cart on tracks

[Federico_992]

Uhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh ... try to count and see if you get the same

good evening mechanicsmg are feder a mechanical passionate student I came across the excel file that concerns the

c(resistente)=t*r+ fn*e It's the limit couple that if you get over it, you start slipping.

I attach here a summary card for a motorization of a cart like your 1000kg porata.

as you can see, setting the usual formulas in excel you can determine that you can make ramps from 0.8m/s^2 and if you try to make a fur more you get that wheels slip.

being a trolley that transfers things from one side to the other, especially in practice you do not go very specifically to deal with (e.g. you neglect the wheel inertia) and you go basically to evaluate two main aspects:
- torque calculation due to wheel deformation and then passing through fv- calculation of the acceleration torque that is discharged as force that is opposed to the tangential force of adherence t and that goes to sum up or remove the force of thrust fright to clarify all doubts:
- the cart weighs 1000kg
- the wheels in total are 4 therefore each wheel has 1000/4=250kg as vertical force
- the limit torque and the torque due to the turning friction determines it on the single generic wheel (motorized or mad that it is)
- the total torque above calculated serves the system, therefore to the 4 wheels but being that the motions are only 2 connected by single motor, it is has that the total torque of the system is to be divided by the number of moving wheels
- therefore the power dissipated by a wheel must be multiplied by the number of motorized wheels and divided by the number of motorizations that are realized (in this case a motor for 2 wheels...in other cases each motor a wheel)View attachment 50283

I've come across this excel file, I can't get out of it to calculate my facc.ruota, it could help me thank you earlier. greetings

 

[meccanicamg]

good evening mechanicsmg are feder a mechanical passionate student I came across the excel file that concerns the

I've come across this excel file, I can't get out of it to calculate my facc.ruota, it could help me thank you earlier. greetings

I see from your profile that you graduated in mechanical engineering, so at least three years you did.
mechanics you did it in at least two courses of which rational mechanics and dynamics.
in both teachings explained the wheel on the ground. They explained to you that there is friction that is that the reaction to the ground is moved forward compared to the shaft of the wheel.
the wheel advances as long as in the contact point there is adherence.
as already said, the moment is done with respect to the wheel centre and the balance equation is determined. Explain what you need and you have equation.

 

[Manny86]

I attach the results for 15t wagon that travels to 10 m/min with wheels diameter 160mm with only one two-wheel motorized axis.
you could do everything with a parallel red axle reducer and a nice ie3 engine braked in alternating current that has double braking torque of the nominal motor, front and back or all around a sensitive contact safety wire.View attachment 51492

Good morning to all, I saw this scheme to determine the various forces/powers. It is not clear to me how you calculated the cra, couple to win to accelerate. what kind of considerations have you made for this data?

Thank you. .

 

[meccanicamg]

Good morning to all, I saw this scheme to determine the various forces/powers. It is not clear to me how you calculated the cra, couple to win to accelerate. what kind of considerations have you made for this data?

Thank you. .

Forgive me but if you read the whole discussion from the beginning you should find it.
the vehicle to accelerate creates inertia force m•a that to return the port to the motor passing from the wheels.

 

[rapa]

Sorry to add... in calculations sometimes it should be considered that the tracks are not perfectly flat but slightly inclined, this involves the weight component as a strong force.

 

[Manny86]

Forgive me but if you read the whole discussion from the beginning you should find it.
the vehicle to accelerate creates inertia force m•a that to return the port to the motor passing from the wheels.

I read all the discussion, I understand very well that in acceleration there is an inertial force of m*a that then goes to the motor shaft. the thing that is not clear to me is the numerical results obtained. It seems to me that there is no correlation between what is defined in theory (m*a summary or subtract the value of the force derived from the tempting friction) and the value in the tables attached. in the example I have reported is indicated 16,7nm that I can not understand how this value turns out.

Sorry to add... in calculations sometimes it should be considered that the tracks are not perfectly flat but slightly inclined, this involves the weight component as a strong force.

Obviously this is another factor to consider in the calculation but in our case, we are probably analyzing an ideal condition with perfectly horizontal plane.

 

[AlexZ72]

on the site of the demag there is dip, catalog, formulas and online program to configure everything.

Could you turn the program link and formulas?
Thank you.

 

[meccanicamg]

Could you turn the program link and formulas?
Thank you.

