Simple Clamping Force / friction question 1

[R3345]

In the picture attached, can somebody please explain if the axial force in each side of the U bolt creates 6000 N on each side of the bar , or is the 6000 N divided between each side?

 

[LittleInch]

IMO divided. Green arrows are 6000N (assuming vertical down arrow is meant to be on the inside surface of the clamp, Assuming this isn't a trick question (currently your support plate isn't attached to anything so will just move with the bar....) red arrow is dependant on friction factor you use between steel and support / U bolt.

My motto: Learn something new every day

Also: There's usually a good reason why everyone does it that way

 

[R3345]

Thanks for the input LittleInch.
So the force to slide the bar out of the U-bolt is 12,000*friction coeff?

That's what I was thinking but I'm not yet convinced. Only 6000 N is applied to the assembly after all.

 

[rb1957]

6000N on each face of the block. but the CoF is different on both sides (should be much lower on the bar side), so Force = 6000*(CoF1+CoF2)

another day in paradise, or is paradise one day closer ?

 

[rb1957]

'cause CoF used above is dependent on the area of contact, no?

another day in paradise, or is paradise one day closer ?

 

[R3345]

No I don't think that's right rb1957.
Coefficient of friction is determined by the materials only.
Friction depends only on the normal forces and the friction coefficient. It is not dependant on the contact area.

 

[rb1957]

it seems pretty reasonable to me that the bar will have a lower CoF on the block than the plate will.

test it ... bolt two bars together around a block, bolt two plates around a block; which is harder to move ?

another day in paradise, or is paradise one day closer ?

 

[CoryPad]

The force is 6000 N multiplied by the number of slip planes, in your case 2. You can learn more about slip critical connections by reading section 5.1 of the Guide to Design Criteria for Bolted and Riveted Joints here:

http://boltcouncil.org/files/2ndEditionGuide.pdf

 

[R3345]

Thanks Cory!

 

[LittleInch]

With U bolts it's not a simple as 6000 x F1 + F2. the bolt tends to deform sideways and bend. Given the very small surface area of the bolt / bar contact point, there is also the not inconsiderable issue of deformation of the steel bar under this sort of contact pressure, creating a small arc like deformation of the bar, be it elastic or plastic deformation.

For locations where axial force is the issue, you really need to have a definable flat area to grip the pipe to avoid these issues or simply clamp the bar between two flat bits of metal.

My motto: Learn something new every day

Also: There's usually a good reason why everyone does it that way

 

[Denial]

I'm with CoryPad on this.[ ] Sliding force[ ]=[ ]2x6000xFrictCoeff.

However there is a possible further complication.[ ] Once slippage begins the bolt assembly might rotate infinitesimally, which would lead to what I believe are known as "binding forces".[ ] It might be this sort of effect that underlies rb1957's feeling that the plate's friction will be greater than the bolt's.

 

[TheTick]

What? No free-body diagram?

I you split the system and diagram it, you will see it must be 6000N on each side (assuming "ideal" bolt that does not bend).

 

[R3345]

Thanks all for clearing up my doubts.

Sliding force = 2x6000xFrictCoeff.
This is precisely what I presented to a customer who laughed and told me I was wrong.

I showed on a fbd how there must be an equal reaction force on the opposite side of the U-bolt, but he maintained that if the tension in the bolt is 6000 , it must be 3000 each side.

Obviously I realise that in reality that assembly would rotate and bind etc. before coming off. I just wanted to clarify the theory.

 

[chicopee]

Yep, the free body diagram as mentioned by TheTick would explain it all. Basic Static problem without getting into all other details such as deformation, slip critical etc., making a mountain out of a mole hill. The pull needed to overcome the static condition is 6000 x static Cof on top + 6000 x static Cof on bottom

 

[FACS]

The friction between the round bar and the plate will not be the same as the friction between the flat plate and the flat bar.

The round U Bolt is made of different material and has a different finish; it will not be the same.

It is [the load * cof] of the flat plates

Plus

the [load * cof] of the flat plate and the round U bolt.



Charlie

http://www.facsco.com

 

[rb1957]

thought so ...

another day in paradise, or is paradise one day closer ?

 

[BrianE22]

I assume you have a load attached to the block? Then the block would have to slip before any load is taken up by the u-bolt on the other side of the plate. So for zero slippage, you can only count on one side of the block producing friction.

 

[R3345]

Thanks Brian E22. That is a good point.

 

[R3345]

So if the load on the u-bolt clamp was eccentric (as in the image attached) how would I calculate the force required for slip?

 

[rb1957]

i'd've thought there'd be some friction for the bar ... less than the plate side, but more than zero.

if you had the plate trapped between two U-bolts there'd be some friction, wouldn't there ?

another day in paradise, or is paradise one day closer ?