10.3 Products of Inertia
10.3 Products of Inertia Example 1, page 1 of 2
1. Determine the product of inertia of the narrow
strip with respect to the x and y axes.
y
x
900 mm
30°
3 mm
y
900 mm
x
1 The product of inertia is defined as
Ixy = xy dA
(1)
Here dA can be taken as an infinitesimal rectangle
with sides ds and 3 mm.
ds
3 mm
10.3 Products of Inertia Example 1, page 2 of 2
dx
2
From geometry
30°
dy
ds = dx/cos 30°
ds
y
dA = (3 mm) ds
= (3/cos 30°) dx
900 mm
y = x tan 30° (equation of the line)
x
30°
Thus Eq 1 becomes
Ixy =
30°
xy dA
(Eq. 1 repeated)
y
(900 mm) cos 30°
ds
900 cos 30°
=
x( x tan 30°)[(3/cos 30°) dx]
900 cos 30°
= 6.31
108 mm4
Ans.
(900 mm) cos 30°
x
10.3 Products of Inertia Example 2, page 1 of 2
2. Determine the product of inertia of the crosshatched
area with respect to the x and y axes.
y
x2 + y2 = 1002
x
100 mm
1
The product of inertia ,
Ixy = xy dA
can be evaluated using double integration. Instead, we will use an
alternative approach based on the equation
Ixy = dIxy
(1)
where dIxy is the product of inertia of an infinitesimal strip.
10.3 Products of Inertia Example 2, page 2 of 2
2 Arbitrarily choosing vertical, rather than horizontal,
strips and applying the parallel-axis theorem to the
strip gives
dIxy = dIx'y' + xelyel dA
y
(2)
where dIx'y' is the product of inertia of the
differential strip about the x y axes (centroidal axes
for the strip).
Because the x axis is an axis of symmetry for the
strip, dIx'y' = 0 and Eq. 2 reduces to
y
x'
dA = y
yel = y/2
x
100 mm
(3)
Because the variable of integration is x, we have to
use the equation of the bounding curve, x2 + y2 =
1002, to obtain y as a function of x :
3 Thus,
Ixy = dIxy
x2
=
Using this expression in Eq. 3 gives, after
simplifying,
100
x(1002 x2)/2 dx
0
= 1.25
2
dIxy = x(100
dx
xel = x
= x(y/2)(y dx)
1002
Centroid of the
infinitesimal
strip
x2 + y2 = 1002
dIxy = xelyel dA
y=
y'
2
x )/2 dx
107 mm4
Ans.
10.3 Products of Inertia Example 3, page 1 of 2
3. Determine the product of inertia of the crosshatched
area with respect to the x and y axes.
y
x = 2y6 50y5
y3 + 100
1.156 m
Scales on the x
and y axes are
not the same.
x
100 m
1
The product of inertia,
Ixy = xy dA
can be evaluated using double integration. An
alternative approach, that will be used here, is to
evaluate
Ixy = dIxy
where dIxy is the product of inertia of an
infinitesimal strip.
(1)
10.3 Products of Inertia Example 3, page 2 of 2
2
Choosing horizontal strips and applying the
parallel-axis theorem to the strip gives
dIxy = dIx'y' + xelyel dA
y
Centroid of the
infinitesimal
strip
y
(2)
where dIx'y' is the product of inertia for the
differential strip about the x y axes.
Because the x axis is an axis of symmetry for the
strip, dIx'y' = 0 and Eq.2 reduces to
dA = x
dy
dy
x
1.156 m
yel = y
dIxy = xelyel dA
x
= (x/2)(y)(x dy)
= x2y dy/2
3
xel = x/2
Replacing x in Eq. 3 by the expression for the
bounding curve gives
dIxy = x2y dy/2
x
(3)
(Eq. 3 repeated)
(2y6 50y5
y3 + 100)
4
Thus the product of inertia is
Ixy = dIxy
1.156
= (2y6 50y5
0
= 1973 m4
y3 + 100)2y dy/2
Ans.
