EMSE 201 — Introduction to Materials Science & Engineering
5 November 2008
Name:
SOLUTION
Department of Materials Science and Engineering
1 of 5
Case Western Reserve University
Test #2 —
75 minutes; 100 points; 5 questions; 5 pages; 10% of course grade
Partial credit
will be given for correct setups and reasoning. Give
units
on numerical answers where ap
propriate. Please write all answers on these pages; use the backs if needed.
1)
A rod, 10.0 cm long, must be able to support a load of 18.0 kN in uniaxial tension along its length. For
each of the four candidate materials listed below,
a)
(6 points)
compute the minimum crosssectional area
of the rod to ensure that its yield strength (or
in the case of silicon nitride, its fracture strength) will not be exceeded under the specified load;
b)
(7 points)
rank the mass
of the resulting rods from 1 (lowest) to 4 (highest);
c)
(7 points)
rank the elongation
of the resulting rods under the specified load from 1 (lowest) to 4
(highest);
Show your work
in the space below. (Use the back if necessary.)
material
yield
strength,
MPa
density,
g cm
–3
modulus of
elasticity,
GPa
cross
sectional
area, cm
2
rank by mass
(1 lowest, 4
highest)
rank by elonga
tion (1 lowest, 4
highest)
annealed 316 stainless steel
205
7.80
193
0.878
4
(68.5 g)
1
(0.106%)
annealed Ti6Al4V
830
4.43
114
0.217
2
(9.61 g)
3
(0.728%)
hotpressed silicon nitride
850*
3.30
304
0.212
1
(6.99 g)
2
(0.280%)
polyetheretherketone
91
1.31
1.10
1.98
3
(25.9 g)
4
(8.27%)
*) fracture strength
For a uniaxial tensile force
F
on a specimen of uniform initial crosssectional area
A
, the en
gineering stress
σ
=
F
/
A
(
2 pts
). Setting the stress equal to the material’s strength,
A
=
18.0 kN/
σ
y
where
σ
y
is the applicable strength parameter (
1 pt
for each correct answer).
The mass
M
of the rod will be given by
AL
ρ
, where
L
is the length (specified as 10.0 cm) and
ρ
is the mass density. Substituting
A
=
F
/
σ
y
gives
M
=
FL
ρ
/
σ
y
(
3 pts
). The masses are
listed in parentheses in the table above. As a ranking was requested, it was necessary only to
compute the ratio
ρ
/
σ
y
and rank the resulting values (
1 pt
for each correct ranking or mass).
The elongation
∆
L
of the rod will be given by
L
ε
=
L
σ
/
E
, where
ε
is the strain and
E
is the
modulus of elasticity. Substituting
σ
=
σ
y
gives
∆
L
=
L
σ
y
/
E
(
3 pts
). As a ranking was re
quested, and
L
is the same for all of the rods, it was necessary only to compute the strain
σ
y
/
E
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 Spring '09
 Materials Science, pts, Tensile strength, Department of Materials Science and Engineering, Materials Science & Engineering

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