10.4 Moments of Inertia About Inclined Axes; Principal Moments
10.4 Moments of Inertia About Inclined Axes; Principal Moments Example 1, page 1 of 3
1. Determine the moments of inertia of the standard rolled-steel
angle section with respect to the u and v axes.
y
0.987 in.
0.5 in.
Ix = 17.40 in4
v
Iy = 6.27 in4
Ixy = 6.08 in.4
6 in.
x
C
45°
u
1.99 in.
0.5 in.
4 in.
10.4 Moments of Inertia About Inclined Axes; Principal Moments Example 1, page 2 of 3
1 The formula for Iu is
Iu =
Ix + Iy
I I
+ x y cos 2  Ixy sin 2
2
2
(1)
We can save some work later, if we calculate and save the expressions
Ix + Iy
=
2
17.40 in4 + 6.27 in4
= 11.835 in4
2
Ix  Iy
=
2
17.40 in4  6.27 in4
= 5.565 in4
2
(2)
and
(3)
Eq. 1 becomes,
11.835 in4, by Eq. 2
6.08 in.4
Ix + Iy
I I
Iu =
+ x y cos 2  Ixy sin 2
2
2
45° (is negative because the x axis
must be rotated clockwise to make it
coincide with the u axis)
5.565 in4, by Eq. 3
= 5.76 in4
Ans.
10.4 Moments of Inertia About Inclined Axes; Principal Moments Example 1, page 3 of 3
2
Similarly, for Iv and Iuv,
11.835 in4, by Eq. 2
Iv =
6.08 in.4
45°
Ix + Iy
I I
 x y cos 2  Ixy sin 2
2
2
5.565 in4, by Eq. 3
Ans.
= 17.92 in4
5.565 in4, by Eq. 3
Ix  Iy
sin 2  Ixy cos 2
Iuv =
2
45°
6.08 in.4
= 5.57 in4
Ans.
10.4 Moments of Inertia About Inclined Axes; Principal Moments Example 2, page 1 of 5
2. Determine the moments of inertia of the crosshatched area with
respect to the u and v axes for a) = 25° and b) = 90°
y
40 mm
v
220 mm
u

x
100 mm
100 mm
20 mm
1
Before we can use the equations for Iu, Iv, and Iuv, we must
determine Ix, Iy, and Ixy. Determining Ixy is easy: the y axis is
an axis of symmetry, so
Ixy = 0
(1)
10.4 Moments of Inertia About Inclined Axes; Principal Moments Example 2, page 2 of 5
2 To find Ix and Iy, consider the crosshatched area to be
the sum of two rectangles and calculate Ix and Iy for
each rectangle
y
y
y
y
y
Centroid
20 mm
20 mm
40 mm
Centroid
220 mm
x
220 mm
=
120 mm
120 mm
+
220 mm
220mm/2 = 110 mm
100 mm
100 mm
20 mm
3
Ix-rectangle-1 = Ix' + d2A
=
(40 mm)(220 mm)3
12
= 1.4197 108 mm4
x
x
x
Rectangle 2
Rectangle 1
I = bh3/12 for
rectangle about
centroidal axis
+ (110 mm)2[(40 mm)(220 mm)
(2)
Iy-rectangle-1
(220 mm)(40 mm)3
=
12
= 0.0117 108 mm4
(3)
10.4 Moments of Inertia About Inclined Axes; Principal Moments Example 2, page 3 of 5
4
Ix-rectangle-2 = Ix' + d2A
=
(240 mm)(40 mm)3
12
+ (220 mm + 20 mm)2(240 mm)(40 mm)
= 5.5424 108 mm4
Iy-rectangle-2 =
=
(4)
bh3
12
(40 mm)(240 mm)3
12
= 0.4608 108 mm4
5
(5)
Adding the results for rectangles 1 and 2 gives
0.4608 108 mm4, by Eq. 5
Iy = Iy-rectangle-1 + Iy-rectangle-2
Ix = Ix-rectangle-1 + Ix-rectangle-2
1.4197 108 mm4, by Eq. 2
5.5424 108 mm4, by Eq. 4
= 6.9621 108 mm4
6
(6)
0.0117 108 mm4, by Eq. 3
= 0.4725 108 mm4
Substitute for Ix, Iy, and Ixy in the equation for Iu, Iv, and Iuv.
