9.1 Centroids by Integration - School of · PDF file9.1 Centroids by Integration Example 1, page 3 of 4 Express the coordinates of the element centroid in terms of the coordinates

9.1 Centroids by Integration

9.1 Centroids by Integration Example 1, page 1 of 4

x

Definition of centroid coordinates

xc = (1)

yc = (2)

where (xel, yel) are the coordinates of the

centroid of the differential area

element dA.

1. Locate the centroid of the plane area shown. Use a

differential element of thickness dx.

1

yel dA

dA

xel dA

dA

y

y = 3x2

2 ft

12 ft


Express the element area in terms of the coordinates of a

point (x, y) on the curve:

dA = height width (of rectangle)

or,

dA = y dx (3)

By choosing an element of width dx, we have also

implicitly chosen x to be the variable of integration.

3

(x, y)

x

x

dx = width

y = height

Locate the differential element so that it extends

from an arbitrary point (x, y) on the curve to an

opposite boundary of the crosshatched region.

y = 3x2

2

y


Express the coordinates of the element

centroid in terms of the coordinates of the

point (x, y) on the curve. The x

coordinate of the element centroid is the

same as the x coordinate of the point on

the curve:

xel = x (4)

Since the centroid of the differential

element is located in the center of the

element, the y coordinate of the element

centroid is

yel = (5)

Since the variable of integration is x, we

now have to express dA and yel in terms of

x (As can be seen from Eq. 4, xel already is

a function of x). The point (x, y) on the

curve satisfies

y = 3x2 (6)

Substituting the expression for y in Eq. 6

into the equations for dA (Eq. 3) and for yel

(Eq. 5) gives

dA = y dx (Eq. 3 repeated)

= 3x2 dx (7)

yel =

= (8)

Evaluate the integral in the denominator

of the equation for xc over the range (from

0 to 2):

dA = 3x2 dx = 8 ft2 (9)

dxx

(x, y)

y = 3x2

y 4 5

6

(xel, yel)

2 ft

y

2

0

y

2

2

y

3x2

2

x


3x2

2

Evaluate the integral in the numerator of the equation for xc

over the range from 0 to 2:

xel dA = x(3x2) dx = 12 ft3 (10)

Evaluate the integral in the numerator of the equation for yc over

the range from 0 to 2:

yel dA = [ ](3x2) dx = 28.8 ft3 (11)

Substitute the results given in Eqs. 9, 10, and 11 into the

definitions for xc and yc:

xc = = 12 8

= 1.5 ft Ans.

yc = = 28.8 8

= 3.6 ft Ans. dA

xel dA

dA

yel dA

2

0

7

2

0


xy = a sin( )2b

2. Locate the centroid of the plane area shown, if a = 3 m

and b = 1 m. Use a differential element of thickness dy.


xc = (1)

yc = (2)



element dA.

xel dA

dA

dA

yel dA

x

x

y

dy

(x, y)

Express the element area in terms of the

coordinates of a point (x, y) on the curve:

dA = width height (of rectangle)

or,

dA = x dy (3)

By choosing an element of width dy, we have

also implicitly chosen y to be the variable of

integration.

Locate the differential element so that it

extends from an arbitrary point (x, y)

on the curve to an opposite boundary

of the crosshatched region.

3

2

1

y = a sin( )2b

x

y

a

b

y

x


2 x

2b ya

12

2bay

b ya

(x, y)

6

x

(xel, yel)

y4 Express the coordinates of the

element centroid in terms of the

coordinates of the point (x, y) on

the curve. Since the centroid of

the differential element is

located in the center of the

element, the x coordinate of the

element centroid is:

xel = (4)

The y coordinate of the element

centroid is the same as the y

coordinate of the point on the

curve:

yel = y (5)

5

2b

3

0

y3

ay

x

y

a

xy = a sin( )2b

2bx

Since the variable of integration is y, we now

have to express dA and xel in terms of y (As

can be seen from Eq. 5, yel already is expressed

as a function of y). The point (x, y) on the

curve satisfies

y = a sin ( )

Solving this equation for x gives

x = ( ) sin-1( ) (6)

Substituting the expression for x in Eq. 6 into

the equations for dA (Eq. 3) and for xel (Eq. 4)

gives

dA = x dy (Eq. 3 repeated)

= [( ) sin-1( )] dy (7)

xel = (Eq. 4 repeated)

= ( ) sin-1( )

= ( ) sin-1( ) (8)

2 x

Substitute a = 3 m and b = 1 m, and evaluate the integral in

the denominator of the equation for xc over the range from 0

to 3 (Use the integral function on your calculator):

dA = ( ) sin-1( ) dy = 1.090 m2 (9)

dy


3

0

3

0

y3

y3

Substitute the results given in Eqs. 9, 10, and 11 into the definitions

for xc and yc:

xc = = 0.2841 1.090

= 0.261 m Ans.

yc = = 2.2500 1.090

= 2.06 m Ans.

