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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
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
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.
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
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:
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
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:
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
Radius of Gyration of an area
Radius of gyration about a point, O
Thus we can
also write:
Same definition for axis y:
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
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:
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:
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
Moment of
Inertia of
common
geometric
shapes
Moment
of Inertia
common
geometric
shapes
Properties of
rolled steel
shapes US
units.
Table 9.13
Properties of
rolled steel
shapes SI
units.
Table 9.13
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.
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
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:
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:
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:
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'
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:
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.
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.
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