Mechanical Engineering Principles

88 MECHANICAL ENGINEERING PRINCIPLES

=

1

2

[

25 x^3

3

−

10 x^4

4

+

x^5

5

] 5

0

125

6

=

1

2

(

25 ( 125 )

3

−

6250

4

+ 625

)

125

6

= 2. 5

Hence the centroid of the area lies at (2.5, 2.5)
(Note from Figure 7.6 that the curve is symmet-
rical aboutx = 2 .5 and thus xcould have been
determined ‘on sight’).

Problem 5. Locate the centroid of the area

enclosed by the curvey= 2 x^2 ,they-axis

and ordinatesy=1andy=4, correct to 3

decimal places.

From Section 7.4,

x=

1

2

∫ 4

1

x^2 dy

∫ 4

1

xdy

=

1

2

∫ 4

1

y

2

dy

∫ 4

1

√

y

2

dy

=

1

2

[

y^2

4

] 4

1

[

2 y^3 /^2

3

√

2

] 4

1

=

15

8

14

3

√

2

= 0. 568

and y=

∫ 4

1

xydy

∫ 4

1

xdy

=

∫ 4

1

√

y

2

(y)dy

14

3

√

2

=

∫ 4

1

y^3 /^2

√

2

dy

14

3

√

2

=

1

√

2

⎡

⎢

⎣

y^5 /^2

5

2

⎤

⎥

⎦

4

1

14

3

√

2

=

2

5

√

2

( 31 )

14

3

√

2

= 2. 657

Hence the position of the centroid is at

(0.568, 2.657).

Now try the following exercise

Exercise 36 Further problems on cent-

roids of simple shapes

1. Determine the position of the centroid of
   a sheet of metal formed by the curve
   y= 4 x−x^2 which lies above thex-axis.
   [(2, 1.6)]


2. Find the coordinates of the centroid of the
   area that lies between the curve

y

x=x−^2

and thex-axis. [(1,−0.4)]

3. Determine the coordinates of the centroid
   of the area formed between the curve
   y= 9 −x^2 and thex-axis. [(0, 3.6)]


4. Determine the centroid of the area lying
   between y = 4 x^2 ,they-axis and the
   ordinatesy=0andy=4.
   [(0.375, 2.40]


5. Find the position of the centroid of the
   area enclosed by the curvey=

√

5 x,the

x-axis and the ordinatex=5.

[(3.0, 1.875)]

6. Sketch the curve y^2 = 9 xbetween the
   limitsx=0andx=4. Determine the
   position of the centroid of this area.
   [(2.4, 0)]

7.7 Second moments of area of regular

sections

Thefirst moment of areaabout a fixed axis of a

lamina of areaA, perpendicular distanceyfrom the

centroid of the lamina is defined asAycubic units.

Thesecond moment of areaof the same lamina

as above is given byAy^2 , i.e. the perpendicular

distance from the centroid of the area to the fixed

axis is squared.

Second moments of areas are usually denoted byI

and have units of mm^4 ,cm^4 , and so on.

Several areas,a 1 ,a 2 ,a 3 ,...at distancesy 1 ,y 2 ,y 3 ,

...from a fixed axis, may be replaced by a single

areaA,whereA=a 1 +a 2 +a 3 +···at distancek

from the axis, such thatAk^2 =

∑

ay^2 .kis called