ME 603 AEP Introduction 1

8/20/2019 ME 603 AEP Introduction 1

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ME 603: Applied Elasticity and Plasticity

INTRODUCTIONINTRODUCTION--11

Prof. S.K.Sahoo


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INTRODUCTION

• Objective – Understand different methods that used to analysestress and strain in solid body.

– Apply various principles to solve problems in a

practical situation and compare its solution with t at o ta ne y so mec an cs approac .

– Analyze stresses inside solid body cause byexternal and internal forces.

– Examine different yield criteria in diverse failuresituations.

– Relate theory of plasticity to manufacturing.


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INTRODUCTION (Cont…)

• Learning Outcome1.0 Can explain the stress, strain, torsion and bending properties.

2.0 Apply the concepts of stress, strain, torsion and bending and deflectionof bar and beam in engineering field

3.0 Calculate and determine the stress, strain and deflection of solid bodythat subjected to external and internal load.

4.0 Enable to design the optimum dimension of the body in a variety of situations where specific properties are required.

5.0 Relates the basic theory of elasticity and plasticity with application of

solid mechanics.6.0 It provides an understanding how the stress-strain characteristics affectultimate failure of materials.

7.0 Able to relate theory of plasticity to design tooling in manufacturinginstead of using ‘thumb rule’.


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TOPICS TO COVER

• Analysis of stress and strain; Notation of stress, Signconvention, Stress tensor, St. Venant principles, Bauschinger effect, Principle of Superposition, Differential equation of Equilibrium, Generalised Hooke’s Law, Compatibility andconstitutive equations; Plane stress and plane strain problems,

• Stress functions; with and without body forces, Applications tos mp e pro ems; pp cat on n rectangu ar an po ar coordinates.

• 3D stress and strain system, Principal stresses and principal

axes, Stress invariants, hydrostatic stress, stress deviator,examples.

• Introduction to complex potentials in two dimensional andaxisymmetric problems; Variational methods; Anisotropicelasticity; Finite deformation elasticity


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TOPICS (Cont…)

• Yield Criteria, Yield surfaces. Deformationand flow theories;

• Theory of plastic constitutive equations;Axis mmetric and s hericall s mmetric

problems;

• Slipline and upper bound theory and

application to simple problems of forgingextrusion, drawing and indentation;

• Introduction to wave propagation in plasticmaterials.


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MappingObjectives/Topics and Outcomes

Analysis of stress and strain, Differential equation of Equilibrium, Compatibility and constitutive equations;Plane stress and plane strain problems

Stress functions; with and without body forces,Applications to simple problems; Application inrectangular and polar coordinates.

Outcome(s)

1.03.0

2.05.0

4.07.0

,stresses and principal axes, Stress invariants,hydrostatic stress, stress deviator, examples.

Theory of plastic constitutive equations; Axisymmetricand spherically symmetric problems

Slipline and upper bound theory and application tosimple problems of forging extrusion, drawing and

indentation

Yield Criteria, Yield surfaces.

Deformation and flow theories4.07.0

4.0

6.0

1.03.0

5.0


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REFERENCES1. Timeshenko & Goodier, Theory of Elasticity - McGraw Hill, 3th

ed. 1982

2. J. Chakrabarty, Applied Plasticity- Springer, New York, 1st ed.,2000

3. Hoffman and Sachs, Theory of Plasticity - McGraw Hill., 2nd ed.

4. Johnson and Mellor, Engineering Plasticity- Van-Nostrand., 1stedition, 1983

5. Computational Elasticity – M Ameen, Narosa Publishing House.

6. Advanced Mechanics of Materials – A P Boresi and R J Schmidt,John Wiley & Sons, Inc.

7. Advanced Mechanics of Solids – L S Srinath, Tata McGraw-Hill


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dx

Example: Find the extension of a uniform cross section bar

subjected to uniformly varying tension due to self weight.

PX= A x (=weight density of bar )d = PX dx / A E;

= PX dx/AE= A x dx/AE

L

0

L

0LX

L

P + dP

P

d

= ( /E) x dx= ( L /2E)

If total weight of bar W= A L = W/AL

=WL/2AE

0

=P/A , / =E , = /L

So, = PL / A E


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F

L

D

d 1

d x

Example: Find the extension of a bar of tapering

cross section from diameter d to D by the load F.

