Engineering Mechanics: STATICS

1
Engineering MechanicsSTATICS

• Anthony Bedford and Wallace Fowler
• SI Edition

Teaching Slides Chapter 8 Moments of Inertia
2
Chapter Outline

• Areas
• Definitions
• Parallel-Axis Theorems
• Rotated Principal Axes
• Masses
• Simple Objects
• Parallel-Axis Theorems

3
8.1 Definition

• Consider an area A in the x-y plane

4
8.1 Definition

• 4 moments of inertia of A are defined
• 1. Moment of inertia about the x axis
• (8.1)
• where y is the y coordinate of the
  differential element
• of area dA
• This moment of inertia is sometimes expressed
  in terms of the radius of gyration about the x
  axis, kx, which is defined by
• (8.2)

5
8.1 Definition

• 2. Moment of inertia about the y axis
• (8.3)
• where x is the x coordinate of the element dA
• The radius of gyration about the y axis, ky,
  is defined by
• (8.4)
• 3. Product of inertia
• (8.5)

6
8.1 Definition

• 4. Polar moment of inertia
• (8.6)
• where r is the radial distance from the origin
  of the coordinate system to dA
• The radius of gyration about the origin, kO,
  is defined by
• (8.7)
• The polar moment of inertia is equal to the
  sum of moments of inertia about the x y axes

7
8.1 Definition

• Substituting the expressions for the moments
  of inertia in terms of the radii of gyration into
  this equation, we obtain
• The dimensions of the moments of inertia of an
  area are (length)4 the radii of gyration have
  dimensions of length
• Notice that the definitions of the moments of
  inertia Ix, Iy JO the radii of gyration imply
  that they have positive values for any area
• They cannot be negative or zero

8
8.1 Definition

• If an area A is symmetric about the x axis, for
  each element dA with coordinates (x, y), there is
  a corresponding element dA with coordinates (x,
  ?y)

9
8.1 Definition

• The contributions of these 2 elements to the
  product of inertia Ixy of the area cancel
• xy dA (?xy) dA 0
• This means that the product of inertia of the
  area is zero
• The same kind of argument can be used for an area
  that is symmetric about the y axis
• If an area is symmetric about either the x or y
  axis, its product of inertia is zero

10
Example 8.1 Moments of Inertia of a Triangular
Area

• Determine Ix, Iy Ixy for the triangular area
  in Fig. 8.3.

Fig. 8.3
11
Example 8.1 Moments of Inertia of a Triangular
Area

• Strategy
• Eq. (8.3) for the moment of inertia about the
  y axis is very similar to the equation for the x
  coordinate of the centroid of an area it can be
  evaluated for this triangular area in exactly in
  the same way by using a differential element of
  area dA in the form of a vertical strip of width
  dx. Then show that Ix Ixy can be evaluated by
  using the same element of area.

12
Example 8.1 Moments of Inertia of a Triangular
Area

• Solution
• Let dA be the vertical strip.
• The equation describing the
• triangular areas upper
• boundary is f(x) (h/b)x,
• so dA f(x) dx (h/b)x dx.
• To integrate over the entire area, we must
• integrate with respect to x from x 0 to x b.

13
Example 8.1 Moments of Inertia of a Triangular
Area

• Solution
• Moment of Inertia About the y Axis
• Moment of Inertia About the x Axis
• 1st, determine the moment of inertia of the strip
  
• dA about the x axis while holding x dx fixed.

