1. Chapter 8 Distributed Forces: Moments of Inertia
CE 102 Statics
Chapter 8
Distributed Forces:
Moments of Inertia

2. Contents Introduction Moments of Inertia of an Area
Moment of Inertia of an Area by Integration
Polar Moment of Inertia
Radius of Gyration of an Area
Sample Problem 8.1
Sample Problem 8.2
Parallel Axis Theorem
Moments of Inertia of Composite Areas
Sample Problem 8.3
Sample Problem 8.4
Product of Inertia
Principal Axes and Principal Moments of Inertia
Sample Problem 8.5
Sample Problem 8.6
Mohr’s Circle for Moments and Products of Inertia
Sample Problem 8.7
Moment of Inertia of a Mass
Parallel Axis Theorem
Moment of Inertia of Thin Plates
Moment of Inertia of a 3D Body by Integration
Moment of Inertia of Common Geometric Shapes
Sample Problem 8.89
Moment of Inertia With Respect to an Arbitrary Axis
Ellipsoid of Inertia. Principle Axes of Axes of Inertia of a Mass

3. Introduction
Previously considered distributed forces which were proportional to the area or volume over which they act.
The resultant was obtained by summing or integrating over the areas or volumes.
The moment of the resultant about any axis was determined by computing the first moments of the areas or volumes about that axis.
Will now consider forces which are proportional to the area or volume over which they act but also vary linearly with distance from a given axis.
It will be shown that the magnitude of the resultant depends on the first moment of the force distribution with respect to the axis.
The point of application of the resultant depends on the second moment of the distribution with respect to the axis.
Current chapter will present methods for computing the moments and products of inertia for areas and masses.

4. Moment of Inertia of an Area
Consider distributed forces whose magnitudes are proportional to the elemental areas on which they act and also vary linearly with the distance of from a given axis.
Example: Consider a beam subjected to pure bending. Internal forces vary linearly with distance from the neutral axis which passes through the section centroid.
Example: Consider the net hydrostatic force on a submerged circular gate.

5. Moment of Inertia of an Area by Integration
Second moments or moments of inertia of an area with respect to the x and y axes,
Evaluation of the integrals is simplified by choosing dA to be a thin strip parallel to one of the coordinate axes.
For a rectangular area,
The formula for rectangular areas may also be applied to strips parallel to the axes,

6. Polar Moment of Inertia
The polar moment of inertia is an important parameter in problems involving torsion of cylindrical shafts and rotations of slabs.
The polar moment of inertia is related to the rectangular moments of inertia,

7. Radius of Gyration of an Area
Consider area A with moment of inertia Ix. Imagine that the area is concentrated in a thin strip parallel to the x axis with equivalent Ix.
kx = radius of gyration with respect to the x axis
Similarly,

8. Sample Problem 8.1 SOLUTION:
A differential strip parallel to the x axis is chosen for dA.
For similar triangles,
Determine the moment of inertia of a triangle with respect to its base.
Integrating dIx from y = 0 to y = h,

9. Sample Problem 8.2 SOLUTION:
An annular differential area element is chosen,
a) Determine the centroidal polar moment of inertia of a circular area by direct integration.
b) Using the result of part a, determine the moment of inertia of a circular area with respect to a diameter.
From symmetry, Ix = Iy,

10. Parallel Axis Theorem
Consider moment of inertia I of an area A with respect to the axis AA’
The axis BB’ passes through the area centroid and is called a centroidal axis.
parallel axis theorem

11. Parallel Axis Theorem
Moment of inertia IT of a circular area with respect to a tangent to the circle,
Moment of inertia of a triangle with respect to a centroidal axis,

12. Moments of Inertia of Composite Areas
The moment of inertia of a composite area A about a given axis is obtained by adding the moments of inertia of the component areas A1, A2, A3, ... , with respect to the same axis.

13. Moments of Inertia of Composite Areas

14. Sample Problem 8.3 SOLUTION:
Determine location of the centroid of composite section with respect to a coordinate system with origin at the centroid of the beam section.
Apply the parallel axis theorem to determine moments of inertia of beam section and plate with respect to composite section centroidal axis.
The strength of a W14x38 rolled steel beam is increased by attaching a plate to its upper flange.
Determine the moment of inertia and radius of gyration with respect to an axis which is parallel to the plate and passes through the centroid of the section.
Calculate the radius of gyration from the moment of inertia of the composite section.