=F/A , / =E , = /L

So, = FL / A E

F

X

d

Bar of Tapering Section:

d1 = d + [(D - d) / L] * X

So, extension of strip of thickness dx is, = F dx / E[(/4){d + [(D - d) / L] * X}2]


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Hence, total extension is,

= 4 F dx /[E {d+kx}2 ]

= - [4F/ E] 1/k [ {1 /(d +kx)}] dx

L

0

L

0

=- [4FL/ E(D-d)] {1/(d+D -d) - 1/d}

= 4FL/( E D d)

Check :- When both diameter is same =dWhen d = D=d =FL/ [( /4)* d2E ] = FL /AE


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PP

X

b2 b1 bx

x

Example: Find the extension of a tapering bar of

uniform thickness t & width varies from b1 to b2.

Bar of Tapering Section:

bx = b1 + [(b2 - b1) / L] * X = b1 + k*x,Where, k = (b2 - b1) / L

= Px / [Et(b1

+ k*X)]


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= P/Et ∫ x / [ (b1 - k*X)],

L = = Px / [Et(b1 - k*X)],

L

0

L

0

0

L

= - P/Etk * loge [ (b1 - k*X)]0 ,

= PLloge(b

1/b

2) / [Et(b

1 – b

2)]


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Example: Find Elongation of a Bar of circular tapering

section due to self weight.from =PL/AE

=Wx*x/(AxE)

now Wx=1/3* AxX

where Wx=Wt.of the bar

x

d

A B

=Wt. density

so = X *x/(3E)so now

L= = X *x/(3E)= /(3E) Xdx= [/3E ] [X2 /2]

= L2/(6E)

L

0

L

0

X

L

0


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Let W=total weight of bar = (1/3)*(/4*d2)L =12W/ (*d2L)

so, =

=2WL/ (*d2E)

=WL/[2*(*d2/4)*E]

=WL /2*A*E


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Example: Find the extension of a bar of uniform strength:(i.e.

stress is constant at all points of the bar) due to self weight.

Down ward force of strip = w*A*dx

w is wt. density

p is stress

Area = A1

dx

L

x

A

B C

DForce = p*A

Force = p*(A+dA)

dxB CA D

comparing force at BC &AD level of strip of thickness dx

p* (A + dA) = p*A + w*A*dx,

dA/A = wdx/p, Integrating logeA = wx/p + C,at x = 0, A = A2 and x = L, A = A1, C = A2

loge(A/A2) = wx/p or A = ewx/p

Area = A2

(where A is cross section area at any level x of bar of uniform strength )


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Example: A rod of length L and cross sectional area A is snugly fitted(no compression

or tension of the spring) between a rigid support at its left and a spring at its right.

The spring has stiffness k. Derive a equation for the stress in the rod due to increase

in temperature of t0C. Thermal expansion coefficient is α.

Strain due to temperature rise: α t

Strain due to compressive force: σ/E

K x = σ A so, x/L = σ A/kL

Both are balanced by strain by spring force = - σA/kL

So, σ/E+ α t = - σA/kL Gives, σ = - (E α t)/(1+EA/kL)

Let P be the balanced force. The free thermal expansion becomes two part,

by elastic force and by spring

- α t L = PL/AE + P/k

Gives, P/A= σ = - (E α t)/(1+EA/kL)


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Example: A material is subjected to tensile stress of σx & σy at right

angle to each other (σx > σy). Find the condition that the resultant

stress makes maximum inclination to its normal.

σxσx

σy

σy

θφ

NR


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Example: A material is subjected to tensile stress of σx & σy at right

angle to each other (σx > σy). Find the condition that the resultant

stress makes maximum inclination to its normal.