It's been a while since I haven't used their portal but it's important for a technician to know how to look after the hit '

Gruppi ruota | Demag

I componenti per la traslazione e i gruppi ruota sono progettati e realizzati come soluzioni modulari per una vasta gamma di applicazioni.

www.demagcranes.com

Demag Designer

Demag Cranes & Components

www.demag-designer.com

 

[AlexZ72]

It's been a while since I haven't used their portal but it's important for a technician to know how to look after the hit '

Gruppi ruota | Demag

I componenti per la traslazione e i gruppi ruota sono progettati e realizzati come soluzioni modulari per una vasta gamma di applicazioni.

www.demagcranes.com

Demag Designer

Demag Cranes & Components

www.demag-designer.com

Thank you for responding. In fact I had already found these web pages of demag but I was interested in the formulas with which you made the excel schemes and on the site demag are not there.
I was able to reproduce the spreadsheet that you posted where couples and powers are drawn, including the limit pair beyond which slip the wheel.

but having read this discussion and the related ones there are some things I did not understand:

1) what is the value of the turning steel friction on steel (I understand the smell of the binding reaction that is then divided by the radius of the wheel).
the baldassini says 0.005 cm, while on wikipedia is 0.0005 m (0.05 cm).
you have indicated as more precise the calculation of niemann (volume I do not have): y=1,075*k/e), but first you indicated the k equal to 13,6 for steel and then you deepened on the niemann and indicated the formula k=q/(d*leff).
I did some accounts with a excel by putting reasonable weights and diameters to the eye and I got that the k of the formula is generally lower than 13,6, while the u is generally between 0.5 and 1 mm therefore would mistake the baldassini.

2) you have also proposed a calculation sheet for the most complete cart that in addition to the torque-power values indicates the contact pressure limit and the minimum permissible diameter of the wheel and here comes the coefficient k (which I did not understand if it is calculated or inserted as given) but does not have the values that would be obtained with the above formula (in one sheet is 13,6 in another 26).
Could you indicate the formulas that are used here?
I place them in the image below, cut from your sheet:

 

[meccanicamg]

Thank you for responding. In fact I had already found these web pages of demag but I was interested in the formulas with which you made the excel schemes and on the site demag are not there.
I was able to reproduce the spreadsheet that you posted where couples and powers are drawn, including the limit pair beyond which slip the wheel.

but having read this discussion and the related ones there are some things I did not understand:

1) what is the value of the turning steel friction on steel (I understand the smell of the binding reaction that is then divided by the radius of the wheel).
the baldassini says 0.005 cm, while on wikipedia is 0.0005 m (0.05 cm).
you have indicated as more precise the calculation of niemann (volume I do not have): y=1,075*k/e), but first you indicated the k equal to 13,6 for steel and then you deepened on the niemann and indicated the formula k=q/(d*leff).
I did some accounts with a excel by putting reasonable weights and diameters to the eye and I got that the k of the formula is generally lower than 13,6, while the u is generally between 0.5 and 1 mm therefore would mistake the baldassini.

2) you have also proposed a calculation sheet for the most complete cart that in addition to the torque-power values indicates the contact pressure limit and the minimum permissible diameter of the wheel and here comes the coefficient k (which I did not understand if it is calculated or inserted as given) but does not have the values that would be obtained with the above formula (in one sheet is 13,6 in another 26).
Could you indicate the formulas that are used here?
I place them in the image below, cut from your sheet:View attachment 69806

formulas related to hardness, pressure and coefficients are derived from the document that I attach to you.

As far as the subject is concerned, we have discussed this with the appropriate updates.
for any doubt, it is necessary to do the tests at variable speed and load to calculate the force of advancement that dissipates the desired friction.

 

[AlexZ72]

formulas related to hardness, pressure and coefficients are derived from the document that I attach to you.

As far as the subject is concerned, we have discussed this with the appropriate updates.
for any doubt, it is necessary to do the tests at variable speed and load to calculate the force of advancement that dissipates the desired friction.

Thank you.

 

[AlexZ72]

formulas related to hardness, pressure and coefficients are derived from the document that I attach to you.

As far as the subject is concerned, we have discussed this with the appropriate updates.
for any doubt, it is necessary to do the tests at variable speed and load to calculate the force of advancement that dissipates the desired friction.

I wanted to ask you a few things.
for my application I checked the wheels with the crane_wheels file... that you have posted and are oversized (it is a project already done and tested that I wanted to verify); Since the c3 parameter is the percentage of time of use, I imagine that the minimum diameter of the wheels is calculated to allow the canons 10,000 hours of operation, is it correct?

instead in the spreadsheet I couldn't find a formula for:k stribeck rolling pressuresigma eff real wheel-binary contact pressuredelta min minimum diameter admissible wheel

just for clarity could you indicate these formulas?
Thank you.

