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Pre-Analysis & Start-Up
Pre-Analysis
Total Deformation
The first
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back-of-the
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envelope
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calculation
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that
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we
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will
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make
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is
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for
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the
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total
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deformation
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of
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the
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crank
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under
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the
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specified
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applied
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load.
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A
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list
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of
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different
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cantilevered
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beam
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loading
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cases
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along
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with
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their
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closed-form
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maximum
...
deflections
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formulas
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can
...
be accessed on this website. Because our beam is loaded at the second hole instead of at the tip, our loading is best represented by the case 2 presented (i.e for a cantilevered beam with a concentrated load, P, at any point). The appropriate formula for the maximum deflection is therefore
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accessed on [this website|http://ruina.tam.cornell.edu/Courses/ME4730/Rand4770Vibrations/BeamFormulas.pdf]. Because our beam is loaded at the second hole instead of at the tip, our loading is best represented by the case 2 presented (i.e for a cantilevered beam with a concentrated load, P, at any point). The appropriate formula for the maximum deflection is therefore {latex} \begin{equation*} \delta_{max} = -\frac{Pa^{2}}{6EI}(3L-a) \end{equation*} {latex} |
where
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"P"
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is
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the
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load,
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"a"
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is
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the
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distance
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from
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the
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support
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to
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the
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load,
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"L"
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is
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the
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distance
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from
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the
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support
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to
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the
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end
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of
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the
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beam,
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"E"
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is
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Young's
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Modulus
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and
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"I"
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is
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the
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moment
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of
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inertia.
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Using
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the
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dimensions
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provided
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below,
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we
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can
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determine
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"a"
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to
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be
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6.69
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in
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and
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"L"
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to
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be
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7.674 in.Â
We also know that "P" is 100 lbs and "E" is 10,000 ksi. The tricky part is to determine the moment of inertia. Recall that for a rectangular cross-section of height h and depth b,Â
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in. [!Screen Shot 2014-06-13 at 1.28.46 PM.png|width=450!|^Screen Shot 2014-06-13 at 1.28.46 PM.png] We also know that "P" is 100 lbs and "E" is 10,000 ksi. The tricky part is to determine the moment of inertia. Recall that for a rectangular cross-section of height h and depth b, {latex} \begin{equation*} I = \frac{1}{12}bh^{3} \end{equation*} {latex} |
where
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"h"
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is
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the
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height
...
and
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"b"
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is
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the
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depth.
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In
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this
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case
...
we
...
know
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that
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the
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depth
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is
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0.375
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in
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but
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what
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should
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we
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do
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about
...
this
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varying
...
height?
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Since
...
height
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varies
...
as
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a
...
function
...
of
...
x,
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the
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moment
...
of
...
inertia
...
also
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varies
...
with
...
x.
...
Finding
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the
...
maximum
...
deflection
...
for
...
a
...
varying
...
moment
...
of
...
inertia
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is
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actually
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very
...
complex.
...
The
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goal
...
here
...
is
...
not
...
necessarily
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to
...
get
...
the
...
exact
...
answer
...
but
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to
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get
...
a
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reasonable
...
idea
...
of
...
what
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we
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should
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expect
...
our
...
ANSYS
...
solution
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to
...
be.
...
One
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simplified
...
approach
...
is
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to
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estimate
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a
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reasonable
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average
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beam
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height
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in
...
order
...
to
...
proceed
...
with
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the
...
moment
...
of
...
inertia
...
calculation.
...
From
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the
...
diagram
...
above,
...
the
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maximum
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height
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is
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2*0.984
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=
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1.968
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in.
...
The
...
minimum
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height
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can
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be
...
approximated
...
as
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2*(1.29-0.984)
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=
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0.612
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in
...
if
...
one
...
subtracts
...
the
...
radius
...
of
...
the
...
left
...
circle
...
from
...
the
...
ellipse
...
(the
...
ellipse
...
makes
...
the
...
curvature).
...
From
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the
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maximum
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and
...
minimum
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heights,
...
we
...
find
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that
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the
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average
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beam
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height
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is
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1.29
...
in.
...
But
...
you
...
and
...