14
Example 8.1 Moments of Inertia of a Triangular
Area

• Solution
• In terms of the element area dAs dx dy

15
Example 8.1 Moments of Inertia of a Triangular
Area

• Solution
• Integrating this expression with respect to x
  from
• x 0 to x b, we obtain the value of Ix for the
  entire
• area

16
Example 8.1 Moments of Inertia of a Triangular
Area

• Solution
• Product of Inertia
• 1st evaluate the product of inertia of the strip
  dA,
• holding x dx fixed

17
Example 8.1 Moments of Inertia of a Triangular
Area

• Solution
• Integrate this expression with respect to x from
• x 0 to x b to obtain the value of Ixy for the
  entire
• area

18
Example 8.1 Moments of Inertia of a Triangular
Area

• Critical Thinking
• This example is chosen so that you can confirm
  that we obtain the results tabulated for a
  triangular area in Appendix B
• Notice that the same procedure can be used to
  obtain the moments of inertia of other areas
  whose boundaries are described by the functions
  of the form y f(x)

19
Example 8.2 Moments of Inertia of a Circular Area

• Determine the moments of inertia radii of
  gyration of the circular area in Fig. 8.4.

Fig. 8.4
Strategy 1st, determine the polar moment of
inertia JO by integrating in terms of polar
coordinates. We know from the symmetry of the
area that Ix Iy since Ix Iy JO, the
moments of inertia of Ix Iy are each equal to ½
JO. We also know from the symmetry of the area
that Ixy 0.
20
Example 8.2 Moments of Inertia of a Circular Area

• Solution
• By letting r change by an amount dr, we obtain an
  
• annular element of area dA 2?r dr.
• The polar moment of inertia is

21
Example 8.2 Moments of Inertia of a Circular Area

• Solution
• And the radius of gyration about O is
• The moments of inertia about the x y axes are
• and the radii of gyration about the x y axes
  are

22
Example 8.2 Moments of Inertia of a Circular Area

• Solution
• The product of inertia is zero
• Critical Thinking
• The symmetry of this example saved us from having
  to integrate to determine Ix, Iy Ixy
• Be alert for symmetry that can shorten your work
• In particular, remember that Ixy 0 if the area
  is symmetric about either the x or the y axis

23
8.2 Parallel-Axis Theorems

• The values of the moments of inertia of an area
  depend on the position of the coordinate system
  relative to the area
• In some situations the moments of inertia of an
  area are known in terms of a particular
  coordinate system but we need their values in
  terms of a different coordinate system
• When the coordinate systems are parallel, the
  desired moments of inertia can be obtained using
  the parallel-axis theorems
• Possible to determine the moments of inertia of a
  composite area when the moments of inertia of its
  parts are known

24
8.2 Parallel-Axis Theorems

• Suppose that we know the moments of inertia of an
  area A in terms of a coordinate system xy with
  its origin at the centroid of the area we wish
  to determine the moments of inertia in terms of a
  parallel coordinate system xy

25
8.2 Parallel-Axis Theorems

• Denote the coordinates of the centroid of A in
  the xy coordinate system by (dx, dy)
  is the distance from the origin of the
  xy coordinate system to the centroid
• In terms of the xy coordinate system, the
  coordinates of the centroid of A are

26
8.2 Parallel-Axis Theorems

• But the origin of xy coordinate system is
  located at the centroid of A, so
• Therefore,
• (8.8)
• Moment of Inertia About the x Axis
• In terms of the xy coordinate system, the moment
  of inertia of A about the x axis is
• (8.9)
• where y is the y coordinate of the element dA
  relative to the xy coordinate system

27
8.2 Parallel-Axis Theorems

• From the figure, y y dy, where y is the
  coordinate of dA relative to the xy coordinate
  system
• Substituting this expression into Eq. (8.9), we
  obtain
• The 1st integral on the right is the moment of
  inertia of A about the x axis
• From Eq. (8.8) the 2nd integral on the right
  equals zero

28
8.2 Parallel-Axis Theorems

• Therefore, we obtain
• (8.10)
• This is a parallel-axis theorem
• It relates the moment of inertia of A about the
  x axis through the centroid to the moment of
  inertia about the parallel axis x

29
8.2 Parallel-Axis Theorems

• Moment of Inertia About the y Axis
• In terms of the xy coordinate system, the moment
  of inertia of A about the y axis is
• From Eq. (8.8), the 2nd integral on the right
  equals zero

