MOMENT OF INERTIA &

POLAR MOMENT OF INERTIA

CLASS NOTES

Moment of inertia is for a cross section.

Larger value reduces bending. Polar

moment of inertia is about the centroid

axis. Larger value reduces torsional twist.

Definition of Moment of Inertia

It is a property of a cross-sectional area to resist bending. Larger is the value of moment of inertia, lesser will be the bending. It is also the second moment of the area of cross-section.

             Symbol

Its symbol is ‘I’.

               Units

Its unit is mm⁴.

Which cross section is best for a beam?

 In case of bending, choose the section for which moment of inertia is largest. For a certain value of the cross section area (say 20 cm²), ‘I’ shape cross section will have the highest moment of inertia as compared to rectangular, square, circular or any other shape of the cross section. Hence I section cross section is best for a beam.

 Equation to find moment of inertia

Mathematically( for a small area dA and its centroid is at a distance of y from xx-axis), then

I_xx = ∫(dA y)y = ∫dA y²

dA y = first moment of the area about x axis

(dA y )y = second moment of the area about x axis

                              Similarly about yy axis,

I_yy = (dA x)x = dA x²

dA x = first moment of the area about y axis

(dA x )x = second moment of the area about y axis

Moment of inertia for standard shapes

(a) For a rectangular cross section, width b and depth d

When x axis is parallel to width b

I_xx = (1/12) bd³

When y axis is parallel to depth d

Iyy = (1/12) db³

(b)

For a square cross section, each side is ‘a’

I_xx =I_yy =  (1/12) a⁴

(c) For a circular solid cross section like a rod

I_xx =I_yy =  (π/64) d⁴

(d) For a circular hollow cross section like a pipe

I_xx =I_yy =  (π/64)[ d_o⁴ –d_i⁴ ]

POLAR MOMENT OF INERTIA, J

Definition

Rotating shafts use it. Thus polar moment of inertia is the resistance against torsional shear. Larger the polar moment of inertia, smaller are torsional shear stress and angle of twist. Thus polar moment of inertia will be applicable to rotating members. The axis of rotation will appear as a point in the cross section of the rotating member. This point in the cross section is ‘Pole’. The moment of inertia about this pole is polar moment of inertia.

Symbol

Its symbol is ‘J’.

Equation for polar moment of inertia

I_zz = ∫(dA r)r =∫dA  r² = ∫dA(x² + y²)= I_xx + I_yy

dA r = first moment of the area about zz axis

(dA r )r = second moment of the area about zz axis

I_zz is polar moment of inertia

The zz axis appears as a point in the cross section. Therefore I_zz is the moment of inertia of the cross section about a point in the cross section. It will be application to the Torsion of shafts since shafts rotate about the axis perpendicular to the cross section.

I_zz= I_xx+ I_yy

^I_zz = J

J = I_zZ

TABLE: Moment & Polar moment of Inertia

Sr. No.	Shape of cross section	Moment Of Inertia	Polar moment of Inertia
1.	Rectangular	Ixx =(1/12) bd3 Here b is parallel to x-axis and d is to the x-axis, Iyy =(1/12) db3 Here d is parallel to y-axis and b is to the y-axis	Not used
2.	Square	Ixx =(1/12) a4= Iyy Iyy =(1/12) a4	Not used
3.	Solid Circular	Ixx=(π/64) d4 = Iyy	J =(π/32) d4 J = Ixx + Iyy
4.	Hollow circular	Firstly Ixx=(π/64)(D4– d4) Secondly Iyy=(π/64)(D4– d4) Ixx = Iyy D is outer diameter d is inner diameter	J = (π/32)(D4– d4) J = Ixx + Iyy

PARALLEL AXIS THEOREM

Parallel axis theorem is to find the moment of inertia. It determines the MOI of a given cross section about a line parallel to the centroid axis.

I_AB = I_NA +Ah²

Where xx is the centroid axis

AB is any line parallel to NA axis

h is the distance between AB and NA

Finding the centroid of the given cross section using Varignon theorem.

Varignon Theorem

Moment of the resultant equals the algebraic sum of moments of the individual forces about the same point.

A y^— =  A₁y₁ +A₂y₂+A₃y₃+A₄Y₄

Application of parallel axis theorem

(i) For the area shown in Fig.1

 

A T-Section with flange width 150 mm, thickness 50 mm, web depth 150 mm & web thickness 50 mm.

 WHILE TAKING MOMENTS FOR FINDING CENTROID.       ALL THE DISTANCES ARE FROM THE TOP EDGE

A₁ =150 x 50= 7500 mm²

Y₁^— = 25 mm

A₂ =150 x 50= 7500 mm²

Y₂^— = 50 + 75 = 125 mm

Using Varignon theorem

A y^— =  A₁y₁ +A₂y₂

A = A1 +A2= 7500+7500 = 15000 mm²

15000 y^— = 7500 x 25 +7500x 125

y^{— =} = 75 mm from the top edge

I _{centroidal axis} =(1/12) b₁d₁³ + A₁h₁² +(1/12) b₂ d₂³ + A₂h₂

 = (1/12) 150 (50)³ + 150x 50 (75-25)² +(1/12) 50 (150)³ + 150x 50 (75-125)²

= 13656x 10⁴ mm⁴

(ii) For the area shown in Fig.2

I SECTION TOP FLANGE 100 x 30, bottom flange 50 x 20 and web as70 x 15 mm

Unsymmetrical section about x-axis

ALL THE DISTANCES ARE FROM THE TOP EDGE

Moment of the resultant equals the algebraic sum of moments of the individual forces about the same point.

A y^— =  A₁y₁ +A₂y₂+A₃y₃

A = A 1 +A 2+A 3= 100 x 30 + 70 x 15 + 50 x 20 = 3000+1050+1000=5050 mm ²

5050 y^— = 3000 x 15 +12050 x 65 + 1000 x 110

y^—  = 44.2 mm From the top edge

I _{centroidal axis} =(1/12) b₁d₁³ + A₁h₁² +(1/12) b₂ d₂³ + A₂h₂² +(1/12) b₃d₃³ + A₃h₃²

= (1/12)100 x 30³+3000(44.2-15)²+(1/12)100 x 30³+3000 x (44.2-65)²+(1/12) 100 x 30³ + 3000 x (44.2-110)²

= 8028850 mm ⁴

Thus parallel axis is very helpful. It can determine the moment of inertia about a line parallel to a centroid axis. Moment of inertia resist bending. Larger its value, lesser will be the bending. Therefore, place the beam to have maximum moment of inertia.

PERPENDICULAR AXIS THEOREM

Fig. Perpendicular Axis Theorem

For concurrent x,y and z axes, perpendicular axis theorem is

I_zz = I_xx + I_yy

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