386 Higher Engineering Mathematics
Table 38.1 Summary 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 with b
bl 3
3
l
√
3
length l, breadth b
(2) Coinciding with l
lb 3
3
b
√
3
(3) Through centroid, parallel to b
bl 3
12
l
√
12
(4) Through centroid, parallel to l
lb 3
12
b
√
12
Triangle
(1) Coinciding with b
bh 3
12
h
√
6
Perpendicular height h,
base b
(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
radius r
plane (i.e. polar axis)
(2) Coinciding with diameter
πr
4
4
r
2
(3) About a tangent
5πr 4
4
√
5
2
r
Semicircle
Coinciding with diameter
πr 4
8
r
2
radius r
Problem 12. Find the second moment of area and
the radius of gyration about axis PP for the
rectangle shown in Fig. 38.19.
P
P
40.0 mm
15.0 mm
25.0 mm
G
G
Figure 38.19
I GG =
lb 3
12
where 1 = 40.0 mm and b = 15.0 mm
Hence I GG =
(40.0)(15.0)
3
12
= 11250 mm
4
From the parallel axis theorem, I PP = I GG + Ad 2 ,
where A = 40.0 × 15.0 = 600 mm 2 and
d = 25.0 +7.5 = 32.5 mm, the perpendicular
distance between GG and PP. Hence,
I PP = 11 250 + (600)(32.5)
2
= 645000 mm
4