10.7 moments of inertia for an area about inclined axes in structural and mechanical design,...

10.7 Moments of Inertia for an Area about Inclined

Axes


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


Axes


Axes

Integrating,

Simplifying using trigonometric identities,

22

22

22

22

sincos2cos

cossin22sin

)sin(cos2cossincossin

cossin2cossin

cossin2sincos

xyyxuv

xyyxv

xyyxu

IIII

IIII

IIII


Axes


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


Axes


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


Axes


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


Axes


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


Axes


Axes


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


Axes


Axes


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


Axes


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.


Axes


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


Axes


AxesView Free Body Diagram

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


Axes


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


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



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



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



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



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



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



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.



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

View Free Body Diagram



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



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



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


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


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



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


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


Example 10.11Determine the mass moment of inertia of the cylinder about the z axis. The density of the material is constant.


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


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


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.


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


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


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


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''

''


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


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


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


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.


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


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


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


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.


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


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


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~


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


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


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


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

10.7 moments of inertia for an area about inclined axes in structural and mechanical design,...

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