And similarly evaluate the integral in the numerator of the equation for yc over the

range from 0 to 3 (Use the integral function on your calculator):

yel dA = y [( sin-1( )] dy = 2.2500 m3 (11)

Similarly, evaluate the integral in the numerator of the equation for xc over the range

from 0 to 3 (Use the integral function on your calculator):

xel dA = [ ] [( sin-1( )] dy = 0.2841 m3 (10)

dA

dA

xel dA

yel dA

7

sin-1 (y/3)


y = 4x5 3x2 + 12x + 1

14 y

y

y

x1 in.

x

y = 4x5 3x2 + 12x + 1y

(x, y)

Locate the differential element so that it extends from

an arbitrary point (x, y) on the curve to an opposite

boundary of the crosshatched region.

2


xc = (1)

yc = (2)



element dA.

3. Locate the centroid of the plane area shown.

1

yel dA

dA

xel dA

dA

Express the element area in terms of the coordinates

of a point (x, y) on the curve:


or,

dA = (14 y) dx (3)

By choosing an element of width dx, we have also


3

dx1 in

13 in.

x

14 in.


y = 4x5 3x2 + 12x + 1

Express the coordinates of the

element centroid in terms of the

coordinates of the point (x, y) on

the curve. The x coordinate of the

element centroid is the same as the

x coordinate of the point on the

curve:

xel = x (4)

Since the centroid of the differential

element is located in the center of

the element, the y coordinate of the

element centroid is:

yel = y +

= 7 + (5)

4Since the variable of integration is x, we now have

to express dA and yel in terms of x (As can be seen

from Eq. 4, xel already is expressed as a function of

x). The point (x, y) on the curve satisfies

y = 4x5 3x2 + 12x + 1 (6)

Substituting the expression for y from Eq. 6 into the

equations for dA (Eq. 3) and for yel (Eq. 5) gives

dA = (14 y) dx (Eq. 3 repeated)

= [14 (4x5 3x2 + 12x + 1)] dx

= ( 4x5 + 3x2 12x + 13) dx (7)

yel = 7 + (Eq. 5 repeated)

= 7 +

= (8)

x

y

x

(x, y)

4x5 3x2 + 12x + 1

4x5 3x2 + 12x + 152

2

y

14 y

(xel, yel)

dx

5

14 y2

14 y

2

y2

y2

14 in


What if we had chosen a differential element dy wide instead of

dx? The figure shows that now the element area is

dA = x dy

Since dy is the variable of integration, we must express x as a

function of y. But x and y are related by

y = 4x5 3x2 + 12x + 1

and this equation is difficult to invert, that is, to solve for x as a

function of y. So it is much easier to use a differential element

dx wide, because then we don't have to solve for x as a function

of y.

The general conclusion to draw is that whether we should choose

a differential element dx or dy wide depends on whether the

equation defining the boundary of the region can be more easily

written as a function of x or as a function of y.

6

dy

x

y = 4x5 3x2 + 12x +1

x

y

7


y = 4x5 3x2 + 12x + 1

x

1 in.

(x, y)

y

dx

And similarly evaluate the integral in the numerator of the equation for yc

over the range from 0 to 1 (Use the integral function on your calculator):

yel dA = [ ]( 4x5 + 3x2 12x + 13) dx

= 69.849 in3 (11)

4x5 3x2 + 12x + 15

Similarly, evaluate the integral in the numerator of the equation for xc

over the range from 0 to 1 (Use the integral function on your calculator):

xel dA = x ( 4x5 + 3x2 12x + 13) dx = 2.679 in3 (10)

Return now to a differential element dx wide.

Evaluate the integral in the denominator of the equation for xc over

the range from 0 to 1 (Use the integral function on your calculator):

dA = ( 4x5 + 3x2 12x + 13) dx = 7.333 in2 (9)

1

0

1

0

11

10

2

8

1

0

9




xc = = 2.679 7.333

= 0.37 in.

yc = = 69.849 7.333

= 9.52 in. dA

12

xel dA

dA

yel dA

Ans.

Ans.


xy = 1

x

y

y

x

(x, y)

x

dy



on the curve to an opposite boundary.


xc = (1)

yc = (2)

where (xel, yel) are the coordinates of the centroid of the

differential area element dA.

2


dA

yel dA

dA

xel dA

xy = 1

0.5 m

2 m

0.5 m

2 m

2 m

0.5 m

x

y


Express the element area in terms of the coordinates of a point

(x, y) on the curve:


or,

dA = x dy (3)

By choosing an element of width dy, we have also implicitly

chosen y to be the variable of integration.

Express the coordinates of the element centroid in terms of the

coordinates of the point (x, y) on the curve. Since the centroid

of the differential element is located in the center of the element,

the x coordinate of the element centroid is

xel = x 2 (4)

The y coordinate of the element is the same as the y coordinate

of the point on the curve:

yel = y (5)

x

x4

dy

y

xy = 1

(x, y)

(xel, yel)

y

3


Evaluate the integral in the denominator of the equation for

xc over the range from 0.5 to 2:

dA = 1 y dy = 1.3863 m2 (9)

Since the variable of integration is y, we now have to

express dA and xel in terms of y (As can be seen from

Eq. 5, yel already is a function of y). The point (x, y) on

the curve satisfies

xy =1

Solving for x gives

x = 1 y (6)

Substituting the expression for x in Eq. 6 into the

equations for dA (Eq. 3) and for xel (Eq. 4) gives

dA = x dy (Eq. 3 repeated)

= 1 y dy (7)

xel = x 2 (Eq. 4 repeated)

=

= 1 2y

(8)

