Vibrations - Rayleigh-Ritz Method Admissible Functions

AI Thread Summary
Choosing an admissible function for a fixed-fixed nonuniform bar involves extending a single function to a family of functions that meet boundary conditions. The proposed function φi(x) = 1 - cos(2πix/L) is appropriate, where i represents the mode number. When calculating stiffness and mass coefficients, the indices i and j should correspond to the mode numbers used in the assumed modes method. For N=4, the indices i and j can be any combination of mode numbers from 1 to 4, and their specific pairing does not affect the outcome as long as they are within the defined range. Understanding these relationships is crucial for accurate modeling in vibration analysis.
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Hi everyone,

I'm having a bit of difficulty choosing an admissible function for a fixed-fixed nonuniform bar.

I chose the function φ(x) = 1 - cos(2πx/L).

But when solving for the the stiffness and mass coefficients:

kij = ∫EA(x)φiφjdx

mij = ∫ρ(x)φiφjdx,

I am not sure where I should have the "i" and "j" in my function.

In an example for a fixed-free beam, the function that was given was:

φ(x) = sin(πx/2L), which was changed to φi(x) = sin[((2i-1)πx)/2L].

Should I choose my function to be φi(x) = 1 - cos(2πix/L)? And why?
 
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Your choice for phi(x) defined only a single function. Your process requires a whole family of functions (i=1,2,3,...), each of which satisfy the boundary conditions. Try to use your original idea, but extend it to a family.
 
Thank you for the quick response Dr. D!

So in that case, could I choose a function (like before):

φi(x) = 1 - cos(2πix/L), i = 1,2,3,...
φj(x) = 1 - cos(2πjx/L), j = 1,2,3,...

Where they would end up being

φ1(x) = 1 - cos(2πx/L)
φ2(x) = 1 - cos(4πx/L)
φ3(x) = 1 - cos(6πx/L)...

Thank you!
 
I think you've got it!
 
Great! Thank you so much!

If I may ask one more slightly related question:

If I am using the Assumed Modes method for N=4, for the stiffness and mass coefficients, do the i and j just need to add up to 4?

For example, I could use:

mij = m1,3
or
mij = m2,2

I'm not sure if that question is clear, as the book did not explain how to get the number of terms. If this is the correct way, does is matter whether I use 1,3 or 2,2?
 
I'm not familiar with your terminology, but N=4 suggests to me that i = 1,2,3,4 and j = 1,2,3,4, so that all the combinations are involved.
 
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