Post on 04-Apr-2015
10.7 Moments of Inertia for an Area about Inclined
Axes
10.7 Moments of Inertia for an Area about Inclined
Axes In structural and mechanical design,
necessary to calculate the moments and product of inertia Iu, Iv and Iuv for an area with respect to a set of inclined u and v axes when the values of θ, Ix, Iy and Ixy are known
Use transformation equations which relate the x, y and u, v coordinates
For moments and product of inertia of dA about the u and v axes,
dAxyyxuvdAdI
dAyxdAudI
dAxydAvdI
xyv
yxu
uv
v
u
)sincos)(sincos(
)sincos(
)sincos(
sincos
sincos
22
22
10.7 Moments of Inertia for an Area about Inclined
Axes
10.7 Moments of Inertia for an Area about Inclined
Axes
Integrating,
Simplifying using trigonometric identities,
22
22
22
22
sincos2cos
cossin22sin
)sin(cos2cossincossin
cossin2cossin
cossin2sincos
xyyxuv
xyyxv
xyyxu
IIII
IIII
IIII
10.7 Moments of Inertia for an Area about Inclined
Axes
10.7 Moments of Inertia for an Area about Inclined
Axes
Polar moment of inertia about the z axis passing through point O is independent of the u and v axes
yxvuO
xyyx
uv
xyyxyx
v
xyyxyx
u
IIIIJ
III
I
IIIII
I
IIIII
I
2cos22sin2
2sin2cos22
2sin2cos22
10.7 Moments of Inertia for an Area about Inclined
Axes
10.7 Moments of Inertia for an Area about Inclined
Axes
Principal Moments of Inertia Iu, Iv and Iuv depend on the angle of
inclination θ of the u, v axes To determine the orientation of these
axes about which the moments of inertia for the area Iu and Iv are maximum and minimum
This particular set of axes is called the principal axes of the area and the corresponding moments of inertia with respect to these axes are called the principal moments of inertia
10.7 Moments of Inertia for an Area about Inclined
Axes
10.7 Moments of Inertia for an Area about Inclined
Axes
Principal Moments of Inertia There is a set of principle axes for every
chosen origin O For the structural and mechanical
design of a member, the origin O is generally located at the cross-sectional area’s centroid
The angle θ = θp defines the orientation of the principal axes for the area. Found by differentiating with respect to θ and setting the result to zero
10.7 Moments of Inertia for an Area about Inclined
Axes
10.7 Moments of Inertia for an Area about Inclined
Axes
Principal Moments of Inertia
Therefore
Equation has 2 roots, θp1 and
θp2 which are 90° apart and
so specify the inclination of the principal axes
2/2tan
02cos22sin2
2
yx
xyp
p
xyyxu
II
I
III
d
dI
10.7 Moments of Inertia for an Area about Inclined
Axes
10.7 Moments of Inertia for an Area about Inclined
Axes
Principal Moments of Inertia
22
2
22
22
22
1
22
11
2/
22cos
2/2sin,
2/
22cos
2/2sin,
xyyxyx
p
xyyx
xypp
xyyxyx
p
xyyx
xypp
IIIII
III
IFor
IIIII
III
IFor
10.7 Moments of Inertia for an Area about Inclined
Axes
10.7 Moments of Inertia for an Area about Inclined
Axes
Principal Moments of Inertia
Depending on the sign chosen, this result gives the maximum or minimum moment of inertia for the area
It can be shown that Iuv = 0, that is, the product of inertia with respect to the principal axes is zero
Any symmetric axis represent a principal axis of inertia for the area
2
2
maxmin 22 xy
yxyx IIIII
I
10.7 Moments of Inertia for an Area about Inclined
Axes
10.7 Moments of Inertia for an Area about Inclined
Axes
Example 10.9Determine the principal moments of
inertia for the beam’s cross-sectional area with respect to an axis passing through the centroid.
