Moment Inertia Integration

10.1 Moments of Inertia by Integration

10.1 Moments of Inertia by Integration Example 1, page 1 of 2

1. Determine the moment of inertia of the rectangle

about the x axis, which passes through the centroid C.

C

b

h/2

x

y

h/2


x

yWe want to evaluate

Ix = y2 dA

where the differential element dA is located a

distance y from the x axis (y must have the same

value throughout dA).

yh/2

h/2

Ix = y2 dA

= y2 (b dy)

by3

3

bh3 Ans.

12

=

=

Limits of integration

dA = area of rectangle

= b dy

1

2

3

4h 2

-h 2

dy

h 2

-h 2

h 2

-h 2

C

b


h

b

y

x

2. Determine the moment of inertia of the rectangle

about its base, which coincides with the x axis.


b

y

x

We want to evaluate

Ix = y2 dA

where the differential element of area dA is located a distance y from the

x-axis (y must have the same value throughout the element dA).

hy

2 dA = area of rectangle

= b dy

3

=

0

h

Ix = y2 dA

= y2 (b dy)

by3

3

= Ans.

Limits of integration4

1

h

0

dy

bh3

3


b

3. Determine the moment of inertia of the

right triangle about the x axis.

x

y

h


x

y

b

x

y

(x,y)

We want to evaluate

Ix = y2 dA

where y has the same value throughout

differential element dA.

Equation of line

y = slope x + intercept

( ) x + h (1) =

1


= x dy

h

Since we will integrate with respect to y,

we must replace x by a function of y.

Solve for x to get

x = ( ) y + b

so,

dA = x dy

= ( y + b) dy

4

5

3

2

dy

b

h

bh

hb


Ix = y2 dA

= y2 ( y + b) dy

= b ( + y2) dy

= b[ + ]

= bh3[ + ]

= Ans.

6

Limits of integration are from

bottom of triangle to top.7

h

0

bh

y3

h

y4

4h

y3

3

h

0

311

4

bh3

12

0

h


x

yx2 y2

a2 b2 = 1+

4. Determine the moments of inertia of the area

bounded by an ellipse about the x and y axes.

b

b

a a


x

y

dy

x

y

We want to evaluate

Ix = y2 dA

where y has the same value throughout the

differential element dA.

x = half the length of the

differential element,

so 2x = entire length

1

(x, y) dA = area of rectangle

= 2x dy

Since we will integrate with

respect to y, we must replace x

by a function of y

Since x locates a point to the right of the y axis, choose

the plus sign:

x = +a 1 (y/b)2 (1)

+ 1=a2 b2

x2 y2

Solve for x to get

x = a 1 (y/b)2

2

4

3

6

5


Evaluate the integral either with a

calculator that does symbolic

integration or use a table of integrals.

To calculate Iy, use symmetry in the following way: in all the

above equations, replace x's by y's, original y's by x's, a's by b's

and original b's by a's. Then the result would be

Iy = Ans.

Limits of integration from

bottom to top of ellipse

8

9

11

10

b

-b

4ab3

a3 b4

7 Using Eq. 1 in the expression for dA gives:

dA = 2x dy

= 2a 1 (y/b)2 dy

Ix = y2 dA

= y2(2a 1 (y/b)2 ) dy

= Ans.


2 ft

5. Determine the moments of inertia of the

crosshatched area about the x and y axes.

y = 4 x2

x

y

4 ft


2 ft

We want to evaluate

I x = y2 dA

where y has the same value throughout the differential element dA.


= x dy

Since we will integrate with respect to y,

we must replace x by a function of y.

Solving for x to get

x = 4 y

Since x locates a point to the right of the y axis, choose the plus sign:

x = + 4 y (1)

1

2

3

4

5

y = 4 x2

(x,y)

y

dy

x

y

4 ft


Using Eq. 1 in the expression for dA gives

dA = x dy

= 4 y dy

6

Ix = y2 dA

= y2 4 y dy

= 19.50 ft4 Ans.

Use the integral function

on your calculator

7

8

9

y = 4 at the top of

the crosshatched area

4

0


2 ft

Next we want to evaluate

Iy = x2 dA

where x has the same value throughout the differential element dA. Thus, we

choose a vertical rectangular strip.


= y dx

= (4 x2) dx

Iy = x2 dA = x2(4 x2 ) dx = 4.27 ft4 Ans.

dx

x

10

11

120

2

yy = 4 x2

x

y

4 ft



crosshatched area about the y axis.

x = 2y6 50y5 y3 + 100

x

y

1.156 m

100 m

Scales on

the x and y axes are

not the same.


100 m

1 We want to evaluate

Iy = x2 dA

where x has the same value throughout the differential

element dA. Thus, we choose a vertical rectangular strip.


= y dx

But this approach won't work because we

can't express y as a function of x.

