Ch07 Distributed Forces Centroids and Centers of Gravity 2

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Vector Mechanics for Engineers: StaticsE i gh t h

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CE 102 Statics

Chapter 7

Distributed Forces:

Centroids and Centers of Gravity





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Contents

Introduction

Center of Gravity of a 2D Body

Centroids and First Moments of Areas

and Lines

Centroids of Common Shapes of Areas

Centroids of Common Shapes of Lines

Composite Plates and Areas

Sample Problem 7.1

Determination of Centroids by

Integration

Sample Problem 7.2

Theorems of Pappus-Guldinus

Sample Problem 7.3

Distributed Loads on Beams

Sample Problem 7.4

Center of Gravity of a 3D Body:Centroid of a Volume

Centroids of Common 3D Shapes

Composite 3D Bodies

Sample Problem 7.5





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Introduction

The earth exerts a gravitational force on each of the particlesforming a body. These forces can be replace by a single

equivalent force equal to the weight of the body and applied

at the center of gravity for the body.

The centroid of an area is analogous to the center of gravity of a body. The concept of the first moment of an

area is used to locate the centroid.

Determination of the area of a surface of revolution and

the volume of a body of revolution are accomplished

with the Theorems of Pappus-Guldinus.





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Center of Gravity of a 2D Body

Center of gravity of a plate

´

§§´§§

!

(!

!(!

dW y

W yW y M

dW x

W xW x M

y

y

Center of gravity of a wire





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Centroids and First Moments of Areas and Lines

x

QdA y A y

y

QdA x A x

dAt x At x

dW xW x

x

y

respect tohmoment witfirst

respect tohmoment witfirst

!

!!

!!!

!

!

´

´

´´

KK

Centroid of an area

´´

´´

!!

!

!

dL y L y

dL x L x

dLa x La x

dW xW x

KK

Centroid of a line





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First Moments of Areas and Lines

An area is symmetric with respect to an axis BB¶

if for every point P there exists a point P ¶ such

that PP ¶ is perpendicular to BB¶ and is divided

into two equal parts by BB¶.

The first moment of an area with respect to a

line of symmetry is zero.

If an area possesses a line of symmetry, its

centroid lies on that axis

If an area possesses two lines of symmetry, its

centroid lies at their intersection.

An area is symmetric with respect to a center O

if for every element dA at ( x,y) there exists an

area dA¶ of equal area at (-x,-y).

The centroid of the area coincides with the

center of symmetry.





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Centroids of Common Shapes of Areas





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Centroids of Common Shapes of Lines





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Composite Plates and Areas

Composite plates

§§§§

!

!

W yW Y

W xW X

Composite area

§§ §§ !! A y AY A x A X





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Sample Problem 7.1

For the plane area shown, determine

the first moments with respect to the

x and y axes and the location of the

centroid.

SOLUTION:

Divide the area into a triangle, rectangle,

and semicircle with a circular cutout.

Compute the coordinates of the area

centroid by dividing the first moments by

the total area.

Find the total area and first moments of

the triangle, rectangle, and semicircle.

Subtract the area and first moment of the

circular cutout.

Calculate the first moments of each area

with respect to the axes.





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Sample Problem 7.1

33

33

mm107.757

mm102.506

v!

v!

y

x

Q

Q Find the total area and first moments of the

triangle, rectangle, and semicircle. Subtract the

area and first moment of the circular cutout.





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Sample Problem 7.1

2333

mm1013.828

mm107.757

v

v!!§

§ A A x X

mm8.54! X

23

33

mm1013.828

mm102.506

v

v!!

§§

A

A yY

mm6.36!Y

Compute the coordinates of the areacentroid by dividing the first moments by

the total area.





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Determination of Centroids by Integration

ydx y

dA y A y

ydx x

dA x A x

el

el

´

´´

´

!

!

!

!

2

? A

? Adx xa y

dA y A y

dx xa xa

dA x A x

el

el

!

!

!

!

´´

´

´

2

¹

º

¸©

ª

¨!

!

