Higher Engineering Mathematics, Sixth Edition

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.