We can save work if we calculate and save the expressions
Ix + Iy
=
2
and
Ix  Iy
=
2
6.9621 108 mm4 + 0.4725 108 mm4
= 3.7173 108 mm4
(8)
2
6.9621 108 mm4  0.4725 108 mm4
= 3.2448 108
2
mm4
(9)
(7)
10.4 Moments of Inertia About Inclined Axes; Principal Moments Example 2, page 4 of 5
7
Part a): Calculate Iu, Iv, and Iuv for = 25°
25°
0, by Eq. 1
Ix + Iy
Ix  Iy
cos 2  Ixy sin 2
+
2
2
Iu =
3.7173 108 mm4, by Eq. 8
3.2448 108 mm4, by Eq. 9
= 5.80 108 mm4
Iv =
Ix + Iy
2
25°
Ix  Iy
cos 2  Ixy sin 2

2
3.7173 108 mm4, by Eq. 8
= 1.632 108 mm4
Iuv
Ans.
0, by Eq. 1
3.2448 108 mm4, by Eq. 9
Ans.
25°
Ix  Iy
sin 2  Ixy cos 2
=
2
3.2448 108 mm4, by Eq. 9
= 2.49 108 mm4
0, by Eq. 1
Ans.
10.4 Moments of Inertia About Inclined Axes; Principal Moments Example 2, page 5 of 5
8
Part b) : 
When the u and y axes coincide, so no transformation
equations are needed. Instead, we have
y, u
Iu = Iy
0.4725 108 mm4, by Eq. 7
= 0.47 108 mm4
Similarly,
90°
v
Ans.
x
Iv = Ix
6.9621 108 mm4, by Eq. 6
= 6.96 108 mm4
Ans.
and
Iuv = Ixy
= 0, by Eq. 1
Ans.
10.4 Moments of Inertia About Inclined Axes; Principal Moments Example 3, page 1 of 4
3. Determine the value of for which the product of inertia of the
crosshatched area with respect to the u and v axes is zero. Calculate
Iu and Iv for this value of and compare Iu and Iv to Imax and Imin
v
y
x2 + y2 = 1002
Ix = 1.9635 x 107 mm4
u
Iy = 1.9635 x 107 mm4

x
Ixy = 1.2500 x 107 mm4
10.4 Moments of Inertia About Inclined Axes; Principal Moments Example 3, page 2 of 4
1 The formula for Iuv is
Iuv =
Ix  Iy
sin 2  Ixy cos 2
2
(1)
We can save some work later, if we calculate and save
Ix  Iy
=
2
1.9635 x 107  1.9635 x 107
= 0
2
(2)
Setting Iuv equal to zero in Eq. 1 gives,
0, by Eq. 2
0 =
Ix+ Iy
sin 2  Ixy cos 2
2
1.2500 x 107 mm4
Solving this equation for  leads to two roots:
= 45°, 45°
Ans.
10.4 Moments of Inertia About Inclined Axes; Principal Moments Example 3, page 3 of 4
2
The formula for Iu is
Ix  Iy
Ix + Iy
cos 2  Ixy sin 2
+
2
2
Iu =
(3)
We can save some work by calculating and saving
Ix + Iy
1.9635  107  1.9635  107 1.9635  107
=
=
2
2
Substituting numerical values in Eq. 3 gives
1.9635  107 mm4
Iu =
1.2500  107 mm4
Ix + Iy
I I
+ x y cos 2  Ixy sin 2
2
2
45°
0, by Eq. 2
= 0.7135  107 mm4
Ans.
Similarly for Iv,
1.9635  107 mm4
Iv =
1.2500  107 mm4
Ix + Iy
I I
 x y cos 2  Ixy sin 2
2
2
45°
0, by Eq. 2
= 3.2135  107 mm4
Ans.
(4)
10.4 Moments of Inertia About Inclined Axes; Principal Moments Example 3, page 4 of 4
Calculate Imax and Imin
3
1.9635  107 mm4
1.2500  107 mm4
 Ix  Iy )2  Ixy2
2
0, by Eq. 2
Imax,min = Ix + Iy 
2
= 3.2135  107 mm4, 0.7135  107 mm4
same as Iv
Same as Iu
Thus Iv = Imax and Iu = Imin. In general if the product of inertia, Iuv, is zero for a
given orientation, then the moments of inertia, Iu and Iv, are the maximum and
minimum moments of inertia possible for any orientation.
