MIT 2.003SC Engineering Dynamics, Fall 2011 View the complete course: ocw.mit.edu/2-003SCF11 Instructor: J. Kim Vandiver License: Creative Commons BY-NC-SA More information at ocw.mit.edu/terms More courses at ocw.mit.edu
It's not important for the point that he was making but at 12:45 the X in the first term of the equation of motion needs to be changed to X "double dot" (the second time derivative). The same applies to the expression at 13:45 where q needs to be changed to q "double dot".
In the final 10-15 minutes we see that with proportional (Rayleigh) damping we obtain 2 parameters in order to approximate the modal damping ratios to the measured ones. How do we measure them in a 2dof system? And what is the point of finding damping ratios this way if we have them measured already?
On 1:01:00 the professor says that "not alway equal to" but as far as I know for any orthogonal matrix u and any matrix c, that will always methamatically result into a diagonal matrix. Am I wrong?
The problem is that the vectors of the modal matrix U are not actually directly orthogonal despite what Prof Vandiver says in the earlier part of the lecture. The way to find the modal matrix is to solve for the eigenvectors of M⁻¹K. Note that if any mass terms are different from each other, M⁻¹K will not be symmetric. Although K is symmetric, M⁻¹ will be populated by the multiplicative inverses of each mass, and so different masses will scale each column of K differently, ruining symmetricity. Since M⁻¹K is not symmetric, then orthogonal eigenvectors are not guaranteed. A concrete example of a non orthogonal modal matrix is shown for an example at 42:58. There is an alternate approach that explains why M and K are diagonalized by the modal matrix (and by extension, why C is not except for a special case). In this alternate approach, x is transformed twice - the first transform being x = M¹ᐟ²q. This transform decouples the modes so that you solve the eigenvectors for M⁻¹ᐟ²KM⁻¹ᐟ² instead (let's call it K̃), which is guaranteed to be symmetric (for any A and any symmetric B, AᵀBA is symmetric, and K must be symmetric, so M⁻¹ᐟ²KM⁻¹ᐟ² is symmetric), which in turn guarantees orthogonality of its eigenvectors. We then normalize those eigenvectors and assemble those eigenvectors into a matrix P. Normalizing the eigenvectors is critical so that P becomes orthogonal - a matrix composed of orthogonal columns is not orthogonal until they are normalized (so that P⁻¹P = I). Now, we can use a second transform (to diagonalize the left side matrices) q = Pr such that the left side of the equation becomes Pr̈ + K̃Pr, then left multiply the system by Pᵀ. Since Pᵀ = P⁻¹, and P is composed of the eigenvectors of K̃, P⁻¹K̃P = Λ by diagonalization, where Λ is a diagonal matrix (specifically, the spectral matrix, where the squared natural frequencies of each mode are the diagonal terms). The left side of the equation becomes Ir̈ + Λr. Note that each term of UᵀMU is a scaled version of a term of I. This is not surprising, since I is the identity matrix. However, UᵀKU is a scaled version of Λ by the same terms. Consequently, any forcing is also scaled this way in his equations versus mine. Both of our methods would thus return the same result. Regardless, the entire reason why I went over this process was to show that the diagonalization process for M and K are unique to M and K. Including a damping matrix that does not commute with K will cause P to not be a basis of eigenvectors for C, which would cause it to not be diagonalized in the r-transform. Similarly, it will also not work using Prof Vandiver's method (UᵀCU).
It is, the mode shape matrix houses the two eigenvectors that are in fact orthogonal to each other. You can't use the dot product to check this but instead the transpose of one of the eigenvectors* mass matrix* the second eigenvector. My answer on matlab was -5.1233*10^-4 which is close enough to 0
This is nice, interesting, and helpful, but the question that lingers for me is that, then, how do you calculate the natural frequencies and mode shapes of a damped MDOF system? Or the natural frequencies and mode shapes are for undamped systems only, which would mean that the natural frequencies and mode shapes of the damped system are those of its undamped version? Thank you.