2

3

y

x = 2y6 50y5 y3 + 100

x

y


4 An alternative approach is to use a horizontal

rectangular strip and employ the equation for the

moment of inertia of a rectangle about its base (BB) :

(1)

B

Bh

b

100 m

1.156 m

yApplying Eq.1 to the differential element

gives the differential moment of inertia.

dIy

(dy)x3

= 3

5

Replacing x by the function of y gives

dIy = x3dy

= (2y6 50y5 y3 + 100)3 dy

Iy = dIy = (2y6 50y5 y3 + 100)3 dy

= 2.72 105 m4 Ans.

Use the integral function on a calculator to evaluate

this integral.

6

7

8

x = 2y6 50y5 y3 + 100

x

x

dy

bh3

3I BB =

=3bh3

31

31

31

0

1.156


y = 3x x2

y = 9 x2


crosshatched area about the x axis.

9 m

3 m

x

y


x

y

We want to evaluate

Ix = y2 dA

where y has the same value throughout the differential

element dA. Thus it appears that we should use a

horizontal rectangular strip.

But using a horizontal strip is awkward we would

have to use three different expressions for dA,

depending on the position of the strip.

2

1

A better approach is to use a vertical strip and then

apply the parallel-axis theorem to the strips.

Parallel-axis theorem

for a general region

Ix = Ic + Ad2d

Centroid

Area A

x

y

For a rectangle in particular,

Ix = bh3 + (b h)d2

= ( h3 + hd2 )b

y

x

y

x

C

d

C

h

b

h

b dx

d

The moment of inertia of a strip of width

b dx is then

dIx = ( h3 + hd2 ) dx (1)

3

121

112

121

C


x

y

(xel,yel)

(x,y1)

(x,y2)

The y-coordinate of the element

centroid is the average of y1 and y2

d = yel = ( y1 + y2 )

Eq. 1 becomes

dIx = ( h3 + hd2 ) dx

= { ( y2 y1 )3 + ( y2 y1 ) [ ( y1 + y2 )]

2} dx

Since the point (x, y2) lies on the curve y = 9 x2, we

can substitute

y2 = 9 x2

in Eq. 2. Similarly, since (x, y1) satisfies y = 3x x2, we

can substitute

y1 = 3x x2

in Eq. 2. Thus Eq. 2 becomes

5 6

h = y2 y1

4

yel

centroid

y = 3x x2

y = 9 x2

dIx = { ( y2 y1 )3 + ( y2 y1 ) [ ( y2 + y1 )]

2} dx

= { [( 9 x2 ) ( 3x x2 )]3 + [( 9 x2 )

( 3x x2 )][ ( 9 x2 ) + ( 3x x2 )]2} dx

= { ( 9 3x )3 + ( 9 3x )[ + x x2 ]2} dxIx = dIx

= { ( 9 3x )3 + (9 3x) ( + x x2 )2 } dx

= 328 m4 Ans.

Enter this expression directly into the

integral function of a calculator.

7

8

x ranges

from 0 to 3.

21 12

1

112 2

1

121 1

2112

29 3

292 2

30

3121

112

122

1

(2)




xy = 1

y = 2x4 m

1 m1 m1/2 m

4 m 1/4 m

y

x


We want to evaluate

Iy = x2 dA


element dA. Thus it appears that we should use

vertical differential strips.

But using a vertical strip is awkwardwe would have to use three different expressions for dA, depending on the location of the strip.

2

1

x

y


3 A better approach is to use a horizontal strip and then

apply the parallel-axis theorem to the strip.

x

y

x

y

y

x

area A

centroid

C

Parallel-axis theorem for a general region

Iy = Ic + Ad2

For a rectangle in particular,

Iy = bh3 + (b h)d2

= ( h3 + hd2 )b

The moment of inertia of a strip of width b dy is then

dIy = ( h3 + hd2 ) dy (1)

C

C

d

d

d

Area = b h

b

h

b dy

h

12

112

112

1


| xel | = | average of x1 and x2 | = | ( x1 + x2 )|d =

h = x2 x1

y

x

xy = 1

y = 2x

2

y

2

y

112 4

y

2y

1

2

y2

yy1

y14

1y

412y

112 2

y1y

y

21y

1y2

1y

2y1

121

y1

12

112

1y

21

121

121

21

y

2

8

7

6

5

Top of region at y = 4

Bottom of

region at y = 1= 2.81 m4 Ans.

