Copyright © 2011 Pearson Education South Asia Pte Ltd Chapter Objectives To discuss the concept of the center of gravity, center of mass, and the centroid

Copyright © 2011 Pearson Education South Asia Pte Ltd

Chapter 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 present a method for finding the resultant of a general distributed loading.

To show how to determine the moment of inertia of an area.


In-Class Activities

1. Reading Quiz

2. Center of Gravity, Center of Mass, and the Centroid of a Body

3. Composite Bodies

4. Resultant of a Distributed Loading

5. Moments of Inertia for Areas

6. Parallel-Axis Theorem for an Area

7. Moments of Inertia for Composite Areas

8. Concept Quiz


READING QUIZ

1) Select the SI units for the Moment of Inertia for an area.

a) m3

b) m4

c) kg·m2

d) kg·m3


READING QUIZ (cont)

2) 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


READING QUIZ (cont)

3) 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


READING QUIZ (cont)

4) 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.


READING QUIZ (cont)

5) 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


READING QUIZ (cont)

6) 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


READING QUIZ (cont)

7) The definition of the Moment of Inertia for an area involves an integral of the form

a) x dA.

b) x2 dA.

c) x2 dm.

d) m dA.


READING QUIZ (cont)

8) The parallel-axis theorem for an area is applied between:

a) An axis passing through its centroid and any corresponding parallel axis.

b) Any two parallel axes.

c) Two horizontal axes only.

d) Two vertical axes only.


READING QUIZ (cont)

9) The Moment of Inertia of a composite area equals the ________ of the MoI of all of its parts.

a) Vector sum

b) Algebraic sum (addition or subtraction)

c) Addition

d) Product


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, CENTER OF MASS, AND THE CENTROID OF A BODY


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, CENTER OF MASS, AND THE CENTROID OF A BODY (cont)


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



Centroid of 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

~,

~;

~



Centroid of 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 of 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

~

;

~

;

