We had an incident where part of a suspended rail fracture suddenly. The fracture face is totally brittle, i.e. no fatigue cracking, and no evidence of obvious pre existing flaws. We think this was due to massive overloading, though no one will own up to what happened. I have been asked to give a figure of the force needed to cause this failure. I am intending to use FEA on the component to determine the load required to produce a certain stress at the failure location. This certain stress I imagine would be a critcal value based on the material properties. Unfortunately I am not experienced in fracture mechanics so am unsure how to proceed. Ideally I would like to put some material properties into a black box and for it to churn out the stress required to break a steel bar with zero flaws. I know there is no such thing as zero flaws in reality but again don't know how to catergorise it when the flaws are microscopic. Any ideas?
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Force required to fracture a rail?
- Thread starter barrind
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- #2
Knowing the configuration would help. I'm assuming that this is normally supposed to be a statically loaded, or slow dynamically loaded (quasi-static) situation. Since you are trying to predict a failure load, you will need to use a non-linear analysis if you are using a general FEA package. You may also want to look at a program called "Franc2D".
It does 2-D fracture mechanics and is pretty easy to understand. There is a 3-D version, but I'm not sure if it is free or not. If this "rail" is a uniform cross section that is long...you can probably make a plane stress assumption and go with the 2-D assessment.
For impact, you will need to go with more of an "event simulation", which several of the general FEA codes have these days.
Garland E. Borowski, PE
Borowski Engineering & Analytical Services, Inc.
Lower Alabama SolidWorks Users Group
It does 2-D fracture mechanics and is pretty easy to understand. There is a 3-D version, but I'm not sure if it is free or not. If this "rail" is a uniform cross section that is long...you can probably make a plane stress assumption and go with the 2-D assessment.
For impact, you will need to go with more of an "event simulation", which several of the general FEA codes have these days.
Garland E. Borowski, PE
Borowski Engineering & Analytical Services, Inc.
Lower Alabama SolidWorks Users Group
if there's no pre-existing flaw and a purely brittle fracture surface, then you probably don't need fracture mechanics ... i think it is as you suspect, a dynamic, impact load on the beam.
what sort of material ? would you have expected a more ductile response ? what about the supports of the beam ? could the beam have sheared ? (was the beam supported rigidly near the load application point ?)
no surprise no-one's owning up ... sounds just like my teenagers !!
what sort of material ? would you have expected a more ductile response ? what about the supports of the beam ? could the beam have sheared ? (was the beam supported rigidly near the load application point ?)
no surprise no-one's owning up ... sounds just like my teenagers !!
To me, it seems the big question is how the material ought to fail. If the material specifiation indicates that it should normally have some ductile behavior, then something else besides just an overload is going on. Perhaps too cold of a temperature, an undetected flaw, a poor weld, etc.
You might also check actual support spacing and rigidity versus designed dimensions.
You might also check actual support spacing and rigidity versus designed dimensions.
MikeHalloran
Mechanical
Can you describe the geometry of the rail, the normal loading condition, and the geometry of the fracture in a little more detail?
Mike Halloran
Pembroke Pines, FL, USA
Mike Halloran
Pembroke Pines, FL, USA
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- #7
Ok, I was hoping me saying it was a rail would keep it simple and still allow for you guys to give me some ideas about sudden material failure. To give it a bit more detail then, the system is basically a suspended C hook used to transport coils of steel round around a despatch bay. The C hook is suspended on a beam via two pins, the beam is then suspened on an overhead rail via two motorised wheels. The beam is basically two channels spaced with two 100mm x 40mm section bars in the centre at either end. Each bar overlaps the channels by about 240mm, and extend out by about 380mm. These bars have a vertical hole to support the wheel mechanism in the extended portion. The location of the failure is the point where the extended part of the bar meets the overlap, i.e. the position of maxiumum bending. Basically the bit which failed is a beam under four point loads.
The failed section is therefore a 100x40mm rectangular cross section. The surface is 95% classic brittle failure with a 5% shear surface where it finally gave. I.e. there is no evidence of a crack before failure.
As I say we do not know for sure where the damage accured. In normal operation forklifts are used to place the coils onto the hook gently. There is no hoist mechanism in the hook drive. We are thinking that either someone managed to drop a coil onto the hook or somehow the weight of the forklift managed to land on the hook.
I know in the classic impact tests a bar is broken by a weight swung into it and the toughness classified by a number which I think is related to the energy of impact, I was hoping that somehow this could be converted into a required stress to cause brittle fracture, maybe combined with the UTS and/or yield.
The UTS and yield of the material are 490MPa and 335Mpa respectively.
What we planning to do in the meantime is derive the load required to reach the UTS at the point of failure and use that as a minimum value to cause failure.
The failed section is therefore a 100x40mm rectangular cross section. The surface is 95% classic brittle failure with a 5% shear surface where it finally gave. I.e. there is no evidence of a crack before failure.
As I say we do not know for sure where the damage accured. In normal operation forklifts are used to place the coils onto the hook gently. There is no hoist mechanism in the hook drive. We are thinking that either someone managed to drop a coil onto the hook or somehow the weight of the forklift managed to land on the hook.
