Clamping force calculation

[RAMA HEGDEParticipant @ramahegde42374]

Hi, I need to calculate the clamping force. Details attached.

[RAMA HEGDEParticipant @ramahegde42374]

[RobinParticipant @robin]

I think we are all wondering what you are pushing against?

The answer depends on the angle of that driving face.

Approx 1 ton

[RAMA HEGDEParticipant @ramahegde42374]

The job is circular bar and it has been pushing against V- block.

[not done it yetParticipant @notdoneityet]

If nothing is moving the opposing forces, by Newton’s First Law of Motion, are equal. Direction is another matter.

One Pascal is one Newton / square metre.

[RedsetterParticipant @redsetter]

I think we are all wondering what are you actually trying to do, and is that really the best way of doing it?

[DC31kParticipant @dc31k]

Since you show a pin joint at the base of the middle arm, there will be no clamping force as you have drawn a mechanism. The horizontal arm will try to move to the left due to the wedging action at the right of it and will result in the piston tilting in the cylinder and all pressure being lost.

To make this statically determinate, you need an encastre joint at the base of the middle leg and to specify the angle at the right hand end of the horizontal arm.

[Dave HalfordParticipant @davehalford22513]

The Vee block will move.

[Michael GilliganParticipant @michaelgilligan61133]

Posted by Dave Halford on 02/05/2020 09:08:39:

The Vee block will move.

.

The Vee block is shown as ‘mechanically earthed’

… Did the Earth move for you today, Dave ?

MichaelG.

[SillyOldDufferModerator @sillyoldduffer]

Here's my attempt at the calculation, done with smath studio (free software) for which I am eternally grateful to Mr Duncan Webster of this Forum. Lots of good things about smath, in this case I use it to finesse the units automatically into consistent Metres/Kilograms/Seconds, and exploit built in values for pi and the gravitational constant (approx 9.81).

Ought to warn I am famously bad at maths! Much prone to silly mistakes, getting the wrong end of the stick and missing bits out. Double check everything, I'm liable to do things like get the lever the wrong way round. (Here I hope the lever reduces force on the job by 4/6…)

My answer: about two thirds of a metric ton. Hope it's right!

Dave

Edited By SillyOldDuffer on 02/05/2020 09:55:38

[SteviegtrParticipant @steviegtr]

If your math is correct then I hope his centre fulcrum is pretty rigid.

Steve.

[Dave HalfordParticipant @davehalford22513]

Posted by Michael Gilligan on 02/05/2020 09:33:32:

Posted by Dave Halford on 02/05/2020 09:08:39:

The Vee block will move.

.

The Vee block is shown as ‘mechanically earthed’

… Did the Earth move for you today, Dave ?

MichaelG.

I can't remember

[duncan webster 1Participant @duncanwebster1]

SOD's sum gives the force acting vertically down, to get the force pushing the round bit into the vee block you have to allow for the angle of the end face. Then it gets involved in friction, which is never a known quantity, and I haven't had my weetabix yet. As Robin says, need to know the angle of the push face, and the exact relationship between the pivots and the bar, if they are not in line it changes the answer

[Rod RenshawParticipant @rodrenshaw28584]

+1 for DC31k's analysis.

The system as drawn will not act as a clamp at all.

The "Centre Fulcrum" at the base of the vertical lever is shown as a pivot and is not rigid at all, so as the pressure in the cylinder rises the horizontal lever will press just a little on the work and then move to the left as there is nothing to restrain it from so doing.

Did I just repeat what DC31k said in less accurate language?

All that wasted maths!

Rod

[not done it yetParticipant @notdoneityet]

Dave(SOD),

Unfortunately it will be accelerating (well, something will) – according to Newton’s Second Law. if accelerating, by definition its velocity cannot be zero (for very long, anyway)!

[duncan webster 1Participant @duncanwebster1]

The base of the pivot point is clearly shown as grounded (slanty lines underneath), it won't move anywhere as long as it doesn't snap. We must assume the base of the cylinder is similarly prevented from moving down, but I think that is a safe assumption.

[SteviegtrParticipant @steviegtr]

Posted by not done it yet on 02/05/2020 12:23:41:

Dave(SOD),

Unfortunately it will be accelerating (well, something will) – according to Newton’s Second Law. if accelerating, by definition its velocity cannot be zero (for very long, anyway)!

Across the workshop floor probably. Not a good method of clamping. Must be a better way than that.

Steve.

[pgk pgkParticipant @pgkpgk17461]

I'd go with SOD but if one wishes to be difficult then it's point contact so the force expressed as pressure per area is infinite. Indeed the piston needn't be round and the whole thing could be end on for a long system….

pgk

[RAMA HEGDEParticipant @ramahegde42374]

Hi,

Here I am giving little more details.

Please find the image.

[duncan webster 1Participant @duncanwebster1]

Ah just had another look at the sketch, I'm in error. Because there is a little link between the lever and the fixed pivot all the side load will be applied to the top of the piston rod, which has no sensible means of resisting it, so it won't work.

Strange how you sometimes look at something and see what ought to be there, not what is there.

[SillyOldDufferModerator @sillyoldduffer]

Posted by duncan webster on 02/05/2020 12:27:01:

The base of the pivot point is clearly shown as grounded (slanty lines underneath), it won't move anywhere as long as it doesn't snap. We must assume the base of the cylinder is similarly prevented from moving down, but I think that is a safe assumption.

 

I agree about safe assumptions!

For completeness I assumed:

1. the title 'Clamping force calculation' indicates the drawing is to identify dimensions rather than being an engineering design.
2. The cylinder & piston is round rather than square.
3. The piston can't tilt.
4. All the fixed points are firmly grounded on Planet Earth. (But the answer in Newtons is OK for other planets!)
5. Friction is zero, ie the piston and joints are all 100% efficient.

But your other post mentions the angle of the lever on the rod as being relevant. I missed that!

So, if the angle at the end of the lever is 20°, what is the force F on the job, and F1 and F2 on the V-Block?

Back at school my mathematical humiliations were only in front of the whole class. Now I'm messing up on the internet!

Dave

PS This post crossed with Rama's clarification.  I guessed 20° correctly!  Amazing.

Edited By SillyOldDuffer on 02/05/2020 13:21:57

Edited By SillyOldDuffer on 02/05/2020 13:24:27

[Neil WyattModerator @neilwyatt]

I suspect this is a college assignment rather than a practical example… but if it gets the grey matter churning…

Neil

[duncan webster 1Participant @duncanwebster1]

Well if it is Rama should just put the answer, 'clamping force zero, stupid design' and leave it at that. What's Rama up to getting us to do his homewrok anyway? When I get back from walking the dog (daily exercise, only once promise!) I'll put something up assuming the lever can't move sideways to show the effect at the slopy end

[duncan webster 1Participant @duncanwebster1]

[Howard LewisParticipant @howardlewis46836]

If this is not a theoretical college question, it seems a needlessly complicated way of clamping a bit of bar in to a V block.

Wouldn't be a lot simpler to replace the linkage with a bell crank.

The clamping force, neglecting friction would be the force exerted by the piston multiplied by the lever ratio of the bell crank.

Or have I missed something?

Howard

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