 

[meccanicamg]

to calculate c3 you need to calculate the intermittent service and then determine c3.
si= intermittent service
Yes.
Yes.
....

in reality the c3 does not have a basis of defined hours but how much it sees the cycle of fatigue regarding its life.

k is the weight that weighs on the single wheel divided the net area of the wheel projection (width angle).


the actual pressure is that of k evaluated with c2 and c3.

the minimum diameter is calculated by reverse formula of actual pressure where the admissible pressure was used in relation to the materials.

 

[AlexZ72]

to calculate c3 you need to calculate the intermittent service and then determine c3.
si= intermittent service
Yes.
Yes.
....

in reality the c3 does not have a basis of defined hours but how much it sees the cycle of fatigue regarding its life.

k is the weight that weighs on the single wheel divided the net area of the wheel projection (width angle).


the actual pressure is that of k evaluated with c2 and c3.

the minimum diameter is calculated by reverse formula of actual pressure where the admissible pressure was used in relation to the materials.

I had speculated that the k was equal to the fract load the projection of the wheel, but the result does not return.
in case I reported above the total mass was 30,000 kg (so on one wheel the load is 30,000 * 9,81 / 4 = 73575 n) and the diameter 300 mm, while the leff =50 mm for which:

= 73575 / (300*50) = 4,9 mpa
while in the above calculation sheet is 13,6 mpa.

for the minimum diameter is clear the calculation, but it is not clear how you get the c2, because the c2 is linked to the diameter of the wheel.
you can get the result by going for attempts: you hypothesize a diameter and you determine the c2 and eventually you change the diameter if you do not support the expected load. but how do you do in the spreadsheet to get the minimum diameter?

 

[meccanicamg]

I had speculated that the k was equal to the fract load the projection of the wheel, but the result does not return.
in case I reported above the total mass was 30,000 kg (so on one wheel the load is 30,000 * 9,81 / 4 = 73575 n) and the diameter 300 mm, while the leff =50 mm for which:

= 73575 / (300*50) = 4,9 mpa
while in the above calculation sheet is 13,6 mpa.

for the minimum diameter is clear the calculation, but it is not clear how you get the c2, because the c2 is linked to the diameter of the wheel.
you can get the result by going for attempts: you hypothesize a diameter and you determine the c2 and eventually you change the diameter if you do not support the expected load. but how do you do in the spreadsheet to get the minimum diameter?

for the minimum diameter see last paragraph previous post.
you have to use pzul in the usual formula, which represents admissible pressure and dmin revenues.

 

[osac]

c(resistente)=t*r+ fn*e It's the limit couple that if you get over it, you start slipping.

I attach here a summary card for a motorization of a cart like your 1000kg porata.

as you can see, setting the usual formulas in excel you can determine that you can make ramps from 0.8m/s^2 and if you try to make a fur more you get that wheels slip.

being a trolley that transfers things from one side to the other, especially in practice you do not go very specifically to deal with (e.g. you neglect the wheel inertia) and you go basically to evaluate two main aspects:
- torque calculation due to wheel deformation and then passing through fv- calculation of the acceleration torque that is discharged as force that is opposed to the tangential force of adherence t and that goes to sum up or remove the force of thrust fright to clarify all doubts:
- the cart weighs 1000kg
- the wheels in total are 4 therefore each wheel has 1000/4=250kg as vertical force
- the limit torque and the torque due to the turning friction determines it on the single generic wheel (motorized or mad that it is)
- the total torque above calculated serves the system, therefore to the 4 wheels but being that the motions are only 2 connected by single motor, it is has that the total torque of the system is to be divided by the number of moving wheels
- therefore the power dissipated by a wheel must be multiplied by the number of motorized wheels and divided by the number of motorizations that are realized (in this case a motor for 2 wheels...in other cases each motor a wheel)View attachment 50283

Good morning I'm struggling with a similar problem. if possible you might know the formulas useful to calculate the “couple to win at the wheel for turning friction” and the “couple to win to accelerate”? I would really help. Thank you in advance

 

[meccanicamg]

read all the explanations written in the various posts carefully.
there are also excel uploaded....when you're looking at this discussion. ...
There's all the information.
you can do a search on the forum and you will also find other material in different discussions.

is your first message.... have you read the rules? Did you show up? Did you look? No.
or so...