Center of Gravity and Centroid 9 Engineering Mechanics: Statics in SI Units, 12e Copyright © 2010 Pearson Education South Asia Pte Ltd

Copyright © 2010 Pearson Education South Asia Pte Ltd

Center of Gravity and Centroid99

Engineering Mechanics: Statics in SI Units, 12e


Chapter Objectives

• Concept of the center of gravity, center of mass, and the centroid

• Determine the location of the center of gravity and centroid for a system of discrete particles and a body of arbitrary shape

• Theorems of Pappus and Guldinus• Method for finding the resultant of a general distributed

loading


Chapter Outline

1. Center of Gravity and Center of Mass for a System of Particles

2. Composite Bodies

3. Theorems of Pappus and Guldinus

4. Resultants of a General Distributed Loading

5. Fluid Pressure


9.1 Center of Gravity and Center of Mass for a System of Particles

Center of Gravity • Locates the resultant weight of a system of particles• Consider system of n particles fixed within a region of

space• The weights of the particles can be replaced by a

single (equivalent) resultant weight having defined point G of application



Center of Gravity • Resultant weight = total weight of n particles

• Sum of moments of weights of all the particles about x, y, z axes = moment of resultant weight about these axes

• Summing moments about the x axis,

• Summing moments about y axis,

nnR

nnR

R

WyWyWyWy

WxWxWxWx

WW

~...~~

~...~~

2211

2211



Center of Gravity • Although the weights do not produce a moment about

z axis, by rotating the coordinate system 90° about x or y axis with the particles fixed in it and summing moments about the x axis,

• Generally,

mmz

zmmy

ymmx

x

WzWzWzWz nnR

~,

~;

~

~...~~2211



Center Mass • Provided acceleration due to gravity g for every

particle is constant, then W = mg

• By comparison, the location of the center of gravity coincides with that of center of mass

• Particles have weight only when under the influence of gravitational attraction, whereas center of mass is independent of gravity

mmz

zmmy

ymmx

x

~,

~;

~



Center Mass • A rigid body is composed of an infinite number of

particles • Consider arbitrary particle having a weight of dW

dW

dWzz

dW

dWyy

dW

dWxx

~;

~;

~



Centroid of a Volume• Consider an object subdivided into volume elements

dV, for location of the centroid,

V

V

V

V

V

V

dV

dVz

zdV

dVy

ydV

dVx

x

~

;

~

;

~



Centroid of an Area• For centroid for surface area of an object, such as

plate and shell, subdivide the area into differential elements dA

A

A

A

A

A

A

dA

dAz

zdA

dAy

ydA

dAx

x

~

;

~

;

~



Centroid of a Line• If the geometry of the object takes the form of a line,

the balance of moments of differential elements dL about each of the coordinate system yields

L

L

L

L

L

L

dL

dLz

zdL

dLy

ydL

dLx

x

~

;

~

;

~


Example 9.1

Locate the centroid of the rod bent into the shape of a parabolic arc.


Example 9.1

Differential element

Located on the curve at the arbitrary point (x, y)

Area and Moment Arms

For differential length of the element dL

Since x = y2 and then dx/dy = 2y

The centroid is located at

yyxx

dyydL

dydydx

dydxdL

~,~

12

1

2

222


Example 9.1

Integrations

m

dyy

dyyy

dL

dLy

y

m

dyy

dyyy

dyy

dyyx

dL

dLx

x

L

L

L

L

574.0479.18484.0

14

14~

410.0479.16063.0

14

14

14

14~

1

02

1

02

1

02

1

022

1

02

1

02


9.2 Composite Bodies

• Consists of a series of connected “simpler” shaped bodies, which may be rectangular, triangular or semicircular