¹ º ¸©

ª¨!

!

´

´

´

´

UU

UU

d r r

dA y A y

d r r

dA x A x

el

el

2

2

2

1sin

3

2

2

1cos

3

2

´´´´ ´´´´ !!!

!!!

dA ydydx ydA y A y

dA xdydx xdA x A x

el

el Double integration to find the first moment

may be avoided by defining dA as a thinrectangle or strip.





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Sample Problem 7.2

Determine by direct integration the

location of the centroid of a parabolic

spandrel.

SOLUTION:

Determine the constant k.

Evaluate the total area.

Using either vertical or horizontal

strips, perform a single integration to

find the first moments.

Evaluate the centroid coordinates.





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Sample Problem 7.2

SOLUTION:

Determine the constant k.

21

21

2

2

2

2

2

yb

a xor x

a

b y

a

bk ak b

xk y

!!

!!

!

Evaluate the total area.

3

30

3

20

2

2

ab

x

a

bdx x

a

bdx y

dA A

aa

!

¼¼½

»

¬¬-

«!!!

!

´´

´





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Sample Problem 7.2

Using vertical strips, perform a single integrationto find the first moments.

1052

2

1

2

44

2

0

5

4

2

0

22

2

2

0

4

2

0

2

2

ab x

a

b

dx xa

bdx y

ydA yQ

ba xab

dx xa

b xdx xydA xQ

a

a

el x

a

a

el y

!¼¼½

»

¬¬-

«!

¹ º

¸©ª

¨!!!

!¼¼½»

¬¬-«!

¹ º

¸©ª

¨!!!

´´´

´´´





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Sample Problem 7.2

Or, using horizontal strips, perform a singleintegration to find the first moments.

10

421

22

2

0

2321

21

21

2

0

22

0

22

abdy yb

aay

dy yb

aa ydy xa ydA yQ

bady yb

aa

dy xa

dy xa xa

dA xQ

b

el x

b

b

el y

!¹ º ¸©

ª¨ !

¹ º

¸©ª

¨!!!

!¹¹ º ¸©©

ª¨ !

!

!!

´

´´´

´

´´´





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Sample Problem 7.2

E

valuate the centroid coordinates.

43

2baab

x

Q A x y

!

!

a x4

3!

103

2abab y

Q A y x

!

!

b y10

3!





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Surface of revolution is generated by rotating a

plane curve about a fixed axis.

Area of a surface of revolution is

equal to the length of the generatingcurve times the distance traveled by

the centroid through the rotation.

L y A T2!





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Body of revolution is generated by rotating a plane

area about a fixed axis.

Volume of a body of revolution is

equal to the generating area timesthe distance traveled by the centroid

through the rotation.

A yV T2!





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Sample Problem 7.3

The outside diameter of a pulley is 0.8

m, and the cross section of its rim is as

shown. Knowing that the pulley is

made of steel and that the density of steel is

determine the mass and weight of the

rim.

33mkg1085.7 v! V

SOLUTION:

Apply the theorem of Pappus-Guldinus

to evaluate the volumes or revolution

for the rectangular rim section and the

inner cutout section.

Multiply by density and accelerationto get the mass and acceleration.





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Sample Problem 7.3

SOLUTION:

Apply the theorem of Pappus-Guldinus

to evaluate the volumes or revolution for

the rectangular rim section and the inner

cutout section.

¹ º

¸©ª

¨vv!! 3393633 mmm10mm1065.7mkg1085.7V m V kg0.60!m

2

sm81.9kg0.60!! mg W N589!W

Multiply by density and acceleration toget the mass and acceleration.





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Distributed Loads on Beams

A distributed load is represented by plotting the load

per unit length, w (N/m) . The total load is equal to

the area under the load curve (dW = wdx).

´´ !!! AdAdxwW L

0

A xdA x AO P

dW xW O P L

!!

!

´

´

0

A distributed load can be replace by a concentratedload with a magnitude equal to the area under the

load curve and a line of action passing through the

area centroid.





