978-0073398242 Appendix B Solution Manual Part 6

subject Type Homework Help
subject Pages 9
subject Words 1210
subject Authors Brian Self, David Mazurek, E. Johnston, Ferdinand Beer, Phillip Cornwell

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page-pf1
PROBLEM B.29 (Continued)
123
() () ()
zz z z
II I I

22
62 22
2.25 1 ft
(950.28 10 lb s /ft) (3.5) in
212 in.








62 2
1(2006.14 10 lb s /ft)(4.75 in.)
18

[(33.322 99.967) (2.784 89.192)

z
page-pf2
PROBLEM B.30*
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.
page-pf3
PROBLEM B.30* (Continued)
1234
222
() () () ()
1(2.76948 kg)[(0.84) (0.21) ] m
12
22
yy y y y
II I I I




2
4

22
4



2
[(0.17302 0.57827) (1.05647 1.87463)

2
2
222
10.84
(10.62713 kg)[(0.84) 6(0.285) ] m (10.62713 kg) m
12 2
zz z z z









2
1(1.80282 kg)(0.285 m)
4



2
[(0.16285 0.48854) (1.05647 1.87463)
(0.03661) (0.03661 1.27207)] kg m


2
(0.65139 2.93110 0.03661 1.30868) kg m
or
2
4.93 kg m
z
I
page-pf4
PROBLEM B.31
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 .)
page-pf5
PROBLEM B.31 (Continued)
(b) 123
2
22
22
() () ()
1(6.31334 kg)(0.08 m)
2
1(0.59188 kg)(0.02 m) (0.59188 kg)(0.04 m)
2
1(0.39458 kg)(0.02 m ) (0.39458 kg)(0.04 m)
2
yy y y
II I I











32
[(20.20269) (0.11838 0.94701)
(0.07892 0.63133)] 10 kg m


32
(20.20269 1.06539 0.71025) 10 kg m

32
20.55783 10 kg m

or
32
20.6 10 kg m
y
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




222 222
1(0.39458 kg)[3(0.02) (0.04) ] m (0.03958 kg)[(0.04) (0.02) ] m
12




32
[(10.94312 2.52534) (0.23675 1.47970)
(0.09207 0.78916)] 10 kg m


32
(13.46846 1.71645 0.88123) 10 kg m

32
14.30368 10 kg m

or
32
14.30 10 kg m
z
I


page-pf6
To the Instructor:
The following formulas for the mass of inertia of a semicylinder are derived at this time for use in the
page-pf7
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM B.32
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
.)
Problem 9.144) for a semicylinder, we have
12345 23
() () () () () where() ()
xx x x x x x x
II I I I I I I
page-pf8
PROBLEM B.32 (Continued)
12345
() () () () ()
y
yy y y y
I
IIIII
15.38406 kg
xI
or
42.9 mm
x
k
20.1837860 kg m
y
I
y
page-pf9
PROBLEM B.33
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 .)
Problem 9.142) for a semicylinder, we have
123
() () ()
xx x x
II I I

page-pfa
PROBLEM B.33 (Continued)
Half-cylinder about base,
2" 2'
2'
2
2
5
1.43925 10
xx
x
IImd
I



22'
2
22
xx
IImd

12 12 12 12



3
4
2.1961 10
x
I

Now combining all components and substituting for ,

344
490 1.26250 10 9.6015 10 2.1961 10
x
I


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