TheRecklessEngineer Posted July 3, 2009 Share Posted July 3, 2009 I only ask this as I know there are some very knowledgeable engineers on here that may be able to point out the error. Our lecturer is currently unavailable. Mods, feel free to remove this if you think it is inappropriate. The LR content is that once I complete this course, I'll be able to afford more random landy parts! This is a mechanics assignment - we had a tensile testing machine and tested a steel sample to destruction. The Young's Modulus (E) of the sample was calculated to be 13GPa, by several independent groups with independent samples. The expected value for steel is 200GPa, and no, the sample was not made by Britpart . Rather than explain our method, this is the equations worked backwards to obtain the force required to extend the sample 0.3mm. Our sample dimensions were 5mm in diameter and 25mm long, broke at an extension of over 4mm, but extended about 0.8mm before permanent deformation. stress = force / area and strain = extension / original length. We also know Young's Modulus (E)= stress / strain. So: E=Fl/xA. and F=ExA/l A=pi x r^2 = pi x (2.5^-3)^2 = 2 x 10^-5 F=(200x10^9 x 0.3x10^-3 x 2x10^-5)/25x10^-3 = 48000N = 4800kg The maths suggests a force of 4.8tonnes. There is no way a round steel bar of 5mm can support a force of 4.8 tonnes - and there is also no way the machine could exert that force. In fact, using this method, the force would have peaked at over 20 tonnes to break the sample. The actual data says the force peaked at a little over 1000kg - sounds much more likely to me - but this in turn leads to a very small value for E. So what is wrong? Did they give us cheese rather than steel, or is our maths flawed somewhere. 16 of us spent about 4 hours trying to crack it last night and got no where. Brownie points issued for anyone who can point out our error. Quote Link to comment Share on other sites More sharing options...
TheRecklessEngineer Posted July 3, 2009 Author Share Posted July 3, 2009 This is the force and extensions graph for the actual data. Looks absolutely classic and the right order of magnitude to me.... Quote Link to comment Share on other sites More sharing options...
bobtail4x4 Posted July 3, 2009 Share Posted July 3, 2009 it all looks correct, are you sure its not britpart steel? Quote Link to comment Share on other sites More sharing options...
q-rover Posted July 3, 2009 Share Posted July 3, 2009 DO you know the quality of the steel sample, did it have a material certificate and C.O.C? Also just because the steel IS certified, doesn't mean it will meet the requirments for it's spec. There has been a big case over here where a company supplied SS piping components (elbows, Tee's, etc) all certified and correct. However they hadn't followed the proper tempering procedure and therefor there have been failures. Looking at the components and certificates doesn't show the problem. And it wasn't a chinese company either. Quote Link to comment Share on other sites More sharing options...
Bush65 Posted July 3, 2009 Share Posted July 3, 2009 From the chart, extension corresponding with 10000N is approx 0.8mm (hard to get exact extension from diagram on my computer screen). Now stress at 10000N is 10000N / (pi * 2.5mm^2) = 509Mpa And corresponding strain is 0.8mm / 25mm = 0.032 Then E is 509MPa / 0.032 = 15.9GPa The stress value indicates the material could be some kind of low alloy steel. If the tensile testing machine was in reasonable calibration, then I question your 25mm value for the gauge length of the test specimen. Edit: was your test specimen, prepared properly, to the standard dimensions? And is 25mm the gauge length for calculating strain? Quote Link to comment Share on other sites More sharing options...
smo Posted July 3, 2009 Share Posted July 3, 2009 4800N is not 4800Kg, its 480Kg! Quote Link to comment Share on other sites More sharing options...
sgnas Posted July 3, 2009 Share Posted July 3, 2009 As the tensile looks fine. It is possible you have material with low ductility. Has it been hardened? Check the harness of the remains. Looks for defects in machining of the sample causing a notch and weakening the sample. Quote Link to comment Share on other sites More sharing options...
DaveSIIA Posted July 3, 2009 Share Posted July 3, 2009 Do you have details (drawing/photo) of the prepared test specimen, showing guage section, markings, etc? A photo could do with some means of ascertaining the scale. Quote Link to comment Share on other sites More sharing options...
Cheesy Posted July 3, 2009 Share Posted July 3, 2009 Are you sure it was actually steel? Quote Link to comment Share on other sites More sharing options...
landroversforever Posted July 3, 2009 Share Posted July 3, 2009 48000N = 4800kg 4800N is not 4800Kg, its 480Kg! I really should know this, we did it this year for my A level engineering ... let me have annother think Quote Link to comment Share on other sites More sharing options...
landroversforever Posted July 3, 2009 Share Posted July 3, 2009 Which type of steel was it? does that make a difference? Quote Link to comment Share on other sites More sharing options...