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Sample Problem 7.4

A beam supports a distributed load as

shown. Determine the equivalent

concentrated load and the reactions atthe supports.

SOLUTION:

The magnitude of the concentrated load

is equal to the total load or the area under

the curve.

The line of action of the concentrated

load passes through the centroid of the

area under the curve.

Determine the support reactions by

summing moments about the beam

ends.





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Sample Problem 7.4

SOLUTION:

The magnitude of the concentrated load is equal to

the total load or the area under the curve.

kN0.18! F

The line of action of the concentrated load passesthrough the centroid of the area under the curve.

kN18

mkN63 ! X m5.3! X





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Sample Problem 7.4

Determine the support reactions by summingmoments about the beam ends.

0m.53kN18m6:0 !!§ y A B M

kN5.10! y B

0m.53m6kN18m6:0 !!§ y B A M

kN5.7! y A





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Center of Gravity of a 3D Body: Centroid of a Volume

Center of gravity G

§ (! jW jW TT

? A jW r jW r

jW r jW r

G

GTTTT

TTTT

v(!v

(v!v

§§

´´ !! dW r W r dW W GTT

R esults are independent of body orientation,

´´´ !!! zdW W z ydW W y xdW W x

´´´ !!! zdV V z ydV V y xdV V x

dV dW V W KK !! and

For homogeneous bodies,





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Centroids of Common 3D Shapes





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Composite 3D Bodies

Moment of the total weight concentrated at thecenter of gravity G is equal to the sum of the

moments of the weights of the component parts.

§§§§§§ !!! W zW Z W yW Y W xW X

For homogeneous bodies,

§§§§§§ !!! V zV Z V yV Y V xV X





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Sample Problem 7.5

Locate the center of gravity of thesteel machine element. The diameter

of each hole is 1 in.

SOLUTION:

Form the machine element from a

rectangular parallelepiped and a

quarter cylinder and then subtracting

two 1-in. diameter cylinders.





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Sample Problem 7.5





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Sample Problem 7.5

34 in.2865in08.3!! §§ V V x X

34 in.2865in5.047!!

§§V V yY

34in.2865in.6181!! §§ V V z Z

in.577.0! X

in.577.0!Y

in.577.0! Z



33

y

x

20 mm 30 mm

Problem 7.6

36 mm

24 mm

Locate the centroid of the plane

area shown.



34

Solving Problems on Your Own

Several points should be emphasized

when solving these types of problems.

Locate the centroid of the plane area

shown.

y

x

20 mm 30 mm

36 mm

24 mm

1. Decide how to construct the given area from common shapes.

2. It is strongly recommended that you construct a table

containing areas or length and the respective coordinates of the centroids.

3. When possible, use symmetry to help locate the centroid .

Problem 7.6



35

Problem 7.6 Solution y

x

24 + 12

20 + 10

10

30

Decide how to construct the given

area from common shapes.C

1

C 2

Dimensions in mm



36


x

24 + 12

20 + 10

10

30

C 1

C 2

Dimensions in mm

C onstruct a table containing areas and

respective coordinates of thecentroids.

A, mm2 x, mm y, mm xA, mm3 yA, mm3

1 20 x 60 =1200 10 30 12,000 36,0002 (1/2) x 30 x 36 =540 30 36 16,200 19,440

7 1740 28,200 55,440



37


x

24 + 12

20 + 10

10

30

C 1

C 2

Dimensions in mm

X7 A = 7 xA

X(1740) = 28,200

Then

or X = 16.21 mm

and Y7 A = 7 yA

Y (1740) = 55,440

or Y = 31.9 mm

A, mm2 x, mm y, mm xA, mm3 yA, mm3

1 20 x 60 =1200 10 30 12,000 36,000

2 (1/2) x 30 x 36 =540 30 36 16,200 19,440

7 1740 28,200 55,440



38

Problem 7.7

The beam AB supports twoconcentrated loads and

rests on soil which exerts a

linearly distributed upward

load as shown. Determine

(a) the distance a for which

w A = 20 kN/m, (b) the

corresponding value w B.

w A w

B

A B

a 0.3 m24 kN 30 kN

1.8 m



39

The beam AB supports two

concentrated loads andrests on soil which exerts a

linearly distributed upward

load as shown. Determine

(a) the distance a for whichw A = 20 kN/m, (b) the

corresponding value w B.

w A w

B

A B

a 0.3 m24 kN 30 kN

1.8 m


1. R eplace the distributed load by a single equivalent force.

The magnitude of this force is equal to the area under the

distributed load curve and its line of action passes through

the centroid of the area.

