1. Problem 9.192 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

2. 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).

3. 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.

4. 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.

5. 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

6. 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

7. 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

8. 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

9. 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

10. 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

11. 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

12. 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