'Allen Key' simulation project by BenLewis


#1

I created a new simulation project called 'Allen Key':

This is a simulation of an allen key being torqued up. Its purpose is to demonstrate the use of remote loads and remote displacements.


More of my public projects can be found here.


#2

This is a simulation of an allen key being torqued up. Its purpose is to demonstrate the use of remote loads and remote displacements.

The ball end is constrained from rotating when it engages with a socket head screw. This is simulated with a remote displacement.

A large moment can be applied to the short end of the key when it is fitted with an extension pipe. This moment is simulated as a remote force.

The applied force is reacted at the right angle bend in the allen key with the operators hand. This is simulated with a deformable remote displacement.

The meshing and solving operations can be run on one core.

The OnShape geometry is available for download from here https://cad.onshape.com/documents/3530c18c4c2f4a34b5b72fa0/w/b41458effd1f464ebf2cef3b/e/374dd0de09f649078890a28b

HAND CALCULATIONS

(excluding notch stress concentrations)

The applied torque is given by:
Torque = Force * Distance
10 Nm = 50 N * 0.2 m

The polar moment of inertia is 877E-9 m^4

The expected shear stress at the narrow cross section, just above the ball, is given by:
Shear Stress = Torque * Radius / Polar Moment of Inertia
342 MPa = 10 Nm * 0.003 m / 877E-9 m^4

Von Mises stress for pure shear is given by:
VM = SQRT (3) * ABS(Shear Stress)
593 MPa = SQRT(3) * (342 MPa)

The stress obtained from simulation is 614 MPa (4% difference).


Torque Boundary Condition
#3


#4

Nice simulation @BenLewis! :smiley:


#5

@BenLewis

Your project seems very similar to the one that I’m currently working on. As an extreme example, I have applied 1 lbf to the end of the long arm approximately 3 inch. The small head size shown in gold on the small arm has hex size of 5/32 inch. Material is A517 Steel, with a tensile yield strength of 100 ksi. The problem is that the model fails with a factor of safety of .7. The max equivalent stress is also shown to be 137 ksi. I’m curious how this is possible from only 1 lbf applied. Any help would be greatly appreciated. Many thanks in advance.


#6

Hi @MannMech,

I’ll take a look at this and get back to you.

Regards, Ben


#7

Hi @MannMech,

I’ve added another simulation to this project. It is for a 5/32" Allen key. From this simulation I get a nominal stress (just above the socket) of around 50 MPa (7.3 ksi) and a peak stress (on the corners of the hex) of 439 MPa (63.7 ksi).

The material yield stress is 689 MPa (100 ksi). Therefore the factor of safety is about 1.6.

Below you can see my hand calculations. There is a good correlation between the simulation results and hand calculations.

The simulation results could be improved by performing a contact analysis (with a socket). Also, the hand calculations could be improved by considering bending (about the z-axis) and shear (across the area of the Allen key just above the socket). But I think excluding these items does not make a big difference to the overall result.

I’m not sure why you are getting a factor of safety of only 0.7. You should check your simulation setup and do your own hand calculations. Some things to check are:

  • Dimensions of model
  • Units (model size, load size, material properties)
  • Loads (size, direction, applied geometry)
  • Constraints (does the model deform as expected)

I hope this helps.

Regards, Ben


#8

Ben lewis,

Sorry for taking so long to get back to you. I experimented around with the
design , and believe I’ve found the source of the problem. My design is a
bit different due to it being telescoping. The hex head sizes now in use
are 12mm(casing), and 8mm, 4mm telescoping bits. To illustrate, if the 4mm
is in the extended position, it will be locked in place with a compressive
spring button (push down on the button and the smaller successive bit can
slide out from the retracted position into a new hole locking in with the
larger encompassing bit). The only real questions I have now is: Will the
max stress occur on the 4mm bit or on one of the holes which lock the 12mm
to the 8mm, or 8mm to the 4mm together? If it is indeed the hole, how would
I go about calculating the stress on the hole due to torsion . Lastly can
bending stress be neglected ? Many thanks for your time and help.

Dimensions :

100mm long arm:
Assume uniform diameter.

On short arm:
40mm length (12mm hex head)
The 8mm bit sticks out 14.375mm from that, and the 4mm bit sticks out 5mm
from that.

12mm-8mm connection:
3mm diameter hole for button.

8mm-4mm connection:
2mm diameter hole for button.

Applied pressure from hand:
50N/750mm^2
5N compression downward

Best regards,

Spenser Mann


#9

Hi @MannMech,

You can apply physical contacts between each of the bodies in your geometry. You can then fix the inner bit and apply a load to the outer bit. The physical contacts will not bond the bodies together, rather, they will prevent one body from penetrating the other. From this analysis you will be able to determine the highest stressed area of your geometry.

If you add your geometry to a SimScale project and share it with me I can show you how to set this up. I will PM you my email address.

Regards, Ben


#10

Is there a simscale project of this anywhere?

I tried to download the geometry from onshape, but it requires a login.


#11

Hi @mhere, the Onshape geometry is public. Try searching for Allen Key - 8mm under public projects.