30
8.2 Parallel-Axis Theorems

• Therefore, the parallel-axis theorem that relates
  the moment of inertia of A about the y axis
  through the centroid to the moment of inertia
  about the parallel axis y is
• (8.11)

31
8.2 Parallel-Axis Theorems

• Product of Inertia
• In terms of the xy coordinate system, the product
  of inertia is
• The 2nd 3rd integrals equal zero from Eq. (8.8)
• The parallel-axis theorem for the product of
  inertia is
  (8.12)

32
8.2 Parallel-Axis Theorems

• Polar Moment of Inertia
• The polar moment of inertia JO Ix Iy
• Summing Eqs. (8.10) (8.11), the parallel-axis
  theorem for the polar moment of inertia is
• (8.13)
• where d is the distance form the origin of the
  xy coordinate system to the origin of the xy
  coordinate system

33
8.2 Parallel-Axis Theorems

• To determine the moments of inertia of a
  composite area
• Suppose that we want to determine the moment of
  inertia about the y axis of the area

34
8.2 Parallel-Axis Theorems

• We can divide it into a triangle, a semicircle
  a circular cutout, denoted as parts 1, 2 3
• By using the parallel-axis theorem for Iy, we can
  determine the moment of inertia of each part
  about the y axis
• E.g. the moment of inertia of part 2 (the
  semicircle) about the y axis is

35
8.2 Parallel-Axis Theorems

• We must determine the values of (Iy)2 (dx)2
• Once this procedure is carried out for each part,
  the moment of inertia of the composite area is
• Notice that the moment of inertia of the circular
  cutout is subtracted

36
8.2 Parallel-Axis Theorems

• Determining a moment of inertia of a composite
  area in terms of a given coordinate system
  involves 3 steps
• 1.Choose the parts try to divide the composite
  area into parts whose moments of inertia you know
  or can easily determine.
• 2.Determine the moments of inertia of the parts
  determine the moment of inertia of each part in
  terms of a parallel coordinate system with its
  origin at the centroid of the part then use the
  parallel-axis theorem to determine the moment of
  inertia in terms of the given coordinate system.

37
8.2 Parallel-Axis Theorems

• 3.Sum the results sum the moments of inertia of
  the parts (or subtract in the case of a cutout)
  to obtain the moment of inertia of the composite
  area.

38
Example 8.3 Demonstration of the Parallel-Axis
Theorems

• The moments of inertia of the rectangular area
  in Fig. 8.8 in terms of the xy coordinate
  system are
• Determine its moment of inertia in terms of
  the xy coordinate system.

Fig. 8.8
39
Example 8.3 Demonstration of the Parallel-Axis
Theorems

• Strategy
• The xy coordinate system has its origin at
  the centroid of the area is parallel to the xy
  coordinate system. Use the parallel-axis theorems
  to determine the moments of inertia of A in terms
  of the xy coordinate system.

40
Example 8.3 Demonstration of the Parallel-Axis
Theorems

• Solution
• The coordinates of the centroid in terms of the
  xy
• coordinate system are dx b/2, dy h/2.
• The moment of inertia about the x axis is
• The moment of inertia about the y axis is
• The product of inertia is

41
Example 8.3 Demonstration of the Parallel-Axis
Theorems

• Solution
• The polar moment of inertia is
• Critical Thinking
• Notice that we could also have determined JO
  using the relation

42
Example 8.3 Demonstration of the Parallel-Axis
Theorems

• Critical Thinking
• This example is designed so that you can confirm
  that the parallel-axis theorems yield the results
  in Appendix B for a rectangular area
• But the same procedure can be used to obtain the
  moments of inertia of the area in terms of any
  coordinate system that is parallel to the xy
  system

43
Example 8.4 Moments of Inertia of a Composite Area

• Determine Ix, kx Ixy for the composite area
  in Fig. 8.9.

