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

AIM:To Create and Analyse the given building in ETABS 2018, 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 data:

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

INTRODUCTION:

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

PROCEDURE:

1. in the first step open the etabs software go to file and click on new model.
2. Model initialisation window will be appeared.
3. Click on the use built in setting and select the indian codes and click ok.
4. Click on the custom grid data and click ok.
5. Now the values as per the given plans are inserted  and our model will be ready.
6. 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.
7. 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.
8. 450 mm x 600 mm beam size is provided for the all the beams.
9. 450 mm x 600 mm beams are provided for the beams at periphery so as to increase the torsional resistance of the building
10. Moment of inertia for beams and columns should be reduced by 0.35% and 0.4% respectively.
11. 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 .
12. Define the stair case slab and provide its thickness as 175 mm( varies from 175-225mm)
13. Draw all the beams using beam command.
14. Draw all the columns using column command
15. In the next step slabs are drawn using slabs command
16. Now we have to decide the location of shear walls and how much should be the length of the shear wall.
17. As per Indian standards the length of the wall should be  5-2% in each direction.
18. The thickness of shear wall is assumed to be 250 mm.
19. Now go to define and define the loads.
20. Next select the beams and floors and add the loads.

   21.AAC load of 3.5 kN/m since the aac blocks are used.

22. The live loads values are got from Is 1875 part 2 for residential buildings.

   24. As per table 1 of IS 875 part 2

   25.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

26. As per clause  number is 875 part2 reduction factor should be applied on the imposed loads.
27. Live load at the roof floor needs to be reduced hence 2 kN/m is applied.
28. Floor finish on all the floors 2 kN/m  is applied.
29. Seismic loads should be define so as per IS 1893 Table no. 8 importance factor should be 1.2
30. Since the building.
31. 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

33. Now define the load combinations
34. 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.
35. And select all the floors and apply diaphragm D1.
36. 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.
37. 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.
38. 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.
39. now go to define Load cases and click on Add new case and Add RSX and Rsy.
40. 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.
41. 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.
42. Now check the model if there are any warning correct it and run the model.

Results  and discussion,

1. Go to display then click show tables and then structural data click ok.
2. Click on show tables , select the Analysis results structure output click ok.
3. 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.
4. 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%.
5. Go to concrete frame design and select the start design or check . and note down the values 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.

4. 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

 

.

Beam 14	Right end	Centre	Left End
Bottom  Reinforcement	672	605	609
required
Dia of bars	20	20	20
Number of bars	2.140127389	1.926751592	1.939490446
Bottom Reinforcement	675	605	605
Provided at Top	2.149681529	1.926751592	1.926751592
	1#24 +2#24 extra is provide	2#20  is provide	2#20  is provide

