MOMENT
OF INERTIA
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6.b
AREA
8.
(tvid - 6.b)
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Also notice that the moments of inertia vary depending on the location of the axis set.
Moment of inertia about x axis:
Moment of inertia about y axis::
Product moment of inertia with respect to xy axis set :
Moment of inertia about x’ axis :
Moment of inertia about y’ axis:
Product moment of inertia with respect to x’ y’ axis set :
8.1 Definitions of moment of inertia of the area in the figure:
8. AREA MOMENT OF INERTIA
x’
y’
x’
y’
Polar moment of inertia with respect to xy axis set :
Figure 8.1
(8.1)
(8.2)
(8.3)
(8.4)
(8.5)
(8.6)
(8.7)
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Find the moment of inertia of the rectangle given in the figure,
a-) with respect to the horizontal x-axis passing through its base, and
b-) with respect to the horizontal xG axis passing through its center of gravity.
Example 8.1:
8. AREA MOMENT OF INERTIA
Solution:
Figure 8.2
Figure 8.3
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Example 8.2:
Strip dA differential area
Solution:
8. AREA MOMENT OF INERTIA
Figure 8.4
Figure 8.5
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8.2 Parallel Axis (Steiner) Theorem:
While the moment of inertia is known about an axis passing through the center of gravity, the moment of inertia can be found about another axis parallel to this axis. As follows:
A: Area, d : It is the perpendicular distance between the x – xG axes
Attention:
There are 2 important conditions for the parallel axis (steiner) theorem to be applied:
1- The axes must be parallel to each other.
2- One axis must pass through the center of gravity.
Similarly;
Moment of inertia about y-axis:
Product moment of inertia with respect to xy axis set :
Polar moment of inertia with respect to xy axis set at point O:
8. AREA MOMENT OF INERTIA
Figure 8.6
(8.8)
(8.9)
(8.10)
(8.11)
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8.3 Moments of Inertia of Fundamental Areas :
quarter circle
Semi-circle
Circle
Note:
Ix and Iy are relative to the axes passing through the circle center (O).
Triangle
Square, rectangle
8. AREA MOMENT OF INERTIA
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Solution:
Example 8.3(2009 final exam)
8. AREA MOMENT OF INERTIA
coordinates of the center of gravity
Figure 8.7
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8. AREA MOMENT OF INERTIA
Solution:
First, let's find the coordinates of the center of gravity G according to the x-y axis set placed in the lower left corner:
Figure 8.8
Figure 8.9
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1
G1
G2
2
3
G3
G
8. AREA MOMENT OF INERTIA
Let's calculate the moments of inertia about the axes passing through the center of gravity:
Product moment of inertia:
0
0
0
Figure 8.10
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r=3m
3m
3m
3m
y
x
8. AREA MOMENT OF INERTIA
Figure 8.11
Question 8.1(*)
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Answers:
Question 8.2
8. AREA MOMENT OF INERTIA
Figure 8.12
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Calculate the coordinates of the center of gravity of the Areas Below and the moments of inertia with respect to the horizontal axis passing through the center of gravity.
(2018 final exam )
(2015- final exam)
8. AREA MOMENT OF INERTIA
Question 8.3
Question 8.4
Question 8.5
Question 8.6
r
Figure 8.13
Figure 8.14
Figure 8.15
Figure 8.16
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v
x
u
y
O
θ
From the definitions of moments of inertia about the x and y axes :
θ
(8.12)
8.4.1 Calculation of moments of inertia according to a different axis set using transformation Relations
8. AREA MOMENT OF INERTIA
Similarly, the moment of inertia about the u-axis is:
(From equations 8.1, 2 and 3)
Figure 8.17
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Similarly, the moment of inertia about the v axis is:
Product moment of inertia with respect to u-v axis set:
We have actually reached our goal with equations (8.12, 13 and 14).
(8.13)
(8.14)
8. AREA MOMENT OF INERTIA
v
x
u
y
O
θ
θ
Figure 8.18
Now we will express these equations in terms of 2θ…>>
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8. AREA MOMENT OF INERTIA
If we use these trigonometric equations in equations 8.12, 8.13, 8.14, we obtain the transformation equations in terms of 2θ as shown below.
(8.15)
(8.16)
(8.17)
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8. AREA MOMENT OF INERTIA
Let's move the first term on the left of equation 8.15 to the right and square both sides:
Let's take the squares of both sides of equation 8.17:
If we add and rearrange both sides of the above equations:
+
=1
We can compare this last equation to a circle equation. Like this:….>>
8.4.2 Mohr Circle for moments of inertia
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The final equation we obtain is:
Equation of a Circle with center (a;0):
So we can express this last equation with a circle
This circle is called Mohr's Circle for Moments of Inertia and is the geometric expression of the moments of inertia with respect to all axes passing through a point.
If the Mohr circle is drawn to scale, the desired moments of inertia can be determined by geometric measurements.
8. AREA MOMENT OF INERTIA
Figure 8.19
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C
O
3- Points M-N are connected. Point C on the x-axis of the MN line is determined. A circle with center C is drawn, passing through M and N.
v
x
u
y
θ
O
where we are
2θ
8.4.3 Drawing of Mohr's circle:
8.4.4
8. AREA MOMENT OF INERTIA
Figure 8.20
Figure 8.21
In reality, if we turn by an angle of θ from the +x axis, we arrive at the +u axis. However, starting from the point (M) where we are on the circle, we turn by an angle of 2θ in the same direction as in reality, and in this case, we arrive at point K on the circle.
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C
O
Where we are
x
y
O
Radius:
D
From the geometry of the circle the following values are found:
B
8.4.5 Principal Axes and Principal Moments of Inertia
(8. 18)
(8.19)
(8.20)
8. AREA MOMENT OF INERTIA
Figure 8.22
Figure 8.23
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8.4.6 ) Principal Axes According to Cross-Section Shapes
8. AREA MOMENT OF INERTIA
G
G
We can make the following determinations for sections where the axis set is placed at the center of gravity.
Figure 8.24
Figure 8.25
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b-)Circular sections: All axes passing through the center of gravity are principal axes. Because there is symmetry with respect to all axes.
8. AREA MOMENT OF INERTIA
G
y
x
G
are different from each other.
(For the strength lesson) Let's keep in mind now: If the direction of the resultant bending moment vector coincides with one of the principal axes, simple bending occurs with respect to the principal inertia axis set. This is a rule valid for all cross-sectional shapes.
Figure 8.26
Figure 8.27
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1
2
Example 8.5
8. AREA MOMENT OF INERTIA
Solution:
a-)
Drawing the Mohr Circle for the moments of inertia.
Figure 8.28
Figure 8.29
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8. AREA MOMENT OF INERTIA
Calculation of moments of inertia:
1
2
0
0
Figure 8.30
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C
M(Ix, Ixy)
N(Iy, -Ixy)
24.9
8.1
−8.1
8.7
8. AREA MOMENT OF INERTIA
b-) Principal axes of inertia and principal moments of inertia
From equation 8.19 :
From equation 8.20 :
O
1-
2-
(is the point we are on the circle.)
3- Points M-N are joined and a circle with center C is drawn through these points.
In these 3 steps we have drawn the Mohr circle.
Figure 8.31
Now let's draw the Mohr circle by following the 3 steps described in article 8.4.3:
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8. AREA MOMENT OF INERTIA
From eq. 8.15:
or from Mohr’s Circle
Minimum principal axis of inertia
u
Maximum principal axis of inertia
C
24.9
8.1
−8.1
8.7
O
B
M(Ix, Ixy)
Figure 8.32
Figure 8.33