2001, W. E. Haisler
Chapter 13: Beam Bending
27
Review of Centroids and Moments of Inertia
Consider a composite cross-section consisting of two separate
areas as shown below. We wish to determine the location of the
centroid of the cross-section.
y
y’
A1
(centroid of A 1)
d1
d2
A2
y’-z’ reference axis
A = A 1+A2
y1
z (centroid of A)
(centroid of A 2 )
y
y2
z’
2001, W. E. Haisler
Chapter 13: Beam Bending
28
Relative to some y'-z' reference axes, define the centroid of
the composite area to be y ,and the centroid of each sub-area
to be yi . We require the "first moment of the area" about the
z' axis in terms of the discrete values yA to be equal to the
integral value  y ' dA so that we write:
A
yA   y ' dA
A

n
n
i 1
i 1
y  Ai   yi Ai
Solving for y gives:
n
 yi Ai
ydA

y A
 i 1
A
n
 Ai
i 1
2001, W. E. Haisler
Chapter 13: Beam Bending
29
Parallel Axis Theorem (or Transfer Theorem)
The moment of inertia about the bending axis (y) is given by
I zz   y 2dA. We can also develop the parallel axis
A
theorem (also called the transfer theorem) for moments of
inertia. Suppose that we know the moment of inertia of an
area about its centroid and wish to determine the moment of
inertia about some other parallel axis. Consider the
following sketch of a rectangular area (Note a rectangular
area is shown for simplicity; however, the area can be any
shape.):
2001, W. E. Haisler
Chapter 13: Beam Bending
30
y
y’
z
A1
(centroid of A1 )
d
1
y’-z’ reference axis
z’
Consider area A1. The moment of inertia of this area about
the z’ axis is defined by
I z ' z '   ( y ')2 dA
A
1
The y and y' coordinates are related by the transformation
y '  y  d1. Substituting this into the above gives
2001, W. E. Haisler
Chapter 13: Beam Bending
31
I z ' z '   ( y ') dA   ( y  d1) dA
2
2
A
1
A
1
  y 2dA   2 yd1dA   (d1)2 dA
A
1
A
1
A
1
 The first term on the right is the moment of inertia about
the z-axis passing through the centroid of area A1:
I zz   y 2dA
1
A
1
 The second term can be written as 2d1  ydA since d1 is a
A
1
constant. However, the term
A1 ydA  0 since the y-z axes
is located at the centroid of A1.
2001, W. E. Haisler
Chapter 13: Beam Bending
32
 The last term is simply (d1)2A1. Consequently we can
write
I z ' z '  I zz  A1d12
1
This last result is called the parallel axis theorem or
transfer theorem. It allows one to determine the moment of
inertia about a parallel axes (z’) in terms of moment of
inertia about the centroidal axis (z) and the distance between
z and z’ (d1).
Now consider a composite body made of n sub-areas Ai such
as that shown below:
2001, W. E. Haisler
y
Chapter 13: Beam Bending
The parallel axis theorem for a
single area can be generalized to
obtain the following expression
for determining the moment of
inertia about the centroidal axis
of the composite body:
33
1
d1
centroid
of body
z
d2
2
d3
n
I zz   y 2 dA   ( I zz  Ai di2 )
A
i 1
i
3
I zz  moment of inertia of body about its centroidal axis
I zz  moment of inertia of area i about its centroidal axis
i
di  distance between centroid of area i and centroidal axis of body
Ai  area i
2001, W. E. Haisler
Chapter 13: Beam Bending
34
Example. Determine the centroid and moment of inertia of
the following composite body.
y
all units are 


in
(mm)


80
10
A1
z’
c1
10
y , d1  ?
c
d2  ?
z
c2
30
A2
40
z
2001, W. E. Haisler
Chapter 13: Beam Bending
35
The centroid of the composite body is labeled “c.” The
centroid of area 1 and 2 is labeled c1 and c2, respectively.
By inspection, the horizontal location of the centroid of the
composite body is 20 cm to the right of the left edge of the
lower area. To determine the vertical location, we use
 Ai yi
y
i
 Ai
.
i
Arbitrarily choose a reference z’ axis to be located at the
centroid of the upper area. Then we write:
2001, W. E. Haisler
Chapter 13: Beam Bending
 yi Ai
36
0(80 x 20)  ( 40)(40 x60)
y

 24mm
(80 x 20)  (40 x60)
 Ai
i
i
Hence the vertical position of the centroid for the composite
body is located 24 mm below the centroid of area 1, or 34
mm below the top of the body.
Now, we determine the moment of inertia about the
centroidal axis of the composite body using the parallel axis
theorem. Knowing the location of the centroid, we know
that d1  24 mm, d 2  16 mm. We will determine the
moment of inertia of each area separately and then sum
them.
2001, W. E. Haisler
Chapter 13: Beam Bending
37
For A1
1 bh3  1 (80  203 )  53.3  103 [mm 4 ]
I z(1)' z '  12
12
(1)
I zz
 I z(1)' z '  A1d12  53,300  (80  20)(24) 2  975,000 [mm 4 ]
For A2
1 40(60)3  720,000
I z(2)

'' z '' 12
[mm 4 ]
(2)
2
2
4
I zz
 I z(2)

A
d

720,000

(40

60)(

16)

1,334,000
[mm
]
2 2
'' z ''
For the Composite Area}
(1)
(2)
I zz  I zz
 I zz
 975,000  1,334,000  2.31106
[mm 4 ]