	COLUMN DESIGN
	COLUMNS ALONG THE  GRID 1B	Column dimension	450*600
1	Longitudinal Reinforcment calculation
a)	Minimum Reinforcment	0.8% of Gross cross sectional Area	2160	mm^2
	As per clause no.26.5.3.1 of Is 456 2000
b)	Maximum Reinforcment	6% of Gross cross sectional Area	15600	mm^2
	As per clause no.26.5.3.1 of Is 456 2000
c)	Maximum Ast along 3 A		4739	mm^2
	As per clause no.26.5.3.1 of Is 456 2000
	Dia of the longitudinal Reinforcment should  12 mm or more
	There for using  16 mm dia bars
	Dia of bar	24
	Area of  1 bar	452.448
	number of bars	10.47413183
	There fore using 8 number of 28 mm dia bars	Ast	4524.48	mm^2
	remaining St	Ast	214.52	mm^2
	2 number #12 extra longtidudinal Reinforcements needs to be provided	Ast	628	mm^2
	There fore  total longitudinal reinforcment provided
	10 no. of 24 mm dia bars	4525
	2no. Of 12 mm dia bars	628
	Ast provided	5152.48	mm^2
	The spacing between the longitudinal bars should not exceed 300mm.
2	Shear Reinforcment calculation
a)	Minimum dia for shear Reinforcment for 32 mm dia	8 mm
	As per clause no 7.4.2 of IS 13290- 2016
b)	Maximum Shear reinforcment from the Etabs out put		665	mm^2
c)	There for using 8 mm dia bars			mm^2
	Dia of bar	8
	Area of  1 bar	50.272
	Special confining reinforcement clause no 8 of IS 13290 2016
	The special cofining reinforcement shall be provided at lo.
	lo should  not be less tham
	larger lateral dimension of the column	600
	1/6 of the clear span of member	554.1666667
	450 mm	450
	There fore lo = 600 mm
	The spacing of shear reinforcment along lo should not be more than
a	1/4 of minimum dimension of beam/column
	minimum dimension of column	150	mm
	1/4th of minimum dimension of column	112.5	mm
b	6 times the dia of smallest longitudinal reinforcement bar
	Dia of smallest longitudinal reinforcement bar	12	mm
	6 times the dia of smallest longitudinal reinforcement bar	72	mm
c	100 mm	100	mm
	There fore  for special confining reinfrocemnet #8mm @100  mm c/cis provided in the lo zone.
	Transverses Reinforcment calculation for the zone upto h/4.
		hc in m	Beam depth	h
	Height of the column  of Plinth level column	1	0.6	0.4
	Height of the column  ofGF column	3.45	0.6	2.85
	Height of the column  of FF column	3.45	0.6	2.85
	Height of the column  of SF column	3.45	0.6	2.85
	Height of the column  of TF column	3.45	0.6	2.85
		h	h/4	approximate
	Plinth level	0.4	0.1	200 mm
	GF	2.85	0.7125	800 mm
	FF	2.85	0.7125	800 mm
	SF	2.85	0.7125	800 mm
	TF	2.85	0.7125	800 mm
	Transverse reinforcement in the splice Zone as per clause number 7.4.of IS 13290 2016
	As per the clause nuo. 7.4.1 of IS 13290 2016 the  ends should have 135  hook with an
	extension of 6d of bars.
	d	32	mm
	6d	192	mm
	As per the clause nuo. 7.4.2 of IS 13290 2016
a	8 mm dia   bars are used in stirrups if the  dia of longitudinal bars is less than or equal to 32 mm
b	The maximum  c/c spacing of stirrups is 300 mm
c	a cross tie should be provided if the dimension of the column exceeds 300 mm.
	There fore upto h/4 #8mm @300  mm c/cis provided in the lo zone.
	As per the clause no. 7.3.2.1 of IS 13290 2016
	100 mm spacng should be provided within this splicing zone
	There fore in the splice zone #8mm @100  mm c/cis provided in the lo zone.
	For the columns terminating in the footing
	as per clause nummber 8.2 of IS 13290 2016.
	special confinement reinforcement should be provided upto 300 mm or larger length inside the footing.
	COLUMNS ALONG THE  GRID 3B	Column dimension	450*650
1	Longitudinal Reinforcment calculation
a)	Minimum Reinforcment	0.8% of Gross cross sectional Area	2340	mm^2
	As per clause no.26.5.3.1 of Is 456 2000
b)	Maximum Reinforcment	6% of Gross cross sectional Area	17550	mm^2
	As per clause no.26.5.3.1 of Is 456 2000
c)	Maximum Ast along 3 A		5022	mm^2
	As per clause no.26.5.3.1 of Is 456 2000
	Dia of the longitudinal Reinforcment should  12 mm or more
	There for using  16 mm dia bars
	Dia of bar	28
	Area of  1 bar	615.832
	number of bars	8.154821445
	There fore using 8 number of 32 mm dia bars	Ast	4926.656
	Remaning Ast	Ast	95.344
	There for using  16 mm dia bars
	Dia of bar	16
	Area of  1 bar	201.088
	number of bars	0.474140675
	There fore using 2 number of 20 mm dia bars	Ast	402.176
	There fore  total longitudinal reinforcment provided
	8 no. of 28 mm dia bars
	2no. Of 16 mm dia bars
	Ast provided	5328.832	mm^2
	The spacing between the longitudinal bars should not exceed 300mm.
2	Shear Reinforcment calculation
a)	Minimum dia for shear Reinforcment for 32 mm dia	8 mm
	As per clause no 7.4.2 of IS 13290- 2016
b)	Maximum Shear reinforcment from the Etabs out put		720.48	mm^2
c)	There for using 8 mm dia bars			mm^2
	Dia of bar	8
	Area of  1 bar	50.272
	Special confining reinforcement clause no 8 of IS 13290 2016
	The special cofining reinforcement shall be provided at lo.
	lo should  not be less tham
	larger lateral dimension of the column	650mm
	1/6 of the clear span of member	554.1666667
	450 mm	450
	There fore lo = 650 mm
	The spacing of shear reinforcment along lo should not be more than
a	1/4 of minimum dimension of beam/column
	minimum dimension of column	450	mm
	1/4th of minimum dimension of column	112.5	mm
b	6 times the dia of smallest longitudinal reinforcement bar
	Dia of smallest longitudinal reinforcement bar	20	mm
	6 times the dia of smallest longitudinal reinforcement bar	120	mm
c	100 mm	100	mm
	There fore  for special confining reinfrocemnet #8mm @100  mm c/cis provided in the lo zone.
	Transverses Reinforcment calculation for the zone upto h/4.
		hc in m	Beam depth	h
	Height of the column  of Plinth level column	1.5	0.5	1
	Height of the column  ofGF column	4	0.5	3.5
	Height of the column  of FF column	3.45	0.5	2.95
	Height of the column  of SF column	3.45	0.5	2.95
	Height of the column  of TF column	3.45	0.5	2.95
		3.45	0.5	2.95
		h	h/4	approximate
	Plinth level	1	0.25	300 mm
	GF	3.5	0.875	900 mm
	FF	2.95	0.7375	900 mm
	SF	2.95	0.7375	900 mm
	TF	2.95	0.7375	900 mm
	Transverse reinforcement in the splice Zone as per clause number 7.4.of IS 13290 2016
	As per the clause nuo. 7.4.1 of IS 13290 2016 the  ends should have 135  hook with an
	extension of 6d of bars.
	d	32	mm
	6d	192	mm
	As per the clause nuo. 7.4.2 of IS 13290 2016
a	8 mm dia   bars are used in stirrups if the  dia of longitudinal bars is less than or equal to 32 mm
b	The maximum  c/c spacing of stirrups is 300 mm
c	a cross tie should be provided if the dimension of the column exceeds 300 mm.
	There fore upto h/4 #8mm @300  mm c/cis provided in the lo zone.
	As per the clause no. 7.3.2.1 of IS 13290 2016
	100 mm spacng should be provided within this splicing zone
	There fore in the splice zone #8mm @100  mm c/cis provided in the lo zone.
	For the columns terminating in the footing
	as per clause nummber 8.2 of IS 13290 2016.
	special confinement reinforcement should be provided upto 300 mm or larger length inside the footing.

