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10.6 Product of Inertia
• Product of Inertia:
I xy = ∫ xy dA
• 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 y A
9- 1
Sample Problem 10.6
Determine the product of inertia of the
right triangle
(a) with respect to the x and y axes and
(b) with respect to centroidal axes
parallel to the x and y axes.
SOLUTION:
• Determine the product of inertia using direct integration with
the parallel axis theorem on vertical differential area strips
• Apply the parallel axis theorem to evaluate the product of
inertia with respect to the centroidal axes.
9- 2
Sample Problem 10.6
(continue)
SOLUTION:
• Determine the product of inertia using direct
integration with the parallel axis theorem on
vertical differential area strips
 x
 x
y = h1 −  dA = y dx = h1 − dx
 b
 b
 x
xel = x
yel = 12 y = 12 h1 − 
 b
Integrating dIx from x = 0 to x = b,
b
I xy = ∫ dI xy = ∫ xel yel dA = ∫ x
0
(12 )h2 1 − bx 

2

2
dx
b
b
 x 2 x3 x 4 
x x 2 x3 
= h 2 ∫  −
+ 2  dx = h  − + 2 
2
b
2
b
0

 4 3b 8b  0
I xy =
1 b 2h 2
24
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Sample Problem 10.6 (continue)
• Apply the parallel axis theorem to evaluate the
product of inertia with respect to the centroidal
axes.
x = 13 b
y = 13 h
With the results from part a,
I xy = I x′′y ′′ + x yA
I x′′y ′′ =
1 b 2h2
24
( )(13 h)(12 bh)
− 13 b
1 b 2h2
I x′′y ′′ = − 72
9- 4
10.7 Moments of Inertia about inclined axis
In structural and mechanical design, it is sometimes necessary to calculate
the moment of inertia with respect to a set of inclined u, v, axes when the
values of θ , Ix, Iy, Ixy are known.
To do this we will use transformation
equations which relates the x, y, and
x’, y’ coordinates.
From Figure, these equations are:
Note: x′ = x cosθ + y sin θ
y ′ = y cosθ − x sin θ
9- 5
Moments of Inertia about inclined axis,, continue
• The change of axes yields
I x′ =
I y′ =
Ix + I y
2
Ix + I y
I x′y′ =
Given
I x = ∫ y 2 dA I y = ∫ x 2 dA
I xy = ∫ xy dA
we wish to determine moments and
product of inertia with respect to
new axes x’ and y’.
2
Ix − I y
2
Ix − I y
cos 2θ − I xy sin 2θ
2
Ix − I y
−
cos 2θ + I xy sin 2θ
2
+
sin 2θ + I xy cos 2θ
• By adding the equations for Ix’ and Ix’ we
can show that the polar moment of inertia
about z axis passing through point O is
independent of the orientation of x’ and y;
Jo = Ix’ + Iy’ = Ix + Iy
• These equations show that Ix’ , Iy’ and Ix’y’
depend on the angle of the inclination, θ,
of the x’, y’ axes.
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Principal Axes and Principal Moments of Inertia
We will now determine the orientation of these axes about which Ix’ , Iy’
are maximum and minimum. This particular axes are called principal axes
By differentiating the first of Eqs. 10-9 with respect to θ and setting the
result to zero. Thus;
dI x '
dθ
Therefore, at θ = θp ;
By substituting for θ in Ix’ , Iy’ and Ix’y’ equations and simplifying ,
we obtain;
9- 7
Summary,,,,,
Ix − I y
…….. 10-9 (a)
−
cos 2θ + I xy sin 2θ
2
2
Ix − I y
=
sin 2θ + I xy cos 2θ
2
…….. 10-9 (a)
I y′ =
I x′y′
Ix + Iy
cos 2θ − I xy sin 2θ
I x′ =
2
Ix + Iy
+
2
Ix − I y
…….. 10-9 (a)
………….. 10-10
…….. 10-11
9- 8
Principal Axes and Principal Moments of Inertia
Squaring Eqs. 10-9a and 10-9c and
adding, it is found that;
(I x ′ − I ave )2
+ I x2′y ′ = R 2
where ;
I ave =
R =
Ix + Iy
2
,
 Ix − I y 
2