• A body can be sectioned or divided into its composite parts

• Accounting for finite number of weights

W

Wzz

W

Wyy

W

Wxx

~~~



Procedure for Analysis

Composite Parts• Divide the body or object into a finite number of

composite parts that have simpler shapes• Treat the hole in composite as an additional

composite part having negative weight or size

Moment Arms• Establish the coordinate axes and determine the

coordinates of the center of gravity or centroid of each part



Procedure for Analysis

Summations• Determine the coordinates of the center of gravity by

applying the center of gravity equations• If an object is symmetrical about an axis, the centroid

of the objects lies on the axis


Example 9.10

Locate the centroid of the plate area.


Solution

Composite Parts

Plate divided into 3 segments.

Area of small rectangle considered “negative”.


Solution

Moment Arm

Location of the centroid for each piece is determined and indicated in the diagram.

Summations

mmA

Ayy

mmA

Axx

22.15.11

14~

348.05.11

4~


9.3 Theorems of Pappus and Guldinus

• A surface area of revolution is generated by revolving a plane curve about a non-intersecting fixed axis in the plane of the curve

• A volume of revolution is generated by revolving a plane area bout a nonintersecting fixed axis in the plane of area



• The theorems of Pappus and Guldinus are used to find the surfaces area and volume of any object of revolution provided the generating curves and areas do not cross the axis they are rotated

Surface Area• Area of a surface of revolution = product of length of

the curve and distance traveled by the centroid in generating the surface area

LrA



Volume• Volume of a body of revolution = product of generating

area and distance traveled by the centroid in generating the volume

ArV


Example 9.12

Show that the surface area of a sphere is A = 4πR2 and its volume V = 4/3 πR3.

Solution

Surface Area

Generated by rotating semi-arc about the x axis

For centroid,

For surface area,

/2Rr

242

2

;~

RRR

A

LrA


Solution

Volume

Generated by rotating semicircular area about the x axis

For centroid,

For volume,

32

34

21

34

2

;~

3/4

RRR

V

ArV

Rr


9.4 Resultant of a General Distributed Loading

Pressure Distribution over a Surface• Consider the flat plate subjected to the loading function

ρ = ρ(x, y) Pa• Determine the force dF acting on the differential area

dA m2 of the plate, located at the differential point (x, y)

dF = [ρ(x, y) N/m2](d A m2)

= [ρ(x, y) d A]N• Entire loading represented as

infinite parallel forces acting on separate differential area dA



Pressure Distribution over a Surface• This system will be simplified to a single resultant force

FR acting through a unique point on the plate



Magnitude of Resultant Force• To determine magnitude of FR, sum the differential

forces dF acting over the plate’s entire surface area dA• Magnitude of resultant force = total volume under the

distributed loading diagram

• Location of Resultant Force is

V

V

A

A

V

V

A

A

dV

ydV

dAyx

dAyxyy

dV

xdV

dAyx

dAyxxx

),(

),(

),(

),(


9.5 Fluid Pressure

• According to Pascal’s law, a fluid at rest creates a pressure ρ at a point that is the same in all directions

• Magnitude of ρ depends on the specific weight γ or mass density ρ of the fluid and the depth z of the point from the fluid surface

ρ = γz = ρgz• Valid for incompressible fluids • Gas are compressible fluids and the above equation

cannot be used


9.5 Fluid Pressure

Flat Plate of Constant Width• Consider flat rectangular plate of constant width

submerged in a liquid having a specific weight γ• Plane of the plate makes an angle with the horizontal

as shown


9.5 Fluid Pressure

Flat Plate of Constant Width• As pressure varies linearly with depth, the distribution

of pressure over the plate’s surface is represented by a trapezoidal volume having an intensity of ρ1= γz1 at depth z1 and ρ2 = γz2 at depth z2

• Magnitude of the resultant force FR = volume of this loading diagram


9.5 Fluid Pressure

Curved Plate of Constant Width• When the submerged plate is curved, the pressure

acting normal to the plate continuously changes direction

• Integration can be used to determine FR and location of center of centroid C or pressure P


