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Section Properties.pdf
Transcript of Section Properties.pdf
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Chapter 2 SECTION PROPERTIES
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Section Properties
Centroid
The centroid of an area is the point about which the area could be balanced if itwas supported from that point. The word is derived from the word center, and itcan be though of as the geometrical center of an area. For three-dimensionalbodies, the term center of gravity, or center of mass, is used to define a similarpoint.
Centroid of Simple Areas
For simple areas, the location of the centroid is easy to visualize.
B
HH/2
B/2
y
xH/3
H
B
xCC
C
D/2
D
CS/2
S/2
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Centroid of Complex Areas
Most complex shapes can be considered to be made up by combining several
simple shapes together.
If the area has an axis of symmetry, the controid will be on that axis. Somecomplex shapes have two axes of symmetry, and therefore the centroid is at theintersection of these two axes.
C C C
CC
C
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Where two axes of symmetry do not occur, the method of composite areas can beused to locate the centroid. For example, consider the following shape. It has avertical axis of symmetry but not a horizontal axis of symmetry. Such areas can
be considered to be a composite of two or more simple areas for which thecentroid can be found by applying the following principle:The product of the total area times the distance to the centroid of the total area isequal to the sum of the products of the area of each component part times thedistance to its centroid, with the distances measured from the same reference axis.
Moment of area
= ydA =(Aiyi)
This can be stated mathematical as
ATY =(Aiyi) Y =(Aiyi)/AT
Where AT =total area of the composite shapeY =distance to the centroid of the composite shape measured from some
reference axisAi=area of one component part of the shapeY i =distance to the centroid of the component part from the reference
axis.
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Example
Find the location of the centroid of following composite area.
Solution
=31.9 mm
T
2
1 ii
A
yAY
=
10605040
551060255040 xxxxY
+
+=
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Chapter 2 SECTION PROPERTIES
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Example
Find the location of centroid of the following T-section.
Solution
Y =(100*10*60+100*10*5) / (100*10+100*10)
=32.5 mm
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Chapter 2 SECTION PROPERTIES
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Chapter 2 SECTION PROPERTIES
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Moment of Inertia (2nd moment of area)
In the study of strength of materials, the property of moment of inertia is an
indication of the stiffness of a particular shape. That is, a shape having a highermoment of inertia would deflect less when subject to bending moments than onehaving a lower moment of inertia.Moment of inertia of simple shapes
Where y =distance from an element of area to the reference axis
A =Area of an element
Example
Determine the moment of inertia, I, of the following area with respect to itscentroidal axis.
dAyI =2
12
)*(
*
3
2/
2/
2
2
bhI
dybyI
dybdA
dAyI
h
h
=
=
=
=
Moment of Inertia, I
In static, it refers to Area moment of Inertia
In dynamic, it refers to Mass moment of Inertia
where I = y2dm
Moment of inertia of a rectangle about centroidal axis:
I =12bd3
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Chapter 2 SECTION PROPERTIES
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Moment of inertia of complex shapes
If the component parts of a composite area all have thesame centroidal axis, thetotal moment of inertia can be found by adding or subtracting the moments ofinertia of the components parts with respect to the centroidal axis.
Example
The cross-section of a beam shown in the following figure has its centroid at theintersection of its axes of symmetry. Compute the moment of inertia of thesection with respect to the horizontal axis x-x.
Ix =I1+I2 +I3I2 = 30*80
3/12 =1.28 x 106 mm4I1 =I3 =30*40
3/12 =0.16 x 106 mm4
Then,
Ix =I1+I2 +I3
=1.6 x 106 mm4
Ix =I1+I2 +I3
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Example
Compute the moment of inertia for the following section with respect to the axis
x-x.
Ix =I1- I2I1
= 504/12 =520.8 x 103 mm4
I2 =*354/64 =73.7 x 103 mm4
Then,
Ix =I1- I2 =447.1 x 103 mm4
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Polar Moment of Inertia (Perpendicular axis theorem)
The moment of inertia for an area relative to a line or axis perpendicular to theplane of the area is called the polar moment of inertia and is denoted by thesymbol J.
The moment of inertia of an area in the xy plane with respect to the z axis is
JZ = r2dA
= (x2
+ y2
)
dA
= x2dA + y2dA
Therefore, (Perpendicular axis theorem)
JZ = Ix + Iy
Expressed in words, this equation states that the polar moment of inertia for anarea with respect to an axis perpendicular to its plane is equal to the sum of themoments of inertia about any two mutually perpendicular axes in its plane that
interest on the polar axis.
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Chapter 2 SECTION PROPERTIES
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Parallel axis theorem
The theorem states that the second moment of an area with respect to any axis is
equal to the second moment of the area with respect to a parallel axis through thecentroid of the area added to the product of the area and the square of the distancebetween the two axes.
The second moment of the area in the following figure about the b axis is
I b = (y + d)2 dA
= y2
dA +2d y dA +d2
dA
=Ic + 2d y dA +d2 A
Where,
2d y dA =0
Therefore, Parallel axis theorem
I b = Ic +Ad2
Where Ic is the moment of inertia about centroidal axis
Id is the distance of Ib from Ic
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Chapter 2 SECTION PROPERTIES
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Example
Compute the moment of inertia for the following inverted T-section with respect
to its centroidal axis.
Solution
Y =(100*10*60+100*10*5)/(100*10+100*10)=32.5 mm
By using the Parallel Axis Theorem,
I =10*1003/12 +100*10*(60-32.5)2+100*103/12 +100*10*(32.5-5)2
=2.35 x 106
mm4
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Chapter 2 SECTION PROPERTIES
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Example
Compute the moment of inertia for the following section with respect to its
centroidal axis.
Solution
Method 1,
We divide the section into 3 small sections and apply the Parallel AxisTheorem,
I =15*3003/12 +2*[215*153/12 +215*15*(322.5-165)2]
=1.94 x 108
mm4
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Chapter 2 SECTION PROPERTIES
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Method 2,
Moment Inertia of the I-section,
I = Iarea 1 Iarea 2 Iarea 3
= 215*3303/12 100*3003/12 100*3003/12
=1.94 x 108 mm4
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Chapter 2 SECTION PROPERTIES
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Example
Compute the moment of inertia for the following section with respect to itscentroidal axis.
Solution
The x-x centroidal axis is shown. It is the axis of symmetry. Only one transferdistance is required (part 1).
Ixx for parts 2 and 4 =50*2003/12
=33.3 x 106 mm4
Ixx for part 3 =150*503/12
=1.56 x 106 mm4
Ixx for parts 1 and 5 =250*503/12 +250*50*(150-25)2
= 197.9 x 106 mm4
Total Ixx =2*33.3 x 106+1.56 x 106 +2*197.9 x 106
=464 x 106 mm4
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Chapter 2 SECTION PROPERTIES
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