The document discusses properties of surfaces, including centre of gravity and moment of inertia. It defines key terms like centre of gravity, centroid, area moment of inertia, radius of gyration, and mass moment of inertia. Methods for calculating these properties are presented for basic shapes like rectangles, triangles, circles, and composite shapes. Theorems like the perpendicular axis theorem and parallel axis theorem are also covered. Examples are provided for determining the moment of inertia of various plane figures and structures.
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CADmantra Technologies pvt. Ltd. is a CAD Training institute specilized in producing quality and high standard education and training. We are providing a perfact institute for the students intersted in CAD courses CADmantra is established by a group of engineers to devlop good training system in the field of CAD/CAM/CAE, these courses are widely accepted worldwide.
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check it out: http://goo.gl/vqNk7m
CADmantra Technologies pvt. Ltd. is a CAD Training institute specilized in producing quality and high standard education and training. We are providing a perfact institute for the students intersted in CAD courses CADmantra is established by a group of engineers to devlop good training system in the field of CAD/CAM/CAE, these courses are widely accepted worldwide.
#catiatraining
#ANSYS #CRE-O
#hypermesh
#Automobileworkshops
#enginedevelopment
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Bending Stresses are important in the design of beams from strength point of view. The present source gives an idea on theory and problems in bending stresses.
CONTENT:
1. Elastic strain energy
2. Strain energy due to gradual loading
3. Strain energy due to sudden loading
4. Strain energy due to impact loading
5. Strain energy due to shock loading
6. Strain energy due to shear loading
7. Strain energy due to bending (flexure)
8. Strain energy due to torsion
9. Examples
When a body is subjected to gradual, sudden or impact load, the body deforms and work is done upon it. If the elastic limit is not exceed, this work is stored in the body. This work done or energy stored in the body is called strain energy.
When a body is subjected to gradual, sudden or impact load, the body deforms and work is done upon it. If the elastic limit is not exceed, this work is stored in the body. This work done or energy stored in the body is called strain energy.
This document gives the class notes of Unit-8: Torsion of circular shafts and elastic stability of columns. Subject: Mechanics of materials.
Syllabus contest is as per VTU, Belagavi, India.
Notes Compiled By: Hareesha N Gowda, Assistant Professor, DSCE, Bengaluru-78.
Bending Stresses are important in the design of beams from strength point of view. The present source gives an idea on theory and problems in bending stresses.
CONTENT:
1. Elastic strain energy
2. Strain energy due to gradual loading
3. Strain energy due to sudden loading
4. Strain energy due to impact loading
5. Strain energy due to shock loading
6. Strain energy due to shear loading
7. Strain energy due to bending (flexure)
8. Strain energy due to torsion
9. Examples
When a body is subjected to gradual, sudden or impact load, the body deforms and work is done upon it. If the elastic limit is not exceed, this work is stored in the body. This work done or energy stored in the body is called strain energy.
When a body is subjected to gradual, sudden or impact load, the body deforms and work is done upon it. If the elastic limit is not exceed, this work is stored in the body. This work done or energy stored in the body is called strain energy.
This document gives the class notes of Unit-8: Torsion of circular shafts and elastic stability of columns. Subject: Mechanics of materials.
Syllabus contest is as per VTU, Belagavi, India.
Notes Compiled By: Hareesha N Gowda, Assistant Professor, DSCE, Bengaluru-78.
Prof. V. V. Nalawade, Notes CGMI with practice numericalVrushali Nalawade
Centre of gravity is a point where the whole weight of the body is assumed to act. i.e., it is a point where entire distribution of gravitational force is supposed to be concentrated
It is generally denoted “G” for all three dimensional rigid bodies.
e.g. Sphere, table , vehicle, dam, human etc
Centroid is a point where the whole area of a plane lamina is assumed to act.
It is a point where the entire length, area & volume is supposed to be concentrated.
It is a geometrical centre of a figure.
It is used for two dimensional figures.
e.g. rectangle, circle, triangle, semicircle
Centroid is a point where the whole area of a plane lamina is assumed to act.
It is a point where the entire length, area & volume is supposed to be concentrated.
