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Calculating the Schmid factor for an FCC single crystal

  1. Mar 4, 2010 #1
    The problem said:
    "Determine the magnitude of the Schmid factor for an FCC single crystal oriented with its [100] direction parallel to the loading axis. "

    The Schmid factor is the

    "cos([tex]\phi[/tex])*cos([tex]\lambda[/tex])"

    term in the equation for resolved shear stress.

    I know that the slip planes for an FCC single crystal are the four {111} planes, and the slip directions are the three <110> directions inside each of the {111} planes.

    To solve for [tex]\phi[/tex], the angle between the <111> direction and the loading direction, <100>, I said that it was equal to

    Cos[tex]^{-1}[/tex]((1*1+1*0+1*0)/[tex]\sqrt{(1^2+1^2+1^2)*(1^2)}[/tex])

    That comes out to 1/[tex]\sqrt{3}[/tex]

    And using the same equation, I found that [tex]\lambda[/tex], the angle between the loading direction, <100>, and the slip direction, <01-1>, is [tex]\pi[/tex]/2

    Multiplying [tex]\phi[/tex] and [tex]\lambda[/tex] should have given me the schmid factor, according to my textbook, but I couldn't get the right answer.

    The answer is .408, but I kept getting something different and I'm not sure why. If some one could explain this to me, I would be very grateful. Thanks
     
  2. jcsd
  3. Mar 4, 2010 #2

    Mapes

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    Hi jbrown110, welcome to PF. Your slip direction doesn't lie within your slip plane. You're going to have a hard time obtaining slip that way. :smile:
     
  4. Mar 5, 2010 #3
    So, basically I picked the one direction that doesn't work for the [111] plane. haha

    I reworked it with the other two possible slip directions and they both gave the right answer.

    Thanks for the reply.
     
  5. Mar 5, 2010 #4

    Mapes

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    You're welcome!
     
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