Center of Mass and Centroids :: Guidelines
Centroids of Lines, Areas, and Volumes
1.Order of Element Selected for Integration
2.Continuity
3.Discarding Higher Order Terms
4.Choice of Coordinates
5.Centroidal Coordinate of Differential Elements
xdL

x
ydL

y
zdL

z
xc dA

x
yc dA

y
z c dA

z
xc dV

x
yc dV

y
z c dV

z
L
A
V
ME101 - Division III
L
A
V
L
A
V
Kaustubh Dasgupta
1
Center of Mass and Centroids
Theorem of Pappus: Area of Revolution
- method for calculating surface area generated by revolving a plane curve
about a non-intersecting axis in the plane of the curve
Surface Area
Area of the ring element: circumference times dL
dA = 2πy dL
If area is revolved
Total area, A  2  y dL
 yL   y dL  A  2 yL
through an angle θ<2π
θ in radians
or
A   yL
y is the y-coordinate of the centroid C for the line
of length L
Generated area is the same as the lateral area of a
right circular cylinder of length L and radius y
Theorem of Pappus can also be used to determine centroid of plane
curves if area created by revolving these figures @ a non-intersecting
axis is known
ME101 - Division III
Kaustubh Dasgupta
2
Center of Mass and Centroids
Theorem of Pappus: Volume of Revolution
- method for calculating volume generated by revolving an area about a
non-intersecting axis in the plane of the area
Volume
Volume of the ring element: circumference times dA
dV = 2πy dA
Total Volume, V  2  y dA
 yA   y dA 
V  2 yA or
If area is revolved
through an angle θ<2π
θ in radians
V   yA
y is the y-coordinate of the centroid C of the
revolved area A
Generated volume is obtained by multiplying the
generating area by the circumference of the circular
path described by its centroid.
Theorems of Pappus can also be used to determine centroid of plane areas if
volume created by revolving these figures @ a non-intersecting axis is known
ME101 - Division III
Kaustubh Dasgupta
3
Area Moments of Inertia
• Previously considered distributed forces which were proportional
to the area or volume over which they act.
- The resultant was obtained by summing or integrating over
the areas or volumes
- The moment of the resultant about any axis was determined
by computing the first moments of the areas or volumes
about that axis
• Will now consider forces which are proportional to the area or
volume over which they act but also vary linearly with distance
from a given axis.
- the magnitude of the resultant depends on the first moment of
the force distribution with respect to the axis.
- The point of application of the resultant depends on the
second moment of the distribution with respect to the axis.
ME101 - Division III
Kaustubh Dasgupta
4
Area Moments of Inertia

• Consider distributed forces F whose magnitudes are
proportional to the elemental areas A on which they
act and also vary linearly with the distance of A
from a given axis.
• Example: Consider a beam subjected to pure bending.
Internal forces vary linearly with distance from the
neutral axis which passes through the section centroid.

F  kyA
R  k  y dA  0
M  k  y 2 dA
 y dA  Qx  first moment
2
 y dA  second moment
First Moment of the whole section about the x-axis =
y A = 0 since the centroid of the section lies on the x-axis.
Second Moment or the Moment of Inertia of the beam
section about x-axis is denoted by Ix and has units of
(length)4 (never –ve)
ME101 - Division III
Kaustubh Dasgupta
5
Area Moments of Inertia by Integration
• Second moments or moments of
inertia of an area with respect to the x
and y axes,
I x   y 2 dA
I y   x 2 dA
• Evaluation of the integrals is simplified
by choosing dA to be a thin strip
parallel to one of the coordinate axes
ME101 - Division III
Kaustubh Dasgupta
6
Area Moments of Inertia by Integration
• For a rectangular area,
h
I x   y dA   y 2bdy  13 bh 3
2
0
• The formula for rectangular areas may also
be applied to strips parallel to the axes,
dI x  13 y 3dx
ME101 - Division III
Kaustubh Dasgupta
dI y  x 2 dA  x 2 y dx
7
Area Moments of Inertia
Polar Moment of Inertia
J 0  I z   r 2 dA
• The polar moment of inertia is an important
parameter in problems involving torsion of
cylindrical shafts and rotations of slabs
• The polar moment of inertia is related to
the rectangular moments of inertia,


J 0  I z   r 2 dA   x 2  y 2 dA   x 2 dA   y 2 dA
 Iy  Ix
Moment of Inertia of an area is purely a mathematical
property of the area and in itself has no physical significance.
ME101 - Division III
Kaustubh Dasgupta
8
Area Moments of Inertia
Radius of Gyration of an Area
• Consider area A with moment of inertia
Ix. Imagine that the area is
concentrated in a thin strip parallel to
the x axis with equivalent Ix.
I
I x  k x2 A
kx  x
A
kx = radius of gyration with respect
to the x axis
• Similarly,
Iy  k A
ky 
2
y
Iy
A
J O  I z  k A  k A kO  k z 
2
O
2
z
kO2  k z2  k x2  k y2
JO
A
Radius of Gyration, k is a measure of distribution of area from a reference axis
Radius of Gyration is different from centroidal distances
ME101 - Division III
Kaustubh Dasgupta
9
Area Moments of Inertia
Example: Determine the moment of inertia of a triangle with respect to its base.
SOLUTION:
• A differential strip parallel to the x axis is chosen for
dA.
dI x  y 2dA
dA  l dy
• For similar triangles,
l h y