15. Sample Problem 8.3 SOLUTION:
Determine location of the centroid of composite section with respect to a coordinate system with origin at the centroid of the beam section.

16. Sample Problem 8.3
Apply the parallel axis theorem to determine moments of inertia of beam section and plate with respect to composite section centroidal axis.
Calculate the radius of gyration from the moment of inertia of the composite section.

17. Sample Problem 8.4 SOLUTION:
Compute the moments of inertia of the bounding rectangle and half-circle with respect to the x axis.
The moment of inertia of the shaded area is obtained by subtracting the moment of inertia of the half-circle from the moment of inertia of the rectangle.
Determine the moment of inertia of the shaded area with respect to the x axis.

18. Sample Problem 8.4 SOLUTION:
Compute the moments of inertia of the bounding rectangle and half-circle with respect to the x axis.
Rectangle:
Half-circle: moment of inertia with respect to AA’,
moment of inertia with respect to x’,
moment of inertia with respect to x,

19. Sample Problem 8.4
The moment of inertia of the shaded area is obtained by subtracting the moment of inertia of the half-circle from the moment of inertia of the rectangle.

20. Product of Inertia Product of Inertia:
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:

21. Principal Axes and Principal Moments of Inertia
Given
we wish to determine moments and product of inertia with respect to new axes x’ and y’.
The change of axes yields
The equations for Ix’ and Ix’y’ are the parametric equations for a circle,
Note:
The equations for Iy’ and Ix’y’ lead to the same circle.

22. Principal Axes and Principal Moments of Inertia
At the points A and B, Ix’y’ = 0 and Ix’ is a maximum and minimum, respectively.
The equation for Qm 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.

23. Sample Problem 8.5 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.
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.

24. Sample Problem 8.5 SOLUTION:
Determine the product of inertia using direct integration with the parallel axis theorem on vertical differential area strips
Integrating dIx from x = 0 to x = b,

25. Sample Problem 8.5
Apply the parallel axis theorem to evaluate the product of inertia with respect to the centroidal axes.
With the results from part a,

26. Sample Problem 8.6 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 ).
For the section shown, the moments of inertia with respect to the x and y axes are Ix = 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.

27. Sample Problem 8.6 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,
Note that the product of inertia with respect to centroidal axes parallel to the xy axes is zero for each rectangle.

28. Sample Problem 8.6
Determine the orientation of the principal axes (Eq. 9.25) and the principal moments of inertia (Eq ).

29. 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,
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.

30. Sample Problem 8.7 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.
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
Based on the circle, evaluate the moments and product of inertia with respect to the x’y’ axes.

31. Sample Problem 8.7 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.

32. Sample Problem 8.7
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 Q = 2(60o) = 120o. The angle that CX’ forms with the x’ axes is f = 120o o = 72.4o.

33. Moment of Inertia of a Mass
Angular acceleration about the axis AA’ of the small mass Dm due to the application of a couple is proportional to r2Dm.
r2Dm = moment of inertia of the mass Dm with respect to the axis AA’
For a body of mass m the resistance to rotation about the axis AA’ is
The radius of gyration for a concentrated mass with equivalent mass moment of inertia is

34. Moment of Inertia of a Mass
Moment of inertia with respect to the y coordinate axis is
Similarly, for the moment of inertia with respect to the x and z axes,
In SI units,
In U.S. customary units,

35. Parallel Axis Theorem
For the rectangular axes with origin at O and parallel centroidal axes,
Generalizing for any axis AA’ and a parallel centroidal axis,

36. Moments of Inertia of Thin Plates
For a thin plate of uniform thickness t and homogeneous material of density r, the mass moment of inertia with respect to axis AA’ contained in the plate is
Similarly, for perpendicular axis BB’ which is also contained in the plate,
For the axis CC’ which is perpendicular to the plate,

37. Moments of Inertia of Thin Plates
For the principal centroidal axes on a rectangular plate,
For centroidal axes on a circular plate,

38. Moments of Inertia of a 3D Body by Integration
Moment of inertia of a homogeneous body is obtained from double or triple integrations of the form
For bodies with two planes of symmetry, the moment of inertia may be obtained from a single integration by choosing thin slabs perpendicular to the planes of symmetry for dm.
The moment of inertia with respect to a particular axis for a composite body may be obtained by adding the moments of inertia with respect to the same axis of the components.

