lifting tong is it statically indeterminate 3

[nighthawk123]

I am trying to figure out the grip force of a tong; however I think it is overconstrained and therefore is statically indeterminate. I've attached a bitmap of the design. Any opinions? Castigliano's theorem? Deflection method?

 

[rb1957]

nice sketch ...
if you're lifting 100 lbs, then you know the load in links AB and AC = (100/2)/sin 27deg. links BDF (and CDE) are three force members; you know the direction of the force at B, and at D (GD is a 2 force member, so the force has to be axial), this means you know the direction of the force at F (thru the intersection of the line of action of the forces at B and D, which would be A). GD and GH are two force members, so the force in these links is axial. then (finally) the lifting finger, EHJ, is a 3 force member, the force at E has a line of action thru A, the force at J is horizontal, and the force at H is thru G; and when you extend these 3 lines of action they should all intersect at the same point.

i would move link GD so that it's between GH and GI (links GI and GH are separated by the thickness of the lifting finger. this may have been what the other posters were noticing, that the mechanism is alittle "twisted".

 

[btrueblood]

Oh, sorry, yes I see what you are saying rb.

Also, don't forget there is also a vertical force acting at J, being the friction force, or mu*Fhorizontal. The friction generated (and/or deformation of the lifted object, but this is statics, not strength of materials and tribology) is what lifts the block, there will be a minimum mu below which you can't lift something.

 

[nighthawk123]

rb / btrue,
thanks for putting the thought into this... where i disagree with the geometric approach from rb1957 is that GD contributes to link BDF as a 2 force member in the y direction, but the contribution from link CDE in the x direction is not included. this moves the intersection away from point A.

On link FIK (or EHJ) the product weight at J is not included. Doing so sends the force vector at an unknown angle (not horizontal), and still leaves the force directions at E&F unknown.

 

[rb1957]

GD is on the plane of symmetry of the part ... the load from CDE and the opposite link offset one another. also GD has only two pins so the load must (assuming no friction moments at the pins) be aligned along GD.

i accept btrue's comment about friction, so that the force vector at J isn't horizontal, but i think it should be close to it. the point of the tongs is to squeeze the thing being lifted. if the thing is too heavy (or if the friction is too low) for the amount of squeeze, the thing will slip out of the tongs. the amount of squeeze force is determined by the force applied at the top of the tong, the geometry of the links is determined by the size of the thing being lifted.

the geometrical approach (for looking at force vectors) is valid, if somewhat old fashioned.

 

[nighthawk123]

thanks again rb1957. can you look at my attached solution and let me know what you think? im kinda stuck on link FIK.

 

[rb1957]

you know the force in GI (consider ptG).

therefore you have only one unknown on FIK.

i think you're doing a "dead lift" at ptA, so that the squeeze force at ptK is determined by the geometry of the links ...

 

[unclesyd]

Once you get this thing going here is short synopsis of what's required to get this gripper in use.

I can't recall the full details but during a an in-house course I took about under the hook lifting and the devices thereof it was mentioned that there is a limit on the amount of multiples that can be used in operation of a gripper. There was also a discussion about having a problem with over center locking when trying to increase the grip by increasing the mechanical advantage.

http://www.pnl.gov/contracts/hoist_rigging/bth_l

ifting.asp

 

[desertfox]

Hi nighthawk123

I think your tongs are statically indeterminate and further more I beleive it won't work, your analysis assumes you have picked the 100lb weight up.
Imagine that the weight is resting on the floor and you place the tongs around it, how do you apply an external force to the tongs,your anaylsis shows a crane hook operating the tongs;therefore the only applied force to the 100lb weight is due to the mass of the tongs as the crane hook moves upward;therefore unless you apply a force greater than that required for sliding friction that block is going nowhere. Assuming friction coefficient to be 0.25 those jaws would require to apply 25lb minimum to prevent the tongs slipping off during lift.

regards

desertfox

 

[hydtools]

desertfox, I think the normal force in your example would be 200 lbs. not 25. Friction force = normal force * friction coefficient.
Friction force = 100/2 = 50
Normal force = 50 * 4 = 200

Ted

 

[unclesyd]

Just for the heck of it. Can you use 2 smaller grabs on a spreader bar. These grabs would the simpler 2 link models.

 

[desertfox]

hi hydtools

Yes your right I was thinking of the block sliding horizontally along the floor thanks for the correction.

desertfox

 

[nighthawk123]

The tong has points that imbed into the product. Therefore a gripping force of 100lb for a 100lb lift would be ok. If I had a product on a larger tong, say 20,000lb, a grip force of 20,000lb on 2 small points would easily imbed into the product, unless picking up something harder, like titanium.

Thanks to everyone's help I feel that a close graphical solution has been attained and the grip force will be quite high, in the neighborhood of 4.5 : 1 (ie 450 lb on a 100lb lift)

I built a scale model of the tong and it has a very high grip force, but maybe too high. It would have been nice to get a working force diagram that I could put into Excel with trig., thereby allowing me to adjust the strap lengths to exactly what they needed to be as well as the proportions of the lower parallel links.

The reason for using this "parallel linkage" design is that I would like to maintain parallel contact with the part throughout a gripping range.