Modelling of 25 storey building with the specified properties using ETABS

Solution.1)

Aim: To model a 25 storey building on ETABS with structural properties as specified below. Ground floor or plinth floor level can be taken at a height of 1.5 metres from the base. Each of the successive 25 storeys has a storey height of 3 metres.

Structural Element Sizes

• Beams : 300 × 400 mm
• Columns : 800 × 800 mm
• Slab : 150 mm thick

Material Properties

• Grade of Concrete : M30
• Grade of Steel Reinforcement Bars : Fe 415

Loading

• Dead Load on beams from infill wall : 10 kN/m (Let us assume the same for roof floor in view of parapet)
• Live load on floor : 3 kN/m2 (Let us assume the same for roof floor)

a) Buildings are assumed to be pinned at the base

b) All columns in all storey are of the same size

c) All beams in all storeys are of the same size

d) Building is located on Soil type-II and in Zone-III. Assume other factors if required.

Carry out the following exercises:

• Check if the translational fundamental natural period of vibration in both the x and y direction is approximately: `T_(x1)` = 4.4 second `T_(y1)` = 4.2 second

This check will help confirm if modelling is done correctly. Some deviations from the above values is possible (± 2%). But any major deviation will suggest an error in modelling.

• Check if the following provision in Table 5 of IS 1893 (part 1) – 2016, on torsional irregularity satisfied:

The natural period corresponding to the torsional mode of oscillation must be less than the two fundamental translational modes of oscillation along each principal plan directions.

• Check if the two provisions of clause 7 in Table 6 (shown below) of IS 1893 (part 1) – 2016, regarding vertical irregularity is satisfied.

• Check if provision on inter-storey drift limitations in clause 7.11.1.1 satisfied. Please note that this check is for Serviceability Limit State and hence is for situation when stucture is subjected to unfactored seismic base shear

• 

Procedure:

• Open ETABS software.

• Click on New Model, a Model Initialization option will pop up, choose the options which are most appropriate. We have selected Indian Design Codes for Steel and Concrete Design as shown in below figure.

• Then a new window will pop up - New mode quick template,in this we have to input the grid dimensions and storey dimensions by choosing the custom grid spacing option as shown in below figure.

• Now, input the grid dimesions as given in the question as shown in the below figure.

• Similarly input storey dimensions.

• Hence, both grid and storey details have been filled.

• Next thing that we need to input is the material property. For this, go to define tab and select material properties and then go to add material option in that window.

• Hence, the Concrete, HYSD500 and HYSD415 materials are added. Now, click ok.

• Now, we need to add the frame members, i.e beams, columns and slabs.
  • Go to Define tab.
  • Section properties.
  • Frame sections (beams and columns)

• Frame properties window opens and in that go to add new material property.
• The reinforcement details and cover to the reinforcement is also provided.

• Similarly, add column to the list by providing the reinforcement details.

• Hence, the Beams and Columns are added to the list is as given below:

• Next we have to add the slab details.
  • For this go to define tab.
  • Section property then slab sections.

• Slab property data dialogue box opens.
• In that edit the slab properties.

• So all the properties of frame members have been added correctly, now we have to place the structural members.
• For this choose quick draw column option from the tools availabe at the left side of the screen and place the columns at all the grid intersections.
• Since columns extends throughout the height of the structure, so care should be taken to check whether columns are assigned to all storeys.

• Next, we have to place the beams in position, for this also quick draw beams option can be chosen.

• Now, we have to create the slabs, for this initially change the all storeys option to similar storeys which may exclude the base.
• Then, choose quick draw floor or wall option to easily draw the floor slabs.

• Next, we have to assign loads to the frame.
  • For this, go to define tab.
  • Then to load patterns.
  • Define load patterns dialogue box appears.
  • Then, apply the loads.

• Now, provide the mass source value to calculate the seismic weight.

• Next, we need to define the load combinations.
  • For this, go to define.
  • Then click on Load combination.
  • Provide the default load combination itself.

• Choose Concrete frame design and click ok.

• Select all the floors.
  • For this ,go to Select.
  • Object type then Floors.
  • Hence all the floors are selected.
  • Then go to assign.
  • Shell loads then Uniform.
  • Then apply 3kN/`(kN)/m^2`.
  • Apply and click on Ok.

• Select all the floors.
  • Select --> Select --> object type --> Beams.
  • Hence all the beams are selected.
• Then go to assign --> Frame loads --> distributed --> load pattern - Brick wall load -->load - 10kN/m -->apply --> ok.

• We have to keep the support as pinned, so for this select all joints by keeping in view the base storey.
• Then Go to assign --> joint --> restraint --> select pinned joint.

• No, go to analyse tab --> check model --> it is checked for any errors and warnings.

• Now, the model can be analysed.
  • For this, go to analysis.
  • Run analysis.
• Hence, analysis has been completed.
• Now, we need to check the details given in the question.
  • So, for this initially go to display tab.
  • Show tables.
  • Analysis result.
  • Structure output.
  • Modal information.
  • Modal periods and frequencies.

• Check if the translational fundamental natural period of vibration in both the x and y direction is as given in the question:

  • 1st mode corresponds to translation in y direction, hence Ty₁ = 2.116 s

  • 2nd mode corresponds to translation in x direction , hence Tx₁ = 1.986 s

• Check if the following provision in Table 5 of IS 1893 (part 1) – 2016, on torsional irregularity satisfied:The natural period corresponding to the torsional mode of oscillation must be less than the two fundamental translational modes of oscillation along each principal plan directions.
  • The natural period corresponding to two translational modes of oscillation is obtained as:
    • Tx₁ = 1.986 s
    • Ty₁ = 2.116 s
  • The natural period corresponding to the torsional mode of oscillation is obtained as:

    • Txy₁ = 1.656 s

  • Hence, it is verified that the natural period corresponding to the torsional mode of oscillation is less than the natural period corresponding to two translational modes of oscillation.
• Now, modal participation ratios have to be taken, for this go to display tab --> Show tables --> Analysis result --> Structure output --> Modal information --> Modal participating mass ratios.

• Check if the two provisions of clause 7 in Table 6 (shown below) of IS 1893 (part 1) – 2016, regarding vertical irregularity is satisfied.
  • Modal mass participation ratios of first three modes in X direction is obtained as obtained are:
  • • 1st = 0.216
    • 2nd = 0.7996
    • 3rd = 0
  • Summing these three we get, 0.216+0.7996+0 = 1.0156 ≈100% > 65% . Hence it satisfies the criteria i.e, the first three modes together contributes atleast 65% mass participation factor in each principal plan direction.
  • Now, the difference between X and Y time period is given by, 2.116-1.986 = 0.130 ≈ 6.14%. Hence, it does not satisfy the criteria of the code i.e the fundamental natural periods of the two buildings in the two principal plan directions are away from each other by at least 10% of the larger value.

Result:

The analysis of the 25 storey building is completed and the specified checks were also carried out. The structure satisfies all the codal provision except the second provisions of clause 7 in Table 6 of IS 1893 (part 1) – 2016, regarding vertical irregularity, which is given as the fundamental natural periods of the two buildings in the two principal plan directions are away from each other by at least 10% of the larger value.