Mechanical Engineering Principles

FIRST AND SECOND MOMENT OF AREAS 97

0.1 m

0.1 m

0.1 m

ab

c d

e f

g h

N A

Figure 7.30

The centroid of this beam will be on the horizontal
axisNA, as shown in Figure 7.30.
The second moment of area of the channel section
is given by:

INA=(Iof rectangleabdcaboutNA)

−(Iof rectangleefhgaboutNA)

=

0. 1 × 0. 23

12

−

0. 08 × 0. 163

12

= 6. 667 × 10 −^5 − 2. 731 × 10 −^5

i.e. INA= 3. 936 × 10 −^5 m^4

Problem 18. Determine the second moment

of area about a horizontal axis passing

through the centroid, for the tee beam shown

in Figure 7.31.

0.2 m

2

1

N A

X X

y Thickness = 0.02 m

0.1 m

Figure 7.31 Tee beam

In this case, we will first need to find the position
of the centroid, i.e. we need to calculate y in
Figure 7.31. There are several methods of achieving
this; the tabular method is as good as any since it
can lead to the use of a spreadsheet. The method
is explained below with the aid of Table 7.2 on
page 98.
First, we divide the tee beam into two rectangles, as
shown in Figure 7.31.
In the first column we refer to each of the two rectan-
gles, namely rectangle (1) and rectangle (2). Thus,
the second row in Table 7.2 refers to rectangle (1)
and the third row to rectangle (2). The fourth row

refers to the summation of each column as appropri-

ate. The second column refers to the areas of each

individual rectangular element,a.

Thus, area of rectangle (1),

a 1 = 0. 1 × 0. 02 = 0 .002 m^2

and area of rectangle (2),

a 2 = 0. 18 × 0. 02 = 0 .0036 m^2

Hence,

∑

a= 0. 002 + 0. 0036 = 0 .0056 m^2

The third column refers to the vertical distance of

the centroid of each individual rectangular element

from the base, namelyXX.

Thus, y 1 = 0. 2 − 0. 01 = 0 .19 m

and y 2 =

0. 18

2

= 0 .09 m

In the fourth column, the product ay is obtained by

multiplying the cells of column 2 with the cells of

column 3,

i.e. a 1 y 1 = 0. 002 × 0. 19 = 3. 8 × 10 −^4 m^3

a 2 y 2 = 0. 0036 × 0. 09 = 3. 24 × 10 −^4 m^3

and

∑

ay= 3. 8 × 10 −^4 + 3. 24 × 10 −^4

= 7. 04 × 10 −^4 m^3

In the fifth column, the productay^2 is obtained by

multiplying the cells of column 3 by the cells of

column 4, i.e.

∑

ay^2 ispartof the second moment

of area of the tee beam aboutXX,

i.e. a 1 y 12 = 0. 19 × 3. 8 × 10 −^4

= 7. 22 × 10 −^5 m^4

a 2 y 22 = 0. 09 × 3. 24 × 10 −^4

= 2. 916 × 10 −^5 m^4

and

∑

ay^2 = 7. 22 × 10 −^5 + 2. 916 × 10 −^5

= 1. 014 × 10 −^4 m^4

In the sixth column, the symbolirefers to the second

moment of area of each individual rectangle about

its own local centroid.

Now i=

bd^3

12

from Table 7.1