We were discussing “Method to determine the area moment of inertia for a hollow rectangular section”, “The
theorem of parallel axis about moment of inertia”, “Area
moment of inertia for the rectangular section about a line passing through the
base” in our previous posts.
Today we will see here the method to
determine the moment of inertia of a circular section with the help of this
post.
Let us consider one circular section,
where we can see that R is the radius and O is the centre of the circular
section as displayed in following figure.
Let us consider one small elementary circular strip of thickens dr and radius r as displayed in following figure.
R = Radius of the circular section
D = Diameter of the circular section
O = Center of the circular section
r = Radius of the small elementary
circular strip
dr = Thickness of the small elementary
circular strip
XX and YY are the axis which is passing
through the center O of the circular section.
ZZ is the axis which is passing through
the center O of the circular section and perpendicular to the plane of paper.
IXX = Moment of inertia
of circular section about XX axis
IYY = Moment of inertia
of circular section about YY axis
IXX = IYY (due
to concept of symmetry)
IZZ = Moment of inertia
of circular section about ZZ axis
Now we will determine the value or
expression for the moment of inertia of circular section about XX axis and also about YY axis
First of all
we will have to find out the moment of inertia of
circular section about ZZ axis and after that we will use the principle of
perpendicular axis i.e. the
perpendicular axis theorem and its proof in order to secure the moment of inertia of circular section about XX axis and also about YY axis.
Let us
determine the moment of inertia of small elementary
circular strip about an axis ZZ which is passing through the center O of the
circular section and perpendicular to the plane of paper.
Area of small elementary circular strip
= 2П * r * dr
Moment of
inertia of small elementary circular strip about
axis ZZ = 2П * r * dr * r2
Moment of
inertia of small elementary circular strip about
axis ZZ = 2П r3 dr
Moment of inertia of the entire circular
section about the axis ZZ will be determined by integrating the above equation
between limit 0 to R and it as displayed here in following figure.
Therefore, moment of inertia of circular
section about ZZ axis, IZZ = ПR4/2
Or we can also say that IZZ =
ПD4/32
Let us use and recall the principle of
perpendicular axis i.e. the
perpendicular axis theorem and its proof in order to secure the moment of inertia of circular section about XX axis and also about YY axis.
IZZ = IXX + IYY
As we have already discussed above that IXX =
IYY
Therefore, moment of inertia of circular
section about XX axis and moment of inertia of circular section about YY axis
could be easily concluded as mentioned here.
IXX = IYY = IZZ
/2
IXX = IYY = ПD4/64
IXX = ПD4/64
IYY = ПD4/64
Do you have any suggestions? Please
write in comment box
Reference:
Strength of material, By R. K. Bansal
Image Courtesy: Google
We will see another important topic i.e.
in the category of strength of material.
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