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Stress tensor transformation and coordinate system rotation

  1. Jan 9, 2014 #1
    1. The problem statement, all variables and given/known data

    Hi, I am not sure if this is the right place for my question but here goes!

    The stress tensor in the Si coordinate system is given below:

    σ’ij = {{-500, 0, 30}, {0, -400, 0}, {30, 0, 200}} MPa

    Calculate the stress tensor in the L coordinate system if: cos-1a33=45°, and X’2 is in the plane defined by X1, X2 and is rotated 60° counterclockwise from X2.

    2. Relevant equations

    The answer the book gives is:

    From the formula

    σ’ij = aikajlσkl, where σ’ij is defined in L,

    (σ’33) = a312σ11+ 2a31a32σ12 + 2a31a33σ13 + a3222 + 2a32a33σ23 + a332σ33

    The direction cosine matrix is given by

    aij = {{0.35,0.61,-0.71}, {-0.87,0.5,0}, {0.35,0.61,0.71}}

    Thus, σ’11 = -150.0 MPa; σ’12 = 92.0 MPa; σ’13 = -225.0 MPa; σ’22 = -450.0 MPa; σ’23 = 31.0 MPa; σ’33 = -100.0 MPa.

    Here is the answer with some explanations and guidance and I’m still not able to reach the same answer as the author of the book I’m following.

    3. The attempt at a solution
    My idea was to just insert the given σij and aij from the answer in the σ’ij = aikajlσkl function but already here I encounter the problem that σ’33 = 0,352*(-500)+2*0,35*0,71*30+0,612*(-400)+0,712*200 = 97,5 MPa, which isn’t that much off, but when they express their answers with one decimal precision I assume it is wrong. I have also calculated the other σ’ij but the error is even greater for them.

    I would very much appreciate help with this, is there an additional step that I am missing?
     
    Last edited: Jan 9, 2014
  2. jcsd
  3. Jan 9, 2014 #2
    Hi Theade88. Welcome to Physics Forums!!

    Are you sure about that direction cosines matrix? It doesn't look right to me. Please show how you got it. My first row is minus yours.

    Chet
     
    Last edited: Jan 9, 2014
  4. Jan 10, 2014 #3
    Hi, Chestermiller. Thank you for replying!

    The cosine matrix is directly taken from the solution that the book gives. What I have tried is to recreate the result using the information which is given in the answer. All of part 2 in my OP is from the solution the book gives to the problem.

    I've been able to recreate the cosine matrix using:

    ry(θ)={{cos(45), 0, -sin(45)}, {0, 1, 0}, {sin(45), 0, cos(45)}}

    rz([itex]\psi[/itex])={{cos(60), sin(60), 0}, {-sin(60), cos(60), 0}, {0, 0, 1}}

    ry x rz = {{cos(60)cos(45), cos(45)sin(60), -sin(45)}, {-sin(60), cos(60), 0}, {sin(45)cos(60), sin(45)sin(60), cos(45)}}

    I don't know if this is the correct cosine matrix, if the answer in the book is wrong, or if I'm missing a step. :)
     
  5. Jan 10, 2014 #4
    I think I must have been the one who made an error in evaluating the cosine matrix. Anyway, I used your cosine matrix to evaluate the primed stresses, and here's what I got:

    σ11'=-124
    σ12'=49
    σ13'=-311
    σ22'=-478
    σ23'=12
    σ33'=-94
     
  6. Jan 10, 2014 #5
    Yeah, that is what I get too!!

    Thank you for your help, for now I will assume that the answer in the book is wrong. :)
     
  7. Jan 10, 2014 #6

    nvn

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    theade88: I used only the given question in section 1 of post 1, and did not look at sections 2 nor 3, nor any other post. Looking only at the question in section 1, I made a quick attempt, without checking my work at all. And I got two possible answers. There are two possible answers for the given question, right? Or am I misinterpreting? The two possible local-to-global rotational transformation matrices are:

    Code (Text):
         [  0.35355 -0.86602  0.35355 ]
    R1 = [  0.61237  0.50000  0.61237 ]
         [ -0.70711  0.00000  0.70711 ]

         [  0.35355 -0.86602 -0.35355 ]
    R2 = [  0.61237  0.50000 -0.61237 ]
         [  0.70711  0.00000  0.70711 ]
    Using R1, and transforming the given stress tensor (σ) from the S to the L coordinate system, I currently obtained the following stress tensor in the L coordinate system.

    σ'11 = -127.5 MPa
    σ'12 = 48.99
    σ'13 = -312.5
    σ'22 = -475.0
    σ'23 = 12.25
    σ'33 = -97.50​

    Using R2 also gave nothing close to the book answer.