 + I xy
2


• At the points A and B, Ix’y’ = 0 and
Ix’ is a maximum and minimum,
respectively.
I max, min = I ave ± R
tan 2θ m = −
2 I xy
Ix − I y
• The equation for θm defines two
angles, 90o apart which correspond to
the principal axes of the area about O.
• Imax and Imin are the principal moments
of inertia of the area about O.
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10.8 Mohr’s Circle for
Moments and Products of Inertia
• The moments and product of inertia for an
area are plotted as shown and used to
construct Mohr’s circle,
I ave =
Ix + Iy
2
 Ix − Iy  2
R = 
 + I xy
 2 
• Mohr’s circle may be used to graphically or
analytically determine the moments and
product of inertia for any other rectangular
axes including the principal axes and
principal moments and products of inertia.
9 - 10
Sample Problem 10.7
For the section shown, the moments of
inertia with respect to the x and y axes
are Ix = 10.38 in4 and Iy = 6.97 in4.
Determine (a) the orientation of the
principal axes of the section about O,
and (b) the values of the principal
moments of inertia about O.
SOLUTION:
• Compute the product of inertia with respect to the xy axes by dividing
the section into three rectangles and applying the parallel axis theorem
to each.
• Determine the orientation of the principal axes (Eq. 9.25) and the
principal moments of inertia (Eq. 9. 27).
9 - 11
Sample Problem 10.7
SOLUTION:
• Compute the product of inertia with respect to the xy axes
by dividing the section into three rectangles.
Apply the parallel axis theorem to each rectangle,
(
I xy = ∑ I x′y′ + x yA
)
Note that the product of inertia with respect to centroidal
axes parallel to the xy axes is zero for each rectangle.
Rectangle Area, in 2
I
II
III
y , in.
x yA, in 4
1.5 − 1.25 + 1.75
− 3.28
1 .5
x , in.
0
0
0
1.5 + 1.25 − 1.75
− 3.28
∑ x yA = −6.56
I xy = ∑ x yA = −6.56 in 4
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Sample Problem 10.7
• Determine the orientation of the principal axes (Eq. 9.25)
and the principal moments of inertia (Eq. 9. 27).
tan 2θ m = −
2 I xy
Ix − I y
=−
2(− 6.56)
= +3.85
10.38 − 6.97
2θ m = 75.4° and 255.4°
θ m = 37.7° and θ m = 127.7°
I x = 10.38 in 4
I max, min =
I y = 6.97 in 4
Ix + I y
2
2
 Ix − I y 
2
 + I xy
± 
 2 
2
I xy = −6.56 in 4
=
10.38 + 6.97
 10.38 − 6.97 
2
± 
 + (− 6.56)
2
2


I a = I max = 15.45 in 4
I b = I min = 1.897 in 4
9 - 13
Sample Problem 10.8
SOLUTION:
• Plot the points (Ix , Ixy) and (Iy ,-Ixy).
Construct Mohr’s circle based on the
circle diameter between the points.
• Based on the circle, determine the
orientation of the principal axes and the
principal moments of inertia.
The moments and product of inertia
with respect to the x and y axes are Ix =
7.24x106 mm4, Iy = 2.61x106 mm4, and
Ixy = -2.54x106 mm4.
• Based on the circle, evaluate the
moments and product of inertia with
respect to the x’y’ axes.
Using Mohr’s circle, determine (a) the
principal axes about O, (b) the values of
the principal moments about O, and (c)
the values of the moments and product
of inertia about the x’ and y’ axes
9 - 14
Sample Problem 10.8
SOLUTION:
• Plot the points (Ix , Ixy) and (Iy ,-Ixy). Construct Mohr’s
circle based on the circle diameter between the points.
OC = I ave =
(
1
2
(I x + I y ) = 4.925 × 106 mm4
)
CD = 12 I x − I y = 2.315 × 106 mm4
R=
6
I x = 7.24 × 10 mm
4
I y = 2.61 × 106 mm 4
I xy = −2.54 × 106 mm 4
(CD )2 + (DX )2
= 3.437 × 106 mm4
• Based on the circle, determine the orientation of the
principal axes and the principal moments of inertia.
tan 2θ m =
DX
= 1.097 2θ m = 47.6°
CD
θ m = 23.8°
I max = OA = I ave + R
I max = 8.36 × 106 mm 4
I min = OB = I ave − R
I min = 1.49 × 106 mm 4
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Sample Problem 10.8
• Based on the circle, evaluate the moments and product
of inertia with respect to the x’y’ axes.
The points X’ and Y’ corresponding to the x’ and y’ axes
are obtained by rotating CX and CY counterclockwise
through an angle Θ = 2(60o) = 120o. The angle that CX’
forms with the x’ axes is φ = 120o - 47.6o = 72.4o.
I x ' = OF = OC + CX ′ cos ϕ = I ave + R cos 72.4o
I x′ = 5.96 × 106 mm 4
I y ' = OG = OC − CY ′ cos ϕ = I ave − R cos 72.4o
I y′ = 3.89 × 106 mm4
I x′y ' = FX ′ = CY ′ sin ϕ = R sin 72.4o
OC = I ave = 4.925 × 106 mm4
R = 3.437 × 106 mm4
I x′y′ = 3.28 × 106 mm4
9 - 16
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