10.3 Products of Inertia Example 4, page 1 of 3
4. Determine the product of inertia of the crosshatched area
with respect to the x and y axes, if a = 3 m and b = 1 m.
y
a
y = a sin(
x
b
x)
2b
10.3 Products of Inertia Example 4, page 2 of 3
y
1
Evaluate the integral
x
Ixy = dIxy
(1)
y'
xel = x/2
dy
dA = x
where dIxy is the product of inertia of a horizontal
infinitesimal strip.
dy
2 Applying the parallel axis theorem to the strip gives
dIxy = dIx'y' + xelyel dA
x'
0, because y is an axis of symmetry of the strip
Centroid of the
infinitesimal
strip
x/2
yel = y
x dy
= xelyel dA
y
= x2y dy/2
x
(2)
The x2 in this expression can be replaced by a function of y by
solving for x from the equation of the bounding curve:
y = a sin(
x)
2b
10.3 Products of Inertia Example 4, page 3 of 3
3 Thus
y
x = (2b/ ) sin-1(y/a)
Eq. 2 becomes
dIxy = x2y dy/2
(Eq. 2 repeated)
= (2b/ ) sin-1(y/a) 2 y/2 dy
Eq. 1 then gives, after substituting a = 3 m and b = 1 m,
Ixy = dIxy
a
x = 2b/ sin-1 (y/a)
3
=
(Eq. 1 repeated)
(2/ ) sin-1(y/3) 2 y/2 dy
0
= 0.669 m4
x
Ans.
10.3 Products of Inertia Example 5, page 1 of 3
5. Determine the product of inertia of the crosshatched
area with respect to the x and y axes.
y
x=4
y2
x = 3y
1m
x
3m
1m
10.3 Products of Inertia Example 5, page 2 of 3
1 Evaluate the integral
Ixy = dIxy
where dIxy is the product of inertia of a
horizontal infinitesimal strip.
Centroid
of the
strip
y
2
dA = (x2
y
x1)
dy
Applying the parallel-axis theorem to the strip
gives
dIxy = dIxy' ' + xel yel dA
0, because y' is an axis of
symmetry of the strip
(xel, yel)
(x1, y)
(x2, y)
(x1 + x2)/2
x
(x2
xel = x1 + (x2
x1)/2
(x2
x1)/2
= (x2 + x1)/2
y
= y(x22
x1)/2
'
x1) dy
= xel yel dA
dy yel = y
x
x1
(x2
x12) dy
(2)
10.3 Products of Inertia Example 5, page 3 of 3
y
y2
x2 = 4
1m
x1 = 3y
x
3
Replacing x1 and x2 in Eq. 2 by functions of y gives
dIxy = y(x22 x12) dy
= y[(4
y2)2
(Eq. 2 repeated)
(3y)2]dy
Thus Eq. 1 becomes
Ixy = dIxy
=
1
y[(4
0
(3)
y2)2
= 3.917 m4
(3y)2] dy
Ans.
10.3 Products of Inertia Example 6, page 1of 2
6. Determine the product of inertia of the crosshatched
area with respect to the x and y axes.
y
y = 10e-x
2
3.68 m
1
x
Ixy = dIxy
1m
1.5 m
y
(1)
where dIxy is the product of inertia of a
vertical infinitesimal strip.
y'
2
x = xel
y = 10e-x
3.68 m
Evaluate the integral
y
x'
2
Centroid
of the
strip
Applying the parallel-axis theorem to the strip gives
dIxy' ' = dIxy + xelyel dA
0, because y' is an axis of
symmetry of the strip
x
= xelyel dA
y dx
yel = y/2
x
dx
dA = y dx
y/2
= xy2 dx/2
(2)
10.3 Products of Inertia Example 6, page 2 of 2
y
y = 10e-x
2
3.68 m
x
1m
1.5 m
3
2
Replacing y in Eq. 2 by y = 10e-x gives
dIxy = xy2 dx/2
(Eq. 2 repeated)
-x 2
10e
Thus the equation for Ixy becomes,
Ixy = dIxy
1.5
2
= x(10e-x )2 dx/2
1
= 1.553 m4
(Eq. 1 repeated)
(Evaluate numerically with a calculator)
Ans.
