Determining the clamping force on a Tube

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The discussion focuses on calculating the clamping force of a clam shell circular clamp securing a hollow tube with an outer diameter of 0.875 inches and an inner diameter of 0.750 inches. Using a torque of 20 lb-in applied to two No. 8-32 screws, the axial force is determined to be 609 lbs, leading to a holding force of 243.6 lbs when accounting for a static friction coefficient of 0.2. The relationship between the holding force and the torque from a weight applied 4 inches from the center of the tube is established, emphasizing the importance of maintaining a net moment of zero to prevent slipping. A recommended safety factor of at least 3 is advised for practical applications.

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mech3d
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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
 
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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)
 
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
 
Could you scan a sketch of your arrangement?
 
Here is a sketch of the question...
001.jpg
 
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.
 

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