4
Show the u and v axes' orientation.
y
v
(Axis of maximum
moment of inertia)
x2 + y2 = 1002
u (Axis of minimum moment of inertia)
45°
x
10.4 Moments of Inertia About Inclined Axes; Principal Moments Example 4, page 1 of 2
4. Determine the principal moments of inertia with respect to all
possible rectangular coordinate systems with their origin at the
centroid C.
y
20 mm
50 mm
C
Ix = 2.2013 x 107 mm4
Iy = 0.9213 x 107 mm4
x
50 mm
20 mm
60 mm
60 mm
10 mm
1
Because the x and y axes are axes of symmetry for the crosshatched area
Ixy = 0
(1)
10.4 Moments of Inertia About Inclined Axes; Principal Moments Example 4, page 2 of 2
2 Apply the equation for the principal moments of inertia.
2.2013  107 mm4
Imax,min =
Ix + Iy
2

0, by Eq. 1
(Ix  Iy)/22 + Ixy2
0.9213  107 mm4
= 2.2013  107 mm4, 0.9213  107 mm4
Ans.
That is, Ix and Iy are principal moments of inertia. In general, you can
immediately recognize if Ix and Iy are principal moments of inertia by noting if
the product of inertia is zero: Ixy = 0 implies that Ix and Iy are principal moments
of inertia.
10.4 Moments of Inertia About Inclined Axes; Principal Moments Example 5, page 1 of 3
5. Determine the principal moments of inertia and
principal axes having their origin at point O.
y
150 mm
15 mm
Ix = 1.1714  107 mm4
30 mm
Iy = 8.9083  107 mm4
Ixy = 2.3971  107 mm4
O
x
10.4 Moments of Inertia About Inclined Axes; Principal Moments Example 5, page 2 of 3
1
Apply the formula for the principal moments of inertia,
1.1714  107 mm4
Imax,min =
2.3971  107 mm4
Ix + Iy 
 Ix  Iy )2  Ixy2

2
2
8.9083  107 mm4
= 9.5908  107 mm4, 4.8892  107 mm4
2
(1)
Apply the formula for the principal directions
2.3971  107 mm4
-1 2Ixy
p = (1/2) tan Ix  I
y
1.1714 107 mm4
8.9083  107 mm4
= 15.8923° and 74.1077°
To determine which p value corresponds to Imax and which to Imin,
substitute 15.8923° into the transformation equation for Ix':
Iu =
1.1714  107 mm4
Ix + Iy
I I
+ x y cos 2  Ixy sin 2
2
2
2.3971  107 mm4
8.9083  107 mm4
= 4.8892  107 mm4
= Imin, by Eq. 1
Ans.
10.4 Moments of Inertia About Inclined Axes; Principal Moments Example 5, page 3 of 3
3
Thus a 15.9° counterclockwise rotation of the
x axis would give the axis for which the
quantity
Iu =  v2 dA
is smaller than the same integral evaluated for
any other orientation of the axis.
y
Axis of minimum
moment of inertia
15.9°
x
O
74.1°
Axis of maximum
moment of inertia
10.4 Moments of Inertia About Inclined Axes; Principal Moments Example 6, page 1 of 4
6. Determine the principal moments of inertia
and principal axes having their origin at point O.
y
6 in.
x
O
4 in.
10.4 Moments of Inertia About Inclined Axes; Principal Moments Example 6, page 2 of 4
y
1
y
Calculate the moment of inertia.
Ix = Ix' + d2A
2 in.
(parallel axis theorem)
(4 in.)(6 in.)3
=
+ (3 in.)2(24 in.2)
12
= 288 in.4
Iy =
x
6 in.
(1)
(6 in.)(4 in.)3
+ (2 in.)2 (24 in.2)
12
= 128 in.4
(2)
3 in.
O
x
4 in.
Ixy = Ix'y' + dxdy A
2
= 0 + (2 in.)(3 in.)(24 in. )
= 144 in.4
Area: A = (4 in.)(6 in.) = 24 in2
(3)
10.4 Moments of Inertia About Inclined Axes; Principal Moments Example 6, page 3 of 4
2
Apply the formulas for the principal moments of inertia.
288 in.4, by Eq. 1
Imax, min=
Ix + Iy
(Ix  Iy)/22 Ixy2

2
128 in.4, by Eq. 2
4
4
 372.73 in. , 43.27 in.
3
144 in.4, By Eq. 3
(4)
Ans.
Apply the formulas for the principal directions
p = 12 tan-1 2Ixy
Ix  Iy
144 in.4, By Eq. 3
288 in.4, by Eq. 1
128 in.4, by Eq. 2
= 30.47° and 59.53°
To determine which p value corresponds to max and which to min,
substitute 30.47° into the transformation equation for Iu:
288 in.4, by Eq. 1
Iu =
Ix + Iy
I I
+ x y cos 2  Ixy sin 2
2
2
4
128 in. , by Eq. 2
144 in.4, By Eq. 3
= 372.73 in.4
= Imax, by Eq. 4
10.4 Moments of Inertia About Inclined Axes; Principal Moments Example 6, page 4 of 4
4 Thus a 30.47° clockwise rotation of the x axis
would give the axis for which the quantity
Iu =  v2 dA
is larger than the same integral evaluated for any
other orientation of the axis.
