Continue to Site

Eng-Tips is the largest engineering community on the Internet

Intelligent Work Forums for Engineering Professionals

  • Congratulations GregLocock on being selected by the Eng-Tips community for having the most helpful posts in the forums last week. Way to Go!

failure of shear pin of gear coupling 2

Status
Not open for further replies.

1234910

Mechanical
Jun 11, 2016
22
Recently we had shear pin failure usedin gear coupling between motor and speed reducer gear box . two pins are used 180 degrees apart , the pin sheared within five months , Pin material is ST - 37 . pin neck dia is 24 mm. what is the shear load ,shear torque , fatigue stength of the pin ,please suggest . Motor power rating - 400 kw .RPM - 306 , twenty four hours running . atthe time of failure motor load was founf nominal, some beach marks are seen on the fractured surface , may be fatigue failure . how to find the life of shear pin .
 
 http://files.engineering.com/getfile.aspx?folder=5849c23b-79ff-4e3a-a0c0-f7184a9f6481&file=DSC01117.JPG
Replies continue below

Recommended for you

Hi 1234910

We really do need more information, any chance of those pictures on the faces of the failed pin and those of the coupling holes?

If this is a coupling you've purchased are you sure it's the correct rating for the job in hand?
Have contacted the gear coupling manufacturer?

Agree with Rolmec on the 90N/mm2 for the allowable shear stress, so the shear stress is exceeded at maximum torque.

Now to do a fatigue analysis you need some data on the load cycling that would cause the pin to fail, you don't appear to have that information however the pin can still fail in fatigue by relative movement between itself and the components it's clamping, thus generating the cyclic loading required.

A rough rule of thumb which avoids fatigue failure is:- if the tensile stress during operation is less than about 40% of the Ultimate tensile strength of the material, then fatigue is not likely to occur.

Why is the pin grooved? It looks to me as the groove would create a stress raiser magnifying the stress in that area which could accelerate the failure.

“Do not worry about your problems with mathematics, I assure you mine are far greater.” Albert Einstein
 
rolmec

what does effective influence of 1.5 means , which you have taken in your calculation .
 
the following pls. to be taken a wee little bit ;-) ... even if we are at a serious issue.... in a serious people's forum ...
script runs
R: "Let's make an agreement: I deliver the background to this 1.5 value, and you deliver some pictures of the surfaces!"
1: "Deal!"
R: "Current design standards (e.g. EC3) rule that in order to cover for the statistical variance of loads, calculation model uncertainties and deviations of load magnitude of unfavorably acting variable loads a partial safety factor is to be applied. The value of 1.5 is basically derived from structural standards, but correlates with applicable mechanical design standards (FEM) for the load case Main Loads."
...
script ends
your turn...
I will stop to push now, promise!

RSVP
 
oh,yes i was thinking about safety factor only ,i needed clarification, for the word effective influence .

thanks and regards
 
hi

Have you got on pictures on the failed surfaces of the pin which we requested earlier?

“Do not worry about your problems with mathematics, I assure you mine are far greater.” Albert Einstein
 
80% chance it is because the drive is misaligned with the motor; 20% chance the shear fuse pin is too small for the load, though it might be exactly the right size to prevent destroying the gearbox or the motor.

This assumes the fit of the pin prevents it from bending significantly under the expected load.

I don't have the correct stress analysis references but this is covered under very short beams under shear, which is not a simplified bending load or pure shear, but will also require evaluation of the contact deformation in the bores the pins are in to get a good approximation of the shear load on the pins. Alternatively, a finite element analysis could be used to determine how much radial displacement is tolerable to get to the desired life.

If these pins don't cost much, replacing them at 4 month intervals may be the least expensive approach. Otherwise find a company that does this sort of alignment.

****

Power = T * omega = 400 kW;
omega = (2*pi radians/revolution)*(306 revolutions/minute)*(1 minute/60 seconds); omega = 32 radians/sec
400 kW = 400,000 n-m/second;
T = Power/omega = (400,000 n-m/second)/(32 radians/second) = 14,000 n-m.

Without knowing the radius the pins are at I can't calculate the shear force and therefore not calculate the nominal shear stress. I expect that the loads from misaligment will be much greater than the nominal loads applied by the motor torque.
 
PCD OF TWO PINS - 450 MM . SO RADIUS = 450/2 MM .
 
3DDAVE
PCD OF TWO PINS - 450 MM . SO RADIUS = 450/2 MM .
 
Based on the .25mm radial flange misalignment shown in your photos, I suspect the fuse pin failed due to a combination of reverse bending cycles and shear. The fuse pins have a close tolerance fit in the flange bushings, the flange faces are closely coupled, the flanges are relatively stiff compared to the fuse pins, so even the modest .25mm radial offset between the flanges would likely produce significant bending stress in the short fuse pin body constrained length.

It would be best to start by correcting the misalignment condition in your drivetrain coupling before considering any other changes.
 
Status
Not open for further replies.

Part and Inventory Search

Sponsor