Moment of Inertia Lecture

7/16/2019 Moment of Inertia Lecture

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IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture AR232 Spring2006

09.06.2013 Dr. Engin Aktaş 1

X.0 SECOND MOMENT OR MOMENT OF INERTIA

OF AN AREA

X.1 Introduction

X.2 Moment of Inertia of an Area

X.3 Moment of Inertia of an Area by Integration

X.4 Polar Moment of Inertia

X.5 Radius of Gyration

X.6 Parallel Axis Theorem

X.7 Moments of Inertia of Composite Areas

X.8 Product of Inertia

X.9 Principal Axes and Principal Moments of Inertia





3.1. 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





3.2. 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 of from 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

dA ydA yk M

QdA ydA yk R Aky F

x

•

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

dA y M

dA y R

A y A p F

x

2





3.3.Moment of Inertia of an Area by Integration• Second moments or moments of inertia of

an area with respect to the x and y axes,

dA x I dA y I y x22

• Evaluation of the integrals is simplified by

choosing d A to be a thin strip parallel to

one of the coordinate axes.

• For a rectangular area,

331

0

22 bhbdy ydA y I h

x

• The formula for rectangular areas may also

be applied to strips parallel to the axes,

dx y xdA xdI dx ydI y x223

31





3.4. Polar Moment of Inertia

• The polar moment of inertia is an important

parameter in problems involving torsion of

cylindrical shafts and rotations of slabs.

dAr J 2

0

• The polar moment of inertia is related to the

rectangular moments of inertia,

x y I I

dA ydA xdA y xdAr J

22222

0





3.5. Radius of Gyration of an Area• Consider area A with moment of inertia

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

the x axis with equivalent I x.

A

I k Ak I x x x x 2

k x = radius of gyration with respectto the x axis

• Similarly,

A

J k Ak J

A

I k Ak I

OOOO

y y y y

2

2

222 y xO k k k





Examples

Determine the moment of

inertia of a triangle with respect

to its base.

SOLUTION:

• A differential strip parallel to the x axis is chosen for

dA.

dyl dAdA ydI x 2

• For similar triangles,

dyh

yhbdA

h

yhbl

h

yh

b

l

• Integrating dI x from y = 0 to y = h,

h

hh

x

y yh

h

b

dy yhyhbdy

h yhb ydA y I

0

43

0

32

0

22

43

12

3bh

I x





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,

r r

OO

O

duuduuudJ J

duudAdAudJ

0

3

0

2

2

22

2

4

2r J O

• From symmetry, I x = I y,

x x y xO I r I I I J 22

2 4

4

4r I I xdiameter





3.6. Parallel Axis Theorem

• Consider moment of inertia I of an area A

with respect to the axis AA’

dA y I 2

•

The axis BB’ passes through the area centroidand is called a centroidal axis.

dAd dA yd dA y

dAd ydA y I

22

22

2

2 Ad I I parallel axis theorem





• Moment of inertia I T of a circular area with

respect to a tangent to the circle,

445

224412

r

r r r Ad I I T

• Moment of inertia of a triangle with respect to a

centroidal axis,

3

361

231

213

1212

2

bh

hbhbh Ad I I

Ad I I

A A B B

B B A A





3.7. Moments of Inertia of Composite Areas• The moment of inertia of a composite area A about a given axis is

obtained by adding the moments of inertia of the component areas

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





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 of composite 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 S





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

A yY A y AY

225 mm

358 mm

20 mm

172 mm

189 mm

12.5095.170011.20SectionBeam

12.50425.76.75Plate

in,in.,in,Section 32

A y A

A y y A mm2mm mm3

4500

7230

11730

189 850.5 x 103

850.5 x 103

2450020225 mmmmmm A

mmmmmm y 1892021358

21

IZMIR INSTITUTE OF TECHNOLOGY D t t f A hit t AR232 S i 2006





• 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

section beam,

mm102.61

51.72189450020225

102.198

51.727230102.160

Ad I I

Y A I I

x x

x x

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

2

46

mm11730

mm1053.82

A

I k x x

mm.1.147 xk

66

plate,section beam, 102.61106.192 x x x I I I

4610254 mm I x

225 mm

358 mm

20 mm

172 mm

189 mm






Example

Determine the moment of inertia

of the shaded area with respect to

the x axis.

