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TRANSCRIPT
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Solid Mechanics - Statics
Chapter 9 Distributed Forces: Moments of
Inertia
Vector Mechanics for Engineers
F.P. Beer, E.R. Johnston, 6th edition
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Second Moment Moment of Inertia
When a beam is subjected to a loading of a pure moment,
internal tension and compression stresses are formed on the
cross-section.
The distributed forces vary
linearly from the neutral axis
and expressed: F = k y A
The resultant of the distributed
forces is:
At the neutral axis the force equals
zero, above and below the neutral
axis the forces have opposite signs
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Second Moment Moment of InertiaThe resultant of the distributed forces is:
The integral above is the first moment Qx of the cross-
section about the neutral axis. The integral over the whole
section is zero (compression = tension).
The moment of a finite F is:
M = yF = k y2A
The total moment of the forces over the
whole section:
The above integral is defined as the second moment or the
moment of inertia about the x-axis. It is denoted Ix and is
the product of y2 by the area, integrated for the entire
section. The moment of inertia is always positive.
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Second Moment Moment of Inertia
We can define the moment of inertia about axis x or axis y
the following way:
These are also called: the rectangular moments of inertia.
If we choose a finite area A = (dx)(dy), the differential
moment of inertia would be:
About x-axis: dIx = y2dA , and
about y-axis: dIy = x2dA
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Second Moment Moment of Inertia
Determine the moment of inertia
about axis-y, using a strip:
dA = (a x) dy
dIx = y2(a x) dy
Determine the moment of inertia
about axis-x, using a strip:
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Determine the moment of inertia
of a rectangle, about the x-axis:
A general concept:
For the strip in the left diagram,
assign: b = dx, and h = y. Thus,
the differential second moment
about the x-axis becomes:
Example: moment of inertia rectangle
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Polar Moment of Inertia
Definition: the polar moment of
inertia is about a point (not about
an axis):
Point O is the pole
From trigonometry we can write:
thus:
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The moment of inertia of area A with
respect to axis x is Ix. If area A can be
concentrated into a thin strip parallel
to axis x, at a distance k from axis x,
and have the same moment of inertia:
Solve for kx:
kx is the radius of gyration
with respect to axis x.
Radius of Gyration of an area
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Radius of Gyration of an area
Radius of gyration about a point, O
Thus we can
also write:
Same definition for axis y:
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Radius of Gyration example
The radius of gyration of the
rectangle with respect to the base is
Note: the centroid is located at y = h/2
while the radius of gyration is located
at k = h /3 = h/1.732 . In other words,
y depends on the first moment, while k
depends on the second moment
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Parallel-axis theoremMoment of inertia with respect
to axis a-a' is defined:
define: y = d + y', where y' is the
distance of dA from the centroid.
d distance of centroid from AA'
The first term is the moment of inertia about BB' , the
second term equals zero, thus,
the Parallel-axis Theorem:
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Parallel-axis theorem The theorem can be extended to the radius of gyration:
Substituting k2A for I (moment of inertia about axis AA',
and k2A for I (moment of inertia about axis BB', the
theorem is expressed:
The theorem can be extended to the moment of
inertia about a point:
or:
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Parallel-axis theorem - examples
Example 2 determine the
moment of inertia of a
triangle about axis BB', and
about axis DD'
Example 1 determine the moment
of inertia of a circle about line T
Ix = r4/4 about the diameter
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Moment of
Inertia of
common
geometric
shapes
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Moment
of Inertia
common
geometric
shapes
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Properties of
rolled steel
shapes US
units.
Table 9.13
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Properties of
rolled steel
shapes SI
units.
Table 9.13
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Product of Inertia
Definition: the following integral is
defined as the product of inertia
about axes x and y:
The product of inertia can be positive,
negative or zero.
When one or both of the x and y axes are
axes of symmetry, the product of inertia
is zero, as seen in the channel example.
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x'-y' is the centroid axes system,
thus the relations: x = x'+x and
y = y'+y.
by definition:
The first term is the product with respect to the centroid
axes. The second and third terms equal zero (first moment
about centroid axes), thus we get:
Parallel axis theorem for Product of Inertia
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Principal axes and principal
moments of InertiaConsider an area A with known moments and product
of inertia
We want to determine the
moments of inertia with
respect to new axes x'-y'
which are rotated by an
angle of with respect to
the original axes.
We can write:
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Principal axes and principal moments of Inertia
Moment of inertia with respect to the new axes:
Plugging in the values for the original moments of inertia:
Similarly we obtain:
trigonometry:
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Principal axes and principal moments of Inertia
Now we can write:
Also, we can write:
This result could be expected from the relation
of the polar moment of inertia:
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Principal axes and principal moments of Inertia
It can be shown that:
define:
The following
are parametric
circle equations
we can write:
Points A and B represent max.
and min. values for Ix'
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Principal axes and principal moments of Inertia
Imin and Imax are 90 apart and
for them we define: principal
axes about point O
Also, at points A and B the
value of the product moment
of inertia Ix'y' is zero, and thus:
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Principal axes and principal moments of Inertia
Summary:
For Ix'y' = 0 we get points A and B which correspond to
Imin and Imax which are 90 apart and are defined principal
moments of inertia, and are related to the principal axes
about point O
When Ix'y' is zero we obtain:
We get two values 2m which
are 180 apart, and thus two m which are 90 apart.
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Mohr circle for the moment of inertiaMohrs circle (Otto Mohr, 1835-1918) is used to illustrate
the relations between the moment and the product of
inertia of a given area.
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Mohr circle for the moment of inertiaIf the moment and the product of an area are known for a
set of rectangular axis, then with Mohr circle we can:
A. Principal axes and
principal moments
B. Moment and product
about any other axes