222

Cos y x y x

n

2

22

2SinSin

y x x y

s

222

22tan

Cos

Sin

y x y x

y x

n

For max. value of φ wrt θ, tanφ should be max. or 0tan d

σxσx

σy

σy

θφ

NR

φσn

τ

02)2(2

22

2cos22

222

SinSinCos

y x y x y x y x y x

22

2cos222

2SinCos y x y x y x

)90tan(cot

22

2222tan

n

y x

y x y x

Sin

Cos

902

y x

y x

1sin

22

1

Value of θ at Maximum value of φ

2sin22

cos2tan

max

max

max

Cos y x y x

y x

)sin1()sin1(

cos)(

cos

sin

maxmax

max

max

max

y x

y x

max

max

sin1

sin1

y

x

y x

y x

maxsin givessolving


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Example: A plate of uniform thickness 1 cm and dimension 3 x 2 cm is acted upon by

the loads shown. Taking E = 2 x 107 N/cm2, determine x yand . Poisson’s ratio is

0.3. 42 kN

y

18 kN 2 cm 18 kN

x

42 kN 3cm

N N cm

180009000 2/

xcm x cm2 1

y N

cm x cm N cm

42000

3 114000 2/

Hooke’s law in two dimensions states that:

x x y E x

x 1 1

2 109000 0 3 14000 240 10

7

6[ ] [ . ( ]

and y y x E x x

1 1

2 10 14000 0 3 9000 565 1076

[ ] [ . ( ]


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Example: A composite bar made up of aluminum and steel is held between two supports. The bars are stress free at 400c. What will be thestresses in the bars when the temp. drops to 200C, if

(a) the supports are unyielding

(b)the supports come nearer to

each other by 0.1 mm.

Take E al =0.7*105 N/mm2 ;al =23.4*10-6 /0C

Steel Aluminum

60cm 30cm

2 cm2 3 cm2

ES=2.1*105 N/mm2 s =11.7*10-6 /0C

Aal=3 cm2 As=2 cm2


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contraction of steel bar s = (s/Es)*Ls =[600/(2.1*105)]* s

contra.of aluminum bar al = (al/Eal)*Lal =[300/(0.7*105)]* al

Free contraction =Ls s t+ LALAlt

=600*11.7*10-6*(40-20)+300*23.4*10-6*(40-20)=0.2808 mm.

Since contraction is checked tensile stresses will be set up. Force beingsame in both As s=Aal al 2 s= 3 al ==> s=1.5 al

(a) When supports are unyielding, s + al = (free contraction)

=[600/(2.1*105)]* s +[300/(0.7*105)]* al =0.2808 mm

s=1.5 al so, al =32.76 N/mm2(tensile) s =49.14 N/mm2(tensile)(b) Supports are yielding: s + al = ( - 0.1mm

so, al =21.09 N/mm2(tensile) s =31.64 N/mm2(tensile)


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Example: A circular section tapered bar is rigidly fixed as shown infigure. If the temperature is raised by 300 C, calculate the maximumstress in the bar. Take E=2*105 N/mm2 ; =12*10-6 /0C

1.0 m

D 2 = 2 0 0 m m

D 1 = 1 0 0 m m

X dX

P P

AB

all c/s.

Free expansion = L t = 1000*12*10-6*30 =0.36 mm

Force P induced will prevent a expansion of 0.36 mm = 4PL/(E*d1*d2) = L t

or P = (/4)*d1*d2 t E=1130400 N

Now Max. stress = P/(least c/s area) =1130400/(.785*1002) = 144MPa


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Elastic Deformation

1. Initial 2. Small load 3. Unload

bondsstretch

return toinitial

Elastic means reversible, ie, No permanent deformation

F F Linear-

elastic Non-Linear-elastic


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Plastic Deformation (Metals)

1. Initial 2. Small load 3. Unload

planesstillsheared

bondsstretch& planesshear

Plastic means permanent.

F

elastic + plastic plastic

F

linear elastic

elastic plastic


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Elastic Recovery During Plastic Deformation

Total Strain, = elastic + plastic= /E + plastic

•Material has permanent strain when deformed plastically and loadis released. The unloading line is parallel to elastic line as E is a

material property. The amount of strain get back is called elasticrecovery.•If stress is reapplied, material again responds elastically at the beginning up to a new yield point that is higher than the originalyield point.