10.7 Moments of Inertia for an Area about Inclined
Axes
10.7 Moments of Inertia for an Area about Inclined
Axes
Solution Moment and product of inertia of the cross-
sectional area with respect to the x, y axes have been computed in the previous examples
Using the angles of inclination of principal axes u and v
Thus,
1.57,9.32
2.1142,8.652
22.22/1060.51090.2
1000.3
2/2tan
1000.31060.51090.2
21
21
99
9
494949
pp
pp
yx
xyp
zyx
II
I
mmImmImmI
10.7 Moments of Inertia for an Area about Inclined
Axes
10.7 Moments of Inertia for an Area about Inclined
Axes
Solution For principal of inertia with respect to
the u and v axes
or
49min
49max
99maxmin
29
299
99
2
2
maxmin
10960.0,1054.7
1029.31025.4
1000.32
1060.51090.2
2
1060.51090.2
22
mmImmI
I
IIIII
I xyyxyx
10.7 Moments of Inertia for an Area about Inclined
Axes
10.7 Moments of Inertia for an Area about Inclined
Axes
10.7 Moments of Inertia for an Area about Inclined Axes10.7 Moments of Inertia for an Area about Inclined Axes
Solution Maximum moment of inertia occurs with
respect to the selected u axis since by inspection, most of the cross-sectional area is farthest away from this axis
Maximum moment of inertia occurs at the u axis since it is located within ±45° of the y axis, which has the largest value of I
10.8 Mohr’s Circle for Moments of Inertia
10.8 Mohr’s Circle for Moments of Inertia
It can be found that
In a given problem, Iu and Iv are variables and Ix, Iy and Ixy are known constants
When this equation is plotted on a set of axes that represent the respective moment of inertia and the product of inertia, the resulting graph represents a circle
222
2
2
2
2
22
RIaI
III
III
I
uvu
xyyx
uvyx
u
10.8 Mohr’s Circle for Moments of Inertia
10.8 Mohr’s Circle for Moments of Inertia
The circle constructed is known as a Mohr’s circle with radius
and center at (a, 0) where
2/
22
2
yx
xyyx
IIa
III
R
10.8 Mohr’s Circle for Moments of Inertia
10.8 Mohr’s Circle for Moments of Inertia
Procedure for AnalysisDetermine Ix, Iy and Ixy Establish the x, y axes for the area, with
the origin located at point P of interest and determine Ix, Iy and Ixy
Construct the Circle Construct a rectangular coordinate system
such that the abscissa represents the moment of inertia I and the ordinate represent the product of inertia Ixy
10.8 Mohr’s Circle for Moments of Inertia
10.8 Mohr’s Circle for Moments of Inertia
Procedure for AnalysisConstruct the Circle Determine center of the circle O, which is
located at a distance (Ix + Iy)/2 from the origin, and plot the reference point a having coordinates (Ix, Ixy)
By definition, Ix is always positive, whereas Ixy will either be positive or negative
Connect the reference point A with the center of the circle and determine distance OA (radius of the circle) by trigonometry
Draw the circle
10.8 Mohr’s Circle for Moments of Inertia
10.8 Mohr’s Circle for Moments of Inertia
Procedure for AnalysisPrincipal of Moments of Inertia Points where the circle intersects the
abscissa give the values of the principle moments of inertia Imin and Imax
Product of inertia will be zero at these points
Principle Axes To find direction of major principal axis,
determine by trigonometry, angle 2θp1, measured from the radius OA to the positive I axis
10.8 Mohr’s Circle for Moments of Inertia
10.8 Mohr’s Circle for Moments of Inertia
Procedure for AnalysisPrinciple Axes This angle represent twice the angle from
the x axis to the area in question to the axis of maximum moment of inertia Imax
Both the angle on the circle, 2θp1, and the angle to the axis on the area, θp1must be measured in the same sense
The axis for the minimum moment of inertia Imin is perpendicular to the axis for Imax
10.8 Mohr’s Circle for Moments of Inertia
10.8 Mohr’s Circle for Moments of Inertia
Example 10.10Using Mohr’s circle, determine the
principle moments of the beam’s cross-sectional
area with respect to an axis passing through the centroid.
