I am currently thinking of a way to lock a piece sliding in and out of a slot, as shown in the diagram supplied, using a clamping knob and M6 bolt pushing onto the inner groove of the linear piece and, essentially, pinching it against the backwall.

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What I would like to estimate is the force applied on the linear piece by the bolt, as well as the force required to prevent it from slipping should Force_D be applied. Here, I'm assuming the linear piece and the structure that surrounds it is aluminium, but the bolt itself is stainless steel, and that the turning moment induced by Force_D is compensated by the structural rigidity of the slot.

Currently, what I've tried to estimate the force applied by tightening the bolt is the Preload Force (and subsequent Clamping Force induced as a reaction) and comparing that with the Friction Force, F_friction = mu_s * F_normal, where mu_s is the static friction coefficient between aluminium and stainless steel (found to be 0.47 after a bit of research). But this kind of goes against my understanding of preload.

For some context, these linear pieces, of which there are four, are adjustable legs to elevate a small chassis (approx. 1.2kg) that's meant to be attached to a robotic arm. The robotic arm needs to push down on the top of the chassis, so the clamping force pinching the legs must be sufficient so that the whole thing doesn't just slide down.

Could anyone give me some advice?

Edit: Chassis and legs overall, the M6 bolt pushes against the sliding piece to pinch it against the backwall, effectively clamping it. But how can I measure F_Bolt?

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    $\begingroup$ estimate the normal force, multiply by friction coefficient, reduce by a generous safety factor. That's it. $\endgroup$
    – Pete W
    Mar 18 at 13:39
  • $\begingroup$ Hello Pete, I suppose that's for the force on the legs. I'm glad it's that simple, but what about for the force exerted by the M6 bolt pushing on the sliding piece? Is that the preload force? $\endgroup$ Mar 19 at 14:17
  • $\begingroup$ The diagram is a little confusing. What is $F_M$? ..... The M6 presumably is the source of additional normal forces, hopefully the dominant source of them. A frictional resistance-force would be generated everywhere where you have contact with a normal force. It's also not practical to model the axial force generated by the screw with any kind of precision (maybe with metal parts in good condition, and lubricant it would be). You have to "overdesign", i.e. have plenty of excess normal force and thus excess friction, to clamp reliably $\endgroup$
    – Pete W
    Mar 19 at 14:26
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    $\begingroup$ Hello, fM is the mass of the chassis the leg is holding up. The M6 bolt is screwed in until it pushes the leg against the structure, effectively clamping it and preventing it from sliding. It's also constrained by friction on all sides against the aluminium structure that houses it. I've added a new diagram to hopefully illustrate this. What I wanted to know is if I can estimate the force applied onto the sliding piece by the M6 screw by using the preload force. $\endgroup$ Mar 22 at 11:32

1 Answer 1


It seems that you are greatly concerned with the possibility of collapse. I suggest that you forget about torque. Drill a small blind hole at the height you want your equipment to sit and use an extended tip machine screw that will fit into that hole. Then there is no way for it to slide up and down. If you want to vary the height, just drill a series of holes at a suitable intervals. To level your platform on an uneven floor, some adjustable feet would do the trick.

Or were you planning on using a torque wrench every time you adjusted the height? You know that isn't going to happen. People will just tighten the bolts until the legs don't move.
This doesn't seem to be an application where torque is critical, unlike, for example, installing Timken bearings.

My analysis of your setup
The bolt presses the linear piece against the frame, which I believe is aluminum. Static friction Al/Al is about 1.05-1.35 (clean and dry) so you have two frictional surfaces.
The preload gives you the clamping force before any additional loads. You also have to take into account any deformation of the aluminum strut. In your set up loading the apparatus will decrease the clamping force on the AL-Al side by (Fd + Fm) sin 15 (minus a load factor of approx 0.1) , but will increase it on the bolt side by the same amount. see for example (This gives an example for forces tending to separate sheets bolted together, but you are more interested in shear forces. critical slip)

The net forces of the strut will be weight minus frictional forces

$F_{net}$ = (Fd+fm)cos15 - final clamping force al/al - final clamping force al/bolt.
If it’s negative you will be fine.


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