Challenge

Project 2_Analyze and Design the RC office building as per IS standard code in TEKLA STRUCTURAL DESIGNER

Aim:-

To model the RC structure analyze and design the model elements as per the IS codal provisions

procedure:

Open a new template in tekla and then go to settings from the home tab

Assign the design codes and units to be considered in the model

Define the structural default settings

Click on ok

step 1:-

To define the construction levels

Click on the construction levels from the model toolbar

Then add the levels in the dialogue box witha a given spacing

Click on ok

Step 2-

To draw horizontal and vertical grids

Click on the grids icoin from the model toolbar

Specify the first point and drag to draw the first grids

Then select on the parallel quick grids to draw the other grid lines with a given spacings

Similarly draw the vertical grids also

step 3:-

To assign the model elements like columns, beams and slabs

Go to manage property sets from the home tab

Click on new members > concrete column

Then define the size of the column and grade of concrete

Click on ok

Now go to model tab and select on the concrete column icon

Select the section from the properties browser

Click on the grid intersections to place the column

Similarly add the properties of the beams and slab using manage property sets and assign them to the model

Now define the properties of the slab and assign slab panels to the model

To add wall panels to the model

Select the wall panels from the model tab

Then assign them to the model

step 4:- Calculation of loads

Calculation of dead loads

Self weight of the slab is directly taken by the tekla model

Dead load of floor finishes

Assume the thickness of the floor finish as 75mm

Unit weight of the concrete is 24 kn/m^3 dead load due to floor finishes= 0.075 x 24 =1.8 kn/m^2

Brick wall loading

Assume the unit weight of the bricks as 20 kn/m^3

Thickness of the exterior wall is 230mm

Thickness of the interior walls is 115mm

Height of the first floor is 3.5m

Height of the second floor is 3.7m

Brick wall load on the ground floor beams(exterior walls) = 3.5x0.23 x 20=16.1 kn/m

Brick wall load on the ground floor beams (interior walls) = 3.5x0.15 x 20=10.5 kn/m

Brick wall load on the first floor

Beams(exterior walls) = 3.7x0.23 x 20-17.02 kn/m

brick wall load on the first floor beams (interior walls) = 3.7x0.15 x 20=11.1 kn/m

Roof loading is assumed as 1.5 kn/m^2 as per IS code provisions

Calculation of live load

Live load are obtained from the IS 875 part 2 based on the type of structure and occupancy

APPLICATION OF LIVE LOAD ON THE SLAB PANELS

Calculation of wind loads

Assume the basic wind speed as 50 m/s

Terrain category as 2

Design life of the structure as 50 years

Assume the structure to be class A

Application of wind loads to the model to define the load cases

Click on loads > load cases and then add the wind load combinations to the dialogue box

Click on ok

Then we need to create frames to assign the wind loads to the mode

Go to model and click on frames and select on the grid to create frames

Then select the corresponding frame to assign the wind loads

Seismic load

Go

To run the analysis

Firstly check with the modifaction factor

Then click on ist order linear analysis from the analyze tab

The analysis has been done but the results view doesnot display any of the combinations in my project because of which couldnot upload the images of deformations, shear force and bending moment

Now we can model the structure by model report

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