Moments of Inertia
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Transcript of Moments of Inertia
Moments of Inertia
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โมเมนต์�ที่��สองของพื้��นที่�� • Stress within the beam varies linearly with the
distance from an axis passing through the centroid C of the beam’s cross-sectional area
σ = kz• For magnitude of the force acting
on the area element dA
dF = σ dA = kz dA
Definition of Moments of Inertia for Areas• Since this force is located a distance z from the y
axis, the moment of dF about the y axis
dM = dF = kz2 dA• Resulting moment of the entire stress distribution
= applied moment M
• Integral represent the moment of inertia of area about the y axis
dAzM 2
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
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,
yxAO
O
IIdArJ
dArdJ
2
2
Parallel Axis Theorem for an Area• 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
• 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'
'
'
• 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
Moments of Inertia for an Area by Integration
Example 10.1Determine 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.
Solution
Part (a)• Differential element chosen, distance y’ from x’
axis• Since dA = b dy’
3
2/
2/
22
12
1
''
bh
dyydAyIh
hAx
Solution
Part (b)• Moment of inertia about an axis passing through
the base of the rectangle obtained by applying parallel axis theorem
32
3
2
3
1
212
1bh
hbhbh
AdII xxb
Solution
Part (c)• For polar moment of inertia about point C
)(12
1
12
1
22
'
3'
bhbh
IIJ
hbI
yxC
y
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Example 10.5
Compute the moment of
inertia of the composite
area about the x axis.
Solution
Composite Parts• Composite area obtained
by subtracting the circle form the rectangle
• Centroid of each area is located in the figure
Solution
Parallel Axis Theorem• Circle
• Rectangle
4623
2'
46224
2'
105.1127515010015010012
1
104.117525254
1
mm
AdII
mm
AdII
yxx
yxx
Solution
Summation• For moment of inertia for the composite area,
46
66
10101
105.112104.11
mm
I x
Example 10.6
Determine the moments
of inertia of the beam’s
cross-sectional area
about the x and y
centroidal axes.
Solution
Composite Parts
• Considered as 3 composite areas A, B, and D
• Centroid of each area is located in the figure
Solution
Parallel Axis Theorem• Rectangle A
4923
2'
4923
2'
1090.125030010010030012
1
10425.120030010030010012
1
mm
AdII
mm
AdII
xyy
yxx
Solution
Parallel Axis Theorem• Rectangle B
493
2'
493
2'
1080.160010012
1
1005.010060012
1
mm
AdII
mm
AdII
xyy
yxx
Solution
Parallel Axis Theorem• Rectangle D
4923
2'
4923
2'
1090.125030010010030012
1
10425.120030010030010012
1
mm
AdII
mm
AdII
xyy
yxx
Solution
Summation• For moment of inertia for the entire cross-
sectional area,
49
999
49
999
1060.5
1090.11080.11090.1
1090.2
10425.11005.010425.1
mm
I
mm
I
y
x