Mechanical Engineering Principles

98 MECHANICAL ENGINEERING PRINCIPLES

Hence, i 1 =

0. 1 × 0. 023

12

= 6. 6 × 10 −^8 m^4

i 2 =

0. 02 × 0. 183

12

= 9. 72 × 10 −^6 m^4

and

∑

i= 6. 6 × 10 −^8 + 9. 72 × 10 −^6

= 9. 786 × 10 −^6 m^4

From the parallel axis theorem:

IXX=

∑

i+

∑

ay^2 ( 7. 1 )

The cross-sectional area of the tee beam

=

∑

a= 0 .0056 m^2 from Table 7.2.

Now the centroidal position, namelyy, is given by:

y=

∑

ay

∑

a

=

7. 04 × 10 −^4

0. 0056

= 0 .1257 m

From equation (7.1),

IXX=

∑

i+

∑

ay^2

= 9. 786 × 10 −^6 + 1. 014 × 10 −^4

i.e. IXX= 1. 112 × 10 −^4 m^4

From the parallel axis theorem:

INA=IXX−(y)^2

∑

a

= 1. 112 × 10 −^4 −( 0. 1257 )^2 × 0. 0056

INA= 2. 27 × 10 −^5 m^4

It should be noted that the least second moment of
area of a section is always about an axis through its
centroid.

Problem 19. (a) Determine the second

moment of area and the radius of gyration

about axisXXfor theI-section shown in

Figure 7.32.

Figure 7.32

(b) Determine the position of the centroid of

theI-section.

(c) Calculate the second moment of area and

radius of gyration about an axisCCthrough

the centroid of the section, parallel to

axisXX.

TheI-section is divided into three rectangles,D,F

andFand their centroids denoted byCD,CEand

CFrespectively.

(a) For rectangle D:

The second moment of area aboutCD(an axis

throughCDparallel toXX)

=

bd^3

12

=

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

12

=18 cm^4

Using the parallel axis theorem:

IXX= 18 +AH^2

Table 7.2

Column 1 2 3 4 5 6

Row 1 Section a y ay ay^2 i

Row 2 (1) 0.002 0.19 3. 8 × 10 −^47. 22 × 10 −^56. 6 × 10 −^8

Row 3 (2) 0.0036 0.09 3. 24 × 10 −^42. 916 × 10 −^59. 72 × 10 −^6

Row 4

∑

0.0056 − 7. 04 × 10 −^41. 014 × 10 −^49. 786 × 10 −^6