Minute 15:45, there are 2 typos: 1- Missing Sqrt, In red: wn = sqrt(k / m) 2- Missing power in the line above: Z0 / y0 = r^2 / sqrt[(1 - r^2)^2 + (2 * zeta * r)^2]
Is this form of the amplification factor specific to the relative coordinate system you have chosen to use? I notice that the numerator in your case is slightly different to what is usually derived when x and y are maintained throughout solving.
There's always a way to use Lagrange's method to find equations of motion. However, in this video, the equation of motion is simple, but its the nature of the solution that we are focused on.
Hi I have a question about the form of ground excitation and particular solution. If the ground excitation is of harmonic/trigonometry identity, the particular solution suggested is also harmonic/trigonometry right? Then how to get the total response of structural model?
It’s not totally random, especially as one moves away from the epicenter. It does has a back-and-forth motion to it which means it has a periodic nature. As such, the ground motion can be approximated using a Fourier Series, which turns it, mathematically, into a harmonic treatment of the problem.
In the previous video, I did not assume that there was any phase shift in the response and as a result, I showed that the amplitude, X, was a complex number and that the phase shift could be determined from this complex number. In this video, I made the assumption that the response already had a phase shift and as a result, the amplitude is purely real. I believe that this is the difference that you are referring to. In fact, they amount to the same thing, but in the previous video I showed step-by-step how to get there. I think if you return to the previous video and go through the math step-by-step, then it will be clear why I am able to treat the problem in this video in this manner as I have (by skipping a couple of steps). I hope I haven't confused you with this. If necessary, apply the technique from the previous video to this problem and you should get to the same result.
I could have done that. It's just a question of choice of coordinate system. Usually this is determined by when if being asked of you in the problem. From my experience, it is usually the relative coordinate system that tends to be the easiest to work with then you can convert to whatever coordinate system you'd like the result in.
Taking the derivative of y with respect to t is the same as multiplying y by iω. So taking the second derivative of y is the same as multiplying y by (iω)^2 = i^2ω^2 = -ω^2. Substituting y_ddot causes the negative (-) sign to change to a positive (+) sign.
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This man's voice is amazing
Minute 15:45, there are 2 typos:
1- Missing Sqrt, In red: wn = sqrt(k / m)
2- Missing power in the line above: Z0 / y0 = r^2 / sqrt[(1 - r^2)^2 + (2 * zeta * r)^2]
Awesome explanation and video once again! ❤️
Nice sort of simulation study.....pls next make a vedio on decoupling of spring damper system......
nice videos, its hard to understand by only reading. This is just beautiful :-)
14.27: OMEGAn = sqrt(K/m), not k/m. Probably just a typo.
Thanks for catching this. You're right, I left out the square root sign.
it would be nice if the chat supported symbols.
@12:30 "Mechanical engineers want to write things in non-dimensional form." So true, dimensionless equations are nice and beautiful to look at.
Amazing explanation
Awesome sauce !
Is this form of the amplification factor specific to the relative coordinate system you have chosen to use? I notice that the numerator in your case is slightly different to what is usually derived when x and y are maintained throughout solving.
Wonderfull explanation thanks.
Can you make me a favor, what drawing app are you using??
The app is called "Paper" by WeTransfer. It is running on a iPad Pro 13 inch and I am using an Apple Pencil.
Thank you bro i was waiting for you btw amazing video thats a SUB🌹
excellent
Thank you
14:00 onwards: this not a zeta, it's a xi . Zeta is smoother as it has just one cusp whereas xi has two.
Thank you. My Greek is definitely not a strong point for me...as you can probably tell.
Is there a way to find the equation of motion using lagrange’s formula instead?
There's always a way to use Lagrange's method to find equations of motion. However, in this video, the equation of motion is simple, but its the nature of the solution that we are focused on.
Hi I have a question about the form of ground excitation and particular solution. If the ground excitation is of harmonic/trigonometry identity, the particular solution suggested is also harmonic/trigonometry right? Then how to get the total response of structural model?
The total response, x, can be found from eqn 2: x = y + z
How did eathquake exhibit harmonic motion.Its random
It’s not totally random, especially as one moves away from the epicenter. It does has a back-and-forth motion to it which means it has a periodic nature. As such, the ground motion can be approximated using a Fourier Series, which turns it, mathematically, into a harmonic treatment of the problem.
Would you be willing to do a video on rotating unbalanced masses and one on whirling of rotating shafts?
I'll add this to my list. Might take me a little while to get there.
@@Freeball99 Thank you!
Sir, if I have a double pendulum and its hitting the surface at a point , can I implement the equation
Not sure I understand the question.
Hi sir, can you send me an email at sudhansumtripathy@gmail.com, I will send you the problem and you can suggest me what to do.
HI, I somehow missed this message from you. You can email me at apf999@gmail.com
Hi Sir, have you received my email
At 12.00, what happened to the e^-itheta of the amplitude Z since in the previous video it is writtien as a port of the amplitude Z
In the previous video, I did not assume that there was any phase shift in the response and as a result, I showed that the amplitude, X, was a complex number and that the phase shift could be determined from this complex number.
In this video, I made the assumption that the response already had a phase shift and as a result, the amplitude is purely real. I believe that this is the difference that you are referring to.
In fact, they amount to the same thing, but in the previous video I showed step-by-step how to get there. I think if you return to the previous video and go through the math step-by-step, then it will be clear why I am able to treat the problem in this video in this manner as I have (by skipping a couple of steps). I hope I haven't confused you with this. If necessary, apply the technique from the previous video to this problem and you should get to the same result.
why didn't you finally write z in terms of x and y
I could have done that. It's just a question of choice of coordinate system. Usually this is determined by when if being asked of you in the problem. From my experience, it is usually the relative coordinate system that tends to be the easiest to work with then you can convert to whatever coordinate system you'd like the result in.
You should be a voice actor
Haha! You should be my agent.
in 07:17 it seems there is a typo error where if we subsuate the value of y double dots the negative sign will not dissper .
Taking the derivative of y with respect to t is the same as multiplying y by iω. So taking the second derivative of y is the same as multiplying y by (iω)^2 = i^2ω^2 = -ω^2. Substituting y_ddot causes the negative (-) sign to change to a positive (+) sign.