1. Problem 9.194 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.)

2. 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
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).

3. 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
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.

4. k = I m Problem 9.194 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
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

5. 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

6. 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

7. 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

8. (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

9. 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

10. 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