ar231 chap08 momentofinertia (1)

8/12/2019 AR231 Chap08 MomentofInertia (1)

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IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture AR231Fall12/13

02.06.2014 Dr. Engin Akta 1

8.0 SECOND MOMENT OR MOMENT OF INERTIA

OF AN AREA

8.1 Introduction

8.2 Moment of Inertia of an Area

8.3 Moment of Inertia of an Area by Integration

8.4 Polar Moment of Inertia

8.5 Radius of Gyration

8.6 Parallel Axis Theorem

8.7 Moments of Inertia of Composite Areas

8.8 Product of Inertia

8.9 Principal Axes and Principal Moments of Inertia


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8.1. Introduction

Forces which are proportional to the area or volume over which they act

but also vary linearly with distance from a given axis.

- the magnitude of the resultant depends on the first moment of the

force distribution with respect to the axis.

- The point of application of the resultant depends on the secondmoment of the distribution with respect to the axis.

Herein methods for computing the moments and products of inertia for

areas and masses will be presented


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8.2. Moment of Inertia of an Area Consider distributed forces whose magnitudes are

proportional to the elemental areas on which they

act and also vary linearly with the distance offrom a given axis.

F

A

A

Example: Consider a beam subjected to pure bending.

Internal forces vary linearly with distance from the

neutral axis which passes through the section centroid.

momentsecond

momentfirst022

dAydAykM

QdAydAykRAkyF

x

Example: Consider the net hydrostatic force on asubmerged circular gate.

dAyM

dAyR

AyApF

x

2


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8.3.Moment of Inertia of an Area by Integration Second momentsor moments of inertiaof

an area with respect to thexandyaxes,

dAxIdAyI yx22

Evaluation of the integrals is simplified by

choosing dA to be a thin strip parallel to

one of the coordinate axes.

For a rectangular area,

331

0

22 bhbdyydAyIh

x

The formula for rectangular areas may also

be applied to strips parallel to the axes,

dxyxdAxdIdxydI yx223

31


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8.4. Polar Moment of Inertia

Thepolar moment of inertiais an important

parameter in problems involving torsion of

cylindrical shafts and rotations of slabs.

dArJ 2

0

The polar moment of inertia is related to the

rectangular moments of inertia,

xy II

dAydAxdAyxdArJ

22222

0


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8.5. Radius of Gyration of an Area Consider areaA with moment of inertia

Ix. Imagine that the area isconcentrated in a thin strip parallel to

thexaxis with equivalentIx.

A

IkAkI xxxx

2

kx= radius of gyrationwith respectto thexaxis

Similarly,

A

JkAkJ

A

IkAkI

OOOO

yyyy

2

2

222yxO kkk


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Examples

Determine the moment of

inertia of a triangle with respect

to its base.

SOLUTION:

A differential strip parallel to thexaxis is chosen for

dA.

dyldAdAydIx 2

For similar triangles,

dyh

yhbdA

h

yhbl

h

yh

b

l

Integrating dIxfromy= 0 toy = h,

h

hh

x

yyh

h

b

dyyhyhbdy

hyhbydAyI

0

43

0

32

0

22

43

12

3bh

Ix


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a) Determine the centroidal polar

moment of inertia of a circular

area by direct integration.

b) Using the result of part a,

determine the moment of inertia

of a circular area with respect to a

diameter.

SOLUTION:

An annular differential area element is chosen,

rr

OO

O

duuduuudJJ

duudAdAudJ

0

3

0

2

2

22

2

4

2rJO

From symmetry,Ix=Iy,

xxyxO IrIIIJ 22

2 4

4

4rII xdiameter


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8.6. Parallel Axis Theorem

Consider moment of inertiaIof an areaA

with respect to the axisAA

dAyI 2

The axisBBpasses through the area centroidand is called a centroidal axis.

dAddAyddAy

dAdydAyI

22

22

2

2AdII parallel axis theorem


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02.06.2014 Dr. Engin Akta 10

Moment of inertiaITof a circular area with

respect to a tangent to the circle,

445

224412

r

rrrAdIIT

Moment of inertia of a triangle with respect to a

centroidal axis,

3

361

231

213

1212

2

bh

hbhbhAdII

AdII

AABB

BBAA


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02.06.2014 Dr. Engin Akta 11

8.7. Moments of Inertia of Composite Areas The moment of inertia of a composite areaAabout a given axis is

obtained by adding the moments of inertia of the component areas

A1,A2,A3, ... , with respect to the same axis.


