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

[salamikorv]

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.

 

[Baluncore]

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.

 

[erobz]

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:

$$v = r_{pulley}~\omega$$

 

[salamikorv]

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?

 

[erobz]

Thanks alot for your reply.
Thats the thing im not sure how i can determine $A_H, A_V$.

Are you familiar with Newton Second Law?

How do i decide the radius for the pulley?

Thats probably a multi faceted question. I would read over this discussion.

https://www.physicsforums.com/threads/how-pulley-diameter-affects-forces.912087/#:~:text=For a given frictional torque at the axle,,against the frictional torque. Apr 23, 2017 #10

 

[salamikorv]

Are you familiar with Newton Second Law?

Thats probably a multi faceted question. I would read over this discussion.

https://www.physicsforums.com/threads/how-pulley-diameter-affects-forces.912087/#:~:text=For a given frictional torque at the axle,,against the frictional torque. Apr 23, 2017 #10

Yea i do, is T=A_V? Not sure what to think of A_H.

 

[erobz]

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. $\sum F_x = 0 , \sum F_y = 0$.

 

[salamikorv]

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?

 

[erobz]

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 $A$.

 

[salamikorv]

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)?

 

[salamikorv]

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?

 

[erobz]

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?

 

[salamikorv]

I think its (approximately) the radial load on the bearing, what do you think?

Ahhh i see, so theres no axial load?

 

[erobz]

Ahhh i see, so theres no axial load?

You mean into/out of the page?

 

[salamikorv]

You mean into/out of the page?

Yes

 

[erobz]

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.

 

[salamikorv]

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

 

[salamikorv]

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?

 

[erobz]

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?

 

[erobz]

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.

 

[salamikorv]

I have to hit the road for a while, someone will figure out what's going on before I get back.

Alright

 

[Baluncore]

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.

 

[erobz]

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

Notice the units on $L_{10}$?

 

[salamikorv]

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

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.

 

[Baluncore]

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.

 

[salamikorv]

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.

 

[Baluncore]

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.

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.

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 ).

 

[salamikorv]

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?

 

[salamikorv]

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.

 

[Baluncore]

Ohh thats PCD, alright then maybe 0.3.

PCD = Pitch Circle Diameter.

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.

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.

 

[Baluncore]

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.

 

[salamikorv]

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.

 

[salamikorv]

I do not know which bearing you are specifying.

Nevermind i had a mistake, its all good.

 

[salamikorv]

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?

 

[Baluncore]

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.

PCD=35mm

You mean more like 300 mm.

 

[salamikorv]

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.You mean more like 300 mm.

Ahh yea thats right hahaha my bad, then i get 6131N with PCD=300mm. I changed the geometry to a circle instead of a square by the way, why cant i have 8 bolts? Does it have to be 2 or 3?

 

[Baluncore]

I changed the baseplate geometry to a circle and not a square. I cant use 8 bolts? Why only 2-3?

You should use 8 bolts, but depending on the direction of the boom load, most will not be doing any real work.
The boom is a lever arm trying to overturn the base.

 

[salamikorv]

You should use 8 bolts, but depending on the direction of the boom load, most will not be doing any real work.
The boom is a lever arm trying to overturn the base.

I didnt understand the part with the boom load, where is that? Could you maybe draw it? Its not the force coming from the lift?

 

[salamikorv]

I got forces here aswell, do i have to do calculation based on them someway to get the forces acting on each bolt maybe and not only the lift load?

 

[Baluncore]

I didnt understand the part with the boom load, where is that? Could you maybe draw it? Its not the force coming from the lift?

The load hanging on the hook is not above the centre of the base.
Extend the line of the rope downwards, then measure the distance from the vertical axis of the base to that line. That is the radius to the load. That is effectively the boom of the crane.

The moment of the load, at that radius, is countered by the base. Some base forces are downwards, but on the side furthest from the load, the forces are upwards. The two or three bolts on that side are under tension.

I got forces here aswell, do i include them in someway?

I have no idea what is fixed and what is moving there.
You must draw and label a better diagram.

 

[erobz]

The load creates a moment $\circlearrowleft$ about the anchor.

The bolts farther away from the axis of rotation ( $\color{red}+$) tend to carry more load. Assuming only 2 of the farthest bolts balance the moment builds in a factor of safety.

You balance the moment from the bolts about the axis of rotation against your moment load. Then you use the proof strength of a desired grade bolt and calculated tensile load to calculate bolt cross-section.

 

[erobz]

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.View attachment 325008

What is going on in this calculation?

Please take a moment to learn how to use Latex to format your calculations so we can clearly see them. These grainy images of chicken scratch are annoying for someone trying to help you to interpret.

1) Why are you cutting the bearing load in each direction in half? Are there two bearings in each sheave?

2) For the $L_{10}$ life the units are millions of revolutions. I tried to point that out, it doesn't appear that it has sunk in.

 

[erobz]

View attachment 325009
I got forces here aswell, do i have to do calculation based on them someway to get the forces acting on each bolt maybe and not only the lift load?

I'm also suspicious of this force balance. Would share how you obtained them using the free body diagram of the boom?

 

[salamikorv]

I'm also suspicious of this force balance. Would share how you obtained them using the free body diagram of the boom?

Sorry for the late reply, theres only bending on that point, 15000Nm. So as you can see, the 20kN is in tension and then theres 30kN for the beam in the middle thats 30kN in compression. Do i use those to calculate the tension on each bolt or?

 

[erobz]

Sorry for the late reply, theres only bending on that point, 15000Nm. So as you can see, the 20kN is in tension and then theres 30kN for the beam in the middle thats 30kN in compression. Do i use those to calculate the tension on each bolt or?View attachment 325315
View attachment 325310View attachment 325311

Where is your free body diagram of the beam?

 

[salamikorv]

Where is your free body diagram of the beam?

My mate is responsible for the calculations of the beam, and he got that, so what i have for the bottomplate is this

 

[erobz]

My mate is responsible for the calculations of the beam, and he got that, so what i have for the bottomplate is thisView attachment 325317

Well, are you not able to do a FBD of the beam? It's always a good idea to check each other, and in this case I think it probably a really good idea.

 

[salamikorv]

Well, are you not able to do a FBD of the beam? It's always a good idea to check each other, and in this case I think it probably a really good idea.

I can but i dont have time to set my mind into doing the calculations for the beams, using s235, finding out the stresses for it, its going to take alot of time its not that easy. Whats important for my part is to determine the right bolts, and he gave me the forces thats acting on the plate and im not sure how to determine what bolts to use since those two forces act towards the middle point. Do you have an idea of what do to with those two forces to determine the right bolts? @Baluncore any ideas? I dont think i can use the formula you gave me before, it cant be that easy.

 

[erobz]

I can but i dont have time to set my mind into doing the calculations for the beams, using s235, finding out the stresses for it, its going to take alot of time its not that easy. Whats important for my part is to determine the right bolts, and he gave me the forces thats acting on the plate and im not sure how to determine what bolts to use since those two forces act towards the middle point. Do you have an idea of what do to with those two forces to determine the right bolts? @Baluncore any ideas? I dont think i can use the formula you gave me before, it cant be that easy.

You must get the correct forces before you can get the correct bolts. I'm not asking you to do the shear and moment diagrams for the beam, just a free body diagram of the forces acting on that beam.

 

[salamikorv]

You must get the correct forces before you can get the correct bolts. I'm not asking you to do the shear and moment diagrams for the beam, just a free body diagram of the forces acting on that beam.

They are correct, the forces he got is correct as our supervisors has checked them for us. If you can assume they are correct, how do i get the bolt tension?