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Thread: WLF shift function

  1. #1

    WLF shift function

    Hello, my friend

    i just have a question about time-temperature supositon. i am modelling material property with Marc mentat, but after i input the reference temperature, and constants. The relaxation curve is a simple line,how can this happen?

    Many thanks!
    ^Q^
    benutella

  2. #2
    Join Date
    2000-02
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    Hello Benutella,

    What stress relaxation function did you specify? Did you use a Prony series? The WLF shift function specifies how the relaxation behavior depends on temperature and a broader time scale, but you still need to specify a stress relaxation function.

    Cheers,
    Jorgen

  3. #3

    i am still lost

    hello,Jorgen

    thanks for your reply!
    but i am still lost. what i want to do is to get the master curve for long-term bebaviour. i use WLF function to specify the shift factor.
    but the result was quite strange
    no idea how to define boundary condition and material property in Marc mentat
    how to get master curve with it?

    Thanks~~~~~~~~~~~~~~!

  4. #4
    Join Date
    2000-02
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    Hello Benutella,
    I am not sure how much I can help you to input the master curve since I currently do not have access to Marc
    Best of luck.
    Jorgen

  5. #5

    shift factor

    hello,Jorgen

    Thanks for your reply!

    just another question about the shift factor. i was using WLF function to calculatethe shift factor, suppose the refference temperature is 30C, the constant is 17.44 and 51.6, the glass transition temperature is 90C

    from all the given, i got the shift factor is -124.6, so i just shift the curve at 30C to left with a value of 124.6, but it seems i shift too much to get a master curve.

    what is wrong? any problem with calculation or something else?
    Thanks in advance!

  6. #6
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    The WLF equation is often written as follows:

    log aT = C1 * (T-T0) / (C2 + T - T0)

    Where C1=17.5 and C2=51.6 K. If I understand you correctly, you have T=30 deg C, and T0 = 90 deg C. Hence, the difference T-T0 is so large that the denominator has switched sign

    You are attempting to apply WLF 60 deg away from the T0, this is more than I would recommend. At 60 deg below Tg you will have very little viscoelastic flow.

  7. #7
    Hello,Jorgen

    It is said that if T<Tg or T<Tg+100C,the WLF function can be used to get the long-term behaviour, i just learned that by reference

    so you mean the temperature difference should be far smaller than that range?
    any suggestion about the application of WLF?

    Thanks a lot!

  8. #8
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    Here's a simple test: if T0=90 deg C, C2=51.6 K, and T=38.4 deg C, then the denominator C2+T-T0=0, and log aT goes to minus infinity and aT=0. Hence, no stress relaxation will occur.

  9. #9
    Hello,Jorgen,

    what does it imply when shift factor has a value with minus infinity? from my understanding, it only suggests that the temperature is lower than reference temperature, so the relaxation curve should be shifted to the left with an infinit value, am i in the right way?

    and also, i made some relaxation tests at 30C, there seems to be relaxation from my experiment result. Theory does not agree with experiment?

    the shift factor must always be positive?

    Thanks again!
    benutella

  10. #10
    Join Date
    2000-02
    Location
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    The factor aT is always a positive number, the term log aT can be either positive or negative. For temperatures above T0, the shift is to the right (log aT > 0); and for temperatures below T0, the shift is to the left (log aT < 0).

    The WLF equation with the constants that you quoted predicts no relaxation at 30 deg C. In your stress relaxation experiments, what strain did you go to? The WLF equation combined with linear viscoelasticity is typically only valid at very small strains. What strains are you trying to simulate?

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