39. Moments of Inertia of Common Geometric Shapes

40. Sample Problem 8.8 SOLUTION:
With the forging divided into a prism and two cylinders, compute the mass and moments of inertia of each component with respect to the xyz axes using the parallel axis theorem.
Add the moments of inertia from the components to determine the total moments of inertia for the forging.
Determine the moments of inertia of the steel forging with respect to the xyz coordinate axes, knowing that the specific weight of steel is 490 lb/ft3.

41. Sample Problem 8.8 SOLUTION: cylinders
Compute the moments of inertia of each component with respect to the xyz axes.
cylinders

42. Sample Problem 8.8 prism (a = 2 in., b = 6 in., c = 2 in.):
Add the moments of inertia from the components to determine the total moments of inertia.

43. Moment of Inertia With Respect to an Arbitrary Axis
IOL = moment of inertia with respect to axis OL
Expressing in terms of the vector components and expanding yields
The definition of the mass products of inertia of a mass is an extension of the definition of product of inertia of an area

44. Ellipsoid of Inertia. Principal Axes of Inertia of a Mass
Assume the moment of inertia of a body has been computed for a large number of axes OL and that point Q is plotted on each axis at a distance
The locus of points Q forms a surface known as the ellipsoid of inertia which defines the moment of inertia of the body for any axis through O.
x’,y’,z’ axes may be chosen which are the principal axes of inertia for which the products of inertia are zero and the moments of inertia are the principal moments of inertia.

45. Problem 8.9 y y2 = mx Determine by direct integration
y1 = kx2 y
Determine by direct integration
the moments of inertia of the
shaded area with respect to
the x and y axes.

46. Solving Problems on Your Own
Determine by direct integration
the moments of inertia of the
shaded area with respect to
the x and y axes.
Problem 8.9 b
y2 = mx a x
y1 = kx2 y
1. Calculate the moments of inertia Ix and Iy. These moments of
inertia are defined by:
Ix = y2 dA and Iy = x2 dA
Where dA is a differential element of area dx dy.
1a. To compute Ix choose dA to be a thin strip parallel to the x
axis. All the the points of the strip are at the same distance y from
the x axis. The moment of inertia dIx of the strip is given by y2 dA.

47. Solving Problems on Your Own
Determine by direct integration
the moments of inertia of the
shaded area with respect to
the x and y axes.
Problem 8.9 b
y2 = mx a x
y1 = kx2 y
1b. To compute Iy choose dA to be a thin strip parallel to the y
axis. All the the points of the strip are at the same distance x from
the y axis. The moment of inertia dIy of the strip is given by x2 dA.
1c. Integrate dIx and dIy over the whole area.

48. m = y1 = k x2: b = k a2 k = y1 = k x2 then: x1 = y1 1/2
Problem 8.9 Solution b
y2 = mx a x
y1 = kx2 y
Determine m and k:
At x = a, y2 = b: b = m a
m =
y1 = k x2: b = k a2
k = b a b a2
Express x in terms of y1 and y2:
y1 = k x then: x1 = y1
y2 = mx then x2 = y2 1
k1/2 m
1/2

49. dA = ( y2 - y1 ) dx = ( m x _ k x2 ) dx
Problem 8.9 Solution y a
To compute Iy choose dA to
be a thin strip parallel to the y
axis. b y2
dA = ( y2 - y1 ) dx
= ( m x _ k x2 ) dx y1 x x dx a
Iy = x2 dA = x2 ( m x _ k x2 ) dx
= ( m x3 _ k x4 ) dx = [ m x4 _ k x5 ]
= m a4 _ k a5 = b a3 a 1 4 1 5 a 1 4 1 5 1 20
Iy = b a3/ 20
Substituting k = , m = b a2 a

50. dA = ( x1 - x2 ) dy = ( y _ y ) dy 1/2 Ix = y2 dA = y2 ( y _ y ) dy
Problem 8.9 Solution y a
To compute Ix choose dA to
be a thin strip parallel to the x
axis. x2 dy b
dA = ( x1 - x2 ) dy
= ( y _ y ) dy 1
k1/2 m
1/2 y x x1
Ix = y2 dA = y2 ( y _ y ) dy
= ( y _ y3 ) dy = [ y _ y4 ]
= b _ b4 = a b3 1
k1/2
1/2 m
5/2 b 2 7
7/2 4 28
Ix = a b3/ 28
Substituting k = , m = b a2 a

