Mechanical Engineering Principles

92 MECHANICAL ENGINEERING PRINCIPLES

Problem 8. Determine the second moment

of area and radius of gyration about axisQQ

of the triangleBCDshown in Figure 7.13.

B

G G

C D

QQ

12.0 cm

8.0 cm6.0 cm

Figure 7.13

Using the parallel axis theorem:IQQ=IGG+AH^2 ,
whereIGGis the second moment of area about the
centroid of the triangle,

i.e.

bh^3

36

=

( 8. 0 )( 12. 0 )^3

36

=384 cm^4 ,

Ais the area of the triangle

=^12 bh=^12 ( 8. 0 )( 12. 0 )

=48 cm^2

andHis the distance between axesGGandQQ

= 6. 0 +

1

3

( 12. 0 )=10 cm

Hence the second moment of area about axisQQ,

IQQ= 384 +( 48 )( 10 )^2

=5184 cm^4

Radius of gyration,

kQQ=

√

IQQ

area

=

√(

5184

48

)

= 10 .4cm

Problem 9. Determine the second moment

of area and radius of gyration of the circle

shown in Figure 7.14 about axisYY.

YY

3.0 cm

GG

r= 2.0 cm

Figure 7.14

In Figure 7.14,

IGG=

πr^4

4

=

π

4

( 2. 0 )^4

= 4 πcm^4

Using the parallel axis theorem,

IYY=IGG+AH^2 ,

where H= 3. 0 + 2. 0 = 5 .0cm

Hence IYY= 4 π+[π( 2. 0 )^2 ]( 5. 0 )^2

= 4 π+l 00 π= 104 π=327 cm^4

Radius of gyration,

kYY=

√

IYY

area

=

√(

104 π

π( 2. 0 )^2

)

=

√

26

= 5 .10 cm

Problem 10. Determine the second moment

of area of an annular section, about its

centroidal axis. The outer diameter of the

annulus isD 2 and its inner diameter isD 1

Second moment of area of annulus about its

centroid,IXX=(IXXof outer circle about its

diameter)−(IXXof inner circle about its diameter)

=

πD 24

64

−

πD^41

64

from Table 7.1

i.e. IXX=

π

64

(D 24 −D^41 )