Contributed by:
Polar moment of inertia, Radius of gyration, Parallel axis theorem, Mass moment of inertia
1.
Distributed Forces: Moments of Inertia
2.
The second moment of area about an axis
is defined as the moment of inertia of area
and is given by
2 2
I x y dA or I y x dA
Note that elemental area of A must be
chosen such that it is parallel to the axis
about which i is being -----
3.
Moments of inertia are important when dealing
with distributed forces that are proportional to
both the area over which the forces are
distributed as well as the distance of that area
from a particular point or axis.
A typical example is beam that is subjected to
pure bending. For such a beam, bending
stresses are linearly distributed across the entire
x-section, such that zero stress exists at the
neutral axis of the x-section and maximum
tension and compression stresses exist at the
outermost force of the section.
4.
Hence for a rectangular x-section
From (c), we observe that y ==> = ky;
but = F/A ==> F = A = KyA
On the elemental area, dA, we have that the
force dF is given by dF = kydA
Hence the resultant F of all the entire elementary
forces dF is obtained by integrated over the
entire x-section, i.e.
F dF kydA k ydA kAy
5.
Because the x-y axes pass through the
centroid of x-section area, y 0 ,
hence F must be equal to zero.
Hence the system of forces dF reduce to a
couple, called the bending moment. Note
that: F dF k ydA
shows that the magnitude of the resultant
force F depends on the first moment of
area . ydA
6.
The magnitude of the bending moment is
obtained by summing up the bending moments
of all the elemental areas along the axis of
interest. Hence,
2
Where, y dA is the second moment of area,
i.e. the moment of inertia about the x-axis.
** To calculate the moment of inertia about a
particular axis, selected a strip of elemental area
parallel to that axis so that all the points forming
that strip are at the same distance from the axis.
7.
Polar Moment of Inertia
Of great importance in problems concerning the
torsion of cylindrical shafts / bars and in
problems dealing with rotation of slabs
For torsion of cylindrical shafts, the shear stress
is proportional to both the area of the x-section
and the distance from longitudinal axis of the
shaft / bar
By definition, the polar moment of inertia Ip is
given by the integral
where r is distance of an elemental area dA from
the longitudinal axis of the shaft.
8.
For a circular x-sectional area of radius r,
determine the moments of inertia Ix and Iy as
well as the polar moment of inertia Ip about the
center O of the circle
To determine the polar moment inertia for the
circular area, use an elemental circular strip as
shown,
9.
Radius of Gyration
The radius of gyration of an area with respect
to an axis is the distance from the axis to a
point where the area is assumed to be
concentrated,
i.e. the moment of inertia of the concentrated
area about the axis is the same as the moment
of inertia of the area about the axis
10.
Radius of Gyration(2)
Consider an arbitrary area A with moment of
inertia Ix about the x-axis. The radius of
gyration about the x-axis, Kx must be such that
11.
Radius of Gyration(3)
Similarly, and
Recalling the relationship , we
see that ==>
12.
Parallel Axis Theorem
You can transfer the moment of Inertia
from one axis to another axis provided that
borh are parallel and one passes through the
centroid of the area.
The Parallel Axis Theorem is used to
compute the moment of inertia of an object
about any axis using the known value of the
moment of inertia about a parallel
centroidal axis.
13.
y y d1 , y d1 y yc dA
y
I x y 2 dA (d1 y) 2 dA d2
2 x x1
I x (d1 2d1 y y2 )dA c xc
2
I x d1 dA 2d1 ydA y2 dA y1
I x d1
2
dA 2 d y dA 2 d1
1
dA
y y
where ydA Is first moment of area about dA x
the centroidal axis
Is moment of inertia
& where y2 dA w.r.t centroidal axis Similarly
2
d1 A 2d1 0 y2 dA
I y x 2 dA ( x d 2 ) 2 dA
2
I x I x c d1 A 2
I y ( x2 2 xd 2 d 2 )dA
2
I y x2 dA 2d 2 xdA d 2 dA
2
I y I yc 2d 2 A x d 2 A, where x = 0
2
I y I yc d 2 A
14.
