Mechanical Engineering Principles

96 MECHANICAL ENGINEERING PRINCIPLES

10. Determine the second moments of areas
    about the given axes for the shapes
    shown in Figure 7.25 (In Figure 7.25(b),
    the circular area is removed.)
    ⎡

⎢

⎢

⎣

IAA=4224 cm^4 ,

IBB=6718 cm^4 ,

ICC=37300 cm^4

⎤

⎥

⎥

⎦

3.0 cm

4.0 cm

9.0 cm

16.0 cm

(a)

4.5 cm

9.0 cm

15.0 cm

10.0 cm

Dia = 7.0 cm

(b)

AAC

B

B

C

Figure 7.25

7.8 Second moment of area for

‘built-up’ sections

The cross-sections of many beams and members of
a framework are in the forms of rolled steel joists
(RSJ’s orI beams), tees, and channel bars, as
shown in Figure 7.26. These shapes usually afford
better bending resistances than solid rectangular or
circular sections.

(a) RSJ (b) Tee beam (c) Channel bar

Figure 7.26 Built-up Sections

Calculation of the second moments of area and the
position of the centroidal, or neutral axes for such
sections are demonstrated in the following worked
problems.

Problem 16. Determine the second moment

of area about a horizontal axis passing

through the centroid, for theIbeam shown

in Figure 7.27.

0.1 m

0.2 m

Thickness = 0.02 m

Figure 7.27

The centroid of this beam will lie on the horizontal

axisNA, as shown in Figure 7.28.

a

N A

b

c

k

d

e f

h

j

l m

0.1 m

0.1 m g

Figure 7.28

The second moment of area of the I beam is

given by:

INA=(Iof rectangleabdc)−(Iof rectangleefhg)

−(Iof rectanglejkml)

Hence, from Table 7.1,

INA=

0. 1 × 0. 23

12

−

0. 04 × 0. 163

12

−^0.^04 ×^0.^16

3

12

= 6. 667 × 10 −^5 − 1. 365 × 10 −^5

− 1. 365 × 10 −^5

i.e. INA= 3. 937 × 10 −^5 m^4

Problem 17. Determine the second moment

of area about a horizontal axis passing

through the centroid, for the channel section

shown in Figure 7.29.

0.2 m

0.1 m

Thickness = 0.02 m

Figure 7.29