Higher Engineering Mathematics, Sixth Edition

390 Higher Engineering Mathematics

For rectangle F:

IXX=

bl^3

3

=

( 15. 0 )( 4. 0 )^3

3

=320cm^4

Total second moment of area for theI-section about

axisXX,

IXX= 3768 + 1267 + 320 =5355cm^4

Total area ofI-section

=( 8. 0 )( 3. 0 )+( 3. 0 )( 7. 0 )+( 15. 0 )( 4. 0 )

=105cm^2.

Radius of gyration,

kXX=

√

IXX

area

=

√(

5355

105

)

=7.14cm

Now try the following exercise

Exercise 152 Further problems on second

moment of areas of regular sections

1. Determine the second moment of area and
   radius of gyration for the rectangle shown in
   Fig. 38.27 about (a) axisAA(b) axisBBand
   (c) axisCC. ⎡

⎣

(a)72cm^4 , 1 .73cm

(b)128cm^4 , 2 .31cm

(c)512cm^4 , 4 .62cm

⎤

⎦

8.0 cm

B

B

C

A A

C

3.0 cm

Figure 38.27

2. Determine the second moment of area and
   radius of gyration for the triangle shown in
   Fig. 38.28 about (a) axisDD(b) axisEEand
   (c) an axis through the centroid of the triangle
   parallel to axisDD.⎡

⎣

(a)729cm^4 , 3 .67cm

(b)2187cm^4 , 6 .36cm

(c)243cm^4 , 2 .12cm

⎤

⎦

12.0 cm

9.0 cm

D D

EE

Figure 38.28

3. For the circle shown in Fig. 38.29, find the
   second moment of area and radius of gyration
   about (a) axisFFand (b) axisHH.
   [
   (a)201cm^4 , 2 .0cm
   (b)1005cm^4 , 4 .47cm

]

r^5

4.0 cm

F

H

H

F

Figure 38.29

4. For the semicircle shown in Fig. 38.30,find the
   second moment of area and radius of gyration
   about axisJJ.
   [3927mm^4 , 5.0mm]

J J

r^5

10.0 mm

Figure 38.30

5. For each of theareas shown in Fig.38.31 deter-
   mine the second moment of area and radius of
   gyration about axisLL, by using the parallel
   axis theorem.
   ⎡
   ⎢
   ⎣

(a)335cm^4 , 4 .73cm

(b)22030cm^4 , 14 .3cm

(c)628cm^4 , 7 .07cm

⎤

⎥

⎦