Extracting mechanical properties from strain-stress plot (real test)

AI Thread Summary
The discussion centers on extracting mechanical properties from a tensile test plot for a 306LN stainless steel sample to improve an elasto-plastic model. The user successfully calculates yield strength and tensile ultimate strength but struggles to reconcile the reported Young's modulus of 187 GPa with their own calculation, which yields only 19.2 GPa. The discrepancy raises questions about potential errors in the lab report, particularly regarding the elastic region of the load versus displacement plot. Other participants suggest that typical strain values for steel at yield should be around 0.002, indicating a possible factor of ten error in the reported modulus. The conversation highlights the need for clarification on the calculation methods used in the lab report.
freddie_mclair
Messages
43
Reaction score
2
Hi everyone, recently I was working out an elasto-plastic model to make some residual stress predictions: see it here.
Recently I got some real tensile tests results from a lab, therefore I would like to extract a better stress-strain curve to plug into my elasto-plastic model using the Ramberg-Osgood equation. The problem is that I cannot get the Young's modulus that they claim to have obtained!

Here's the tensile test plot:
Tensile-test.jpg


The stress-strain curve I would like to extract corresponds to the plots on the bottom part of the image (ultimate stress around 7.5kN).
The geometry of the samples is the following:
Sample-dimensions.jpg


The results reported by the lab are the following:
  • ##\mbox{- Yield strength: } S_y=480\mbox{ MPa}##
    • from the plot we get ##6{kN}/(\pi 2^2)=477.5\mbox{ MPa}##. OK here!
  • ##\mbox{- Tensile ultimate strength: } S_u=615\mbox{ MPa}##
    • from the plot we get around ##7.5{kN}/(\pi 2^2)=597\mbox{ MPa}##. OK here too!
  • ##\mbox{- Elongation: } 54\%##
    • from the plot, we see that it has been elongated up to around 12mm, which, if we consider the length of the sample to be 24mm, it corresponds to these 50%. Also fine!
  • ##\mbox{- Young's modulus: } E=187\mbox{ GPa}##
    • I unable to get this result... from the plot, the displacement corresponding to the Yield strength (##@6\mbox{ kN}##), ##S_y=480\mbox{ MPa}## is around 0.6mm, which then leads to a strain of ##(24.6-24)/24=0.025##. Consequentially, this gives me an order of magnitude lower for Young's modulus: ##480/0.025=19.2\cdot10^3\mbox{ MPa}##, i.e., ##19.2\mbox{ GPa}##.
What am I doing wrong here?
Thanks in advance!
 
Engineering news on Phys.org
Where did the 187 come from?
 
Chestermiller said:
Where did the 187 come from?
This is the value declared from the lab report (it's not mentioned how it was calculated).
I was just trying to calculate it (and cross-check it) by myself based on the plot results.

Ah, this is a 306LN stainless steel.
 
Maybe there's a typo, and a decimal point was omitted?
 
No, for sure that it is not a typo. Also, for stainless steel, 190GPa are the "normal" values... I just don't get what is happening in the elastic part of this Load VS Displacement plot.

They have done other tensile tests on similar samples, and the values are also similar.
 
Typical strain values for steel at yield might be in the order of
0.002 rather than .025. Something funny with the decimal point here. Factor of 10. Something is amiss.
 
Thread 'Turbocharging carbureted petrol 2 stroke engines'
Hi everyone, online I ve seen some images about 2 stroke carbureted turbo (motorcycle derivation engine). Now.. In the past in this forum some members spoke about turbocharging 2 stroke but not in sufficient detail. The intake and the exhaust are open at the same time and there are no valves like a 4 stroke. But if you search online you can find carbureted 2stroke turbo sled or the Am6 turbo. The question is: Is really possible turbocharge a 2 stroke carburated(NOT EFI)petrol engine and...
I need some assistance with calculating hp requirements for moving a load. - The 4000lb load is resting on ball bearing rails so friction is effectively zero and will be covered by my added power contingencies. Load: 4000lbs Distance to travel: 10 meters. Time to Travel: 7.5 seconds Need to accelerate the load from a stop to a nominal speed then decelerate coming to a stop. My power delivery method will be a gearmotor driving a gear rack. - I suspect the pinion gear to be about 3-4in in...
Back
Top