integration

386 INTEGRAL CALCULUS
I PP = AkPP 2 , from which,
k PP =
√ √ (645000 )
IPP
area = = 32.79 mm
600
Problem 14. Determine the second moment of
area and radius of gyration of the circle shown
in Fig. 38.21 about axis YY.
Problem 13. Determine the second moment of
area and radius of gyration about axis QQ of the
triangle BCD shown in Fig. 38.20.
G
r = 2.0 cm
G
B
12.0 cm
G
G
3.0 cm
C
D
8.0 cm 6.0 cm
Y
Figure 38.21
Y
Figure 38.20
Q
Using the parallel axis theorem: I QQ = I GG + Ad 2 ,
where I GG is the second moment of area about the
centroid of the triangle,
i.e. bh3
36 = (8.0)(12.0)3
36
A is the area of the triangle,
= 384 cm 4 ,
=
2 1 bh = 2 1 (8.0)(12.0) = 48 cm2
and d is the distance between axes GG and QQ,
= 6.0 + 1 3
(12.0) = 10 cm.
Hence the second moment of area about axis QQ,
Q
In Fig. 38.21, I GG = πr4
4 = π 4 (2.0)4 = 4π cm 4 .
Using the parallel axis theorem, I YY = I GG + Ad 2 ,
where d = 3.0 + 2.0 = 5.0 cm.
Hence I YY = 4π + [π(2.0) 2 ](5.0) 2
= 4π + 100π = 104π = 327 cm 4 .
Radius of gyration,
√ √ ( )
IYY 104π
k YY =
area = 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 axis XX.
G
10.0 mm
G
I QQ = 384 + (48)(10) 2 = 5184 cm 4 .
B
B
Radius of gyration,
15.0 mm
k QQ =
√ √ (5184 )
IQQ
area = = 10.4 cm
48
X
Figure 38.22
X