Stress-Strain Notes

8/11/2019 Stress-Strain Notes

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Fundamentals of mechanics andstrength of materials

Ashis Mallick ME 13101 (Solid Mechanics)Indian School of Mines

shis allick

Department of Mechanical EngineeringIndian School of Mines


LectureLecture PlanPlan1. The principles and basic concepts of mechanics of materials 12. Stress strain behavior of engineering materials 23. Concept of stress and strain field 14

Stress strain transformation Hooks law and compatibility conditions Mohrs circle representation for plane stress and plane strain

Thermal stresses and strain Volumetric stress and strain4. Stresses in pressure vessels: thin, thick and compound cylinders 65. Beam Analysis: 8

Deflection in beams Statically indeterminate beam analysis Strain energy concept for structural members Stresses in beam

6. Torsion of a circular members and thin walled walled tubes 57. Springs : Helical and leaf spring 48. Failure theories: Buckling of columns 6

Total contact Hour (App.) 46


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Reading MaterialsReading Materials

1 . Elements of Strength of Materials by S.P Timoshenko & D.H. Young East West Press 2. Strength of Materials (Part 1 & 2) by S.P. Timoshenko CBS Publication3. Mechanics of Materials by J. M. Gere Brooks / Cole

Introduction of Continuum MechanicsIntroduction of Continuum Mechanics


A steel ball dropped in air, oil, on a cotton wad and on a steel block behavesdifferently

What will happen ?


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Containing air

Containing oil

Containing cotton

marble accelerates downward rapidly

marble moves downward relatively slowly


Containing steel block

Why these materials behave differently?

How would we describe this difference in behaviour?

it hardly moves

Atomic distance Air > Oil > Cotton > Steel block

Materials are composed by atoms


o escr e e e av our o ese our ma er a s we cou eg n w escr ng the atoms and their arrangement.

1 m 1m 1m cube contains approximately 30 billion atoms

Largest computer can handle only 10 billion atoms

Change the approach

Treat all materials as continuous media

Advantages of this approach

No need to track the large numbers of atoms that compose our body.

Body can treat using continuous variables measuring the force, elongation etc.

Changes of continuous variables can describe by partial differential equations.


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In the process of moving from atomic to continuous description of matter we loss many informations.

Disadvantages of this approach


distance between atoms

structures of atoms etc.

We need to explicitly build in this information for us to be able to describe the body under consideration. This is called constitutive information.

It is an object which is not liquid or gas. Is it a sufficient definition or scientificdefinition of solid?


SolidSolid

c en c e n on:A solid is an object that has three dimensions (length, width and height),in which the molecules vibrate about fixed positions and cannot migrateto other positions in the substance. Unlike a gas or liquid, a solid has afixed shape, and unlike a gas, a solid has a fixed volume. In most solids(with exceptions such as glass), the molecules are arranged in crystallattices of various sizes.

It has definite size and shape. It cant flow like liquid or gas. Molecules are tightly packed as compare to liquid or gas.

Examples: spheres, cubes, pyramids, cylinders etc.


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Mechanics is the study of deformation and motion of bodies under applied forces .MechanicsMechanics

Deal mechanics problem:

Describe the bod


the location of the body at various times and be able to track it along its motion known as kinematics .

Ensure that some basic balance laws are satisfiedmass of the body is conserved

the energy of the entire system is conserved and

Newtons laws of motion are being satisfied

Description of the actual material in the form of constitutive equations satisfy the second law of thermodynamics

satisfy other symmetry principles which are particular to the material under consideration

finally, we need to consider the actual geometry of the problem and combine all the above information results in a boundary value problem.

Vector and Tensor AnalysisVector and Tensor Analysis

We start to lay the framework for a three dimensional description of thecontinuum. The language of three dimensional continuum mechanics is vectorand tensor al ebra and calculus .


Notation

Direct notation Scalars are denoted by lower and upper case light face letters (a, b, , , A, B) Vectors are denoted by lower case bold face letters (a , b , ) Tensors are represented by upper case bold face letters (A, B, )

Index notation

All represented by light face with appropriated number of subscripts (e.g scalars a , b , c; vectors a i , b i , ci ; tensors Aij , Bij , C ij ).


