978-0077687342 Appendix B Part 3

subject Type Homework Help
subject Pages 14
subject Words 1166
subject Authors Brian Self, E. Johnston, Ferdinand Beer, Phillip Cornwell

Unlock document.

This document is partially blurred.
Unlock all pages and 1 million more documents.
Get Access
page-pf1
PROBLEM B.25
A 2-mm thick piece of sheet steel 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 each of the coordinate axes.
SOLUTION
z
page-pf2
PROBLEM B.25 (Continued)
z
page-pf3
PROBLEM B.26
A 2-mm thick piece of sheet steel is cut and bent into the
machine component shown. Knowing that the density of steel is
7850 kg/m
3
, determine the mass moment of inertia of the
component with respect to each of the coordinate axes.
SOLUTION
x
page-pf4
PROBLEM B.26 (Continued)
z
page-pf5
PROBLEM B.27
A subassembly for a model airplane is fabricated from three pieces of 1.5-
mm plywood. Neglecting the mass of the adhesive used to assemble the
three pieces, determine the mass moment of inertia of the subassembly
with respect to each of the coordinate axes. (The density of the plywood
is
3
780 kg/m .)
SOLUTION
x
page-pf6
PROBLEM B.27 (Continued)
z
page-pf7
PROBLEM B.28
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
18
v
For composite area:
page-pf8
PROBLEM B.28 (Continued)
Plate 2 in x-z plane:
Note: By Eqn. 9.38,
y
z
z
page-pf9
PROBLEM B.29
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
62
2006.14 10 lb s /ft

Using Figure 9.28 for components 1 and 2 and the equations derived above for component 3, we have
123
62 2
() () ()
1(3325.97 10 lb s /ft)(3.5 in.)
xx x x
II I I


page-pfa
PROBLEM B.29 (Continued)
62 2 22
1(2006.14 10 lb s /ft)[(4.75) (2) ] in
18
 
y
page-pfb
PROBLEM B.29 (Continued)
123
62 2 22
() () ()
1(3325.97 10 lb s /ft)[(2.25) (3.5) ] in
zz z z
II I I

z
page-pfc
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.
SOLUTION
34
2
1.80282 kg


Using Figure 9.28 for component 1 and the equations derived above for components 2 through 4, we have
34
() ()
xx
II
x
page-pfd
PROBLEM B.30* (Continued)
1234
222
() () () ()
1(2.76948 kg)[(0.84) (0.21) ] m
yy y y y
II I I I
y
1234
() () () ()
zz z z z
II I I I
1(1.80282 kg)(0.285 m)

z
page-pfe
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 .)
SOLUTION
First compute the mass of each component. We have
ST
mV
3
Using Figure 9.28 and the parallel-axis theorem, we have
x
page-pff
PROBLEM B.31 (Continued)
(b) 123
() () ()
yy y y
II I I
[(20.20269) (0.11838 0.94701)

y
(c) 123
() () ()

zz z z
II I I
1(0.59188 kg)[3(0.02) (0.06) ] m (0.59188 kg)[(0.04) (0.03) ] m

z
page-pf10
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.
page-pf11
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
.)
SOLUTION
x
page-pf12
PROBLEM B.32 (Continued)
y
page-pf13
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 .)
SOLUTION
2"
12 12 12 12
x




page-pf14
PROBLEM B.33 (Continued)
Half-cylinder about base,
2" 2'
2
2
53
0.7639
xx
IImd



x

Trusted by Thousands of
Students

Here are what students say about us.

Copyright ©2022 All rights reserved. | CoursePaper is not sponsored or endorsed by any college or university.