I know in the classic impact tests a bar is broken by a weight swung into it and the toughness classified by a number which I think is related to the energy of impact, I was hoping that somehow this could be converted into a required stress to cause brittle fracture, maybe combined with the UTS and/or yield.
The UTS and yield of the material are 490MPa and 335Mpa respectively.
What we planning to do in the meantime is derive the load required to reach the UTS at the point of failure and use that as a minimum value to cause failure.
MikeHalloran
Mechanical
Ah. Using the word 'rail' confused me.
The section sounds rather delicate for the service and loading condition that you have described.
That, and the brittle failure, suggests that it was intended to do just as it did, i.e. fail when the coil is dropped on it, perhaps to minimize risk of injury downstream, or to avoid pulling down the tramrail or the building.
Mike Halloran
Pembroke Pines, FL, USA
The section sounds rather delicate for the service and loading condition that you have described.
That, and the brittle failure, suggests that it was intended to do just as it did, i.e. fail when the coil is dropped on it, perhaps to minimize risk of injury downstream, or to avoid pulling down the tramrail or the building.
Mike Halloran
Pembroke Pines, FL, USA
sounds a little like a dog's breakfast ... probably something modified time and again ... anyways, so the rectangular bar attaches to the back-to-back channels with 2 fasteners (to support the moment when the load is offset). I'm surprised that any material with your strength (either hi-strength Al alloy or low strength steel) doesn't exhibit ductile failure ... a little surprised that your detailled description doesn't say which material it is (I'm betting steel, probably annealled).
I guess you can do the P/A (well, M/Z) calcs yourself. Without a ductile failure I wouldn't use plastic bending (like Cozzone) to calc the strength of the bar. i figure something like 32,000 Nm of moment on the section, maybe 10,000N applied about 1/3m away ... still I'd've expected ductile failure in bending. How does this compare with the weight of your steel coils ? or maybe a forklift ??
Another question, for when you do some calcs ... why didn't the fasteners fail (even in bearing ?); i'd've expected these would shear before the bar would break. how did the rest of the structure react when the bar broke ... kicked like a mule ?
I guess you can do the P/A (well, M/Z) calcs yourself. Without a ductile failure I wouldn't use plastic bending (like Cozzone) to calc the strength of the bar. i figure something like 32,000 Nm of moment on the section, maybe 10,000N applied about 1/3m away ... still I'd've expected ductile failure in bending. How does this compare with the weight of your steel coils ? or maybe a forklift ??
Another question, for when you do some calcs ... why didn't the fasteners fail (even in bearing ?); i'd've expected these would shear before the bar would break. how did the rest of the structure react when the bar broke ... kicked like a mule ?
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- #10
The exact material is BS EN 10025 1993 S355 J2G3. The coils are about 2,200Kg and a forklift about 10,000Kg. As I said though we have no way of knowing if these were dropped or what happened, that is part of the mystery. When I look at the drawing, if it was going to fail anywhere due to excessive load, it would fail at the point it has. I am relying on our materials guys to say if it was partially ductile before failure and they say it was fully brittle. Anyway thanks for the replies so far, I am coming to the conlusion that it is not simply a case of equating a stress at the geomtery back to a known material property as it seems to depend greatly on how dynamicly the load was applied.
Hi barrind
Quickly reading your post if the beam failed in bending then you could work to find what static load applied gradually would cause the beam to go completely plastic. Having obtained this figure a shock load applied without impact the beam would fail at twice the static load.
A beam failing due to an impact load would require the height from which the coil was dropped.
However if you use the shock load figure without impact it may give you a minimum failure load that the beam saw.
regards
desertfox
desertfox
Quickly reading your post if the beam failed in bending then you could work to find what static load applied gradually would cause the beam to go completely plastic. Having obtained this figure a shock load applied without impact the beam would fail at twice the static load.
A beam failing due to an impact load would require the height from which the coil was dropped.
However if you use the shock load figure without impact it may give you a minimum failure load that the beam saw.
regards
desertfox
desertfox
are my numbers way off ?? (wouldn't be the 1st time)
if the steel coils weigh 2,200kg, that'd be about 22,000N, somewhat more than my 10,000N static failure load ...
490 = M*6/(40*100^2), M = 32E6Nmm = 32E3Nm ... no, looks about right ??
in plain english (i don't know your BS standards; not that they're BS) i think your material is low strength, low alloy steel, yes?
if the steel coils weigh 2,200kg, that'd be about 22,000N, somewhat more than my 10,000N static failure load ...
490 = M*6/(40*100^2), M = 32E6Nmm = 32E3Nm ... no, looks about right ??
in plain english (i don't know your BS standards; not that they're BS) i think your material is low strength, low alloy steel, yes?
desertfox,
By shock load without impact I am taking that to mean a 'suddenly applied load'. If stress when the beam goes plastic is found would not the static load to produce this stress need to be divided by 2 if the load was suddenly applied?
thanks
By shock load without impact I am taking that to mean a 'suddenly applied load'. If stress when the beam goes plastic is found would not the static load to produce this stress need to be divided by 2 if the load was suddenly applied?
thanks
Since the material failed in fracture i.e. no indication of yielding, it can be classified as brittle. (It may be wise to find the brinell hardness and determine this absolutely) Therefore you can apply some failure theories to find the stress at which the beam fractured. Brittle materials generally have higher compressive strengths than tensile so both must be checked. S355 J2G3 is a structural steel. I don't know any impact failure theories but you may be able to get an idea of the stress from static loading theories.
