moment of inertia by gp capt nc chattopadhyay. what is moment of inertia? it is the moment required...
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MOMENT OF INERTIA
BY GP CAPT NC CHATTOPADHYAY
WHAT IS MOMENT OF INERTIA?
IT IS THE MOMENT REQUIRED BY A SOLID BODY TO OVERCOME IT’S RESISTANCE TO ROTATION
IT IS RESISTANCE OF BENDING MOMENT OF A BEAM
IT IS THE SECOND MOMENT OF MASS (mr2) OR SECOND MOMENT OF AREA (Ar2)
IT’S UNIT IS m4 OR kgm2
PERPENDICULAR AXIS THEOREM
The moment of inertia of a plane area about an axis normal to the plane is equal to the sum of the moments of inertia about any two mutually perpendicular axes lying in the plane and passing through the given axis. Moment of Inertia: Iz = Ix+Iy
PARALLEL (TRANSFER)AXIS THEOREM
THE MOMENT OF AREA OF AN OBJECT ABOUT ANY AXIS PARALLEL TO THE CENTROIDAL AXIS IS THE SUM OF MI ABOUT IT’S CENTRODAL AXIS AND THE PRODUCT OF AREA WITH THE SQUARE OF DISTANCE OF CG FROM THE REF AXIS
IXX= IG+Ad2
A is the cross-sectional area. : is the perpendicuar distance between the centroidal axis and the parallel axis.
Moment of Inertia - Parallel Axis Theorem
Parallel axis theorem:
Consider the moment of inertia Ix of an area A with respect to an axis AA’. Denote by y the distance from an element of area dA to AA’.
2xI y dA
Moment of Inertia - Parallel Axis Theorem
Consider an axis BB’ parallel to AA’ through the centroid C of the area, known as the centroidal axis. The equation of the moment inertia becomes
22x
2 22
I y dA y d dA
y dA y dA d dA
Moment of Inertia - Parallel Axis Theorem
The first integral is the moment of inertia about the centroid.
0
0
y A y dA y
y dA
2xI y dA
The second component is the first moment area about the centroid
Moment of Inertia - Parallel Axis Theorem
Modify the equation obtained with the parallel axis theorem.
2x 2I y dA y dA 2
2x
d dA
I d A
Example – Moment of Inertia
Compute the moment of inertia in the x about the AA` plane.
AA`
Example – Moment of Inertia
Compute the moment of inertia in the x about the AA` plane.
AA`
h b2 2
x
Area 0 0
h3 3
03 3
I y dA y dxdy
y bhb
Example – Moment of Inertia
From earlier lecture, the moment of inertia about the centroid
h/22 2
x
Area h/2
h/23 3
h/23 12
I y dA y bdy
y bhb
Example – Moment of Inertia
Using the parallel axis theorem
2x x
233
3
4
12 2 12
3
I I d A
bh hbh bh
bh
AA`
Parallel Axis - Why?
Recall that the method of finding centroids of composite bodies? -
Follow a Table technique
How would you be able to find the moment of inertia of the body. Use a similar technique, table method, to find the moment of inertia of the body.
Parallel Axis - Why?
Use a similar technique, table method, to find the moment of inertia of the body.
Bodies Ai yi yi*Ai Ii di=yi-ybar di2Ai
i i
i
y Ay
A
2x x
i
2
xi i i
I I d A
I y y A
Moment of Inertia
Use a set of standard tables:
Example - Moment of Inertia
Find the moment of inertia of the body, Ix and the radius of gyration, kx (rx)
I THINK…. I CAN THINK….
Example - Moment of Inertia
Set up the reference axis at AB and find the centroidBodies Ai yi yi*Ai Ii di=yi-ybar di
2Ai
1 18 1 182 18 5 90
36 108
3i i
2i
108 in3.0 in.
36 in
y Ay
A
Example - Moment of Inertia
From the table to find the moment of inertia
2
x xi i i
4 4 460 in 144 in 204 in
I I y y A
Bodies A i y i y i*A i Ii di=y i-ybar di2A i
1 18 1 18 6 -2 722 18 5 90 54 2 72
36 108 60 144
ybar 3 in.
I 204 in4
Example - Moment of Inertia
Compute the radius of gyration, rx.
4x
x 2
204 in
36 in2.38 in.
Ir
A
Example - Moment of Inertia
Find the moment of inertia of the body, Ix and the radius of gyration, kx (rx)
Example - Moment of Inertia
The components of the two bodies and subtract the center area from the total area.
Bodies Ai yi yi*Ai Ii di=yi-ybar di2Ai
1 21600 90 1944000 58320000 0 02 -9600 90 -864000 -11520000 0 0
12000 1080000 46800000 0
ybar 90 mm
I 46800000 mm4
Example - Moment of Inertia
Compute the radius of gyration, rx.
4x
x 2
46800000 mm
12000 mm62.45 mm
Ir
A
PROBLEM….
FIND THE MI OF A CIRCULAR LAMINA OF RADIUS “r“?