Recall that FCC and BCC metals are common and these crystal structures feature d

Question

Recall that FCC and BCC metals are common and these crystal structures feature different slip systems (slip plane and slip direction combinations), The purpose of this question is to help you derive the angles relevant to slip in these cubic systems. Although you can find these angles listed in a variety of resources, you are encouraged to solve for them so you can further your understanding of cubic geometry and prepare yourself for the exam. In the next lecture, we'll learn that the angles featured in this analysis are common to all cubic systems, but the angles are used in different ways when calculations pertaining to slip for different crystal structures. For now, we can just focus on the fact that we are interested in a cubic structure. so we solve this generically here and apply what we learn later YouT want to remember or record the angles that you establish on this homework. Consider the following cubic unit cell: [0 0 1] [0 01] [10 1] (1 1 1] 1 1 1)

Explanation / Answer

Blue vector, A = [1 0 1]

|A| = (12 + 02 + 12)0.5 = 20.5

Red vector, B = [1 1 1]

|B| = (12 + 12 + 12)0.5 = 30.5

C = [0 0 1]

|C| = 1

D = [1 0 0]

|D| = 1

1)

is the angle between A and C

A.C = |A||C| cos

1.0 + 0.0 + 1.1 = 20.5 * 1 * cos

cos = 1 / 20.5

= 45°

2)

is the angle between A and D

A.D = |A||D| cos

1.1 + 0.0 + 1.0 = 20.5 * 1 * cos

cos = 1 / 20.5

= 45°

3)

is the angle between B and C

B.C = |B||C| cos

1.0 + 1.0 + 1.1 = 30.5 * 1 * cos

cos = 1 / 30.5

= 54.7°

Angle (90° - ) = 35.3° is the angle between B and [1 1 0]

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