Young's Modulus of Elasticity Problems -

  1. Young's Modulus of Elasticity Problems - URGENT

    1. The problem statement, all variables and given/known data

    Calculate Young's Modulus of Elasticity for the three materials using the graphs only.

    Stress (N/m2) Strain
    70 0.06
    100 0.11
    150 0.22
    200 0.35
    220 0.48


    2. Relevant equations

    ε = Stress / Strain

    3. The attempt at a solution

    I believed that to calculate Young's Modulus was a case of stress over strain but surely it should give the same value throughout the elastic region to be a constant. The stress and strain results provided must be correct as they were already provided.

    70 / 0.06 = 1166.66
    100 / 0.11 = 909.09
    150 / 0.22 = 681.82
    200 / 0.35 = 571.43
    220 / 0.48 = 458.33

    So my main question is why are these so different, am i applying an incorrect formulae. Please note i was only using the formulae because i wanted to check i was getting it right and wasn't sure how to use the graphs to obtain the value.

    How do a derive the youngs modulus from a Stress-Strain graph?

    Please help as i only have 1 more day before this is due in.

    Many thanks, Joe
     
  2. jcsd
  3. Not my field but I found this which suggests the curve isn't allways straight...

    [​IMG]

    Presumably the Youngs modulus normally quoted is the slope at the bottom left in the elastic region.

    Perhaps wait for other suggestions.
     
  4. Do you know what the material is? What units for strain?..

    I ask because a youngs modulus of 1000 N/m2 seems quite low..

    For example this page quotes that of rubber as 0.01 * 109 to 0.1 * 109 N/m2

    http://www.engineeringtoolbox.com/young-modulus-d_417.html
     
  5. That is the lowest values that were pre-given, the other 2 materials are

    Stress Strain E
    50000000 0.00128 39062500000
    100000000 0.0016 62500000000
    150000000 0.00187 80213903743
    200000000 0.00213 93896713615
    250000000 0.00227 1.10132E+11
    300000000 0.00253 1.18577E+11
    350000000 0.00339 1.03245E+11
    375000000 0.00427 87822014052
    400000000 0.00707 56577086280


    Stress Strain E
    100000000 0.00109 91743119266
    200000000 0.00145 1.37931E+11
    300000000 0.00182 1.64835E+11
    400000000 0.00215 1.86047E+11
    470000000 0.00236 1.99153E+11
    460000000 0.00273 1.68498E+11
    460000000 0.00291 1.58076E+11
    500000000 0.00318 1.57233E+11
    550000000 0.00382 1.43979E+11
    600000000 0.00436 1.37615E+11



    The idea of the assignment is to calculate the modulus of E and then research into what materials they could be. I'm going to presume the one i have already given would be something such as glass, which would explain why its so different.

    I'm meant to calculate the E from the graphs that i have anyway, how do i go about this?

    I thought modulus was meant to be a consistant number across the whole elastic region so why does it change??

    Thanks
     
  6. SteamKing

    SteamKing 8,382
    Staff Emeritus
    Science Advisor
    Homework Helper

    Not every material has a perfectly elastic region in its stress-strain curve.
     
  7. So what value would i use for the elasticity since there so different, would i have to take an average?
     
  8. Chestermiller

    Staff: Mentor

    Show us what the stress-strain graphs look like. (Strain = abscissa, Strain = ordinate). Seeing the graphs by eyeball should tell a lot.

    Chet
     
  9. SteamKing

    SteamKing 8,382
    Staff Emeritus
    Science Advisor
    Homework Helper

    strain = abscissa, stress = ordinate
     
  10. Chestermiller

    Staff: Mentor

    Hi. Thanks for the correction.

    Chet
     
  11. Chestermiller

    Staff: Mentor

    Please include the strain and stress 0,0 on these arithmetic scale plots. Also, maybe you can provide plots of log(stress) vs. log(strain) so we can see the behavior expanded at small strains (of course without the point 0,0 on the log-log plot). Please also show the data points on the plots.

    These plots that you have shown so far are already very revealing.
     
  12. Thankyou, my initial task was to calculate the CSA which i did at 0.00002m2. I was pre-given Load and Extension data for 2 materials and given Stress-Strain data for the third. I was tasked to calculate the stress-strain data for all three and then plot the relevant graphs.

    Material X

    TEST DATA FOR X
    LOAD..EXT...CSA.........Stress (F/A)..Strain........Orig Length E (Stress / strain)
    1000..0.48...0.00002....50000000.....0.00128............375...39062500000
    2000..0.6....0.00002.....100000000...0.0016..............375...62500000000
    3000..0.7....0.00002.....150000000...0.001866667......375...80357142857
    4000..0.8....0.00002.....200000000...0.002133333......375...93750000000
    5000..0.85..0.00002......250000000..0.002266667......375....1.10294E+11
    6000..0.95..0.00002.....300000000..0.002533333.......375....1.18421E+11
    7000..1.27..0.00002.....350000000..0.003386667.......375....1.03346E+11
    7500..1.6....0.00002.....375000000..0.004266667.......375....87890625000
    8000..2.65..0.00002.....400000000..0.007066667........375....56603773585
    8500..5.5...0.00002......425000000..0.014666667.......375....28977272727
    8300..7......0.00002......415000000..0.018666667.......375....22232142857
    8000..7.9....0.00002.....400000000..0.021066667.......375.....18987341772
    7000..8.7....0.00002.....350000000..0.0232...............375.....15086206897
    6100..9.1....0.00002.....305000000..0.024266667......375......12568681319