10.8 Mohr’s Circle for Moments of Inertia
10.8 Mohr’s Circle for Moments of Inertia
SolutionDetermine Ix, Iy and Ixy Moments of inertia and the product of inertia
have been determined in previous examples
Construct the Circle Center of circle, O, lies from the origin, at a
distance
25.42/)60.590.2(2/
1000.3
1060.51090.2
49
4949
yx
xy
yx
II
mmI
mmImmI
10.8 Mohr’s Circle for Moments of Inertia
10.8 Mohr’s Circle for Moments of Inertia
Solution With reference point A (2.90, -3.00)
connected to point O, radius OA is determined using Pythagorean theorem
Principal Moments of Inertia Circle intersects I axis at
points (7.54, 0) and (0.960, 0)
29.3
00.335.1 22
OA
10.8 Mohr’s Circle for Moments of Inertia
10.8 Mohr’s Circle for Moments of Inertia
Solution
Principal Axes Angle 2θp1 is determined from
the circle by measuring CCW from OA to the direction of the positive I axis
2.11429.300.3
sin180
sin1802
10960.01054.7
1
11
49min
49max
OA
BA
mmImmI
p
10.8 Mohr’s Circle for Moments of Inertia
10.8 Mohr’s Circle for Moments of Inertia
SolutionThe principal axis for Imax = 7.54(109)
mm4 is therefore orientated at an angle θp1 = 57.1°, measured CCW from the positive x axisto the positive u axis
v axis is perpendicular to this axis
10.9 Mass Moment of Inertia10.9 Mass Moment of Inertia
Mass moment of inertia of a body is the property that measures the resistance of the body to angular acceleration
Mass moment of inertia is defined as the integral of the second moment about an axis of all the elements of mass dm which compose the body
Example Consider rigid body
10.9 Mass Moment of Inertia10.9 Mass Moment of Inertia
For body’s moment of inertia about the z axis,
Here, the moment arm r is the perpendicular distance from the axis to the arbitrary element dm
Since the formulation involves r, the value of I is unique for each axis z about which it is computed
The axis that is generally chosen for analysis, passes through the body’s mass center G
mdmrI 2
10.9 Mass Moment of Inertia10.9 Mass Moment of Inertia
Moment of inertia computed about this axis will be defined as IG
Mass moment of inertia is always positive
If the body consists of material having a variable density ρ = ρ(x, y, z), the element mass dm of the body may be expressed as dm = ρ dV
Using volume element for integration,V
dVrI 2
10.9 Mass Moment of Inertia10.9 Mass Moment of Inertia In the special case of ρ being a constant,
When element volume chosen for integration has differential sizes in all 3 directions, dV = dx dy dz
Moment of inertia of the body determined by triple integration
Simplify the process to single integration by choosing an element volume with a differential size or thickness in 1 direction such as shell or disk elements
VdVrI 2
Procedure for Analysis Consider only symmetric bodies having
surfaces which are generated by revolving a curve about an axis
Shell Element For a shell element having height z, radius
y and thickness dy, volume dV = (2πy)(z)dy
10.9 Mass Moment of Inertia10.9 Mass Moment of Inertia
10.9 Mass Moment of Inertia10.9 Mass Moment of Inertia
Procedure for AnalysisShell Element Use this element to determine the moment
of inertia Iz of the body about the z axis since the entire element, due to its thinness, lies at the same perpendicular distance r = y from the z axis
Disk Element For disk element having radius y, thickness
dz, volume dV = (πy2) dz
10.9 Mass Moment of Inertia10.9 Mass Moment of Inertia
Procedure for AnalysisDisk Element Element is finite in the radial direction
and consequently, its parts do not lie at the same radial distance r from the z axis
To perform integration using this element, determine the moment of inertia of the element about the z axis and then integrate this result
10.9 Mass Moment of Inertia10.9 Mass Moment of Inertia
Example 10.11Determine the mass moment of inertia of the cylinder about the z axis. The density of the material is constant.
