harmonically excited vibrations
Post on 19-Dec-2015
19 Views
Preview:
DESCRIPTION
TRANSCRIPT
Harmonically Excited Vibrations
ME-304 â Mechanical Vibrations
ð ð¥ + ðð¥ = ð¹ cosðð¡
Undamped system under Harmonic Excitation:
ðð ðððð€ ð¡âð âðððððððð¢ð ð ððð¢ð¡ððð ð€ðð¡â ð¹ = 0 :ð¥â ð¡ = A1 cosððð¡ + ðŽ2 sinððð¡
ððððŠ ð€ððŠð ð¡ð ðððð ð¡âð ðððð¡ððð¢ððð ð ððð¢ð¡ððð.ððâ²ðð ð¢ð ð ð¡âð ððð¡âðð ðð ð¢ðððð¡ððððððð ððððððððððð¡ð ð¥ð ð¡ = ð cosðð¡
ððâ²ðð ððð¢ð ð¡âðð ðððð ðð ð¡ð ð¡âð ðžðð ððð ð ðððððððŠððð¡ðð ððððððððð¡ððð¡ððð ð¡ð€ððð:âð2ðð + ðð = ð¹
ð¿ =ð
ð âððððð ð =
ð
ð âðððððððð
ððð¡ðð ð ððð¢ð¡ðððð ðð ð¥ ð¡ = ð¥â ð¡ + ð¥ð ð¡ = A1 cosððð¡ + ðŽ2 sinððð¡ + X cosðð¡
ðð ððð ðŒð¶ð : ð¥ 0 = ð¥ð; ð¥ 0 = ð£ðð€ð ððð ðððð ðŽ1ððð ðŽ2
ðŽ1 = ð¥ð â ð
ðŽ2 =ð£ððð
ððð¡ðð ð ððð¢ð¡ðððð ðð
ð¥ ð¡ = (ð¥ð â ð) cosððð¡ +ð£ððð
sinððð¡ + X cosðð¡
ð€âððð ð =ð¹
ð â ðð2
ð
ð¿ð ð¡=
1
1 â ð2
ð·ððððð ð ð¡ðð¡ðð ððððððð¡ððð: ð¿ð ð¡ =ð¹
ð
ððð ððððð¢ððððŠ ððð¡ðð: ð =ð
ðð
This term is called the amplification factor, amplitude ratio, magnification factor or simply gain
Gain is a function of frequency ratio
In Phase Response(Frequency ratio, r < 1)
Out of Phase Response(Frequency ratio, r > 1)
Resonance(Frequency Ratio, r = 1)
ðŽð ð ð¢ðððð ð§ððð ðððð¡ððð ðððððð¡ðððð : ð¥ 0 = 0 ; ð¥ 0 = 0
ððð¡ðð ð ððð¢ð¡ðððð ðð ð¥ ð¡ = X (cosðð¡ â cosððð¡)
ð·ððððð ð¿ð ð¡ =ð¹
ðððð ð =
ð
ðð
ð€âððð ð =ð¹
ð âðð2
ð¥(ð¡) =ð¿ð ð¡
1 âððð
2 (cosðð¡ â cosððð¡ )
ð¥(ð¡) =ð¿ð ð¡
1 âððð
2 (cosðð¡ â cosððð¡ )
ðŽð¡ ð = 1,ð = ðð; ðâð ððð ðððð ð ðð ð¡âð ð ðŠð ð¡ðð ðð ððð£ðð ððŠð¡âð ððððð€ ððð¢ðð¡ððð, ððððððð ð¢ðððððððð
We can use L'HÃŽpital's rule:
Beating Phenomenon(Frequency Ratio, r close to 1)
We found the response of the system with zero initial conditions to be:
Using trigonometric identities, we can rewrite:
ð¥(ð¡) =ð¿ð ð¡
1 âððð
2 (cosðð¡ â cosððð¡ )
When the difference between the driving frequency and natural frequency is small. We define a small quantity epsilon, ϵ :
Rewriting the response of the system:
ð¥(ð¡) =ð¿ð ð¡
1 âððð
2 (cosðð¡ â cosððð¡ )
ð¥(ð¡) =ðð2ð¿ ð ð¡
ðð2 â ð2
(2 sin(ðð + ð
2ð¡) sin(
ðð â ð
2ð¡) )
ð¥(ð¡) =
ððÃð¹ð
4ðð(2 sin(
ðð + ð
2ð¡) sin(
ðð âð
2ð¡) )
ð¥(ð¡) =ð¹
2ððð(sinðð¡ ð ðð ðð¡ )
ððððð ð ðð ððð¡âðð ð ðððð: sin ðð¡ ð€ððð âðð£ðð ððððð ð¡ððð ðððððð
ðŽðð ð ðð ðð¢ðâ ðððððð, sinðð¡ ð€ððð âðð£ðð ð ðððð ð¡ððð ðððððð
ð ð¥ + ð ð¥ + ðð¥ = ð¹ cosðð¡
Damped System with Harmonic Force
Many ways to get the particular solution.
