Transcript Stress and strain - TATI University College

INTRODUCTION
Name : Mohamad Redhwan Abd Aziz
Post
: Lecturer @ DEAN CENTER OF HND STUDIES
Subject : Solid Mechanics
Code : BME 2033
Room : CENTER OF HND STUDIES OFFICE
H/P No. : 019-2579663
W/SITE : Http://tatiuc.edu.my/redhwan
email : [email protected]
INTRODUCTION (Cont..)
 Monday
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11-1pm (L13/Mech lab)
Tutorial
Lab
 Tuesday
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Lecture
10-12pm (L9)
INTRODUCTION (Cont..)
 BME
2023 SOLID MECHANICS
 3 CREDIT HOURS
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2 Hours for lecture
2 hours for lab/tutorial
 Assessment/marks
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distribution
Quizzes/Assignments
Test
Lab report
Final Exam
- 10%
- 30%
- 20%
- 40%
INTRODUCTION (Cont…)
 Objective
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Understand different methods that used to
analyse stress and strain in solid body
Apply various principles to solve problems in
solid mechanics
Analyse forces of solid body cause by
external force
Analyse the result of solid mechanics
experiments
INTRODUCTION (Cont…)
 Learning
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Outcome
Apply the concepts of stress, strain, torsion and bending and
deflection of bar and beam in engineering field
Explain the stress, strain, torsion and bending
Calculate and determine the stress, strain and bending of solid
body that subjected to external and internal load
Use solid mechanic apparatus and analyse the experiments
result
Work in-group that relates the basic theory with application of
solid mechanics
TOPICS
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Topic cover
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Stress and strain
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Introduction to stress and strain, stress strain diagram
Elasticity and plasticity and Hooke’s law
Shear Stress and Shear strain
Load and stress limit
Axial force and deflection of body
Torsion
• Introduction, round bar torsion, non-uniform torsion.
• Relation between Young’s Modulus E,  and G
• Power transmission on round bar
TOPICS (Cont…)
 Topic
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cover
Shear Force and bending moment
• Introduction, types of beam and load
• Shear force and bending moment
• Relation between load, shear force and bending moment
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Bending Stress
• Introduction, Simple bending theory
• Area of 2nd moment, parallel axis theorem
• Deflection of composite beam
TOPICS (Cont…)
 Topic
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cover
Non Simetric Bending
• Introduction, non-simetric bending
• Product of 2nd moment area, determination of stress
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Shear Stress in beam
• Introduction, Stream of shear force
• Shear stress and shear strain in edge beam
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Deflection of Beam
• Introduction
• Equation of elastic curve, slope equation and integral
deflection
• Statically indeterminate Beams and shaft
REFERENCES
1.
2.
3.
James M. Gere (2006) “ Mechanics of Materials”. 6th
Edition, Thompson
R.C. Hibbeler (2003) “ Mechanics of Materials”. 5th
Edition, Prentice Hall
Raymond Parnes (2001), “Solid Mechanics in
Engineering”. John Willey and Son
Stress and strain
DIRECT STRESS
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
When a force is applied to an elastic body, the body deforms. The
way in which the body deforms depends upon the type of force
applied to it.
Compression force makes the body shorter.
A tensile force makes the body longer
Tensile and compressive forces are called DIRECT FORCES
Stress is the force per unit area upon which it acts.
Force F
Stress   

Area
A
….. Unit is Pascal (Pa) or N / m 2
( Simbol – Sigma)
Note: Most of engineering fields used kPa, MPa, GPa.
DIRECT STRAIN ,

In each case, a force F produces a deformation x. In engineering, we
usually change this force into stress and the deformation into strain
and we define these as follows:
Strain is the deformation per unit of the original length.
x
Strain   
L
The
symbol

called EPSILON
Strain has no unit’s since it is a ratio of length to length. Most
engineering materials do not stretch very mush before they become
damages, so strain values are very small figures. It is quite normal to
change small numbers in to the exponent for 10-6( micro strain).
MODULUS OF ELASTICITY (E)
•Elastic materials always spring back into shape when released.
They also obey HOOKE’s LAW.
•This is the law of spring which states that deformation is directly
proportional to the force. F/x = stiffness = kN/m
•The stiffness is different for the different material and different sizes of the
material. We may eliminate the size by using stress and strain instead of
force and deformation:
•If F and x is refer to the direct stress and strain , then
F  A
x  L
F A
and
hence

x L
FL 

Ax 
•The stiffness is now in terms of stress and strain only and this
constant is called the MODULUS of ELASTICITY (E)
E
FL 

