PROBLEM 9.138
A section of sheet steel 0.03 in. thick is cut and bent into the
sheet metal machine component shown. Determine the mass
moment of inertia of the component with respect to each of the
coordinate axes. (The specific weight of steel is 490 lb/ft3.)
SOLUTION
First consider one triangular plate with local axes u-v. We have
332
2
490 lb/ft 1 4.5 in. 0.03 in.
1 ft 7.1332 10 lb s /ft
2 12 in./ft 12 in./ft
32.2 ft/s
mV tA
g
mass area area
m
I
tI I
A
PROBLEM 9.138 (Continued)
Plate 2 in x-z plane:
Note: By Eqn. 9.38,
66
‘
62
55.728 10 396.29 10
452.02 10 lb ft s
yuv
III
I
PROBLEM 9.139
A framing anchor is formed of 0.05-in.-thick galvanized steel. Determine the
mass moment of inertia of the anchor with respect to each of the coordinate
axes. (The specific weight of galvanized steel is
3
470 lb/ft .)
SOLUTION
First compute the mass of each component. We have
G.S.
mV tA
g
Then
3
32
12
470 lb/ft 1 ft
0.05 in. (2.25 3.5) in 12 in.
32.2 ft/s
m
PROBLEM 9.139 (Continued)
62 2 22
22 2
62 2
1(2006.14 10 lb s /ft)[(4.75) (2) ] in
18
21 1 ft
(2006.14 10 lb s /ft) 4.75 2 in
33 12 in.
62 2
22
62 2 2
1(2006.14 10 lb s /ft)(2 in.)
18
11 ft
(2006.14 10 lb s /ft) (2.25) 2 in
3 12 in.
PROBLEM 9.139 (Continued)
123
62 2 22
22 2
62 2
() () ()
1(3325.97 10 lb s /ft)[(2.25) (3.5) ] in
12
2.25 3.5 1 ft
(3325.97 10 lb s /ft) in
22 12 in.
zz z z
II I I
62 2
1(950.28 10 lb s /ft)(2.25 in.)
12
PROBLEM 9.140*
A farmer constructs a trough by welding a rectangular piece
of 2-mm-thick sheet steel to half of a steel drum. Knowing
that the density of steel is
3
7850 kg/m
and that the thickness
of the walls of the drum is 1.8 mm, determine the mass
moment of inertia of the trough with respect to each of the
coordinate axes. Neglect the mass of the welds.
SOLUTION
First compute the mass of each component. We have
ST ST
mV tA
Then
32
1(7850 kg/m )(0.002 m)(0.84 0.21) m
2.76948 kg
m
PROBLEM 9.140* (Continued)
1234
222
22
2
() () () ()
1(2.76948 kg)[(0.84) (0.21) ] m
12
0.84 0.21
(2.76948 kg) 0.285 m
22
yy y y y
II I I I
1234
2
2
2
222
() () () ()
10.84
(2.76948 kg)(0.84 m) (2.76948 kg) m
12 2
10.84
(10.62713 kg)[(0.84) 6(0.285) ] m (10.62713 kg) m
12 2
zz z z z
II I I I
PROBLEM 9.141
The machine element shown is fabricated from steel.
Determine the mass moment of inertia of the assembly with
respect to (a) the x axis, (b) the y axis, (c) the z axis. (The
density of steel is
3
7850 kg/m .)
SOLUTION
First compute the mass of each component. We have
ST
mV
Then
32
1(7850 kg/m )( (0.08 m) (0.04 m)]
6.31334 kg
m
PROBLEM 9.141 (Continued)
(b) 123
2
22
() () ()
1(6.31334 kg)(0.08 m)
2
1(0.59188 kg)(0.02 m) (0.59188 kg)(0.04 m)
2
yy y y
II I I
(c) 123
222 2
() () ()
1(6.31334 kg)[3(0.08) (0.04) ] m (6.31334 kg)(0.02 m)
12
zz z z
II I I
222 222
1(0.59188 kg)[3(0.02) (0.06) ] m (0.59188 kg)[(0.04) (0.03) ] m
12
To the Instructor:
The following formulas for the mass of inertia of a semicylinder are derived at this time for use in the
solutions of Problems 9.142 through 9.145.