b
h
l b
h y
h
dA  b
h y
dy
h
• Integrating dIx from y = 0 to y = h,


h y
bh 2
I x   y dA   y b
dy   hy  y 3 dy
h
h0
0
2
h
2
h
b  y3 y 4 
 h  
h 3
4
0
ME101 - Division III
Kaustubh Dasgupta
bh3
I x
12
10
Area Moments of Inertia
Example: a) Determine the centroidal polar moment of inertia of a circular area
by direct integration.
b) Using the result of part a, determine the moment of inertia of a circular area with
respect to a diameter.
SOLUTION:
• An annular differential area element is chosen,
dA  2 u du
dJ O  u 2dA
r
r
J O   dJ O   u 2 u du   2  u 3du
2
0
0
JO 

2
r4
• From symmetry, Ix = Iy,
JO  I x  I y  2I x

2
r 4  2I x
I diameter  I x 
ME101 - Division III
Kaustubh Dasgupta

4
11
r4
Area Moments of Inertia
Parallel Axis Theorem
• Consider moment of inertia I of an area A
with respect to the axis AA’
I   y 2 dA
• The axis BB’ passes through the area centroid
and is called a centroidal axis.
Parallel Axis theorem:
MI @ any axis =
MI @ centroidal axis + Ad2
The two axes should be
parallel to each other.
ME101 - Division III
II 
  yy 22dA
dA 
   yy 
 dd 22 dA
dA


22 dA  2d  ydA  d 22  dA


y


 y dA  2d y dA  d dA



• Second term = 0 since centroid lies on BB’
(∫y’dA = ycA, and yc = 0
I  I  Ad 2
Kaustubh Dasgupta
Parallel Axis theorem
12
Area Moments of Inertia
Parallel Axis Theorem
• Moment of inertia IT of a circular area with
respect to a tangent to the circle,
 
I T  I  Ad 2  14  r 4   r 2 r 2
 54  r 4
• Moment of inertia of a triangle with respect to a
centroidal axis,
I AA  I BB  Ad 2
I BB  I AA  Ad
2
1 bh 3
 12
 
2
1
1
 2 bh 3 h
1 bh 3
 36
• The moment of inertia of a composite area A about a given axis is obtained by adding the
moments of inertia of the component areas A1, A2, A3, ... , with respect to the same axis.
ME101 - Division III
Kaustubh Dasgupta
13
Area Moments of Inertia: Standard MIs
ME101 - Division III
Kaustubh Dasgupta
14
Area Moments of Inertia
Example:
Determine the moment of inertia of the shaded area with respect to the x axis.
SOLUTION:
• Compute the moments of inertia of the
bounding rectangle and half-circle with
respect to the x axis.
• The moment of inertia of the shaded area is
obtained by subtracting the moment of
inertia of the half-circle from the moment
of inertia of the rectangle.
ME101 - Division III
Kaustubh Dasgupta
15
Area Moments of Inertia
Example: Solution
SOLUTION:
• Compute the moments of inertia of the bounding
rectangle and half-circle with respect to the x axis.
Rectangle:
I x  13 bh3 
4r 490
a

 38.2 mm
3
3
b  120 - a  81.8 mm
A  12 r 2  12  90
2
 12.72  10 mm
3
2
1
3
2401203  138.2 106 mm 4
Half-circle:
moment of inertia with respect to AA’,
I AA  18 r 4  18  904  25.76  106 mm4
Moment of inertia with respect to x’,

 

I x  I AA  Aa 2  25.76 10 6  12.72 103 38.2
2
 7.20 10 6 mm 4
moment of inertia with respect to x,


I x  I x  Ab 2  7.20 10 6  12.72 103 81.8
2
 92.3 10 6 mm 4
ME101 - Division III
Kaustubh Dasgupta
16
Area Moments of Inertia
Example: Solution
• The moment of inertia of the shaded area is obtained by
subtracting the moment of inertia of the half-circle from
the moment of inertia of the rectangle.
Ix

138.2  106 mm4

92.3  106 mm4
I x  45.9  106 mm4
ME101 - Division III
Kaustubh Dasgupta
17
Area Moments of Inertia
Products of Inertia: for problems involving unsymmetrical cross-sections
and in calculation of MI about rotated axes. It may be +ve, -ve, or zero
• Product of Inertia of area A w.r.t. x-y axes:
I xy   xy dA
x and y are the coordinates of the element of area dA=xy
• When the x axis, the y axis, or both are an
axis of symmetry, the product of inertia is
zero.
• Parallel axis theorem for products of inertia:
I xy  I xy  x yA
- Ixy
+ Ixy
Quadrants
+ Ixy
ME101 - Division III
Kaustubh Dasgupta
- Ixy
18