It is a geometrical centre of a figure.
It is used for two dimensional figures.
e.g. rectangle, circle, triangle, semicircle
Centroid is a point where the whole area of a plane lamina is assumed to act.
It is a point where the entire length, area & volume is supposed to be concentrated.
It is a geometrical centre of a figure.
It is used for two dimensional figures.
e.g. rectangle, circle, triangle, semicircle
Mathematical Analysis of Spherical Triangle (Spherical Trigonometry by H.C. R...Harish Chandra Rajpoot
All the important parameters of a spherical triangle have been derived by Mr H.C. Rajpoot by using simple geometry & trigonometry. All the articles (formula) are very practical & simple to apply in case of a spherical triangle to calculate all its important parameters such as solid angle, covered surface area, interior angles etc. & also useful for calculating all the parameters of the corresponding plane triangle obtained by joining all the vertices of a spherical triangle by straight lines. These formula can also be used to calculate all the parameters of the right pyramid obtained by joining all the vertices of a spherical triangle to the center of sphere such as normal height, angle between the consecutive lateral edges, area of base etc.
Operation “Blue Star” is the only event in the history of Independent India where the state went into war with its own people. Even after about 40 years it is not clear if it was culmination of states anger over people of the region, a political game of power or start of dictatorial chapter in the democratic setup.
The people of Punjab felt alienated from main stream due to denial of their just demands during a long democratic struggle since independence. As it happen all over the word, it led to militant struggle with great loss of lives of military, police and civilian personnel. Killing of Indira Gandhi and massacre of innocent Sikhs in Delhi and other India cities was also associated with this movement.
Francesca Gottschalk - How can education support child empowerment.pptxEduSkills OECD
Francesca Gottschalk from the OECD’s Centre for Educational Research and Innovation presents at the Ask an Expert Webinar: How can education support child empowerment?
Unit 8 - Information and Communication Technology (Paper I).pdfThiyagu K
This slides describes the basic concepts of ICT, basics of Email, Emerging Technology and Digital Initiatives in Education. This presentations aligns with the UGC Paper I syllabus.
Embracing GenAI - A Strategic ImperativePeter Windle
Artificial Intelligence (AI) technologies such as Generative AI, Image Generators and Large Language Models have had a dramatic impact on teaching, learning and assessment over the past 18 months. The most immediate threat AI posed was to Academic Integrity with Higher Education Institutes (HEIs) focusing their efforts on combating the use of GenAI in assessment. Guidelines were developed for staff and students, policies put in place too. Innovative educators have forged paths in the use of Generative AI for teaching, learning and assessments leading to pockets of transformation springing up across HEIs, often with little or no top-down guidance, support or direction.
This Gasta posits a strategic approach to integrating AI into HEIs to prepare staff, students and the curriculum for an evolving world and workplace. We will highlight the advantages of working with these technologies beyond the realm of teaching, learning and assessment by considering prompt engineering skills, industry impact, curriculum changes, and the need for staff upskilling. In contrast, not engaging strategically with Generative AI poses risks, including falling behind peers, missed opportunities and failing to ensure our graduates remain employable. The rapid evolution of AI technologies necessitates a proactive and strategic approach if we are to remain relevant.
Welcome to TechSoup New Member Orientation and Q&A (May 2024).pdfTechSoup
In this webinar you will learn how your organization can access TechSoup's wide variety of product discount and donation programs. From hardware to software, we'll give you a tour of the tools available to help your nonprofit with productivity, collaboration, financial management, donor tracking, security, and more.
Synthetic Fiber Construction in lab .pptxPavel ( NSTU)
Synthetic fiber production is a fascinating and complex field that blends chemistry, engineering, and environmental science. By understanding these aspects, students can gain a comprehensive view of synthetic fiber production, its impact on society and the environment, and the potential for future innovations. Synthetic fibers play a crucial role in modern society, impacting various aspects of daily life, industry, and the environment. ynthetic fibers are integral to modern life, offering a range of benefits from cost-effectiveness and versatility to innovative applications and performance characteristics. While they pose environmental challenges, ongoing research and development aim to create more sustainable and eco-friendly alternatives. Understanding the importance of synthetic fibers helps in appreciating their role in the economy, industry, and daily life, while also emphasizing the need for sustainable practices and innovation.