Ax 
• A graph of stress against strain will be straight line with
gradient of E. The units of E are the same as the unit of
stress.
ULTIMATE TENSILE STRESS
•If a material is stretched until it breaks, the tensile stress has
reached the absolute limit and this stress level is called the
ultimate tensile stress.
STRESS STRAIN DIAGRAM
STRESS STRAIN DIAGRAM
Elastic behaviour
The curve is straight line trough out most of the region
Stress is proportional with strain
Material to be linearly elastic
Proportional limit
The upper limit to linear line
The material still respond elastically
The curve tend to bend and flatten out
Elastic limit
Upon reaching this point, if load is remove, the
specimen still return to original shape
STRESS STRAIN DIAGRAM
Yielding
 A Slight increase in stress above the elastic limit will
result in breakdown of the material and cause it to
deform permanently.
This behaviour is called yielding
The stress that cause = YIELD STRESS@YIELD
POINT
Plastic deformation
Once yield point is reached, the specimen will
elongate (Strain) without any increase in load
Material in this state = perfectly plastic
STRESS STRAIN DIAGRAM
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STRAIN HARDENING
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When yielding has ended, further load applied, resulting in a curve
that rises continuously
Become flat when reached ULTIMATE STRESS
The rise in the curve = STRAIN HARDENING
While specimen is elongating, its cross sectional will decrease
The decrease is fairly uniform
NECKING
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At the ultimate stress, the cross sectional area begins its localised
region of specimen
it is caused by slip planes formed within material
Actual strain produced by shear strain
As a result, “neck” tend to form
Smaller area can only carry lesser load, hence curve donward
Specimen break at FRACTURE STRESS
SHEAR STRESS

•Shear force is a force applied sideways on the material (transversely
loaded).
When a pair of shears cut a material
When a material is punched
When a beam has a transverse load
Shear stress is the force per unit area carrying the load. This
means the cross sectional area of the material being cut, the
beam and pin.
F
•Shear stress,  
A
and symbol is called Tau
The sign convention for shear force and stress is based on how it
shears the materials as shown below.
SHEAR STRAIN

The force causes the material to deform as shown. The shear strain
is defined as the ratio of the distance deformed to the height
x
L
.

Since this is a very small angle ,
Shear strain
x
 
L
( symbol called
Gamma)
we can say that :
MODULUS OF RIGIDITY (G)
•If we conduct an experiment and measure x for various values of F,
we would find that if the material is elastic, it behave like spring and
so long as we do not damage the material by using too big force,
the graph of F and x is straight line as shown.
The gradient of the graph is constant so
F
 cons tan t
x
and this is the spring stiffness of the block in N/m.
•If we divide F by area A and x by the height L, the relationship is
still a constant and we get
•If we divide F by area A and x by the height L, the relationship is
still a constant and we get
F
A  F x L  FL  con tant
x
A x Ax
L
Where
F

A
x
 
L
then
F L FL 
x 
  con tant
A x Ax 
This constant will have a special value for each elastic material
and is called the Modulus of Rigidity (G).


G
ULTIMATE SHEAR STRESS
If a material is sheared beyond a certain limit and it becomes
permanently distorted and does not spring all the way back to its
original shape, the elastic limit has been exceeded.
If the material stressed to the limit so that it parts into two, the
ultimate limit has been reached.

The ultimate shear stress has symbol
and this value is used
to calculate the force needed by shears and punches.
DOUBLE SHEAR
Consider a pin joint with a support on both ends as shown. This
is called CLEVIS and CLEVIS PIN
 By balance of force, the force in the two supports is F/2 each
The area sheared is twice the cross section of the pin
So it takes twice as much force to break the pin as for a case of
single shear
Double shear arrangements doubles the maximum force
allowed in the pin
LOAD AND STRESS LIMIT
DESIGN CONSIDERATION
Will help engineers with their important task in Designing
structural/machine that is SAFE and ECONOMICALLY perform for
a specified function
DETERMINATION OF ULTIMATE STRENGTH
An important element to be considered by a designer is how the
material that has been selected will behave under a load
This is determined by performing specific test (e.g. Tensile test)
ULTIMATE FORCE (PU)= The largest force that may be applied
to the specimen is reached, and the specimen either breaks or
begins to carry less load
ULTIMATE NORMAL STRESS
(U) = ULTIMATE FORCE(PU) /AREA
ALLOWABLE LOAD / ALLOWABLE STRESS
Max load that a structural member/machine component will be allowed
to carry under normal conditions of utilisation is considerably smaller
than the ultimate load
This smaller load = Allowable load / Working load / Design load
Only a fraction of ultimate load capacity of the member is utilised when
allowable load is applied
The remaining portion of the load-carrying capacity of the member is
kept in reserve to assure its safe performance
The ratio of the ultimate load/allowable load is used to define FACTOR
OF SAFETY
FACTOR OF SAFETY = ULTIMATE LOAD/ALLOWABLE LOAD
@
FACTOR OF SAFETY = ULTIMATE STRESS/ALLOWABLE STRESS
SELECTION OF F.S.
1. Variations that may occur in the properties of the member under
considerations
2. The number of loading that may be expected during the life of the
structural/machine
3. The type of loading that are planned for in the design, or that may
occur in the future
4. The type of failure that may occur
5. Uncertainty due to the methods of analysis
6. Deterioration that may occur in the future because of poor
maintenance / because of unpreventable natural causes
7. The importance of a given member to the integrity of the whole
structure
WORKED EXAMPLE 8
0.6 m
SOLUTION
SOLUTION
SELF ASSESSMENT NO. 5
AXIAL FORCE & DEFLECTION OF BODY
Deformations of members under axial loading
If the resulting axial stress does not exceed the proportional limit of
the material, Hooke’s Law may be applied   E
Then deformation (x / ) can be written as
FL

AE
WORKED EXAMPLE 9
0.4 m
WORKED EXAMPLE 9
WORKED EXAMPLE 9
SELF ASSESSMENT NO. 6