Moments of Inertia
ความหมายของโมเมนต์ของพื้นที่
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ตัดคาน
โมเมนต์ที่สองของพื้นที่
• 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
M   z 2 dA
• Integral represent the moment of inertia of area
about the y axis
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
•
dI x  y 2 dA
dI y  x 2 dA
For entire area, moments of
inertia are given by
I x   y 2 dA
A
I y   x 2 dA
A
Moment of Inertia
• Formulate the second moment of dA about the
pole O or z axis
• This is known as the polar axis
•
dJO  r 2 dA
where r is perpendicular from the pole (z axis)
to the element dA
Polar moment of inertia for entire area,
J O   r 2 dA  I x  I y
A
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
dI x   y ' d y  dA
2
• For entire area
I x    y ' d y  dA
2
A
  y '2 dA  2d y  y ' dA  d y2  dA
A
A
A
• First integral represent the moment of inertia of the area
about the centroidal axis
• Second integral = 0 since x’ passes through the area’s
centroid C
 y ' dA  y  dA  0;
y0
• Third integral represents the total area A
• Similarly
I x  I x  Ad y2
I y  I y  Ad x2
• For polar moment of inertia about an axis perpendicular
to the x-y plane and passing through pole O (z axis)
J O  J C  Ad 2
Moments of Inertia for an Area by
Integration
Example 10.1
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.
Solution
Part (a)
• Differential element chosen, distance y’ from x’
axis
• Since dA = b dy’
I x   y ' dA  
2
A
1
3

bh
12
h/2
h / 2
y '2 dy
Solution
Part (b)
• Moment of inertia about an axis passing through
the base of the rectangle obtained by applying
parallel axis theorem
I xb  I x  Ad 2
2
1
1 3
h
3

bh  bh   bh
12
3
2
Solution
Part (c)
• For polar moment of inertia about point C
1
3
I y' 
hb
12
JC  I x  I y'
1
2
2

bh( h  b )
12
โมเมนต์ อนิ เนอร์ เชียของพืน้ ที่ประกอบ
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
I x  I x '  Ad y2
 
1
4
2
2
  25   25 75  11.4 10 6 mm4
4
• Rectangle
I x  I x '  Ad y2
 
1
1001503  100150752  112.5 106 mm4

12
Solution
Summation
• For moment of inertia for the composite area,
 
 
I x  11.4 10  112.5 10
6
 
 101 10 mm
6
4
6
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
I x  I x '  Ad y2
 
1
3
2
100300  100300200  1.425 109 mm4

12
I y  I y '  Ad x2
 
1
3
2
9
4








300 100  100 300 250  1.90 10 mm
12
Solution
Parallel Axis Theorem
• Rectangle B
I x  I x '  Ad y2
 
1
3
600100  0.05 109 mm4

12
I y  I y '  Ad x2
 
1
3
100600  1.80 109 mm4

12
Solution
Parallel Axis Theorem
• Rectangle D
I x  I x '  Ad y2
 
1
3
2
100300  100300200  1.425 109 mm4

12
I y  I y '  Ad x2
 
1
3001003  1003002502  1.90 109 mm4

12
Solution
Summation
• For moment of inertia for the entire crosssectional area,
 
 
 
 2.9010 mm
I  1.9010   1.8010   1.9010 
 5.6010 mm
I x  1.425 10  0.05 10  1.425 10
9
9
9
4
9
9
y
9
4
9
9