# What is the maximum torque a cylinder can withstand without turning when clamped by 2 set screws?

I am trying to calculate the maximum torque a cylindrical body can withstand when clamped by 2 set screws. The cylinder is placed in a housing with a sliding fit against housing surface, and the clamping force produced by the set screws are represented by F1 and F2.

I would like to verify that my below analysis is correct, would appreciate any comments:

For example, if we apply 20Nm torque to the cylinder, the force tangent to the surface of the cylinder is T=FR, 20Nm=F0.05m, F=400N.

Using a formula for estimated force produced from an M4 screw torqued to 1Nm, the force from each set screw is estimated to be ~850N. Assuming the coefficient of static friction between the set screw (steel) and the cylinder (brass) to be 0.5, the friction force from the set screws is 850N as shown above, and the cylinder should not turn.

• Probably why the bicycle handlebars have serrations where they are clamped... Sep 26, 2023 at 9:22
• your calcs seem right, but experience says that set screws are a poor method for preventing rotation. This is why keys exist. Sep 26, 2023 at 15:14
• They are nearly always used in conjunction with a keyway. If two setscrews are used, they go at 90 degrees to each other - not 180. Typically, one setscrew acts on the key itself. If it's just a little plastic blower fan, then the plastic is the weak point. Sep 26, 2023 at 18:52
• Ah, tiger guy beat me to it. I wrote my comment this morning but apparently didn't post it. Sep 26, 2023 at 18:53
• Thanks for the insights, esp about keyways used in conjuction with the set screws! Sep 27, 2023 at 10:08