51. Problem 8.10 For the 5 x 3 x -in. angle cross
section shown, use Mohr’s circle
to determine (a) the moments of
inertia and the product of inertia
with respect to new centroidal
axes obtained by rotating the x
and y axes 30o clockwise, (b) the
orientation of the principal axes
through the centroid and the
corresponding values of the
moments of inertia. 1 2
3 in
5 in
0.5 in
1.75 in x
0.5 in
0.75 in y
L5x3x 1 2

52. Solving Problems on Your Ownx
0.5 in
0.75 in
L5x3x 1 2
Solving Problems on Your Own
For the 5 x 3 x -in. angle cross section
shown, use Mohr’s circle to determine
(a) the moments of inertia and the
product of inertia with respect to new
centroidal axes obtained by rotating
the x and y axes 30o clockwise, (b) the
orientation of the principal axes through
the centroid and the corresponding
values of the moments of inertia. 1 2
1. Draw Mohr’s circle. Mohr’s circle is completely defined by the
quantities R and IAVE which represent, respectively, the radius of
the circle and the distance from the origin O to the center C of the
circle. These quantities can be obtained if the moments and
product of inertia are known for a given orientation of the axes.

53. Solving Problems on Your Ownx
0.5 in
0.75 in
L5x3x 1 2
Solving Problems on Your Own
For the 5 x 3 x -in. angle cross section
shown, use Mohr’s circle to determine
(a) the moments of inertia and the
product of inertia with respect to new
centroidal axes obtained by rotating
the x and y axes 30o clockwise, (b) the
orientation of the principal axes through
the centroid and the corresponding
values of the moments of inertia. 1 2
2. Use the Mohr’s circle to determine moments of inertia of
rotated axes. As the coordinate axes x-y are
rotated through an angle q, the associated
rotation of the diameter of Mohr’s circle is
equal to 2q in the same sense
(clockwise or counterclockwise). x y 2q
Ix , Iy
Ixy x’ y’

54. Solving Problems on Your Ownx
0.5 in
0.75 in
L5x3x 1 2
Solving Problems on Your Own
For the 5 x 3 x -in. angle cross section
shown, use Mohr’s circle to determine
(a) the moments of inertia and the
product of inertia with respect to new
centroidal axes obtained by rotating
the x and y axes 30o clockwise, (b) the
orientation of the principal axes through
the centroid and the corresponding
values of the moments of inertia. 1 2
2. Use the Mohr’s circle to determine the orientation of principal
axes, and the principal moments of inertia. Points A and B where
the circle intersects the horizontal axis
represent the principal moments of inertia.
The orientation of the principal axes is
determined by the angle 2qm.
Ixy
2qm x B A
Ix, Iy y

55. Ix = 9.45 in4, Iy = 2.58 in4 Ixy = (Ixy)1 + (Ixy)2
Problem Solution
3 in
5 in
0.5 in
1.75 in x
0.5 in
0.75 in
L5x3x 1 2
From Fig. 9.13A:
Ix = 9.45 in4, Iy = 2.58 in4
Product of inertia Ixy:
Ixy = (Ixy)1 + (Ixy)2
For each rectangle Ixy = Ix’y’ + x y A
and Ix’y’ = 0 (symmetry)
Thus: Ixy = S x y A y
0.5 in 2
A, in2 x, in y in x y A, in4
Ixy = S x y A = -2.81
1 in x
1.5 in 1
0.75 in

56. OC = Iave = (Ix + Iy) OC = (9.45 + 2.58) = 6.02 in4
Draw Mohr’s circle.
Problem Solution
The Mohr’s circle is defined by the
diameter XY where X(9.45, -2.81)
and Y(2.58, 2.81).
The coordinate of the center C and
the radius R are calculated by:
OC = Iave = (Ix + Iy)
OC = ( ) = 6.02 in4
R = [ (Ix - Iy)]2 + Ixy
R = [ ( )]2 + (-2.82)2
R = 4.44 in4 1 2 1 2 1 2
X (9.45, -2.81)
Ix, Iy (in4)
Ixy (in4)
Y (2.58, +2.81) C O
2qm F 1 2
tan 2qm = = FX CF
2.81
qm = 19.6o

