Applies To |
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Product(s): |
STAAD.Pro | ||

Version(s): |
ALL | ||

Environment: |
ALL | ||

Area: |
Analysis | ||

Subarea: |
Rayleigh Frequency calculation | ||

Original Author: |
Biswatosh Purkayastha, Bentley Technical Support Group | ||

Question:

I have modeled a sample cantilever column and applied a seismic joint weight of 10 KN at the tip but after performing the seismic analysis per IBC 2012 code, I am getting the Rayleigh Time period (T) calculated by Staad as 1.795 sec for seismic load generation, whereas on invoking the CALCULATE RAYLIEGH FREQUENCY in the seismic load case, Staad reports different Rayleigh Time period of 1.26 sec.

What is the reason for this difference?

Answer:

In case of Static seismic analysis, Staad needs the fundamental natural frequency or the natural time period to determine the design spectral acceleration which is a multiplication factor of Seismic Weight.

This fundamental time period is determined by the maximum of the approximate method as outlined in the clause 12.8.2 of IBC 2012 code or the Rayleigh frequency which is internally determined by Staad if the user has not specified natural frequency by PX and PZ parameters.

Now as there is no user input in the model, Staad has to determine the Rayleigh frequency internally before determining the base shear.

For determining the Rayleigh frequency, Staad needs the loading information to know the shape function but before determining the base shear Staad could not determine the pseudo static lateral loading information, so the question is which loading is considered by Staad to calculate the Rayleigh frequency.

The answer is Staad takes the predefined Seismic Weight information before applying them laterally at the defined nodes to determine the shape function and the corresponding Rayleigh frequency.

However, in another instance where you have specified the instruction under IBC load case to calculate Rayleigh frequency by the command CALCULATE RAYLIEGH FREQUENCY, Staad has already determined the base shear and hence the program already has the information of the pseudo static lateral load. So, this time Staad determines the Rayleigh frequency from the new shape function corresponding to the new lateral loading information.

In your case, the equivalent lateral load (pseudo static lateral ) is lesser than the corresponding seismic weight as the base shear is lesser than the total seismic weight and hence the Rayleigh frequencies in two cases vary.

Cantilever column has the seismic weight only defined at the tip with the magnitude of 10 KN. Now, while computing the Rayleigh frequency internally before determining the base shear, the seismic weight of 10 KN is assumed to be acting laterally at the same node.

Staad determines the shape function which resembles the fundamental mode shape and then the Rayleigh frequency and the corresponding Time Period of (T=1.795 sec).

Again, the Rayleigh Time period calculated against the user instructed CALCULATE RAYLEIGHT FREQUENCY is (1/0.78831 = 1.26 sec).

Invoke a command PRINT LOAD DATA just after the PERFORM ANLAYSIS command to know the pseudo lateral forces determined by Staad.

After analysis , this lateral force value at the tip of the column is reported as 4.992 KN.

So, let us create another load case (Load case -2) and apply the lateral load same as the pseudo static lateral forces and then invoke CALCULATE RAYLEIGHT FREQUENCY command.

After performing the analysis, the Rayleigh frequency calculated by the program is 0.788 and the corresponding time period of (1/0.788 = 1.26 sec) which is same as the time period reported in the seismic load case.

So, in the nutshell, if we see the basic equation of the Rayleigh frequency calculation, it interprets that higher the applied loading , more the displacement (y) and higher the Rayleigh Time period.

=>T=(1/ω) ∝ √(y)

=> (T_{1}/T_{2})=√(y_{2}/y_{1})

So if the internal calculation of Rayleigh time period by Staad against Load 10 KN for the seismic load generation is 1.795 then the Rayleigh Time period against the load 4.99 KN is √[(4.99/10)x1.795]= 1.26 sec ( considering a linear elastic analysis where displacement is proportional to applied load).