AE304_fall09_HW2_solution

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Unformatted text preview: 
 AE304
–
Aircraft
Structures
I
–
Fall
2009
 Due:
Wednesday,
September
9,
2009
 H omework 2 - S olution Exercise
1:
 The
state
of
stress
at
a
critical
point
of
an
aircraft
component
is
given
by

 
 The
tensile
yield
strength
of
the
material
is
82
MPa
and
the
factor
of
safety
is
1.2.
Determine
whether
failure
 takes
place
at
the
point,
according
to
(a)
maximum
shearing
stress
theory
and
(b)
maximum
energy
distortion
 theory
 Solution:
 The
applied
stresses:

 
 The
invariant:
 
 
 The
equation
and
the
corresponding
roots
or
principal
stresses:
 
 
 a.
Maximum
shearing
stress
(Tresca)
theory:

 
 
 b.
maximum
energy
of
distortion
(Von
Mises)
theory:
 ∴

Based
on
Tresca
theory,
failure
takes
place
at
this
point
 
 ∴

Based
on
Von
Mises
theory,
failure
takes
place
at
this
point
 Exercise
2:
 Simple
tension
and
compression
tests
on
a
brittle
material
reveal
that
failure
occurs
by
fracture
at
σut
=
260
 MPa
and
σuc
=
420
MPa,
repectively.
In
actual
application,
the
material
is
subjected
to
perpendicular
tensile
 and
compressive
stresses,
σ1
and
σ2,
respectively,
such
that
σ1/σ2
=
‐1/4.
Determine
the
limiting
values
of
σ1
 and
σ2
according
to
the
Coulomb‐Mohr
theory.
 Solution:
 The
principal
stresses:
 
 According
the
Coulomb‐Mohr
theory,
yielding
occurs
when
 
 Then,
 
 Hence
the
limiting
values
for
the
principal
stresses
are:
 
 
 Exercise
3:
 The
100‐mm
diameter
bar
shown
in
Figure
1
is
made
of
steel
that
has
a
yield
stress
σY
=
420
MPa.
The
free
end
 of
 bar
 is
 subjected
 to
 loads
 2
 P
 in
 x‐direction
 and
 5
 P
 in
 z‐direction.
 Determine
 the
 magnitude
 of
 P
 that
 initiates
yielding
at
point
A,
based
on
(a)
the
distortional
energy
density
criterion
and
(b)
maximum
shearing
 stress
criterion,
if
the
factor
of
safety
is
1.5.
 
 Solution:
 Loads
at
the
fixed
end:
 a. Normal
force:
Fx
=
2
P
 b. Shear
force:
Fz
=
5
P
 c. Moment
about
x‐axis
or
Torque:

 Mx
=
T
=
2
P
 d. Moment
about
y‐axis:
My
=
‐
0.6
P
 e. Moment
about
z‐axis:
Mz
=
‐
0.8
P
 
 
 Figure
1
 
 
 
 At
the
point
A,
top
surface:
 Applied
stresses
at
point
A:
 
 
 The
principal
stresses
at
point
A:
 
 
 Evaluation
at
point
A:
 a. based
on
maximum
shearing
stress
(Tresca)
theory:

 
 
 b. based
on
maximum
energy
of
distortion
(Von
Mises)
theory:
 
 Exercise
4:
 A
2024‐T851
aluminum
alloy
frame
with
an
edge
crack
supports
a
concentrated
load
as
shown
in
Figure
2.
 Determine
the
magnitude
of
the
fracture
load
P
based
on
a
safety
factor
of
n=1.5
for
crack
length
of
4
mm.
 The
dimensions
are
w
=
50
mm,
d
=
125
mm,
and
t=
25
mm.
The
fracture
toughness
of
the
material
is
KIC
=
23
 MPa√m.
 
 Figure
2
 Solution:
 
 The
normal
stresses
are
due
to
normal
force
and
bending
moment,
hence
 
 The
geometry
factor
for
each
stress
with

a/w
=
0.08
:
 
 
 The
stress
intensity
factor
at
fracture
is
equal
to
the
fracture
toughness,
hence
 
 
 ∴
 
 
 ...
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This note was uploaded on 04/12/2011 for the course AE 304 taught by Professor Lestari during the Fall '10 term at Prescott College.

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