Read| The latest issue of The Challenger is here! We are thrilled to announce that our school paper has qualified for the NATIONAL SCHOOLS PRESS CONFERENCE (NSPC) 2024. Thank you for your unwavering support and trust. Dive into the stories that made us stand out!
2024.06.01 Introducing a competency framework for languag learning materials ...Sandy Millin
http://sandymillin.wordpress.com/iateflwebinar2024
Published classroom materials form the basis of syllabuses, drive teacher professional development, and have a potentially huge influence on learners, teachers and education systems. All teachers also create their own materials, whether a few sentences on a blackboard, a highly-structured fully-realised online course, or anything in between. Despite this, the knowledge and skills needed to create effective language learning materials are rarely part of teacher training, and are mostly learnt by trial and error.
Knowledge and skills frameworks, generally called competency frameworks, for ELT teachers, trainers and managers have existed for a few years now. However, until I created one for my MA dissertation, there wasn’t one drawing together what we need to know and do to be able to effectively produce language learning materials.
This webinar will introduce you to my framework, highlighting the key competencies I identified from my research. It will also show how anybody involved in language teaching (any language, not just English!), teacher training, managing schools or developing language learning materials can benefit from using the framework.
Acetabularia Information For Class 9 .docxvaibhavrinwa19
Acetabularia acetabulum is a single-celled green alga that in its vegetative state is morphologically differentiated into a basal rhizoid and an axially elongated stalk, which bears whorls of branching hairs. The single diploid nucleus resides in the rhizoid.
2. 2.1 CENTRE OF GRAVITY
Centre of gravity(C.G) of a body is defined as the
point through which the entire weight of the body
acts. A body has only one centre of gravity for all
its positions.
Centroid is defined as the point at which the total
area of a plane figure is assumed to be
concentrated. It is the CG of a plane figure
CG of a rectangle is the point where the
diagonals meet
CG of a triangle is the point where the medians
meet
CG of a circle is at its centre
3. 2.1 CENTRE OF GRAVITY
Consider a plane figure of total area A composed
of a number of small areas a1, a2, a3,...., an
Let x1= Distance of CG of a1 from axis of
reference OY
x2= Distance of CG of a2 from axis of reference
OY
x3= Distance of CG of a1 from axis of reference
OY
;
;
xn= Distance of CG of an from axis of reference
OY
4. 2.1 CENTRE OF GRAVITY
The moments of all areas about axis OY= a1x1 +
a2x2 + a3x3+ ....... + anxn
Let G be the centre of gravity of the total area
about the axis OY whose distance from OY is Xc
5. 2.2 CENTRE OF GRAVITY OF
PLANE FIGURES BY METHOD OF
MOMENTS
Moment of total area about OY= AXc
Sum of moments of small areas about axis OY
must be equal to the moment of total area about
OY.
Hence a1x1 + a2x2 + a3x3+ ....... + anxn= AXc
Hence Xc= (a1x1 + a2x2 + a3x3+ ....... + anxn)/A
Similarly
6. 2.2 CENTRE OF GRAVITY OF
PLANE FIGURES BY METHOD OF
INTEGRATION
If the areas are large in number (that is i tends to
infinity), then the summations in the above
equations can be replaced by integration. And
when the number of split up areas are large, the
size of component areas will be small, hence a
can be replaced by dA in the above equation.
Hence
Xc=∫xdA/∫dA
Yc=∫ydA/∫dA
7. 2.3 IMPORTANT NOTES TO
CALCULATE CENTRE OF
GRAVITY
Axes about which moments of areas are taken, is
known as axis of reference.
Axis of reference of plane figures is generally
taken as the lowest line of the figure for
determining Axis of reference of plane figures is
generally taken as the lowest line of the figure for
determining c and left line of the figure for
calculation Xc
If the given section is symmetrical about X-X or YY, CG will lie in the line of symmetry
8. 2.2 AREA MOMENT OF INERTIA
Consider an area A as shown. Let dA be an elemental area
of the area A with coordinates x and y. The term Σx²dA is
called the moment of inertia of area A about y axis and
term Σy²dA is called the moment of inertia of area A about
x-axis.
