# Will This Friction Clamp Hold the Weight?

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• NoelFa
NoelFa
TL;DR Summary
Friction Clamp Conundrum
Hi guys, I am trying to design a brake/locking mechanism to hold up a weight, here is some background; I have a weight 18kg, 170mm away from the pivot point on an arm. A brake compresses the arm at the pivot point. This consist of a M10 bolt compressing a surface area of 1500mm^2 on each side of the arm. The torque I apply to the bolt/brake is 5 N/m. The CoF is 0.3. Will the brake hold up the weight? I am struggling to find the correct formula for this calc online. I know the weight will produce a torque around the pivot point of approx 30Nm. The clamping force I think would then be Torque/(CoF x bolt diameter) = 2.5 Nm. Is this correct?

On a different not why does a taper help with these kind of friction brakes? I've seen it mentioned online but I'm not sure why. If the contact surface was at a 15° angle what would this do to the calculation?

Welcome to PF.

How do you apply the torque to the bolt brake, or nut?
We need a picture or a diagram. Please attach one to your next post.

A standard M-10 bolt has a pitch of 1.5 mm.
The effective thread diameter is; 10 - 1.5 = 8.5 mm.
Thread circumference = 8.5 * Pi = 26.7 mm
When you apply 5 Nm to the nut, the axial force on the bolt is greater.
5 Nm * 17.8 * (1000 / 4.25) = 20.944 kN.

Thanks for the feedback. Does the below illustration help?

What are the inside and outside diameters, of the friction pad annulus?

Pressure is not important in friction, force is important. The area of the annulus is not important, but the average radius is. In this case, we need to know that the force is distributed evenly, radially over the annulus.

If rather than a pancake, your friction brake had a coaxial sleeve arrangement, then there would be an advantage in a tapered bolt as one of the friction surfaces.

Understood. Lets assume the inner diameter is 12 mm and the outer is 40. How do we calculate the force in the case of flat 'brake pads' on the arm?

And then what changes in the calculation when you go to a conical bake pad like shown below? I know that this set up will give a better braking output but I'm not sure how to calculate this

NoelFa said:
Understood. Lets assume the inner diameter is 12 mm and the outer is 40. How do we calculate the force in the case of flat 'brake pads' on the arm?
ID = 12 mm; IR = 6 mm. OD = 40 mm; OR = 20 mm. Average radius = 13 mm.

From earlier; bolt axial force = 5 Nm * 17.8 * (1000 / 4.25) = 20.944 kN.
The two surfaces are in parallel; With CofF = 0.3
2 * 0.3 * 20.944 kN = 12.57 kN, acting at 13 mm.
The lever arm is; 13.0 : 170.0 = 0.0764526 ;
That can support; 12.57 kN * 13.0 / 170 = 961 N.
That would be 961 / 9.8 = 98.1 kg.
NoelFa said:
And then what changes in the calculation when you go to a conical bake pad like shown below?
The effective radius will be the radius in the middle of the taper. There is only one surface with a single taper. With the opposed double taper you show, there are two surfaces.

The taper is a wedge with angle α, that converts bolt axial force into a radial force, in proportion to; run/rise = 1 / tan( α ).

If the taper angle is too small, the taper will self-lock, so not always release when you loosen the nut.
That critical angle; α ≈ atan( CofF ) ;
For CofF = 0.3, we get; atan( 0.3 ) ≈ 16.7° wedge angle.
That is also the angle, above which, a wedge or a tapered pin will be ejected.

NoelFa said:
TL;DR Summary: Friction Clamp Conundrum

I am trying to design a brake/locking mechanism to hold up a weight,
Do you need the mechanism to be infinitely adjustable in angle? Or can you have it lock every 10 degrees or so? You can get a much stronger locking mechanism if you can use a gear-faces as locking surfaces ("Hirth Joint")...

https://www.hirth-joint.com/hirth-joint/

Lnewqban
@Baluncore I came at this from a slightly different angle, using the following calculator to get started (https://www.engineersedge.com/calculators/torque_calc.htm):

Arm/Lever torque around pivot point: 18 (kg) x 0.17 (mm) x 9.81 (m/s2) = 30 Nm

Bolt Clamp force: Bolt torque 5 (Nm) / Bolt diameter 0.01 (m) x CofF 0.2 = 2500 N

Number of braking faces = 2

Torque from braking mech: Clamping Force x CofF x average radius x number of braking faces = 2500 x 0.2 x 0.013 x 2 = 13 Nm
Torque required: 30 Nm

Therefore I would need to increase bolt torque or play around with some of the other figures.

Assuming this is correct I'm still not sure how to turn the calculation into a taper angle calc as you have mentioned. Can you run through the step by step here? Really appreciate all the help!!

You use the general approximation of torque; Tn = Kn * Faxial * Dp
Where Dp is the pitch diameter, and the nut factor; Kn = 0.2 ;
Bolt M10-1.5 has; Dp = 10.0 - ( 2 * 1.5 * 0.8660 ) = 7.402 mm ;
Then Faxial = Tn / ( Kn * Dp ) = 5 / ( 0.2 * 0.007402) = 3377.5 N.
Using CoF = 0.3; the friction disc torque = 26.34 Nm
Compared to the lever arm torque = 30.02 Nm

When you replace the flat pancake clamp with the α conical clamp, there is an advantage of 1 / Tan( α ) ;
For a cone of; α = 30°, that is a factor of 1.732
The radius of the friction surface is still; ( Ro + Ri ) / 2
With data from post #5, r = ( 30 + 40 ) / 4 = 17.5 mm.
The friction cone torque becomes = 61.42 Nm
Compared to the lever arm torque = 30.02 Nm

NoelFa said:
And then what changes in the calculation when you go to a conical bake pad like shown below?
There is a problem with the design of your conical clamp in post #5.
You have not specified which surfaces slide and which are keyed.

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