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Example

The strength of a W360x57 rolled steel

beam is increased by attaching a

225x20 mm plate to its upper flange.

Determine the moment of inertia and

radius of gyration with respect to an

axis which is parallel to the plate and

passes through the centroid of the

section.

SOLUTION:

Determine location of the centroid ofcomposite section with respect to a

coordinate system with origin at the

centroid of the beam section.

Apply the parallel axis theorem todetermine moments of inertia of beam

section and plate with respect to

composite section centroidal axis.

Calculate the radius of gyration from the

moment of inertia of the compositesection.

225 mm

358 mm

20 mm

172 mm

f /


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SOLUTION:

Determine location of the centroid of composite sectionwith respect to a coordinate system with origin at the

centroid of the beam section.

mm.51.7211730

1050.58 3

A

AyYAyAY

225 mm

358 mm

20 mm

172 mm

189 mm

12.5095.170011.20SectionBeam

12.50425.76.75Plate

in,in.,in,Section 32

AyA

AyyA mm2 mm mm3

4500

7230

11730

189 850.5 x 103

850.5 x 103

2450020225 mmmmmmA

mmmmmmy 1892021358

21

IZMIR INSTITUTE OF TECHNOLOGY D t t f A hit t AR231 F ll12/13


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02.06.2014 Dr. Engin Akta 14

Apply the parallel axis theorem to determine moments of

inertia of beam section and plate with respect to composite

section centroidal axis.

46

23

1212

plate,

4

262

sectionbeam,

mm102.61

51.72189450020225

102.198

51.727230102.160

AdII

YAII

xx

xx

Calculate the radius of gyration from the moment of inertiaof the composite section.

2

46

mm11730

mm1053.82

A

Ik xx

mm.1.147xk

66

plate,sectionbeam, 102.61106.192 xxx III

4610254 mmIx

225 mm

358 mm

20 mm

172 mm

189 mm



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02.06.2014 Dr. Engin Akta 15

Example

Determine the moment of inertia

of the shaded area with respect to

thexaxis.

SOLUTION:

Compute the moments of inertia of the

bounding rectangle and half-circle with

respect to thex axis.

The moment of inertia of the shaded area is

obtained by subtracting the moment ofinertia of the half-circle from the moment

of inertia of the rectangle.



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02.06.2014 Dr. Engin Akta 16

SOLUTION:

Compute the moments of inertia of the bounding

rectangle and half-circle with respect to thex axis.

Rectangle:

46313

31 mm102.138120240 bhIx

Half-circle:moment of inertia with respect toAA,

464814

81 mm1076.2590 rI AA

23

2

2

12

2

1

mm1072.12

90

mm81.8a-120b

mm2.383

904

3

4

rA

ra

moment of inertia with respect tox,

46

362

mm1020.7

1072.121076.25

AaII AAx

moment of inertia with respect tox,

46

2362

mm103.92

8.811072.121020.7

AbII xx



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02.06.2014 Dr. Engin Akta 17

The moment of inertia of the shaded area is obtained by

subtracting the moment of inertia of the half-circle fromthe moment of inertia of the rectangle.

46mm109.45 xI

xI 46mm102.138 46mm103.92

IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture AR231 Fall12/13


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02.06.2014 Dr. Engin Akta 18

8.8. Product of Inertia Product of Inertia:

dAxyIxy

When thexaxis, theyaxis, or both are an

axis of symmetry, the product of inertia iszero.

Parallel axis theorem for products of inertia:

AyxII xyxy



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02.06.2014 Dr. Engin Akta 19

8.9. Principal Axes and Principal Moments of Inertia

Given

dAxyI

dAxIdAyI

xy

yx22

we wish to determine momentsand product of inertia with

respect to new axesxandy.