10.9 Mass Moment of Inertia10.9 Mass Moment of Inertia
SolutionShell Element For volume of the element,
For mass,
Since the entire element lies at the same distance r from the z axis, for the moment of inertia of the element,
32 2
2
2
hrdmrdI
drrhdVdm
drhrdV
z
10.9 Mass Moment of Inertia10.9 Mass Moment of Inertia
Solution Integrating over entire region of the
cylinder,
For the mass of the cylinder
So that2
2
0
4
0
32
2
1
2
22
mRI
hRrdrhdmm
hRdrrhdmrI
z
R
m
R
mz
10.9 Mass Moment of Inertia10.9 Mass Moment of Inertia
Example 10.12A solid is formed by revolving the shaded area about the y axis. If the density of the material is 5 Mg/m3, determine the mass moment of inertia about the y axis.
10.9 Mass Moment of Inertia10.9 Mass Moment of Inertia
SolutionDisk Element Element intersects the curve at the arbitrary
point (x, y) and has a mass dm = ρ dV = ρ (πx2)dy
Although all portions of the element are not located at the same distance from the y axis, it is still possible to determine the moment of inertia dIy about the y axis
10.9 Mass Moment of Inertia10.9 Mass Moment of Inertia
Solution In the previous example, it is shown that
the moment of inertia for a cylinder is I = ½ mR2
Since the height of the cylinder is not involved, apply the about equation for a disk
For moment of inertia for the entire solid,
1
0
2281
0
4
222
.873.873.02
5
2
5
2
1)(
2
1
mkgmMgdyydyxI
xdyxxdmdI
y
y
10.9 Mass Moment of Inertia10.9 Mass Moment of Inertia
Parallel Axis Theorem If the moment of inertia of the body
about an axis passing through the body’s mass center is known, the moment of inertia about any other parallel axis may be determined by using parallel axis theorem
Considering the body where the z’ axis passes through the mass center G, whereas the corresponding parallel z axis lie at a constant distance d away
10.9 Mass Moment of Inertia10.9 Mass Moment of Inertia
Parallel Axis Theorem Selecting the differential mass element dm,
which is located at point (x’, y’) and using Pythagorean theorem,
r 2 = (d + x’)2 + y’2
For moment of inertia of body about the z axis,
First integral represent IG
mmm
mm
dmddmxddmyx
dmyxddmrI
222
222
'2''
''
10.9 Mass Moment of Inertia10.9 Mass Moment of Inertia
Parallel Axis Theorem Second integral = 0 since the z’ axis
passes through the body’s center of mass
Third integral represents the total mass m of the body
For moment of inertia about the z axis, I = IG + md2
10.9 Mass Moment of Inertia10.9 Mass Moment of Inertia
Radius of Gyration For moment of inertia expressed using
k, radius of gyration,
Note the similarity between the definition of k in this formulae and r in the equation dI = r2 dm which defines the moment of inertia of an elemental mass dm of the body about an axis
m
IkormkI 2
10.9 Mass Moment of Inertia10.9 Mass Moment of Inertia
Composite Bodies If a body is constructed from a number
of simple shapes such as disks, spheres, and rods, the moment of inertia of the body about any axis z can be determined by adding algebraically the moments of inertia of all the composite shapes computed about the z axis
Parallel axis theorem is needed if the center of mass of each composite part does not lie on the z axis
10.9 Mass Moment of Inertia10.9 Mass Moment of Inertia
Example 10.13If the plate has a density of 8000kg/m3 and athickness of 10mm, determine its mass moment of inertia about an axis perpendicular to the page and passing through point O.