TheRecklessEngineer Posted July 3, 2009 Author Share Posted July 3, 2009 Thanks guys, This is the best photo I have of the specimen: I can confirm it is 25.05mm long (between the lumpy bits on the end) and 5.05mm in diameter. I actually went and checked it this morning, and the machine. The data is correct, and it appears to be steel - feels like steel, slightly rusty and magnetic. I could not see any notches/dents or cracks in the unused samples. We have all come up with a figure of around 13GPa, using different samples and (slightly) differing methods - this is what is so odd. What really gets me though is the maths above suggests that if this sample did have E=200GPa, then it would take nearly 20,000kg to break it. Anyone snapped a half shaft recently? (I know it's in a different direction, but a similar theory holds). It shouldn't take this force to snap a 5mm steel rod - and certainly the machine is not capable of applying it. I get the feeling that there is something fundamentally wrong with the maths here somehow...the data 'feels' correct, but the numbers don't work. Quote Link to comment Share on other sites More sharing options...
bishbosh Posted July 3, 2009 Share Posted July 3, 2009 You can only apply those formulae in the linear elastic portion of the graph. All you need is the gradient of the "straight" part of the line. Which from your best fit curve is about 11,000N/mm2 So either your equipment is faulty or your lecturer is testing you with a fake sample.... Or.... my brain might be slightly befuddled as I am 23 hours out of a general anaesthetic and popping some nice happy pills. Quote Link to comment Share on other sites More sharing options...
TheRecklessEngineer Posted July 3, 2009 Author Share Posted July 3, 2009 Thanks Bish - yes, we mostly did the calculations from the gradient of the straight portion. What I really don't understand though, are the massive forces required when you work the equations backwards. I.e. to extend the sample by a mere 0.3mm you need a force of nearly 5000kg - which to me sounds ludicrous! Quote Link to comment Share on other sites More sharing options...
bishbosh Posted July 3, 2009 Share Posted July 3, 2009 How much necking was there at failure? Your calculated stress will vary a fair bit on a sample that small. Quote Link to comment Share on other sites More sharing options...
Night Train Posted July 3, 2009 Share Posted July 3, 2009 I clicked on this to test my memory. Sadly it has failed the test with regards to the maths! I last did this experiment a quarter century ago and we were comparing the effects of temperature on the properties of the steel samples. Temperature had a large effect on the outcomes of the tests based on lab certified samples. Also we found that notches and surface damage due to corrosion also had an effect. The tests we did covered tensile stress through extension of the sample and also impact testing by placing the sample in a carrage and releasing a weighted 'axe' to impact it. We found that even with the lab certified samples there was variation. Quote Link to comment Share on other sites More sharing options...
Fatboy Posted July 3, 2009 Share Posted July 3, 2009 As a basic cross-check. Can you do a surface hardness measurement to see how it correlates with your results? Clicky Linky Thing Quote Link to comment Share on other sites More sharing options...
Bush65 Posted July 3, 2009 Share Posted July 3, 2009 As the tensile looks fine. It is possible you have material with low ductility.Has it been hardened? Check the harness of the remains. Looks for defects in machining of the sample causing a notch and weakening the sample. Ductility affects the elongation to failure, not the Young's Modulus. Quote Link to comment Share on other sites More sharing options...
Bush65 Posted July 3, 2009 Share Posted July 3, 2009 ... The data is correct, ... But something appears wrong with the results presented in your graph of load vs extension. Possibly the units. Quote Link to comment Share on other sites More sharing options...
TheRecklessEngineer Posted July 4, 2009 Author Share Posted July 4, 2009 Sod it. I have more of a life than wondering what on earth can be wrong with this. It will make my discussion of the experiment all the more interesting. Something along the lines of: "All the numbers are correct, therefore my value for pi must be wrong. I suspect some sort of relativistic effect causing a change in the value for pi. This would also explain the difficulty a large majority of my course has in turning an accurate diameter on the lathes next door" Or: "I must have sine flu as the trig simply does not work for me" Quote Link to comment Share on other sites More sharing options...
