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"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

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# Calculating the Schmid factor for an FCC single crystal

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