1/y2

5

2 m

6

0.5

2

0.5 m

y

xy = 1

dy

x


2 m

(x, 2)

y

(x, y)

xy = 1

x

10

2

0.5

0.5

2

Evaluate the integral in the numerator of the equation for yc

over the range from 0.5 to 2:

yel dA = y ( 1 y ) dy = 1.5 m3 (11)


over the range from 0.5 to 2:

xel dA = ( 1 2y

)( 1 y ) dy = 0.7500 m3 (10)



xc = = 0.7500 1.3863

= 0.541 m Ans.

yc = = 1.5 1.3863

= 1.082 m Ans. dA

dA

yel dA

xel dA

9

8

7 What if we had chosen a vertical element rather

than a horizontal element? The figure below

shows that we would have to define the element

areas in two equations, one for the region where y

is constant (y = 2 m), and one for the region where

y varies. We would also have to evaluate two

integrals, one over each region. Thus using a

vertical element rather than a horizontal element

would almost double the amount of work required.

Nevertheless, it would give us the correct answer.


6 m

2 m

1 m

4 m

y2 y1

y1

x

(x, y2)

(x, y1)Express the element area in terms of the



or,

dA = (y2 y1) dx (3)

By choosing an element of width dx, we have

also implicitly chosen x to be the variable of

integration.


extends from an arbitrary point on the

lower curve to a point directly above on

the upper curve.


xc = (1)

yc = (2)


centroid of the differential area element dA.

3

21


dA

yel dA

dA

xel dA

x

y =

y

y = x2 +

x(13 x)

6

14 11x

3

dx

x

y


y2 + y1

7x2 + 35x 28

14 11x

x(13 x)

7x2 + 35x 28

[x(13 x)/6] + [x2 + (14 11x)/3]

12(5x2 9x + 28)

Since the variable of integration is x, we now have to express

dA and yel in terms of x (As can be seen from Eq. 4, xel already

is a function of x). The y coordinates of the top and bottom of

the element satisfy

y2 =

y1 = x2 +

y2 y1 = [ ] [x2 + ]

= (6)

Substituting the expression for y in Eq. 6 into the equations for

dA (Eq. 3)

dA = (y2 y1) dx (Eq. 3 repeated)

= dx (7)

and for yel (Eq. 5) gives

yel = (Eq. 5 repeated)

=

= (8)

(x, y2)dx

(xel, yel)

x = x el

(x, y1)

y2 y1

2

x

2

Express the coordinates of the element centroid in terms

of the coordinates of the point (x, y) on the curve. The

x coordinate of the element is the same as the x

coordinate of the point on the curve:

xel = x (4)

Since the centroid of the differential element is located

in the center of the element, the y coordinate of the

element centroid is average of the y coordinates of the

points on the top and bottom curves.

yel = (5)

4

y2 + y1

y

2

6

5

6

2

6

6

3

314 11xx(13 x)

y1

y2 y1


Evaluate the integral in the numerator of the equation for yc

over the range from 1 to 4:

yel dA = [ ][ ] dx

= 17.0625 m3 (11)

Evaluate the integral in the numerator of the equation for xc over the

range from 1 to 4:

xel dA = x [ ] dx = 13.1250 m3 (10)

(5x2 9x + 28)



xc = = 13.125 5.25

= 2.50 m Ans.

yc = = 17.0625 5.25

= 3.25 m Ans.

xel dA

dA

yel dA

dA

9

4

1

7x2 + 35x 2812 6

7x2 + 35x 28

7x2 + 35x 28

Evaluate the integral in the denominator of the equation for xc over


dA = dx = 5.2500 m2 (9)

8

7

4

1

6

1

4

6

6

14 11xy = x2 +3

dx

x

x(13 x)y =

6

y

4 m

1 m


x = 3yx = 4 y2

y

x




or,

dA = (x2 x1) dy (3)

By choosing an element of height dy, we

have also implicitly chosen y to be the

variable of integration.





xc = (1)

yc = (2)

where (xel, yel) are the coordinates of

the centroid of the differential area

element dA.

32

1


dA

yel dA

dA

xel dA

(x1, y)

xdy

(x2, y)

yx = 4 y2

x = 3y

1 m

3 m 1 m

y

x1 x2 x1



express dA and xel in terms of y (As can be seen from

Eq. 5, yel already is a function of y). The point (x, y)

on the curve satisfies

x1 = 3y

x2 = 4 y2

x2 x1 = (4 y2) (3y)

= y2 3y + 4 (6)

Substituting the expression for x2 x1 in Eq. 6 into the

equations for dA (Eq. 3) gives

dA = (x2 x1) dy (Eq. 3 repeated)

= ( y2 3y + 4) dy (7)

and for xel (Eq. 4) gives


=

= (8)

2

(4 y2) + (3y)

x2 x1

y

2

x y2 + 3y + 4

2

2

Express the coordinates of the element centroid in terms

of the coordinates of the point (x, y) on the curve. Since

the centroid of the differential element is located in the

center of the element, the x coordinate of the element

centroid is

xel = + x1

= (4)

The y coordinate of the element is the same as the y

coordinate of the point on the curve:

yel = y (5)

x2 x1

x2 x1

x = 3y

y

2

2

(x1, y)

4

x = 4 y2

(xel, yel)

