es11!10!2 student version

Lecture 10

3

x

y A

F=kyA

AkyF

From mechanics of materials:

The resultant force is:

ydAkkydAR

The resultant moment is:

dAykdAkyM22

First moment of area

Second moment of area / moment of inertia

xdA

y

y

dAyI x2

x

dAxI y2

Moment of inertia about the x-axis:

Moment of inertia about the y-axis:

Geometric property and depends on its reference axis. the smallest value occurs at the axis passing through the centroid.

It measures the ability of cross-section to resist bending. the larger the moment of inertia the less bending will occur

x

Which has the largest moment of inertia about the x-axis?

Which is harder to rotate about the x-axis?

It measures the resistance to rotation about an axis

Choosing a differential element

x

dy

x2

y

1x dx

y1

y2

= 2

= 2

METHOD 1

dyxxy

dAyI

y

y

x

12

2

2

2

1

Choosing a strip such that the distance of all parts of the strip to the axis is constant. (Strip is parallel to axis)

x

dy

x2

y

1x

y1

y2

dyxxxx

dAxI

x

x

ely

12

2

21

2

2

1 2

NOTE: x or y here is not the y-coordinate of the centroid! In this derivation, all x or y in the strip for the term x2 or y2 should be at the same distance from the axis

METHOD 2

xdx 1y

y 2y

x1 x2

Similarly

dxyyx

dAxI

x

x

y

12

2

2

2

1

METHOD 2

y

b

x

h

Find the moment of inertia of a rectangle with respect to one of its base

dy

y

dAyI x2

dybyh

02

When using the form y2dA, all parts of the strip must have the same distance to the axis

3

3

1bh

x

dxydI x3

3

1

y

y dy

dx

Recall that the moment of inertia of a rectangle with respect to its base is (1/3)bh3

The moment of inertia of a differential strip perpendicular to and touching the axis is then

METHOD 3

OR, if the strip is not touching

dxyydI x ]3

1

3

1[ 3132

x

dx 1y

y 2y

x1 x2

xdA

y

y

r

dArJO2

xPolar moment of inertia wrt pole O

O

dAyxJO22

dAydAxJO22

yxO IIJ

It measures the resistance to rotation about a point

xy

O

x

y

kx

Concentrating the area into a strip such that it has the same moment of inertia wrt x-axis as the original,

AkI xx2

A

Ik xx Radius of gyration wrt y-axis

xy

Ox

y

ky

Similarly, for the y-axis

Iy = ky2A ky =

Iy

ARadius of gyration wrt x-axis

xy

O

Similarly, concentrating the area into a ring such that it has the same polar moment of inertia as the original,

AkJ oO2

A

Jk Oo

ko

x

y

222

yxo kkk

Find the polar moment of inertia of a circle wrt its center

r r duuuduuodJoJ 0 0

32)2(2

y

du

x O

u

r

ududA 2

dAudJO

2

4

2rJ

O

Determine the moment of inertia with respect to the x and y axes.

0

0.5

1

1.5

2

2.5

0 1 2 3 4 5

x = ky2

y = 0.25 x

Find the moment of inertia of the shaded area about the x-axis.

x

y

100 mm

y2 = kx

100 mm

150 mm

dAyIA

x 2

461040 mmxI x

x

y

y

y2 = kx

y

x = 150

Find k.

At x = 150, y = 100, thus k = 200/3 and the equation of the curve is y2 = (200/3)x

Using horizontal strips, we can use the form:

Thus,

= 2 150

3

2002

100

100

= 2 2 1

100

100

x

y

y

y2 = (200/3)x

x

x = 150

Alternatively, the moment of inertia of the top half is the same as the moment of inertia of the bottom half. The moment of inertia dIx of the differential strip is:

Thus,

=1

3

200

3

1/23

150

0

=1

323

= 20106 4

Since we only got the moment of inertia of the top half, we multiply it to two:

= 40106 4

Find the moment of

inertia of the shaded

area about its vertical

centroidal axis.

x

y

y = 2 sin x

0

21

Reference: Beer, F. B., Johnson, E. R., and Eisenberg, E. R., 2006. Vector Mechanics for Engineers: Statics. 9th Ed. McGraw-Hill.

x

y

x

y

b y

x

a

C

d

r

z

dA y

x

z

Inertia of the shaded area about the x axis is

dAyIA

X 2

' )'(

Since y = y + b, then

dAbyIA

X 2

')(

AAA

dAbydAbdAy 22 2

2

'AbII XX

Assume x is a centroidal axis

Parallel-axis theorem always involves one centroidal axis!

x

y

x

y

b y

x

a

C

d

r

z

dA y

x

z

Similarly

2

'AaII yy

Since

thenIIJ yxz ,'''

)()( 22'

baAIIJ yxz

JO ' = JO +Ad2

Since integration is a summation of small areas, composite areas can be used

n

i

iTotalAxisSpecified IIIII ...321)(

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.