2. When possible, complex distributed loads should be

divided into common shape areas.

Problem 7.7



40

Problem 7.7 Solution

w B

A B

a 0.3 m24 kN 30 kN

20 kN/m

C

0.6 m 0.6 m

R I R II

We have RI= (1.8 m)(20 kN/m) = 18 kN

1

2

RII= (1.8 m)(w

BkN/m) = 0.9 w

BkN

1

2

R eplace the distributed

load by a pair of

equivalent forces.



41


w B

A B

a 0.3 m24 kN 30 kN

C

0.6 m 0.6 m

R I = 18 kN

R II

= 0.9 w B

kN

(a) 7 M C = 0: (1.2 - a)m x 24 kN - 0.6 m x 18 kN

- 0.3m x 30 kN = 0

or a = 0.375 m

(b) 7 F y = 0: -24 kN + 18 kN + (0.9 w B

) kN - 30 kN= 0

or w B

= 40 kN/m

+

+



42

Problem 7.8

y

x

0.75 in z

1 in2 in2 in

3 in

2 in2 in

r = 1.25 in

r = 1.25 in

For the machine element

shown, locate the z coordinateof the center of gravity.



43


y

x

0.75 in z

1 in2 in2 in

3 in

2 in2 in

r = 1.25 in

r = 1.25 in

X 7V = 7 x V Y 7V = 7 y V Z 7V = 7 z V

where X, Y, Z and x, y, z are the coordinates of the centroid of the

body and the components, respectively.

For the machine elementshown, locate the z coordinate

of the center of gravity.

Problem 7.8

Determine the center of

gravity of composite body.For a homogeneous body

the center of gravity coincides

with the centroid of its volume. For this case the center of gravity

can be determined by

8 S



44


x

0.75 in z

1 in2 in2 in

3 in

2 in2 in

r = 1.25 in

r = 1.25 in

Determine the center of gravity

of composite body.

First assume that the machine

element is homogeneous so

that its center of gravity will

coincide with the centroid of

the corresponding volume.

y

x

z

I

IIIII

IVV

Divide the body into

five common shapes.



45

y

x

z

I

IIIII

IVV

y

x

0.75 in z

1 in2 in2 in

3 in

2 in2 in

r = 1.25 in

r = 1.25 in

V, in3 z, in. z V, in4

I (4)(0.75)(7) = 21 3.5 73.5

II (T/2)(2)2 (0.75) = 4.7124 7+ [(4)(2)/(3T)] = 7.8488 36.987

III -T(11.25)2

(0.75)= -3.6816 7 -25.771IV (1)(2)(4) = 8 2 16

V -(T/2)(1.25)2 (1) = -2.4533 2 -4.9088

7 27.576 95.807

Z 7V = 7 z V : Z (27.576 in3 ) = 95.807 in4 Z = 3.47 in



46

Problem 7.9

Locate the centroid of the volume

obtained by rotating the shaded area

about the x axis.

h

a

x

y y = kx1/3



47


1. When possible, use symmetry to help locate the centroid .

Locate the centroid of the volume

obtained by rotating the shaded area

about the x axis.

h

a

y = kx1/3

x

y

The procedure for locating the

centroids of volumes by direct

integration can be simplified:

2. If possible, identify an element of volume dV which produces

a single or double integral , which are easier to compute.

3. After setting up an expression for dV , integrate and determine

the centroid.

Problem 7.9



48


x

y

z x

dx

y = kx1/3

r

U se symmetry to help locate the

centroid . Symmetry implies

y = 0 z = 0

I dentify an element of volume dV

which produces a single or double integral.

