chapter 7 and 8 centroids and inertia

8/3/2019 CHAPTER 7 and 8 Centroids and Inertia

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STATICS

Chapter 7/8 The Centroid andCenters of Gravity


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Learning Objectives

Understand the concepts of the centroid,center of mass, and moment of inertia.

Determine the centroid of any geometry.

Compute the moments of inertia about axesthrough the centroid.

Transform the moment of inertia between twoparallel axes using the parallel axis theorem.


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Center of Gravity (COG)

Definition--The Point located at anobjects average position of the weight

In other words. The center of an

objects weightSymmetrical objects, like a baseball

the C of G would be in the exact centerof object

However other oddly shaped objects will findCOG in any number of positions, dependingon weight distribution

COG


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C.O.G.

When objects rotate freely they mustrotate about an axis through the COG

Basically treat the object as if all its weightis concentrated at that one pt.


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C.O.G. --BalancingFor an object to balance,

and not topple support

must be directly below

C.O.G.


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Where C.O.G. is located

Generally found in the

middle of all the weightDoes not even have to be

within, the object itselfEx. boomerang

Will be located towardone side of an objectwhere most of its mass is

focusedEx. Weebles

COG

gravity


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Balancing Stuff

Again, all that has to happen to

balance, is for a support to be directlybeneath COG


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Introduction

The earth exerts a gravitational force on each of the particles

forming a body. These forces can be replaced 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 thefirst moment of an

area is used to locate the centroid.


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The center of mass of a system of masses is the point where thesystem can be balanced in a uniform gravitational field


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Center of Gravity of a 2D Body

Center of gravity of a plate

W

Wyy

WyWyM

WWxx

WxWxM

aa

y

aa

y


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mrI2 mass moment of inertia


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The Centroid

The centroid is a point that locates the geometric centerof an object.

The position of the centroid depends only on the objects

geometry (or its physical shape) and is independent ofdensity, mass, weight, and other such properties.

The average position along different coordinate axes

locates the centroid of an arbitrary object.


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The Centroid of an Area

We can divide the object into anumber of very small finite elementsA1, A2, An.

In this particular case, each smallsquare grid represents one finite area.

Let the coordinates of these areas be(x1, y1), (x2, y2), , (xn, yn).

The coordinates x1 and y1 extend tothe center of the finite area.

Now, the centroid is given by

i i

i

i

i

x A

x =A

i i

i

i

i

y A

y =A


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The Centroid

The calculations will result in thelocation of centroid C.

Because point C is at the center ofthe rectangle, the results intuitivelymake sense.

Consider the moment due to thefinite areas (instead of the forces)about two lines (AA and BB)parallel to the x- and y-axespassing through the centroid.

Because the rectangle issymmetric about these two lines,the net moment will be zero.


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The Centroid

Centroid always lieson the line ofsymmetry.

For a doublysymmetric section(where there are two

lines of symmetry),the centroid lies at theintersection of thelines of symmetry.


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Functional Symmetry

The area is symmetricabout line BB, its

centroid must lie onthis line.

The area is notsymmetric about lineAA.


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Functional Symmetry

The four holes are equidistantfrom line AA, and the momentsfrom the two holes on the topof line AA counteract that ofthe two bottom holes.

Even though the area is notphysically symmetric about lineAA, functionally line AA can beviewed as the line ofsymmetry.

Therefore, the centroid lies onthe intersection of the twolines.


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The Centroid

The calculation of the centroid for a compositesection requires the following three steps:Divide the composite geometry into simple geometries

for which the positions of the centroid are known or canbe determined easily.

Determine the centroid and area of individualcomponents.

Apply the equation to determine the centroid location.


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Example 1

Determine thecentroid of thecomposite section.


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Example 1

Step 1: Divide thecomposite section intosimple geometries

The composite geometry

can be divided into threeparts:

two positive areas

one negative area(circular cutout).


