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Materials Science Help - Slip Systems & Resolved Shear Stress

  1. Mar 21, 2012 #1
    Hi everyone,

    I've taken a course at uni that may be a little bit over my head, and am pushing through an assignment for it at the moment in an attempt to do well. I've hit a bit of a roadblock on one question though, so am hoping that someone kind here can help me out! The question is structured like this:

    1) As we know, slip in FCC crystals can occur on 12 independent {hkl}<uvw> slip systems.

    i) Write down all systems.

    I have {111}<110>, and there are 4 planes in this system, each with 3 directions in each plane <-1,1,0>, hence the 12. Is this right? This seems fine, but the real question I have issues with is below.

    ii) A tensile load of 60kN is applied to a cylindrical FCC single crystal of iridium of diameter 10mm. If the centre of rotation of the cylindrical sample and the applied load are parallel to the [1,-2,1] direction, determine the resolved shear stress acting on the (1,-1,-1) plane in the [0,-1,-1],[1,0,1] and [1,-1,0] directions.

    I've attempted to go about the problem by using:

    Shear stress (t) is:
    t = (F/A)*cos(phi)*cos(lambda)

    The F is the force applied, given to us as 60kN. The A is the cross sectional area, which is easy to calculate for a cylinder with diameter of 10mm using pi*r^2, taking r to be 5mm.

    I can also calculate cos(phi), the angle between the tensile axis (the axis upon which the force is applied, which I think is [1,-2,1], and the normal to this, which I am unsure of. I can also calculate cos(lambda), which is the angle between the tensile axis, again, which is think is [1,-2,1] and the slip direction, although I am also not sure of the slip direction. I assume it is one of the directions given.

    My main issue stems from the miller index co-ordinates - i'm not sure how I should be using all 5 to calculate this. When I'm given a tensile axis coordinate, and a system of two coordinates I can work it out fine, but with 5 seperate co-ordinates I'm quite confused.

    Any help would be appreciated, and my apologies for the long post.
  2. jcsd
  3. Mar 22, 2012 #2
    The tensile direction is [1-21], you are right about that.
    The normal to the slip plane is exactly the same miller index as the plane itself. e.g., the normal to plane (1-1-1) is the direction [1-1-1] (note the different brackets). The angle between [1-21] (tensile axis) and [1-1-1] (normal to the slip plane) is cos[itex]\phi[/itex].

    There are three possible slip directions [0,-1,-1],[1,0,1] and [1,-1,0]. So there are three possible values of cos λ, depending on which direction you pair up with the tensile direction [1-21] (and find the angle), and therefore, 3 values of the resolved shear stress.
    At no point do you need to use all 5 data together.
  4. Mar 22, 2012 #3
    Firstly thanks for the reply Bavid.

    But yes I see now. So if there are three possible slip directions, and therefore three possible values for cos(λ), and therefore three values of the RSS, how will I know which is the correct RSS? Or is the question asking for the RSS in each direction, so there is expected to be three different solutions?

    So I guess the method would be to find cos(ϕ) and the area first, as they're constant through all cases, as is the given applied force of 60kN. Then find cos(λ) for the first case, find the RSS for that first case, sub in for RSS, find the RSS for the first case, then move on to finding the cos(λ) and RSS for the 2nd case, and so on for the 3rd case?
  5. Mar 22, 2012 #4
    Given the state of loading, you will have one value of rss for each slip system. You are only concerned with 3 out of the 12 slip systems here, that's why there are 3 values of rss. If the LARGEST rss value (in magnitude) is larger than the critical rss, slip will initiate on that slip system.
  6. Mar 22, 2012 #5
    for your question, merely listing the 3 rss values seems to be enough.
  7. Mar 23, 2012 #6
    Thanks for the replies Bavid. This is the working i've gone through, I think something may be wrong though, I'm getting oddly small values for the RSS.

    First, F = 60kN.
    A = pi*5^2 = 25*pi.

