Mechanical Engineering Principles

FIRST AND SECOND MOMENT OF AREAS 99

where A=( 8. 0 )( 3. 0 )=24 cm^2

and H= 12 .5cm

Hence IXX= 18 + 24 ( 12. 5 )^2 =3768 cm^4

For rectangle E:

The second moment of area aboutCE(an axis

throughCEparallel toXX)

=

bd^3

12

=

( 3. 0 )( 7. 0 )^3

12

= 85 .75 cm^4

Using the parallel axis theorem:

IXX= 85. 75 +( 7. 0 )( 3. 0 )( 7. 5 )^2

=1267 cm^4

For rectangle F:

IXX=

bd^3

3

=

( 15. 0 )( 4. 0 )^3

3

=320 cm^4

Total second moment of area for the I-

section about axisXX,

IXX= 3768 + 1267 + 320 =5355 cm^4

Total area of I-section = ( 8. 0 )( 3. 0 )+

( 3. 0 )( 7. 0 )+( 15. 0 )( 4. 0 )=105 cm^2

Radius of gyration,

kXX=

√

IXX

area

=

√(

5355

105

)

= 7 .14 cm

(b) The centroid of theI-section will lie on the
axis of symmetry, shown asSSin Figure 7.32.
Using a tabular approach:

Part Area Distance of centroid Moment

(acm^2 ) fromXX aboutXX

(i.e.ycm) (i.e.aycm^3 )

D2412.5 300

E217.5 157.5

F602.0 120

∑

a=A= 105

∑

ay= 577. 5

Ay=

∑

ay, from which,

y=

∑

ay

A

=

577. 5

105

= 5 .5cm

Thus the centroid is positioned on the axis

of symmetry 5.5 cm from axisXX.

(c) From the parallel axis theorem:

IXX=ICC+AH^2

i.e. 5355=ICC+( 105 )( 5. 5 )^2

=ICC+ 3176

from which,second moment of area about

axisCC,

ICC= 5355 − 3176 =2179 cm^4

Radius of gyration,

kCC=

√

ICC

area

=

√

2179

105

= 4 .56 cm

Now try the following exercise

Exercise 38 Further problems on second

moment of area of ‘built-up’

sections

Determine the second moments of area about a

horizontal axis, passing through the centroids,

for the ‘built-up’ sections shown below. All

dimensions are in mm and all the thicknesses

are 2 mm.

1. Figure 7.33 [17329 mm^4 ]

r

30

Figure 7.33

2. Figure 7.34 [37272 mm^4 ]

40

30

Figure 7.34