Project 2_Design & Analysis of a Hostel building at Guwahati, Assam
Q1. ) In this project, participants will get to experience the entire design process from modelling, analysis, design, and preparation of construction drawings. The architectural drawings of a G+3 RCC building is provided. The following information on the building is to be used:
- Location – Guwahati, Assam
- Function – Hostel block in an academic institute


The walls will be made of AAC blocks. Other loadings are to be calculated as per the architectural drawings.
The buildings rests on type 3 soil. Piles of length 10 meters and 500 mm diameter with a load-carrying capacity of 280 KN is to be used.
The ground floor level will be at a height of 2 meters above the existing site level to avoid floodwater enter the building.
M25 grade of concrete with Fe500 for longitudinal reinforcement and Fe 415 for shear reinforcement is to be used.
The following design IS codes will be required for reference:
- IS 875 part 1
- IS 875 part 2
- IS 13920
- IS 1893
- IS 2911 part 2
- IS 456
- SP 34
Students will have to produce the following:
- Etabs model for the two blocks on either side of the expansion gap
- Foundation details illustrating the pile & pile cap reinforcement
- Tie beam details connecting the pile caps
- Column reinforcement
- Framing plan for typical floor
- Beam reinforcements for ground floor
AIM :- The structural model needs to be created and analysed using the specified building in ETABS 2018 in line with IS 1893 and IS 13920. To design the specifics for the pile cap and pile reinforcement, To create the ground floor frame plan, the column reinforcements, the tie beam linking the pile caps, and the ground floor beam detailing.
provided data:
AAC blocks are used to build the wall. Pile strength: 280 kN, M25 steel, Fe 500, and soil type 3.
INTRODUCTIONS:-
- Etabs is the name of the program that creates a structural model of a particular structure.
- The grid spacing values from the included structural designs should be noted first.
- After building the structural model, run it next.
- Take notes on the ETABS result numbers after you analyze the results.
- The SAFE 2016 software is then used to design the foundations using the model.
PROCEDURE:-
- Open the Etab software, then select File > File > New Model from the menu.
- A window for initializing the model will appear.
- Select the Indian codes by clicking the use built-in setting, then click "OK."
- Select the custom grid data, then select OK.
- As soon as the values specified in the plans are inserted, our model will be complete.
- The next step is to define the material. Go to define, select material properties, and then specify the concrete and rebar materials.
- Properties need to be defined in the subsequent step section. Add the beam and column specification by going to define and clicking on section properties.
- 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 .
- Describe the stair case slab and specify that it is 175 mm thick (ranges from 175-225 mm).
- Using the beam command, draw every beam.
- Use the column command to draw every column.
- The slabs command is used to draw slabs in the subsequent step.
- We now need to decide where the shear walls will be placed and how long they should be.
- According to Indian standards, the wall's length should be 5-2% in each direction.
- It is assumed that the shear wall is 250 mm thick.
- Go ahead and define the loads now.
- The loads are then added after choosing the beams and floors.
- 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.
- Hell for the proportion of jarring walls visit the base level, then select reaction When ex and ey values are available, divide them by two. A minimum of 90% should be the total base shear. Add the shear walls now in the proper directions.























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| 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
- 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 go 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
















