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 .
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failure of shear pin of gear coupling 2
- Thread starter 1234910
- Start date
Predictions require more information than you have supplied and are very affected by detailed circumstances that sometimes show up as empirical factors with wide variations in results. I suggest finding the analysis that was done to design the joint and see what the prediction was and whether there are any differences between those assumptions and what was delivered.
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Hi
A picture of the inside faces of the pin would be useful, also what are those marks on the pin half on the left of the picture?
Also a picture of the coupling might be good so we can see whether the hole in it as elongated or not.
We need to know what the torque value was during running and whether it's a fluctuating torque or constant.
My guess is this:- the taper pin as elongated the hole that it was inserted in over time, this as allowed slight movement between pin and coupling which creates bending on the pin as well as the shear stress and the pin eventually fails in fatigue due to bending, if you look at that picture the pin is bent.
“Do not worry about your problems with mathematics, I assure you mine are far greater.” Albert Einstein
A picture of the inside faces of the pin would be useful, also what are those marks on the pin half on the left of the picture?
Also a picture of the coupling might be good so we can see whether the hole in it as elongated or not.
We need to know what the torque value was during running and whether it's a fluctuating torque or constant.
My guess is this:- the taper pin as elongated the hole that it was inserted in over time, this as allowed slight movement between pin and coupling which creates bending on the pin as well as the shear stress and the pin eventually fails in fatigue due to bending, if you look at that picture the pin is bent.
“Do not worry about your problems with mathematics, I assure you mine are far greater.” Albert Einstein
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- #7
I will send you the photographs of inside face of pin as well as that of coupling ,can't get the torque value as the drive system shows only motor load ,which was nominal at that time . as you have written,pin is bent , from where ?
i am also interested in the mathematical part also . please suggest ,regarding shear load , shear torque ,fatigue strength of ST - 37 material also its life cycle ,shall we go for endurance strength or shear stress for the life of pin .
i am also interested in the mathematical part also . please suggest ,regarding shear load , shear torque ,fatigue strength of ST - 37 material also its life cycle ,shall we go for endurance strength or shear stress for the life of pin .
- Thread starter
- #8
Hi 1234910
Without an knowledge of the load or torque fluctuation we cannot predict how long the pin will last for, all we can do is take the maximum torque and show the pin should not fail under this maximum load but this doesn't prevent a fatigue failure and I would have thought that somebody would have already done the maximum torque load on the pin at the design stage.
I got the impression that the pin was slightly bent on the photograph, maybe it's how the picture as been taken.
“Do not worry about your problems with mathematics, I assure you mine are far greater.” Albert Einstein
Without an knowledge of the load or torque fluctuation we cannot predict how long the pin will last for, all we can do is take the maximum torque and show the pin should not fail under this maximum load but this doesn't prevent a fatigue failure and I would have thought that somebody would have already done the maximum torque load on the pin at the design stage.
I got the impression that the pin was slightly bent on the photograph, maybe it's how the picture as been taken.
“Do not worry about your problems with mathematics, I assure you mine are far greater.” Albert Einstein
- Thread starter
- #10
Motor torque = 9550x KW X SF /RPM
= 9550 X 400 X 1.5 /306
= 18725 Nm
Since Max Torque = 110 % of mean torque ( as set from elect side )
SO MAX torque - 1.1 x 18725
- 20597.5 Nm
Shear load on pin = T X 2/D
= 20597.5 X2X/.395 Since PCD of gear coupling = 395 mm
= 104291.13 N
= 104.29113 KN
ie , design load of pin = 104.29113 KN
BREAKING LOAD OF PIN - stress x araa of notch , now this stress we shall take endurance stress of shear stress of the material ST -37
- DO SUGGEST
= 9550 X 400 X 1.5 /306
= 18725 Nm
Since Max Torque = 110 % of mean torque ( as set from elect side )
SO MAX torque - 1.1 x 18725
- 20597.5 Nm
Shear load on pin = T X 2/D
= 20597.5 X2X/.395 Since PCD of gear coupling = 395 mm
= 104291.13 N
= 104.29113 KN
ie , design load of pin = 104.29113 KN
BREAKING LOAD OF PIN - stress x araa of notch , now this stress we shall take endurance stress of shear stress of the material ST -37
- DO SUGGEST
- Thread starter
- #11
if we take endurance strength of material for ST -37 material ,it is = 222 N/ sqmm .