Strategy This area can be divided into 2
rectangles. Use the parallel-axis theorems to
determine Ix Ixy for each rectangle in terms
of the xy coordinate system sum the results
for the rectangles to determine Ix Ixy for the
composite area. Then use Eq. (8.2) to determine
the radius of gyration kx for the composite area.
44
Example 8.4 Moments of Inertia of a Composite Area

• Solution
• Choose the Parts
• Determine the moments of inertia by dividing the
• area in 2 rectangular parts 1 2

45
Example 8.4 Moments of Inertia of a Composite Area

• Solution
• Determine the Moments of Inertia of the Parts
• For each part, introduce a coordinate system xy
  
• with its origin at the centroid of the part

46
Example 8.4 Moments of Inertia of a Composite Area

• Solution
• Use the parallel-axis theorem to determine the
• moment of inertia of each part about the x axis
• (Table 8.1)

Table 8.1 Determining the moments of inertia of
the parts about the x axis
47
Example 8.4 Moments of Inertia of a Composite Area

• Solution
• Sum the Results
• The moment of inertia of the composite area
  about
• the x axis is
• The sum of the areas is A A1 A2 6 m2,
• so the radius of gyration about the x axis is

48
Example 8.4 Moments of Inertia of a Composite Area

• Solution
• Repeating this procedure, determine Ixy for each
• part in Table 8.2
• The product of inertia of the composite area is

Table 8.2 Determining the products of inertia of
the parts of the xy coordinate system
49
Example 8.4 Moments of Inertia of a Composite Area

• Critical Thinking
• The moments of inertia you obtain do not depend
  on how you divide a composite area into parts
  you will often have a choice of convenient ways
  to divide a given area.

50
Example 8.5 Moments of Inertia of a Composite Area

• Determine Iy ky for the composite area in
  Fig. 8.10.

51
Example 8.5 Moments of Inertia of a Composite Area

• Strategy
• Divide the area into a rectangle without the
  semicircular cutout, a semicircle without the
  semicircular cutout a circular cutout. Use a
  parallel-axis theorem to determine Iy for each
  part in terms of the xy coordinate system. Then,
  determine Iy for the composite area by adding the
  values of the rectangle semicircle
  subtracting the circular cutout. Then use Eq.
  (8.4) to determine the radius of gyration ky for
  the composite area.

52
Example 8.5 Moments of Inertia of a Composite Area

• Solution
• Choose the Parts
• Divide the area into a rectangle,
• a semicircle the circular cutout,
• calling them parts 1, 2 3,
• respectively

53
Example 8.5 Moments of Inertia of a Composite Area

• Solution
• Determine the Moments of Inertia of the Parts
• The moments of inertia of the parts in terms of
  the
• xy coordinate systems location of the
  centroid
• of the semicircular part are summarized in Table
  8.3.
• Use the parallel-axis theorem to determine the
• moment of inertia of each part about the y axis.

54
Example 8.5 Moments of Inertia of a Composite Area

• Solution
• Sum the Results
• The moment of inertia of the composite area
  about
• the y axis is
• The total area is

55
Example 8.5 Moments of Inertia of a Composite Area

• Solution
• So the radius of gyration about the y axis is
• Critical Thinking
• Integration is an additive process, which is why
  the moments of inertia of composite areas can be
  determined by adding (or in the case of a cutout,
  subtracting) the moments of inertia of the parts

56
Example 8.5 Moments of Inertia of a Composite Area

• Critical Thinking
• But the radii of gyration of composite areas
  cannot be determined by adding or subtracting the
  radii of gyration of the parts
• This can be seen from the equations relating the
  moments of inertia, radii of gyration area
• For this example, we can demonstrate it
  numerically the operation
• does not yield the correct radius of gyration
  of the composite area.

