mechanical exercises applied to machines

[Fulvio Romano]

Thank you. I'll be studying your answer calmly.
As for the three hydraulic cylinders, why don't I consider them as members?
Perhaps because they are motive members and do not add or remove anything from the mechanism itself?

because they are actuators, and not film members.
If you had an electric motor on the joint, would you also consider the stator and rotor as members?

I repeat, I find it wrong to consider the cylinders as members of the cinematic structure. If, however, your professor is of other opinion, he might resent and consider my point of view a "error".
check well.

 

[snaroz]

Yes, your answer convinces me.
therefore, recapitulating, removing the cylinders the members are in total 6.
cinematic pairs, all lower and rotary, are also 6, in particular:
1) boom-house;
2) stunga-house;
3) stunga-tripower;
4) tripower-boom;
5) stick-tripower;
6) stick bucket

for the calculation of degrees of freedom I apply the formula of the grubler, which says that f=3(l-1)-2a-b, where "l" is the number of members, "a" the number of lower cinematic pairs and "b" of the higher ones.
inserting the data, it is found that the mechanism has 3 degrees of freedom.
Is that right?
thank you very much for the availability!

 

[Vortexsheet]

thanks fulvio,
I solved by downloading mingw where I found the missing dll.
snaroz in the meantime - I don't know if you already did - identify the degrees of freedom - you see to eye - the mechanism is slow. the motive organs you can simply model them with glifi and course. but the denavit-hartenberg method is not used for spatial mechanisms?
I tried freemat, really made me a good impression! I copied a matlab code and I didn't have to change anything. .
practically for educational purposes (except for simulink) is equivalent to matlab?

edit: just three degrees of freedom. What are you with?

 

[Fulvio Romano]

therefore, recapitulating, removing the cylinders the members are in total 6.
cinematic pairs, all lower and rotary, are also 6,

Right.

in particolare:
1) boom-house;
2) stunga-house;
3) stunga-tripower;
4) tripower-boom;5) stick-tripower;6) Stick bucket

the stick is not directly connected to the tripower. what is missing is the stick-boom.

for the calculation of degrees of freedom I apply the formula of the grubler, which says that f=3(l-1)-2a-b, where "l" is the number of members, "a" the number of lower cinematic pairs and "b" of the higher ones.
inserting the data, it is found that the mechanism has 3 degrees of freedom.
Is that right?

Look, just the three g.d.l. that, removed from the three cylinders, make the structure not labile...:wink:

 

[Fulvio Romano]

but the denavit-hartenberg method is not used for spatial mechanisms?

not necessarily. d-h is a recurring method that allows you to solve any cinematic structure in a methodical way, so with less chance of wronging.
the method was invented to solve the articulated quadrilaterals of trains. is normally used in robotics.
the d-h method then allows to write in an always recurring and methodical way, the structure jacobiano.

I tried freemat, really made me a good impression! I copied a matlab code and I didn't have to change anything. .
practically for educational purposes (except for simulink) is equivalent to matlab?

I would also say for non-didactic purposes. . .

 

[snaroz]

Right.

the stick is not directly connected to the tripower. what is missing is the stick-boom.

Yes thank you, I was wrong to write:finger:

Look, just the three g.d.l. that, removed from the three cylinders, make the structure not labile...:wink:

I understand....so applying the grubler formula is correct? the number of degrees of freedom of the mechanism could not have been calculated even by analytical means?

 

[Vortexsheet]

not necessarily. d-h is a recurring method that allows you to solve any cinematic structure in a methodical way, so with less chance of wronging.
the method was invented to solve the articulated quadrilaterals of trains. is normally used in robotics.
the d-h method then allows to write in an always recurring and methodical way, the structure jacobiano.

Very interesting! any applications? Some exercise I can have fun with. .

 

[Fulvio Romano]

I understand....so applying the grubler formula is correct? the number of degrees of freedom of the mechanism could not have been calculated even by analytical means?

What do you mean by analytical?
in the plan each member has 3g.d.l., each rotary couple takes two, they remain 6. remove the three gdl representing a rigid rounding in the plane, three remain. the grubler formula does exactly this reasoning, nothing more.

