It includes explanation of stress-strain diagram and properties deduce from stress-strain diagram and working of UTM machine

2. DEPARTMENT OF MECHANICAL ENGINEERING
MAY 10, 2020
TIME : 11:00 AM TO 12:30 PM
URL :
EMAIL : hodmech@meu.edu.in
CONTACT / WHATSAPP
An
Online Workshop
On
“Demonstration of Mechanial Behaviour of Material
Using a Practical Approach”
For Class 10th & 12th Students

3. An
Online Workshop
on
“Demonstration of Mechanical Behavior of Materials Using a
Practical Approach”
By:- Dr. Gaurav Gugliani
(Ph.D. IIT, B.H.U., M.Tech. N.I.T., Rourkela, B.E. R.G.T.U.)
Mr. Pawan Bhawsar (B.Tech., Mechanical )
Assistant Professor, Department of Mechanical Engineering
Mandsaur University, Mandsaur, M.P.

4. Outline of the Presentation
 Introduction to Mechanical Engineering
 Introduction to Mechanical Properties
 Introduction to Stress-Strain and its Types
 Material Properties
 Elastic Modulii
 Practice Questions
 Stress- Strain Diagram
 Introduction to Universal Testing Machine
 Types of Test perform on UTM
 Practice Exercise
 Introduction to Department of Mechanical Engineering
 Overview of the Placements, Labs, Industrial Visits

5. What is Mechanical Engineering?
Mechanical engineering is a discipline of engineering that applies the principles
of physics and materials science for analysis, design, manufacturing, and
maintenance of mechanical systems. It is the branch of engineering that involves
the production and usage of heat and mechanical power for the design,
production, and operation of machines and tools. It is one of the oldest and
broadest engineering disciplines.
Why study Mechanical Engineering?
If you’re interested in the design, development, installation, operation or
maintenance of just about anything that has moveable parts then mechanical
engineering could be the programme of study for you

6. Scope of Mechanical Engineering

7. Mechanical properties that are important to a design engineer differ from
those that are of interest to the manufacturing engineer.
 In design, mechanical properties such as elastic modulus and yield strength
are important in order to resist permanent deformation under applied
stresses. Thus, the focus is on the elastic properties.
 In manufacturing, the goal is to apply stresses that exceed the yield strength
of the material so as to deform it to the required shape. Thus, the focus is on
the plastic properties.
Introduction

8. • Ductility: Ability of a material to deform under tensile load; high % elongation and
% reduction of area indicate ductility.
• Brittleness: material failure with little deformation; low % elongation and %
reduction of area indicate brittleness.
Elasticity: By virtue of which certain material returns back to their original position
After the removal of the external forces is ‘elasticity’
Plasticity: is the opposite of elastic deformation and is accepted as unrecoverable
strain. Plastic deformation is retained even after the relaxation of the applied
stress.
Strength: when an external forces act on body, the body tends to undergo some
deformation due to cohesion between the molecules ,the body resist deformation .this
resistances by which material of body opposes the deformation is known as ‘strength of
material’
Material Properties:

9. The forces of resistances per unit area offered by a body against
deformation is known as ‘stress’
Mathematical written as:
Unit of Stress is:
Pascal OR Newton/m2
OR dyne/cm2 OR Kgf/m2
( )
( )
( )
Force F
Stress
Area A
 
Introduction to Stress

10. Types of Stress

11. When a body is subjected to some external forces ,their is some
change of dimension of the body
The rate of change of dimension of the body to the original
dimension is known as ‘strain’
Mathematical Expression:
Change in dimension
Strain ( )
Original dimension
 
Introduction to Strain
Units of strain is no units
Here the units of length is meter so
meter/meter gets cancel

12. Tensile Strain: When we apply a tensile force on a body its length increases.
The ratio of increase in length to original length
Compressive Strain: When we apply a compressive force on a body its length
decreases. The ratio of decreases in length to original length
0
0
0
-
Tensile Strain ( )
= original length
New Length
t
L L
L
where L
L
 

