We were discussing method to determine the area moment of inertia for a hollow rectangular section, area moment of inertia for the rectangular section about a line passing through the base  and basic concepts of mass moment of inertia in our previous posts.

Today we will see here the method to determine the mass 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 width dr and radius r as displayed in following figure.

R = Radius of the circular section
D = Diameter of the circular section
T = Uniform thickness of the circular section
O = Centre of the circular section

A= Area of the entire circular section (Π.R2)
V= Volume of the entire circular section
V= A .t = Π.R2 t
M= Mass of the entire circular section
M= Density x Volume of the entire circular section
M= ρ Π.R2

(Im) XX = Mass moment of inertia of the circular section about X-axis
(Im) YY = Mass moment of inertia of the circular section about Y-axis
(Im) ZZ = Mass moment of inertia of circular section about ZZ axis
(Im) XX = (Im) YY (due to concept of symmetry)

r = Radius of the small elementary circular strip
dr = Width of the small elementary circular strip

XX and YY are the axis which is passing through the centre O of the circular section.

ZZ is the axis which is passing through the centre O of the circular section and perpendicular to the plane of paper.

### Now we will determine the value or expression for the mass moment of inertia of circular section about X-axis and also about Y-axis

First of all we will have to find out the mass 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 mass moment of inertia of circular section about X- axis and also about Y-axis.

Let us determine the mass moment of inertia of small elementary circular strip about an axis ZZ which is passing through the centre O of the circular section and perpendicular to the plane of paper.

Area of small elementary circular strip, dA = 2П * r * dr

Volume of the elementary circular strip, dV= 2П * r * dr * t
Volume of the elementary circular strip, dV= 2П r dr t

Mass of the elementary circular strip, dm = ρ 2П r dr t
Mass of the elementary circular strip, dm = ρ 2П t. r dr

Mass moment of inertia of small elementary circular strip about Z-axis = ρ 2П t. r dr * r2
Mass moment of inertia of small elementary circular strip about Z-axis = ρt 2П r3 dr

Mass 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 will be displayed here in following figure.

Therefore, mass moment of inertia of circular section about ZZ axis, (Im) ZZ = Πρt. R4/2
Or we can also say that IZZ = ρΠ R2t x R2/2
IZZ = M R2/2

Let us use and recall the principle of perpendicular axis i.e. the perpendicular axis theorem and its proof in order to secure the mass moment of inertia of entire circular section about XX axis and also about YY axis.

(Im)ZZ = (Im) XX + (Im) YY

As we have already mentioned above that (Im) XX = (Im) YY

(Im) XX = (1/2) (Im) ZZ
(Im) XX = M R2/4

Therefore, mass moment of inertia of circular section about X-axis and mass moment of inertia of circular section about Y-axis could be easily concluded as mentioned here.

### (Im) YY = M R2/4

Do you have any suggestions? Please write in comment box

### Reference:

Strength of material, By R. K. Bansal