Help with a bolt torque for this vice please.

This is a Chinese vice I bought to manufacture multiple pieces in one op. Its awesome actually and cost US$650 plus freight.

Build quality, I give it 85% but I am fussy fussy. It took a while to learn how to bring it into Fusion, and it has a couple of variable points with the wedges that need to be setup if I am machining more than 2 pieces. Anyway, I am very happy with it.

Question though, and the manufacturer has no idea, they just copied it from somewhere.

The way this vice works is the wedges are pulled down with a M16 bolt. This down force on the wedge, pushes the face to clamp the workpiece. I guess its obvious to you guys who use these types.

I do understand there is a mechanical advantage in the ramp of the M16 thread, and also there will be an advantage in the wedge yes? So if the wedge was 45deg it would be no mechanical advantage?

But today, first time its happened, the 2.5mm cut-out on the top of my jaws, actually sheared off the bottom corner of my 6061 workpiece. Both of them actually different sides. It was especially cold today but I do pull these bolts up quite tight. The force must be huge.

So my question is if I was to use a torque wrench what would I pull these bolts down to? I use up to 14mm end mills and a 45mm facing mill so I do get paranoid a workpiece might move as has happened once or twice. Can someone advise on that? Vice image attached.

Hi Boydage - a coarse thread iso M14 bolt class 8.8 pulled up to 121Nm will have a load on it of about 44kN (4485kgf or 9900lbf) This is 65% of its yield strength. If you can tell me the angle of the wedge I can calculate the clamping force... or see attached maths Cheers Peter

Hi. Awesome. Thanks

Ok sorry I might have been incorrect its a M16 bolt. The angle I measured is 15deg on the angle. I looked at the maths and paused on the tan as I am unsure how to work that one out. What do you think?

Also, if I was to be able to work out the clamping force, I wonder what is appropriate to hold a workpiece? 500kg? 1000? I use the 2.5mm slot most times, and sometimes the workpiece might be 100mm x 80mm x 50mmH which is not super large but does need to be held.

Thoughts?

Hi Boy - to get technical I suppose its relative to cutting forces. So if you know the max loads your machine can apply this will give you a guide. Say you have a piece of metal in a vice and your typical coefficient of friction for an oiled surface is 0.1 then you need 10x the cutting force to grip it with. Say your machine can push 100kgf? (some VMC's can push 3T) then you need 1000kgf to resist that cutting force without the part slipping. That's why some fixtures have little teeth on it then its independent of friction. 15degs --- clamping force 1/(tan15deg) is 3.7 so for every 1kgf you get 3.7kgf clamping force per 1kgf bolt force. The diagram I just sent is wrong. I wrote Fc=F x tan(theta) but it should have been Fc=F/tan(theta) opps... Peter