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