y
Axis of minimum
moment of inertia
59.53°
x
O
30.47°
Axis of maximum
moment of inertia
10.4 Moments of Inertia About Inclined Axes; Principal Moments Example 7, page 1 of 3
7. Use Mohr's circle to determine the principal moments of
inertia and principal axes having their origin at the centroid C
of the standard rolled-steel channel section.
y
15.3 mm
Ix = 32.6  106 mm4
Iy = 1.14  106 mm4
127 mm
C
x
1
127 mm
Because the x axis is an
axis of symmetry, Ixy = 0
10.4 Moments of Inertia About Inclined Axes; Principal Moments Example 7, page 2 of 3
2 Draw the I and Ixy axes.
Ixy
3 Plot the point corresponding to the x axis:
(Ix, Ixy) = (32.6  106, 0).
R
X
I
C
4
Plot the point C at the center of Mohr's circle:
(Iaverage, 0) = ((Ix + Iy)/2, 0)
5
Calculate the radius:
= ((32.6  106 + 1.14 x 106)/2, 0)
R = 32.6 x 106  16.87 x 106
= (16.87  106, 0)
= 15.73  106
10.4 Moments of Inertia About Inclined Axes; Principal Moments Example 7, page 3 of 3
6 Draw Mohr's circle.
R = 15.73 106
Ixy
7
Calculate Imax.
Imax = 16.87  106 + 15.73  106
Imin
C
Imax
= 32.6  106 mm4
I
8
Calculate Imin.
Imin = 16.87  106  15.73  106
= 1.14  106 mm4
16.87 106
9
Principal axes: Because Ix is the same
as Imax and Iy is the same as Imin, the x
and y axes are the principal axes.
Ans.
10.4 Moments of Inertia About Inclined Axes; Principal Moments Example 8, page 1 of 4
8. Use Mohr's circle to determine the principal moments of
inertia and principal axes having their origin at the centroid C
of the standard rolled-steel angle section.
y
0.987 in.
0.5 in.
Ix = 17.40 in4
Iy = 6.27 in4
Ixy = 6.08 in.4
6 in.
x
C
1.99 in.
0.5 in.
4 in.
10.4 Moments of Inertia About Inclined Axes; Principal Moments Example 8, page 2 of 4
1 Draw the I and Ixy axes.
Ixy
4
Calculate the radius.
17.40
R=
= 8.242
I
C
R
6.08
2
X
11.835
17.40 11.835 = 5.565
3 Plot the point C at the center of Mohr's circle:
(Iaverage, 0) = ((17.40 + 6.27)/2 , 0)
= (11.835, 0)
(6.08)2 + (5.565)2
Plot the point corresponding to the x axis:
(Ix, Ixy) = (17.40, 6.08).
10.4 Moments of Inertia About Inclined Axes; Principal Moments Example 8, page 3 of 4
5
Draw Mohr's circle.
6 Calculate Imax.
Ixy
Imax = 11.835 + 8.242
R = 8.242
= 20.1 in.4
C
R
2
I
R
R
6.08
X
11.835
Ans.
5.565
8 Determine the orientation of the Imax axis.
= (1/2) tan-1( 6.08 )
5.665
= 23.8° (Counterclockwise rotation of the x
axis gives the axis of Imax.)
7
Calculate Imin.
Imin = 11.835  8.242
= 3.59 in.4
Ans.
10.4 Moments of Inertia About Inclined Axes; Principal Moments Example 8, page 4 of 4
9
Sketch the principal axes.
y
Axis of minimum
moment of inertia
0.987 in.
Axis of maximum
moment of inertia
23.8°
x
C
1.99 in.
10.4 Moments of Inertia About Inclined Axes; Principal Moments Example 9, page 1 of 4
9. Use Mohr's circle to determine the principal moments of inertia
and principal axes having their origin at the centroid C.
105 mm
y
15 m
Ix = 6.7245  106 mm4
52.5 mm
Iy = 6.3520  106 mm4
80 mm
C
x
15 mm
7.5 mm
80 mm
15 mm
Ixy = 5.1300  106 mm4
10.4 Moments of Inertia About Inclined Axes; Principal Moments Example 9, page 2 of 4
1 Draw the I and Ixy axes.
2
Ixy
Plot the point corresponding to the x axis:
(Ix, Ixy) = (6.7245  106, 5.1300  106).
