Given force, need to determine what bearing to use for a crane

In summary: I don't know. There probably is in reality, but it's not a part of the "zeroth order" model. @Baluncore suggests it's about ##450~\rm{N}##. Perhaps they can weigh in on what is causing it, and adjust the model. Also, like I said, this is a baseline model. Friction and dynamics should be considered to improve...
  • #1
salamikorv
36
2
Hey!
I want to determine what bearings to use for this manual crane and the only thing i know is that it should be able to lift a wieght of 1000kg at max and that the winch should pull 1.24 m of cable/min. Im looking to find the force P to then determine C, and i think P is only the radial force so 10kN but i need to show that its true right? So i tried to draw a Free Body Diagram (Its FBD not PBD as in the picture) but im not sure what to do now.
Image (18).jpeg
Image (19).jpeg
 
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  • #2
Welcome to PF.

As a rough guide, the base of those lifting arms is usually a thick wall steel tube, about 300 mm long, 100 mm diameter.

It must handle a vertical weight of about 1250 kg, which can be carried on a small diameter bump with grease, in the end of the tube.

The side thrust at top and bottom of the tube is about 450 kg. That side force can be carried on a short polymer sleeve bushing at each end of the tube.

To prevent dirt and water from falling into the bearings, make the female sleeve rotate with the arm, while the male pintle is anchored to the base.
 
  • #3
salamikorv said:
Hey!
I want to determine what bearings to use for this manual crane and the only thing i know is that it should be able to lift a wieght of 1000kg at max and that the winch should pull 1.24 m of cable/min. Im looking to find the force P to then determine C, and i think P is only the radial force so 10kN but i need to show that its true right? So i tried to draw a Free Body Diagram (Its FBD not PBD as in the picture) but im not sure what to do now.
View attachment 324839View attachment 324840
Can you determine the reaction forces ##A_H, A_V##?

If you have a linear velocity ##v = 1.24 ~\rm{\frac{m}{min}}## you relate that to the angular velocity via:

$$v = r_{pulley}~\omega$$
 
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  • #4
erobz said:
Can you determine the reaction forces ##A_H, A_V##?

If you have a linear velocity ##v = 1.24 ~\rm{\frac{m}{min}}## you relate that to the angular velocity via:

$$v = r_{pulley}~\omega$$
Thanks alot for your reply.
Thats the thing im not sure how i can determine ##A_H, A_V##. How do i decide the radius for the pulley?
 
  • #5
  • #7
salamikorv said:
Yea i do, is T=A_V? Not sure what to think of A_H.
You have two directions to consider. Because the acceleration of point ##A## is zero, forces are balanced in each direction.

Apply Newtons Law for each direction independently i.e. ##\sum F_x = 0 , \sum F_y = 0 ##.
 
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  • #8
erobz said:
You have two directions to consider. Because the acceleration of point ##A## is zero, forces are balanced in each direction.

Apply Newtons Law for each direction independently i.e. ##\sum F_x = 0 , \sum F_y = 0 ##.
A_V=T and A_H=-T? Correct?
 
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  • #9
salamikorv said:
Its A_H=-T?
Ok, so you have the reaction components. Now you need to add them vectorially to find the magnitude of the reaction at ##A##.
 
  • #10
erobz said:
Ok, so you have the reaction components. Now you need to add them vectorially to find the magnitude of the reaction at ##A##.
Are you talking about the resultant force? So sqrt(A_V^2+A_H^2)?
 
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  • #11
salamikorv said:
Are you talking about the resultant force? So sqrt(A_V^2+A_H^2)?
What can i do with the resultant force here? Wait, is that the force P im looking for?
 
  • #12
salamikorv said:
What can i do with the resultant force here? Wait, is that the force P im looking for?

I think its (approximately) the radial load on the bearing, what do you think?
 
  • #13
erobz said:
I think its (approximately) the radial load on the bearing, what do you think?
Ahhh i see, so theres no axial load?
 
  • #14
salamikorv said:
Ahhh i see, so theres no axial load?
You mean into/out of the page?
 
  • #15
erobz said:
You mean into/out of the page?
Yes
 
  • #16
salamikorv said:
Yes
I don't know. There probably is in reality, but it's not a part of the "zeroth order" model. @Baluncore suggests it's about ##450~\rm{N}##. Perhaps they can weigh in on what is causing it, and adjust the model. Also, like I said, this is a baseline model. Friction and dynamics should be considered to improve it.
 
  • #17
erobz said:
I don't know. There probably is reality, but it's not a part of the "zeroth order" model. @Baluncore suggests its about ##450~\rm{N}##. Perhaps they can weigh in on what is causing it?
Would be great if i could get an explanation to whats causing it to have 450N as the axial load
 
  • #18
But alright until someone explains that lets assume its only the radial force, then that gives me P=sqrt((1000*9,82)^2+(1000*9,82)^2)=13,888kN. And lets say i want almost infinite life so L10=10^4 revolutions and a=3 for ball bearings, so that gives me the dynamic load capacity to be C=299 kN which is pretty high, have i done something wrong?
 
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  • #19
salamikorv said:
But alright until someone explains that lets assume its only the radial force, then that gives me P=14,142kN. And lets say i want infinite life so L10=10^6 revolutions and a=3 for ball bearings, so that gives me the dynamic load capacity to be C=1414213 N which isnt even in the SKF catalouge, what have i done wrong?

Are you Euro ( or whatever) where you use the comma as a decimal point?
 
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  • #20
erobz said:
Are you Euro ( or whatever) where you use the comma as a decimal point?
I have to hit the road for a while, someone will figure out what's going on before I get back.
 
