clamped plate resistance to tangential forces

[pedro113]

Hello,

How do you calculate clamped down plate resistance to rotational force?
Let's say you have 400mm x 400mm aluminum plate clamped to a workbench (also aluminum) with 5KN clamping force in the middle.
How to you calculate the max tangential force to the plate side before the plate starts rotating along the axis?

Cheers

 

[dik]

I use a program similar to the attached... see if it uploads... extract it and change the*.x to *.exe. No bugs, malware, etc. My system is pretty secure and if you look at my profile... I'm reasonably serious...

Rather than think climate change and the corona virus as science, think of it as the wrath of God. Feel any better?

-Dik

 

[3DDave]

It varies a lot with the flatness of the bench and the plate. The most likely resistance case is to integrate over the area of the clamp face, which may produce a rather small amount.

 

[onatirec]

I would also suggest integrating over the clamp area.

However, I'd likely localize the clamp area to a 30 conical conical degree frustum. You'd integrate the pressure of this area x distance from center of rotation to figure the moment it could withstand - then set equal to your tangential moment (force x distance) and solve for the force.

 

[IRstuff]

Seems to me that friction would be the big unknown

TTFN (ta ta for now)
I can do absolutely anything. I'm an expert!

https://www.youtube.com/watch?v=BKorP55Aqvg

faq731-376 forum1529 Entire Forum list

http://www.eng-tips.com/forumlist.cfm

 

[saplanti]

I would not rely on the friction on the system that you described. I understand that you are trying to do something on the bench which may have lots of solutions. As you described we cannot visualise what you are trying to achieve.
Or you may use additional locaters on the perimeter of the plate against the torsional forces.

 

[rb1957]

clamp the plate at two points, or provide a stop (so the plate will bear against the stop and the clamp is resisting shear, not torque).

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

 

[Tmoose]

How many lb-ft resistance do you need ?
Will the force be smoothly applied, or more of an impact ?

By "inspection" of non-existent dwgs and sketches I'd guess 50 lb-feet at most, until or unless the aluminum galls and seizes.
As others said, the introduction of lubricant on purpose or by accident likely will create a lazy susan.

I don't think I saw the thickness of the plate and the workbench, or the fastener size, if the clamping will be applied with a thru bolt.
If the clamping force is applied with a lower jaw of a velociraptor with the teeth digging into the aluminum, the plates' friction may not count for much.

For plate thickness less than 13 mm ~ .5 inch and with typical manufacturing flatness and surface finish I'd expect the clamping force to be mainly applied to an area with diameter just 2 or 3 times the bolt Ø.
If on the other hand the concave faces of the plates were assembled together, so there initially is a gap ~ 1 mm new th center, the clamping force would be applied out at a much larger diameter, and the resistance to rotating would be proportionally larger as well.

The expectation of successful "calculating" requires some level of realism in the assumptions.
Just turning to page 212 or using calculator at

http://www.whatever

may be disappointing.

 

[rb1957]

instead of clamping, can you put two fasteners through a plate ? that would be much more quantifiable than friction.

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