We use the parallel axis theorem to get the moment of inertia about a non-centroidal axis

Find the moment of inertia of the composite area about the x-axis.

x

0.9 m

1.5 m

1.8 m

Components:

21 xxx III 2

iixixi dAII where

2 2

222 dAII xx

1 2

111 dAII xx

PARALLEL AXIS THEOREM

0.9 m

1.5 m

1.8 m

Solution

468585.4 mI x ANSWER:

423

2 025.2)75.0)(5.1)(8.1(12

)5.1)(8.1(mI x

423

1 66085.22

)8.1)(9.0)(8.1(

36

)9.0)(8.1(mI x

23

1 )8.1(236

bhbhI x

23

2 )75.0(12

bhbh

I x

21 xxx III Solution

Determine the moment of inertia of the area with respect to the x-axis.

y 75 mm

50 mm

100 mm

150 mm

60 mm

x

The moment of inertia of the figure is the moment of inertia of the large rectangle minus the moment of inertia of the small triangle and the semicircle.

y 75 mm

50 mm

100 mm

150 mm

60 mm

x

d2

d1

1

2

3

= 1 2 3

1 =1

3225 1503 = 253.125 1064

2 =1

375 503 = 3.125 1064

To get I3, the form Ix1 = r4/8 is the moment of

inertia about x1 and is not a centroidal axis of the semicircle, thus we cannot use this in the parallel axis theorem to get Ix of the semicircle directly. We can, however find the moment of inertia with respect to the centroid, Ix,centroid first

1 = + 12

60 4

8= +

60 2

2

4 60

3

2

y 75 mm

50 mm

100 mm

150 mm

r = 60 mm x1

Xcentroid of semicircle

x

d2

d1 2

= 1.42245 106

Using the parallel axis theorem again to get the moment of inertia of the semicircle wrt x.

3 = + 22

3 = 1.42245 106 + 60 2

2150

4 60

3

2

3 = 89.12388 1064

= 253.125 106 3.125 1064

89.12388 1064

Thus, the total moment of inertia is

= 160.88 1064

y 75 mm

50 mm

100 mm

150 mm

60 mm

x

d2

d1

1

2

3

1. Beer, F. B., Johnson, E. R., and Eisenberg, E. R., 2006. Vector Mechanics for Engineers: Statics. 9th Ed. McGraw-Hill.

Determine the moment of inertia of the shaded area about the x-axis.

x

y

100 mm

y2 = 100 - x

dAyIA

x

2

Ix = y2(

0

10

100- (100- y2 ))dy

dyyIx

10

0

4

431020 mmxIxANSWER:

x

y

y

y2 = 100 - x

Determine the radius of gyration of the shaded area with respect to y-axis.

100 mm

y

100 mm

Components:

21 yyyT III

2

iiyiyi dAII where

623

1 1067.163

100

236x

bhbhI y


64

2 1027.398

)100(xI y

2

111 dAII yy

yy II 2

1

2

100 mm

y

100 mm

Solution

21 yyyT III

61094.55 xI yT

21 AAAT T

yT

A

I

yk Radius of Gyration

32

1071.252

)100(

2

)100)(200(xAT

mmx

xky 65.46

1071.25

1094.553

6

ANSWER: mmky 65.46

Solution

Obtain the moment of inertia of the composite area about the y-axis

y

75 mm

50 mm

100 mm

150 mm

60 mm

Components:

321 yyyyT IIII

2

iiyiyi dAII where


1

2

3

2111 dAII yy

yy II 2

2111

311

112

dhbhb

I y

2222

322

212

dhbhb

I y

23

24

328

drr

I y

2

333 dAII yy

y

75 mm

50 mm

100 mm

150 mm

60 mm

623

1 1075.168)75)(150)(150(12

)150)(150(xI y

623

2 101875.267)5.187)(100)(75(12

)75)(100(xI y

6224

3 10447.25)60(2

)60(

8

)60(xI y

46321 105.410 mmxIIII yyyyT

ANSWER: 46105.410 mmxI yT

Solution

Find the moment of inertia of the shaded area about: a) y-axis b) x-axis

x

y

75 mm

x2 = 25y

x

x2 = 25y

y

x x=75, y=225

Using horizontal strips:

a) Moment of inertia about the x-axis, Ix

x

x2 = 25y

y

y

x=75, y=225 =

2

= 2 25 1/2 0

225

0

Alternatively, we can use vertical strips:

= 1

323 1

313

=1

3 2253

2

25

3

75

0

= 244.06106 4

= 244.06106 4

x

x2 = 25y

y

x x=75, y=225

Using vertical strips:

b) Moment of inertia about the y-axis, Iy

= 2

= 2 225

2

25

75

0

Alternatively, we can use horizontal strips:

= 1

323 1

313

=1

3 25 1/2

3 03

225

0

= 12.66106 4

x

x2 = 25y

y

y

x=75, y=225

= 12.66106 4

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