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MASSACHUSETTS
INSTITUTE
OF
TECHNOLOGY
DEPARTMENT
OF
MECHANICAL
ENGINEERING
CAMBRIDGE,
MASSACHUSETTS
02139
2.002
MECHANICS
AND
MATERIALS
II
HOMEWORK
SOLUTION
NO.
5
Distributed
:
Wednesday,
April
10,
2003
Due
:
Wednesday,
April
16
,
2003
Problem
1
(
30
points
)
The
critical
stress
intensity
factor,
K
IC
,
is
given
by
the
inequality
K
IC
≥
Qσ
∞
√
πa
(1)
Solving
for
the
critical
crack
length,
a
crit
we
get
2
1
±
K
IC
a
≤
a
crit
≡
(2)
π
Qσ
∞
For
an
edge
crack
(whose
length
is
less
than
or
equal
to
0.65
times
the
plate’s
width,
w
),
the
conﬁguration
correction
factor,
Q
,
is
1
.
12
Q
=
²
1
−
0
.
7
³
a
´
1
.
5
µ
3
.
25
(3)
w
a
We
are
told
that
the
the
plate
is
large
,
which
implies
that
w
1.
It
is
also
explicitly
stated
that
Q
=
1
.
12
(4)
We
are
also
told
to
use
a
design
stress
equal
to
0
.
6
times
the
minimum
yield
strength.
σ
∞
=
0
.
6
σ
y
min
(5)
Substituting
equations
4
and
5
into
equation
2
we
get
1
±
K
IC
2
a
crit
=
(6)
π
0
.
672
σ
y
min
1
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View Full Document We
are
asked
to
compare
the
critical
crack
sizes
for
four
diﬀerent
grades
of
steel.
The
values
of
σ
y
min
and
K
IC
for
these
steels
are
given
in
the
second
and
third
columns
of
table
1.
The
fourth
column
of
the
table
contains
the
critical
crack
length
as
computed
by
equation
6.
Steel
(ASTM
Grade)
σ
y
min
[
ksi
]
K
IC
at
−
60
o
F,
[
ksi
√
in
]
a
crit
[
in
]
A36
36
60
1.958
A441
50
53
0.792
A572
50
100
2.82
A514
100
60
0.254
Table
1:
Material
properties
and
Critical
crack
length
for
available
steels
Based
on
table
1,
the
steel
that
would
most
likely
be
selected
is
A572.
A572
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This note was uploaded on 02/23/2012 for the course MECHANICAL 2.002 taught by Professor Davidparks during the Spring '04 term at MIT.
 Spring '04
 DavidParks
 Mechanical Engineering

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