Engineering Mechanics: Statics Chapter 9: Center of Gravity and Centroid Chapter 9: Center of Gravity and Centroid

Engineering Mechanics: StaticsEngineering Mechanics: Statics

Chapter 9: Center of Gravity and

Centroid

Chapter 9: Center of Gravity and

Centroid

Chapter ObjectivesChapter Objectives

To discuss the concept of the center of gravity, center of mass, and the centroid.

To show how to determine the location of the center of gravity and centroid for a system of discrete particles and a body of arbitrary shape.

To use the theorems of Pappus and Guldinus for finding the area and volume for a surface of revolution.

To present a method for finding the resultant of a general distributed loading and show how it applies to finding the resultant of a fluid.

Chapter OutlineChapter Outline

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

Center of Gravity and Center of Mass and Centroid for a Body

Composite BodiesTheorems of Pappus and GuldinusResultant of a General Distributed

LoadingFluid Pressure

9.1 Center of Gravity and Center of Mass for a System of Particles9.1 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

comprise a system of parallel forces which 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 of Gravity Where represent the coordinates of the

center of gravity G of the system of particles, represent the coordinates of each particle in the

system and represent the resultant sum of the weights of all the particles in the system.

These equations represent a balance between the sum of the moments of the weights of each particle and the moment of resultant weight for the system.

zyx ,,

W

zyx ~,~,~


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

~,

~;

~

9.2 Center of Gravity and Center of Mass and Centroid for a Body9.2 Center of Gravity and Center of Mass and Centroid for a Body

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

~;

~;

~

9.2 Center of Gravity and Center of Mass and Centroid

for a Body

9.2 Center of Gravity and Center of Mass and Centroid

for a BodyCenter Mass γ represents the specific weight and dW =

γdV

V

V

V

V

V

V

dV

dVz

zdV

dVy

ydV

dVx

x

~

;

~

;

~


Center of Mass Density ρ, or mass per unit volume, is

related to γ by γ = ρg, where g = acceleration due to gravity

Substitute this relationship into this equation to determine the body’s center of mass

V

V

V

V

V

V

dV

dVz

zdV

dVy

ydV

dVx

x

~

;

~

;

~


Centroid Defines the geometric center of object Its location can be determined from

equations used to determine the body’s center of gravity or center of mass

If the material composing a body is uniform or homogenous, the density or specific weight will be constant throughout the body

The following formulas are independent of the body’s weight and depend on the body’s geometry


CentroidVolume 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

~

;

~

;

~


CentroidArea 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

~

;

~

;

~


CentroidLine If the geometry of the object such as a thin

rod or wire, 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

~

;

~

;

~


Line Choose a coordinate system that

simplifies as much as possible the equation used to describe the object’s boundary

Example Polar coordinates are appropriate for

area with circular boundaries


Symmetry The centroids of some shapes may be

partially or completely specified by using conditions of symmetry

In cases where the shape has an axis of symmetry, the centroid of the shape must lie along the line

Example Centroid C must lie along the

y axis since for every element length dL, it lies in the middle

Symmetry For total moment of all the elements

about the axis of symmetry will therefore be cancelled

In cases where a shape has 2 or 3 axes of symmetry, the centroid lies at the intersection of these axes


00~ xdLx


Procedure for AnalysisDifferential Element Select an appropriate coordinate system,

specify the coordinate axes, and choose an differential element for integration

For lines, the element dL is represented as a differential line segment

For areas, the element dA is generally a rectangular having a finite length and differential width


Procedure for AnalysisDifferential Element For volumes, the element dV is either a

circular disk having a finite radius and differential thickness, or a shell having a finite length and radius and a differential thickness

Locate the element at an arbitrary point (x, y, z) on the curve that defines the shape


Procedure for AnalysisSize and Moment Arms Express the length dL, area dA or

volume dV of the element in terms of the curve used to define the geometric shape

Determine the coordinates or moment arms for the centroid of the center of gravity of the element


Procedure for AnalysisIntegrations Substitute the formations and dL, dA and

dV into the appropriate equations and perform integrations

Express the function in the integrand and in terms of the same variable as the differential thickness of the element

The limits of integrals are defined from the two extreme locations of the element’s differential thickness so that entire area is covered during integration


Example 9.1Locate the centroid of the rod bent into the shape of a parabolic arc.


SolutionDifferential 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


SolutionIntegrations

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


Example 9.2Locate the centroid of the circular wire segment.


SolutionDifferential element A differential circular arc is selected This element intersects the curve at (R,

θ)Length and Moment Arms For differential length of the element

For centroid,

sin~cos~ RyRx

RddL



R

dR

dR

dR

dRR

dL

dLy

y

R

dR

dR

dR

dRR

dL

dLx

x

L

L

L

L

2

sinsin~

2

coscos~

2/

0

2/

02

2/

0

2/

0

2/

0

2/

02

2/

0

2/

0


Example 9.3Determine the distance from the x axis to the centroid of the area of the triangle


SolutionDifferential element Consider a rectangular element having

thickness dy which intersects the boundary at (x, y)

Length and Moment Arms For area of the element

Centroid is located y distance from the x axis

yy

dyyhhb

xdydA

~



32161

~

2

0

0

h

bh

bh

dyyhhb

dyyhhby

dA

dAy

yh

h

A

A


Example 9.4Locate the centroid

for the area of a quarter circle.