2cos2sin2

2sin2cos22

2sin2cos22

xyyxyx

xyyxyx

y

xyyxyx

x

IIII

IIIII

I

IIIII

I

The change of axes yields

The equations forIxandIxyare the

parametric equations for a circle,

2

222

22 xy

yxyxave

yxavex

III

RII

I

RIII

The equations forIyandIxy lead to the

same circle.

sincos

sincos

xyy

yxx

Note:



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02.06.2014 Dr. Engin Akta 20

2

222

22 xy

yxyxave

yxavex

III

RII

I

RIII

At the pointsAandB, Ixy= 0 andIxis

a maximum and minimum, respectively.RII aveminmax,

yx

xym

II

I

2

2tan

ImaxandIminare theprincipal momentsof inertiaof the area about O.

The equation for Qmdefines two

angles, 90oapart which correspond to

theprincipal axesof the area about O.



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02.06.2014 Dr. Engin Akta 21

Example

Determine the product of inertia of

the right triangle (a)with respect

to thexandyaxes and

(b)with respect to centroidal axes

parallel to thexandyaxes.

SOLUTION:

Determine the product of inertia using

direct integration with the parallel axis

theorem on vertical differential area strips

Apply the parallel axis theorem to

evaluate the product of inertia with respect

to the centroidal axes.



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02.06.2014 Dr. Engin Akta 22

SOLUTION:

Determine the product of inertia using direct integrationwith the parallel axis theorem on vertical differential

area strips

b

xhyyxx

dxb

xhdxydA

b

xhy

elel 1

11

2121

Integrating dIxfromx= 0 tox= b,

bb

b

elelxyxy

b

x

b

xxhdx

b

x

b

xxh

dxb

xhxdAyxdII

02

4322

02

322

0

22

21

83422

1

22

24

1 hbIxy



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02.06.2014 Dr. Engin Akta 23

Apply the parallel axis theorem to evaluate the

product of inertia with respect to the centroidal axes.

hybx31

31

With the results from part a,

bhhbhbI

AyxII

yx

yxxy

21

31

3122

241

22

721 hbI yx



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02.06.2014 Dr. Engin Akta 24

For the section shown, the moments of

inertia with respect to thexandyaxes

areIx= 10.38 in4andIy= 6.97 in4.

Determine (a) the orientation of the

principal axes of the section about O,

and (b) the values of the principal

moments of inertia about O.

SOLUTION:

Compute the product of inertia with

respect to thexyaxes by dividing the

section into three rectangles and applying

the parallel axis theorem to each.

Determine the orientation of theprincipal axes (Eq. 9.25) and the

principal moments of inertia (Eq. 9. 27).

Example



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SOLUTION:

Compute the product of inertia with respect to thexyaxesby dividing the section into three rectangles.

276985813849295.445.31988

000988

13849295.445.31988

m,mm.,mm.,mmArea,Rectangle 42

AyxIII

II

I

mAyxyx

Apply the parallel axis theorem to each rectangle,

AyxII yxxy

Note that the product of inertia with respect to centroidal

axes parallel to thexy axes is zero for each rectangle.

42770000mmAyxIxy

102 mm

76 mm

13 mm

13 mm

76 mm

31.5 mm

44.5 mm

44.5 mm

31.5 mm



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Determine the orientation of the principal axes by

46

46

46

mm1077.2

mm1093.2

mm1042.4

xy

y

x

I

I

I

254.9and9.742

72.31093.21042.4

1077.2222tan66

6

m

yx

xy

mII

I

5.127and5.37 mm

262

6666

2

2

minmax,

1077.22

1093.21042.4

2

1093.21042.4

22

xy

yxyx IIIIII

4

min

4

max

mm07.8

mm54.6

II

II

b

a

yx

xym

II

I

2

2tan

RII aveminmax,the principal moments of

inertia by

2

222

22 xy

yxyxave

yxavex

III

RII

I

RIII

m = 127.5o

m = 37.5o

ar231 chap08 momentofinertia (1)

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