10.9 Mass Moment of Inertia10.9 Mass Moment of Inertia
SolutionThe plate consists of 2 composite parts,
the 250mm radius disk minus the 125mm radius disk
Moment of inertia about O is determined by computing the moment of inertia of each of these parts about O and then algebraically adding the results
10.9 Mass Moment of Inertia10.9 Mass Moment of Inertia
SolutionDisk For moment of inertia of a disk about an
axis perpendicular to the plane of the disk,
Mass center of the disk is located 0.25m from point O
222
22
2
2
.473.125.071.1525.071.152
12
1
71.1501.025.08000
2
1
mkg
dmrmI
kgVm
mrI
ddddO
ddd
G
10.9 Mass Moment of Inertia10.9 Mass Moment of Inertia
SolutionHole
For moment of inertia of plate about point O,
2
222
22
2
.20.1276.0473.1
.276.025.093.3125.093.32
12
1
93.301.0125.08000
mkg
III
mkg
dmrmI
kgVm
hOdOO
hhhhO
hhh
10.9 Mass Moment of Inertia10.9 Mass Moment of Inertia
Example 10.14The pendulum consists of two thin robs each having a mass of 100kg. Determine the pendulum’s mass moment of inertia about an axis passing through (a) the pin at point O, and (b) the mass center G of
the pendulum.
10.9 Mass Moment of Inertia10.9 Mass Moment of Inertia
SolutionPart (a) For moment of inertia of rod OA about an axis
perpendicular to the page and passing through the end point O of the rob,
Hence,
Using parallel axis theorem,
22222
2
222
2
.3005.11003100121
121
121
.300310031
31
31
mkgmdmlI
mlI
mkgmlI
mlI
OOA
G
OOA
O
10.9 Mass Moment of Inertia10.9 Mass Moment of Inertia
Solution For rod BC,
For moment of inertia of pendulum about O,
2
2
2222
.1275975300
.975
3100310012
1
12
1
mkgI
mkg
mdmlI
O
OBC
10.9 Mass Moment of Inertia10.9 Mass Moment of Inertia
SolutionPart (b) Mass center G will be located relative to pin at
O For mass center,
Mass of inertia IG may be computed in the same manner as IO, which requires successive applications of the parallel axis theorem in order to transfer the moments of inertias of rod OA and BC to G
mkgkg
kgmkgm
m
myy 25.2
100100
)100(3)100(5.1~
10.9 Mass Moment of Inertia10.9 Mass Moment of Inertia
Solution Apply the parallel axis theorem for IO,
2
22
2
.5.262
25.2200.125
;
mkgI
Imkg
mdII
G
G
GO
Chapter Summary Chapter Summary
Area Moment of Inertia Represent second moment of area about
an axis Frequently used in equations related to
strength and stability of structural members or mechanical elements
If the area shape is irregular, a differential element must be selected and integration over the entire area must be performed
Tabular values of the moment of inertia of common shapes about their centroidal axis are available
Chapter Summary Chapter Summary
Area Moment of Inertia To determine moment of inertia of these
shapes about some other axis, parallel axis theorem must be used
If an area is a composite of these shapes, its moment of inertia = sum of the moments of inertia of each of its parts
Product of Inertia Determine location of an axis about which
the moment of inertia for the area is a maximum or minimum
Chapter Summary Chapter Summary
Product of Inertia If the product of inertia for an area is known
about its x’, y’ axes, then its value can be determined about any x, y axes using the parallel axis theorem for product of inertia
Principal Moments of Inertia Provided moments of inertia are known,
formulas or Mohr’s circle can be used to determine the maximum or minimum or principal moments of inertia for the area, as well as orientation of the principal axes of inertia
Chapter Summary Chapter Summary
Mass Moments of Inertia Measures resistance to change in its rotation Second moment of the mass elements of the
body about an axis For bodies having axial symmetry, determine
using wither disk or shell elements Mass moment of inertia of a composite body
is determined using tabular values of its composite shapes along with the parallel axis theorem
Chapter ReviewChapter Review
Chapter ReviewChapter Review
Chapter ReviewChapter Review
Chapter ReviewChapter Review