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
- 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
- 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
- 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
- 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
- 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
⎤
⎥
⎦