2- handouts lecture-44 45

COURSE:COURSE: CE 201 (STATICS)CE 201 (STATICS)

LECTURE NO.:LECTURE NO.: 44 & 4544 & 45

FACULTY:FACULTY: DR. SHAMSHAD AHMADDR. SHAMSHAD AHMAD

DEPARTMENT:DEPARTMENT: CIVIL ENGINEERINGCIVIL ENGINEERING

UNIVERSITY:UNIVERSITY: KING FAHD UNIVERSITY OF PETROLEUM KING FAHD UNIVERSITY OF PETROLEUM & MINERALS, DHAHRAN, SAUDI ARABIA& MINERALS, DHAHRAN, SAUDI ARABIA

TEXT BOOK:TEXT BOOK: ENGINEERING MECHANICSENGINEERING MECHANICS--STATICS STATICS by R.C. HIBBELER, PRENTICE HALLby R.C. HIBBELER, PRENTICE HALL

LECTURE NO. 44 & 45LECTURE NO. 44 & 45MOMENT OF INERTIAMOMENT OF INERTIA

Objectives:Objectives: To develop a method for determining the moment To develop a method for determining the moment

of inertia for an areaof inertia for an area

To explain the ParallelTo explain the Parallel--Axis TheoremAxis Theorem

To show how to determine the moment of inertia To show how to determine the moment of inertia for composite areasfor composite areas

MOMENT OF INERTIA FOR AREAS MOMENT OF INERTIA FOR AREAS

Moment of inertia for an area is determined by taking integral ofthe second moment of area about an axis, as follows.

2

2

2

x A

y A

o x yA

I y dA

I x dA

J r dA I I

=

=

= = +

Where:

Ix = moment of inertia about x-axis Iy = moment of inertia about y-axis Jo = polar moment of inertia

PARALLEL AXIS THEOREMPARALLEL AXIS THEOREM

The parallel-axis theorem is useful for determination ofthe moment of inertia for composite bodies.

For an area, as shown in the figure below, the parallel-axis theorem may be stated as:

2'

2'

2

x x y

y y x

o C

I I Ad

I I Ad

J J Ad

= +

= +

= +

PROBLEM SOLVING: PROBLEM SOLVING: Example # 1Example # 1Determine the moment of inertia of the triangular area about the y-axis.

PROBLEM SOLVING: PROBLEM SOLVING: Example # 1Example # 1

Let us consider a differential element of thickness dx at a distance x from the y-axis, as shown in the following figure:

dA=ydx= (b-x)dxhb

( )

( )

2 2

3 42 3

4 4 4

3

3 4

3 4 12

Ans.12

b

yo

bb

o o

y

hI x dA x b x dxb

h h bx xbx x dxb b

h b b hbb b

hbI

= =

= =

= =

=


Determine the moment of inertia ofthe shaded area about the y-axis.

PROBLEM SOLVING: PROBLEM SOLVING: Example # 2Example # 2y

x

dxx

h

b

y

y'=h-y

y

x

dxx

h

b

y

y'=h-y

( )

'

22

2 22

( )dA y dx h y dxhdA h x dxb

hdA b x dxb

= =

=

=

( ) ( )2 2 2 2 2 2 42 22 3 5 5 5

2 2

3

3 5 3 5

2 Ans.15

b b

yo o

b

o

y

h hI x dA x b x dx b x x dxb b

h b x x h b bb b

b hI

= = =

= =

=

PROBLEM SOLVING: PROBLEM SOLVING: Example # 3Example # 3Locate the centroid x of the abovecomposite section. Then find the moment of inertia of the compositearea about y-axis passing through thecentroid (i.e., about y-axis).


1in

4in

4in

2in

1in

B

A

x

yy'

x'

x

Centroid1in

4in

4in

2in

1in

B

A

x

yy'

x'

x

Centroid

The x coordinate of the centroid (i.e., x ) can be determined as:Rectangle A(in2) (in)x xA A 4 2 = 8 1 8 B 8 2 = 16 4 64 A = 24 x A = 72

72 3 in24

xAxA

= = =

For rectangle A: ( ) ( )'

32 44 2 4 2 2 34.66 in

12yI = + =

For rectangle B: 3

2 4'

2 8 8 2 1 101.33 in12y

I = + =

Summation: 4' 34.66 101.33 136 inyI = + = Ans.


