Some applications of integration 389
Problem 18. Determine correct to 3 significant
figures, the second moment of area about axisXX
for the composite area shown in Fig. 38.25.
4.0 cm
1.0 cm 1.0 cm
8.0 cm
CT
X X
2.0 cm
6.0 cm
2.0 cm
T T
Figure 38.25
For the semicircle,
IXX=
πr^4
8
=
π( 4. 0 )^4
8
= 100 .5cm^4
For the rectangle,
IXX=
bl^3
3
=
( 6. 0 )( 8. 0 )^3
3
=1024cm^4
For the triangle, about axisTTthrough centroidCT,
ITT=
bh^3
36
=
( 10 )( 6. 0 )^3
36
=60cm^4
By the parallel axis theorem, the second moment of area
of the triangle about axisXX
= 60 +
[ 1
2 (^10 )(^6.^0 )
][
8. 0 +^13 ( 6. 0 )
] 2
=3060cm^4.
Total second moment of area aboutXX
= 100. 5 + 1024 + 3060
= 4184. 5
=4180cm^4 ,correct to 3 significant figures.
Problem 19. Determine the second moment of
area and the radius of gyration about axisXXfor the
I-section shown in Fig. 38.26.
CF
CE
CD
3.0 cm
7.0 cm
4.0 cm
3.0 cm
y
CC
X X
S
S
15.0 cm
8.0 cm
Figure 38.26
The I-section is divided into three rectangles,D,E
andFand their centroids denoted byCD,CEandCF
respectively.
For rectangle D:
The second moment of area aboutCD(an axis through
CDparallel toXX)
=
bl^3
12
=
( 8. 0 )( 3. 0 )^3
12
=18cm^4
Using the parallel axis theorem:
IXX= 18 +Ad^2
whereA=( 8. 0 )( 3. 0 )=24cm^2 andd= 12 .5cm
HenceIXX= 18 + 24 ( 12. 5 )^2 =3768cm^4.
For rectangle E:
The second moment of area aboutCE(an axis through
CEparallel toXX)
=
bl^3
12
=
( 3. 0 )( 7. 0 )^3
12
= 85 .75cm^4
Using the parallel axis theorem:
IXX= 85. 75 +( 7. 0 )( 3. 0 )( 7. 5 )^2 =1267cm^4.