Enter this expression directly into the

integral function of a calculator

Iy = dIxy

= [ ( + )3 + ( + ) ( )2 ] dy

dIy = { ( x2 x1 )3 + ( x2 x1 )[ ( x2 + x1 )]

2 } dy

= { ( ( ))3 + ( ( ))[ ( + ( ) )]2 } dy

= { ( + )3 + ( + )[ ]2 } dy

Since the point (x1,y) lies on the line y = 2x,

we can solve for x and substitute

x1 =

in Eq. 2. Similarly, since (x2,y) lies on the

curve xy = 1, we can substitute

x2 =

in Eq. 2. Thus Eq. 2 becomes

4

Centroid of

strip

From the figure, we see that Eq. 1 can be written as

dIy = { h3 + hd2} dy (Eq. 1 repeated)

= { ( x2 x1 )3 + ( x2 x1 )[ ( x2 + x1 )]

2 } dy (2)

(x2,y)(x1,y)

(xel,yel)

C



crosshatched area about the x axis.

y

x

y = 10e x2

x2ey = 10

1 m

1.5 m


1

2

3 Ix for region D

No integration needed.

Use the formula for the moment of inertia of a

rectangle about a centroidal axis.

4

b

h/2

h/2

So

=

=

C

IxD

12

bh3

Ic = (1)

12

33.1915 m4 (2)

x

y

3.6788 m

3.6788 m

1 m

(1 m)(2 3.6788 m)3

y(1) = 10e = 3.6788 m(1)2

1 m

We want to evaluate

Ix = y2dA

Because of the shape of the region, the integral has to

be evaluated over two sub-regions, D and E.

Region E

Region D


5 Ix for region E. Because the region is

symmetrical about the x axis, we can save

work by using vertical rectangular strips and

by applying Eq. 1 to these strips:

Ic = bh3

dIxE = (dx)(2y)3

= (dx)[2( )]3

= dx

Thus, with the aid of the integral function on a

calculator, we have

IxE = dIx

= e dx

= 4.7985 m4 (3)

Adding the results for regions D and E gives

Ix = IxD + IxE

= 33.1915 + 4.7985

= 38.0 m4 Ans.

by Eq. 2 by Eq. 3

6121

112

112

20003

1

1.5

32000

or,

x

y

dx

1 m

x

1.5 m

10e x2

x2ey = 10

3x2e

3x2


1 in.

2 in.

4 in.

x

y

y = x2

y = 3x



3 in.


1

2

3 Integrate over region B.

y

x

x

1 in.

(x,y2)

(x,y1)

y = 3x

y = x2


= ( y2 y1 ) dx

= ( 3x x2 ) dx

4

5 IyB

= x2 dA

= x2(3x x2) dx

= 0.5500 in4 (2)

1

0

dx

y

x

y2 y1

We want to evaluate

Iy = x2dA (1)


element dA.

Given the shape of the area, we have to evaluate the

integral in Eq.1 over two sub-regions, B and C.

Region C

Region B


x

y

6 Integrate over region C.

4 in.

x

y2 y1

dx

2 in.

(x,y1)

y = x2

(x,y2)

Write the equation of the line: =

Solving gives

y = x + 2

dA = Area of rectangle

= ( y2 y1 ) dx

= [(x + 2) x2 ] dx

IyC

= x2 dA

= x2( x + 2 x2 ) dx

= 2.2167 in4 (3)

Adding the results for regions B and C gives

Iy = IyB

+ IyC

= 0.5500 + 2.2167 = 2.77 in4 Ans.

by Eq. 2 by Eq. 3

7

8

9

10

2 1

x 1 y 3

4 3

1

2

1 in.

3 in.


Area of quarter circle

4a2 4a

3

4a4

9

34a

a

11. Given that the centroid C of the area bounded by a

quarter-circle lies a distance 4a/(3 above the base of the

quarter-circle, determine the moment of inertia of the cross

hatched area about an axis xc through the centroid.

y yc

xc

x

x2 + y2 = a2

C

We want to evaluate

Ixc = yc2 dA

but because the equation of the circle is given in

terms of x and y instead of xc and yc, it is easier to

evaluate

Ix y2 dA

and then use the parallel-axis theorem:

Ix = Ixc + Ad2

= Ixc + ( )2

Solving gives

Ixc = Ix

Thus now we need to calculate Ix.

1


x2 + y2 = a2

x

y2

y

x

(x,y)

dA = Area of rectangle

= x dy

Solve x2 + y2 = a2

to get

x = a2 y2

Substitute into the equation for dA

dA = x dy

= a2 y2 dy

Integrate

Ix = y2 dA

= y2 a2 y2 dy

= (2)

a

0

Use the result given by Eq. 2 in Eq. 1:

IxC = Ix (Eq. 1 repeated)

( ) a4 Ans.

Evaluate the integral with a

calculator that does symbolic

integration or use a table.

3

4

5

6

7

dy

16

4a4

9

94a4 4

9a4

16

a

a4

16

Date post:	08-Apr-2016
Category:	Documents
Upload:	tsuki-zombina
View:	111 times
Download:	2 times

Moment Inertia Integration

Documents