57. Ix’ = OE = OC _ CE = 6.02 _ 4.44 cos 80.7o Ix’ = 5.30 in4y
Problem Solution y’
Use the Mohr’s circle to determine
moments of inertia of rotated axes.
(a) The coordinates of X’ and Y’ give the
moments and product of inertia with
respect to the x’y’ axes. x
30o
Ix’ = OE = OC _ CE = 6.02 _ 4.44 cos 80.7o
Ix’ = 5.30 in4
Iy’ = OD = OC + CD = cos 80.7o
Iy’ = 6.73 in4 x’
Ixy (in4)
6.02 in4 Y’ Y
R = 4.44 in4 E D O C
Ix, Iy (in4)
Ix’y’ = EX’ = _ 4.44 sin 80.7o
80.7o X
39.3o
Ix’y’ = _ 4.38 in4 X’
2q = 60o

58. y
Problem Solution b
Use the Mohr’s circle to determine
the orientation of principal axes, and
the principal moments of inertia. a
qm = 19.6o
(b) The principal axes are obtained by
rotating the xy axes through angle qm. x
tan 2qm = = FX CF
2.81
qm = 19.6o
X (9.45, -2.81)
Ix, Iy (in4)
Ixy (in4)
Y (2.58, +2.81) C O
2qm F
The corresponding moments of inertia
are Imax and Imin : _ _
Imax, min = OC + R =
Imin
Imax
Imax = in4
Imin = in4
a axis corresponds to Imax
b axis corresponds to Imin
R = 4.44 in4

59. Problem 8.11 A section of sheet steel 2 mm y
thick is cut and bent into the
machine component shown.
Knowing that the density of
steel is 7850 kg/m3,
determine the mass moment
of inertia of the component
with respect to (a) the x axis,
(b) the y axis, (c) the z axis. x y z
120 mm
150 mm

60. Solving Problems on Your Ownx y z
120 mm
150 mm
Solving Problems on Your Own
A section of sheet steel 2 mm
thick is cut and bent into the
machine component shown.
Knowing that the density of
steel is 7850 kg/m3,
determine the mass moment
of inertia of the component
with respect to (a) the x axis,
(b) the y axis, (c) the z axis.
1. Compute the mass moments of inertia of a composite body with
respect to a given axis.
1a. Divide the body into sections. The sections should have a
simple shape for which the centroid and moments of inertia can
be easily determined (e.g. from Fig in the book).

61. Solving Problems on Your Own A section of sheet steel 2 mm x y z
120 mm
150 mm
Solving Problems on Your Own
A section of sheet steel 2 mm
thick is cut and bent into the
machine component shown.
Knowing that the density of
steel is 7850 kg/m3,
determine the mass moment
of inertia of the component
with respect to (a) the x axis,
(b) the y axis, (c) the z axis.
1b. Compute the mass moment of inertia of each section. The
moment of inertia of a section with respect to the given axis is
determined by using the parallel-axis theorem:
I = I + m d2
Where I is the moment of inertia of the section about its own
centroidal axis, I is the moment of inertia of the section about the
given axis, d is the distance between the two axes, and m is the
section’s mass.

62. Solving Problems on Your Ownx y z
120 mm
150 mm
Solving Problems on Your Own
A section of sheet steel 2 mm
thick is cut and bent into the
machine component shown.
Knowing that the density of
steel is 7850 kg/m3,
determine the mass moment
of inertia of the component
with respect to (a) the x axis,
(b) the y axis, (c) the z axis.
1c. Compute the mass moment of inertia of the whole body. The
moment of inertia of the whole body is determined by adding the
moments of inertia of all the sections.

