ENME 382 Section 0201
Homework 8
Due in class: April 26, 2010
1.
a)
The critical resolved shear stress for copper is 0.48 MPa (70 psi).
Determine the maximum
possible yield strength for a single crystal of Cu pulled in tension.
Solution
In order to determine the maximum 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.
bii) On the basis of these resolved shear stress values, which slip system(s) is (are) most favorably
oriented?
Solution
(a)
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
[
]
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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.
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 Spring '10
 shang
 Tensile strength, 4 m, 2 mM, 4 mm, 1.0 j, 0.79 mm

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