1
ENSC 3313, Spring 2005
TEST 2 - KEY
Name_________________________
Answer all 10 questions, show all work, 2points each
1.
Visualize your tensile specimen of lab 1! A test, such as you did, gave:
0.1% offset load
= 2,000 lb
Peak load
= 6,000 lb
Extension at fracture = 0.4 inches
Calculate the 4 property values
you would expect to find in handbooks and use in design.
Answer
You should recollect, diameter = 0.5", hence area 0.2 in
2
, length = 2", hence elongation/2 = strain
Thus,
i) Elastic Modulus =
(
)
psi
10
10
001
.
0
2
.
0
2000
6
×
=
ii) Yield Stress =
psi
000
,
10
2
.
0
2000
=
iii) Tensile Strength =
psi
000
,
30
2
.
0
000
,
6
=
iv) % Elongation (2") =
(
)
%
20
100
2
4
.
0
=
2.
Shown is the plane (110) and the direction [
11
1
] in a BCC iron unit cell.
Is this a possible slip system?
If
so, calculate the resolved shear stress if 10,000 psi tensile stress is applied in the [010] direction.
Yes, this is a slip system since the direction lies in the
plane:
0
0
1
1
1
1
1
w
l
v
k
u
h
=
⋅
+
⋅
+
⋅
−
=
⋅
+
⋅
+
⋅
The resolved shear stress is given by the
formula:
λ
φ
σ
τ
cos
cos
=
r
The angle between the plane normal [110] and the stress
direction [010] is 45
o
so cos
φ
=
2
1
The cosine of the angle between the direction [
11
1
] and
[010] is:
3
1
cos
=
λ
Hence,
psi
080
,
4
6
000
,
10
r
=
=
τ
x
y
z
φ
λ
4R
R
3
4

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