07 cross-sectional properties

7/30/2019 07 Cross-Sectional Properties

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7 Properties of an Area

KL2103 Class 01

Semester I 2009/2010

Introduction

We have been able to determine the internal forcesin statically determinate structures subjected to

given loads. The internal forces will be the sameregardless of the materials (steel, concrete, wood,etc.) or the size of the structural members.

The strength of a structure, however, will depend onthe material properties and cross-sectional

properties of its members.

Knowing the properties of a cross-sectional area isthe first step to analyze the strength of a structure.


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Cross-Sectional Area

When we cut a member of a structure and look in

the direction perpendicular to the section line, we

will see the cross-sectional area of the member.

A B C

First Moments of an Area

x

A

y

A

Q y dA

Q x dA

=

=


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Centroid of an Area

A

A

x dA Ax

y dA Ay

=

=

x

y

Q Ay

Q Ax

=

=

First Moments of an Area

The first moments could be positive, zero, or

negative, depending upon the position of the

coordinate axes.

The first moment of an area with respect to an axis

that passes the centroid of the area is zero.


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Composite Area

1 2

1 1 2 2

n

x

A A A A

x n n

i i

Q y dA y dA y dA y dA

Q A y A y A y

A y

= = + + +

= + + +

=

;i i i i

i i

A x A yx y

A A= =

Second Moments / Moments of Inertia

Polar moment of inertia

Product of inertia

2 2

x y

A A

I y dA I x dA= =

2

O

A

I r dA=

xy

A

I xy dA=


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Example 1

Determine the first and second moment about the

x andy axes for the rectangular section shown.

x

y

b

h

Example 1

2

0 0 0

2 2

0 0 0

2

2 2

h b h

x

A

h b h

y

A

bhQ y dA y dx dy by dy

b b hQ x dA x dx dy dy

= = = =

= = = =

First moments

32 2 2

0 0 0

3 32 2

0 0 0

3

3 3

h b h

x

A

h b h

y

A

bhI y dA y dx dy by dy

b b hI x dA x dx dy dy

= = = =

= = = =

Second moments


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Example 2

Determine the moment of inertia about the x andy

axes that pass through the centroid of the

rectangular section.

x

y

b

h C

Example 2

2 2 2

2 2 2

2 2 2

2 2 2

32 2 2

3 32 2

12

12 12

h b h

h b h

h b h

h b h

x

A

y

A

bh

I y dA y dx dy by dy

b b hI x dA x dx dy dy

= = = =

= = = =


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Moments of Inertiafor Some Common Sections

3

3

12

12

x

y

bhI

b hI

=

=

3

3

36

36

x

y

bhI

b hI

=

=

2 2

72xy

b hI =

x

y

C

b

h

x

y

C

b

h

y

C x

4

4

4

64

x y

rI I

D

= =

=

Moments of Inertia

Moments of inertia are always positive.

Product of inertia Ixy

can be positive, zero, or

negative, depending upon the position of the

coordinate axes.

Ixy will be zero ifeither x or y is an axis of

symmetry.


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Parallel-Axis Theorem

0

0

0 0

2

2

2

x x x

y y y

O C

xy x y x y

I I d A

I I d A

I I d AI I d d A

= +

= +

= += +

Example 3

Determine the moments of inertia and product of

inertia for the angle section shown about the x and

y axes that pass through the centroid of the area.

x

y

5

C

1

1

7


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Example 3

( )( )( ) ( )( )( )( )( ) ( ) ( )

( )( )( ) ( )( )( )( ) ( ) ( ) ( )

1 8 0.5 5 1 3.51.65"

1 8 5 1

1 8 4 5 1 0.52.65"

1 8 5 1

x

y

+= =

+

+= =

+

Location of the centroid

x

y

C

y

x

Moments of inertia

( )( ) ( ) ( )( )

( )( ) ( ) ( )( )

3 2

3 2

4

11 8 4 2.65 1 8

12

15 1 0.5 2.65 5 1

12

80.78 in.

xI = +

+ +

=

( ) ( ) ( ) ( )( )

( ) ( ) ( ) ( )( )

3 2

3 2

4

11 8 0.5 1.65 1 8

12

15 1 3.5 1.65 5 1

12

38.78 in.

yI = +

+ +

=

Product of inertia( )( )( )( )

( )( )( )( )4

0 4 2.65 0.5 1.65 1 8

0 0.5 2.65 3.5 1.65 5 1

32.31 in.

xyI = +

+ +

=

Moments of Inertia about Inclined Axes


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Moments of Inertia about Inclined Axes

'

'

' '

cos2 sin 22 2

cos2 sin 22 2

sin 2 cos22

x y x y

x xy

x y x y

y xy

x y

x y xy

I I I II I

I I I II I

I II I

+ = +

+ = +

= +

Example 4

Determine the moments of inertia and product of

inertia for the angle section shown about the x

andy axes given.

1

y

5

C

1

7

x

60o


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Example 4

( )

( )

( )

'

4

'

4

' '

4

80.78 38.78 80.78 38.78cos120 32.31 sin120

2 2

77.25 in.

80.78 38.78 80.78 38.78cos120 32.31 sin120

2 2

42.30 in.

80.78 38.78 sin120 32.31 cos1202

34.34 in.

o o

x

o o

y

o ox y

I

I

I

+ = +

=

+ = +

=

= +

=

Principal Moments of Inertia

2

2

max,min2 2

x y x y

xy

I I I I

I I

+

= +

2tan2

xy

p

y x

I

I I =

Principal axes


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Example 5

Determine the principal moments of inertia for thesection shown about the axes that pass through thecentroid of the area, and the rotation angles that

produce the principal axes.

x

y

5

C

1

1

7

Example 5

( )

( )

22

1

4

22

2

4

80.78 38.78 80.78 38.7832.31

2 2

98.31 in.

80.78 38.78 80.78 38.7832.31

2 2

21.24 in.

I

I

+ = + +

=

+ = +

=

( )1

1

2

2 32.311tan

2 38.78 80.78

28.5

118.5

p

o

p

o

p

=

=

=


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Example 5

1

2

5

C

1

1

7

4

1 '

4

2 '

28.5 98.31 in.

118.5 21.24 in.

o

p x

o

p x

I

I

= =

= =

28.5o

Mohrs Circle

A graphical way to transform Ix, Iy, andIxy into Ix,

Iy, andIxy and the principal moments of inertia.

y Establish a coordinate system with horizontal axis Iand

vertical axis Ixy.

y Plot point O on Iaxis at a distance (Ix + Iy)/2 from the

origin. This is the center of the Mohrs circle.

y Plot point A (Ix, Ixy).

y Draw the circle with O as the center andOA as the radius

of the circle.


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EF

Mohrs Circle

Point Crepresents (Ix, Ixy) if the x-y axes are

rotated with an angle counterclockwise.

Point Eis the maximum moment of inertia Imax or

I1.

Point Fis the minimum moment of inertia Imin orI2.

The angle is the rotation angle forx-axis to obtain

the principal axis.


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Example 6

Use Mohrs circle to reproduce the results of

Examples 4 and 5 for the section shown.

x

y

5

C

1

1

7

Example 6

A

O80.78

32.31

98.31I

21.2457o

120o

C

38.78

34.34

77.25

42.30

Ixy

0 50 in.410 20 30 40

07 cross-sectional properties

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