Projefct 1_Comparative study of different storey buildings for Seismic forces

PROJECT 1:

AIM: Comparative study of different storey buildings for Seismic forces

The table shows different properties of different buildings

CREATION OF MODELS

BUILDING A

• Number of storeys=2
• Number of Bays in X direction=4
• Number of bays in Y direction=3
• size of column=400mmx400mm
• size of beam=300mmx400mm
• grade of concrete=M30
• Grade of steel reinforcement bars=Fe415
• Height of floors=3m
• height of plinth=1.5m
• Slab thickness=150mm

STEP1-

• Open Etabs->select the type of IS codes you use->click OK
• Enter the custom grids plan
• enter the number of grids and their spacing

STEP2

• Enter the story data according to the requirements
• plint 1.5m and 2 storeys of 3m-3m

• The Builing structure is formed according to the data given

STEP3

• Go to Define->material properties
• Now we define properties of materials
• Add new material->We use property if M30 and Fe415

STEP4

• Now we define frame properties->define ->section properties->frame sections
• We add column and beam sections

STEP5

• Now Slab property will be added
• go to define->section property->slab sections

STEP6

• Now assign the beams,columns and slabs to the building
• using the quick draw property
• use quick draw column->Select the property as C400x400
• drag the cursor from one corner to other diagonally

• similarly select qucikdraq beams->select B300x400 from the property
• drag the cursor diagonally to apply ot all beam

• similarly for quick draw beam/floor

• Now go to the base->select the base points
• Assign->Joints->restraints
• JOints will be assigned

STEP7

• Now Loads will be assigned->Define->load patterns
• assign dead,brickwalls,lives and seismic loads
• Modify lateral loads of the seismic forces

STEP8

• Assigning live load on the floors and walls
• Go to select->object types->select floors
• Go to assign ->shell loads->uniform load
• Give live load of 3KN

• Assigning brickwall load on the beams
• Go to select ->object types->beams(all stories)
• enter the load o 10Kn

STEP9

• Now we enter the mass source
• Define->mass source
• live dead,brickwall and live=0.25 for live load on floors<= 3

• Now add load combinations
• define->load combinations->add default combination->concrete frame designs->rewrite

• Now we add diaphragm
• Select all the floors
• Assign->Shell->Diaphragm
• select D1 combinations and click OK

STEP10

• Go to analyze->check model
• In this case no error message showed up
• Now run the model ,press F5
• Click on deformed shapes and view the shapes on different cases

STEP11

• checking thr table for th results of the structure
• Display->show tables
• analysis results->structure output->modal information->modal mass participating ratios

Building A

Builidng B

BUilidng C

Column X oriented

BUilidng D

• Column Y Axis oriented
• go to view panel->set view options
• Uncheck the floors &beams
• Select ->object types->Columns
• Assign->frame->local axes
• and set the angle to 90degree to rotate the column

Building E

Building F

• copy save the E building
• rename it and save as F buildign file
• Go to dislplay options->uncheck the floors and beams
• Select ->properties->frame sections
• select the C400x400 column //It'll select only the C400x400 column
• Go to assign->frame ->section property
• select the C600x600 //It'll replace the Previous assigned column with C600x600
• cehck for all live and walls loads
• run the check for errors and run the analysis

Building G

• copy the F file and open it
• save as new fil->add extra 15 storeys to it
• View ->buiding view limits->set the top story and lower storey where column is to be changes
• set display units->uncheck beams and columns to hide them from view
• select the columns ->assign the 400x400,600x600,800x800 columns accordin the floors required
• analysi the result

Building H

• copy and save the G building file ->rename it
• got to section properties->create a column of 800x800
• select ->object types->columns
• Assign->frames->section property->800x800 columns
• analyze the frame and run it

TESTING FACTOR INFLUENCING NATURAL PERIOD OF BUILDING

Comparison1- EFFECT OF STIFFNESS ON T

• FOR BUILDING E & F
• time period for building E is 1.731sec and of F is 1.735sec
• due to change in the size of columns along the height of E,there is signinficant difference in time period of building E
• Less column size along the height causes the reduction in the stiffness of the columns ,And we know the More the stiffness the less the time period.
• The building F column provides more stiffness compared to E dur to the same column size along the height;therefore this sgnificant time difference can be seen
• FOR BUILDING G & H
• G time period is 4.17sec and H time period is 4.416sec
• The building H has more stiffness because of Same 800x800 column along full height wheres as the G building columns sizes varies along with height.

BOTH F AND H BUILDING ARE STIFFER THAN E & G ,BUT ALSO HAS MORE MASS ,THEREFORE THE NATURAL TIME PERIOD INCREASES

Building J Imposed load 10%more than that of H

• Open the H building file->save it as J building file
• Now select->object types->floors
• Assign->shell loads->Uniform loads
• give live load 10% more than that of H building ;i.e. 3.3KN/m^2
• Select the beams and assign wall loads of 10
• save the file->check model->run analysis

Building K Imosed load 20%more than H building

• Open the H building file->save it as K building file
• Now select->object types->floors
• Assign->shell loads->Uniform loads
• give live load 10% more than that of H building ;i.e. 3.6KN/m^2
• Select the beams and assign wall loads of 10
• save the file->check model->run analysis

EFFECT OF MASS ON T

• Time period of H is 4.416 sec
• time period of J is 4.425 sec
• time period of K is 4.435 sec
• From the above result it is proved that on increasing the mass of the buildign floor the time period increases
• Building H has 3KN/m^2 of mass on each floor, J has 10% more and K has 20% more than H
• With column having same stiffness but with the mass increasing ,makes it flexible,which causes the increse in time period H < J < K

Comparison3 -EFFECT OF BUILDING HEIGHT ON T

• Building A has 2 storeys and time period of 0.5 sec
• Building B has 5 storeys and time period of 1.036 sec
• Building F has 10 storeys and time period of 1.735 sec
• Buildign H has 25 storeys and time period of 4.416 sec
• As the height increases the,Its mass increases ,but the overall stiffness decreases therefore the results shows the increase in time period of taller buildigns.
• The fundamental natural period of: 25 storeys=3.14sec,10storeys=1.35sec,5 storeys=0.89sec and 2 storeys =0.45sec

Comparison4 -EFFECT OF COLUMN ORIENTATION ON T

• Building B has square columns and time period of 1.036 sec
• Building C has rectangular columns and time period of 1.145 sec
• Building D has rectangular columns and time period of 1.12 sec
• FOR RECTANGULAR COLUMNS ,Lateral stiffness along longer direction is more,therefore the natural period of building along the longer direction of column is less.
• Building C has X oriented column therefore the time period is more on Y direction i.e. 1.145 sec and on X direction is always less that Y
• Building D has Y oriented column therefore the time period is more on X direction i.e. 1.12 sec and on Y direction is always less that X