Types of Strain

13. Volumetric Strain: When we apply a force on all sides of a body its volume will
decrease. The ratio of change in volume to the original volume.
Shear Strain: When we apply a tangential force on a body there is an angular
displacement which measures the shear strain.
0
0
0
-
Volumetric Strain ( )
V = original Volume
V New Volume
v
V V
V
where
 

Shear Strain ( ) tan( )s
d
h
  

14. Hooke’s law
Hooke’s law, states that when a material load within the elastic limit
the stress is directly proportional to the strain produced by the stress .
This means the ratio of the stress to the corresponding strain is a
constant with in the elastic limit this constant is known as ‘modulus of
elasticity’ (or) ‘modulus of rigidity’(or) ‘elastic moduli’
Its units is newton /meter square
Relationship Between Stress and Strain

15. Modulus of Elasticity (or) Young’s Modulus
The ratio of tensile stress (or) compressive stress to
corresponding strain is a constant, this ratio is known as
‘young’s modulus’ it is denoted by ‘E’
0F L
E
A

 

 

Units: Gpa =109Newton/m2

16. Modulus of rigidity (or) shear modulus
The ratio of the shear stress to the corresponding shear strain with in the
elastic limit is known as ‘modulus of rigidity’ (or) ‘shear modulus’
It is denoted by C ,G or N
Bulk Modulus
The ratio of the Normal stress acting on a body from all direction to the
corresponding Volumetric strain with in the elastic limit is known as ‘Bulk
Modulus’
It is denoted by K
/
tan( )s
F A
G

 
 
n
v
K




17. Relationship between Young Modulus, Modulus of Rigidity
and Bulk Modulus
Young Modulus and Modulus of Rigidity
Young Modulus and Bulk Modulus
3 (1 2 )E K  
2 (1 )E G  

18. Materials Modulus of Elastic (GPa)
1. Steel 200 to 220
2. Cast iron 100 to 160
3. Copper 90 to 110
4. Brass 80 to 90
5. Aluminum 60 to 80
6. Timber 10

19. Practice Question # 1
Ex:1 Two wires are made of the same material and have the same volume. The first
wire has cross-sectional area A and the second wire has cross-sectional area 3A. If the
length of the first wire is increased by on applying a load Force F, how much
force is needed to stretch the second wire by the same amount.
a. 4F
b. 9F
c. F
d. 6F
l
Solution: We know that Volume= Area *Length
As two wires have same Volume :
According to Hooke’s Law
1 2
1 1 2 2
1
2 1 1 1
2
3
3 3 3 9
Stress Stress
E
Strain Strain
F L F L
A l A l
l
F F F F
l
 
   
    
   
 

 
    
1 2
1 23
1
3
2
V V
A l A l
l
l

  


20. Practice Question # 2
Ex 2. Which is more elastic Steel or Rubber ?
Answer: Steel is more elastic than rubber.
Explanation: Modulus of Elastic of Steel: 200-250 GPa
Modulus of rubber: 0.01-0.1 GP
If you elongate steel and rubber both to double its length, the force
required to elongate steel is much higher than that of rubber.

21. Practice Question # 3
Ex 2. Which is more elastic Steel or Diamond ?
Modulus of Elastic of Steel: 200-250 GPa
Modulus of Diamond: 1200 GP
Answer: Diamond cannot be drawn into a thin wire. It
is a brittle material it break because of its Hardness.

22. Poisson’s Ratio or Poisson Coefficient (µ):
It is the ratio of transverse contraction strain to longitudinal extension strain in
the direction of stretching force. Tensile deformation is considered positive
and compressive deformation is considered negative
Strain Energy
Strain energy is defined as the energy stored in a body
due to deformation. The strain energy stored in the
body is equal to the work done by the applied load
in stretching the body, which is equal to the shaded area
Of load extension curved.
Lateral Strain
Longitudnal Strain
  
2
1
* *
2 2
Strain energy Load Extension V
E

 

23. Practice Question # 4
Ex 4: A Gradual force of 15000N applied to a metal rod and it
gets 0.01 m shorter as a result. How much strain energy stored in
the rod?
a. 75J
b. 150J
c. 625J
d. 0.05J
Solution:
15000
0.01
1 1
* *15000*0.01 75
2 2
F N
x
Strain Energy stored workdone by the force Area of a triangle
base height Joule