Another thing you won't know is the factor of safety that the designer used to create your track. This too can be estimated at 2.0 to 2.5 depending on the reliability requirements of the component.
You'll have to first model the element and find the principal stresses using formulas or Mohr's Circle.
If both principal stresses have the same sign, compressive or tensile, you can use the MAXIMUM NORMAL STRESS THEORY where failure is assumed to occur if the largest tensile principal stress is greater than the ultimate tensile strength OR if the larger principal compressive stress is larger than the ultimate compressive stress.
sigma 1 & sigma 2 = principal stresses
Failure criterion is: sigma 1, sigma 2 > Ultimate Stress / Factor of Safety
COULOMB-MOHR THEORY is used when the principal stresses have opposite signs (compressive and tensile).
Failure criterion is: (FS*sigma1)/Sut + (FS*sigma2)/Suc > 1
The least conservative theory is the MODIFIED MOHR THEORY which you should be able to find in a good machine design book I just don't have it in fromt of me.
Another thing you won't know is the factor of safety that the designer used to create your track. This too can be estimated at 2.0 to 2.5 depending on the reliability requirements of the component.
You'll have to first model the element and find the principal stresses using formulas or Mohr's Circle.
If both principal stresses have the same sign, compressive or tensile, you can use the MAXIMUM NORMAL STRESS THEORY where failure is assumed to occur if the largest tensile principal stress is greater than the ultimate tensile strength OR if the larger principal compressive stress is larger than the ultimate compressive stress.
sigma 1 & sigma 2 = principal stresses
Failure criterion is: sigma 1, sigma 2 > Ultimate Stress / Factor of Safety
COULOMB-MOHR THEORY is used when the principal stresses have opposite signs (compressive and tensile).
Failure criterion is: (FS*sigma1)/Sut + (FS*sigma2)/Suc > 1
The least conservative theory is the MODIFIED MOHR THEORY which you should be able to find in a good machine design book I just don't have it in fromt of me.
Hi barrind,
Thinking further about your failure, wouldn't your equipment have had to undergo a load test before it was put into service? load tests are usually two and a half times the normal service load so assuming that your equipment would have had to undergo one of these tests you will have a static load figure available.
Half of this figure could be taken as a shock load without impact, therefore whatever shock load or impact load the equipment saw it would have to have been in excess of this value.
regards,
desertfox
Thinking further about your failure, wouldn't your equipment have had to undergo a load test before it was put into service? load tests are usually two and a half times the normal service load so assuming that your equipment would have had to undergo one of these tests you will have a static load figure available.
Half of this figure could be taken as a shock load without impact, therefore whatever shock load or impact load the equipment saw it would have to have been in excess of this value.
regards,
desertfox
Sounds like a ductile to brittle/Fast-Fracture thing maybe- what was the metal temperature?
Check Charpy's on the material. Also check compliance with ASME B30.20 "Below-the-Hook Lifting Devices" or similar standards
There's some good easy-to-understand info on Fast Fracture in Ashby& Jones "Engineering Materials" (PergamonPress,1980)
Check Charpy's on the material. Also check compliance with ASME B30.20 "Below-the-Hook Lifting Devices" or similar standards
There's some good easy-to-understand info on Fast Fracture in Ashby& Jones "Engineering Materials" (PergamonPress,1980)
- Thread starter
- #18
Thanks for all your replies. We decided to get a second opinion on the failure from a different set of met lab guys. Lo and behold they did indeed find evidence of fatigue cracking at two of the corners of the section. Quite small but flaws none the less. They now reccomend we do three charpy tests at -20 deg C and compare with the spec of the steel. A couple of fracture toughness tests at operating temperature +10 deg C and then they say it is relatively straight forward to evaluate the stress required to fracture assuming static and impact lifts, I assume by measuring the size of the flaw at failure. FE can then be used to equate the actual load. My idea of using the UTS is not valid as there was no apparent plastic defomation before failure hence the bar failed below UTS. Some very useful tips have been given above though, so thanks for that.
assuming that your material is low allow, low strength, steel, it's surprising that a small flaw would produce a brittle failure, given that the steel should be quite ductile.
charpy tests will confirm that the material is in spec. ... i suspect that (like the uts failure condition) the stress required is going to be much higher than you'd expect with static loading.
this leaves impact loading, and at least a story get try a get the truth about what happened out of the guys that were there.
did my numbers above make any sense ? 'cause it looked like you were overloading the bar ??
charpy tests will confirm that the material is in spec. ... i suspect that (like the uts failure condition) the stress required is going to be much higher than you'd expect with static loading.
this leaves impact loading, and at least a story get try a get the truth about what happened out of the guys that were there.
did my numbers above make any sense ? 'cause it looked like you were overloading the bar ??
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