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Material pointA body size vanishingly small compared with the size of the areasin which it moves. It is thus endowed with a certain geometricpoint mass


Perfectly rigid body(indeformable), it is the body which points do not change withinthe forces

System of material pointsIt arises from the breakdown of the body in the infinitely growingnumber of material points forming a continuum material.

Solid mechanics is concerned with the stressing, deformation and failure of solid materials and structures.

FCD = 25 kN and FCB = 20 kN

Engineering Mechanics to Solid MechanicsEngineering Mechanics to Solid Mechanics


The interatomic distance between two neighboring atoms will increase in CD. The rod will reach a new equilibrium. Thus, we can write:

1

N

i CDi

F F Resultant internal force = external applied force


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Consider an element of continuous (no voids) and cohesive (no cracks, breaks anddefects) material subjected to a number of externally applied loads as shown in

Concept of stress at a pointConcept of stress at a point

0lim A

F A

. .

Stress : Stress is the intensity of the internal resisting force (due to external loading) on a specific plane passing through a point.


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Qn

Q7

Q6Qn

Q7

Q6


Q1

Q5 Q5Q

Q n

n

Q2 Q34

Q1

Q2 Q3Q40

lim A

Q A

Normal and Shear Stress:

Now lets resolve the force F in normal and tangential direction of the acting


area. e n ens y o e orce or orce per un area ac ng norma y o sec on A is called Normal Stress, (sigma), and it is expressed as:

0

lim nn A

F

A

If this stress pulls on the area it is referred as Tensile Stress and defined as Positive. If it pushes on the area it is called Compressive Stress and defined as Negative.

0( ) lim t t A

F A

The intensity or force per unit area acting tangentially to A is called Shear Stress, (tau) , and it is expressed as:


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To begin, we only look at beams that carry tensile or compressive loads and whichare long and slender. Such beams can then be assumed to carry a constant stress,and the Eq. for stress is simplified to:

F F

Average Normal


We call this either Average Normal Stress or Uniform Uniaxial Stress.

F A

Units of StressThe units in the SI system is the Newton per square meter or Pascal, i.e. : Pa = N/m 2

In engineering, Pa seems too small, so we usually use: Kilo Pascal KPa (=Pa10 3 ) e.g. 20,000Pa=20kPaMega Pascal MPa (=Pa10 6) e.g. 20,000,000Pa=20MPaGiga Pascal GPa (=Pa10 9) e.g. 20,000,000,000Pa=20GPa

F F

F = 80 kN

A 100 mm 2

= 800 N/mm 2

= 800 N/mm 2


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Whenever a body is subjected to external force, its shape and size will be changed. These changes are referred as deformations. Due to these deformations the body may be either elongate ( positive) or shortened (negative) as shown in Fig.

Concept of strain at a pointConcept of strain at a point

F4

F

P

Q

SPP

Q

S

F1F2

F3

n

0limS

S S S

Normal StrainThe elongation (+ve) or contrac on ( ve) of a body per unit length is termed Strain.


verage orma ra nIf the stress in the body is everywhere constant, in other words, the deformation is uniform in the material (e.g. uniform uniaxial tension or compression) then the strain can be measured by its average value:

i.e. the chan e in len th of the bod over its ori inal len th,

deformed originalave

original

L L L L L


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StressStress--Strain relationship and Hooks lawStrain relationship and Hooks law


1

ln(1 )true e


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(a) Ductile (Cup cone fracture)

Very ductileExam: Pb, Au

Moderate ductileExam: Al

BrittleExam: Ceramic, glass at room temp

(b) Brittle fracture


Ductile FractureDuctile Fracture


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1.2.2: Calculate the compressive stress c inthe circular piston rod when a force P = 40 N isapplied to the brake pedal. Assume that theline of action of the force P is parallel to thepiston rod, which has diameter 5 mm. Also, the


other dimensions shown in the figure ( 50 mmand 225 mm) are measured perpendicular tothe line of action of the force P. Ans.: 11.2

1.8.2 A steel pipe having yield stress = 270Mpa is to carry an axial compressive load P =1200 kN. A factor of safety of 1.8 against yieldingis to be used. If the thickness t of the pipe is tobe one ei ht of its outer diameter, what is the


maximum required outer diameter d min ? Ans. 153


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1.8.4: Two bars of rectangular cross section (thickness t = 15 mm) are connected by a bolt in the manner shownin the Fig. The allowable shear stress in the bolt is 90MPa and allowable bearing stress between the bolt andthe bars is 150 Mpa.