Cheesy Posted July 5, 2009 Share Posted July 5, 2009 Thanks guys,This is the best photo I have of the specimen: I can confirm it is 25.05mm long (between the lumpy bits on the end) and 5.05mm in diameter. I actually went and checked it this morning, and the machine. The data is correct, and it appears to be steel - feels like steel, slightly rusty and magnetic. I could not see any notches/dents or cracks in the unused samples. We have all come up with a figure of around 13GPa, using different samples and (slightly) differing methods - this is what is so odd. What really gets me though is the maths above suggests that if this sample did have E=200GPa, then it would take nearly 20,000kg to break it. Anyone snapped a half shaft recently? (I know it's in a different direction, but a similar theory holds). It shouldn't take this force to snap a 5mm steel rod - and certainly the machine is not capable of applying it. I get the feeling that there is something fundamentally wrong with the maths here somehow...the data 'feels' correct, but the numbers don't work. Your first bit of maths is pretty much right... to get 0.3mm extension would require 49kN however it does not mean that the material will withstand this stress, infact if you check back this would require the sample to have a yeild strength of around 2520MPa which it doesnt. From your plot the UTS and Yeild could be steel, the numbers look about right however the extension/strain do not look right for a carbon steel, these plots typically show a very distinctive discontinuous yeild, your plot looks much more like an FCC structured metal (stainless steel or aluminium for example). Now im guessing that you didnt have a strain gauge/displacement gauge on the sample when you tested it and the recorded extention was from the head displacement? If this is the case Im betting that you have a 'soft' tensile test machine, and you extention measurement includes all of the deformation in the test equipment as well, it is also likely that it doesnt have the resolution to measure the extension accuratly. For instance if you sample is infact steel with a modulus of 210GPa the extension at yield would be around 0.07mm. If you did the test again with a smaller say 2mm sample you will probably get better results Quote Link to comment Share on other sites More sharing options...
TheRecklessEngineer Posted July 5, 2009 Author Share Posted July 5, 2009 Cheesy, Thanks for the reply. I like the theory that the machine may be deforming - tbh I didn't much like the look of it anyway. And no, we didn't have a strain gauge. I did consider that the extension was in the wrong order of magnitude. However, the broken sample when put back together is 3.75mm longer than it started. This suggested to me that the extension was about right. It is also interesting what you say about the plots looking incorrect for a carbon steel. Our lecturer is a prime example of "If you can't, then teach". I would not put it past him to have confused the samples with something else. We have spoken to him since these figures arose - he was very surprised that we did not get the right answer. He confirmed that assuming our maths is correct we won't be marked down. Quote Link to comment Share on other sites More sharing options...
Cheesy Posted July 5, 2009 Share Posted July 5, 2009 Cheesy,Thanks for the reply. I like the theory that the machine may be deforming - tbh I didn't much like the look of it anyway. And no, we didn't have a strain gauge. I did consider that the extension was in the wrong order of magnitude. However, the broken sample when put back together is 3.75mm longer than it started. This suggested to me that the extension was about right. It is also interesting what you say about the plots looking incorrect for a carbon steel. Our lecturer is a prime example of "If you can't, then teach". I would not put it past him to have confused the samples with something else. We have spoken to him since these figures arose - he was very surprised that we did not get the right answer. He confirmed that assuming our maths is correct we won't be marked down. Its not a theory, it is deforming, it just depends on how much.... Thats why bigger/more high tec machines will have a seperate device to measure the elongation while the sample is in the linear elastic region, this eliminates the machine stiffness from the measurements, The samples probably were mild steel but they were too big for the machine you were using, reducing the calculated modulus. You will probably get marks for mentioning machine stiffness in your report as well..... Quote Link to comment Share on other sites More sharing options...
TheRecklessEngineer Posted July 6, 2009 Author Share Posted July 6, 2009 Thanks cheesy - I've handed in the report this morning. We did a number of different experiments, and to be honest, the quality was diabolical. Our most accurate measuring device for anything over 30mm was a wooden metre rule - really not too useful! All the experiments have now had a slating, not one result was within 10% of the expected value. I hope next time they are set up with a little more effort towards obtaining accurate results. Quote Link to comment Share on other sites More sharing options...
A Twig Posted July 7, 2009 Share Posted July 7, 2009 Could it be something to do with the lumps on the ends of the sample? The lumps have a differing area to the actual bit you're testing, so with Pressure/Area = Force, then with the bigger area of the top bits, it will give the impression of requiring more Force? Maybe? I'm not sure, I do electronics! Quote Link to comment Share on other sites More sharing options...
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