Higher Engineering Mathematics

386 INTEGRAL CALCULUS

IPP=Ak^2 PP, from which,

kPP=

√

IPP

area

=

√(

645000

600

)

=32.79 mm

Problem 13. Determine the second moment of

area and radius of gyration about axisQQof the

triangleBCDshown in Fig. 38.20.

B

G G

CD

QQ

12.0 cm

8.0 cm 6.0 cm

Figure 38.20

Using the parallel axis theorem:IQQ=IGG+Ad^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

anddis the distance between axesGGandQQ,

= 6. 0 +^13 (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

)

=10.4 cm

Problem 14. Determine the second moment of

area and radius of gyration of the circle shown

in Fig. 38.21 about axisYY.

YY

3.0 cm

GG

r = 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 .0 cm.

Hence IYY = 4 π+[π(2.0)^2 ](5.0)^2

= 4 π+ 100 π= 104 π=327 cm^4.

Radius of gyration,

kYY=

√

IYY

area

=

√(

104 π

π(2.0)^2

)

=

√

26 =5.10 cm

Problem 15. Determine the second moment of

area and radius of gyration for the semicircle

shown in Fig. 38.22 about axisXX.

G G

B B

XX

15.0 mm

10.0 mm

Figure 38.22