chapter 7 and 8 centroids and inertia
TRANSCRIPT
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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