If the tensile load P = 31 kN, what is the minimum required diameter d min of the bolt? Ans: 14.8

2.2.12 The horizontal rigid beam ABCD issupported by vertical bars BE and CF and isloaded by vertical forces P1 = 400 kN and P2 =360 kN acting at points A and D, respectively.Bars BE and CE are made of steel ( E = 200 GPa)


BE = ,mm 2 and ACF = 9,280 mm 2 . The distancebetween the various points on the bars areshown in Fig. Determine the verticaldisplacement A and D of points A and D,respectively.

A = 0.2 mm and D = 0.88 mm


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Shear strains measure changes of angles as the material distort in response to shear stress.To define shear strains it is necessary to look at two directions that form the plane that undergoes shear distortion. Therefore a one dimensional view is insufficient to describe

Shear strainsShear strains

what happens. It takes two to shear.

Poisson's ratioPoisson's ratio


Poissons ratio is defined as ratio of lateral strain to axial strain :

The ve sign is introduced for convenience so that comes out positive. For structural

materials

lies in

the

range

0.0

<

0.5.

For

most

metals

0.250.35.

For

concrete

and ceramics, 0.10. For cork 0. For rubber, 0.5 to 3 places. A material for which = 0.5 is called incompressible.


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Imagine that threetension tests, labeled (1), (2) and (3) respectively, areconducted along x, y and z, respectively.


Strain to Stress Relationship:Strain to Stress Relationship:

(1) Pulling the material by applying xx along x will produce following normal strains:

1 1 1( ) , ( ) , ( ) xx xx xx

xx yy zz E E E

(2) Pulling the material by applying yy along y will produce following normal strains:

yy yy yy , , yy zz xx E E E

(3) Pulling the material by applying zz along z will produce following normal strains:

3 3 3( ) , ( ) , ( ) zz zz zz

zz xx yy E E E

In the general case the cube is subjected to combined normal stresses xx , yy and zz . Since we assumed that the material is linearly elastic, the combined strainscan be obtained by superposition of the foregoing results:

y yy1 2 3( ) ( ) ( )

yy xx zz xx xx xx xx


x

xx

zz

1 2 3( ) ( ) ( ) yy xx zz

yy yy yy yy E E E

1 2 3( ) ( ) ( ) yy xx zz

zz zz zz zz E E E

The shear strains and shear stresses are related by the shear modulus as:

, , . xy yx yz zy zx xz xy yx yz zy zx xzG G G G G G


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The complete strain to stress relationship can be expressed in the matrix form as:

1 0 0 0

1 0 0 0

E E E


1 0 0 0

10 0 0 0

yy

zz

xy

yz

zx

E E E

E E E

G

0

10 0 0 0 0

10 0 0 0 0

yy

zz

xy

yz

zxG

ij ijkl klS

(1 ) ,(1 2 )(1 ) xx xx yy zz

E

E , xy yx xy yxG G


Stress to Strain Relationship:Stress to Strain Relationship:

,(1 2 )(1 ) yy xx yy zz

(1 ) ,(1 2 )(1 ) zz xx yy zz

E

,

. yz zy yz zy

zx xz zx xzG G

(1 ) 0 0 0

(1 ) 0 0 0

xx

yy

E E E

E E E

xx

yy

E With

(1 ) 0 0 0

0

zz

xy

yz

zx

E E E

0 0 0 0

0 0 0 0 0

0 0 0 0 0

zz

xy

yz

zx

G

G

G

(1 2 )(1 )

ij ijkl klC


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Volumetric strainVolumetric strain

( )( )( )V a a b b c c abc

The volumetric strain is the unit change in volume due to a deformation or due to the effect of volumetric stress.