Moments of Inertia of Composite
Areas
1. Divide/Split the composite areas into regular geometric shapes whose
moments of inertia are known/provided or can easily be determined
2. Determine the moments of inertia of each constutuent part about its
centoidal axis (which must be parallel to the axis of interest) and use the
parallel-axis theorem to compute the moment of inertia about the axis of
interest.
3. The momeent of intertia of the composite area is the algebraic sum of
the constituent parts/areas talking care to treat holes as negative
quantities
4. Tabulating your calculationsmay be helpful in keeping track of things:
eg. Given an area, calculate moments of inertia I x and Iy Jo and Ixy. We
know that the following expressions are valid:
2 2
I x I x d y A; I y I y d x A; I xy I xy d x d y A; J o J o d 2 A
15.
Mass Moment of Inertia
In dynamics, we know that accelerations of an object as a
result of a force F acting on it is given by a = F/m ; where m is
the mass of the body. Hence acceleratoin of a body is
dependant on its mass.
The angular acceleration or rotational acceleration of an object
about an axis is dependant on a quantity called the Mass
Moment of Inertia, which by definition is the second
moment of mass about an axis, ie
2 Observe that the unit for mass
I o r dm moment of inertia is kg•m2
m
Where r is the perpendicular distance from the axis to the
elemental mass dm and Io is the mass moment of inertia about
the z-axis.
16.
Mass Moment of Inertia of a Slender Bar
Given: A straight slender prismatic bar of mass m, length L and
cross-sectional area A and density
Req’d: Mass moment of inertia about a perpendicular axis through
its center of mass.
solution
Determine the location of the center or mass.
x
dm
x dx L
xdm
x but dm dV Adx CONTINUED
m
1
m dm Adx dL
0
17.
2 L
L
x
xdm x( Adx) A xdx 0
2 oL
x
m AL AL L 2
dm=dV= Adr
o r dr
L
L L
2 2 r3 2
2
2 A L3 L3
I 0 r dm r Adr A r dr A
L 3
L 3 8 8
2
2
AL3 AL( L2 ) 1 2 1 2 m Where Ix is the moment of
( V ) L mL I x inertia of area of unit about
12 12 12 12 L axis perpendicular to length
Hence the mass moment of inertia of a slender bar/rod about an axis through its
1 2
center of mass is given by Io= 12 mL
18.
Mass Moments of Inertia of Thin Plates
Consider a homogeneous flat plate of density , mass m, and uniform thickness T
which is shown below:
y By definition the mas moment of inertia
about the z-axis I Z axis r 2 dm
For the elemental area dA, its volume
dV=TdA and its mass dm = dV= TdA
hence I z axis r 2 TdA T r 2 dA
y r m
x I z axis TJ 0 since T constant
A
x
observe that :
I x axis y 2 dm y 2 TdA T y 2 dA TI x
m
m
Also I y axis x 2 dm T x 2 dA Iy
m A
I z axix I x axix I y axix
19.
Hence, the mass moment of inertia of a thin homogenious
plate of uniform thickness can be expressed in terms of the
moments of inertia of the crossectional area up the plate; or in
terms of the radii of gyration as shown below:
m 2 Note these Relationships:
I x axis J o mk p
A If you know the moment of
inertia of the x-sectional
m 2 area of a plate, you can
I y axis I x mk x
A obtain its mass moment of
inertia, provied that the
m 2 plate is thin, has uniform
I z axis I y mk y thickness and is made from
A homogenous material.
20.
Parallel Axis Theorem for Mass Moment
of Inertia
This theorem is used to computer the mass moment of inertia
of an object bout any asxis which is parallel to an axis that
passes through the centre of mass of the object. The theorem
is of the same form as that for moment of inertia of area:
Io = Ic + d2m
Where Ic is the mas moment of inertia about an axis that
passes he centre of mass, m is the mass of the body and di is
the distance between the centroidal axis and a parallel axis
through point O
+1-315-636-4485
+91 98151-74050 
