natural period.....

[kantiman]

dear friends

i hv solve one RCC building in Etabs considering dynamic effects using response spectra method an i m getting natural period 2.37 sec an base shear 1319 kN. An i hv solved same problem manually also i got natural period 0.97 sec and base shear 1320 kN but , the problem s that natural period by manually an etabs r not matching plz some one help me i m attaching etab file also.
these r the details of building
Live load : 4.0 kN/m2 at typical floor
: 1.5 kN/m2 on terrace
Floor finish : 1.0 kN/m2
Water proofing : 2.0 kN/m2
Terrace finish : 1.0 kN/m2
Zone III : Z= 0.16
Earthquake load : As per IS-1893 (Part 1) - 2002
Depth of foundation below ground : 2.5 m
Type of soil : Type II, Medium as per IS:1893
Storey height : Typical floor: 5 m, GF: 3.4 m
Floors : G.F. + 5 upper floors.
Plinth level : 0.6 m
Walls : 230 mm thick brick masonry walls
only at periphery.

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[highcivil]

pls upload with e2k file. I can not see your project.

Sincerely...

[kantiman]

pls upload with e2k file. I can not see your project.

Sincerely...

thanks for reply....... can u tell me what s that e2k file.

[Grunf]

@kantiman: "natural period 2.37 sec an base shear 1319 kN" and/ or "natural period 0.97 sec and base shear 1320 kN" looks to suspicious to me... I have only experience with Eurocode and for two buildings similar to your described above that I have designed, I've got periods in interval from 0,30-0,45 sec and base shear 10 times your base shear (actually about 14000kN)

for prelimenary calculation of natural period use this two formulas:
1. T = 0,06 × Nf (where Nf = number of floors) in your case it is T = 0,06 × 5 = 0,30 sec
2. T = 0,022 × Hb (where Hb is height of the building from the ground level to the top of the building) in your case it is T = 0,022 × 5 × 3,4 = 0,374 sec

you should check the design for both periods - 0,30 sec and 0,374 sec as well.

concerning the base shear:
It depend on a lot of parameters (according to Eurocode) but in short I can tell you that you can get base shear force using this formula (just prelimenary and use with caution!):

Fb = ag × M
Where ag is ground acceleration in [m/s2], and M is mass of the complete building with live load in it times the corresponding factors in [kg] (e.g. according to EC (just prelimenary!) 100% of dead load + 30% of imposed live load)

PLEASE NOTE: above described formulas are only for evaluation/ prelimenary calculations!


You can find Eurocode 8 posted here:

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so you can do the check of the construction on your own...

My best regards.
G

[pontiac]

It seems there is a problem in your etabs model. In my opinion there may be column(s) and beam(s) that their nodes are not jointed ( I mean they are not working properly,load etc. is not transferred ) or at the base level your restraint definitions are not valid.But your findings quite strange because for there are many simple formulations in the litareture and codes to check first period and for such a building your results seems wrong.. For example up to 10 storey building you can use 0.1N (N=highest storey number)
for 6 floor 0.1*6=0.6 sec can be expected.
regards
pontiac

[highcivil]

thanks for reply....... can u tell me what s that e2k file.

I can use different version of Etabs so i can not open your documents. İf you save as e2k format (as picture below) anybody will open it.

[Badis]

dear friends

i hv solve one RCC building in Etabs considering dynamic effects using response spectra method an i m getting natural period 2.37 sec an base shear 1319 kN. An i hv solved same problem manually also i got natural period 0.97 sec and base shear 1320 kN but , the problem s that natural period by manually an etabs r not matching plz some one help me i m attaching etab file also.

Hi kantiman

the period that Etabs gave you is the analytical period based on eigenvalues ((K-w~2M)=0), while the period that you calculate manually is based on the approximate formula given by your seismic design code (I guess Ct*h~3/4) that usually gives conservative value.

I roughly checked your model and it seem that every thing is ok. your numerical period is too long may be because of the story height, notice also that you didn't take into account the stiffness of the brick masonry walls which may decrease the period.

but which is strange is that with two different periods (numerical and empirical) you have the same base shear?!

[chigozie]

kaniman wrote:-
...but , the problem s that natural period by manually an etabs r not matching plz some one help me i m attaching etab file also.
these r the details of building
Live load : 4.0 kN/m2 at typical floor
: 1.5 kN/m2 on terrace
Floor finish : 1.0 kN/m2
Water proofing : 2.0 kN/m2
Terrace finish : 1.0 kN/m2
Zone III : Z= 0.16
....
Comment:-

I think that you confused some of the data i.e. you were putting the wrong data at the wrong places e.g.:-
· Live load: 4.0 kN/m2 at typical floor: 1.5 kN/m2 on terrace INSTEAD of Live load: 1.5kN/m2 at typical floor: 4.0 kN/m2 on terrace. ( the live load for the terrace is usually greater than the live load for the typical floor and it is usually in the range of 4.0 kN/m2)
· Storey height: Typical floor: 5 m, GF: 3.4 m Floors: G.F. + 5 upper floors. INSTEAD of Storey height: Typical floor: 3.4m, GF (ground floor): 5m, Floors: G.F (ground floor). + 5 upper floors. (It seems strange that the upper floors would be higher than the ground floor.It is also strange that 5 floor each will be of height 5m).
· The weight for Water proofing material of 2.0 kN/m2 is too much (it should not be greater than 1.0 kN/m2. The floor finish of 1.0 kN/m2 also seems to be too much unless you have internal partitions that have not been counted for in your weight estimate. Be careful in the choosing the loads as overloading could lead to uneconomic structure and could result in wrong forces and wrong moment at the wrong place-thus unsafe design/structure.
If what I assumed is what you meant, then the fundamental period of the structures (T) could be calculated as CH^0.75 (H = height of structure = 2.5 + 5 + 5 x 3.4 = 24.5m) If you are using brick masonry walls
only at periphery, that means that your base shear is to be resisted by moment resisting frames (brick masonry walls do not contribute to the resistance and I assumed moment resisting frames are in reinforced concrete) as such, C in the above equation = 0.075 for which the period T = 0.075 x 24.5^0.75 = 0.82s
I am not conversant with the code that you are using and the nomenclatures. So, could you give written and technical description of your soil type, the exposure condition the peak ground acceleration and other relevant technical details (or their equivalent in Euro code) as I could not digest the information that you gave which is particular to the code that you are using.
Regards
Teddy

[zipatton]

Why not dear Rafik ? if we are on the stair of the spectrum then it is the same base shear !
Zipatt

[Badis]

Why not dear Rafik ? if we are on the stair of the spectrum then it is the same base shear !
Zipatt

Yes dear zipatton, you are right, for periods of the plateau region of the response spectrum we have same acceleration (and base shear). But between period of 0.97sec and 2.37sec we can not be in the plateau region.