1316 KN

Pile capacity = 280KN

No of piles = 1316/280 = 4.7 = Assume 6 piles

For the N4 & N1 grid pile cap-

Axial load = 544 KN

Pile capacity = 280KN

No of piles = 544/280 = 1.94 = Assume 2 piles

For the N3 & N2 grid pile cap-

Axial load = 691 KN

Pile capacity = 280KN

No of piles = 691/280 = 2.46 = Assume 4 piles

Define materials

Define slab properties.

After that using draw point option drop the points for footing for each column.

Now, go to draw option and draft points on footing for the piles to be developed.
Next we need to provide pile, using the above used option and material property as pile500.