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Stress-Strain Diagram

4

2

3

5

Elastic

Region Plastic

Region

Strain

Hardening

ultimatetensile

strengthnecking

yieldstrength

UTS

y

Elastic

limit Rate of increase of

Rate of increase of strength is less thanrate of decrease of strength due todecrease of diameter

Strain ( ) ( L/Lo)

rac ure

Elastic regionslope =Young’s (elastic) modulusyield strength

Plastic regionultimate tensile strengthstrain hardeningfracture

εEσ

ε

σE

Proportionallimit

Offset

more than rateof decrease of strength due todecrease of diameter

= 0.002

= 0.2%


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Strain /Work Hardening

• Curve fit to the stress-strain response:

• An increase in y due to plastic deformation.

large

hardening y1

- Locking up of grains => increase in strength- Under plastic strain, grains slipping along boundaries

- We can see this in the true-stress-strain curve also

T K T n

“true” stress( F / A) “true” strain: ln( L/ Lo)

hardening exponent:n = 0.15 (some steels)to n = 0.5 (some coppers)

hardening y0

Applications:- Cold rolling, forging: part isstronger than casting


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• Ductility may be expressed as either percent elongation (%

plastic strain at fracture) or percent reduction in area.• %AR > %EL is ossible if internal voids form in neck.

Ductility, %ELDuctility is a measure of the plastic

deformation that has been sustained at fracture.Measures how much the material can be stretched before fracture

L f Ao

A f Lo

Engineering tensile strain,

Engineeringtensilestress,

smaller %EL(brittle if %EL5%)

100% x A

A A AR

o

f o

100% xl

l l EL

o

o f

A material that suffers very little plastic deformation is brittle.

High ductility: platinum, steel, copper Good ductility: aluminum

Low ductility (brittle): chalk, glass, graphite


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• Stress at which noticeable plastic deformation has occurred.

It is difficult to distinct this point.Hence, many cases stress value at p = 0.002 offsetis taken for this purpose.Stress at this strain value is also called Proof stress.

Yield Strength, y

tensile stress,

y

for 250 mm sample, y is at

= 0.002 = z / z

z = 0.5 mm

engineering strain,

p = 0.002 y = yield strength


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Example: Calculate deflection if the proof stress is applied and

then partially removed to a stress, MPa) E is the Young’smodulus (200 Gpa).

Yield Plastic

Failure

proof stress

Stress

Strainx


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Example: Calculate deflection if the proof stress is applied and

then partially removed to a stress, MPa) E is the Young’smodulus (200 Gpa).

Yield

0.2% proof stress

Plastic

Failure

Stress

Strain0.2%

0.002 /E

The sample is loaded up to the 0.2% proof stress and then unloaded to a stress, kPa)the strain x = 0.2% + /E = 0.002 + 200/200000= 0.002+0.001 = 0.003


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Ideal Stress Strain Curves


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Typical Stress-Strain Curves for Materials

Rubber

Aluminium alloy

Cast Iron


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Tensile Strength, TS

y

F = fracture or

ultimate

strength

r i n g

TS

s s

• Maximum stress on engineering stress-strain curve.

• Metals: occurs when noticeable necking starts.• Polymers: occurs when polymer backbone chains are

aligned and about to break.

strain

Typical response of a metal

Neck – actsas stressconcentrator

e n g i n e

s t r

engineering strain


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• Energy to break or Energy absorbed breaking a unit volume of material• Approximately, the area under the true stress-strain curve.

Toughness

Tensile

small toughness (ceramics)

large toughness (metals)fracture

Brittle fracture: elastic energyDuctile fracture: elastic + plastic energy

very small toughness(unreinforced polymers)

Tensile strain,

stress,

fracture

fracture


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Toughness Again….

• The area under the wholestress/strain curve is theenergy (per unit volume)needed to make it fail.

• But you get some energy back

in elastic recovery (the black

.• The remaining area is the

energy absorbed by thematerial in failing. This is onemeasure of the “toughness”

of the material.

S t r e s s ,

Strain,

X

O

Elastic

recovery


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• Energy to break or Energy absorbed breaking a unit volume of material• Approximately, by the area under the true stress-strain curve.