I know, it's not that I don't know, I remember it very well... but if there were someone who doesn't remember it... it's not that you can remember the difference between upper and lower film couple? :redface:

 

[snaroz]

a cinematic couple is a set of two members connected through two cinematic elements so that each member has compared to the other a relative motion.
Moreover, a cinematic couple is said to be lower if the contact between the two cinematic elements is superficial, superior if of linear type or dots.

 

[snaroz]

What do you mean by analytical?

I have seen it on the book of rational mechanics, in a few words mathematically describe the constraints to which each member of the mechanism is subjected and solve a certain system of equations.

 

[Fulvio Romano]

Very interesting! any applications? Some exercise I can have fun with. .

boh...for now I can post an extract from the book "Industrial Robotics" of Sicilian brown with which I had the fortune to do thesis. robotics students call it "the bible."

It's 30 pages of 400... 7.5%... the limit not to take complaints is 15%, right?

 

[Vortexsheet]

Of course you can do it, but in the particular case of the plan mechanisms things are simplified. if you want to start from zero you should consider the bond of flatness,
You should check that this bond and the 'cerniera' constraints constitute a set of non-regular constraints (dini's theorem). ergo applies grubler, justifying these simplifications takes much more time and a certain rigour - other risks of producing false motivations

 

[snaroz]

Okay, thank you.
I attached the representation of the mechanism according to graph theory, okay?
other thing. the number of independent circuits l, applying the formula l=j-l+1, where j is the number of cinema pairs and that of members, I come 1. Okay?
Thank you!

 

[Vortexsheet]

boh...for now I can post an extract from the book "Industrial Robotics" of Sicilian brown with which I had the fortune to do thesis. robotics students call it "the bible."

It's 30 pages of 400... 7.5%... the limit not to take complaints is 15%, right?

Thank you! :finger: Well you could always get a half complaint:biggrin:

 

[snaroz]

hi, the text explicitly says to consider the pistons of the hydraulic cylinders as motive members, so they are members to all effects or wrong?
analyzing the mechanism only "on one side", if the above cylinders are also considered, 6 members are added and also the following cinematic pairs:
1) prismatic couple inside-outside boom cylinder;
"""""""""""""""""""""""""""""""""""""""""""""""""""""""""""" stick cylinder;
""""""""""""""""""""""""""""""""""""""""""""""""""""bucket roll cylinder;
4) boom cylinder-house (rotoidale);
5) cylinder-tripower boom (rotoidal);
6) stick cylinder-boom
7) stick cylinder-stick
8) bucket roll cylinder-bucket;
9) bucket roll cylinder-tripower.
therefore total members become 12 and pairs 15. applying the grubler formula the degrees of freedom, of course, remain always 3. Is that okay?
Thank you!

 

[snaroz]

hi, then, indicated with h1, h2, a and b respectively the stunga hook point at the house, the boom hook point at the house, the tripower coupling point at the boom and the stunga coupling point at the tripower, I considered the articulated quadrilateral formed by the h1h2, h1b, h2a, ab. I replaced therefore the rods with vectors and I set the equation
I then fixed a reference system and considered the values of the angles of inclination of the rods regarding the x axis.
I also assumed that the vector modules, i.e. the lengths of the rods, were known, and that the angles of inclination of the h2h1 and h1b vectors were also known (i.e. I fixed the inclination of the stunga and the passing segment for h1 and h2). so I got a system of two equations in the two unknowns that are two angles of inclination.
under these conditions, so that it can close the mechanism, it is obvious that that system must have only one solution.
If what I have written now is correct, how do I proceed?
Thank you!

 

[snaroz]

If I'm not wrong I should write four closing equations, then solve a "system" of eight equations or wrong? What would there be to implement in matlab?
Thank you!

 

[snaroz]

writing the four closing equations and considering the hydraulic cylinders as variable vectors (their length is obviously not constant), I got a system of four vector equations in six unknowns. Now, according to what's written on the book, such a stuff can be solved with the newton-raphson method. However, I do not think I should do this, but do I have to implement these equations in matlab so that the software can solve them?
Thank you!

 

[Fulvio Romano]

Sorry snaroz, but my family is increasing by 33.3% at these times... I can't follow the speech.

 

[snaroz]

in what sense:tongue:?
Are you "reproduced"?
Well if it's so good!! !

in any case my question is simple. After setting four closing equations, I get a system of eight scaling equations. What do I do? Do I have to fix this? What does matlab have to do with it?
Thank you!