How to calculate the Torque for a Set screw

In summary, the O-rings provide friction to keep the sleeve against the set screws. The purpose of the O-rings is unknown, but they are needed to keep the sleeve against the set screws. The set screws are tightened to rotate the sleeve against the O-rings. The stack up dimensions of the inner and outer assemblies are not guaranteed to match, so a gap is left in the casing.
  • #1
psugumaran
3
0
Hi,
I am working on a problem where I have to find the tightening torque required for rotating the outer sleeve. The Outer sleeve is rotated by human hand ( One hand operation) hence the sleeve torque feel is determined by the set screw.

Problem statement: Please refer the image. O-ring 1 at the spacer will compress the Outer sleeve against O-ring 2 by means of tightening the set screws in the collar. The spacer is movable (floating position) and collar is fastened with shaft (fixed position). When the set screw is tighten, the spacer moves and compress the O-ring (1) against outer sleeve.

I have to calculate, How much torque is required for set screw to tighten so that sleeve will be rotated against the O-rings (1&2). Consider it is a dry condition
Set screw = 3 nos (M3)
O-ring material = Viton
Sleeve material = Aluminum 6061-T6
Maximum Torque applied by human hand (One hand) = 9000 N-mm

Thanks in advance.
upload_2017-7-7_10-11-20.png
 

Attachments

  • upload_2017-7-7_9-39-7.png
    upload_2017-7-7_9-39-7.png
    13.3 KB · Views: 748
Engineering news on Phys.org
  • #2
psugumaran said:
Maximum Torque applied by human hand (One hand) = 9000 N-mm
Where does the figure 9000 N⋅mm come from?
 
  • #3
A bit off point, but, if these o-rings are for providing mechanical friction only, unless there is a chemical resistance issue, Viton is poor material for that type of application. A 60 or 90 Duro NPR (Buna N) would be a better choice for contact surface durability.
 
  • #4
I doubt any calculation will be very accurate but will need the diameter of the O-rings and the coefficient of friction which seems hard to find.
 
Last edited:
  • Like
Likes Nidum
  • #5
Rubber materials have a coefficient of friction that is highly dependent on water or oil as lubricant. If the surfaces are dry, rotation of the sleeve will damage the O-rings. You need a surface with a predictable coefficient of friction that can slide without damage. I expect that when the O-ring is dry, the spacer will slide on the steel set-screw ends rather than the O-ring slide on the sleeve.

What is the purpose of the O-rings? Why do you need to rotate the sleeve by hand?
 
  • Like
Likes CWatters
  • #6
@psugumaran : Sorry but use of three separate and non locked screws adjusted individually to position an essentially floating collar to compress an O ring by an unspecified amount is not good engineering .

Use of torque values on the adjusting screws to set up the assembly even approximately correctly is in any case going to be problematic .

In this kind of assembly the proper quantities to specify are the dimensions which control the compression of the O ring .

In components which nominally touch then the compression is controlled by the ring groove dimensions .

In components which have some separation then compression is additionally controlled by the separation distance .

Ideally the components are dimensioned so that they can just be put together and then the O ring compression gets set correctly without any adjustment at all .

If it is absolutely necessary to have adjustment then a gap dimension between components can be given and set to correct value during assembly using feeler gauges .
 
Last edited:
  • #7
Purely as a matter of interest the Bristol Siddeley/Rolls Royce Olympus engines of series 101 , 201 , 301 and early 593 all had a designed in gap in the outer casing . With so many component dimensions involved the stack up dimensions of inner and outer assemblies could not be guaranteed to match exactly . So a deliberate gap was left in the casing which was measured and then a precision ring specially manufactured to close the gap .
 

1. What is torque and why is it important in set screw calculations?

Torque is a measure of rotational force that is applied to an object. In set screw calculations, torque is important because it is used to determine the amount of clamping force that the set screw can produce on the object it is securing.

2. How do I calculate torque for a set screw?

The formula for calculating torque for a set screw is T = F x D, where T is torque, F is the force applied to the set screw, and D is the distance between the force and the axis of rotation. This formula assumes a perpendicular force to the axis of rotation.

3. What units are used to measure torque in set screw calculations?

Torque is typically measured in units of pound-force inches (lb-in) or newton-meters (N-m) in set screw calculations. These units can be converted using the conversion factor 1 lb-in = 0.11298 N-m.

4. How does the thread pitch of a set screw affect torque calculations?

The thread pitch of a set screw can affect the amount of torque required to secure an object. A finer thread pitch will require more torque to produce the same amount of clamping force as a coarser thread pitch. It is important to consider the thread pitch when calculating torque for a set screw.

5. Are there any other factors that can affect the torque calculation for a set screw?

Yes, there are other factors that can affect the torque calculation for a set screw, such as the material of the set screw and the material of the object being secured. Friction between the threads of the set screw and the object can also impact the required torque. It is important to consider all of these factors when calculating torque for a set screw.

Similar threads

Replies
8
Views
2K
  • Mechanical Engineering
Replies
9
Views
2K
Replies
2
Views
763
  • Engineering and Comp Sci Homework Help
Replies
13
Views
5K
  • Sci-Fi Writing and World Building
Replies
21
Views
1K
  • Mechanical Engineering
Replies
1
Views
4K
  • Engineering and Comp Sci Homework Help
Replies
3
Views
1K
Replies
9
Views
4K
  • Special and General Relativity
8
Replies
264
Views
28K
  • Introductory Physics Homework Help
Replies
22
Views
12K
Back
Top