replacement masses by dynamic equivalence:biella

[Ing Italy]

Hello.

I have a question to ask and I thank in advance who will answer me: as you will know when to introduce the dynamic equivalence is added to the moment of inertia of the biella (and also in other cases) a moment of fictitious inertia (can possibly also be negative and does not correspond to the real mass distribution in the biella) so that the scheme to two masses more fictitious moment jo uguagli dynamically the real biella.bene, as can be noticed in the link to page 67 when I go to write the
.
♪

I wonder why you use only the fictitious moment and not the real moment of inertia since the moment of inertia of the equivalent system is the same as the royal biella and therefore =m(a) a^2 + m(b) b^2 +jo therefore in theory as obstacle to rotation there must also be the moment of inertia created by the two masses of substitution in a and b. so the question is: why use only the fictitious moment Thank you very much who will give me a hand, how you will understand from the time the exam is approaching.saluti

http://www.scribd.com/doc/38055075/48/metodo-delle-masse-di-sostituzione

 

[Ste_d]

but you have obtained that j0=j(biella)-m(biella)*ab from the equivalence with the biella. therefore the replacement body has moment of baricentric inertia j0 and when you write the second cardinal regarding the center of gravity it is right that only the moment of baricentric inertia appears.
the relationship j(biella)=m(a) a^2 + m(b) b^2 +jo the impotence to say that the 2 bodies have the same moment of baricentric inertia.

 

[Ing Italy]

excuse but when you have the equivalent system practically you have an auction with two masses at the ends and you want to write at the time of baricentric inertia in direction z, so it would seem logical to say that in a system with two masses the moment of inertia is given by the sum of the moment of inertia given by the two masses, then as it also introduces a moment of inertia phytizio I will submerge this moment of additional inertia.


if the equivalent system and the royal biella have the same moment of inertia to be equivalent dynamically they will have to have uquale rotational inertia, if I in rotational inertia put only the moment of fictitious inertia no longer have the same pair of inertia, no?

 

[Ste_d]

I think once you've got yourself j0 according to the parameters of the real biella you're good... that is the moment of inertia of the equivalent body therefore including rotational inertia of the masses (also expressed according to the mass of the biella).
Also on my notes the pair is expressed with j0, it is not a mistake so... But I've never been too careful. I can't give any more explanations, wait for more opinions!

 

[Ing Italy]

I think once you've got yourself j0 according to the parameters of the real biella you're good... that is the moment of inertia of the equivalent body therefore including rotational inertia of the masses (also expressed according to the mass of the biella).
Also on my notes the pair is expressed with j0, it is not a mistake so... But I've never been too careful. I can't give any more explanations, wait for more opinions!

ahahah, look at most of the times the things I go to ask the prof are things for which neither did they ever happen,unfortunately or fortunately I am an incredible pygnolo,cmq thanks for having responded colleague =)

 

[Ing Italy]

I would have hoped to have other opinions, so to all

 

[paulpaul]

Bye!
when you take the equivalent biella considering only the two masses placed in the two centers, you must consider only the j0 fictitious moment. your equivalent body is formed only by the two masses and by the moment of fictitious inertia.

to have the equivalence between the motion plane of a body and that of a system of masses to it equivalent the latter must have:

1. same mass;
2. same moment of inertia;
3. same product of inertia;
4. static moments (they are two because you have two axes) null;

5 equations ---> 5 unknown (5 fictitious masses)

1 allows you to have the same kinetic energy duses for translation motions. 2 and 3 same kinetic energy for rotation motions. 4 same center.

for the girls, when they want to calculate the actions of inertia of these in general they put the masses all on a same axis (and one of the 4 falls), remaining 4 masses. If you then consider the center of the axes as the origin of the axes, you also fall 3 and therefore you need only 3 masses.

However, since it is convenient to have only two masses m1 and m2 (one with pure translatory motion the other of pure rotary motion) because of these points you know the accelerations with the formulas of the tavellism, it is advisable to replace the 3 masses with 2 masses and a moment of inertia (the unknowns are always 3 because you always have 3 equations!!!).

So you have

1. m1 + m2 = m (real biella mass)
2. m1*a1^2+m2*a2^2+j0 = j (real biella inertia)
4. m1*a1 = m2*a2

a1 and a2 are the distances of the two centers from the center (the latter you revenues from the cad). j you also revenues from the cad. and then consider m1, m2, j0 for inertia actions!

I carry out the balance of the inertia forces of my compressors by applying this method equal (on xls) to calculate the actions on the biella!

 

[Ing Italy]

Hello paul!

your comments are always very clear and meaningful. Thank you for contributing to this discussion. ..unfortunately in this period I have several things to do and I can't follow this forum and in fact I wanted to answer you also regarding the post on the load losses but I can't find the right time. I just wanted to let you know that I really appreciate it, hello and good day!