(x2, y)

dy

x2 x1

5

x2 x1x1


Substitute the results given in Eqs. 9, 10, and 11

into the definitions for xc and yc:

xc = = 5.2667 2.1667

= 2.43 m Ans.

yc = = 0.75 2.1667

= 0.346 Ans.


the range from 0 to 1 (Use the integral function on your calculator):

yel dA = y( y2 3y + 4) dy = 0.75 m3 (11)

Evaluate the integral in the numerator of the equation for xc over the range

from 0 to 1 (Use the integral function on your calculator):

xel dA = [ ] [ y2 3y + 4] dy = 5.2667 m3 (10) y2 + 3y + 4


xc over the range from 0 to 1 (Use the integral function on

your calculator):

dA = ( y2 3y + 4) dy = 2.1667 m2 (9)

8

7

2

6

dA

9

x = 3y

xel dA

dA

yel dA

y

x

x = 4 y2

1

0

0

1

0

1

1 m


h

b

x

y

y

x




or,

dA = y dx (3)

By choosing an element of width dx, we

have also implicitly chosen x to be the






xc = (1)

yc = (2)



element dA

3

2

1


differential element of thickness dx.

dA

yel dA

dA

xel dA

(x, y)

dx

y = h b

x

y = h b

x

x

y


dxx

(xel, yel)

yy = x

(x, y)y

2

y

2

h

2b

b

h

b

h

h

b

x

y

Since the variable of integration is x, we now have to express dA and

yel in terms of x (As can be seen from Eq. 4, xel already is a function

of x). The point (x, y) on the curve satisfies

y = x (6)

Substituting the expression for y in Eq. 6 into the equations for dA

(Eq. 3) and for yel (Eq. 5) gives

dA = ( x) dx (7)

yel = x (8)


coordinates of the point (x, y) on the curve. The x coordinate

of the element is the same as the x coordinate of the point on

the curve:

xel = x (4)

Since the centroid of the differential element is located in the

center of the element, the y coordinate of the element centroid

is

yel = (5)

5

4


dA

xel dA

yel dA

dA

6

9

b2h/3bh/2

bh2/6bh/2

dx

x

y

b


xc over the range from 0 to b:

dA = h b

x dx = bh 2 (9)

Evaluate the integral in the numerator of the equation for

xc over the range from 0 to b:

xel dA = x( h b

x) dx = (10)


yc over the range from 0 to b:

yel dA = ( h 2b

x)( h b

x) dx = (11)bh2b

0

8

7

b

0

b2h3

0

b

6



xc = = = 2b 3

Ans.

yc = = = h 3

Ans.


a x

a




or,

dA = (a x) dy (3)

By choosing an element of width dy, we have

also implicitly chosen y to be the variable of

integration.


extends from an arbitrary point (x, y) on

the curve to an opposite boundary of the

crosshatched region.


xc = (1)

yc = (2)



element dA.

3

21


differential element of thickness dy.

dA

yel dA

dA

xel dA

(x, y)

(a, y)

dy

x = a[1 ( )2]

y

x

x

y

x = a[1 ( )2]

y

b

y

b

b

a

y

x


y

dy

a x2

x

(a, y)

x = a[1 ( )2]y

b

y

b

a + x

2

a + x

2

y

by

b

a[2 (y/b)2]2

y

Since the variable of integration is y, we now have to express dA and

xel in terms of y (As can be seen from Eq. 5, yel already is a function

of y). The point (x, y) is on the curve so it satisfies

x = a[1 ( )2] (6)

Substituting the expression for x in Eq. 6 into the equation for dA

(Eq. 3) gives

dA = (a x dy (Eq. 3 repeated)

= (a a[1 ( )2] dy

= a( )2 dy (7)

2

a + a[1 (y/b)2]

Substituting the expression for x in Eq. 6 into

the equation for xel (Eq. 4) gives


=

= (8)


coordinates of the point (x, y) on the curve. Since the centroid

of the differential element is located in the center of the

element, the x coordinate of the element centroid is

xel = + x

= (4)

The y coordinate of the element is the same as the y coordinate

of the point on the curve:

yel = y (5)

(a x)

5

4

2

(xel, yel)

(a x)

(x, y)

x

a


7a



xc = = = Ans.

yc = = = Ans. yel dA

xel dA

9

dA

dA 10

y

dy

x

y

b 3ab

ab/3

(7/30)a2 b

ab2 /4

ab/3 4

3b

30

7

b


over the range from 0 to b:

xel dA = [ ][a( )2] dy

= a2b (10)


the range from 0 to b:

yel dA = y[a )2] dy = (11)

Evaluate the integral in the denominator of the equation for x

c over the range from 0 to b:

dA = a( )2 dy = (9)

8

6

7

[2a (a/b2) y2]

0

b

0

b

b

2

y

0

b

4ab2

b

y

9.1 Centroids by Integration Example 9, page 1 of 3 +

xc

9. A sign is made of 0.5 in. thick steel plate in the shape shown.

Determine the reactions at supports B and C.

x = 50 + (10) sin

Specific weight of steel = 490 lb/ft3

Free-body diagram of plate

The weight acts through the

centroid of the plate.

Equilibrium equations for the plate

Fx = 0: Bx + Cx = 0 (1)

Fy = 0: By W = 0 (2)

MB = 0: (72 in.)Cx xcW = 0 (3)

Calculate the weight of the plate

by calculating the area. Also

calculate the distance xc to the

center of gravity.

