Higher Engineering Mathematics, Sixth Edition

Some applications of integration 387

IPP=Ak^2 PP, from which,

kPP=

√

IPP

area

=

√(

645000

600

)

=32.79mm

Problem 13. Determine the second moment of

area and radius of gyration about axisQQof the

triangleBC Dshown in Fig. 38.20.

QQ

C D

6.0 cm

12.0 cm

8.0 cm

G

B

G

Figure 38.20

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

i.e.

bh^3

36

=

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

36

=384cm^4 ,

Ais the area of the triangle,

=^12 bh=^12 ( 8. 0 )( 12. 0 )=48cm^2

anddis the distance between axesGGandQQ,

= 6. 0 +^13 ( 12. 0 )=10cm.

Hence the second moment of area about axisQQ,

IQQ= 384 +( 48 )( 10 )^2 =5184cm^4

Radius of gyration,

kQQ=

√

IQQ

area

=

√(

5184

48

)

=10.4cm

Problem 14. Determine the second moment of

area and radius of gyration of the circle shown in

Fig. 38.21 about axisYY.

Y

3.0 cm

Y

G G

r 5 2.0 cm

Figure 38.21

In Fig. 38.21,IGG=

πr^4

4

=

π

4

( 2. 0 )^4 = 4 πcm^4.

Using the parallel axis theorem, IYY=IGG+Ad^2 ,

whered= 3. 0 + 2. 0 = 5 .0cm.

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

= 4 π+ 100 π= 104 π=327cm^4

Radius of gyration,

kYY=

√

IYY

area

=

√(

104 π

π( 2. 0 )^2

)

=

√

26 =5.10cm

Problem 15. Determine the second moment of

area and radius of gyration for the semicircle shown

in Fig. 38.22 about axisXX.

X X

G G

B B

10.0 mm

15.0 mm

Figure 38.22