 
  

24. It is a tool for understanding material behavior under load.
A stress strain diagram help engineers to select the right
materials for specific loading conditions. Or
It is a graph that represents how a part behaves under an
increasing load, and used by engineers when selecting
materials for specific designs.
Stress-Strain Diagram

25. •Elastic region: This portion is generally represented as a linear
relationship between stress and strain. If the load is released the specimen
will return to its original dimensions.
•Plastic region: In this portion, the
specimen begins to yield.
The maximum strength of the
specimen occurs in this zone.
The specimen endures some
permanent deformation that
remains after the load is released.
•Rupture: The point at which a
specimen breaks into two parts.
A stress-strain diagram generally contains three
regions:

27. Elastic Limit: The limit in which the material will return its original shape
when the load is removed.
• Yield Point: Yield point is the point at which the material will have an
appreciable elongation OR a slight increase in stress above the elastic limit will
result in permanent deformation. This behavior is called yielding for ductile
materials (In Engineering, the transition from elastic behavior to plastic
behavior).
• Upper yield point: which corresponds to the load reached just before yield starts.
• Lower yield point: which corresponding to the load required maintain yield.
Lower yield point should be used to determine the yield strength of the material.
Explanation of Stress-strain Curve

28. Yield strength: you can draw a line parallel to the initial linear portion, and where
this line intersects the curve is the yield point. The Y coordinate of this point is the
yield strength.
Ultimate Strength: The maximum ordinate in the stress-strain diagram is the
ultimate strength or tensile strength. This is the maximum load the specimen
sustains during the test.
• Necking: After the ultimate stress, the cross sectional area begin to decrease in a
region of the specimen because of local instability. This phenomenon is known as
necking. After necking has been beginning, we note that rupture occurs at an angle
of 45 degree with the original surface of the specimen. This indicates that shear
stresses are responsible for failure of the ductile materials.
Cont…

29. • Rapture or Fracture: The specimen spilt into two or more pieces or break into
parts.
• Rapture Strength: is the strength of the material at rupture. This is also known as
the breaking strength.
• Strain Hardening: The stress must be increased to keep elongating the specimen,
until the maximum value has been reached. This is due to a property of the material
known as strain hardening.
Cont…

30. • Brittle materials: such as cast iron, glass, concrete and carbon fiber
(composite materials) are characterized by the fact that rupture occurs
without any prior change in the rate of deformation.
• These do not have a yield point and do not strain harden, which means
that ultimate strength and breaking strength are at same point.
Stress Strain Diagrams for Brittle Materials

31. Reading of Stress-strain Curve


• Slope is high High Young
Modulus High Stiffness
• Highest Ultimate Stress
Strongest
• No Plastic Deformation Brittle
• Stong + Stiff + Brittle Hard
Material
• Slope is Low Low Young
Modulus More Flexible
• Lowest Ultimate Stress Weakest
• For low Stress, high strain More
Elastic
• Shaded area is high High Strain
Energy High Toughest
• Large Plastic Deformation More Plastic
• Tough+Plastic Ductile OR Malleable ct

32. UNIVERSAL TESTING MACHINE (UTM)
A universal testing machine (UTM), also
known as a universal tester, materials
testing machine or materials test frame,
is used to test the tensile strength and
compressive strength of materials . It is
named after the fact that it can perform
many standard tensile and compression
tests on materials, components, and
structures.

