Mechanical Engineering Principles

90 MECHANICAL ENGINEERING PRINCIPLES

P G

C

x

b

P d G

2

d

2

dx

Figure 7.9

From the parallel axis theorem

IPP=IGG+(bd)

(

d

2

) 2

i.e.

bd^3

3

=IGG+

bd^3

4

from which,

IGG=

bd^3

3

−

bd^3

4

=

bd^3

12

Perpendicular axis theorem

In Figure 7.10, axesOX,OYandOZare mutually
perpendicular. IfOXandOYlie in the plane of area
Athen the perpendicular axis theorem states:

IOZ=IOX+IOY

A summary of derived standard results for the sec-
ond moment of area and radius of gyration of regular
sections are listed in Table 7.1, on page 91.

Area

Figure 7.10

The second moment of area of ahollow cross-
section, such as that of a tube, can be obtained by
subtracting the second moment of area of the hole
about its centroid from the second moment of area
of the outer circumference about its centroid. This
is demonstrated in worked problems 10, 12 and 13
following.

Problem 6. Determine the second moment

of area and the radius of gyration about axes

AA,BBandCCfor the rectangle shown in

Figure 7.11.

A

B

C

b= 4.0 cm

d= 12.0 cm A

B

C

Figure 7.11

From Table 7.1, the second moment of area about

axisAA,

IAA=

bd^3

3

=

( 4. 0 )( 12. 0 )^3

3

=2304 cm^4

Radius of gyration,kAA=√d

3

=^12 √.^0

3

= 6 .93 cm

Similarly, IBB=

db^3

3

=

( 12. 0 )( 4. 0 )^3

3

=256 cm^4

and kBB=

b

√

3

=

4. 0

√

3

= 2 .31 cm

The second moment of area about the centroid of a

rectangle isbd

3

12 when the axis through the centroid

is parallel with the breadthb. In this case, the axis

CCis parallel with the lengthd

Hence ICC=

db^3

12

=

( 12. 0 )( 4. 0 )^3

12

=64 cm^4

and kCC=

b

√

12

=

4. 0

√

12

= 1 .15 cm

Problem 7. Find the second moment of

area and the radius of gyration about axisPP

for the rectangle shown in Figure 7.12.

40.0 mm

15.0 mm

G

25.0 mm

G

PP

Figure 7.12