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ENMA 300
Homework 8
Solutions
1a) The critical resolved shear stress for copper is 0.48 MPa (70 psi).
Determine the
minimum
possible
yield strength for a single crystal of Cu pulled in tension.
In order to determine the minimum possible yield strength for a single crystal of Cu pulled in tension, we
simply employ Equation 8.5 as
σ
y
= 2
τ
crss
= (2)(0.48 MPa) = 0.96 MPa
(140 psi)
bi)
A single crystal of a metal that has the BCC crystal structure is oriented such that a tensile stress is
applied in the [100] direction.
If the magnitude of this stress is 4.0 MPa, compute the resolved shear
stress in the
[1
1 1]
direction on each of the (110), (011), and
(10
1 )
planes.
This part of the problem asks, for a BCC metal, that we compute the resolved shear stress in the
[1
1 1] direction on
each of the (110), (011), and
(10
1 ) planes.
In order to solve this problem it is necessary to employ Equation 8.2,
which means that we first need to solve for the
for angles
λ
and
φ
for the three slip systems.
For each of these three slip systems, the
λ
will be the same—i.e., the angle between the direction of the
applied stress, [100] and the slip direction,
[1
1 1].
This angle
λ
may be determined using Equation 8.6
λ =
cos

1
u
1
u
2
+
v
1
v
2
+
w
1
w
2
u
1
2
+
v
1
2
+
w
1
2
(
)
u
2
2
+
v
2
2
+
w
2
2
(
)
where (for [100])
u
1
= 1,
v
1
= 0,
w
1
= 0, and (for
[1
1 1])
u
2
= 1,
v
2
= –1,
w
2
= 1.
Therefore,
λ
is determined as
λ =
cos

1
(1)(1)
+
(0)(

1)
+
(0)(1)
(1)
2
+
(0)
2
+
(0)
2
[
]
(1)
2
+
(

1)
2
+
(1)
2
[
]
=
cos

1
1
3
=
54.7
°
Let us now determine
φ
for the angle between the direction of the applied tensile stress—i.e., the [100] direction—
and the normal to the (110) slip plane—i.e., the [110] direction.
Again, using Equation 8.6 where
u
1
= 1,
v
1
= 0,
w
1
= 0 (for [100]), and
u
2
= 1,
v
2
= 1,
w
2
= 0 (for [110]),
φ
is equal to
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[100]

[110]
=
cos

1
(1)(1)
+
(0)(1)
+
(0)(0)
(1)
2
+
(0)
2
+
(0)
2
[
]
(1)
2
+
(1)
2
+
(0)
2
[
]
=
cos

1
1
2
=
45
°
Now, using Equation 8.2
τ
R
= σ
cos
φ
cos
λ
we solve for the resolved shear stress for this slip system as
τ
R
(110)

[1
1 1]
=
(4.0 MPa) cos (45
°
) cos(54.7
°
)
[
]
=
(4.0 MPa) (0.707)(0.578)
=
1.63 MPa
Now, we must determine the value of
φ
for the (011)–
[1
1 1] slip system—that is, the angle between the
direction of the applied stress, [100], and the normal to the (011) plane—i.e., the [011] direction.
Again using
Equation 8.6
λ
[100]

[ 011]
=
cos

1
(1)(0)
+
(0)(1)
+
(0)(1)
(1)
2
+
(0)
2
+
(0)
2
[
]
(0)
2
+
(1)
2
+
(1)
2
[
]
=
cos

1
(0)
=
90
°
Thus, the resolved shear stress for this (011)–
[1
1 1] slip system is
τ
R
(011)

[1
1 1]
=
(4.0 MPa) cos (90
°
) cos (54.7
°
)
[
]
=
(4.0 MPa) (0)(0.578)
=
0 MPa
And, finally, it is necessary to determine the value of
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 Spring '11
 AlSheikhly

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