Choose as the element of volume a disk or radius r and

thickness dx . Then

dV = Tr 2 dx xel

= x

S



49


x

y

z x

dx

y = kx1/3

r

I dentify an element of volume dV

which produces a single or

double integral.

dV = Tr 2 dx xel

= x

Now r = kx 1/3 so that

dV = Tk 2 x2/3dx

At x = h, y = a : a = kh1/3 or k = a/h1/3

Then dV = T x2/3dxa2

h2/3

P bl 7 9 S l ti



50


x

y

z x

dx

y = kx1/3

r

dV = T x2/3

dxa2

h2/3

I ntegrate and determine the centroid .

´0

h

V = T x2/3dxa2

h2/3

´

= T x5/3a2

h2/335[ ]

0

h

= Ta2h3

5

´0

h

xel

dV = x (T x2/3 dx) = T [ x8/3 ]a2

h2/3a2

h2/338

Also

= Ta2h23

8

P bl 7 9 S l ti



51


x

y

z x

dx

y = kx1/3

r

I ntegrate and determine the centroid .

´

V = Ta2

h

3

5

xel

dV = Ta2h23

8

Now xV = xdV :´ 3

8 x ( Ta2h) = Ta2h23

5

x = h5

8

y = 0 z = 0

P bl 7 10



52

Problem 7.10

The square gate AB is held in theposition shown by hinges along its

top edge A and by a shear pin at B.

For a depth of water d = 3.5 ft,

determine the force exerted on the

gate by the shear pin.

A

B

d

1.8 ft

30o

Problem 7 10



53


The square gate AB is held in the

position shown by hinges along itstop edge A and by a shear pin at B.




A

B

d

1.8 ft

30o

Assuming the submerged body has a width b, the load per unit

length is w = b Vgh, where h is the distance below the surface of

the fluid.

1. First, determine the pressure distribution acting perpendicular

the surface of the submerged body. The pressure distribution

will be either triangular or trapezoidal.

Problem 7.10

Problem 7 10



54


The square gate

AB is held in theposition shown by hinges along its

top edge A and by a shear pin at B.




A

B

d

1.8 ft

30o

2. R eplace the pressure distribution with a resultant force, and

construct the free-body diagram.

3. Write the equations of static equilibrium for the problem, and

solve them.

Problem 7.10

Problem 7 10 Solution



55


A

B

Determine the pressure distribution

acting perpendicular the surface of thesubmerged body.

1.7 ft

(1.8 ft) cos 30o

P A

P B

P A

= 1.7 V g

P B = (1.7 + 1.8 cos 30o) V g

Problem 7 10 SolutionA



56


A

B

(1.8 ft) cos 30o

R eplace the pressure

distribution with a

resultant force, and construct the free-body

diagram.

A y

A x

F B

1.7 V g

(1.7 + 1.8 cos 30o) V g

L AB

/3

L AB

/3

L AB

/3

P1

P2

The force of the water

on the gate is

P = Ap = A( Vgh)1

21

2

P 1 = (1.8 ft)2(62.4 lb/ft3)(1.7 ft) = 171.85 lb1

2

P 2 = (1.8 ft)2(62.4 lb/ft3)(1.7 + 1.8 cos 30o)ft = 329.43 lb1

2

Problem 7 10 SolutionA



57


A

B

(1.8 ft) cos 30o

A y

A x

F B

1.7 V g

(1.7 + 1.8 cos 30o) V g

L AB

/3

L AB

/3

L AB

/3

P1

P2

P 1 = 171.85 lb P 2 = 329.43 lb

Write the equations of

static equilibrium for the

problem, and solve them.

7 M A = 0:

( L A B) P 1 + ( L A B) P 213

23

- L A B F B = 0

+

(171.85 lb) + (329.43 lb) - F B = 013

2

3 F B = 276.90 lb

30o

FB = 277 lb

Documents

Ch07 Distributed Forces Centroids and Centers of Gravity 2