Than breaking load on pin = stress x notch area
= 222 x 3.14 /4 x 24 x 24 since notch dia is 24 mm
= 81388.8 N
=81.3888 KN
and design load is =104.29113 KN
SO 81388.8/104291.13 X100 = 78.04%
IE, PIN CAN CARRY 78.04 % of design load before failure .
is it ok , please check ,or something to be added .
next how to find life of shear pin ( no of cycles before failure ) . and its fatigue strength .using Goodsman equation or using SN curve. please help .
Than breaking load on pin = stress x notch area
= 222 x 3.14 /4 x 24 x 24 since notch dia is 24 mm
= 81388.8 N
=81.3888 KN
and design load is =104.29113 KN
SO 81388.8/104291.13 X100 = 78.04%
IE, PIN CAN CARRY 78.04 % of design load before failure .
is it ok , please check ,or something to be added .
next how to find life of shear pin ( no of cycles before failure ) . and its fatigue strength .using Goodsman equation or using SN curve. please help .
I agree with others that the pin appears to have bent. You might also consider that there was some bending stress produced on the pin due to slight misalignment of the inner/outer bores at installation.
Were the coupling pin bores match machined to ensure precise alignment?
Were the coupling pin bores match machined to ensure precise alignment?
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Hello,
as per the given:
A) loading of one section
Given nominal torque & geometrics etc.: 9550*400*2/(306*0,395*452,4) = 140 N/mm²
B) loadbearing capac. of material (~Eurocode 3 / general)
Given fy with 235 N/mm² & γF to 1,5 & shear: 235/(1,5*sqrt(3)) approx. 90 N/mm²
Concluding:
One section would not suffice to transmit the full load. It is necessary to consider the second pin. The load distribution between these is statically indeterminate, as a rule of thumb (refer parallel keys) we assume an effective influence of 1,5. So:
90*1,5 = 135 N/mm² which is approximately equal to the load. At a give or take something philosophy, one could consider that this is ok.
So it is necessary to go into detail, and here there's not enough database to get a stringent conclusion.
Let's try:
a) Usually, to make up for defects within the material, EC3 & Co lower the loadbearing capac. of material by γM of 1,1
b) you state a possible overloading of 1,1 over the nominal
90*1,5/1,1=123 N/mm² whereas 140*1,1= 154 N/mm².
This is not soundly acceptable.
From your picture:
one could conclude that the shear pin was fixed on one side but was able to move on the other side, which would have induced a bending load (however small) onto the section in question.
Prediction (pls. allow me... ;-)
The fracture shall have started at a single point due to some bending effect and shall have developed as a fatigue crack over a longer period, as the base material is quite ductile. The final crack might show a bending to shear overlay of patterns.
remaining questions
- views on the fracture surfaces
- Check of 400 kW @ 306 rpm and/or info about drive setup & configuration
- derivation of Sf 1,5 in your calc: if factor of safety, where did you take it from and why was this value applied?
- Why was HRC test applied to this low fu / high ductility material? EN 18265 even does not give correlation values at HRC for this range of fu. Perhaps HRB testing is to be preferred?
Regards
R.
RSVP
as per the given:
A) loading of one section
Given nominal torque & geometrics etc.: 9550*400*2/(306*0,395*452,4) = 140 N/mm²
B) loadbearing capac. of material (~Eurocode 3 / general)
Given fy with 235 N/mm² & γF to 1,5 & shear: 235/(1,5*sqrt(3)) approx. 90 N/mm²
Concluding:
One section would not suffice to transmit the full load. It is necessary to consider the second pin. The load distribution between these is statically indeterminate, as a rule of thumb (refer parallel keys) we assume an effective influence of 1,5. So:
90*1,5 = 135 N/mm² which is approximately equal to the load. At a give or take something philosophy, one could consider that this is ok.
So it is necessary to go into detail, and here there's not enough database to get a stringent conclusion.