Ixx= Σy²dA & Iyy= Σx²dA
When dA is very small, mathematically Σy²dA= ∫y²dA
Σx²dA= ∫x²dA
9. 2.4 AREA MOMENT OF
INERTIA
Ixx= ∫y²dA & Iyy= ∫x²dA
If r is the distance of an area dA (which is a part
of area A) from an axis AB, then the sum of terms
r²dA (ie Σ r²dA) to cover the entire area is called
the moment of inertia of the area A about axis AB
or second moment of area of area A about axis
AB.
Moment of inertia is a fourth dimensional term as
it is obtained by multiplying area by distance
squared. Hence the SI unit is m^4
10. 2.5 RADIUS OF GYRATION
Mathematical term defined by the expression, k=√
(I/A) is called radius of gyration.
I=Ak²
Hence the radius of gyration can be considered a
that distance at which the complete area is
squeezed and kept as a strip of negligible width
such that there is no change in the moment of
inertia
11. 2.6 THEOREMS OF MOMENT OF
INERTIA
1) Perpendicular axis theorem:-
The area moment of inertia about an axis
perpendicular to its plane at any point is equal to
the sum of moment of inertia about two mutually
perpendicular axes through the point O and lying
in the same plane of area.
Polar moment of inertia is defined as the product of
area and square of distance between CG from
the axis of reference perpendicular to the area.
12. 2.6 THEOREMS OF MOMENT OF
INERTIA
Proof:- Consider an infinitesimal elemental area
dA with co-ordinates (x,y).
Ixx=∫y²dA ;Iyy=∫x²dA; Izz=∫r²dA
r²= x² + y²
Hence r² x dA= x² dA + y² x dA
Hence Izz= Ixx + Iyy
13. 2.6 THEOREMS OF MOMENT OF
INERTIA
Parallel axis theorem:- Moment of inertia about
any axis in the plane of an area is equal to sum of
moment of inertia about a parallel centroidal axis
and the product of area and square of the
distance between the two parallel axes
Referring to the figure given below, the theorem
means IAB= IGG + Ayc²
14. 2.6 THEOREMS OF MOMENT OF
INERTIA
IAB Moment of inertia about axis AB
IGG Moment of inertia about centroidal axis GG
parallel to AB
A The area of plane figure given.
Yc The distance between the axis AB and the
parallel centroidal axis GG
Proof:- Consider an elemental strip dA whose CG
is at a distance y from centroidal axis G-G
15. 2.6 THEOREMS OF MOMENT OF
INERTIA
IGG= ∫y² dA
IAB= ∫(y + yc)² dA
IAB= ∫(y² + yc² + 2yyc) dA
y is a variable and yc is a constant; hence ∫y² dA=
IGG and ∫yc² dA= Ayc².
∫y dA/ A=distance of centroid from the axis of
reference= 0 as GG is passing through the
centroid.
Hence IAB= IGG + Ayc²
16. 2.7 DETERMINATION OF MOMENT
OF INERTIA
1) Moment of inertia of a rectangular section
about the centroidal axis in the plane of
section
Consider a rectangular section of length b and
depth d. Let X-X be the horizontal axis passing
through the centroid of the area.
17. 2.7 DETERMINATION OF MOMENT
OF INERTIA
Consider a rectangular elemental strip whose CG
is at a distance y from the horizontal centroidal
axis.
Area of the strip dA= b x dy
Area moment of inertia of strip about the
centroidal axis,
Ixx =bd³/12
Similarly Iyy =db³/12
18. 2.7 DETERMINATION OF MOMENT
OF INERTIA
2) Moment of inertia of a rectangular section
about an axis passing through the base of the
rectangle
Consider a rectangular section of length b and
depth d. Let AB be the horizontal axis passing
through the base of the rectangle.
