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Strain tensor on a Triangle

  1. Sep 13, 2009 #1
    1. The problem statement, all variables and given/known data
    Strain tensor E = [0.02 0 0; 0 0.01 0; 0 0 0.03] (no shear strains).

    A triangle consists of points A, B, and C, each on axis X1, X2, and X3 respectively. The lengths OA = OB = OC, and D is the midpoint of AC. The direction cosines of AC are (1/sqrt(2), 0, -1/sqrt(2)) and those of BD are (-1/sqrt(6), sqrt(2)/sqrt(3),-1/sqrt(6)). Find:

    A) The elongation of AC
    B) The change of initial right angle BDA

    9a65va.jpg

    2. Relevant equations

    A^2 + B^2 = C^2 for right triangle

    3. The attempt at a solution

    Part A --

    Drawing the triangle OAC (in the plane X1-X2), we see that the initial angles CAO and ACO are 45 degrees, while COA is a right angle. The deformation at point A is AO(0.02) while the deformation at point C is CO(0.03). Since AO=CO, we can write the change as:

    A'C' - AC = sqrt( (1.02AO)^2 + (1.03AO)^2 ) - sqrt ( 2 AO^2 ) = 0.0354 AO

    Part B --

    Here I am a bit stuck. It seems to me that it should still be a right angle, but this is probably not true. I know I should have to use the direction cosines for BD. Any advice or hints?
     
  2. jcsd
  3. Sep 13, 2009 #2
    Ah, I had some insight. So the reason its not a right angle anymore is because C' is farther away from C than A' is from A. (The midpoint of AC has moved closer to C!) Still kinda lost about how to use this information though.
     
  4. Sep 13, 2009 #3
    I have decided a game plan:

    Try to solve cos(theta) = dot(B'D', A'C') / |B'D'|*|A'C'|

    I have solved for |A'C'| already. I think I can solve for |B'D'| without much difficulty. Then I need to determine the direction cosines for B'D' and A'C'. Finally, plug in and solve.

    Still fishing for help. I'm particularly confused about how to solve for the direction cosines.
     
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