04-23-2010, 11:42 AM
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
...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