33. Schematic Diagram of Tensile testing
in UTM

35. Tensile Testing
• Uses an UTM to apply measured force to an test specimen. The amount of
extension can be measured and graphed.
• Variables such as strain, stress, elasticity, tensile strength, ductility and
shear strength can be gauged.
• Test specimens can be round or flat.
Why are metals Tested?
• Ensure quality
• Test properties
• Prevent failure in use
• Make informed choices in using material
Tensile Test

36. Determination of following properties using tensile test
• STRENGTH
• DUCTILITY
• ELASTICITY
• STIFFNESS
• MALLEABILTY
• MODULUS of TOUGHNESS
• MODULUS of RESILIENCE

37. • Aerospace Industry
• Shear and tensile strength testing of fasteners
e.g. Bolts, nuts and screws
• Textiles Industry
• 'Pull-off' characteristics of buttons, stitched-on
decorations, press studs, zip fasteners, hookand-
loop fasteners
Applications:

38.  Compressive strength is the capacity of a material or structure to
withstand axially directed pushing forces.
 When the limit of compressive strength is reached, brittle materials are crushed.
COMPRESSIVE STRENGTH
COMPRESSIVE TEST
Before Loading After Loading

39. In engineering mechanics, flexure or bending characterizes the behavior of a
slender structural element subjected to an external load applied perpendicularly
to a longitudinal axis of the element.
FLEXURAL TEST
WHY PERFORM A FLEXURE TEST?
A flexure test produces tensile stress in the convex side of the specimen and compression
stress in the concave side. This creates an area of shear stress along the midline. To
ensure the primary failure comes from tensile or compression stress the shear stress must
be minimized. This is done by controlling the span to depth ratio; the length of the outer
span divided by the height (depth) of the specimen.
For most materials S/d=16 is acceptable.
Some materials require S/d=32 to 64 to
keep the shear stress low enough.

40. In this test a specimen with rectangular or flat cross-section is placed on two
parallel supporting pins. The loading force is applied in the middle by means
loading pin.
3-POINT BEND TEST
Determination of following properties using 3-point bending test
 Modulus of elasticity in bending,
 Flexural stress,
 Flexural strain
 Flexural stress-strain response of the material.

41. Calculation:

42. Department of Mechanical
Engineering
Mandsaur University

43. Teaching Staff
of
ce
Sr.No Name of Faculty Qualification Total Year of
Experience
1 Dr. Ravi Kumar Ph.D , M.Tech , B.Tech (Mech.) 23+
2 Mr. Sanjeev singh Chauhan ME (IE & M) , BE (Mech.) 23+
3 Dr. Haider Hussain Ph.D , ME(CAD/CAM), Me(Mech.) 12+
4 Dr. Gaurav Guglinai Ph.D. IIT, B.H.U., M.Tech. N.I.T.,
Rourkela, B.E. R.G.T.U
3+
5 Mr. Ashish Sethiya M.Tech (Energy Mgmt. , BE (Mech.) 8+
6 Mr. Abdul Gani M.Tech (Thermal Engg.) , B.Tech (Mech.) 5+
7 Mr. Yuvraj Singh Jhala ME (Energy) , BE (Mech.) 2
8 Mr. Abhishek Rohilla ME (Prodction Engg.) , BE (Mech.) 8+
9 Mr. Pushpraj Singh Rathore ME (Thermal Engg.) , BE (Mech.) 7+
10 Mr. Pawan Bhawsar M.S , BE (Mech.) 5
11 Mr. Shailendra Singh Chauhan BE (Mech.) 7+
12 Mr. Kalpit Bhargav Diploma (Mech.) 9+

44. About The Department
Lab
Lab

45. Fluid Mechanics Lab
Computer Aided Design Lab
LABORATORY

46. 3-D Printer

47. KOM/DOM Lab

48. CNC Machine Lab
Mechanical Workshop

49. Internal Combustion Engine Lab
HMT LabRAC Lab

50. Research Achievements By
Faculties of ME Department

51. Placements
Our department student are placed in following companies:-
• L & T Infotech
• Modtech Machine PVT. LTD. Ahmedabad
• Bridgestone
• IPCA LAB.
• Maruti Suzuki
• Eicher Motor
• RentEase international LLP, Navi Mumbai
• Macleods Pharma.
• Plasto plastic Pvt. Ltd.
• Cipla
• RAPS Automotovie
• SPM Autocomp system pvt. Ltd.
• Khodiyar group
Note:- Number of students palced 100 %

