MOMENT OF INERTIA & POLAR MOMENT OF INERTIA CLASS NOTES
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 mm^{4}.
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 cm^{2}), ‘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 y^{2}
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 x^{2}
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) bd^{3}
When y axis is parallel to depth d
Iyy = (1/12) db^{3}
(b)
For a square cross section, each side is ‘a’
I_{xx} =I_{yy} = (1/12) a^{4}
(c) For a circular solid cross section like a rod
I_{xx} =I_{yy} = (π/64) d^{4}
(d) For a circular hollow cross section like a pipe
I_{xx} =I_{yy} = (π/64)[ d_{o}^{4} –d_{i}^{4} ]
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 r^{2} = ∫dA(x^{2} + y^{2})= 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 ofcross section |
MomentOf Inertia |
Polar moment of Inertia |
1. |
Rectangular |
I_{xx} =(1/12) bd^{3}Here b is parallel to x-axis and d is to the x-axis,I_{yy} =(1/12) db^{3}Here d is parallel to y-axis and b is to the y-axis |
Not used |
2. |
Square |
I_{xx =}(1/12) a^{4}= I_{yy}I_{yy =}(1/12) a^{4} |
Not used |
3. |
Solid Circular |
I_{xx=}(π/64) d^{4} = I_{yy} |
J =(π/32) d^{4}J = I_{xx} + I_{yy} |
4. |
Hollow circular |
FirstlyI_{xx}=(π/64)(D^{4}– d^{4})SecondlyI_{yy}=(π/64)(D^{4}– d^{4})I_{xx} = I_{yy}D is outer diameterd is inner diameter |
J = (π/32)(D^{4}– d^{4})J = I_{xx} + I_{yy} |
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} +Ah^{2}
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^{— }= A_{1}y_{1} +A_{2}y_{2}+A_{3}y_{3}+A_{4}Y_{4}
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
A_{1} =150 x 50= 7500 mm^{2}
Y_{1}^{—} = 25 mm
A_{2} =150 x 50= 7500 mm^{2}
Y_{2}^{—} = 50 + 75 = 125 mm
Using Varignon theorem
A y^{— }= A_{1}y_{1} +A_{2}y_{2}
A = A1 +A2= 7500+7500 = 15000 mm^{2}
15000 y^{—} = 7500 x 25 +7500x 125
y^{— =} = 75 mm from the top edge
I_{ centroidal axis} =(1/12) b_{1}d_{1}^{3} + A_{1}h_{1}^{2} +(1/12) b_{2} d_{2}^{3} + A_{2}h_{2}
= (1/12) 150 (50)^{3} + 150x 50 (75-25)^{2} +(1/12) 50 (150)^{3} + 150x 50 (75-125)^{2}
= 13656x 10^{4} mm^{4}
(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^{— }= A_{1}y_{1} +A_{2}y_{2}+A_{3}y_{3}
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) b_{1}d_{1}^{3} + A_{1}h_{1}^{2} +(1/12) b_{2} d_{2}^{3} + A_{2}h_{2}^{2} +(1/12) b_{3}d_{3}^{3} + A_{3}h_{3}^{2}
= (1/12)100 x 30^{3}+3000(44.2-15)^{2}+(1/12)100 x 30^{3}+3000 x (44.2-65)^{2}+(1/12) 100 x 30^{3} + 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}
https://www.mesubjects.net/wp-admin/post.php?post=3454&action=edit Bending stresses Class notes