SolutionMethod 1Differential element Use polar coordinates for circular boundary Triangular element intersects at point (R,θ)Length and Moment Arms For area of the element


sin32~cos

32~

2)cos)((

21 2

RyRx

dR

RRdA



34

sin32

2

2sin

32~

34

cos32

2

2cos

32~

2/

0

2/

0

2/

0

2

2/

0

2

2/

0

2/

0

2/

0

2

2/

0

2

R

d

dR

dR

dR

R

dA

dAy

y

R

d

dR

dR

dR

R

dA

dAx

x

A

A

A

A


SolutionMethod 2Differential element Circular arc element

having thickness of dr Element intersects the

axes at point (r,0) and (r, π/2)


SolutionArea and Moment Arms For area of the element


/2~/2~

4/2

ryrx

drrdA



34

242

422~

34

242

422~

0

02

0

0

0

02

0

0

R

drr

drr

drr

drrr

dA

dAy

y

R

drr

drr

drr

drrr

dA

dAx

x

R

R

R

R

A

A

R

R

R

R

A

A


Example 9.5Locate the centroid of the area.


SolutionMethod 1Differential element Differential element of thickness dx Element intersects curve at point (x, y),

height yArea and Moment Arms For area of the element

For centroid2/~~ yyxx

ydxdA



m

dxx

dxxx

dxy

dxyy

dA

dAy

y

m

dxx

dxx

dxy

dxxy

dA

dAx

x

A

A

A

A

3.0333.0100.0

)2/()2/(~

75.0333.0250.0

~

1

02

1

022

1

0

1

0

1

02

1

03

1

0

1

0


SolutionMethod 2Differential element Differential element

of thickness dy Element intersects

curve at point (x, y) Length = (1 – x)




yy

xxxx

dyxdA

~2

12

1~

1



m

dyy

dyyy

dxx

dxxy

dA

dAy

y

m

dyy

dyy

dxx

dxxx

dA

dAx

x

A

A

A

A

3.0333.0100.0

11

1~

75.0333.0250.0

1

121

1

12/1~

1

0

1

02/3

1

0

1

0

1

0

1

01

0

1

0


Example 9.6Locate the centroid of the shaded are bounded by the two curves

y = x and y = x2.


SolutionMethod 1Differential element Differential element of thickness dx Intersects curve at point (x1, y1) and (x2, y2),

height yArea and Moment Arms For area of the element

For centroidxx

dxyydA

~

)( 12



m

dxxx

dxxxx

dxyy

dxyyx

dA

dAx

x

A

A

5.0

61

121

~

1

02

1

02

1

0 12

1

0 12


SolutionMethod 2Differential element Differential

element of thickness dy

Element intersects curve at point (x1, y1) and (x2, y2)

Length = (x1 – x2)




22~ 2121

2

21

xxxxxx

dyxxdA



m

dxyy

dyyy

dxyy

dyyyyy

dxxx

dxxxxx

dA

dAx

x

A

A

5.0

61

121

21

2/2/~

1

0

1

0

2

1

0

1

01

0 21

1

0 2121


Example 9.7Locate the centroid for the paraboloid of revolution, which is generated by revolving the shaded area about the y axis.


SolutionMethod 1Differential element Element in the shape of a thin disk, thickness

dy, radius z dA is always perpendicular to the axis of

revolution Intersects at point (0, y, z) Area and Moment Arms For volume of the element dyzdV 2


Solution For centroid

Integrations

mm

dyy

dyy

dyz

dyzy

dV

dVy

y

yy

V

V

7.66

100

100~

~

100

0

100

0

2

100

0

2

100

0

2


SolutionMethod 2Differential element Volume element in the form of

thin cylindrical shell, thickness of dz

dA is taken parallel to the axis of revolution

Element intersects the axes at point (0, y, z) and radius r = z




2/1002/100~

10022

yyyy

dzyzrdA



m

dzzz

dzzz

dzyz

dzyzy

dV

dVy

y

V

V

7.66

101002

1010

1002

10022/100~

100

0

22

100

0

444

100

0

100

0


Example 9.8Determine the location of the center of mass of the cylinder if its density varies directly with its distance from the base ρ = 200z kg/m3.


SolutionFor reasons of material symmetry

Differential element Disk element of radius 0.5m and thickness dz

since density is constant for given value of z Located along z axis at point (0, 0, z) Area and Moment Arms For volume of the element

dzdV

yx

25.0

0

View Free Body Diagram


Solution For centroid

Integrations

mzdz

dzz

dzz

dzzz

dV

dVz

z

zz

V

V

667.0

5.0)200(

5.0)200(~

~

~

1

0

1

02

1

0

2

1

0

2


Solution Not possible to use a

shell element for integration since the density of the material composing the shell would vary along the shell’s height and hence the location of the element cannot be specified

Documents

Engineering Mechanics: Statics Chapter 9: Center of Gravity and Centroid Chapter 9: Center of Gravity and Centroid