Let's try:
a) Usually, to make up for defects within the material, EC3 & Co lower the loadbearing capac. of material by γM of 1,1
b) you state a possible overloading of 1,1 over the nominal
90*1,5/1,1=123 N/mm² whereas 140*1,1= 154 N/mm².
This is not soundly acceptable.
From your picture:
one could conclude that the shear pin was fixed on one side but was able to move on the other side, which would have induced a bending load (however small) onto the section in question.
Prediction (pls. allow me... ;-)
The fracture shall have started at a single point due to some bending effect and shall have developed as a fatigue crack over a longer period, as the base material is quite ductile. The final crack might show a bending to shear overlay of patterns.
remaining questions
- views on the fracture surfaces
- Check of 400 kW @ 306 rpm and/or info about drive setup & configuration
- derivation of Sf 1,5 in your calc: if factor of safety, where did you take it from and why was this value applied?
- Why was HRC test applied to this low fu / high ductility material? EN 18265 even does not give correlation values at HRC for this range of fu. Perhaps HRB testing is to be preferred?
Regards
R.
RSVP
- Thread starter
- #14
Given nominal torque & geometrics etc.: 9550*400*2/(306*0,395*452,4) = 140 N/mm²
What does 452.4 inidcates i nthe above equation ? do suggest and what does 140 N/sqmm indicates ?
Given fy with 235 N/mm² & γF to 1,5 & shear: 235/(1,5*sqrt(3)) approx. 90 N/mm²
give the specific formula ? for th e above equation?
What does 452.4 inidcates i nthe above equation ? do suggest and what does 140 N/sqmm indicates ?
Given fy with 235 N/mm² & γF to 1,5 & shear: 235/(1,5*sqrt(3)) approx. 90 N/mm²
give the specific formula ? for th e above equation?
- Thread starter
- #15
LittleInch
Petroleum
What exactly was this gear coupling driving for 24 hrs/day?
The thing missing from this discussion for me was the degree and number of cycles of any shock loadings or high stresses.
Remember - More details = better answers
Also: If you get a response it's polite to respond to it.
The thing missing from this discussion for me was the degree and number of cycles of any shock loadings or high stresses.
Remember - More details = better answers
Also: If you get a response it's polite to respond to it.
a) 452,4 [mm²] from PI/4*24*24
b) 140 N/mm² is the shear load on one section of 24 mm dia, given nominal loading
c) loadcarrying ability in shear is 1/sqrt(3) smaller than that in tension etc.
d) ? = question mark to indicate request to the OP for input, not verified knowledge (..couldn't be sure, too far away ;-) )
e) It would be very kind to have (some) of these questions from the people around answered, esp. to fatigue issues & the fracture surface pictures. Like some kind of compensation for a good boy, see? ;-)
Thank you in advance
RSVP
b) 140 N/mm² is the shear load on one section of 24 mm dia, given nominal loading
c) loadcarrying ability in shear is 1/sqrt(3) smaller than that in tension etc.
d) ? = question mark to indicate request to the OP for input, not verified knowledge (..couldn't be sure, too far away ;-) )
e) It would be very kind to have (some) of these questions from the people around answered, esp. to fatigue issues & the fracture surface pictures. Like some kind of compensation for a good boy, see? ;-)
Thank you in advance
RSVP
Is the gearbox and coupling a commercial product?
How is the motor coupling half retained to the shaft.
Is this your type of coupling?
A 1 inch nominal pin suggests a shaft diameter at least 5 inches diameter
The acceptable methods of fitting large gear type couplings are-
- slip fit coupling hub with keys and setscrews for mild applications
- interference cylindrical fit between coupling hub ID and shaft OD with key.
- key with shallow taper fit which creates an interference fit when the hub is "drawn up" the tapered shaft
How is the motor coupling half retained to the shaft.
Is this your type of coupling?
A 1 inch nominal pin suggests a shaft diameter at least 5 inches diameter
The acceptable methods of fitting large gear type couplings are-
- slip fit coupling hub with keys and setscrews for mild applications
- interference cylindrical fit between coupling hub ID and shaft OD with key.
- key with shallow taper fit which creates an interference fit when the hub is "drawn up" the tapered shaft
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