Recent content by mech3d

Here is a sketch of the question...

It is as simple as that? According to the calculator the axial force is 609 lbs. Assuming 0.2 for the coefficient of friction we have: 609 lb X 0.2 = 121.8 lbs, therefore with 2 screws X 121.8 lbs = 243.6 lbs holding force. So if I have 2.5 lbs of weight 4" from the center of the tube...

Given the following: 1. A hollow tube 0.875" O.D. with a 0.750" I.D. 2. A clam shell circular clamp 0.875" wide. 3. Two No. 8-32 screws securing the 2 halves of the clamp onto the tube. 4. Screws are torqued to 20 lb-in. I am looking for how much weight that can be supported by the...