  • #21
erobz said:
I have to hit the road for a while, someone will figure out what's going on before I get back.
Alright
 
  • #22
salamikorv said:
Would be great if i could get an explanation to whats causing it to have 450N as the axial load
Time for me to wake up.
I was summarising with regard to the rotation bearing at the base.

The winch wire passes over pulleys, the minimum radius of the pulley is determined by the winch wire details.
The radial force applied to the bearing of a guide-pulley could be from zero to twice the wire tension, depending on the change in wire direction over that pulley; Sin(2Θ).

There is little axial force on the guide pulley support bearings.
The winch drum is another issue, with alignment problems.

We need a better diagram of the machine, with identification labels on components. We need a specific question about one part of the mechanism.
 
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  • #23
salamikorv said:
But alright until someone explains that lets assume its only the radial force, then that gives me P=sqrt((1000*9,82)^2+(1000*9,82)^2)=13,888kN. And lets say i want almost infinite life so L10=10^4 revolutions and a=3 for ball bearings, so that gives me the dynamic load capacity to be C=299 kN which is pretty high, have i done something wrong?
As far as I can tell from the SKF website:

Bearing Rating Life
1681474348711.png


Notice the units on ##L_{10}##?
 
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  • #24
erobz said:
As far as I can tell from the SKF website:

Bearing Rating LifeView attachment 324876

Notice the units on ##L_{10}##?
Sorry for the late reply. Yea ive been using these formulas quite alot now, thanks. How would i do with the bolts at the bottom? I want to choose what bolts to use for the bottomplate, here:d
nynyny.png

Each bolts is evenly spaced out, i couldnt draw it so good but i think you get the point. I need a force thats acting on that plane right, but what force? I only know the force from the lift which is directed "into" the plane/bottomplate.
 
  • #25
Will the baseplate fail by bending?
How thick is the steel baseplate?
How is the vertical mast-tube attached to the baseplate?
What foundation is the baseplate bolted to?

Bolt tension is determined by mode of baseplate failure and:
1. Hook radius from axis.
2. Baseplate bolt Pitch Circle Diameter.
3. Number of bolts.

Lift_Arm.png
 
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  • #26
Baluncore said:
Will the baseplate fail by bending?
How thick is the steel baseplate?
How is the vertical mast-tube attached to the baseplate?
What foundation is the baseplate bolted to?

Bolt tension is determined by mode of baseplate failure and:
1. Hook radius from axis.
2. Baseplate bolt Pitch Circle Diameter.
3. Number of bolts.

View attachment 324997
1. Im not sure.
2. I think ill use 30mm in thickness.
3. Through welding.
4. I was thinking to the floor.
5. The hook radius would be 1.4 meters.
6. Is that the length of each side of the baseplate? 0,5 meters.
7. 8 bolts i think would be enough, placed like in the picture.
 
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  • #27
salamikorv said:
2. I think ill use 30mm in thickness.
I think that will be more than sufficient. I would guess closer to 12 mm.
But I would use diagonal gussets to stiffen the plate, like shown in my picture.

salamikorv said:
6. Is that the length of each side of the baseplate? 0,5 meters.
No. The 4 closer of those bolts lie on that diameter of a circle.

salamikorv said:
7. 8 bolts i think would be enough, placed like in the picture.
Assume only two bolts carry all the tension. n = 2.

Bolt tension = load * radius / ( n * PCD ).
 
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  • #28
Ohh thats PCD, alright then maybe 0.3. How can i use the bolt tension to determine what bolts to use, and is the load in the formula 1000*9.81?
 
  • #29
Another thing, this is for the bearing, i got these calculations and i ended up with 0,7 revolutions....
And the text part is saying "choosing a ball bearing diameter of 35mm from SKF, bearing 16003 with C = 6,37kN" What should i do to fix this and get atleast a million revolutions or higher.
Image (20).jpeg
 
  • #30
salamikorv said:
Ohh thats PCD, alright then maybe 0.3.
PCD = Pitch Circle Diameter.

salamikorv said:
How can i use the bolt tension to determine what bolts to use, and is the load in the formula 1000*9.81?
Load =1000 * 9.81 newton.

Look at manufacturer's bolt specifications.

salamikorv said:
4. I was thinking to the floor.
You need to consider how the bolts are held in the floor, so they do not pull out.
 
  • #31
salamikorv said:
Another thing, this is for the bearing, i got these calculations and i ended up with 0,7 revolutions....
I do not know which bearing you are specifying.
 
  • #32
Baluncore said:
PCD = Pitch Circle Diameter.Load =1000 * 9.81 newton.

Look at manufacturer's bolt specifications.You need to consider how the bolts are held in the floor, so they do not pull out.
Thanks alot i will come back if something is off with the bolts. I'll calculate in a moment.
 
  • #33
Baluncore said:
I do not know which bearing you are specifying.
Nevermind i had a mistake, its all good.
 
  • #34
Baluncore said:
PCD = Pitch Circle Diameter.Load =1000 * 9.81 newton.

Look at manufacturer's bolt specifications.You need to consider how the bolts are held in the floor, so they do not pull out.
For n=8, PCD=35mm, r=1,5m, and the load=1000*9,81, i get 52,5kN, is that the tension that every bolt have? Isnt that super high? Thats the force on all 8 bolts seperately?
 
  • #35
salamikorv said:
Thats the force on all 8 bolts seperately?
But only two or three bolts are holding down the side of the base furthest from the hook, where that moment can be of any real advantage.

salamikorv said:
PCD=35mm
You mean more like 300 mm.
 

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