Determine the moment of inertia of thefollowing composite section about the x-axis.

PROBLEM SOLVING: PROBLEM SOLVING: Example # 4Example # 4 y

x

y'

x'C

x = 68mm

1

2

3

80mm

80mm

y = 80mm

80mm80mm

60mm20mm

48mm

12mm

y

x

y'

x'C

x = 68mm

1

2

3

80mm

80mm

y = 80mm

80mm80mm

60mm20mm

48mm

12mm

The x coordinate of the centroid (i.e., x ) can be determined as:

( )mmx ( )3mmxASegment A (mm2)1 40 80 = 3200 20 640002 160 40 = 6400 80 5120003 160 40 = 6400 80 512000

2 3 = 16000 mm ; 1088000 mmA xA = 1088000 68 mm16000

xAxA

= = =

PROBLEM SOLVING: PROBLEM SOLVING: Example # 4Example # 4 y

x

y'

x'C

x = 68mm

1

2

3

80mm

80mm

y = 80mm

80mm80mm

60mm20mm

48mm

12mm

y

x

y'

x'C

x = 68mm

1

2

3

80mm

80mm

y = 80mm

80mm80mm

60mm20mm

48mm

12mm

Summation 6 4

'6 4

'

1706666.66 2 23893333.33 49.49 10 mm A

7799466.66 2 14574933.33 36.94 10 mm Ax

y

I

I

= + =

= + =

Rectangle 1: 3

4'

32 4

'

40 80 1706666.66 mm12

80 40 80 40 48 7799466.66 mm12

x

y

I

I

= =

= + =

Rectangle 2: 3

2 4'

32 4

'

160 40 160 40 60 23893333.33 mm12

40 160 160 40 12 14574933.33 mm .12

x

y

I

I

= + =

= + =

Rectangle 3: 3

2 4'

32 4

'

160 40 160 40 60 23893333.33 mm12

40 160 160 40 12 14574933.33 mm .12

x

y

I

I

= + =

= + =

PROBLEM SOLVING: PROBLEM SOLVING: Example # 5Example # 5Determine the moment of inertia of thefollowing shaded area about the x and yaxes.


( )4 2'x x yI in Ad= +I ( )4 2'y y xI in Ad= +ISegment A (in2) Ix (in4) Iy (in4)dy

(in)dx

(in)1 18 54 13.5 7 1.5 936 54.02 9 18 4.5 6 4.0 342 148.53 24 32 72 2 3.0 128 288.0

4 12.566

12.566

12.566 4 3.0 213.628 126.663

Ans. Ix= 1192.37 in4 Iy= 364.83 in4

PROBLEM SOLVING: PROBLEM SOLVING: Example # 6Example # 6Determine the moment of inertia of thefollowing composite section about the x

axis. y =154.4 mm


1

2

3

50mm

d y3=

60.6

mm

100m

m

d y2=

64.4

mm

y=154.4mm

15m

m

150mm

d y1=

146.

9mm

150m

m

15mm

x

1

2

3

50mm

d y3=

60.6

mm

100m

m

d y2=

64.4

mm

y=154.4mm

15m

m

150mm

d y1=

146.

9mm

150m

m

15mm

x

32 6 4

'150 15Segment 1: 150 15 146.9 48.596 10 mm

12xI = + =

32 6 4

'15 150Segment 2: + 15 150 64.4 = 13.550 10 mm

12xI =

( )4 2 2 6 4'Segment 3: 50 + (50) 60.6 = 33.751 10 mm4xI =

Summation: Ix = 95.898 106 mm4 Ans.

Multiple Choice Problems1. The moment of inertia of the composite section (shown in the following

figure) about the x-axis is

(a) 288 in4 (b) 928 in4 (c) 2016 in4 (d) 5760 in4

Ans: (d) Feedback:

3 342 28 12 6 12 5760 in8 12 6 6 12 2

12 12xI = + =+ +

Multiple Choice Problems

2. The moment of inertia of the composite section (shown in the followingfigure) about the x-axis is

(a) 290.66 in4 (b) 190.66 in4 (c) 90.66 in4 (d) 85.33 in4 Ans: (a) Feedback:

3 3

42 2'

2 8 8 2 290.66 in2 8 2.5 8 2 2.512 12x

I = + =+ +

6.5 iny =

2- handouts lecture-44 45

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