Force required to fracture a rail?
Force required to fracture a rail?
(OP)
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?





RE: Force required to fracture a rail?
RE: Force required to fracture a rail?
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
RE: Force required to fracture a rail?
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 !!
RE: Force required to fracture a rail?
You might also check actual support spacing and rigidity versus designed dimensions.
RE: Force required to fracture a rail?
Mike Halloran
Pembroke Pines, FL, USA
RE: Force required to fracture a rail?
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.
RE: Force required to fracture a rail?
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
RE: Force required to fracture a rail?
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 ?
RE: Force required to fracture a rail?
RE: Force required to fracture a rail?
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
RE: Force required to fracture a rail?
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?
RE: Force required to fracture a rail?
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
RE: Force required to fracture a rail?
Yes your exactly right I should have said half the static load my mistake.
I posted in my lunch hour with very little time to spare.
thanks
desertfox
RE: Force required to fracture a rail?
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.
RE: Force required to fracture a rail?
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
RE: Force required to fracture a rail?
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)
RE: Force required to fracture a rail?
RE: Force required to fracture a rail?
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 ??