Properties of Area
Centroid
1st Moment of area
2nd Moment of area
Section Modulus
CENTROID OF AREAS
• Centroid of an area is the point at which the total area
may be considered to be situated for calculation
purposes.
• Corresponds to the centre of gravity of a lamina of the
same shape as the area
• Often possible to deduce the centroid by SYMMETRY of
the area.
• Need to know position of the centroid of a section as
bending occurs with compression above and tension
below this axis.
• Distance from centroid to axis of rotation
(x or y) is 1st moment of area /total area
1st Moment of Area
B
Moment of Force =
Fxd
F
d
A
Likewise;
First Moment of Area
about the line CD =
D
d
C
Area
A
G = centroid of area
Axd
CENTROID OF AREAS
y
Total
Area A
x
G
x
x
Elemental area
y
a
y
x
y
 ax
a
 ay
a
1st moment of Area - Example
30
60
30
Find centroid of
the composite
beam section
shown
50
7
15
35
65
20 Dia
1st moment of Area – Example (Ans)
distance to centroid
Area Ref breadth
1
2
3
50
35
28
hole
-10
width
area
30
30
60
ΣA =
1st Moments of area
1500
1050
1680
from Oy
x
15
105
60
From Ox
y
25
18
14
About Oy
Ax
22500
110250
100800
About Ox
Ay
37500
18900
23520
-314
65
15
-20420
-4712
213130
75208
Σ Ax =
3916
x=
y=
Σ Ax
ΣA
Σ Ay
ΣA
=
213130
=
54
=
19
3916
=
75208
3916
= Σ Ay
2nd Moment of Area
• A property of area used in many engineering
calculations (e.g. stress in beams)
• Elemental
Elemental area a
Second Moment of Area
about the line CD = I
D
x
C
I 
ax
2
Standard Results for I
• Using differential calculus we can formulate standard
solutions, eg:
b
• Rectangle about its base
bd 3
d
I
• Rectangle about its centre
b
d
O

3
3
bd
I CC  12
• For more complicated shapes can use compound areas and
parallel axes theorem
• Or, easier, use tables from steel joist manufacturers
Example / Exercise
• Loaded Timber beam has max BM of 5 kNm, find stress in the section.
100
300
5 kNm
Section
compression
tension
Stress block
M
I
BMD
I = bd3 / 12
= 100 x 3003 mm4
12
= 102 x 33 x 1003
12
= 27 x 102 x 106
12
= 2.25 x 108 mm4
f
E


y
R
f
My

Hence
I
Hence f = 5 x 103 x 103 Nmm x 150 mm
2.25 x 108 mm4
= 750 x 106
225 x 106
= 3.33 N/mm2