1. Using undetermined coefficientsa. Assume a particular solution of the form:
b. Plug it back in to the EOM:
c. Use trigonometric identities to expand cos and sin terms:
d. Equate the coefficients of cos(Ït) and sin(Ït) in above equation
e. Solve for α and X
f. Solution may be expressed in terms of dimensionless numbers
ð ððð ð
2. Using Complex Form
ð ð¥ + ð ð¥ + ðð¥ = ð¹ðððð¡
ðð ððð ðððððð ððð¡ ð âððððððð ððððððð ðð¢ððð¡ðððð¢ð ððð ððððððð¥ ððð¡ðð¡ððð:ðŒð ð€ð ð¢ð ð ð ððððððð ðð¢ððð¡ððð ðð ð¡âð ðððð ð¹ðððð¡ , ð¡âð ðððððððð¡ ðð ð¹ðððð¡ ðð ð¹ cosðð¡ . ðð ð¡âð ðððð ðððð¡ ðð ð¡âð ð ððð¢ð¡ððð ð€ððððððððð ðððð ð¡ð ð¡âð ð ððð¢ð¡ððð ð€ðð¡â ððððððð ðð¢ððð¡ððð ð¹ cosðð¡
ðŽð ð ð¢ðð ð ðððð¡ððð¢ððð ð ððð¢ð¡ððð ðð ð¡âð ðððð:
ð¥ ð = ððððð¡
ð·ðððððððð¡ððð¡ð ð¡ð€ððð ððð ðð¢ð¡ ðð¡ ðððð ðð ð¡âð ðžðð:
âðð2ððððð¡ + ðððððððð¡ + ðððððð¡ = ð¹ðððð¡
ð ð âðð2 + ððð = ð¹
ð ð âðð2 + ððð = ð¹
This is called the âMechanical impedanceâ of the system: ð ðð = ð âðð2 + ððð
ð =ð¹
ð âðð2 + ððð
ð =ð¹
ð âðð2 + ðððÃð âðð2 â ððð
ð âðð2 â ððð
There is an imaginary term in the denominator. We can get rid of that and separate the real and imaginary parts by multiplying and dividing by the complex conjugate of the denominator.