57
Design Example 8.6 Beam Design

• The equal areas in Fig. 8.11 are candidates
  for the cross-section of a beam. (A beam with the
  2nd cross section shown is called an I-beam.)
  compare their moments of inertia about the x axis.

58
Design Example 8.6 Beam Design

• Strategy
• Obtain the moment of inertia of the square
  cross section from Appendix B. Divide the I-beam
  into 3 rectangles use the parallel-axis theorem
  to determine its moment of inertia by the same
  procedure used in Examples 8.4 8.5.

59
Design Example 8.6 Beam Design

• Solution
• Square Cross Section
• From Appendix B, the moment of inertia of the
• square cross section about the x axis is
• I-Beam Cross Section
• Divide the area into the rectangular
• parts shown

60
Design Example 8.6 Beam Design

• Solution
• Introducing the coordinate system xy with their
  
• origins at the centroids of the parts

61
Design Example 8.6 Beam Design

• Solution
• Use the parallel-axis theorem to determine the
• moments of inertia about the x axis (Table 8.4)

62
Design Example 8.6 Beam Design

• Solution
• Their sum is
• The moment of inertia of the I-beam about the x
• axis is 3.06 times that of the square cross
  section
• of equal area.

63
Design Example 8.6 Beam Design

• Design Issues
• A beam is a bar of material that supports lateral
  loads, meaning loads perpendicular to the axis of
  the bar

• 2 common types of beams
• Simply supported beam a beam with pinned ends
• Cantilever beam a beam with a single, built-in
  support

64
Design Example 8.6 Beam Design

• Design Issues
• The lateral loads on a beam cause it to bend it
  must be stiff or resistant to bending to support
  them
• It is shown in mechanics of materials that a
  beams resistance to bending depends directly on
  the moment of inertia of its cross-sectional area

65
Design Example 8.6 Beam Design

• Design Issues
• Consider the beam
• The cross section is symmetric about the y axis
  the origin of the coordinate system is placed at
  its centroid

• If the beam consists of a homogeneous structural
  material such as steel it is subjected to
  couples at the ends, it bends into a circular arc
  of radius
• where Ix is the moment of inertia of the
  beam cross
• section about the x axis

66
Design Example 8.6 Beam Design

• Design Issues
• The modulus of elasticity E has different
  values for different materials
• (The equation holds only if M is small enough
  so that the beam returns to its original shape
  when the couples are removed)
• Thus, the amount the beam bends for a given value
  of M depends on the material the moment of its
  inertia of its cross section
• Increasing Ix increases the value of R, which
  means the resistance of the beam to bending is
  increased

67
Design Example 8.6 Beam Design

• Design Issues
• This explains in large part the cross sections of
  many of the beams you see in use
• E.g. in highway overpasses in frames of
  buildings
• They are configured to increase their moments of
  inertia

68
Design Example 8.6 Beam Design

• Design Issues
• The cross sections in the figure all have the
  same area
• (the numbers are the ratios of the moment of
  inertia Ix to the value of Ix for the solid
  square cross section)

69
Design Example 8.6 Beam Design

• Design Issues
• However, configuring the cross section of a beam
  to increase its moment of inertia can be carried
  too far
• The box beam in the figure has a value of Ix
  that is 4 times as large as a solid square beam
  of the same cross-sectional area but its walls
  are so thin that they may buckle

70
Design Example 8.6 Beam Design

• Design Issues
• The stiffness implied by the beams large moment
  of inertia is not realized because it becomes
  geometrically unstable
• 1 solution used by engineers to achieve a large
  moment of inertia in a relatively light beam
  while avoiding failure due to buckling is to
  stabilize its walls by filling the beam with a
  light material such as honeycombed metal or
  foamed plastic

71
8.3 Rotated Principal Axes

• Suppose Fig. a is the cross section of a
  cantilever beam
• If you apply a vertical force to the end of the
  beam, a larger vertical deflection results if the
  cross section is oriented as shown in Fig. b than
  if it is oriented as shown in Fig. c