9.5 Fluid Pressure

Flat Plate of Variable Width• Consider the pressure distribution acting on the surface

of a submerged plate having a variable width• Since uniform pressure ρ = γz (force/area) acts on dA,

the magnitude of the differential force dF

dF = dV = ρ dA = γz(xdy’)


9.5 Fluid Pressure

Flat Plate of Variable Width• Centroid V defines the point which FR acts

• The center of pressure which lies on the surface of the plate just below C has the coordinates P defined by the equations

• This point should not be mistaken for centroid of the plate’s area

A VR VdVdAF

V

V

V

V

dV

dVyy

dV

dVxx

'~'

~


Example 9.14

Determine the magnitude and location of the resultant hydrostatic force acting on the submerged rectangular plate AB. The plate has a width of 1.5m; ρw = 1000kg/m3.


Solution

The water pressures at depth A and B are

For intensities of the load at A and B,

kPamsmmkggz

kPamsmmkggz

BwB

AwA

05.49)5)(/81.9)(/1000(

62.19)2)(/81.9)(/1000(23

23

mkNkPambw

mkNkPambw

BB

AA

/58.73)05.49)(5.1(

/43.29)62.19)(5.1(


Solution

For magnitude of the resultant force FR created by the distributed load.

This force acts through the centroid of the area,

measured upwards from B

N

trapezoidofareaFR

5.154)6.734.29)(3(2

1

mh 29.1)3(58.7343.29

58.73)43.29(2

3

1


Solution

Same results can be obtained by considering two components of FR defined by the triangle and rectangle.

Each force acts through its associated centroid and has a magnitude of

Hence

kNmmkNF

kNmmkNF

t 2.66)3)(/15.44(

3.88)3)(/43.29(Re

kNkNkNFFF RR 5.1542.663.88Re


Solution

Location of FR is determined by summing moments about B

mh

h

MM BBR

29.1

)1(2.66)5.1(3.88)5.154(

;


QUIZ

1.The _________ is the point defining the geometric center of an object.

A)Center of gravity B) Center of mass

C)Centroid D) None of the above

2. To study problems concerned with the motion of matter under the influence of forces, i.e., dynamics, it is necessary to locate a point called ________.

A) Center of gravity B) Center of mass

C) Centroid D) None of the above


QUIZ

3. If a vertical rectangular strip is chosen as the differential element, then all the variables, including the integral limit, should be in terms of _____ .

A) x B) y

C) z D) Any of the above.

4. If a vertical rectangular strip is chosen, then what are the values of x and y?

A) (x , y) B) (x / 2 , y / 2)

C) (x , 0) D) (x , y / 2)

~ ~


QUIZ

5. A composite body in this section refers to a body made of ____.

A) Carbon fibers and an epoxy matrix

B) Steel and concrete

C) A collection of “simple” shaped parts or holes

D) A collection of “complex” shaped parts or holes

6. The composite method for determining the location of the center of gravity of a composite body requires _______.

A) Integration B) Differentiation

C) Simple arithmetic D) All of the above.


QUIZ

7. Based on the typical centroid information, what are the minimum number of pieces you will have to consider for determining the centroid of the area shown at the right?

A)1 B) 2 C) 3 D) 4

8. A storage box is tilted up to clean the rug underneath the box. It is tilted up by pulling the handle C, with edge A remaining on the ground. What is the maximum angle of tilt (measured between bottom AB and the ground) possible before the box tips over?

A) 30° B) 45 ° C) 60 ° D) 90 °


QUIZ

7. What are the min number of pieces you will have to consider for determining the centroid of the area?

A)1 B) 2 C) 3 D) 4

8. A storage box is tilted up by pulling C, with edge A remaining on the ground. What is the max angle of tilt possible before the box tips over?

A) 30° B) 45 ° C) 60 ° D) 90 °

3cm 1 cm

1 cm

3cm

30º

G

C

AB


QUIZ

9. For determining the centroid, what is the min number of pieces you can use?

A) Two B) Three

C) Four D) Five

10. For determining the centroid of the area,what are the coordinates (x, y ) of the centroid of square DEFG?

A) (1, 1) m B) (1.25, 1.25) m

C) (0.5, 0.5 ) m D) (1.5, 1.5) m

2cm 2cm

2cm

4cm

x

y

A1m

1m

y

E

F G

CB x

1m 1m

D

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Center of Gravity and Centroid 9 Engineering Mechanics: Statics in SI Units, 12e Copyright © 2010 Pearson Education South Asia Pte Ltd