6.7245 106
X
R
5.1300 106
C
I
6.5382 106
6.7245 106 6.5382 106
= 0.1863 106
3 Plot the point C at the center of Mohr's circle:
(Iaverage, 0) = ((Ix + Iy)/2, 0)
=
(6.7245  106 + 6.3520  106 , 0)
2
= (6.5382  106, 0)
4
Calculate the radius.
R=
(5.1300  106)2 + (0.1863  106)2
= 5.1334  106
10.4 Moments of Inertia About Inclined Axes; Principal Moments Example 9, page 3 of 4
5
Draw Mohr's circle.
6 Calculate Imax.
Ixy
Imax = 6.5382  106 + 5.1334  106
X
5.1300 106
= 11.67  106 mm4
Ans.
R
R
C
2
R
I
R = 5.1334 106
6.5382 106
6
= (1/2) tan-1 5.1300  106
0.1863  10
0.1863 106
7
= 44.0° (Clockwise rotation of the
x axis gives the axis of Imax.)
Calculate Imin.
Imin = 6.5382  106  5.1334  106
= 1.40  106 mm4
8 Determine the orientation of the Imax axis.
Ans.
10.4 Moments of Inertia About Inclined Axes; Principal Moments Example 9, page 4 of 4
9
Sketch the principal axes.
y
Axis of minimum moment of inertia
x
C
44.0°
Axis of maximum moment of inertia
10.4 Moments of Inertia About Inclined Axes; Principal Moments Example 10, page 1 of 5
10. Use Mohr's circle to determine the principal moments of
inertia and principal axes having their origin at point O
y
2 in.
2 in.
2 in.
2 in.
2 in.
2 in.
O
x
10.4 Moments of Inertia About Inclined Axes; Principal Moments Example 10, page 2 of 5
1 Consider the crosshatched area to be the difference
between a large square and a small square.
y
y
y
2 in.
2 in.
2 in.
y
3 in.
y
2 in.
2 in.
3 in.

=
x
6 in.
3 in.
2 in.
2 in.
x
x
O
Square 1
2 Calculate the moment of inertia of square 1 about the x
axis (the moment about the y axis will be the same, by
symmetry):
(6 in.)(6 in.)
+ (3 in.)2[(6 in.)(6 in.)]
12
(1)
3 in.
x
Square 2
3 Ixy-square-1 = Ix'y' + dxdy A
= 324 in4
3
= 432 in4
O
= 0 + (3 in.)(3 in.)(6 in.)(6 in.)
Ix-square-1 = Ix'-square-1 + d2A
=
x
2 in.
(2)
10.4 Moments of Inertia About Inclined Axes; Principal Moments Example 10, page 3 of 5
y
4 Calculate the moment of inertia of square 2.
3 in.
Ix-square-2 = Ix'-square-2 + d2A
=
(2 in.)(2 in.)3
+ (3 in.)2(2 in.)(2 in.)
12
= 37.333 in.4
(3)
2 in.
O
= 0 + (3 in.)(3 in.)[(2 in.)(2 in.)]
= 36 in4
(4)
5 Calculate the moment of inertia for the composite square.
= 394.667 in.4
6
37.333 in.4, by Eq. 3
Square 2
Ixy = Ixy-square-1  Ixy-square-2
= 288 in.4
(5)
(6)
3 in.
x
324 in.4, by Eq. 2
Iy = Ix, by symmetry
= 394.667 in.4
x
2 in.
Ixy-square-2 = Ix'y' + dxdy A
432 in.4, by Eq. 1
Ix = Ix-square-1  Ix-square-2
y
36 in.4, by Eq. 4
(7)
10.4 Moments of Inertia About Inclined Axes; Principal Moments Example 10, page 4 of 5
7
Draw the I and Ixy axes
Ixy
8
394.667
(Ix, Ixy) = (394.667, 288)
X
R
288
C
I
9
10 Calculate the radius
R = 288
Plot the point corresponding to the x axis:
Plot the point C at the center of Mohr's circle:
(Iaverage, 0) = ((Ix + Iy)/2, 0)
= ((394.667 + 394.667)/2, 0)
= (394.667, 0)
10.4 Moments of Inertia About Inclined Axes; Principal Moments Example 10, page 5 of 5
R = 288
11 Calculate Imax and Imin
Ixy
Imax = 394.667 + 288
X
Ans.
= 683 in.4
Imin = 394.667  288
C
Imin
Imax
Ans.
= 106.7 in.4
I
12 Clockwise rotation of the x axis 90°/2 = 45° gives the
axis of Imax.
y
394.667
Ixy
Axis of
minimum
moment of
inertia
X
90°
Imin
C
Imax
I
O
x
45°
Axis of maximum moment of inertia