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

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To determine the appropriate bearings for a manual crane capable of lifting 1000 kg, the discussion focuses on calculating the radial force (P) and the dynamic load capacity (C) needed for the bearings. The radial force is estimated at 10 kN, and participants suggest using a Free Body Diagram to analyze forces, including potential axial loads. There is a debate about the correct calculations for bearing life and load capacity, with some participants noting discrepancies in the results obtained. Additionally, considerations for the baseplate and bolt specifications are discussed, emphasizing the importance of understanding the load distribution and the role of the boom in the crane's mechanics. Overall, accurate calculations and a clear understanding of the forces involved are crucial for selecting the right components for the crane.
  • #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.
 
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  • #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.
 
  • #36
Baluncore said:
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?
 
  • #37
salamikorv said:
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.
 
  • #38
Baluncore said:
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?
 
  • #39
Image (21).jpeg

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?
 
  • #40
salamikorv said:
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.

salamikorv said:
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.
 
  • #41
The load creates a moment ## \circlearrowleft## about the anchor.

1681737787954.png


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.
 
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  • #42
salamikorv said:
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.
 
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  • #43
salamikorv said:
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?
 
  • #44
erobz said:
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?
crane.png

342339288_971008814075326_1670727586082600693_n.jpg
342384676_181770108078580_4037401906670205964_n.jpg
 
  • #45
salamikorv said:
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?
 
  • #46
erobz said:
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
kn.png
 
  • #47
salamikorv said:
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.
 
  • #48
erobz said:
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.
 
  • #49
salamikorv said:
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.
 
  • #50
erobz said:
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?
 
  • #51
salamikorv said:
the forces he got is correct as our supervisors has checked them for us
What supervisors? Is this for a schoolwork project? Or are you tasked with doing this in real life at work?
 
  • #52
berkeman said:
What supervisors? Is this for a schoolwork project? Or are you tasked with doing this in real life at work?
I'm sensing this is a school project.
 
  • #53
berkeman said:
What supervisors? Is this for a schoolwork project? Or are you tasked with doing this in real life at work?
Yea its a school project, no its not for real life work. We get help from the studentassistents in the lectures all the time. They give us direct methods that we can use to make our parts, thats how i determined the bearings because its thanks to them. But ive been sick this week, still i am, so i havent been able to go to school and ask for help and its been difficult to understand how i can get the tension for the bolts when i have two forces directed in two different directions.
 
  • #54
Okay, thanks for the info. I'll move your thread to the schoolwork forums then. You will still get the same good help there.
 
  • #55
berkeman said:
Okay, thanks for the info. I'll move your thread to the schoolwork forums then. You will still get the same good help there.
Oh alright. @erobz can you still help?
 
  • #56
1682180509966.png


Maybe I'm missing something or inferring too much about dimensions. Lets try a sanity check to see if that is indeed the loading on your base. Apply the loads your supervisors tell you are correct. Is the beam in equilibrium?
 
  • #57
erobz said:
View attachment 325318

Maybe I'm missing something or inferring too much about dimensions. Lets try a sanity check to see if that is indeed the loading on your base. Apply the loads your supervisors tell you are correct. Is the beam in equilibrium?
Yes it is.
 
  • #58
salamikorv said:
Yes it is.
what are the missing dimensions?
 
  • #59
erobz said:
what are the missing dimensions?
1.5m to the left, i mean thats the length from the left end of the beam to the point where you drew the upward force, and 0.75m is the length thats left
 
  • #60
salamikorv said:
1.5m to the left and 0.75m to the right
Explain how its in static equilibrium with an unbalanced horizontal force component from the support and the tension on the pulley?
 

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