IAB= IGG + A x Yc²= bd³/12 + bd x (d/2)²
IAB= bd³/3
19. 2.7 DETERMINATION OF MOMENT
OF INERTIA
3) Moment of inertia of a hollow rectangular
section about the centroidal axis in the plane
of section
Moment of inertia of main section about X-X axis=
b2d2³/12
Moment of inertia of the cut out section about X-X
axis= b1d1³/12
20. 2.7 DETERMINATION OF MOMENT
OF INERTIA
Moment of inertia of the hollow rectangular
section about X-X axis= (b2d2³/12- b1d1³/12)
Similarly Moment of inertia of the hollow
rectangular section about Y-Y axis= (d2b2³/12d1b1³/12)
Moment of inertia of the hollow rectangular
section about any axis= ( MI of outer rectangular
section about the axis - MI of cut-out rectangular
section about the axis )
21. 2.7 DETERMINATION OF MOMENT
OF INERTIA
4) Moment of inertia of a circular section
passing through the centre and lying in the
plane of the figure
Consider an elementary circular ring of radius r
and thickness dr. Area of the circular ring =2πrdr
The moment of inertia about an axis passing
through the centre O of the circle and
perpendicular to the plane of area,
Izz= πR^4/2= πD^4/32
According to perpendicular axis
theorem, Ixx=Iyy= ½ Izz= πD^4/64
22. 2.7 DETERMINATION OF MOMENT
OF INERTIA
5) Moment of inertia of a hollow circular
section
Moment of inertia of outer circle about X-X axis=
πD^4/64
Moment of inertia of the cut-out circle= πd^4/64
Moment of inertia of the hollow circular section
about X-X axis Ixx= π x (D^4 – d^4) /64
Similarly Iyy= π x (D^4 – d^4) /64
23. 2.7 DETERMINATION OF MOMENT
OF INERTIA
6) Moment of inertia of a semicircular area
INN= Moment of inertia of the semicircular lamina
about an axis passing through the centre of the
semicircle=1/2 x Moment of inertia of a circular
lamina about an axis passing through the centre
and lying in the plane of the figure= πD^4/128
24. 2.7 DETERMINATION OF MOMENT
OF INERTIA
Moment of inertia of the semicircle about an axis
passing through the CG of the semicircle= Inn + A
x Yc²
Ixx= πD^4/128 + πD²/8 x (2D/3π)²= 0.11R^4
25. 2.7 DETERMINATION OF MOMENT
OF INERTIA
7) Moment of inertia of a triangular section
about its base
Consider an elemental strip DE of the triangle at
a distance y from the vertex opposite to the base
of the triangle. From the above figure , DE/b= (dy)/d
DE=b (1- y/d)
Area of the elemental strip dA= DE x dy = b (1-
y/d) x dy
Moment of inertia of the area about N-N,
INN =bd³/12
26. 2.7 DETERMINATION OF MOMENT
OF INERTIA
8) Moment of inertia of a triangle about the
centroidal axis parallel to the base of the
triangle
INN =bd³/12=Igg + Ayc²
bd³/12 - bd/2 x d²/9= Igg
Igg= bd³/36
27. 2.8 MASS MOMENT OF
INERTIA
Consider a body of mass M lying in the XY plane.
Let
X= distance of CG of the body from OY axis
Y= distance of CG of the body from OX axis
Moment of mass M about OY axis= Mx
28. 2.8 MASS MOMENT OF
INERTIA
Second moment of mass M about OY axis= Mx²
Second Moment of mass is known as mass
moment of inertia
Mass moment of inertia(IM) about an axis is
hence defined as the product of mass of a body
and the square of its perpendicular distance from
the axis of reference.
29. 2.8 MASS MOMENT OF
INERTIA
Suppose the body is split up into small masses
m1, m2, m3,....., mn. Let the distance of CGs of
masses be at distances r1, r2, r3,.....rn from an
axis of reference. Then mass moment of inertia
about that axis is given by
IM= Σmi x ri²
If the small masses are large in number, then the
summation in above equation can be replaced by
integral,
Im= ∫r²dm
Physical meaning of mass moment of inertia:- It is
the resistance of a rotating body against the
change in angular velocity
30. 2.9 MASS MOMENT OF INERTIA OF A
RECTANGULAR PLATE ABOUT CENTROIDAL
AXIS PARALLEL TO THE BASE OF THE PLATE
Consider a rectangular plate of width b, depth d
and thickness t composed of a material of density
ρ.
