Higher Engineering Mathematics, Sixth Edition

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