Importing the software  Safe 2016.

1. Go to file click on export and select the safe v12.fk.
2. 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,
3. Goto SAFE 2016 and fo to file click import and select the file and save the file as FDB.
4. 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:

1. Number of piles=Total Axial load/pile capacity=(1.5DL+1.5LL)/280= numbers
2. 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.

3. 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.
4. 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.
5. Select the values .
6. Go to define section properties and define pile caps ,pile Dia , and column.
7. Add new , defined pile cap pile 500 as Mat , and  Column as Stiff.
8. 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.
9. 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.
10. Select the pile caps and changed the rebar material as Fe 500 .
11. Next go to draw and draw point and a window will open up there draw the points where piles are to be drawn.
12. 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.
13. Now we have to define all the piles  as spring for that go to define.
14. select soil spring properties add new property there click on compression only and add the calculation value of K=EA?L.
15. Select all the piles and go to assign and select support data and select spring properties and assign it.
16. Go to define and select the define beam and define the tie beams whose thickness should be as of pile caps.
17. Check weather the rebar material is Fe 500 under design preferences.
18. Go to define and define the load combinations sls and uls.
19. Go to design and design preferences.
20. Save the model and run the Analysis.
21. Deformed shapes showed that deformations are negligible.
22. Assign add the design strips.
23. Run the analysis.
24. Deformed shape all the values are less than 25 mm hence ok.
25. Now go to display and click on the punching shear the values we have go are less than 1 hence ok.
26. Now go to display and select slab design.
27. All the reinforcement details will be available
28. 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.

29. 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.