10.3 Products of Inertia Example 7, page 1 of 4
7. Determine the product of inertia of the beam's
cross-sectional area with respect to axes passing through the
centroid C.
105 mm
y
15 m
52.5 mm
80 mm
C
x
15 mm
7.5 mm
80 mm
15 mm
10.3 Products of Inertia Example 7, page 2 of 4
1
Express the area as the sum of three rectangles.
All lengths are in millimeters.
105
15
y, y
y
52.5
52.5
15/2 = 45
y
y
80
=
15
C
x
C
7.5
x
+
15
80/2 + 7.5 = 47.5
105
x
80
Rectangle 2
80
y
105/2 - 15/2 = 45
y
15
15
x, x
C
Centroids of
rectangles
Rectangle 1
x
80
+
80/2 + 7.5 = 47.5
x
C
15
Rectangle 3
10.3 Products of Inertia Example 7, page 3 of 4
2
y
y
Apply the parallel-axis theorem to rectangle 1:
45 mm
C
Ixy1 = Ix'y' + dxdy A
0, because the x' and y' axes are axes
of symmetry of the rectangle
x
47.5 mm
x
80 mm
= 0 + ( 45 mm)( 47.5 mm)[(80 mm)(15 mm)]
= 2.5650
106 mm4
y, y
(1)
15 mm
3
Similarly for rectangle 2:
15 mm
Ixy2 = Ix'y' + dxdy A
105 mm
= 0 + (0)(0)A
y
y
=0
4
x, x
C
45 mm
(2)
Rectangle 3:
80 mm
Ixy3 = Ix'y' + dxdy A
x
47.5 mm
= 0 + (45 mm)(47.5 mm)[(80 mm)(15 mm)]
= 2.5650
106 mm4
(3)
x
C
15 mm
10.3 Products of Inertia Example 7, page 4 of 4
5
Add the Ixy terms from Eqs. 1, 2, and 3:
Ixy = Ixy1 + Ixy2 + Ixy3
= 2.565
= 5.13
106 mm4 + 0 + 2.5650
106 mm4
106 mm4
Ans.
10.3 Products of Inertia Example 8, page 1 of 4
8. Determine the product of inertia of the beam's
cross-sectional area with respect to axes passing through
the centroid C.
y
1 in.
11 in.
6 in.
5 in.
x
8 in.
1 in.
1 in.
6 in.
C
5 in.
11 in.
10.3 Products of Inertia Example 8, page 2 of 4
1
Express the area as the sum of three rectangles.
y
All lengths are in inches.
1 in.
11 in.
y, y
y y
6 in.
6 in.
5 in.
6 11/2 = 0.5
x
x
8 in.
1 in.
C
=
8
C
+
5 1/2 = 4.5
5 in.
x
1 in.
11 in.
x, x
C
1
1
11
Rectangle 2
Rectangle 1
11 in.
y
1 in.
x
y
+
5 1/2 = 4.5
6 11/2 = 0.5
C
Rectangle 3
x
10.3 Products of Inertia Example 8, page 3 of 4
2
Apply the parallel-axis theorem to rectangle 1:
y y
0.5 in.
Ixy1 = Ix'y' + dxdyA
x
C
0, because the x' and y' axes are axes
of symmetry of the rectangle
4.5 in.
x
= 0 + ( 0.5 in.)( 4.5 in)[(1 in.)(11 in.)]
4
= 24.75 in.
(1)
1 in.
11 in.
Rectangle 1
3
Similarly for rectangle 2:
y, y
Ixy2 = Ix'y' + dxdyA
= 0 + (0)(0)A
=0
(2)
8 in.
x, x
C
1 in.
Rectangle 2
10.3 Products of Inertia Example 8, page 4 of 4
4
Rectangle 3:
11 in.
y
1 in.
x
Ixy3 = Ix'y' + dxdyA
= 0 + (0.5 in.)(4.5 in.)[(11 in.)(1 in.)]
