Eighth Edition - Welcome to APL100 - APL100apl100.wdfiles.com/local--files/notes/Inertia chapter...
Transcript of Eighth Edition - Welcome to APL100 - APL100apl100.wdfiles.com/local--files/notes/Inertia chapter...
VECTOR MECHANICS FOR ENGINEERS:
STATICS
Eighth Edition
Ferdinand P. Beer
E. Russell Johnston, Jr.
Lecture Notes:
J. Walt Oler
Texas Tech University
CHAPTER
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9 Distributed Forces:
Moments of Inertia
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9 - 2
Moment of Inertia of a Mass
• Angular acceleration about the axis AA’ of the
small mass m due to the application of a
couple is proportional to r2 m.
r2 m = moment of inertia of the
mass m with respect to the
axis AA’
• For a body of mass m the resistance to rotation
about the axis AA’ is
inertiaofmomentmassdmr
mrmrmrI
2
23
22
21
• The radius of gyration for a concentrated mass
with equivalent mass moment of inertia is
m
IkmkI 2
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9 - 3
Moment of Inertia of a Mass
• Moment of inertia with respect to the y coordinate
axis is
dmxzdmrI y222
• Similarly, for the moment of inertia with respect to
the x and z axes,
dmyxI
dmzyI
z
x
22
22
• In SI units, 22 mkgdmrI
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I is the Inertia matrix which has a size of 3 x3 and is
symmetric
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Vector Mechanics for Engineers: Statics
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© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
Vector Mechanics for Engineers: Statics
Eig
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© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
Vector Mechanics for Engineers: Statics
Eig
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© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
Vector Mechanics for Engineers: Statics
Eig
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© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
Vector Mechanics for Engineers: Statics
Eig
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© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
Vector Mechanics for Engineers: Statics
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•Since r is independent of inclination of
the coordinate axes and depends only on
the position of the origin. Therefore
sum of moment of inertia at a point in
space for a given body is an invariant
with respect to rotation of axes.
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Mirror Symmetry & mixed moments of Inertia
• Ixz of each half gives the contribution of same magnitude
but opposite sign.
• This conclusion is also true for Ixy .
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• Similarly Ixy = 0
• If two axes form a plane of symmetry for the mass
distribution of a body, the products of inertia having as an
index the coordinate that is normal to the plane of
symmetry will be zero.
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Body of Revolution
9 - 16
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Body of Revolution
• Let z axis coincide with the axis of symmetry.
This is true all possible xy axis formed by rotating
the z axis at O.
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Radius of gyration
• kx, ky and kz are radius of gyration.
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9 - 19
Parallel Axis Theorem
•For the
rectangular axes
with origin at O
and parallel
centroidal axes
x’y’z’,
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9 - 20
The coordinates of any point are (x,y,z)
xbar, ybar, zbar (the quantities with bar
on them give the coordinates of CM
from O
x’, y’, z’ are distances of any point
fromCM
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9 - 21
2 2
x
2 2
I y z dm
y y z z dm
x
2 2 2 2
I
y z dm 2y y dm 2z z dm y z dm
22 zymII xx
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9 - 22
x
2 2 2 2
I
y z dm 2y y dm 2z z dm y z dm
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9 - 23
y dm 0 z dm 0
Definition of CM
Ycm and Zcm are the coordinates of CM from origin. Here CM is
at the origin itself.