Toughness

Tensile

small toughness (ceramics)

large toughness (metals)fracture

Brittle fracture: elastic energyDuctile fracture: elastic + plastic energy

very small toughness(unreinforced polymers)

Tensile strain,

stress,

fracture

fracture


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Resilience, U r and Elastic energy

• Ability of a material to store energy

– Energy stored best in elastic region

• If you load up a material in its elastic region, tosome stressthen the area under the line is a measure of the

energy you used to do it. This area is actuallythe ener er unit volume of material in the

If we assume a linear stress-straincurve this simplifies to y yr 2

1U

y

d U r

0

sample. It is called Elastic energy.• This energy is stored in the material and will

be released if you unload it.


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FatigueFracture/failure of a material subjected cyclic stresses

• Fatigue properties are shown onS-N diagrams.

• When the stress is reduced belowthe endurance limit , fatiguefailures do not occur for anynumber of cycles.

• A member may fail due to fatigueat stress levels significantly belowthe ultimate strength if subjectedto many loading cycles.


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Non-Linear Elasticity, Hysteresis

• In some materials (e.g. some polymers) the stress/strain line iscurved in the elastic region…

• …and sometimes the loading and .

• In that case E is not constant…

• …and some energy is lost, (given by

the area between the lines). This iscalled hysteresis.


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Residual Stresses

• When a single structural element is loaded uniformly

beyond its yield stress and then unloaded, it is permanentlydeformed but all stresses disappear. This is not the generalresult.

• Residual stresses will remain in a structure after loading

and unloading if

• Residual stresses also result from the uneven heating or cooling of structures or structural elements

- only part of the structure undergoes plastic deformation

- different parts of the structure undergo different plasticdeformations


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Bauschinger Effect

• The Bauschinger effect refers toa decrease in the compressiveyield stress due to work hardening in tension.

• It can also refer to a decrease in

Tension stress-strain curve

Unload

work hardening in compression.

• Work hardening can be used to

increase the yield strength of amaterial, but it does so at thecost of a lower yield stress inthe reversed direction of loading.

stress strain curve

in compression

Actual compression

stress-strain curve

following tensile work

hardening

c c


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Principle of Superposition

• It states that the effects of several actions taking placesimultaneously can be reproduced exactly by addingthe effect of each action separately.

• The principle is general and has wide applications

• (i) The structure is elastic• (ii) The stress-strain relationship is linear

• (iii) The deformations are small.


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Saint-Venant’s Principle

• Loads transmitted through rigid plates result in uniform distributionof stress and strain.

• Concentrated loads result in largestresses in the vicinity of the loadapplication point.

• Saint-Venant’s Principle:Stress distribution may be assumedindependent of the mode of loadapplication except in the immediatevicinity of load application points.

• Stress and strain distributions become uniform at a relatively shortdistance from the load application

points.


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Saint-Venant’s Principle

The Stress, Strain and Displacement Fields Due to TwoDifferent Statically Equivalent Force Distributions onParts of the Body Far Away From the Loading Points AreApproximately the Same.

x

P

x

P/2 P/2

y y

xy y

x

xy y

x

Stresses Approximately Equal


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Stress - Strain Terminology


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Engineering Stress

• Shear stress, :

Area, A

F t

F s

F • Tensile stress, :

Area, A

F t

Stress has units:

N/m2 or kgf/cm2 or Pa

F t F

F s

= F s

Ao

original area

before loading

F t

= F t

Ao2m

N=


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• Tensile strain: • Lateral strain:

Engineering Strain

Lo

LL

w o

/2

Low o

• Shear strain:

90º

90º -y

x = = tan = x /y

L/2


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When pure shear acts on an element, the

element deforms into a rhombic shape.

For convenience the element is rotated by anangle and represented as shown.

A’ y ’

Shear strain…

Ξ

τ x

A

xD C

For small angles = → tan = AA’

AD(radians)


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A pure shear strain is produced in torsion.

