Determining the clamping force on a Tube

[mech3d]

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 clamp (before twisting loose) when the weight is 4" from the center of the tube and connected perpendicular to the clamp. Assuming a rigid connection between the weight and the clamp.

Thanks for your help.
Mech3D

 

[edgepflow]

First compute the clamping force of the screws with this link:

http://www.engineersedge.com/calculators/torque_calc.htm

Then compute the holding force by multiplying by the coefficient of static friction (say 0.2 for a reasonable but safe number - but reaserch this for your actual materials)

 

[mech3d]

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, then the induced torque is then 10 in-lbs from the center or from the outer surface of the tube 2.5 lbs X 3.563" = 8.9 in-lbs?

If this is correct, then what is the relationship between the holding force and the torque from the weight? How can I determine let's say the safety factor?

Thanks Again

 

[edgepflow]

Could you scan a sketch of your arrangement?

 

[mech3d]

Here is a sketch of the question...

 

[edgepflow]

With this arrangement, you can see that the net moment must be zero to avoid slipping. Hence,

F_clamping X Radius_tube = F_applied X Larm

or,

F_applied = F_clamping X (Radius_tube / Larm)

Thus, the force you may apply at the end, F_applied, is less than the clamping force by the ratio:

Radius_tube / Larm = 0.4375 / 4 = 0.109

Test your setup slowly (add weight slowly) and use a generous safety factor (al least 3) on your clamping force you figured earlier.