Hi. I know this is an old question but let me try to help others if not you. A) If want to solve by hand and the system is lightly damped, just eliminate the damping from the EoM and you should be fine. B) If you really want to solve the problem if the damping, keep in mind it would be very time-consuming. Basically, you would have a term [C]*dot{x} in your equation of motion, where the dot{x} is the x first time derivative. Due to that term, you'll get pairs of conjugated complex poles as roots of the Characteristic Equation for each mode shape. Your eigenvalues and eivenvectors will be complex. B1) Although I believe there is more than one way to solve this kind of problem, the most used one is transforming the EoM to the State-Space form. Let's assume you're solving a 2 DoFs system. By doing the transform to state-space you're converting the set of two 2nd order differential equations to a set of four first order differential equations, by defining a state variable 'z'. Then you can write dot{z} = Az whereas A is a 4x4 matrix derived from the oritinal equations of motion. B2) Then you can solve the Characteristic Equation: as for the undamped case, for each egenvalue you'll find the corresponding eigenvector. B3) Formulate the general solution writing it as a combination of exponential functions based on the egenvalues and eigenvectors. B4) Aplly the initial conditions to solve for the constants in the general form. The state-space form is also very useful to solve damped problems in Matlab.
I am having question regarding continuous system, plz help , when continuous system/MDOFs is vibrating, how do we comes to know which modes are significant and till that mode sum to be taken?
+Chirag Palan Engineering intuition, thats what my professor told me. For simple problems you first just visualize the modes in your head and see if you can figure out breaking points, which can be done for simple shapes till like the fifth mode. Otherwise you need a analysis to figure out which modes are important.
in earthquake engineering the fundamental modes are those who are capable of 'activating' at least 80-90 of the total mass of the system when you add their particular contribution for each analysed direction . some approaches take into account those modes who move/activate at least 1%- 5%.
Course materials including the suggested readings can be found in the full course site ocw.mit.edu/2-003SCF11. The readings are in the two following textbooks: Hibbeler: Engineering Mechanics: Dynamics www.amazon.com/exec/obidos/ASIN/0136077919/ref=nosim/mitopencourse-20 and Williams: Fundamentals of Applied Dynamics www.amazon.com/exec/obidos/ASIN/0471109371/ref=nosim/mitopencourse-20.
For my previous education I used the book "Modal Testing: A Practitioner's Guide" from Peter Avitabile. The book helped me with questions and practical tips for my custom vibration measurement services. Some of these are shown on my channel.
The problem with a mathematical description of physics is that most people understand the physics before they understand the mathematics and when teaching it, you basically made it really difficult. You learn it in school and then you forget whereas a physical understanding is reinforced through physical observation of reality and most people don't forget just as we don't forget what it felt like to fall on our ass.
People fell on their arses for millennia before Newton and analysts following him, and had zero idea how any of this worked. They thought arrows were kept in the air by motions of the air. So, no.
DOUBT, (Need help...) we know natural modes are linearly independent to each other (Orthogonal), then why the dot product of the natural modes is not zero in most of the videos (including this video). Is it just a calculation mistake by the professor or my above mentioned concept is wrong????? Any help will be appreciated ...
That inappropriate behavior of the students is the result of "democracy" in class, where you can come in what ever clothes you want, you can sit how ever you want, and at the end you can do what ever you want... Disrespect the knowledge and world will fall apart
it looks unfortunate that at the end of class, students do not care to let the teacher first come out of class before they leave the class room. This is slowly visible in other countries such as India also. The lecture is remarkable for its content and presentation.
LOL if u really think that you haven't attended many engineering classes, this guy does a really good approach in both senses compared to many other proffessors that don't even understand physics behind this kind of problems
My sentiments precisely! I had this 'experiment' with rope at grade seven. I've stopped watching when he missed double dots in the equation. Is it really MIT?
Anna B. I would say he used a 'simple' experiment to make the math sensible. lol If you try to work this stuff out, there's a lot of confusion of how terms turn out and it makes you feel very uneasy. If you demonstrate whatever problem you have with a simple experiment regardless of the level of education, it takes that uneasy feeling away slightly. Usually, I try to imagine how motion occurs all at once, but that's not always sufficient.
Alex Poulin prolly just low quality bait. Surely if someone was taught this so early on and understood it they wouldn't look up introductory lectures on it right?