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Example 1

Step II: Determine the centroid and the area ofindividual component

Part Dimensions Area (sq. in) x y

Area 1 24 8 3 5

Area 2 10

6 60 9 5

Area 3 2 radius 4 10 5


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Example 1

Step III: Determine the centroid location

i ix A i iy A

iA

55.434

i ix A438.34

i iy A277.17

Part Dimensions Area (sq.in)

x y (in3) (in3)

Area 1 24 8 3 5 24 40

Area 2 106 60 9 5 540 300

Area 3 2 radius 4 10 5 40 20


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Example 1

i i

i

i

i

x A

x =

A

i i

i

i

i

y A

y =

A

438.34

x = 7.91in55.434

277.17

y = 5.00in55.43


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Centroids of CommonShapes of Areas


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Locate the centroid of the composite area. Seta= 150 , b= 450 , d = 70 , h = 210 .

xbar = a = 150 mm

Triangle (area 1)

ybar = 2/3 h =2/3(210)=140730-140= 590 mm

Hollow Tube (area 2)

by symmetry

ybar = (d+b) = ( 70 + 450)=260

xbar = a = 150

Hole (area 3)

xbar = a =15=ybar = b/2 = 450/2 = 225

PART AREA XBAR YBAR XA YA


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AREA 1 31,500 150 590 4725000 18585000

AREA 2 156,000 150 260 23400000 40560000

AREA 3 60,800 150 225 -9120000 -13680000

SUM 126,700 19,005,000 45,465,000

XBAR 150

YBAR 358.839779


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a1

a2 a3

a4


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MOMENTS OF INERTIA

-IS AN ABTRACT, MATHEMATICAL CONCEPT, IS NOT SOMETHINGYOU CAN SEE OR TOUCH

-IS ABSOLUTELY CRUCIAL TO UNDERSTANDING STRENGTHIN MATERIALS

-THE MOMENT OF INERTIA OF A COMPONENTS CROSS SECTION HASA DIRECT RELATIONSHIP TO ITS STRENGTH.

HOW WELL IT RESISTS:BENDINGBUCKLINGTORSION


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MOMENT OF INERTIA

-MOMENTS OF INERTIA ARE ALWAYSCOMPUTED RELATIVE TO A SPECIFICAXIS

THERE WILL BE 2 DIFFERENTRECTANGULAR MOMENTS OF INERTIA

THE MOMENT RELATIVE TO AXISX-X, CALLED IX

THE MOMENT RELATIVE TO AXISY-Y, CALLED IY 2ayI

X

2

axIY IS SOMETIMES CALLED THE SECOND MOMENT OF AREA


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LETS SAY BEAM 1 = 3 X 3BEAM 2 = 4 X 4 OD

3.5 X 3.5 IDAREA BEAM 1= 9 IN2

AREA BEAM 2 = 4 X 4 3.5 X 3.5=3.75 IN2

4

33

75.6

12

)3(3

12

inbh

I 82.812

150256

12

3

11

3

dbbd

I


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1in diameter 2 in OD1.625 in ID

A = 3.14 in2A=1.067 in2

2

44

049.64

)1(14.3

64in

dI

0667.64

)( 441

ddI


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TRANSFER FORMULA

-USED TO FIND THE MOMENT OF INERTIA OF AN AREA ABOUT

A NONCENTROIDAL AXIS, PARALLEL TO THE CENTROIDAL AXIS

I = IO + ad2

COMPOSITE SECTIONS


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COMPOSITE SECTIONS


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DISTANCE FROM REFERENCE AXISTO CENTROIDAL AXIS


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R di f G i f A


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Radius of Gyration of an Area

Consider areaA with moment of inertia

Ix. Imagine that the area is

concentrated in a thin strip parallel to

thex axis with equivalentIx.

A

IrArI x

xxx 2

rx = radius of gyration with respectto thex axis

similarly

larger r = greater resistance to buckling


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The steel columns shown below all have areas of 3-1/8 in2. The safe loads for an 8 ft length are

shown. The only difference between them is the way in which the cross-sectional area isdistributed about the centroid.