= 24.75 in.4
5
(3)
Add the Ixy terms from Eqs. 1, 2, and 3:
4.5 in.
0.5 in.
C
Rectangle 3
Ixy = Ixy1 + Ixy2 + Ixy3
= 24.75 in.4 + 0 + 24.75 in.4
= 49.50 in.4
y
Ans.
x
10.3 Products of Inertia Example 9, page 1 of 3
9. The cross section of a standard rolled-steel angle is shown.
Determine the product of inertia of the section with respect to x
and y axes passing through the centroid C. Make the
simplifying assumption that all corners are square.
y
0.987 in.
0.5 in.
6 in.
x
C
1.99 in.
0.5 in.
4 in.
10.3 Products of Inertia Example 9, page 2 of 3
Express the area as the sum of two rectangles.
1
All lengths are in inches.
y
0.987 in.
y
y
0.5 in.
0.987
0.5/2 = 0.737
=
y
x
6 in.
6 in.
6/2
x
C
1.99 = 1.01
3.5/2 + 0.5 0.987
= 1.263
y
+
1.99
x
x
C
0.5/2 = 1.74
C
1.99 in.
x
0.5 in.
4 in.
0.5
Rectangle 1
Centroids
of rectangles
3.5
Rectangle 2
10.3 Products of Inertia Example 9, page 3 of 3
2
y
y
Apply the parallel-axis theorem to rectangle 1 .
Rectangle 1:
Ixy1 = Ix'y' + dxdy A
0.737 in.
0, because the x' and y' axes are axes
of symmetry of the rectangle
x
= 0 + ( 0.737 in.)(1.01 in.)[(0.5 in.)(6 in.)]
6 in.
1.01 in.
= 2.233 in.4
3
(1)
x
C
Rectangle 2 :
Ixy2 = Ix'y' + dxdy A
= 0 + (1.263 in.)( 1.74 in.)[(3.5 in.)(0.5 in.)]
0.5 in.
(2)
Rectangle 1
= 3.846 in.4
y
4
Add the Ixy terms from Eqs. 1 and 2:
1.263 in.
x
Ixy = Ixy1 + Ixy2
=
y
C
4
1.74 in.
4
2.233 in. + ( 3.846 in. )
= 6.08 in.4
Rectangle 2
x
Ans.
3.5 in.
10.3 Products of Inertia Example 10, page 1 of 4
10. Determine the product of inertia of the crosshatched
area with respect to the x and y axes.
y
150 mm
15 mm
30 mm
x
10.3 Products of Inertia Example 10, page 2 of 4
1
y
Express the area as a rectangle minus a circle
plus a semicircle.
y
75 mm
150 mm
150 mm
y
15 mm
x
60 mm
30 mm
=
30 mm
x
y
y
Radius
= 15 mm
x
150 mm
2 A table of properties of planar regions
gives the information shown below.
x
Centroid
y
x
Centroid location
2
A=
r
30 mm
r
2
C
Semicircular area
y
150 mm
4r
3
x
+
y
4(30 mm)
= 12.7324 mm
3
30 mm
x
30 mm
x
10.3 Products of Inertia Example 10, page 3 of 4
y
3 Apply the parallel-axis theorem to the rectangle.
150 mm
Ixy-rectangle = Ix'y' + dxdy A
y
0, because the x' and
y' axes are axes of
symmetry
75 mm
= 0 + (75 mm)(30 mm) (150 mm)(60 mm)
= 2.0250
107 mm4
x
60 mm
30 mm
(1)
x
4
Similarly for the circle,
Ixy-circle = Ix'y' + dxdy A
= 0 + (150 mm)(30 mm)
0.3181
107 mm4
15 mm)2
(2)
y
y
150 mm
Radius = 15 mm
x
30 mm
10.3 Products of Inertia Example 10, page 4 of 4
5 Similarly for the semicircle,
y
y
150 mm
Ixy-semicircle = Ix'y' + dxdy A
= 0 + (150 mm + 12.7324 mm)(30 mm)
30 mm)2/2
12.7324 mm
= 0.6902
7
4
10 mm
(3)
30 mm
x
30 mm
x
6
Add the Ixy values for the rectangle and semicircle and subtract the circle.