B

’

=

AB

AA’

A’ A

Shear strain…

A A’

L

θ: Angle of twist of radial line AB to position A’B

r: radius of cross-sectional area

=r

=Lr θ


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Torsion test : Modulus of rigidity : Shear stress

L

A

A’

T

T

LT

T

D

d

T = Applied torque,J = Polar moment of inertiaJ = r 2 dA

Cylindrical shell: J = D4

d4

)/32R= Outer radiusr = radius of consideration

Shear strain = = r /LAngle of twist: = TL/GJShear stress: = Tr/J

Maximum shear stress = max = TR/J

G: Modulus of rigidity

= G

= =


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Tensile test : Young’s modulus : Tensile strength

Final

Necking

In the linear elastic range: Hooke’s law: /e = E or, e

E : Young’s modulus

rac ure


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Shear strength and Tensile strength

[approximate relation between shear and tensile strengths]

Material Tensile-Relation Yield-Relation

Ultimate Tensile Strength = Su Ultimate Shear Strength = SsuTensile Yield Strength = Syp Shear yield point = Ssyp

Wrought Steel & alloy steel Ssu ≈ 0.75 x Su Ssyp = Approx 0,58 x Syp

Ductile Iron Ssu ≈ 0.90 x Su Ssyp = Approx 0,75 x Syp

Cast Iron Ssu ≈ 1.3 x Su -

Copper & alloys Ssu ≈ [0.6-0.9] x Su -

Aluminum & alloys Ssu ≈ 0.65 xSu Ssyp = Approx 0,55 x Syp


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Thermal Stresses

• A temperature change results in a change in length or thermal strain. There is no stress associated with thethermal strain unless the elongation is restrained bythe supports.

PL

• Treat the additional support as redundant and applythe principle of superposition.

coef.expansionthermal

AE

P T

0

0

AE

PL LT

P T

• The thermal deformation and the deformation fromthe redundant support must be compatible.

T E A

P

T AE P P T

0


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Poisson's ratio,• Poisson's ratio, nu :

L L

ametals: ~ 0.33ceramics: ~ 0.25 polymers: ~ 0.40

Ratio of lateral to axial strain calledPoisson's ratio .

F

Ao

/2

Units:: dimensionless

> 0.50 Volume decrease with tensile stress

Volume increase with compressive stress

-

L/2

owo

Limit of

-1 < ν ≤ 0.5


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Volume of cube is (assume the extensions are small)

Let consider the volume change for a cube

and since it follows that ≤0.5

Change in volume

0


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• Elastic Shearmodulus, G :

G

= G

Other Elastic Properties and Relationship

simpletorsiontest

M

M

•

• Special relations for isotropic materials:

2(1 )

E G

3(1 2 )

E K

modulus, K:

pressuretest: Init.

vol =V o.Vol chg.= V

P P P = - K V

V o

V

K V o


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Relationship between Elastic Modulus (E) and Bulk Modulus, K

x x y z

x y z

x

E

F o r h y d r o s t a t ic s t r e s s

i e

E E

1

12 1 2

( )

,

. .

y z

v x y z

v

v

v

S im ila rly a n d a r e e a c h E

V olu m e t ric st ra in

E

E

B u l k M o d u l u s K V olu m e tric o r h yd ro st a t ic s t re ss

V olu m e tr ic st ra in

i e E K a n d K E

1 2

31 2

3 1 2

3 1 23 1 2

,

,

. .


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Example: A bar made of mild steel has the dimensions shown. If an axialforce of P is applied to the bar, determine the change in its dimensionsafter applying the load. The material is in elastic zone.

E= 200 Gpa

= P/A Pa100.161080 6

3

P z

z = / E

Lz = L * z

x = y = - zLx = L * xLy = L * y

..

mm/m108010200

100.16 69

6

st

z z

E

(Ans)m1205.11080 6z z z L

m/m6.25108032.0 6 z st y x v

(Ans)m28.105.0106.25

(Ans)m56.21.0106.25

6

6

y y y

x x x

L

L


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Normal & Shear components of stress 2D Case

xy

yy

xy

xx

xy

xx n x x

x y

y x y

y

xy

yy

2cos2sin

2

2sin2cos22

xy

y x

xy

y x y x

n

Considering the conditions for equilibriumof a triangular/ prismatic element


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Principal StressesPrincipal stresses: Maximum and minimum normal stresses

Principal planes : Planes on which the principal stresses act

2sin2cos22

xy

y x y x

n

0 d n

2cos22sin2 x y x

: p The angle defines the orientation of the principal planes.

y x

xy p

22tan

- Principal directions are orthogonal to each other -- No Shear stress along principal directions


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2

2

22 xy

y x y x

n I

Principal Stresses…..

