Lecture Strength of Materials I: Chapter 5 - PhD. Tran Minh Tu

STRENGTH OF MATERIALS

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TRAN MINH TU - University of Civil Engineering,
Giai Phong Str. 55, Hai Ba Trung Dist. Hanoi, Vietnam

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CHAPTER

5
Geometric Properties of an Area
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Contents
5.1. Introduction

5.2. First moment of area
5.3. Moment of inertia for an area
5.4. Moment of inertia for some simple areas

5.5. Parallel - axis theorem
5.6. Examples

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5.1. Introduction

Dimension, shape?

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5.2. First Moment of Area
5.2.1. Definition
• The first moment of a plane A about
the x- and y-axes are defined as

Sx 



( A)

ydA

Sy 



xdA

( A)

• Value: positive, negative or zero

• Dimension: [L3]; Unit: m3, cm3,...
• Centroidal axes: are axes, which first moment of a plane A about them
is zero
5.2.2. The centroid of an area

• The centroid C of the area is defined as the point in the xy-plane that
has the coordinates
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5.2. First Moment of Area
xC 

Sy
A

Sx
yC 
A

yC

• If the origin of the xy-coordinate system
is the centroid of the area then Sx=Sy=0

C

xC

• Whenever the area has an axis of
symmetry, the centroid of the area will lie
on that axis
• If the area can be subdivided in to simple geometric shapes
(rectangles, circles, etc., then
n

Sx   S
i 1

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n

i
x

S y   S yi
i 1

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5.2. First Moment of Area
y

5.2.3. The centroid of composite area
yC1

n

xC 

Sy
A



x
i 1
n

Ci

Ai
i

y

Ci

i 1

C3
x

xC1

n

Ai

n

A
i 1

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C2

A
i 1

Sx
yC 

A

C1

i

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5.3. Moment of Inertia for an Area
5.3.1. Moment of inertia

Ix 



y 2dA

Iy 



x 2dA

( A)

( A)

5.3.2. Polar moment of inertia

Ip 



 2 dA  I x  I y

( A)

5.3.3. Product of inertia

I xy 

 xydA

( A)

• The value of moment of inertia and polar
moment of inertia always positive, but the
product of inertia can be positive, negative,
or zero
• Dimension: [L4]; Unit: m4, cm4,...

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5.3. Moment of Inertia for an Area
- The product of inertia Ixy for an area will be zero if either the x or the y
axis is an axis of symmetry for the area
- The area with hole, then the hole’s
area is given by minus sign.
- The composite areas:
n

Sx   S
i 1

i 1

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n

S y   S yi
i 1

n

Ix   I

i
x

n

i
x

I y   I yi
i 1

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5.4. Moment of Inertia for some simple areas
•

Rectangular

hb3
Iy 
12

Ip 

R
2

Ix  I y 

•

y
h

bh3
Ix 
12
•
Circle
4

y



 R4
4

D
32


x

4

 0,1D 4

 D4
64

x
b

 0,05D

4

D

Triangular

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h

bh3

Ix 
12

x
b

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5.5. Paralell-axis Theorem
• In the xy coordinates, an area
has geometric properties: Sx, Sy,
Ix, Iy, Ixy.
• In the uv coordinates: O'u//Ox,
O'v//Oy và:

u  xb

v ya

• Geometric properties of an area
in the coordinates O'uv are:

Su  S x  a. A

Sv  S y  b. A
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Iu  I x  2aS x  a 2 A

I v  I y  2bS y  b2 A
Iuv  I xy  aS y  bS x  abA

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5.5. Paralell-axis Theorem
If O go through centroid C, then:

Iu  I x  a 2 A
I v  I y  b2 A

Iuv  I xy  abA

C

C

. Radius of gyration
The radius of gyration of an area about the x and y axes, and the point
O are defined as

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Iy
Ix
rx 
; ry 
A
A

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5.5. Paralell-axis Theorem

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5.5. Paralell-axis Theorem

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Example 5.1
Problem 5.6.1. An area with the shape
and the dimension as shown in the
figure. Determine the principal moment
of inertia for area .
Solution
Choosing the primary
coordinates x0y0 as shows in the figure.
Divide the composite area to 2 simple
2
areas 1

y0

1. Determine the centroid:

1

- xC=0 (y0 – axis of symmetry)
2

x0
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Example 5.1
- Draw the principal coordinates Cxy

y0

- The Principal moment of inertia for an area:
1
2
x
0

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Example 5.2
Problem 5.2.

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Example 5.2

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Example 5.3

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Example 5.3

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THANK YOU FOR
ATTENTION !

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