Mechanical Engineering Principles

94 MECHANICAL ENGINEERING PRINCIPLES

d

Figure 7.17

The lamina is shown in Figure 7.17.

From the perpendicular axis theorem:

IZZ=IXX+IYY

IXX=

db^3

3

=

( 40 )( 15 )^3

3

=45000 mm^4

and IYY=

bd^3

3

=

( 15 )( 40 )^3

3

=320000 mm^4

Hence IZZ= 45000 + 320000

=365000 mm^4 or 36 .5cm^4

Radius of gyration,

kZZ=

√

IZZ

area

=

√(

365000

( 40 )( 15 )

)

= 24 .7mmor 2 .47 cm

Problem 15. Determine correct to 3

significant figures, the second moment of

area about axisXXfor the composite area

shown in Figure 7.18.

X X

1.0 cm

8.0 cm

6.0 cm

T T

2.0 cm

CT

4.0 cm

1.0 cm

2.0 cm

Figure 7.18

For the semicircle,

IXX=

πr^4

8

=

π( 4. 0 )^4

8

= 100 .5cm^4

For the rectangle,

IXX=

bd^3

3

=

( 6. 0 )( 8. 0 )^3

3

=1024 cm^4

For the triangle, about axisTTthrough centroidCT,

ITT=

bh^3

36

=

( 10 )( 6. 0 )^3

36

=60 cm^4

By the parallel axis theorem, the second moment of

area of the triangle about axisXX

= 60 +[^12 ( 10 )( 6. 0 )][8. 0 +^13 ( 6. 0 )]^2

=3060 cm^4

Total second moment of area aboutXX

= 100. 5 + 1024 + 3060 = 4184. 5

=4180 cm^4 ,

correct to 3 significant figures.

Now try the following exercise

Exercise 37 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 Figure 7.19 about (a) axisAA
   (b) axisBBand (c) axisCC
   ⎡

⎢

⎢

⎣

(a) 72 cm^4 ,1.73cm

(b) 128 cm^4 ,2.31cm

(c) 512 cm^4 ,4.62cm

⎤

⎥

⎥

⎦

Figure 7.19