Determining the clamping force on a Tube

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Discussion Overview

The discussion revolves around determining the clamping force on a hollow tube secured by a clam shell circular clamp, specifically how much weight the clamp can support before loosening. The context includes calculations related to torque, clamping force, and safety factors, with a focus on mechanical engineering principles.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • Mech3D presents a scenario involving a hollow tube and a clam shell clamp, seeking to understand the weight it can support based on the torque applied to the screws.
  • One participant suggests calculating the clamping force using a torque calculator and multiplying by a coefficient of static friction, proposing a value of 0.2 as a reasonable estimate.
  • Another participant calculates an axial force of 609 lbs from the torque and applies the friction coefficient to derive a holding force of 243.6 lbs, questioning the relationship between holding force and torque.
  • A later reply introduces a formula relating clamping force and applied force, emphasizing the need for the net moment to be zero to prevent slipping.
  • Participants discuss the importance of testing the setup slowly and recommend using a safety factor of at least 3 on the clamping force.

Areas of Agreement / Disagreement

Participants present various calculations and approaches, but there is no consensus on the exact relationship between holding force and torque, nor on the appropriate safety factor, indicating multiple competing views remain.

Contextual Notes

Assumptions include the rigidity of the connection between the weight and the clamp, as well as the choice of the coefficient of friction. The discussion does not resolve the exact values for safety factors or the implications of different coefficients of friction based on material properties.

Who May Find This Useful

Mechanical engineers, students studying mechanics, and individuals interested in clamping force calculations and torque applications may find this discussion relevant.

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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