SOLUTION:

• Compute the moments of inertia of the

bounding rectangle and half-circle with

respect to the x axis.

• The moment of inertia of the shaded area is

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

of inertia of the rectangle.






SOLUTION:

• Compute the moments of inertia of the bounding

rectangle and half-circle with respect to the x axis.

Rectangle:

46

313

31 mm102.138120240 bh I x

Half-circle:moment of inertia with respect to AA’,

464

814

81 mm1076.2590 r I A A

23

2

2

12

2

1

mm1072.12

90

mm81.8a-120 b

mm2.383

904

3

4

r A

r a

moment of inertia with respect to x’,

46

362

mm1020.7

1072.121076.25

Aa I I A A x

moment of inertia with respect to x,

46

2362

mm103.92

8.811072.121020.7

Ab I I x x






• 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 x I

x I 46mm102.138 46mm103.92






3.8. Product of Inertia• Product of Inertia:

dA xy I xy

• When the x axis, the y axis, or both are an

axis of symmetry, the product of inertia iszero.

• Parallel axis theorem for products of inertia:

A y x I I xy xy






3.9. Principal Axes and Principal Moments of Inertia

Given

dA xy I

dA x I dA y I

xy

y x22

we wish to determine momentsand product of inertia with

respect to new axes x’ and y’.

2cos2sin2

2sin2cos22

2sin2cos22

xy y x y x

xy y x y x

y

xy y x y x

x

I I I I

I I I I I

I

I I I I I

I

• The change of axes yields

• The equations for I x’ and I x’y’ are the

parametric equations for a circle,

2

222

22xy

y x y xave

y xave x

I I I

R I I

I

R I I I

• The equations for I y’ and I x’y’ lead to the

same circle.

sincos

sincos

x y y

y x x

Note:






2

222

22xy

y x y xave

y xave x

I I I

R I I

I

R I I I

• At the points A and B, I x’y’ = 0 and I x’ is

a maximum and minimum, respectively. R I I ave minmax,

y x

xym

I I

I

22tan

•

I max and I min are the principal momentsof inertia of the area about O.

• The equation for Q m defines two

angles, 90o apart which correspond to

the principal axes of the area about O.






Example

Determine the product of inertia of

the right triangle (a) with respect

to the x and y axes and

(b) with respect to centroidal axes

parallel to the x and y axes.

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.






SOLUTION:

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

area strips

b

xh y y x x

dxb

xhdx ydA

b

xh y

el el 1

11

2121

Integrating dI x from x = 0 to x = b,

bb

b

el el xy xy

b

x

b

x xhdx

b

x

b

x xh

dxb

xh xdA y xdI I

02

4322

02

322

0

22

21

83422

1

22

24

1 hb I xy






• Apply the parallel axis theorem to evaluate the

product of inertia with respect to the centroidal axes.

h yb x31

31

With the results from part a,

bhhbhb I

A y x I I

y x

y x xy

21

31

3122

241

22

721 hb I y x






For the section shown, the moments of

inertia with respect to the x and y axes

are I x = 10.38 in4 and I y = 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 the xy axes by dividing the

section into three rectangles and applying

the parallel axis theorem to each.

•

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

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

Example






SOLUTION:

•

Compute the product of inertia with respect to the xy axes by dividing the section into three rectangles.

276985813849295.445.31988

000988

13849295.445.31988

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

A y x III

II

I

m A y x y x

Apply the parallel axis theorem to each rectangle,

A y x I I y x xy

Note that the product of inertia with respect to centroidal

axes parallel to the xy axes is zero for each rectangle.

42770000mm A y x I xy

102 mm

76 mm

13 mm

13 mm

76 mm

31.5 mm

44.5 mm

44.5 mm

31.5 mm





• 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

y x

xy

m I I

I

5.127 and 5.37 mm

26

26666

2

2

minmax,

1077.22

1093.21042.4

2

1093.21042.4

22

xy

y x y x I I I I I I

4

min

4

max

mm07.8

mm54.6

II

I I

b

a

y x

xym

I I

I

22tan

R I I ave minmax,

the principal moments of

inertia by

2

222

22xy

y x y xave

y xave x

I I I

R I I

I

R I I I

m = 127.5o

m = 37.5o

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Moment of Inertia Lecture