Ing94
Guest
Sorry guys I can't do a problem (five mechanical expert).
Is there anyone willing to help me solve it?? Thank you very much in advance
Is there anyone willing to help me solve it?? Thank you very much in advance
lucrezio
Guest
si
:smile:
:smile:
PierArg
Guest
Hi, if you haven't introduced yet, I suggest you do it here.
http://www.cad3d.it/forum1/showthread.php?t=187&page=136regarding your problem try to attach something or at least describe it.
radio
Guest
- to read the rules.
-present yourself.
-expose the problem so that it is understandable to all, even using sketches, patterns etc...
n.b. if you want to have more chances of finding someone to help you, describe how you plan to deal with the problem, where you can solve and where you are in trouble.
Usually questions like "does me exercise?" do not get answers.
Bye.
PierArg
Guest
I repeat the new proposal/ initiative launched by exatem.
from now on we will not respond with "resolving answers" but with further "intelligent questions" that make the student reason.
for more details:http://it.wikipedia.org/wiki/metodo_socratico
Ing94
Guest
I have attached the drawing given me by the school. It's a vertical milling head, and I need to check the two shafts and the two bearings.
data:
spindle power= 0.5kw
yield = 0,95
spindle rotation speed= 350 rpm
toothed wheel module= 2mm
the problem that when I go to check the trees and I design my usual beam as a sketch I can't see where the f force that I need to solve is. I do not ask to solve it but simply a council from more experienced people of me, even because I can solve it but I can not see the force in newton where it is located.
Thank you very much to all.
Attachments
PierArg
Guest
Let's start with the first "response/demand".
Do you have any idea how the object you posted works?
if you try to describe it (you will see that while you do it you will get yourself some clarifications...or even the answer)
Do you have any idea how the object you posted works?
if you try to describe it (you will see that while you do it you will get yourself some clarifications...or even the answer)
Ing94
Guest
being a spindle obviously turns. you will put the tool that will then work the piece. inside I noticed 2 conical toothed wheels that gear.
lucrezio
Guest
schools are over.
How come you want to solve this problem right now that you graduated?
PierArg
Guest
thanks to the loop! :tongue::
Of course the spindle turns... a bit like saying that the wheel rolls!
try to get down even more in detail.
however well for the geared wheels!
suppose to start from the application, that is the milling that turns: now describe the bike until you reach the horizontal shaft.
I apologize if I answer you with questions but it is a new method that we think we adopt in the forum (especially for students :biggrin
: make reason instead of providing the answers immediately.lucrezio
Guest
all forces go halfway through the dense band.
in the case of cylindrical wheels you only have tangential and radial forces; in the case of conical wheels you have tangential, radial and axial components.
PierArg
Guest
You should not have answered:tongue:. He was coming.
Ing94
Guest
I didn't expect all these questions.thanks to the loop! :tongue::
Of course the spindle turns... a bit like saying that the wheel rolls!
try to get down even more in detail.
however well for the geared wheels!
suppose to start from the application, that is the milling that turns: now describe the bike until you reach the horizontal shaft.
I apologize if I answer you with questions but it is a new method that we think we adopt in the forum (especially for students :biggrin
I'm telling you, because I looked at it, but I couldn't figure out where the force was. However I think that there is a motor that rotates the horizontal shaft and that consequently through the gearing of the two wheels dense rough conical also the vertical shaft (however I am not graduated I still have to do fifth being of 94).
PierArg
Guest
Ing94
Guest
I'm gonna have strength on my teeth, being the geared wheels? ? ?
MBT
Guest
Of course. . .
If not how do you exchange the time to pass on?
Besides, in addition to tangential forces, you will also have radials (see pressure angle) and axial (if the gear is helical)
If not how do you exchange the time to pass on?
Besides, in addition to tangential forces, you will also have radials (see pressure angle) and axial (if the gear is helical)
PierArg
Guest
now you can consider what wrote lucrezio :biggrin::finger:
lucrezio
Guest
I would do exactly the opposite: that is the horizontal shaft that turns and transmits power to the spindle.
as he knows the power that must come out (and the performance) the power (and therefore the strength) that enters and from there proceeds to cascade.
is much more linear.
At least in my opinion.
lucrezio
Guest
Yes, but you have the same radial, tangential and axial forces.
Don't get confused by that "right-headed" you wrote.
You tell me what you're doing with such a problem at school locked up and graduation?