72
8.3 Rotated Principal Axes

• The minimum vertical deflection results when the
  beams cross section is oriented so that the
  moment of inertia Ix is a maximum (Fig. d)
• In many engineering applications you must
  determine moments of inertia of areas with
  various angular orientations relative to a
  coordinate system also determine the
  orientation for which the value of the moment of
  inertia is a maximum or minimum

73
8.3 Rotated Principal Axes

• Rotated Axes
• Consider an area A, a coordinate system xy a
  2nd coordinate system xy that is rotated
  through an angle ? relative to the xy coordinate
  system

74
8.3 Rotated Principal Axes

• Suppose that we know the moments of inertia of A
  in terms of the xy coordinate system
• The objective is to determine the moments of
  inertia in terms of the xy coordinate system
• In terms of the radial distance r to a
  differential element of area dA the angle ?,
  the coordinates of dA in the xy coordinate system
  are
• (8.14)
• (8.15)

75
8.3 Rotated Principal Axes

• The coordinates of dA in the xy coordinate
  system are
• (8.16)
• (8.17)
• In Eqs. (8.16) (8.17), we use identities for
  the cosine sine of the difference of 2 angles

76
8.3 Rotated Principal Axes

• By substituting Eqs. (8.14) (8.15) into Eqs.
  (8.16) (8.17), we obtain equations relating the
  coordinates of dA in the 2 coordinate systems
• (8.18)
• (8.19)
• We can use these expressions to derive relations
  between the moments of inertia of A in terms of
  the xy xy coordinate systems

77
8.3 Rotated Principal Axes

• Moment of Inertia About the x Axis
• From this equation we obtain
• (8.20)

78
8.3 Rotated Principal Axes

• Moment of Inertia About the y Axis
• This equation gives us the result
• (8.21)

79
8.3 Rotated Principal Axes

• Product of Inertia
• In terms of the xy coordinate system, the
  product of the inertia of A is
• (8.22)
• Polar Moment of Inertia
• From Eqs. (8.20) (8.21), the polar moment of
  inertia in terms of the xy coordinate system
  is
• Thus the value of the polar moment of inertia is
  unchanged by a rotation of the coordinate system

80
8.3 Rotated Principal Axes

• Principal Axes
• Consider the following question for what values
  of ? is the moment of inertia Ix a maximum or
  minimum?
• Use the identities

81
8.3 Rotated Principal Axes

• With these expressions, we can write Eqs.
  (8.20)?(8.22) in the forms
• (8.23)
• (8.24)
• (8.25)
• Denoting a value of ? at which Ix is a maximum
  or minimum by ?P

82
8.3 Rotated Principal Axes

• To determine ?P, evaluate the derivative of Eq.
  (8.23) with respect to 2? equate it to zero,
  obtaining
• (8.26)
• If we set the derivative of Eq. (8.24) with
  respect to 2? equal to zero to determine a
  value of ? for which Iy is a maximum or minimum,
  we again obtain Eq. (8.26)

83
8.3 Rotated Principal Axes

• The 2nd derivatives of Ix Iy with respect to
  2? are opposite in sign
• which means that at an angle ?P for which Ix
  is a maximum, Iy is a minimum and at an angle
  ?P for which Ix is a minimum, Iy is a maximum

84
8.3 Rotated Principal Axes

• A rotated coordinate system xy that is oriented
  so that Ix Iy have maximum or minimum values
  is called a set of principal axes of the area A
• The corresponding moments of inertia Ix Iy are
  called the principal moments of inertia
• Because the tangent is a periodic function, Eq.
  (8.26) does not yield a unique solution for the
  angle ?P
• However, it does determine the orientation of the
  principal axes within an arbitrary multiple of 90

85
8.3 Rotated Principal Axes

• Observe in the figure that if 2?0 is a solution
  of Eq. (8.26), then 2?0 n(180) is also a
  solution for any integer n

