Calculating the Schmid factor for an FCC single crystal

[jbrown110]

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(\phi)*cos(\lambda)"

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 \phi, the angle between the <111> direction and the loading direction, <100>, I said that it was equal to

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

That comes out to 1/\sqrt{3}

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

Multiplying \phi and \lambda 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

 

[Mapes]

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.

 

[jbrown110]

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.

 

[Mapes]

You're welcome!