Calculations for a lifting tool for a bar

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SUMMARY

The discussion focuses on the design and calculations for a lifting tool intended for a 10-meter long, 10mm diameter steel bar with a flared head. Key calculations involve determining the lifting force using the formula P=F/A, where F is the weight of the bar multiplied by 9.81, and A is the area under the flared head. The design considerations include minimizing oscillation during lifting and ensuring the tool can effectively clamp the bar without causing deformation. Recommendations include using a collet mechanism to enhance grip and manage stress concentrations effectively.

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  • Understanding of Finite Element Analysis (FEA) principles
  • Knowledge of mechanical stress calculations, specifically F=ma and P=F/A
  • Familiarity with oscillation dynamics in long, flexible structures
  • Experience with lifting mechanisms and their design considerations
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  • Explore the design and application of collet mechanisms in lifting systems
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Engineers, mechanical designers, and safety professionals involved in the design and implementation of lifting tools for long and flexible materials, particularly in industrial settings.

oblong-pea
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TL;DR
Is the analysis for this lifting tool as simple as?
I'm designing a lifting tool for a bar with a flared head. The bar is say 10m long and 10mm in diameter with a head on the top that is 20mm in diameter and 10mm thick. There is a fillet from the head to the bar of 2mm (for reducing stress concentrations).
The bar will be lifted vertically and then transported to the side. Therefore i'm not expecting transverse forces such as torsion or shear.
I'm trying to do some FEA and hand calcs on designs for lifting tools and the below seems too simple?

I tend to second guess myself but is it as simple as: P=F/A
Where F = the weight of the bar x 9.81 for a force
and A = The area under the flared head which is engaged with a lifting tool?

Much appreciated
 
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A bar that is 10 meters long and 10 mm in diameter is a very flexible and hard to manipulate bar.

Steel bar?
How will your lifting tool attach to the vertical bar?
Will the vertical stroke of the tool be more than 10 meters?
How fast and jerky the vertical and horizontal movements will be?
 
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Lnewqban said:
A bar that is 10 meters long and 10 mm in diameter is a very flexible and hard to manipulate bar.

Steel bar?
How will your lifting tool attach to the vertical bar?
Will the vertical stroke of the tool be more than 10 meters?
How fast and jerky the vertical and horizontal movements will be?
Thanks for the reply

It will be steel, and I'm planning to use a system that goes around the disc/ flared head and clamps underneath (lifting from the underside of the disc) and lifts up from there. I'm aiming to make it something that connects to a crane or similar system so it's hoisted up; where the stroke would be more than 10m, possibly 12m giving 2m clearance.

The bar is in a guide tube, so shouldn't jerk when be removed until its clear of this guide tube, however its indoors so there won't be any wind considerations.
 
Once out of the guide, the bar will start sideways movement on its own, even in zero wind conditions.
It is a natural oscillation that becomes an S-shaped oscillation with one or more nodes.
All depends on the natural oscillation frequency of the bar.

In that case, I recommend you grabbing the head in such a way that it is free to tilt in all directions.
That would eliminate any deformation due to transferred moment from those oscillations.
A way of restraining those oscillation by means of cords manipulated by a worker, or other way, is also recommendable.
 
oblong-pea said:
The bar is say 10m long and 10mm in diameter with a head on the top that is 20mm in diameter and 10mm thick. There is a fillet from the head to the bar of 2mm (for reducing stress concentrations).
I assume the bar remains vertical throughout the lifting operation.

The tool could be made to operate like a cam, or a tapered collet, where the force applied against the bottom of the head is multiplied mechanically, to pinch the bar below the head, with a greater force than is applied by the head.
 
Thanks for the replies and guidance, really appreciated.

I am still struggling to clarify how to calculate the forces on the head though. If I were to use a collet or similar and lift from under the head, is the stress simply F/A? And is it purely compression on the under side?
Many thanks
 
Lifting by compression against the underside of the head becomes tension in the bar. The tension will be focussed on the fillet below the head. The failure mode would probably be by shearing along a conical surface that intersects the fillet. Once the stress reaches the bar shank, it will be F/A.

The advantage of a collet is that it clamps onto the shank of the bar, using the head as a position reference. The head only needs to carry the forces that activate the collet during the attachment. The head does not then need a step with a fillet, it only requires a conical taper that is steeper than the collet taper. It is easier to upset and spread the end of the bar than it is to form a flat head step. The length of the collet can be made sufficient, such that the tension in the rod rises gradually to F/A.
The critical included angle of the collet taper is the arc tangent of the friction coefficient between the bar and collet. A steeper collet taper will release easily, while a more gentle taper will lock, requiring a reverse impact to break the grip, that may have safety implications.
 

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