From Figure 9.28:
Symmetry and the definition of the mass moment of inertia
2
()Irdm
imply
semicylinder cylinder
1
() ()
2
II
2
sc cyl
11
() 22
x
Ima
and
22
sc sc cyl
11
() () (3 )
212
yz
II maL
PROBLEM 9.142
Determine the mass moments of inertia and the radii of
gyration of the steel machine element shown with respect
to the x and y axes. (The density of steel is 7850 kg/m
3
.)
SOLUTION
First compute the mass of each component. We have
ST
mV
Then
33
1
(7850 kg/m )(0.24 0.04 0.14) m
10.5504 kg
m
PROBLEM 9.142 (Continued)
12345
() () () () ()
y
yy y y y
I
II II I
where 234 5
() () () |()|
yyy y
III I
Then 222
1(10.5504 kg)(0.24 0.14 ) m
12
y
I
PROBLEM 9.143
Determine the mass moment of inertia of the steel machine
element shown with respect to the x axis. (The density of
steel is
3
490 lb/ft .)
SOLUTION
First compute the mass of each component. We have
2
490 lb s ;
32.2 ft mV
g
Using Figure 9.28 for components 1 and 3 and the equation derived above (before the solution to
Problem 9.142) for a semicylinder, we have
123
() () ()
xx x x
II I I
PROBLEM 9.143 (Continued)
Half-cylinder about base,
2″ 2′
2′
2′
2
2
53
5
0.7639
3.3489 10 4.7124 10 12
1.43925 10
xx
x
x
IImd
I
I
Now combining all components and substituting for ,
344
490 1.26250 10 9.6015 10 2.1961 10
32.2
x
I
or
32
30.5 10 lb ft s
x
I
PROBLEM 9.144
Determine the mass moment of inertia of the steel machine
element shown with respect to the yaxis. (The density of
steel is
3
490 lb/ft .)
SOLUTION
First compute the mass of each component. We have
2
490 lb s ;
32.2 ft mV
g
Then
3
13
3.6 1.2 6 ft
12
m
PROBLEM 9.144 (Continued)
22‘
2
2
22 2
33
4.7124 10 1.8 1.6 5.2
3 4.7124 10
12 12 12 12
yy
y
IImd
I
PROBLEM 9.145
Determine the mass moment of inertia of the steel fixture
shown with respect to (a) the x axis, (b) the y axis, (c) the
z axis. (The density of steel is 7850 kg/m
3
.)
SOLUTION
First compute the mass of each component. We have
ST
mV
Then
33
1
7850 kg/m (0.08 0.05 0.160) m
5.02400 kg
m
PROBLEM 9.145 (Continued)
(b) 123
22
222 2
() () ()
10.080.16
(5.02400 kg)[(0.08) (0.16) ] m (5.02400 kg) m
12 2 2
yy y y
II I I
22
222 2
10.080.07
(1.67048 kg)[(0.08) (0.07) ] m (1.67048 kg) 0.05 m
12 2 2
To the instructor:
The following formulas for the mass moment of inertia of wires are derived or summarized at this time
for use in the solutions of Problems 9.146 through 9.148.
Slender Rod
2
1
0 (Figure 9.28)
12
xyz
IIImL
2
1(Sample Problem 9.9)
3
yz
II mL
Semicircle
Following the above arguments for a circle, We have
22
1
2
xz y
II ma Ima
PROBLEM 9.146
Aluminum wire with a weight per unit length of 0.033 lb/ft is used to
form the circle and the straight members of the figure shown.
Determine the mass moment of inertia of the assembly with respect to
each of the coordinate axes.
SOLUTION
First compute the mass of each component. We have
AL
1WW
mL
ggL
Then 12
32
2345 2
2
11 ft
0.033 lb/ft (2 16 in.) 12 in.
32.2 ft/s
8.5857 10 lb s /ft
11 ft
0.033 lb/ft 8 in. 12 in.
32.2 ft/s
0.6832 lb s /ft
m
mmmm
Using the equations given above and the parallel-axis theorem, we have