OR

II I 2

2

22 xy

y x y x

n II


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Maximum Shear Stresses

22

212

2

max

xy

y x

1 1

2 1

4

4

s p

s p

The angle defines the orientation of the maximum shear stress planes.

- Maximum shear stress directions are orthogonal to each other -- Maximum Shear stress direction is 450 to principal directions


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Plane (2D) Strain

2cos2

2sin22

2sin2

2cos22

11

1

xy y x y x

xy y x y x

x

Normal and shear strain on a plane making angle with y-axis

Differentiating wrt ,

22

22

222

222

xy y x y x II

xy y x y x

I

Principal Strains:

tan 2 xy P x y

we have:

M h ’ Ci l f Pl St


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Mohr’s Circle for Plane Stress

• Mohr’s circle for plane stress may be appliedwith simple geometric considerations to

estimate graphically the principal stresses,max. shear stress, normal & shear stress on adefined plane.

• For a known state of plane stress plot the points P and P’ and construct thecircle centered at C .

xy y x ,,

2

2

22 xy

y x y x

a ve R

• The principal stresses are obtained at Q & Q’.

y x

xy

pa ve II I R

22tan&,

•CP is the x-axis and CP’ is y-axis•Angle doubles and rotate in opposite direction

• To find stress at an angle with respect to the yaxis, a new line CB may be drawn at an angle2 with respect to C P location point B on circle

gives the stress values.

I t ti t d M h ’ Ci l


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Instruction to draw Mohr’s Circle

1. Identify the stresses σx, σy, and τxy = τyx with the proper sign.2. Draw a set of σn - τ coordinate axes with σn being positive to the right

and τ being positive in the upward direction. Choose an appropriate scale for theeach axis. Locate the σx ( point L) and σy (point M) on σn axis.

3. Plot the stresses on the x- face of the element in this coordinate system ( point P ). Repeat the process for the y -face ( point P’ ).

4. Draw a line between the two point P and P’. The point where this linecrosses the σn axis establishes the center of the circle.

5. Draw the complete circle. Radius gives value of maximum shear stress. 6. The line from the center of the circle to point P identifies the x axis or

reference axis for angle measurements (i.e. θ = 0). Note: The angle between the reference axis and the σn axis is equal to 2θ p, (

angle PCQ) one of the principal plane angle.7. The circle cross the σn axis is at point Q and Q’. OQ represents first

principal stress σI and OQ’ represents second principal stress σII. The anglemade by CQ’ with CP gives other principal plane angle 2θ p +.

8. To find stress at an angle with respect to the y axis, a new line CB may be drawn at an angle 2 with respect to C P location point B on circle gives the stress values, ie, BN= σn and BS= τ

M h ’ Ci l f Pl St


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Mohr’s Circle for Plane Stress

• Mohr’s circle for centric axial loading:

0, xy y x A

P

A

P xy y x

2

• Mohr’s circle for torsional loading:

J

Tc xy y x 0 0 xy y x

J

Tc

M h ’ Ci l f Pl St i


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Mohr’s Circle for Plane Strain

• The equations for the transformation of plane strain are of the same form as theequations for the transformation of planestress - Mohr’s circle techniques apply.

• Abscissa for the center C and radius R ,22

222

xy y x y x

a ve R

• Principal axes of strain and principal strains,

R R a vea ve

y x

xy p

minmax

2tan

22max 2 xy y x R

• Maximum in-plane shearing strain,

St i Thi W ll d P V l


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Stresses in Thin-Walled Pressure Vessels• Cylindrical vessel with principal stresses

1 = hoop stress

2 = longitudinal stress

pr

xr p xt F z

1 220

• Hoop stress:

t

21

2

2

2

2

2

20

t

pr

r prt F x

• Longitudinal stress:

St i Thi W ll d P V l


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Stresses in Thin-Walled Pressure Vessels

• Points A and B correspond to hoop stress, 1,and longitudinal stress, 2

• Maximum in-plane shearing stress:

t

pr

42

12) planeinmax(

• Maximum out-of-plane shearing stresscorresponds to a 45o rotation of the planestress element around a longitudinal axis

t

pr

22max

Stresses in Thin Walled Pressure Vessels


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Stresses in Thin-Walled Pressure Vessels

• Spherical pressure vessel:

t

pr

221

• Mohr’s circle for in-planetransformations reduces to a point

0 plane)-max(in

• Maximum out-of-plane shearingstress

t

pr

412

1max

True Stress () and Strain ()


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True Stress () and Strain ()

i A

P

0

ln

0 L

L

L

dL L

L

Ai → instantaneous area P → applied load

d → strain increament

P s 0 L L Le i

The definitions of true stress and true strain are based oninstantaneous values of area (Ai) and length (Li).