    Find cos(∅) first, as it's constant (angle between tensile axis and normal):

    Angle between [1,-2,1] and [1,-1,-1] is:

    cos(∅) = ((1*1)+(-2*-1)+(1*-1))/(sqrt((1^2+(-2)^2+1^2))*sqrt(1^2+(-1)^2+(-1)^2))
    cos(∅) = 2/(3*sqrt(2))

    So now for the first case, we look at the angle between [1,-2,1] and [0,-1,1]:

    cos(λ) = ((1*0)+(-2*-1)+(1*1))/(sqrt((1^2+(-2)^2+1^2))*sqrt(0^2+(1)^2+(1)^2))
    cos(λ) = sqrt(3)/2

    So subbing in to find the RSS for this case:

    τ = (F/A)*cos(∅)*cos(λ)
    τ = (60/25*pi) * 2/(3*sqrt(2)) * sqrt(3)/2
    τ = 0.311Mpa

    Is this the correct method? I repeated the same calculations for cases 2 and 3, and got 0.207Mpa for case 2 and 0.311Mpa again for case 3. The worry comes from the next part, which asks which slip system is most likely to glide first.

    I know that it would be the system with the highest RSS, but two systems have the same RSS which I don't think is correct, hence I might have an issue somewhere. Either that or I haven't noticed that with one of the cases, the angle between one of the directions and the tensile axis is 90 degrees, meaning cos of that angle is 0, meaning no RSS at all.
  8. Mar 23, 2012 #7
    1. Your method is correct but you need to have consistent units. Using Pa as the standard, the applied stress is 60e3 / (pi * 5e-3 * 5e-3) = 763 943 727Pa. The values you are getting for rss are in units kN/(mm)^2 which equals 10^9 Pa! In fact, actually τ = 0.311* 10^9 Pa for the case below. Clearly, it is not as small as you imagined.

    2. It is perfectly possible for two slipsystems to have the same rss. This can happen if the slip plane is common and the slip directions make equal angles with the tensile axis. Also, note that in this case, the tensile axis lies in one of the slip planes. It is obviously normal to the plane's normal direction and the rss is 0 for all slip directions in that plane.

    3. If two slipsystems have the same rss (and that rss is the highest) , you have glide on both systems simultaneously.
  9. Mar 24, 2012 #8
    Thanks for the reply again Bavid. Have fixed up my units and am getting a far more reasonable answer now. This is the only point that i'm still a little lost on, which of the cases will have a rss of 0? Breaking it down from what I see:

    [1,-2,1] for tensile axis, [0,-1,-1] for slip dir, not normal to each other so rss non-zero.
    [1,-2,1] for tensile axis, [1,0,1] for slip dir, not normal to each other so rss non-zero.
    [1,-2,1] for tensile axis, [1,-1,0] for slip dir, not normal to each other so rss non-zero.

    None of my cos(λ) angles are 90/0 degrees either, so surely none of the slip directions in this have a rss of 0?

    Which case am I messing it up with? :s
    Last edited: Mar 24, 2012
  10. Mar 24, 2012 #9
    [1-21] is normal to the [111] direction. [111] is normal to the (111) plane. Therefore the [1-21] direction is in the (111) plane.
  11. Mar 24, 2012 #10
    Ah yes I see, but that doesn't directly impact any slip systems listed does it?
  12. Mar 24, 2012 #11
    It impacts all 3 slip systems that lies on the plane (111)
  13. Mar 24, 2012 #12
    for the combinations you show:

    [1,-2,1] for tensile axis, [0,-1,-1] for slip dir, not normal to each other so rss non-zero.
    [1,-2,1] for tensile axis, [1,0,1] for slip dir, not normal to each other so rss non-zero.
    [1,-2,1] for tensile axis, [1,-1,0] for slip dir, not normal to each other so rss non-zero.
    cos lambda is never zero, but cos phi is always zero. hence, the rss is zero
  14. Mar 24, 2012 #13
    Wait, so now all my values for the RSS should be zero...? Above you indicated my calculations were correct :S

