AAE 590, CRN 65011, Spring 2013
Homework #2; Due on Thursday, January 24
Solve the following problems. This HW is for your own benefit and is necessary to properly learn the
material. It is expected that each student put forward an honest effort in solving each problem and
contributes to every part of the HW. Please place all names on the HW. Please ask any questions during
class or office hours.
Problem 1:
Plastic deformation is achieved by slipping of atoms between closed-packed planes along
some low index directions. Show your work!
a) In an FCC crystal with lattice parameter a:
i) Determine the distances between neighboring atoms along [100], [110], and
[111] directions. Which is the smallest?
Solution:
1
Along [100] , a
Along [110],
a2 + a2
2
a
=
a=
= 0.707a
2
2
2
Along [111],
a 2 + a 2 + a 2 = 3a = 1.7329a
Smallest distance is along [110]
ii) Determine the distances between the neighboring (100), (110), and (111)
planes. Which is the largest?
Solution:
a
2
Along [100],
2
Along [110],
Along [111],
a2 + a2
2
a
=
a=
= 0.355a
2× 2
4
2 2
a2 + a2 + a2
3
a
=
a=
= 0.577a
3
3
3
(3 planes exist between 2 corner atoms along this direction)
Largest distance is along [111]
iii) From this analysis, what can you infer about the slip systems in an FCC
crystal?
Solution:
In an FCC slip system, slip planes are {111} having largest spacing and
slip directions are <110> having smallest distance.
b) In a BCC crystal with lattice parameter a:
i) Determine the distances between neighboring atoms along [100], [110], and
[111] directions. Which is the smallest?
Solution:
3
Along [100], a
Along [110],
a 2 + a 2 = 2a = 1.41a
Along [111],
a2 + a2 + a2
3
=
a = 0.866a
2
2
Smallest is along [111]
ii) Determine the distances between the neighboring (100), (110), and (111)
planes. Which is the largest?
Solution:
4
Along [100],
a
= 0.5a
2
Along [110] ,
Along [111],
a2 + a2
2
a
=
a=
= 0.707a
2
2
2
a2 + a2 + a2
3
a
=
a=
= 0.288a
6
2×3
2× 3
(3 planes exist between one corner atom and the central atom along this direction)
Largest is along [110]
iii) From this analysis, what can you infer about the slip systems in a BCC
crystal?
Solution:
In an BCC slip system, slip planes are {1 1 0} having largest spacing, and slip
directions are <111> having smallest distance.
Problem 2:
Visualization of slip in an FCC crystal:
a) Draw the atoms of an FCC crystal projected on the (111) plane. (Hint to help you
visual, the following website will help: http://stokes.byu.edu/fcc.htm )
Solution:
5
b) In the FCC crystal, there is a stacking sequence according to the position relative to
the projected plane. Please label your atoms in part A, with the associated stacking
sequence (A,B,C,A’)
Solution :
c) Draw a slip vector in the [110] direction (Hint it connects ‘B’ atom to ‘B’ atom).
Solution:
d) Draw a pair of slip vectors in the [211] direction (connecting ‘B’ atom to ‘C’ atom)
and in the [12 1 ] direction (connecting ‘C’ atom to ‘B’ atom)
Solution :
6
Problem 3:
Crystallography Geometry (this is mostly number crunching, so feel free to use a tool
such as Matlab to cut down on the work):
a) What is the angle between the (111) and (1 1 1 ) planes?
Solution:
Taking dot product of the two normal directions,
[111].[1 1 1 ] = 12 + 12 + 12 (−1) 2 + 12 + (−1) 2 cos α
⇒ cos α =
(1)(−1) + (1)(1) + (1)(−1)
12 + 12 + 12 (−1) 2 + 12 + (−1) 2
=−
1
3
⇒ α = 109.47 0
b) What is the angle between the [110] and [211] directions?
Solution:
Taking dot product of the two directions,
[110].[211] = 12 + 12 + 0 2 2 2 + 12 + 12 cos α
(1)(2) + (1)(1) + (0)(1)
3
⇒ cos α =
=
2
2
2
2
2
2
2 6
1 +1 + 0 2 +1 +1
⇒ α = 30 0
c) What is the angle between the [211] and [12 1 ] directions?
Solution:
Taking dot product of the two directions,
[211].[12 1 ] = 2 2 + 12 + 12 12 + 2 2 + (−1) 2 cos α
⇒ cos α =
(2)(1) + (1)(2) + (1)(−1)
2 2 + 12 + 12 12 + 2 2 + (−1) 2
=
3
6 6
⇒ α = 60 0
d) What is the direction normal to the [ 211] and [01 1 ] directions?
Solution:
Taking cross product of the two directions,
[211] × [011] = [022]
So, the normal direction is (0 1 1)
e) What is the direction normal to the [33 2] and [1 1 0] directions?
Solution:
Taking cross product of the two directions,
[33 2] × [1 1 0] = [ 2 2 6]
So, the normal direction is [ 1 1 2]
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f) What is the line of intersection between the (113) and (111) planes?
Solution:
Taking cross product of the two normal directions,
[113] × [111] = [220]
So, the intersecting line is [ 1 10]
g) What is the line of intersection between the (113) and (1 1 1) planes?
Solution:
Taking cross product of the two normal directions,
[113] × [1 1 1] = [42 2]
So, the intersecting line is [21 1 ]
Problem 4:
Elastic Modulus – Model 316L stainless steel as a cubic material. Based off the uniaxial
σ-ε data for single crystals in the [111], [001], and [123] orientations, calculate the elastic
modulii in the [112], [101], and [113] directions.
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Solution:
We have,
1
E[ hkl ]
=
=
1
E<001>
1
E<001>
− 3(
− 3(
1
E<001>
1
E<001>
−
−
1
E<111>
1
E<111>
)
)(α 2 β 2 + α 2γ 2 + β 2γ 2 )
k 2 h 2 + k 2l 2 + h 2l 2
(k 2 + h 2 + l 2 ) 2
Where α, β and γ are the direction cosines of the [h k l] direction and the [1 0 0], [0 1 0]
and [0 0 1] directions respectively.
From the given data plot, we have,
E[111] = 264.087 GPa and E[ 001] = 100.014 GPa ( calculated from the plots )
1
E[112]
=
1
1
1
(12 )(12 ) + (12 )(2 2 ) + (12 )(2 2 )
− 3(
−
)
100.014
100.014 264.087
(12 + 12 + 2 2 ) 2
∴ E[112] = 187.28GPa
1
E[101]
1
1
1
(0 2 )(12 ) + (0 2 )(12 ) + (12 )(12 )
=
− 3(
−
)
100.014
100.014 264.087
(12 + 0 2 + 12 ) 2
∴ E[101] = 187.28GPa
1
E[113]
=
1
1
1
(12 )(12 ) + (12 )(32 ) + (12 )(32 )
− 3(
−
)
100.014
100.014 264.087
(12 + 12 + 32 ) 2
∴ E[113] = 141.397GPa
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