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mechanical resistance toothed wheel

  • Thread starter Thread starter marco-v94
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marco-v94

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Hello everyone, I have an important question. I have a gear with two toothed module 1 wheels having a 90 teeth and an 114.vorrei know how to calculate mechanical resistance on the primitive diameter. the material is c40 with induction tempering.
thanks in advance to all those who can help me.
 
on the site khk, which is a Japanese gear manufacturer, you can download the program, or if you do not want to download it, simuli purchase of the two wheels as yours and see the loads that can hold.
 
What tooth width do you have? straight or helical teeth? rectified teeth or only cut of creator? What are the incoming spin numbers? in what tolerance are the gears?
 
What tooth width do you have? straight or helical teeth? rectified teeth or only cut of creator? What are the incoming spin numbers? in what tolerance are the gears?
the band size is 15mm. The teeth are straight. Now I'll explain the case better. I have a magnetic plate of 150kg (on which there are objects of 200kg) to whose center is connected, through a tree, the wheel of 114. to it is connected the wheel by 90.turning the wheel by 90 you have to tilt the plane by 50°. I wanted to know, when it is in the minimum and maximum angle position of the plate, the mechanical resistance between the two wheels. so I actually don't have a number of laps that can serve. As for tolerances, I don't know them.
 
from a spannometric calculation with the khk program

wheel z90:
permissible tangential force 102 kg
transmission torque 4.61 kgm

wheel z114
permitted tangential force 94 kg
Transmissible torque 5.39 kgm

try to make a verification, if you get back the numbers, the better would be a supervision of the good meccanimg!
 
Maybe if you put a sketch of work that you have to make things easy!
nb: the numbers I gave you refer to the calculation of bending teeth, I did not consider wear as you have zero velocity.
 
with z1=90 and z2=114 band 15mm material c45 I put for 5000 hours in input 0.023kw to 1rpm i.e. 219nm.
rectified wheels and made well.
Then it depends on what you want to do.
 
with z1=90 and z2=114 band 15mm material c45 I put for 5000 hours in input 0.023kw to 1rpm i.e. 219nm.
rectified wheels and made well.
Then it depends on what you want to do.
I'm sorry, but I'm not so clear about your answer. If I can get a sketch of everything in the day.
 
I'm sorry, but I'm not so clear about your answer. If I can get a sketch of everything in the day.
I simply geared the two straight teeth wheels and applied the rotation of 1 rpm in the input and I tried to raise the torque and power so as long as the bending check was at the limit.
I assumed that you knew the strength your system requires and therefore the strength it serves under the conditions you mentioned. I gave you the value "admissible" or otherwise bearable.

if instead the wheels are in c45 reclaimed but not tempered you only have 120nm that you can transmit as pair on the input wheel z90.
 
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I simply geared the two straight teeth wheels and applied the rotation of 1 rpm in the input and I tried to raise the torque and power so as long as the bending check was at the limit.
I assumed that you knew the strength your system requires and therefore the strength it serves under the conditions you mentioned. I gave you the value "admissible" or otherwise bearable.
Unfortunately the data I have are only the weight of the plate with what goes on, the type of toothed wheels and the maximum angle to which you have to rotate the plate. I wanted to know if those wheels resist mechanically during rotation. As I would also need some calculations, could you give me some formula to use?
 
for a static application I would say that you can use the formula of lewis that we wrote in different posts.

theory lewis post #6.
calcolo reducer.

you will certainly get much lower values than mine, at least one order of magnitude.

the basic formulation with the theory of lewis considers the breaking load that already in itself is doubtful since considering a steel with the hardened teeth there is to consider how many actual hrc there are and from that deducing the breaking load.

it is assumed that at least 50hrc there are and therefore we will have a breaking load about 1650mpa.
If you have taken wheels in the catalogue, ask the supplier to give you the applicable strength and strength value.

with this insert the data into the formula and determine the bearable couple.

this number revenues from the force that you generate from your system (which you have to post pattern, and tell us in what direction it is turned to understand where it is to consider gravity).
 
then applying what written in the last post I get that for the pinion I can have no more than 82nm.Screenshot_20200214_221052.webpwhile for the wheel can not bear more than 104nm.Screenshot_20200214_221116.webpclearly it is necessary to calculate the primitive diameter as m•z and to determine the force divides the pair by half primitive diameter.
but I repeat, we need to see what has been declared.
Besides reading and rereading I have doubts and I did not understand exactly what and how it tilts and therefore I can not make continuous reasoning on the load.

we have 150kg table and 200kg load that tilts by 50° but how much is the orthogonal distance of the resulting weight force? that distance multiplied the result of the sum of the loads makes the couple that will have to endure the hypothetical wheel from 114 teeth.
 
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This is a scheme to make you understand better how it is done. in a few words you want to rotate the plan of 50° (differently if clockwise or anticlockwise) by pressing a rotation to the wheel with z=90. I hope it is clear now15817296407514698640371603793357.webp
 
if the center of the toothed wheel z114 is in the center of the plate of 150 kg, in theory, if it turns slowly, neglecting the inertia, not from torque.
so the only pair is given by the 200 kg plate, so please, to make an exact calculation, also placed the size of this plate.
 
I forgot... placed also the distance of the baricentro plate from 200 kg regarding the axis of the toothed wheel. .
I'm sorry. So, what school are you doing?
 
I forgot... placed also the distance of the baricentro plate from 200 kg regarding the axis of the toothed wheel. .
I'm sorry. So, what school are you doing?
the 200 kg plate has the same size as the lower one, I designed them different to make you understand that they are two distinct things. However I do mechanical engineering, but let's say this is more of a personal project I'm working on
 
Sorry, but you didn't answer my questions.
Now I know that I'm going to draw many ires. but it doesn't seem possible to me, I graduated from the technical institute in the very distant 1976, but already in third if as an exercise of mechanics I couldn't set a simple calculation like yours, rightly they gave me a nice four and goodbye. I don't think I am a genius. .
But if we're at this point in engineering school today... We need to cry!
Excuse the vent! :
 
Sorry, but you didn't answer my questions.
Now I know that I'm going to draw many ires. but it doesn't seem possible to me, I graduated from the technical institute in the very distant 1976, but already in third if as an exercise of mechanics I couldn't set a simple calculation like yours, rightly they gave me a nice four and goodbye. I don't think I am a genius. .
But if we're at this point in engineering school today... We need to cry!
Excuse the vent! :
In truth I had already done the calculations only that I considered as a condition that the weight was applied at the end of the plate and were voltage values to the tooth really exaggerated (about 6000 mpa) against the admissible voltage values of 500 mpa. so the wheel, according to my calculations, would not have rectified which seemed strange to me. That's why I wanted advice from who knew more.
 
I had the doubt that it was the system. If the weight falls in the center of gravity you will have to pair at constant speed nothing. You only have to win the frictions. If you have a transient then you will have the system inertia to win as a couple.
if the resultant falls out of the rotation axis you will have equal strength per arm.

with the helical wheels you will have greater load capacity due to the fact that the total replenishment factor is greater.
 
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