B

C

C

B

Weight, W

+

+

Definition of centroid coordinate

xc = (1)

where xel is the coordinate of the

differential element dA.

dA

xel dA

1

2

34

y24

Cx

By

Bx

y

x

x

5

x = 50 + (10) sin y24

50 in.

72 in.

72 in.


y

(xel, yel)

76

y24

2

y24

x = 50 + (10) sin y

24

x 2

x

Since the variable of integration is y, we now have to express

dA and xel in terms of y (As can be seen from the equation of

the curve, x already is a function of y). Therefore

dA = x dy

= [50 + (10) sin dy (7)

xel = x 2

=

= 25 + 5 sin (8)

50 + (10) sin( y/24)

Express the element area in terms of the coordinates of a

point (x, y) on the curve:


or,

dA = x dy (4)

By choosing an element of width dy, we have also implicitly

chosen y to be the variable of integration.

y

Express the coordinate of the x-element centroid

in terms of the coordinates of the point (x, y) on

the curve. Since the centroid of the differential

element is located in the center of the element, the

x coordinate of the element centroid is

xel = x 2 (5)

8

x9




(x, y)dy


y24

y24

Ans.

Ans.

Ans.

xc = = 99,439

3753

= 26.5 in.

Substitute the results given in Eqs. 9 and 10 into the

definition for xc:

Thus the weight is

W = area thickness specific weight

= (3753 in2) (0.5 in.) (490 lb/ft3) ( 1 ft 12 in.

)3

= 532.1 lb

Substituting xc = 26.5 in. and W = 532.1 lb in the

equilibrium equations, Eqs. 1, 2, and 3, and solving gives

Bx = 196 lb

By = 532 lb

Cx = 196 lb


xc over the range from 0 to 72 (Use the integral function

on your calculator):

xel dA = [25 + 5 sin ][50 + (10) sin ]dy

= 99,439 in3 (10)

12

72 in.

x

xel dA

dA

x = 50 + (10) sin

y

24y 13

Evaluate the integral in the denominator of the equation

for xc over the range from 0 to 72 (Use the integral

function on your calculator):

dA = [50 + (10) sin dy = 3753 in2

(9)

10

72

0

y24

11

72

0


y = 2x2

(xel, yel)

dL

dx

dydL

x

y

Since dy/dx is slightly easier to compute than

dx/dy, express dL in terms of dy/dx and dx:

dL = [(dx)2 + (dy)2]

= [1 + ( dy dx

)2] (dx)2

= [1 + ( dy dx

)2] dx (5)

By expressing the length dL in terms of dx, we

have also implicitly chosen x to be the variable

of integration.

Locate the differential element at an

arbitrary point (x, y) on the curve and

express the centroidal coordinates of

the element in terms of x and y.

xel = x (3)

yel = y (4)


xc = (1)

yc = (2)


the centroid of the differential length

element dL.

3

2

1

10. Locate the centroid of the wire shown.

dL

yel dL

dL

xel dL

x

y

y = 2x23 m

18 m y

x



xc over the range from 0 to 3:

dL = [1 + (4x)2] dx = 18.46 m (10)

Since the variable of integration is x, we now have to express

dL and yel in terms of x (As shown in Eq. 3, xel already is

expressed as a function of x, since xel = x). The point (x, y) on

the curve satisfies:

y = 2x2 (6)

Thus

dy dx

= 4x (7)

Substituting these expressions for y and dy/dx into the

equations for yel (Eq. 4) and dL (Eq. 5) gives

yel = y

= 2x2 (8)

dL = [1 + ( dy dx

)2] dx

= [1 + (4x)2] dx (9)

0

3

5

4

x

dL

(x, y)

y

y = 2x2

3 m


Evaluate the integral in the numerator of the equation for xc over the range

from 0 to 3:

xel dL = x [1 + (4x)2] dx = 36.36 m2 (11)



yel dL = (2x2) [1 + (4x)2] dx = 163.11 m2 (12)

Substitute the results given in Eqs. 10, 11, and 12 into the definitions for xc

and yc:

xc = = 36.36 18.46

= 1.97 m Ans.

yc = = 163.11 18.46

= 8.84 m Ans.

6

dL

yel dL

xel dL

dL

8

7

0

3

0

3


x = 300[1 ( )4]

y

x

dL

(xel, yel)


xc = (1)

yc = (2)


centroid of the differential length element dL.

Locate the differential element at an arbitrary

point (x, y) on the curve and express the

centroidal coordinates of the element in terms

of x and y.

xel = x (3)

yel = y (4)

2

dL

dL

yel dL

xel dL

1

11. Locate the centroid of the wire shown.

x

y

y200

x = 300[1 ( )4]200y

300 mm

200 mm

y

x


x = 300[1 ( )4]200y

200y

4y3

2004

y200

dy

x

ydx

dxdydx

dy

dxdy

dy

Since dx/dy is slightly easier to compute than dy/dx, express

dL in terms of dx/dy and dy:

dL = [(dx)2 + (dy)2]

= [( )2 + 1] (dy)2

= [( )2 + 1] dy (5)

By expressing the length dL in terms of dy, we have also

implicitly chosen y to be the variable of integration.