ð = ð¹ð âðð2
ð âðð2 2 + ðð 2â ð
ðð
ð âðð2 2 + ðð 2
ð ðð ðð ð¡âð ðððð ð â ðð
ðððð ð¡ð ððð€ððð¡ð ð ðð ð¡ðððð ðð ð ðððððð¡ð¢ðð ððð ðâðð ð ððððð:
ð = ðŽðâðð, ð€âððð ðŽ = ð2 + ð2 ððð ð = tanâ1ð
ð
ðŽ = ð¹ð âðð2
ð âðð2 2 + ðð 2
2
+ðð
ð âðð2 2 + ðð 2
2
ðŽ =ð¹
ð âðð2 2 + ðð 2 ð = tanâ1ðð
ð âðð2
ð¥ ð = ððððð¡ = ðŽðð(ðð¡âð) =ð¹
ð âðð2 2 + ðð 2ðð(ðð¡âð)
Use these results to rewrite the particular solution of the system:
ð¥ ð =ð¹
ð âðð2 2 + ðð 2cos ðð¡ â ð
The real part of the particular solution is the same as what we got before:
The equation,
can rewritten in terms of dimensionless numbers:
ð
ð¿ð ð¡=
1
1 â ð2 + ð2ððâ¡ ð»(ðð)
This term is called the âFrequency Response Functionâ
ð =ð¹
ð âðð2 + ððð
ð =ð¹/ð
1 âðð2
ð+ðððð
ððŠ ð¢ð ððð ð =ð
ðð; ðð =
ð
ð; ð¿ð ð¡ =
ð¹
ðððð ð =
ð
2 ðð
Total Solution:
ðŒ
Using ICs, we can find unknowns
Note that only the particular solution doesnât decay exponentially, whereas the homogenous part will decay and die out eventually. So the particular part is often called the âSteady Stateâ solution, and the homogenous part is called the âtransientâ solution
f. Solution may be expressed in terms of dimensionless numbers
ð =ð¹
ð âðð2 2 + ðð 2
ð =ð¹/ð
1 âððð2
2+ 2
ð
2 ððð
ðð
2
ð
ð¿ð ð¡=
1
1 â ð2 2 + 2ðð 2
ðððððððððŠ ð = tanâ1ðð
ð âðð2= tanâ1
2ðð
1 â ð2
This expression gives us the Gain of the system as a function of âfrequency ratioâ r and damping ratio ζThis is plotted on the next slide, for different damping ratios
Base Excitation
Lets look at the response of the system when the base of a mass-spring-damper system undergoes harmonic motion.
The Equation of Motion of the system (using the free-body diagram) looks like:
ð ð¥ = âð ð¥ â ðŠ â ð( ð¥ â ðŠ)
ð¹ =ð ð¥
ð ð¥ + ð ð¥ + ðð¥ = ð ðŠ + ððŠ
ðŒð ðŠ ð¡ = ð sinðð¡
ð ð¥ + ð ð¥ + ðð¥ = ððð cosðð¡ + ðð sinðð¡
ðâð ððððððð ðð¢ððð¡ððð: ððð cosðð¡ + ðð sinðð¡ððð ðð rewritten in the fom A sin(ðð¡ â ðŒ)
A sin(ðð¡ â ðŒ) = ððð cosðð¡ + ðð sinðð¡ðŽ sinðð¡ cos ðŒ â ðŽ sin ðŒ cosðð¡ = ððð cosðð¡ + ðð sinðð¡
Must find A and α, by comparing the terms on either side of the equation
ðŽ sin ðŒ = âððððŽ cos ðŒ = ðð
ðŽ = ð ðð 2 + ð2
ðŒ = tanâ1 âðð
ð= tanâ1(â2ðð)
ð ð¥ + ð ð¥ + ðð¥ = ðŽ sin(ðð¡ â ðŒ)
And the equation of motion is:
This shows that giving excitation of the base is equivalent to applying a harmonic force of magnitude A.
Weâve already solved this for a slightly different forcing function, we donât really need to solve this again. We can just borrow results we derived before with a few modifications.