Ixy = Ixy-rectangle + Ixy-semicircle
= 2.0250
= 2.40
Ixy-circle
107 mm4 + 0.6902
107 mm4
107 mm4
0.3181
107 mm4
Ans.
10.3 Products of Inertia Example 11, page 1 of 6
11. Determine the product of inertia of the crosshatched
region with respect to the x and y axes.
y
280 mm 210 mm
40 mm
50 mm
x
210 mm
330 mm
10.3 Products of Inertia Example 11, page 2 of 6
y
y
1 Express the area as a rectangle minus a triangle.
330 mm/2 = 165 mm
y
x
280 mm
=
280 mm 210 mm
x
330 mm
280 mm/2 = 140 mm
y
40 mm
50 mm
y
50 mm + 210 mm/3
= 120 mm
x
210 mm
40 mm + 210 mm/3
= 110 mm
330 mm
210 mm
x
x
210 mm
10.3 Products of Inertia Example 11, page 3 of 6
2
Apply the parallel-axis theorem to the rectangle.
y
y
Ixy-rectangle= Ix'y' + dxdy A
165 mm
0, because the x' and y' axes are
axes of symmetry
= 0 + (165 mm)(140 mm) (280 mm)(330 mm)
= 21.3444
108 mm4
x
280 mm
(1)
140 mm
x
330 mm
10.3 Products of Inertia Example 11, page 4 of 6
3
Apply the parallel-axis theorem to the triangle. Note that Ix'y'
0.
y
y
120 mm
Ixy-triangle = Ix'y' + dxdy A
A = (1/2)(210 mm)(210 mm)
= 22050 mm2
= Ix'y' + (120 mm)(110 mm)(22050 mm2)
= Ix'y' + 2.9106
108 mm4
(2)
210 mm
x
110 mm
4
Compute Ix'y' by integration. The integration will be
simpler if we first find Ix''y'' and then use the
parallel-axis theorem. That is, by the parallel axis
theorem,
x
210 mm
y
Ix''y'' = Ix'y' + dx'dy'A
y
210 mm/3 = 70 mm
Thus
Ix'y' = Ix''y'' dx'dy'A
= Ix''y'' (70mm)(70mm)(22050 mm2)
= Ix''y'' 1.0804
108 mm4
210 mm
x
(3)
210 mm/3 = 70 mm
x
210 mm
10.3 Products of Inertia Example 11, page 5 of 6
5
Integrate by using vertical strips. Applying the parallel-axis
theorem to the vertical strip gives
y
dIx''y'' = dIxcyc + xel'' yel" dA
= 0 + x''(y''/2)(y'' dx'')
yc xel = x
Equation of line:
y'' = x'' + 210
Centroid of
infinitesimal strip
= x''( x'' + 210)2 dx''/2
210 mm
Thus
xc y
Ix''y'' =
dIx''y''
x
dx
210 mm
210
=
x'' ( x'' + 210)2 dx''/2
0
= 0.8103
108 mm4
(4)
yel = y /2
10.3 Products of Inertia Example 11, page 6 of 6
6
Substitute in Eq. 3:
Ix'y' = Ix''y''
108 mm4
1.0804
0.8103
(Eq. 3 repeated)
108, by Eq. 4
108 mm4
= 0.2701
(5)
Substitute this result in Eq. 2:
Ixy-triangle = Ix'y' + 2.9106
0.2701
= 2.6405
108 mm4
(Eq. 2 repeated)
108 mm4, by Eq. 5
108 mm4
(6)
The product of inertia for the entire area is the difference between
Ixy-rectangle and Ixy-triangle:
Ixy = Ixy-rectangle
21.3444
Ixy-triangle
108 mm4, by Eq.1
= 1.870
109 mm4
2.6405
108, by Eq. 6
Ans.