31. RECTANGULAR PLATE ABOUT
CENTROIDAL AXIS PARALLEL TO THE
BASE OF THE PLATE
Mass of the plate= ρ x b x d x t
Consider an elementary rectangular strip of width
b, depth dy and thickness t at a distance y from
the centroidal axis X-X . The area of the strip
is, dA= b x dy
Mass of the strip, dm= Volume of the strip x
density= Thickness x Area of the strip x density=t
x b x dy x ρ
Mass moment of inertia of the strip = y² dm= y²x
(t x b x dy x ρ)
32. RECTANGULAR PLATE ABOUT
CENTROIDAL AXIS PARALLEL TO THE
BASE OF THE PLATE
Mass moment of inertia of the entire mass about
XX axis= ∫ y² dm = ∫y²x (t x b x dy x ρ)
=bt ρ x d³/12= ρ x t x b d³/12= Density x thickness
x Moment of inertia of the rectangular section
about the centroidal axis parallel to the base
bt ρ x d= Mass of the rectangular plate= M
33. RECTANGULAR PLATE ABOUT
CENTROIDAL AXIS PARALLEL TO THE
BASE OF THE PLATE
Hence moment of inertia of a rectangular plate
about the centroidal axis parallel to its base
Imxx= Md²/12
Moment of inertia of a rectangular plate about the
centroidal axis perpendicular to its base, Imyy=
Mb²/12
Moment of inertia of a hollow rectangular plate of
outer section dimensions B, D and inner section
dimensions b, d is given by the equation, Imxx=
1/12 (MD² - md²); where M is the mass of outer
section and m is the mass of cut-out section.
34. 2.9 MASS MOMENT OF INERTIA OF A
RECTANGULAR PLATE ABOUT AN AXIS
PASSING THROUGH ITS BASE
Consider a rectangular plate of width b, depth d
and thickness t composed of a material of density
ρ.
Mass of the plate= ρ x b x d x t
Consider an elementary rectangular strip of width
b, depth dy and thickness t at a distance y from
the base AB . The area of the strip is, dA= b x dy
Mass of the strip, dm= Volume of the strip x
density= Thickness x Area of the strip x density=t
x b x dy x ρ
35. 2.9 MASS MOMENT OF INERTIA OF A
RECTANGULAR PLATE ABOUT AN AXIS
PASSING THROUGH ITS BASE
Mass moment of inertia of the strip = y² dm= y²x
(t x b x dy x ρ)
Mass moment of inertia of the entire mass about
XX axis= ∫ y² dm = ∫y²x (t x b x dy x ρ)
=bt ρ x d³/3= ρ x t x b d³/3= Density x thickness x
Moment of inertia of the rectangular section about
the centroidal axis parallel to the base
36. 2.9 MASS MOMENT OF INERTIA OF A
RECTANGULAR PLATE ABOUT AN AXIS
PASSING THROUGH ITS BASE
bt ρ x d= Mass of the rectangular plate= M
Hence moment of inertia of a rectangular plate
about a horizontal axis passing through to its
base Imxx= Md²/3
Moment of inertia of a rectangular plate about the
centroidal axis perpendicular to its base and
passing through the vertical side, Imyy= Mb²/3
37. 2.10 MASS MOMENT OF INERTIA
OF A CIRCULAR PLATE
Consider a circular plate of radius R, thickness
t, with O as its centre. Consider an elementary
circular ring of radius r and width dr as shown.