    Material Y

    TEST DATA FOR Y
    LOAD...EXT...CSA......... Stress.........Strain..........Orig Length....E (Stress / strain
    2000....0.6...0.00002.....100000000..0.001090909.......550...91666666667
    4000....0.8...0.00002.....200000000..0.001454545.......550....1.375E+11
    6000...1......0.00002.....300000000..0.001818182.......550...1.65E+11
    8000...1.18..0.00002.....400000000..0.002145455.......550....1.86441E+11
    9400....1.3...0.00002.....470000000..0.002363636.......550...1.98846E+11
    9200...1.5...0.00002.....460000000..0.002727273.......550...1.68667E+11
    9200...1.6...0.00002.....460000000..0.002909091.......550....1.58125E+11
    10000.1.75..0.00002.....500000000..0.003181818.......550....1.57143E+11
    11000.2.1...0.00002.....550000000..0.003818182.......550...1.44048E+11
    12000.2.4...0.00002.....600000000..0.004363636.......550...1.375E+11
    14000.5.....0.00002.....700000000..0.009090909.......550....77000000000
    12000.7.2..0.00002.....600000000..0.013090909......550....45833333333
    10000.7.25.0.00002.....500000000..0.013181818.......550...37931034483
    8000...7.27..0.00002.....400000000..0.013218182.......550....30261348006

    For Material Z only the Stress-Strain data was given already

    TEST DATA FOR Z
    Stress...Strain.......CSA................................E (Stress / strain)
    70...... 0.06..........0.00002........................... 1166.666667
    100.... 0.11 ..........0.00002...........................909.0909091
    150.... 0.22 ..........0.00002..........................681.8181818
    200.... 0.35 ..........0.00002..........................571.4285714
    220.... 0.48..........0.00002...........................458.3333333

    You have asked for 0,0 data but i was not given any data close to that amount. Should i still include this on the graphs and just start the line from thin air essentially.

    Is my use of formulae correct. Should i change the scale data on my graphs?
     
    Last edited: Apr 23, 2013
  13. Chestermiller

    Staff: Mentor

    The point 0,0 is known to be an exact point on your graph. If there is no applied stress, then there is no strain, and vice versa. I asked for the results on a log-log plot because it will increase the resolution in the region of small strains where the stress-strain behavior is expected to be linear. On a log-log plot, the linear stress-strain region should have a slope of 45 degrees (assuming equal spacing of the decades in the horizontal and vertical axes). If you are using a graphics package, just change the scales from arithmetic to logarithmic. And if you are switching to log scales, make sure you omit the point 0,0.

    Chet
     
  14. So your suggesting the graph starts in mid air? I'm not sure what you mean but i'm confident i have to use arithmetic graphs and not log. Do i just use the gradient to determine Young's Modulus? So Change in Y axis / Change in X axis?

    I have very little time left to get this done, i need to get a value and find the material.

    Thanks, Joe
     
  15. Chestermiller

    Staff: Mentor

    I'm not suggesting that the graph starts in mid air. I'm suggesting that there is an exact point on the graph at 0,0. How much was the strain before any stress was applied? Zero, right?

    If you don't want to use log scales, then you need to zoom in on the data at small strains and stresses. To do this, omit the data points that are well beyond the linear region, and, by doing so, expand the scales on both axes. Then you will get a better eyeball view of what is happening at small strains.

    Chet
     
  16. Chestermiller

    Staff: Mentor

    I plotted up your data for two of the cases, and, in both cases, the data is offset along the strain axis. Otherwise, the data is pretty linear in the region of small strains, and you should be able to get a pretty accurate determination of the Young's modulus. I don't know why your data is offset along the strain axis.

    Chet
     
  17. Thankyou, all the Load and extension data was provided. The area i believe to be correct and i'm confident my calculations are correct so any idea what to do now?
     
  18. Chestermiller

    Staff: Mentor

    The graphs are "screaming" that something is wrong with the strains. This must mean that something was wrong with the extension measurements. This must somehow be related to the zero point of the extension measurements.
     
  19. I've double checked them and i can't see where its wrong. The extension was pre-given in mm and the Original length of sample was also pre-given in mm.

    Strain = Change of Length / Original Length

    For Material X

    Original Sample Length = 375mm

    Ext (mm) / Original Sample Length (mm)
    0.48 / 375 = 0.00128


    Is this correct?
     
  20. Chestermiller

    Staff: Mentor

    Yes, it looks correct. But, as best I can tell, the extensions look off (too large) by about 0.38mm (and the strains look off by about 0.001, at least for this case) . The only thing I can recommend is that you take the slope of the stress vs strain graph in the region of small strains. You will find that the data are fit very nicely by a straight line. The slope is what I would report as the Young's modulus. For this case, I would estimate the slope to be about 2 x 1011.
     
Know someone interested in this topic? Share a link to this question via email, Google+, Twitter, or Facebook

0
Draft saved Draft deleted