The combination, v = xx + yy + zz is called the volumetric strain, or dilatation. Thenegative of v is known as the condensation.

V abc(1 )(1 )(1 ) 1 x z x y y z

When a body is subjected to the mutually perpendicular like and equal direct stress orvolumetric stress (e.g hydrostatic pressure), the ratio of direct stress to the correspondingvolumetric strain if found to be constant for a given material when the deformation iswithin a certain limit. The ratio is known as bulk modulus and is usually denoted by K.


Bulk modulusBulk modulus

*** Volumetric stress causes change in volume

If Poissons ratio approaches 0.5 , which happens for near incompressible materials, K .

Hydrostatic pressureVolumetric strain

Volumtric stressVolumetric strain

K

V

pK

dV V

dV

3(1 2 )

E K


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Determine the elongation of a conical bar under the action of its own weight(shown in Fig.) if the length of the bar is l , the diameter of the base is d and theweight per unit of the volume of the material is .

d 2d l

dy

y

l

4 32 3

24 3 xd y

V l

3 y

d dy E

2l l ydy

0


Principle of SuperpositionPrinciple of Superposition

P P P PA A

L1L2

L3

A2E1 E2 E3


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400 mm x

y z = 140 MPaA circular diameter d = 220mm is madeon a unstressed aluminum plate. Forcesacting on the plate cause normalstresses x and z as shown in Fig.

20 mm

z

x = 80 MPa220 mm

A B

C

Take E = 70 GPa and = 1/3.

Determine:(a) change in length in dia AB(b) change in length in dia CD(c) change in thickness and(d) change in volume of the plate.


Thermal stressesThermal stresses

Metal expand on heating and contract on cooling. If the free expansion orcontraction is prevented by an object then stresses will be developed in theheated or contracted material. The stresses are due to chan e in tem erature.

1T ol l T t o T o

l lT

l

T E T

Thermal strain:

Thermal stress:


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M T T E

( ) E T

1D Hookes law with thermal effect1D Hookes law with thermal effectTotal strain = mechanical strain + thermal strain


Prob.: A steel tube 2.4 cm external diameter and 1.8 cm internal diameter encloses acopper rod 1.5 cm diameter to which it is rigidly joined at each end. If, at a temperatureof 10 0C there is no longitudinal stress calculate the stresses in the rod and tube whenthe temperature is raised to 200 0C. Before heat

Final position

Es = 2.1(10) 6 Pa, s = 11(10) 6 / 0C

Ec = 1(10) 6 Pa, s = 18(10) 6 / 0C

Free expansion


One of the most common problems in mechanics of materials involves transformation of axes. For instance, we may know the stresses acting on xy planes, but are really more interested in the stresses acting on planes oriented at, say, to the x axis as seen in Fig.

Transformation Stresses and strainTransformation Stresses and strain

, occur, or is the angle at which two pieces of lumber are glued together in a \scarf" joint. We seek a means to transform the stresses to these new xy planes.

x'

y'


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=


, Shear force V tangential to the inclined plane, V = P sin

2max

coscos (1 cos 2 )

/ cos 2 x

xinclined

N P P A A A

max

sinsin cos sin 2

/ cos 2 2 x x

inclined

V P P A A A

Shear stress to the inclined plane:


y

xy

y

xy

x

x


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2 2cos sin 2 sin cos x x y xy

2 2cos sin cos sin xy y x xy x'

y y'

xy A


x

A

y( A sin )

x( A cos )

xy( A sin )

xy( A cos )

Explanation:


cos(2 ) sin(2 )2 2

x y x y x xy

The transformation equation for plane stress conditions:

cos(2 ) sin(2 )2 2

x y x y y xy

sin(2 ) cos(2 )2

x y xy xy

The summation of normal stress at any orientation: x y x y


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Principal stresses: The maximum and minimum normal stresses ( 1 and 2) areknown as the principal stresses. To find the principal stresses, we mustdifferentiate the transformation equations.