Final

NeckingFracture

• In engineering/nominal stress/strain since we divide by

original area, A0 and original length L0.

L

dLd →

0 00

engg strain L/Lo true strain ln(L/Lo)

t r u e s t r e s s P / A i

e n g g s t r e s s

P / A

o

fracture

fracture

• nce vo ume rema ns constant n t e p ast cregion of the test, true strain can be expressed as

D

D

D

D

A

A

L

L 02

00

0

ln2lnlnln

AL L A 00

DL L D 00

If, specimen is a round rod and D indicate

diameter

Comparison between “Engineering” and “True” quantities


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e)( ε 1ln

e) s( 1

Comparison between Engineering and True quantities

00 0 i i

0

From volume constancy A L =A L i

i

A L

A L

Valid tillnecking starts

Let A0, L0 are original dimensions and Ai and Li are dimensionsat a particular instant.

0

0 0 0

1 1 (1 )i i

i

A L L P s s s e

A A L L

0

ln

0 L

L

L

dL L

L

i A

P

0

ln 1 1 ln(1+e) L

L

Comparison between true strain and engineering strain

True strain () 0.01 0.10 0.20 0.50 1.0 2.0 3.0 4.0Engineering strain (e) 0.01 0.105 0.22 0.65 1.72 6.39 19.09 53.6

Note that for strains of about 0.4, ‘true’ and ‘engineering’ strains can be assumed to be equal. At large strains the deviations between the values are large.

As we shall see that during the tension test localized plastic deformation occurs after some strain (called necking). This leads to inhomogeneity in the stress across thelength of the sample and under such circumstances true stress should be used.

T t t t i


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True stress–true strain curves• A typical true stress–true strain curve is typically approximated

by the equation.

K n

“true” stress ( P / Ai) “true” strain: ln( L/ Lo)

hardening exponentn = 0.15 (some steels)to n = 0.5 (some coppers)

strength coefficient

• Note that ‘n’ is always positive and that the slope decreaseswith increasing strain.

YYield StressYf flow stress

K strength coefficient

n K


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Example: A cylinder pressure vessel 10 m long has closedends, a wall thickness of 5 mm, and a diameter at mid-

thickness of 3 mm. If the vessel is filled with air to a pressureof 2 MPa, how much do the length, diameter, and wallthickness change, and in each case state whether the change isan increase or a decrease. The vessel is made of a steel havingelastic modulus E = 200,000 MPa and the Poisson’s ratio =

0.3. Ne lect an effects associated with the details of how the ends are attached.

z-axis is normal to the surface


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The ratio of radius to thickness, r/t, is such that it is reasonable to employ the thin walledtube assumption, and the resulting plane stress equations.Denoting the pressure as p, we have

M Pamm

mm M Pa

t

pr

M Pamm

mm M Pa

t

pr

y

x

6005

)1500)(2(

300)5(2

)1500)(2(

2

z

0 z

The value of varies from -p on the insidewall to zero on the outside, and for a thinwalled tube is everywhere sufficiently smallthat can be used. Substitute thesestresses, and the known E and v into Hooke’sLaw, which gives 410*00.6 x

310*55.2 y

310*35.1 z

These strains are related to the changes in length , circumference, diameter , and thickness , as follows:

Substituting the strains from above and the known dimensions gives

Thus, there are small increases in length and diameter, and a tinydecrease in the wall thickness.

L

)( d d t

L

L x

d

d

d

d y

)(

t

t z

mm L 6 mmd 65.7 310*75.6 t

Example


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Example

Answer:


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Example


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Example

Answer:

Example


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Example

Answer:


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Example


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Example

Answer:


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Documents

ME 603 AEP Introduction 1