Calculation of Seismic Load generation by Rayleigh frequency.

Question: I have modeled a sample cantilever columncommunities.bentley.com/.../Rayleigh.STDn and applied a seismic joint weight 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 fro this difference?

Answer:

The simple 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).

So there is a difference in two scenario. The difference arises because of the fact that in the latter case, the Rayleigh time frequency is calculated from the pseudo static lateral forces distributed from the base shear and the base shear is lesser than the total seismic weight.

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

After analysis , these 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)

=> (T₁/T₂)=√(y₂/y₁)

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).