One way of doing this, is by comparing with the solution we got using complex notation
ð ð + ð ð + ðð = ððððð ð ð + ð ð + ðð = ðšðð(ððâð¶)
ð ð = ð¿ðððð
ð ð =ð
ð âððð ð + ðð ððð(ððâð)
Assume particular solution: Assume particular solution:
ð ð = ð¿ðð(ððâð¶)
ð ð =ðš
ð âððð ð + ðð ððð(ððâð¶âðð)
ðâð ððððððð ðð¢ððð¡ððð ð¹ cosðð¡ ðð ðððððððð ððŠ ð¡âð ðððð ðððð¡ ðð ð¹ ðððð¡
ðâð ððððððð ðð¢ððð¡ððð ðŽ sin(ðð¡ â ðŒ) ðð ðððððððð ððŠ ð¡âð ðððððððððŠ ðððð¡ ðððŽðð(ðð¡âðŒ)
Solution in complex form: Solution in complex form:
ððððð ð = ðððâððð
ð âðððððððð ðð = ðððâð
ðð
ð âððð
The real part of this solution represents the solution for the forcing function F cos ðð¡
The imaginary part of this solutionrepresents the solution for the forcingfunction ðŽ sin(ðð¡ â ðŒ)
Base ExcitationHarmonic Excitation
ð¥ ð =ð¹
ð â ðð2 2 + ðð 2ðð(ðð¡âð) ð¥ ð =
ðŽ
ð â ðð2 2 + ðð 2ðð(ðð¡âðŒâð1)
ð
ð¿ð ð¡=
1
1 â ð2 2 + 2ðð 2
Here, we were trying to compare the âstatic displacementâ ÎŽst with the magnitude of the particular solution, (or steady state solution), X
ð¥ ð =ð ð2 + ðð 2
ð â ðð2 2 + ðð 2ðð(ðð¡âð)
Weâre trying to compare the magnitude of the input displacement, Y and the magnitude of the particular solution, X
ð
ð=
ð2 + ðð 2
ð â ðð2 2 + ðð 2
ð
ð=
1 + 2ðð 2
1 â ð2 2 + 2ðð 2
We can rewrite this in terms of frequency ratio and damping ratio:
ð€âððð ð = ðŒ + ð1
This term is called displacement transmissibility:
Base Excitation â Force Transmissibility
In base excitation, a force F is transmitted to the base or support due to the reactions from the spring and the dashpot. This force can be determined as:
ð¹ = ð ð¥ â ðŠ + ð ð¥ â ðŠ = âð ð¥
ð¥ ð =ð ð2 + ðð 2
ð âðð2 2 + ðð 2ðð(ðð¡âð)
Consider the particular solution:
ð¥ ð = ð sin(ðð¡ â ð)
ð€âððð ð =ð ð2 + ðð 2
ð â ðð2 2 + ðð 2
ð¹ = ðð2X sin ðð¡ â ð
ð¹ = ðð2ð ð2 + ðð 2
ð âðð2 2 + ðð 2sin ðð¡ â ð
The maximum value of this force FT is given by:
ð¹ð = ðð2ð ð2 + ðð 2
ð âðð2 2 + ðð 2
ð¹ðð= ðð2
ð2 + ðð 2
ð âðð2 2 + ðð 2
ð¹ðð= ðð2
1 + 2ðð 2
1 â ð2 2 + 2ðð 2
Lets try to rewrite this in terms of frequency ratio and damping ratio:
ð¹ððð
=ð
ðð2
1 + 2ðð 2
1 â ð2 2 + 2ðð 2
ð¹ððð
= ð21 + 2ðð 2
1 â ð2 2 + 2ðð 2
This term is called the âforce transmissibilityâ
Rotating Unbalance
Rotating Unbalance
Rotating Unbalance
ð ð¥ + ð ð¥ + ðð¥ = moeð2 sinðð¡
ð¹ =ð ð¥
ð¥ ð =ðððð
2
ð â ðð2 2 + ðð 2ðð(ðð¡âð)
ð€âððð ð = tanâ1ðð
ð â ðð2
This is effectively the same problem as before, with F= moeÏ2
ð¥ ð = ð sin(ðð¡ â ð)
ð =ðððð
2
ð âðð2 2 + ðð 2
Time to make this non-dimensional. Rewrite in terms of zeta and r
ð =ðððð
2
ð âðð2 2 + ðð 2 ð =
ðððð2
ð
1 â ð2 2 + 2ðð 2
ð =
ðððð2
ðÃðð
1 â ð2 2 + 2ðð 2
ðð
ððð=
ð2 Ãðð
1 â ð2 2 + 2ðð 2
ðð
ððð=
ð2
1 â ð2 2 + 2ðð 2
This term is the âdimensionless displacement magnitudeâ of the system.
The force F is transmitted to the foundation due to the rotating unbalanced force is given by:
ð¹ ð¡ = ðð¥ ð¡ + ð ð¥(ð¡)
ð¹ð = ðððð2
1 + 2ðð 2
1 â ð2 2 + 2ðð 2
The maximum value of this force FT is given by:
top related