386 Higher Engineering Mathematics
Table 38.1Summary of standard results of the second moments of areas of regular sections
Shape Position of axis Second moment Radius of
of area,I gyration,k
Rectangle (1) Coinciding withb
bl^3
3
l
√
3
lengthl, breadthb
(2) Coinciding withl
lb^3
3
b
√
3
(3) Through centroid, parallel tob
bl^3
12
l
√
12
(4) Through centroid, parallel tol
lb^3
12
b
√
12
Triangle (1) Coinciding withb
bh^3
12
h
√
6
Perpendicular heighth,
baseb (2) Through centroid, parallel to base
bh^3
36
h
√
18
(3) Through vertex, parallel to base
bh^3
4
h
√
2
Circle (1) Through centre, perpendicular to
πr^4
2
r
√
2
radiusr plane (i.e. polar axis)
(2) Coinciding with diameter
πr^4
4
r
2
(3) About a tangent
5 πr^4
4
√
5
2
r
Semicircle Coinciding with diameter
πr^4
8
r
2
radiusr
Problem 12. Find the second moment of area and
the radius of gyration about axisPPfor the
rectangle shown in Fig. 38.19.
P P
40.0 mm
15.0 mm
25.0 mm
G G
Figure 38.19
IGG=
lb^3
12
where 1= 40 .0mmandb= 15 .0mm
HenceIGG=
( 40. 0 )( 15. 0 )^3
12
=11250mm^4
From the parallel axis theorem, IPP=IGG+Ad^2 ,
whereA= 40. 0 × 15. 0 =600mm^2 and
d= 25. 0 + 7. 5 = 32 .5mm, the perpendicular
distance betweenGGandPP. Hence,
IPP= 11250 +( 600 )( 32. 5 )^2
=645000mm^4