Area of the circular ring, dA= (2πr)dr
Volume of the circular ring, dV=(2πr)dr x t
Mass of the circular ring, dm=(2πr)dr x t x ρ
38. 2.10 MASS MOMENT OF INERTIA
OF A CIRCULAR PLATE
Mass moment of inertia of the circular ring about
axis perpendicular to the plane of the
figure, Imzz= ∫r²dm=
Imzz= πR^4t ρ/2
Imzz=Imxx=1/2 x Imzz= πR^4t ρ/4
Imxx= πD^4t ρ/64
Mass of the circular disc, M= Area of the disc x
thickness of the disc x density= πD²/4 x t x ρ
Imxx= Md²/16
39. 2.11 MASS MOMENT OF INERTIA
OF A HOLLOW CIRCULAR
CYLINDER
Let
R2 Outer radius of the cylinder
R1 Inner radius of the cylinder
L Length of cylinder
M Mass of cylinder
ρDensity of the material of cylinder
M= π x (R2² - R1²) x L x ρ
40. 2.11 MASS MOMENT OF INERTIA
OF A HOLLOW CIRCULAR
CYLINDER
Area of the circular ring, dA= (2πr)dr
Volume of the circular ring, dV=(2πr)dr x L
Mass of the circular ring, dm=(2πr)dr x L x ρ
Mass moment of inertia of the circular ring about
axis perpendicular to the plane of the
figure, Imzz= ∫r²dm
41. 2.11 MASS MOMENT OF INERTIA
OF A HOLLOW CIRCULAR
CYLINDER
=πρL x (R2^4 – R1^4)/2
= πx (R2² - R1²) ρL x (R2² + R1²)/2=M x(R2² +
R1²)/2
According to perpendicular axis theorem,
Imxx=Imyy=Imzz/2= M x (R2² + R1²)/4
42. 2.12 MASS MOMENT OF INERTIA
OF A UNIFORM ROD
Consider a uniform thin rod AB of length L. Let m
be the mass per unit length of the rod and M be
the mass of the rod. Hence M=mL
Consider a strip of length dx at a distance x from
the Y-Y axis of reference.
43. 2.12 MASS MOMENT OF INERTIA
OF A UNIFORM ROD
Mass of the strip dm=mdx
Mass moment of inertia about axis YY,Imyy=
∫x²dm
=mL³/3=ML²/3
Hence Imyy= ML²/3
44. 2.13 PRODUCT OF AREA
Consider a small elemental area dA of a body of
area A.
Moment of area dA about X-axis= y.dA
Moment of y.dA about Y-axis=x.(ydA)
xydA is known as the product of inertia(Ixy) of
area dA about x-axis and y-axis.∫ xydA is known
as the product of inertia of the area A about x-axis
and y-axis.
Ixy=∫ xydA
Hence product of area is obtained if an elemental
area is multiplied by the product of its coordinates and is integrated over an area.
45. 2.13 PRODUCT OF AREA
The product of inertia can also be written
mathematically as
Ixy= Σ(xiyiAi)= x1y1A1 + x2y2A2 + .......+xnynAn
Where (xi,yi) are the co-ordinates of the CG of
area Ai.
Notes:
Product of inertia can be positive (if both x and y
have the same sign) or negative (if both x and y
have different signs)
If the area is symmetrical w.r.t one or both the
axis, then the product of inertia will be zero
The product of inertia w.r.t the centroidal axis will
46. 2.14 PRINCIPAL AXES
Principal axes are the axes about which the
product of inertia is zero.
Consider a body of area A. Consider a small area
dA. The product of inertia of the total area A w.r.t
x and y axes is given by
47. 2.14 PRINCIPAL AXES
Let now the axes be rotated by 90° in CCW
direction keeping the total area in the same
position. Let X1 and Y1 be the new axes. The coordinates of the area dA w.r.t new axes are x’ and
y’.
Hence the product of inertia w.r.t new
axes=∫(x’y’dA);
Ixy=∫xydA
Let now the axes be rotated by 90° in CCW
direction keeping the total area in the same
position. Let X1 and Y1 be the new axes. The coordinates of the area dA w.r.t new axes are x’ and
y’.
Hence the product of inertia w.r.t new
48. 2.14 PRINCIPAL AXES
Comparing the two configurations, x’= y and y’=-x
Hence ∫(x’y’dA)= -∫(xydA).
That means, the product of area changes its sign
from positive to negative while rotating the axes
through an angle of 90° in CCW direction. Hence
it is also possible that on rotating the axes
through a certain angle, the product of inertia
becomes zero. The new axes about which
product of inertia is zero is called principal axes.