cmy dm y
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9 - 24
Parallel Axis Theorem
22 zymII xx
22
22
yxmII
xzmII
zz
yy
• Generalizing for any axis AA’ and a
parallel centroidal axis,
2mdII
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Parallel axis for product moments
9 - 25
'xy x yI I mx y
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9 - 26
Moments of Inertia of Thin Plates
• For a thin plate of uniform thickness t and homogeneous
material of density , the mass moment of inertia with
respect to axis AA’ contained in the plate is
areaAA
AA
It
dArtdmrI
,
22
• Similarly, for perpendicular axis BB’ which is also
contained in the plate,
areaBBBB ItI ,
• For the axis CC’ which is perpendicular to the plate,
BBAA
areaBBareaAAareaCCC
II
IItJtI ,,,
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9 - 27
t is mass per unit area
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9 - 28
Moments of Inertia of Thin Plates
• For the principal centroidal axes on a rectangular plate,
21213
121
, mabatItI areaAAAA
21213
121
, mbabtItI areaBBBB
22121
,, bamIII massBBmassAACC
• For centroidal axes on a circular plate,
2414
41
, mrrtItII areaAABBAA
221 mrIII BBAACC
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9 - 29
Moments of Inertia of a 3D Body by Integration
• Moment of inertia of a homogeneous body
is obtained from double or triple
integrations of the form
dVrI 2
• For bodies with two planes of symmetry,
the moment of inertia may be obtained
from a single integration by choosing thin
slabs perpendicular to the planes of
symmetry for dm.
• The moment of inertia with respect to a
particular axis for a composite body may
be obtained by adding the moments of
inertia with respect to the same axis of the
components.
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Vector Mechanics for Engineers: Statics
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© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
Vector Mechanics for Engineers: Statics
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Vector Mechanics for Engineers: Statics
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Part b
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Part c
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9 - 35
Moments of Inertia of Common Geometric Shapes
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9 - 36
Moments of Inertia of Common Geometric Shapes
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Sample Problem 9.12
9 - 37
Determine the moments of inertia of
the steel forging with respect to the
xyz coordinate axes, knowing that
the density of steel is 7850 kg/m3.
SOLUTION:
• With the forging divided into a prism and
two cylinders, compute the mass and
moments of inertia of each component
with respect to the xyz axes using the
parallel axis theorem.
• Add the moments of inertia from the
components to determine the total moments
of inertia for the forging.
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Sample Problem 9.12
9 - 38
Determine the moments of inertia of the steel
forging with respect to the xyz coordinate
axes, knowing that the density of steel is 7850
kg/m3.
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Sample Problem 9.12
9 - 39
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Sample Problem 9.12
9 - 40
•With the forging divided into a prism
and two cylinders, compute the mass
and moments of inertia of each
component with respect to the xyz axes
using the parallel axis theorem.
•Add the moments of inertia from the
components to determine the total
moments of inertia for the forging.
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Sample Problem 9.12
9 - 41
kg161
)kg/m7850)(m1047301(
m104731
m)0750(m)0250(
:cylindereach
334-
34-
2
.
.m
.
..V
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Sample Problem 9.12
9 - 42
23
2
1000502
100025
21
22
21
mkg102363
161m161
.
..
ymmaIx
cylinders :mm50,mm5.62,mm75,mm25 yxLa
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Sample Problem 9.12
9 - 43
23
2
10005622
1000752
100025
121
222
121
mkg102565
1613161
3
.
..
xmLamI
.
y
cylinders
:mm50,mm5.62,mm75,mm25 yxLa
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Sample Problem 9.12
9 - 44
23
2
1000502
10005622
1000752
100025
121
2222
121
kg.m101568
1613161
3
.
..
yxmLamI
.
z
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Sample Problem 9.12
9 - 45
prism (a = 50 mm., b =
150 mm, c = 50 mm):
23
2
1000502
1000150
12122
121
mkg 10125.6
kg94.2cbmII zx
23
2
1000502
100050
12122
121
m kg102251
kg942
.
.acmI y
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Sample Problem 9.12
9 - 46
• Add the moments of inertia from the components
to determine the total moments of inertia.
33 1026332101256 ..Ix
23 mkg106512.Ix
33 1025652102251 ..I y
23 mkg107411.I y
33 1015682101256 ..Iz
23 mkg104422.Iz
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Moment of Inertia With Respect to an Arbitrary Axis
kk is the arbitrary axis
r is the vector from
origin to point (x,y,z)
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Transformation of product MI
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© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
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Find Iz’z’ and Ix’z’
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