3

dL

dx

(xel, yel)

y

dL

x


express dL and xel in terms of y (As shown in Eq. 4, yel

already is expressed as a function of y, since yel = y). The

coordinates of the point (x, y) on the curve satisfy

x = 300[1 ( )4] (6)

Thus

= 300[ ]

= 7.5 10-7 y3 (7)

Substituting these expressions for x and dx/dy into the

equations for xel (Eq. 3) and dL (Eq. 5) gives

xel = x

= 300[1 ( )4] (8)

dL = [( )2 + 1] dy

= [( .5 10-7 y3)2 + 1] dy (9)

4



the range from 0 to 200 (Use the integral function on your

calculator):

yel dL = y [( .5 10-7 y3)2 + 1] dy = 54 861.7 m2 (12)

Evaluate the integral in the numerator of the equation for xc over the


xel dL = 300[1 ( )4] [( .5 10-7 y3)2 + 1] dy

= 75 209.6 m2 (11)

Evaluate the integral in the denominator of the equation for xc over the


dL = [( .5 10-7 y3)2 + 1] dy = 407.4 mm (10)

6

7

5

Substitute the results given in Eqs. 10, 11, and 12 into

the definitions for xc and yc:

xc = = 75 209.6407.4

= 184.6 mm Ans.

yc = = 54 861.7 407.4

= 134.7 mm Ans.

dL

yel dL

xel dL

dL

8

y

dL

(x, y)

x

x = 300[1 ( )4]200y

y200

200

0

200

0

200

0

200 mm


12. The rod is bent into the shape of a circular arc.

Determine the reactions at the support A.

If L represents the length

of the rod, the weight W

is (0.2 lb/ft)L.

The weight of the rod acts

through the centroid (center of

gravity), located by xc.

Equations of equilibrium for the rod.

Fx = 0: Ax = 0 (1)

Therefore Ax = 0 Ans.

Fy = 0: Ay (0.2)L = 0 (2)

MA = 0: MA (0.2 lb/ft)(L)(xc) = 0 (3)

The length L of the rod can be calculated without

using an integral:

L = (angle in radians) (radius)

= (180 20 )( /180°) 3 ft

= 8.378 ft (4)

Substituting L = 8.378 ft into Eq. 2 and solving

gives

Ay = 1.68 lb Ans.

3 ft

20°

20°3 ft

xc

y

x

MA

0.2 lb/ft

0.2 lb/ft

+

+

2

3

4

5

6

W

Free-body diagram of rod1

Ay

Ax

A

+


Express the element length in terms of the polar

coordinate angle

dL = (3 ft) d (6)

To solve for the moment MA in Eq. 3 we have to

calculate the horizontal coordinate of the centroid:

xc = (5)

where xel is the horizontal coordinate of the centroid

of the differential length element dL.

xel dL

Locate the differential

element at an arbitrary point

on the curve.

x(3 ft) cos

dL

A

y 8

d

3 ft

(xel, yel)dL

20°

We do not need to evaluate the integral in the

denominator of Eq. 5 for xc, since we already know

that

dL = L = 8.378 ft (8)

by Eq. 4.

Express the horizontal coordinate of the element in

terms of .

xel = 3 ft + (3 ft) cos (7)

11

10

9

7

3 ft

xe1


To evaluate the integral in the numerator in the

expression for xc, we note the limits of

integration.

y

20° 180°

radians

xel dL = (3 + 3 cos )(3)d = 22.055 ft2 (9)

Substitute the results given in Eq. 8 and 9 into the

definition of xC:

xc = = 22.055 8.378

= 2.632 ft (10)

Substituting this result in Eq. 3 and solving for MA gives

MA = 4.41 lb·ft Ans.

x

dL

xel dL

13

18020

12


2


xc = (1)

yc = (2)

where (xel, yel) are the coordinates of the centroid of the

differential length element dL.

Because of symmetry about the y-axis, the centroid must lie

on the y-axis, that is,

xc = 0 Ans.

1

yel dL

dL

xel dL

dL

Approximate equation of centerline:

y = 639.9 ft (68.78 ft) cosh[(0.01003 ft-1)x]

y

x

625 ft

299 ft 299 ft

13. a) Locate the centroid of the Gateway Arch in St.

Louis, Missouri, USA. b) During the pre-dawn hours of

September 14, 1992, John C. Vincent of New Orleans,

Louisiana, USA, climbed up the outside of the Arch to the

top by using suction cups and then parachuted to the

ground. Estimate the length of his climb.

To avoid cluttering the equations with so many digits, define

a = 639.9 ft (3)

b = 68.78 ft (4)

c = 0.01003 ft-1 (5)

Then the equation of the arch becomes

y = 639.9 ft (68.78 ft) cosh[(0.01003 ft-1)x]

= a b cosh(cx) (6)


dL

x

Since the variable of integration is x, we now have to

express dL and yel in terms of x. The point (x, y) on the

curve satisfies

y = a b cosh (cx) (9)

Thus

= bc sinh (cx) (10)

Substituting these expressions for y and into Eq. 7 for

yel and Eq. 8 for dL gives

yel = y

= a b cosh (cx) (11)

dL = [ 1 + ( )2 ] dx

= [1 + ( bc sinh (cx))2] dx (12)

Locate the differential element at a general point (x, y)

on the curve and express the y-centroidal coordinate of

the element in terms of the point on the curve.