63. Divide the body into sections.
Problem Solution x y z
120 mm
150 mm
Divide the body into sections. x y z x’ y’ z’
y’’
x’’
z’’ 2 1 3

64. m2 = r V2 = (7850 kg/m3)(0.002 m)(0.150 m)(0.120 m) = 0.2826 kgx y z
120 mm
150 mm
Problem Solution x y z x’ y’ z’
y’’
x’’
z’’ 2 1 3
Computation of Masses:
Section 1: m1 = r V1 = (7850 kg/m3)(0.002 m)(0.300 m)2 = kg
Section 2:
m2 = r V2 = (7850 kg/m3)(0.002 m)(0.150 m)(0.120 m) = kg
Section 3: m3 = m2 = kg

65. x y z
120 mm
150 mm
Problem Solution x y z x’ y’ z’
y’’
x’’
z’’ 2 1 3
Compute the moment of inertia of each section.
(a) Mass moment of inertia with respect to the x axis.
Section 1: (Ix)1 = (1.413) (0.30)2 = 1.06 x 10-2 kg . m2 1 12
Section 2: (Ix)2 = (Ix’)2 + m d2
(Ix)2 = (0.2826) (0.120) 2 + (0.2826)( )
(Ix)2 = 7.71 x 10-3 kg . m2 1 12
Section 3: (Ix)3 = (Ix)2 = 7.71 x 10-3 kg . m2

66. x y z
120 mm
150 mm
Problem Solution x y z x’ y’ z’
y’’
x’’
z’’ 2 1 3
Compute the moment of inertia of the whole area.
For the whole body:
Ix = (Ix)1 + (Ix)2 + (Ix)3
Ix = x x x 10-3 = 2.60 x 10-2 kg . m2
Ix = 26.0 x 10-3 kg . m2

67. x y z
120 mm
150 mm
Problem Solution x y z x’ y’ z’
y’’
x’’
z’’ 2 1 3
Compute the moment of inertia of each section.
(b) Mass moment of inertia with respect to the y axis.
Section 1: (Iy)1 = (1.413) ( ) = 2.12 x 10-2 kg . m2 1 12
Section 2: (Ix)2 = (Ix’)2 + m d2
(Iy)2 = (0.2826) (0.150)2 + (0.2826)( )
(Iy)2 = 8.48 x 10-3 kg . m2 1 12
Section 3: (Iy)3 = (Iy)2 = 8.48 x 10-3 kg . m2

68. x y z
120 mm
150 mm
Problem Solution x y z x’ y’ z’
y’’
x’’
z’’ 2 1 3
Compute the moment of inertia of the whole area.
For the whole body:
Iy = (Iy)1 + (Iy)2 + (Iy)3
Iy = 2.12 x x x 10-3 = 3.82 x 10-2 kg . m2
Iy = 38.2 x 10-3 kg . m2

69. x y z
120 mm
150 mm
Problem Solution x y z x’ y’ z’
y’’
x’’
z’’ 2 1 3
Compute the moment of inertia of each section.
(c) Mass moment of inertia with respect to the z axis.
Section 1: (Iz)1 = (1.413) (0.30)2 = x 10-2 kg . m2 1 12
Section 2: (Iz)2 = (Iz’)2 + m d2
(Iz)2 = (0.2826) ( ) + (0.2826)( )
(Iz)2 = 3.48 x 10-3 kg . m2 1 12
Section 3: (Iz)3 = (Iz)2 = 3.48 x 10-3 kg . m2

70. x y z
120 mm
150 mm
Problem Solution x y z x’ y’ z’
y’’
x’’
z’’ 2 1 3
Compute the moment of inertia of the whole area.
For the whole body:
Iz = (Iz)1 + (Iz)2 + (Iz)3
Iz = 1.06 x x x 10-3 = x 10-2 kg . m2
Iz = x 10-3 kg . m2

71. Problem 8.12 y Determine the moment of inertia and the radius of
15 mm x y z
40 mm
50 mm
60 mm
45 mm
38 mm
Determine the moment of
inertia and the radius of
gyration of the steel machine
element shown with respect to
the x axis. (The density of
steel is 7850 kg/m3.)

72. Solving Problems on Your Own
Determine the moment of
inertia and the radius of
gyration of the steel machine
element shown with respect to
the x axis. (The density of
steel is 7850 kg/m3.)
Problem 8.12
15 mm x y z
40 mm
50 mm
60 mm
45 mm
38 mm
1. Compute the mass moments of inertia of a composite body with
respect to a given axis.
1a. Divide the body into sections. The sections should have a
simple shape for which the centroid and moments of inertia can
be easily determined (e.g. from Fig in the book).