52. Placements
Highest Package 12.4 LAKHS

53. Autocad Workshop
By:- Mr. Sudhanshu Shukla
From Engg. CAD Institute , Indore

54. Omkareshwar Hydal Power Project

55. JK cement works, Mangrol , Nimbahera
(Rajasthan)

56. Diesel Locomotive Shed Ratlam
(W.R. Divison)

57. Industrial Visit ToNTPC , Singaji

59. Guest Lectures

60. Lectures By TPO

61. Industrial Visit to Sanchi Milk Plant located at Jaggakhedi
Mandsaur(13/3/2018) & Gandhi Sagar Hydel Power
Station located at Gandhisagar, Mandsaur (17/3/2018)

62. Aptitude Test & Technical test
(27/2/2018 & 28/2/2018)

63. " Four Wheeler (automotivevehicles)"
(2/5/2018 to 3/5/2018)
Resource Agency : Ram Motors Mandsaur (M.P.)
Resource person: Mr Gopal Mali

64. "Wheel alignment and balancing“(1/5/2018)
• Resource agency: Prabhukripa wheel alignment
center (Ramtekri) Mandsaur

65. Career in government sector

66. Career in government sector

67. INDIAN RAILWAYS
• There are many jobs in Indian Railways for Mechanical Engineers the recruitment for which is
conducted by Railway Recruitment Board. The list of some important position for Mechanical
Engineers are:
Indian Railway Services for Mechanical Engineers like:
• Junior Engineer
• Junior Engineer
• Site Engineer
• Quality Engineer
• Signal Processing Engineer
• Indian Railway Store Service
• Central Power Engineering Service
They can also try in Railways through Railway Recruitment examination. Generally JE and
SSE group C posts are filled through open competitive examinations. In order to become a
Railway JE, one should have the Diploma in Engineering certificate and to become an SSE,

68. INDIAN ARMY, INDIAN NAVY &
INDIAN AIR FORCE
• Mechanical Engineers can also serve their by getting into Indian Navy, Indian Air force
and Indian Army. For this, the Government of India conducts various exams such as:
• UES: This is a University Entry Scheme exam and only final year students are eligible to
apply.CDS: Union Public Service Commission (UPSC) conducts Combined Defense
Services (CDS) examination twice a year for the recruitment of officers into Indian
Military Academy, Officers Training Academy, Indian Naval Academy and Indian Air
ForceAcademy.
• CDS: Union Public Service Commission (UPSC) conducts Combined Defense Services
(CDS) examination twice a year for the recruitment of officers into Indian Military
Academy, Officers Training Academy, Indian Naval Academy and Indian Air Force
Academy.TES: This is Technical Entry Scheme (TES) and only B.Tech applicants can
apply.
• TES: This is Technical Entry Scheme (TES) and only B.Tech applicants can
apply.GDOC: B.Tech students can apply for this exam, which is Ground Duty Officers
Course exam.
• GDOC: B.Tech students can apply for this exam, which is Ground Duty Officers
Course exam.

69. ENGINEERING SERVICES
EXAMINATION
• Every Single Year Union Public Service Commission (UPSC) conducts Engineering Services
Examination, which is the most prestigious examination, held in India. This is also considered
as one of the toughest examinations in India and in the world as well.
Mechanical Engineers are shortlisted for this exam for various services listed below:
• Border Roads Engineering Service (Border Roads Organisation)
• Border Roads Engineering Service (Border Roads Organisation)
• Indian Supply Service (Directorate General of Supply and Disposals)
• Central Engineering Service (Ministry of Road Transport & Highways)
• Indian Railway Stores Service
• Central Electrical & Mechanical Engineering Service (Central Public Works Department)
• Central Power Engineering Service (Central ElectricityAuthority)
• Indian Ordinance Factories Service (IOFS)
• Indian Railway Service of Mechanical Engineers
• Central Water Engineering Service (Central Water Commission)
• Corps of Electrical and Mechanical Engineers (Indian Army)
• Indian Railway Stores Service
• Indian Naval Stores Service (Indian Navy)
• Indian Naval Armament Service (Directorate of NavalArmament)
• Indian Inspection Service (Directorate General of Supply and Disposals)

70. Recreational Moments:
Cultural Program
Some Moments of Annual Fest

71. Recreational Moments: Sports