yel = y (7)

3

5

dy

dx

dy

dx

dy

dx

dx

dy

dx

dy

y

Since dy/dx is easier to compute than dx/dy, express dL in

terms of dy/dx and dx:

dL = [(dx)2 + (dy)2]

= [ 1 + ( )2 ] (dx)2

= [1 + ( )2] dx (8)

By expressing the length dL in terms of dx, we have also


dx

dL dy

4

y

y = a b cosh (cx)


y = a b cosh (cx)y

299 ft

x

dL

6 Evaluate the integral in the denominator of the equation for yc over the

range from 299 ft to +299 ft (Use the integral function on a calculator):

dL = [1 + ( bc sinh (cx))2] dx = 1,476 ft (13)

Similarly evaluating the numerator of the equation for yc gives

yel dL = (a b cosh (cx)) [1 + ( bc sinh (cx))2] dx

= 439,685 ft2 (14)

Substitute the results given in Eqs. 13 and 14 into the definitions for yc.

yc = = = 298 ft Ans.

Since Mr. Vincent only climbed half of the total arc length, we must

divide the length given in Eq. 13 by 2:

Mr. Vincent's climb = = 738 ft Ans.

(Actually he climbed a little more than 738 ft, since he climbed on the

outside surface, not on the centerline of the cross section.)

yel dL

dL

299

-299

1,476

2

439,6851,476

-299

299

299 ft



xc = (1)

yc = (2)

zc = (3)

where (xel, yel) are the coordinates of the centroid of

the differential element dV.

By symmetry, the centroid must lie on the y-axis, so

xc = 0 Ans.

zc = 0 Ans.

The centroidal coordinate yc remains to be calculated.

14. Locate the centroid of the cone shown.

zel dV

dV

dV

dV

yel dV

1

xel dV

z

Radius = 2 m

Ox

y

3 m


(0, yel, 0)

z

dy

3 The boundary of the disk

intersects the xy plane at an

arbitrary point P(x, y, 0).

The y coordinate of the

element centroid equals the

y coordinate of the point P

in the xy plane.

yel = y (5)

x

P(x, y, 0)

O

r

4

Use a differential element in the form of a

disk of thickness dy. The volume of the

disk is

dV = (area of circular base)

thickness of disk

= r2 dy (4)

By choosing an element of width dy, we

have also implicitly chosen y to be the


B

y

A

2

y


y


express dV and x in terms of y.

By similar triangles OCP and OBA, we have

=

Thus

x = (7)

Substituting x = into Eq. 6 gives

r = (8)

Thus the volume of the differential element in Eq. 4

becomes

dV = r2 dy

= )2 dy (9)

AB2 m

7 Evaluate the integral in the denominator of the

equation for yc over the range from 0 to 3:

dV = )2 dy = 4 m3 (10)0

3

P (x, y, 0)

y

z

Ox

C

dy

r

x O

The distance from the y axis to

the boundary of the disk is x, so

r = x (6)

y

5

A

CP

B

3 m

6

y3

2y

32y

3

3

2y

2y

3

x2

3 m

2 m

x

2y

3


Evaluate the integral in the numerator of the equation for yc:

yel dV = y [ )2] dy = 9 m4 (11)

Using the results given by Eqs. 10 and 11 in the definition of yc yields

yc = = = 2.25 m Ans. yel dV

dV 4

9

8

0

3 2y

3


Definition of centroidal coordinates

xc = (1)

yc = (2)

zc = (3)

where (xel, yel, zel) are the coordinates of the centroid

of the differential volume-element dV.

Because of symmetry, xc = yc = zc, so we only have

to compute one centroidal distance. Let's arbitrarily

choose to compute zc.

15. Locate the centroid of the volume shown.

yel dV

dV

1

xel dV

dV

yel dV

dV

z

y

x

One-eighth of a

sphere of radius "a"

a


4

y

4

The boundary of the quarter disk

intersects the yz plane at an

arbitrary point P(0, y, z).

The differential element is one-quarter of a disk of thickness dz.

The volume of the disk is

dV = (one-fourth of the area of a circle) thickness of disk

= dz (4)

Note that by choosing an element of width dz, we have also

implicitly chosen z to be the variable of integration.

3

2

(xel, yel, zel)

z

r

dz

x

The z coordinate of the element

centroid equals the z-coordinate of

the point P in the yz-plane.

zel = z (5)

P(0, y, z)z

r2


The distance from the z axis to the

boundary of the disk is y so

r = y (6)

a

z

yr

z

y

5 Since the variable of integration is z, we now have to express y

and dV in terms of z. Because point P lies on a circle of radius

"a" in the yz plane , y and z must satisfy the equation of a

circle

y2 + z2 = a2

Solving this for y gives

y = a2 z2 (7)

Substituting y = a2 z2 into Eq. 6 gives

r = y

= a2 z2 (8)

Thus the volume of the differential element in Eq. 4 becomes

dV = dz

= dz

= dz (9)

dz

x

a2 z2)4

a2 z2 2

4

4

r2

P (0, y, z)

6

[ ]


dz

y

x

Evaluate the integral in the numerator of the equation

for zc over the range from 0 to a:

zel dV = z[ ] dz = (11)

Using the results given by Eqs. 10 and 11 in the

definition of zc yields

zc = = = 3a 8

Ans.