Principal stressesPrincipal stresses

Principal planes: The planes on which only principal stresses are acting called principal planes.

cos(2 ) sin(2 )2 2

x y x y x xy

Transformation stress:

( 2sin 2 ) (2 cos 2 ) 02

x y x xy

d d

2tan 2 xy p x y

Here, p s are the principal angles associated with principal stresses

There are two values of 2 p in the range 0 360, with values differing by 180.There are two values of p in the range 0 180, with values differing by 90.So, the planes on which the principal stresses act are mutually perpendicular.


2tan 2 xy p

x y

2

xyR

2

2 2

2 x y

xy R

cos2 x y p

( x y ) /2sin2 xy p R

Putting the value of sin(2 p) and cos(2 p) in the transformation eq., the principal stress obtained:

2

21

x y x y x

Explanation:


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Exam. 7.1 (Gere): An element in plane stress is subjected to stresses are x = 16000psi, y = 6000 psi, and xy = yx = 4000 psi (Fig.2). Determine the stresses acting on anelement at an angle = 450.

6000 si

xy = 4000 psi

16000 psi


Mohrs circle for plane stressMohrs circle for plane stress

Mohrs circle is named after the famous German civil engineer OttoChristian Mohr (1835 1918), who developed the circle in 1882 for

The transformation equations for plane stress can be represented in

graphical form by a plot known as Mohrs Circle.This graphical representation is extremely useful because it enables

you to visualize the relationships between the normal and shearstresses acting on various inclined planes at a point in a stressed body.

.

Using Mohrs Circle you can also calculate principal stresses,maximum shear stresses and stresses on inclined planes.

Mohrs is also valid for quantities of a similar of mathematical nature,including strains and moment inertia.


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MhorsMhors circle for plane stresscircle for plane stressyy

xy yy

cos(2 ) sin(2 )2 2

x y x y x xy

xx

xy xx

sin(2 ) cos(2 )2

x y xy xy

2 2 2( ) x ave xy R ( xa)2 + y2 = R2

This is the equation of a circle in standard algebraic form. The coordinatesare x1 and x1y1 the radius is R and the centre of circle has coordinates x =

Explanation:

ave and xy = 0

Mohrs circle can be constructed in a variety of ways, depending upon which stresses are known and which are to be found

Let us assume that we know the stresses x , y xy acting on the x and ylanes of an element in lane stress Fi .a


The above information is sufficient to construct the circle

Then, with the circle drawn, we can determine the stresses x ,

y

xyacting on an inclined element (Fig . b)

yy

x yy

xx

xy

+ xx


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Consider the normal stress x positive to the right and the shear stress xypositive downward (Fig.). The advantage of plotting shear stress positivedownward is that the angle 2 will be positive when counter clockwise,which agrees with the positive direction of 2


xx

xy

+

A

B

B ( y , xy ) at = 90

D ( y , xy ) at +900

max

xx

xy yy

A ( x , xy ) at = 0

1 2

2 p

2C ( x , xy ) at ave

R R

Exam. 7.4 (Gere): At a point on the surface of a pressurized cylinder, the material issubjected to biaxial stresses x = 90 MPa and y = 20 MPa, as shown on the stresselement on Fig.1. Using Mohrs circle, determine the stresses acting on an elementinclined at an angle = 300. (consider only the in plane stresses, and show the resultson a sketch of a ro erl oriented element.


Xx= 90 MPa

A

B

yy = 20 MPa


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Exam. 7.5 (Gere): An element in plane stress at the surface of a large machine issubjected to stresses x = 15000 psi, y = 5000 psi, and xy = 4000 psi (Fig.). UsingMohrs circle, determine the following quantities: (a) the stresses acting on anelement inclined at an angle = 400. (b) the principal stresses, and (c) the max. Shearstresses. (Consider only the in plane stresses, and show the results on a sketch of aproperly oriented element.)

(a)

p1 = .

(b)

s= 25.7 0

(c)


Exam. 7.6 (Gere): At a point on the surface of a generator shaft the stresses are x =50 MPa, y = 10 MPa, and xy = 40 Mpa (Fig.2). Using Mohrs circle, determine the

following quantities: (a) the stresses acting on an element inclined at an angle = 400.(b) the principal stresses, and (c) the max. Shear stresses. (Consider only the in planestresses, and show the results on a sketch of a ro erl oriented element.

xy

A

B

50 MPa

10 MPa

a

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