    Do you agree that there are two possible answers? Or not? Also, did you make a typographic mistake by putting a prime mark on the given stress tensor in section 1? The L coordinate system uses the prime mark, but not the S coordinate system.
     
  8. Jan 10, 2014 #7
    Is it possible that R2 is for a left hand coordinate system?
     
  9. Jan 10, 2014 #8

    nvn

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    Chestermiller: No, I thought R1 and R2 are both for a right-handed coordinate system.
     
  10. Jan 11, 2014 #9
    Aaaah. You're right. There are 2 sets of primed axes that satisfy the problem specifications. Here's how it plays out: The y' axis is unique, and there is only one y' axis. But there are two z' axes that satisfy the specifications. The z' axis must lie on the surface of a cone surrounding the z axis at a cone angle of 45 degrees, but it also must lie in a plane perpendicular to the y' axis. This plane cuts the cone along two different lines. So there are two different possibilities for the z' axis. This automatically leads to two corresponding possibilities for the x' axis.

    Chet
     
    Last edited: Jan 11, 2014
  11. Jan 11, 2014 #10
    Yes there is a typo, the Si stress tensor matrix should not be primed.

    I think that you have used -60° instead of 60° which is the counter clockwise direction in trigonometry. Your point of it being two cosine direction matrices that meets the problem specification, I really don't know. In my opinion, which often is flawed, the rotations which complies with the books best is a counter clockwise rotation around the X'3 axis → X2∠X'2=60° and a counter clockwise rotation around the X'2 axis X3∠X'3 = 45°. In this specific order the new X'2 axis will be in the X1, X2 plane since the X'3 axis will have a perpendicular angle to the plane at the moment of rotation. After this rotation a counter clockwise rotation of 45° is performed around the X'2 axis which gives cos-1(a33)=45°. Any other order or direction will not fulfill the specified information from the problem statement?!

    Since the rotation order goes from right to left this means:

    ry(45°) * rz(60°) = ...
     
    Last edited: Jan 11, 2014
  12. Jan 11, 2014 #11
    Did you read my post #9?

    Chet
     
  13. Jan 11, 2014 #12
    Hm, now when i read it again i realize that every angle around the positive X3 axis is positive therefore if X'3 is mirrored through the X3 X'2 plane the specifications is still met. So you two are right there is two solutions to the cosine matrix.

    ry(45°) * rz(60°)
    and
    r(-45°) * rz(60°)
     
  14. Jan 11, 2014 #13
    Now the question is "does either of these two transformations give stress tensor components that agree with your book?"
     
  15. Jan 11, 2014 #14

    nvn

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    Chestermiller: I touched on that in the second to last paragraph of post 6. Neither of these transformations give anything close to the book answer.
     
    Last edited: Jan 11, 2014
  16. Jan 11, 2014 #15

    nvn

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    theade88: Agreed. In posts 1 and 3, you are using C programming language, two-dimensional matrix initialization notation, where, for braces inside of braces, the inner braces mean matrix rows, not columns. This is an excellent convention.

    Therefore, we see your rotational transformation matrix in post 1, which you called aij, is the global-to-local rotational transformation matrix, whereas my rotational transformation matrices in post 6 are local-to-global rotational transformation matrices, which I called R1 and R2. Anyway, they are merely the transpose of each other.

    And we see in posts 3, 10, and 12, your ry and rz nomenclature again denotes global-to-local rotational transformation matrices. Therefore, in that case,

    Code (Text):
            [  cos(θ) 0 -sin(θ) ]
    ry(θ) = [    0    1    0    ]
            [  sin(θ) 0  cos(θ) ]

            [  cos(θ) sin(θ) 0  ]
    rz(θ) = [ -sin(θ) cos(θ) 0  ]
            [    0      0    1  ]
    Therefore, ry(45 deg)*rz(60 deg) =

    Code (Text):
    [  0.35355  0.61237 -0.70711 ]
    [ -0.86602  0.50000  0.00000 ]
    [  0.35355  0.61237  0.70711 ]
    We see this matches R1T. Good.

    For the second possible solution, ry(-45 deg)*rz(60 deg) =

    Code (Text):
    [  0.35355  0.61237  0.70711 ]
    [ -0.86602  0.50000  0.00000 ]
    [ -0.35355 -0.61237  0.70711 ]
    We see this indeed matches R2T. Therefore, theade88, we see that your post 12 is correct. We also see, all of your matrices in post 3 are exactly correct, except for one minor typographic mistake. The left-hand side of your third equation in post 3 should be ry*rz (i.e., ry dot rz), not ry cross rz.
     
    Last edited: Jan 11, 2014
  17. Jan 12, 2014 #16
    Ahh, I didn't know that it makes a difference which symbol you use for multiplication of matrices. :) Anyway, I'm still not reaching the same answer as the book.

    Thank you both very much for all the help!
     
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