86
8.3 Rotated Principal Axes

• The resulting orientations of the xy coordinate
  system are

87
8.3 Rotated Principal Axes

• Determining the principal axes principal
  moments of inertia of an area involves 3 steps
• 1.Determine Ix, Iy Ixy you must determine the
  moments of the area in terms of the xy coordinate
  system.
• 2.Determine ?P solve Eq. (8.26) to determine
  the orientation of the principal axes within an
  arbitrary multiple of 90.
• 3.Calculate Ix Iy once you have chosen the
  orientation of the principal axes, you can use
  Eqs. (8.20) (8.21) or Eqs. (8.23) (8.24) to
  determine the principal moments of inertia.

88
Example 8.7 Determining Principal Axes Moments
of Inertia

• Determine a set of principal axes the
  corresponding principal moments of inertia for
  the triangular area in Fig. 8.20.

Fig. 8.20
Strategy Obtain the moments of inertia of
the triangular area from Appendix B. Then use Eq.
(8.26) to determine the orientation of the
principal axes evaluate the principal moments
of inertia with Eqs. (8.23) (8.24).
89
Example 8.7 Determining Principal Axes Moments
of Inertia

• Solution
• Determine Ix, Iy Ixy
• The moments of inertia of the triangular area are

90
Example 8.7 Determining Principal Axes Moments
of Inertia

• Solution
• Determine ?P
• From Eq. (8.26),
• and we obtain ?P 21.4.
• The principal axes corresponding
• to this value of ?P are

91
Example 8.7 Determining Principal Axes Moments
of Inertia

• Solution
• Calculate Ix Iy
• Substituting ?P 21.4 into Eqs. (8.23)
  (8.24),
• we obtain (with moments of inertia in m4)

92
Example 8.7 Determining Principal Axes Moments
of Inertia

• Critical Thinking
• The product of inertia corresponding to a set of
  principal axes is zero
• In this example, substituting ?P 21.4 into Eq.
  (8.25) confirms that Ixy 0

93
Example 8.8 Rotated Principal Axes

• The moments of inertia of the area in Fig. 8.21
  in
• terms of the xy coordinate system shown are
• Ix 22 m4, Iy 10 m4 Ixy 6 m4.
• (a) Determine Ix, Iy Ixy for ? 30.
• (b) Determine a set of principal axes the
• corresponding principal moments of inertia.

Fig. 8.21
94
Example 8.8 Rotated Principal Axes

• Strategy
• (a) Determine the moments of inertia in terms of
• the xy coordinate system by substituting
• ? 30 into Eqs. (8.23)(8.25).
• (b) The orientation of the principal axes is
• determined by solving Eq. (8.26) for . Once
  has
• been determined, the moments of inertia
  about
• the principal axes can be determined from
  Eqs.
• (8.23) (8.24).

95
Example 8.8 Rotated Principal Axes

• Solution
• (a) Determine Ix, Iy Ixy
• By setting ? 30 into Eqs. (8.23)(8.25), we
  obtain
• (with moments of inertia in m4)

96
Example 8.8 Rotated Principal Axes

• Solution
• (b) Determine ?P
• Substitute the moments of inertia in terms of the
  
• xy coordinate system into Eq. (8.26), yielding
• Thus, ?P ? 22.5

97
Example 8.8 Rotated Principal Axes

• Solution
• The principal axes corresponding to
• this value of ?P are

Calculate Ix Iy Substitute ?P ? 22.5 into
Eqs. (8.23) (8.24), obtaining the principal
moments of inertia Ix 24.5 m4, Iy 7.5 m4
98
Example 8.8 Rotated Principal Axes

• Critical Thinking
• Remember that the orientation of the principal
  axes is only determined within an arbitrary
  multiple of 90
• In this example, we chose to designate the axes
  as the positive x y axes but any of the 4
  choices in this figure is equally valid