~



• 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

~~~

COMPOSITE BODIES


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

COMPOSITE BODIES (cont)



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

COMPOSITE BODIES (cont)


EXAMPLE 1


Locate the centroid of the plate area.


EXAMPLE 1 (cont)


Composite Parts• Plate divided into 3 segments.• Area of small rectangle considered “negative”.

Solution


EXAMPLE 1 (cont)


SolutionMoment 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~


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](dA m2)

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

infinite parallel forces acting on separate differential area dA

RESULTANT OF A DISTRIBUTED LOADING


Pressure Distribution over a Surface• This system will be simplified to a single resultant force FR acting through a unique point on the plate

RESULTANT OF A DISTRIBUTED LOADING (cont)


Magnitude of Resultant Force

• To determine magnitude of FR, sum up 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

),(

),(

),(

),(

RESULTANT OF A DISTRIBUTED LOADING (cont)


• Centroid for an area is determined by the first moment of an area about an axis

• Second moment of an area is referred as the moment of inertia

• Moment of inertia of an area originates whenever one relates the normal stress σ or force per unit area

MOMENTS OF INERTIA FOR AREASPlease refer to the website for the animation: Moment of Inertia


Moment of Inertia • Consider area A lying in the x-y plane• Be definition, moments of inertia of the differential

plane area dA about the x and y axes

• For entire area, moments of

inertia are given by

Ay

Ax

yx

dAxI

dAyI

dAxdIdAydI

2

2

22

MOMENTS OF INERTIA FOR AREAS (cont)


Moment of Inertia • Formulate the second moment of dA about the pole O

or z axis• This is known as the polar axis

where r is perpendicular from the pole (z axis) to the element dA

• Polar moment of inertia for entire area,

dArdJO2

yxAO IIdArJ 2

MOMENTS OF INERTIA FOR AREAS (cont)


• For moment of inertia of an area known about an axis passing through its centroid, determine the moment of inertia of area about a corresponding parallel axis using the parallel axis theorem

• Consider moment of inertia of the shaded area

• A differential element dA is located at an arbitrary distance y’ from the centroidal x’ axis

PARALLEL AXIS THEOREM FOR AN AREA


• The fixed distance between the parallel x and x’ axes is defined as dy

• For moment of inertia of dA about x axis

• For entire area

• First integral represent the moment of inertia of the area about the centroidal axis

)

)

AyAyA

A yx

yx

dAddAyddAy

dAdyI

dAdydI

22

2

2

'2'

'

'

PARALLEL AXIS THEOREM FOR AN AREA (cont)


• Second integral = 0 since x’ passes through the area’s centroid C

• Third integral represents the total area A

• Similarly

• For polar moment of inertia about an axis perpendicular to the x-y plane and passing through pole O (z axis)

2

2

2

0;0'

AdJJ

AdII

AdII

ydAydAy

CO

xyy

yxx

PARALLEL AXIS THEOREM FOR AN AREA (cont)


EXAMPLE 2


Determine the moment of inertia for the rectangular area with respect to (a) the centroidal x’ axis, (b) the axis xb passing through the base of the rectangular, and (c) the pole or z’ axis perpendicular to the x’-y’ plane and passing through the centroid C.


EXAMPLE 2 (cont)


Part (a)• Differential element chosen, distance y’ from x’ axis.• Since dA = b dy’,

Part (b)• By applying parallel axis theorem,

32/

2/

22/

2/

22

12

1')'('' bhdyybdyydAyI

h

h

h

hAx

32

32

3

1

212

1bh

hbhbhAdII xxb

Solution


EXAMPLE 2 (cont)


Solution

Part (c)• For polar moment of inertia about point C,

)(12

112

1

22'

3'

bhbhIIJ

hbI

yxC

y


• Composite area consist of a series of connected simpler parts or shapes

• Moment of inertia of the composite area = algebraic sum of the moments of inertia of all its parts

Procedure for AnalysisComposite Parts• Divide area into its composite parts and indicate the

centroid of each part to the reference axisParallel Axis Theorem• Moment of inertia of each part is determined about its

centroidal axis

MOMENTS OF INERTIA FOR COMPOSITE AREAS



Parallel Axis Theorem• When centroidal axis does not coincide with the

reference axis, the parallel axis theorem is used

Summation• Moment of inertia of the entire area about the

reference axis is determined by summing the results of its composite parts

MOMENTS OF INERTIA FOR COMPOSITE AREAS (cont)


EXAMPLE 3


Compute the moment of inertia of the composite area about the x axis.


EXAMPLE 3 (cont)


Composite Parts• Composite area obtained by subtracting the circle form the rectangle.• Centroid of each area is located in the figure below.

Solution


EXAMPLE 3 (cont)


Solution

Parallel Axis Theorem

Circle

Rectangle

) ) ) ) 46224

2'

104.117525254

1mm

AdII yxx

) ) ) ) ) ) 4623

2'

105.1127515010015010012

1mm

AdII yxx


EXAMPLE 3 (cont)


Solution

Summation

For moment of inertia for the composite area,

) ) ) 46

66

10101

105.112104.11

mm

I x


CONCEPT QUIZ

1) A pipe is subjected to a bending moment as shown. Which property of the pipe will result in lower stress (assuming a constant cross-sectional area)?

a) Smaller Ix

b) Smaller Iy

c) Larger Ix

d) Larger Iy

My

M

Pipe section

x


CONCEPT QUIZ (cont)

2) In the figure to the right, what is the differential moment of inertia of the element with respect to the y-axis (dIy)?

a) x2 ydx

b) (1/12)x3dy

c) y2 x dy

d) (1/3)ydy x

y=x3

x,y

y


CONCEPT QUIZ (cont)

3) For the area A, we know the centroid’s (C) location, area, distances between the four parallel axes, and the MoI about axis 1. We can determine the MoI about axis 2 by applying the parallel axis theorem ___ .

a) Directly between the axes 1 and 2.

b) Between axes 1 and 3 and then between the axes 3 and 2.

c) Between axes 1 and 4 and then axes 4 and 2.

d) None of the above.

d3

d2

d1

432

1

A

•C

Axis


CONCEPT QUIZ (cont)

4) For the same case, consider the MoI about each of the four axes. About which axis will the MoI be the smallest number?

a) Axis 1

b) Axis 2

c) Axis 3

d) Axis 4

e) Can not tell.

d3

d2

d1

4

3

2

1

A

•C

Axis


CONCEPT QUIZ (cont)

5) For the given area, the Moment of Inertia about axis 1 is 200 cm4 . What is the MoI about axis 3?

a) 90 cm4

b) 110 cm4

c) 60 cm4

d) 40 cm4

A=10 cm2

•C

d2

d1

3

2

1

•C

d1 = d2 = 2 cm


CONCEPT QUIZ (cont)

6) The Moment of Inertia of the rectangle about the x-axis equals:

a) 8 cm4.

b) 56 cm4 .

c) 24 cm4 .

d) 26 cm4 .

2cm

2cm

3cm

x


CONCEPT QUIZ (cont)

7) 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.


CONCEPT QUIZ (cont)

8) 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)

~ ~


CONCEPT QUIZ (cont)

9) 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°


CONCEPT QUIZ (cont)

10) 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

3cm1 cm

1 cm

3cm


CONCEPT QUIZ (cont)

11) 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°30º

G

C

AB


CONCEPT QUIZ (cont)

12) For determining the centroid, what is the minimum number of pieces you can use?

a) Two

b) Three

c) Four

d) Five

2cm 2cm

2cm

4cm

x

y


CONCEPT QUIZ (cont)

13) 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

A1m

1m

y

E

FG

CB x

1m 1m

D

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Copyright © 2011 Pearson Education South Asia Pte Ltd Chapter Objectives To discuss the concept of the center of gravity, center of mass, and the centroid