I
...
know
...
that
...
greater
...
deflection
...
arises
...
from
...
the
...
thinner
...
part
...
of
...
the
...
crank
...
and
...
so
...
using
...
the
...
average
...
beam
...
height
...
will
...
likely
...
undershoot
...
the
...
actual
...
maximum
...
deflection.
...
So
...
let
...
us
...
take
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a
...
value
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for
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h
...
that
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is
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slightly
...
below
...
the
...
average
...
beam
...
height,
...
say
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1
...
in.
...
We
...
now
...
have
...
all
...
variables
...
needed
...
to
...
find
...
the
...
maximum
...
deflection.
...
Using
...
the
...
equations
...
above,
...
we
...
estimate
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the
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maximum
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deflection
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to
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be
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0.039
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inches.
...
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{latex}$\sigma_x${latex} |
...
...
the
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height
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of
...
the
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cross-section
...
Now
...
we
...
wish
...
to
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do
...
hand-calculations
...
to
...
predict
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the
...
stress
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in
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the
...
x-component
...
across
...
the
...
height
...
of
...
the
...
cross-section
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at
...
the
...
thinnest
...
part
...
of
...
the
...
crank
...
(i.e
...
the
...
middle).
...
Because
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{latex}$\sigma_x${latex} |
...
...
linear
...
along
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the
...
height
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of
...
the
...
cross-section,
...
and
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because
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from
...
symmetry,
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the
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value
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at
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the
...
top
...
of
...
the
...
cross-section
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is
...
the
...
same
...
as
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the
...
value
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at
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the
...
bottom
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of
...
the
...
cross-section,
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one
...
can
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simply
...
calculate
...
the
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{latex}$\sigma_x${latex} |
...
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the
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top
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to
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compare
...
with
...
ANSYS.
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This
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value
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is
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the
...
only
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real
...
unknown
...
because
...
stress
...
at
...
other
...
heights
...
are
...
directly
...
dependent
...
on
...
that
...
value
...
and
...
can
...
be
...
found
...
quite
...
easily
...
by
...
interpolation.
...
So
...
let's
...
calculate
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{latex}$\sigma_x${latex} |
...
...
the
...
top
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of
...
the
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cross
...
section.
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Dividing
...
the
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minimum
...
height
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found
...
above
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by
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two,
...
we
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find
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the
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coordinate
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of
...
this
...
point
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to
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be
...
(3.345",
...
0.306")
...
with
...
the
...
origin
...
at
...
the
...
left
...
hole
...
center.
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{latex} \begin{equation*} M=-P*(a-x)=-(100)*(6.69-3.345) = -334.5 lb*inches \end{equation*} \begin{equation*} I = \frac{1}{12}bh^{3} = \frac{1}{12}(0.375)(0.612)^{3} = 0.00716 in^{4} \end{equation*} \begin{equation*} \sigma_{x} = \frac{M*y}{I} = \frac{-334.5*0.306}{0.00716} = -14295.7 psi \end{equation*} {latex} |
We
...
now
...
have
...
a
...
good
...
idea
...
of
...
what
...
to
...
expect
...
from
...
our
...
ANSYS
...
results.
...
We
...
will
...
come
...
back
...
to
...
these
...
hand-calculation
...
results
...
in
...
the
...
verification
...
and
...
validation
...
step.
...
Let's
...
now
...
solve
...
this
...
problem
...
using
...
ANSYS!
...
Start-Up
The following video shows how to launch ANSYS Workbench and choose the appropriate analysis system (which, under the hood, sets the governing equations that one will be solving). The video also shows how to add a new material to the material list for this project. We'll later assign our material to the model in the Physics Setup step.
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h3. The following video shows how to launch ANSYS Workbench and choose the appropriate analysis system (which, under the hood, sets the governing equations that one will be solving). The video also shows how to add a new material to the material list for this project. We'll later assign our material to the model in the [Physics Setup|Bike Crank - Physics Setup] step. \\ {html} <iframe width="600" height="338" src="//www.youtube.com/embed/GqxfyBzfAeY?rel=0" frameborder="0" allowfullscreen></iframe> {html} \\ [*Go to Step |
...