73. Solving Problems on Your Own
15 mm x y z
40 mm
50 mm
60 mm
45 mm
38 mm
Solving Problems on Your Own
Determine the moment of
inertia and the radius of
gyration of the steel machine
element shown with respect to
the x axis. (The density of
steel is 7850 kg/m3.)
Problem 8.12
1b. Compute the mass moment of inertia of each section. The
moment of inertia of a section with respect to the given axis is
determined by using the parallel-axis theorem:
I = I + m d2
Where I is the moment of inertia of the section about its own
centroidal axis, I is the moment of inertia of the section about the
given axis, d is the distance between the two axes, and m is the
section’s mass.

74. k = I m Problem 8.12 Solving Problems on Your Own y
15 mm x y z
40 mm
50 mm
60 mm
45 mm
38 mm
Solving Problems on Your Own
Determine the moment of
inertia and the radius of
gyration of the steel machine
element shown with respect to
the x axis. (The density of
steel is 7850 kg/m3.)
Problem 8.12
1c. Compute the mass moment of inertia of the whole body. The
moment of inertia of the whole body is determined by adding the
moments of inertia of all the sections.
2. Compute the radius of gyration. The radius of gyration k of a
body is defined by:
k = I m

75. Divide the body into sections.
15 mm x y z
40 mm
50 mm
60 mm
45 mm
38 mm
Problem Solution
Divide the body into sections. x y z x y z x z y

76. m = r V V = (0.15 m)(0.09 m)(0.03 m) V = 4.05 x 10-4 m3y z
40 mm
50 mm
60 mm
45 mm
38 mm
Problem Solution
Compute the mass moment
of inertia of each section.
Rectangular prism:
m = r V
V = (0.15 m)(0.09 m)(0.03 m)
V = 4.05 x m3
m = (7850 kg/m3)(4.05x10-4 m3)
m = 3.18 kg x y z
(Ix) = ( Ix’) + m d2
(Ix) = (3.18 kg)[(0.15 m)2+(0.03 m)2]
+ (3.18 kg) (0.075 m)2
(Ix) = x kg . m2 1 12 x’
75 mm

77. 15 mm x y z
40 mm
50 mm
60 mm
45 mm
38 mm
Problem Solution
Semicircular cylinder:
m = r V
V = p (0.045 m)2 (0.04 m)
V = 1.27 x 10-4 m3
m = (7850 kg/m3)(1.27x10-4 m3)
m = 1.0 kg 1 2 x y z x’
130 mm
15 mm
x’’ 4r 3p
4.45
= 19.1 mm =
(centroidal axis) C
40 mm

78. (1.0 kg) [3 (0.045 m)2 + (0.04 m)2] = Ix’’ + (1.0 kg)(0.0191 m)2
Problem Solution x y z x’
130 mm
15 mm
x’’ 4r 3p
4.45
= 19.1 mm =
(centroidal axis) C
40 mm
Ix’ = Ix’’ + m d2
(1.0 kg) [3 (0.045 m)2 + (0.04 m)2] = Ix’’ + (1.0 kg)( m)2
Ix’’ = x kg . m2
Ix = ( Ix’’) + m d2 = x kg . m2
+ (1.0 kg)[(0.13 m)2 + (0.015 m m)2]
Ix = x kg . m2 1 12

79. 15 mm x y z
40 mm
50 mm
60 mm
45 mm
38 mm
Problem Solution
Circular cylindrical hole:
m = r V
V = p (0.038 m)2 (0.03 m)
V = x 10-4 m3
m = (7850 kg/m3)(1.36x10-4 m3)
m = kg
(Ix) = ( Ix’) + m d2
(Ix) = (1.07 kg) [ 3 (0.038 m)2
+ (0.03 m)2 ]
+ (1.07 kg) (0.06 m)2
(Ix) = x kg . m2 x z y x’
60 mm 1 12

80. 15 mm x y z
40 mm
50 mm
60 mm
45 mm
38 mm
Problem Solution
Compute the mass moment
of inertia of the whole body.
For the entire body, adding the
values obtained:
Ix = x x x 10-3
Ix = 3.81 x kg . m2
Compute the radius of gyration.
The mass of the entire body:
m = = 3.11 kg
Radius of gyration:
k = =
k = m I m
3.81 x kg . m2
3.11 kg