    I have my cos(phi) as the angle between [1,-2,1] and [1,-1,-1], which is:

    cos(∅) = ((1*1)+(-2*-1)+(1*-1))/(sqrt((1^2+(-2)^2+1^2))*sqrt(1^2+(-1)^2+(-1)^2))
    cos(∅) = 2/(3*sqrt(2))

    I can't see how this is zero. If cos(∅) was zero and then the RSS was zero for all three cases, the question wouldn't really make sense at all either.
    Last edited: Mar 24, 2012
  15. Mar 24, 2012 #14
    Sorry about that, I think the word "always" confused you. I should say: cos phi is zero for the slip system (hkl)[uvw] iff the tensile axis lies in the plane (hkl), regardless of what [uvw] is.
    when you say "[1,-2,1] for tensile axis, [1,-1,0] for slip dir, not normal to each other so rss non-zero." , you are only considering cos lambda. cos phi for this case is 1*1 +1*-2+1*1=0, and that's why rss is 0. You are perfectly right about [1-21] and [1-1-1]: cos phi is not zero.
    Remember, a slip system is (hkl)[uvw], where (hkl) is the plane, and [uvw] is a direction IN that plane. If you merely evaluate cos lambda, your calculation is incomplete.
  16. Mar 24, 2012 #15
    Ahh I see now, so I need to be recalculating cos(phi) for every slip system, I had just assumed that it was a constant throughout all systems. That'll mean some of my other RSS values will be slightly off as I used an incorrect cos(phi), but no issue, it can be easily fixed.

    The only issue is see that is still confusing me is this:

    Here you've multiplied incorrectly I think, you have a 1 instead of a -1, and 1*1 instead of 1*0 at the end - either that or I have learnt it the wrong way. Should it not be given by:

    (1*1) + (-2 * -1) + (1*0) = 3

    Also, am I correct in saying that slip in fcc crystals can occur on 12 {hkl}<uvw> systems, and those systems are {111}<110>, where there are 4 planes, and each plane has its 3 directional vectors within it, <0,-1,-1>,<1,0,1> and <1,1,0>?

    Thanks for all your help Bavid.
    Last edited: Mar 24, 2012
  17. Mar 25, 2012 #16
    You seem to have learnt it the wrong way. cos phi is calculated from the PLANE and the tensile axis, no role of the slip direction there. The plane is (111) and the tensile axis is [1-21]. Please remember what cos phi and cos lambda mean. They are different.
    You are correct about the slip systems.
  18. Mar 26, 2012 #17
    Hi Bavid,

    I think you might be mistaken, I've double checked on a few different sites/textbooks and all seem to say this:

    Or worded such as:

    Pulled from these nice sites:
    http://oregonstate.edu/instruct/engr322/Homework/Previous/S10/ENGR322HW3.html [Broken]

    Or am I *still* getting very confused? Sorry for posting so much. I guess my big issue is I have 3 values for the RSS for the 3 slip directions, but two of the slip directions have the same RSS, when I think one of the duplicate values is meant to be zero.
    Last edited by a moderator: May 5, 2017
  19. Mar 26, 2012 #18
    I said "cos phi is calculated from the PLANE and the tensile axis, no role of the slip direction there." I assumed you know that the slip plane's normal is the exact same miller indices as the slip plane itself, only that the the two are written using different brackets. All I meant was that only information regarding the slip plane (i.e., its normal) and the tensile axis are needed to calculate cos phi, the slip direction does not figure. The resources you cite here are saying the same thing as I am.
    And two slip systems can have the same value of rss, if they have the same slip plane (cos phi is same for the two) and the slip directions are equally inclined to the tensile axis (cos lambda is also the same for the two). Why do you *think* one of the duplicate values are meant to be zero?
    If you are still confused, I would suggest you read a text book with these ideas in mind; some of the statements I make here NEED visual demonstration, for which this forum is probably not the best place.
  20. Mar 29, 2012 #19
    Ah yes I see Bavid, it's all good now. I've gone back to the basics and reworked everything, and am now getting three nice RSS values, all which seem intrinsically correct.

    Thanks for all your help, it's much appreciated!
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