Thus from symmetry,

xc = yc = 3a 8

Ans.

1640

zel dV

9

dV

a4/16

a3/6

Evaluate the integral in the denominator of the

equation for yc over the range from 0 to a:

dV = dz = (10)a2 z2)

8

a

0

7

a a3

a2 z2)

4 6

a4

z

a


x

x = a[1 ( z b

)2]

y

z

16. Determine the x coordinate of the centroid of the solid

shown. The solid consists of the portion of the solid of

revolution bounded by the xz and yz planes.

(This curve is rotated about the

x-axis to generate the solid.)

b

a

x = a[1 ( z b

)2]z

P(x, 0, z)

r

dx

y

The differential element is one-half of a disk of thickness dx.

The volume of the disk is

dV = (one-half of the area of a circle) thickness of disk

= dx (2)

By choosing an element of width dx, we have also implicitly

chosen x to be the variable of integration.

The boundary of the half disk intersects the xz

plane at an arbitrary point P(x, 0, z) on the

generating curve.

x

3

Definition of centroidal coordinates

xc = (1)

where xel is the coordinate of the centroid

of the differential volume-element dV.

2

r2

2

dV

1

xel dV

x


The distance from the x-axis to the

boundary of the half-disk is z so

r = z (4)

r

zP (x, 0, z)

5

x

z

4

(xel, yel, 0)

The x coordinate of the element centroid

equals the x coordinate of the point P in

the xy plane.

xel = x (3)

y

Since the variable of integration is x, we now have to

express z and dV in terms of x. Because point P lies

on the generating curve, x and z must satisfy the

equation of that curve

x = a[1 ( z b

)2]

Solving this for z gives

z = b 1 ( x a ) (5)

Substituting z = b [1 ( x a )] into Eq. 4 gives

r = z

= b 1 ( x a ) (6)

Thus the volume of the differential element in Eq. 2

becomes

dV = dx

= dx

= dx (7)

x

r2

b 1 (x/a) ]2

b2 x/a)2

2

2

6


adx

y

Evaluate the integral in the numerator of the equation for xc over

the range from 0 to a:

xel dV = x[ ] dx = (9)a2b2

Using the results given by Eqs. 8 and 9 in the definition of xc

yields

xc = = = a 3

Ans. dV

xel dV

9

ab2/4

a2b2/12

b2 x/a)

2 12

Evaluate the integral in the denominator of the equation for xc

over the range from 0 to a:

dV = dx = (8)b2 x/a)

8

x

7

2ab2

40

a

0

a

z

P(x, 0, z)


x

y

z

a a

b

b

17. Locate the centroid of the pyramid shown.

xel dV

yel dV

dV

dV

dV

zel dV


xc = (1)

yc = (2)

zc = (3)

where ( xel, yel) are the coordinates of the

centroid of the differential volume-element

dV.

By symmetry, the centroid must lie on the

y-axis, so

x c = 0 Ans.

z c = 0 Ans.

y c remains to be calculated.

1

h


The y coordinate of the element centroid equals

the y coordinate of the point P in the xy plane.

yel = y (5)

5

Q(0, y, z)

z

b

a

B

O

ydy C

The boundary of the differential element

intersects the xy and xz planes at arbitrary points

P and Q on the sloping lines in those planes.

4

x

2 Because of symmetry, we need consider

only one-fourth of the pyramid.

A

(xel, yel, zel)

P(x, y, 0)

y

The differential element is a rectangular box of

thickness dy. The volume of the box is

dV o = (area of base) thickness

= (xz) dy (4)

By choosing an element of width dy, we have also

implicitly chosen y to be the variable of integration.

3

h


y

O

z

a

Cx

h

x

h y

P(x, y, 0)

z

B b

zQ(0, y, z)

xO

y

h

A

y

6

A

h y

y

7

h y

h

y

h

x a

Because the variable of integration is y, we now have to dexpress dV in terms of y. Because point P lies on the line AC in the xy plane, similar triangles can be used to derive an equation relating x and y:

= Solving for x gives

x = a(1 ) (6)

Because point Q lies on the line AB in the yz plane, similar triangles can be used to derive an equation relating z and y:

= Solving for x gives

z = b(1 ) (7)

bz

h

y

h

h y


Substituting Eqs. 6 and 7 in Eq. 4 and multiplying by

4 to get the volume of the whole pyramid (not just one

fourth) gives the volume element in terms of y.

dV = 4(dV o)

= 4(xz)dy

= 4[a(1 )] [b(1 ] dy

= 4ab(1 )2 dy (8)

z

y

h

x

Evaluate the integral in the denominator of the equation for yc

over the range from 0 to h:

dV = 4 ab(1 )2 dy = (9)

Evaluate the integral in the numerator of the equation for yc:

yel dV = 4 y[ab(1 )2]dy = (10)

Using the results given by Eqs. 9 and 10 in the definition of yc

yields

yc = = = h 4 Ans.

3abh2

yel dV abh2/34abh/3 dV

9

8

y

h

34abh

h

y

h

y

h

y y

h

h

0

h

0

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9.1 Centroids by Integration - School of · PDF file9.1 Centroids by Integration Example 1, page 3 of 4 Express the coordinates of the element centroid in terms of the coordinates