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Polar Moment of Inertia

Thepolar moment of inertia is an important

parameter in problems involving torsion ofcylindrical shafts and rotations of slabs.

dArJ 20

The polar moment of inertia is related to the

rectangular moments of inertia,

xy

IIJ

dAydAxdAyxdArJ

0

22222

0


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Calculate RectangularMoments of inertia

Transfer Inertia tocomposite centroid


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Radius of Gyration

Polar Inertia


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Example 3

Locate the centroid ofthe line whoseequation is

with x ranging from 0to 1

2y = 1 - x


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Example 3

2

2 2 yL x + y 1 x

x

dd d d d

d

y=-2x

x

d

d

2

L 1 4 xd x d


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Example 3

L

L

x L

x =L

d

d

L

L

y L

y =L

d

d

12

0

1

2

0

x 1 4x x

x =

1 4x x

d

d

1

2 2

0

1

2

0

1 x 1 4x x

y =

1 4x x

d

d

x = 0.8667 y = 0.2861


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The Center of Mass

The center of mass is a point that locates theaverage position of the mass of an object.

For an object with uniform density, it coincideswith the centroid.

It is often called the center of gravity because

the gravitational pull on an object can berepresented as a concentrated force acting atthis point.


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The Center of Mass

The equation for finding the center of mass of a volume takes theform of

For a three-dimensional surface of uniform thickness and density,the center of mass coincides with the centroid of the surface.

The same concepts can be used to determine the center of mass of

a line. The equation takes the form of

m

m

x m

x =m

d

d

m

m

y m

y =m

d

d

m

m

m

z =m

zd

d

A

A

x A

x =A

d

d

A

A

y A

y =A

d

d

A

A

z A

z =A

d

d

L

L

x L

x =L

d

d

L

L

y L

y =L

d

d

L

L

L

z =L

zd

d


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The Moment of Inertia

2

A

y AxI d

2

A

x AyI d

2

A

r AOJ d

Moment of inertia

about x-axis:

Moment of inertia

about y-axis:

Polar moment ofinertia:

Product of inertia:A

x y AxyI d

Ag

IR

The moment of inertia is sometimes expressed in terms of theradius of gyration. The radius of gyration determines how thearea is distributed around the centroid.


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Example 4

Determine themoments of inertiaabout the x- and y-

axes. Also, determinethe polar moment ofinertia.


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Example 4

d d

2 22 2 2

d dA

2 2

y A y b. y b y yxI d d d

3bd

12xI

3db

12yI

O x yJ I I 2 2bd

b +d12

OJ


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Parallel Axis Theorem

2

A

2

y

A

2 2

y y

A A A

2y y

A A

y A

y +d A

y A 2 y d A d A

2d y A d A

x

x

x

I d

I d

d d d

I d d

2yAdx xI I

2xAdy yI I

In the second term, is equal to zero as the x-axis

passes through the centroid.

2AdO CJ J


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Example 5

Determine themoments of inertiaabout the x- and y-

axes about thecentroid.

Also, determine the

polar moment ofinertia.


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Example 5


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Example 5

PartDimen

-sions

Area

(sq.

in)

x y (in3) (in3) Ix Iy dx dy

Area

124 8 3 5 24 40 10.67 2.67 4.91 0 192.86 0

Area

2106 60 9 5 540 300 180 500 1.09 0 71.286 0

Area

3

2

radius 4 10 5 40 20 -0.785 -0.785 2.09 0 -54.89 0

Summation 55.43 438.34 189.89 501.89 209.26

i ix A i iy A

2

xAd

2yAd

277.17


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Example 5

2yAdx xI I

2xAdy yI I

4189.89 inxI 4711.15 inxI

O x yJ I I

4

901.04 inOJ

chapter 7 and 8 centroids and inertia

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