Mechanics of ε Materials σ Laboratory σcourses.washington.edu/me354/reading/Notes_Jenkins.pdf ·...

165
University of Washington Mechanics of Materials Laboratory σ τ ε σ P P ME 354 σ τ τ τ σ τ τ τ σ x xy xz xy y yz xz yz z Course website: http://swhite.me.washington.edu/~jenkinsm/me354/ Department of Mechanical Engineering University of Washington Seattle, Washington Michael G. Jenkins Associate Professor, Mechanical Engineering TEL: 206-685-7061; FAX: 206-685-8047; e-mail: [email protected] Rev 2.0 01 January 2001

Transcript of Mechanics of ε Materials σ Laboratory σcourses.washington.edu/me354/reading/Notes_Jenkins.pdf ·...

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University of Washington

Mechanics of Materials Laboratory σ

τ

ε

σ

P

P

ME 354σ τ ττ σ ττ τ σ

x xy xz

xy y yz

xz yz z

Course website: http://swhite.me.washington.edu/~jenkinsm/me354/

Department of Mechanical EngineeringUniversity of Washington

Seattle, Washington

Michael G. JenkinsAssociate Professor, Mechanical Engineering

TEL: 206-685-7061; FAX: 206-685-8047; e-mail: [email protected]

Rev 2.001 January 2001

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Mechanical Engineering 354Mechanics of Materials Laboratory

TABLE OF CONTENTS

Topic Page

1. Introduction ....................................................................................................1.1Laboratory ProcedureLaboratory Report

2. Stress/Strain/Constitutive Relations...........................................................2.1StressStrainConstitutive Relations

3. Beams: Stress, Strain, Deflection (Lab Exercise) ....................................3.1StrainStress: Normal and ShearDeflections

4. Beams: Curved, Composite, Unsymmetrical Bending (Lab Exercise) 4.1Curved BeamsUnsymmetrical BendingComposite Beams

5. Mechanical Properties and Performance of Materials(Tension, torsion, hardness, impact tests) (Lab Exercise).......................5.1

Mechanical TestingTensionHardnessTorsionImpactPlasticity Relations

6. Stress Concentration and Stress Raisers (Lab Exercise) .......................6.1Stress Concentration FactorsEffects of Stress RaisersExperimental Techniques

7. Fracture and Effects of Cracks (Lab Exercise)...........................................7.1

8. Time Dependent Behaviour: Creep Deformation (Lab Exercise)........8.1

9. Time Dependent Behaviour: Cyclic Fatigue (Lab Exercise)..................9.1

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Topic Page

10. Compression and Buckling of Columns (Lab Exercise)........................10.1

11. Structures: Complex Stresses and Deflections (Lab Exercise)...........11.1Failure CriteriaCombined StressesTypes of Engineering Structures

12. Pressure Vessels: Combined Loading (Lab Exercise) ........................12.1Thin-Walled Pressure VesselsThick-Walled Pressure Vessels

13. References...................................................................................................R.1

Appendix A: Lab Report Format .............................................................A.1

Appendix B: Lab Exercise Handouts......................................................B.1

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1.1

MECHANICS OF MATERIALS LABORATORY NOTES

1. INTRODUCTION

Mechanics of Materials is generally the name applied to a discipline in which the stress,

strain and deflections of loaded structural elements are considered. This set of notes presents

the laboratory aspects of this subject.

For nearly all design work it is necessary to know something of the elastic and, often,

plastic properties of the material to be used. While these properties are often available from

handbooks, sometimes particular properties of less common materials are needed, in which

case the engineer must perform his own tests. The performance of these typical tests in this

laboratory will give a better feeling for the significance of the various material properties and for

the accuracy with which these quantities can be determined.

The various sections of these notes are concerned with review of the subject of

mechanics of materials and related material properties including the laboratory application of

these principles to a simple structures. In various exercises, the stresses, strains and

deflections of a both simply-supported straight and curved beams are measured and

compared to analytical predictions. In another exercise, selected mechanical properties and

performance of representative engineering materials are measured using standardized test

methods and quantitatively compared to handbook values. The effects of stress

concentrations are the focus of another exercise in which photoelasticity is used to determine

stress raisers for comparisons to values obtained from compendiums. Fracture mechanics and

crack interactions are examined in a study of the load carrying reduction of cracks in components.

Time-dependent behaviour is evaluated through measurement and analysis of creep

deformation and cyclic fatigue failures. Structural instabilities such as column buckling are

compared to material strength in assessing engineering failures. Complex structures are

analyzed through experimental measurements and both simple and complex analytical

methods to assess the implications of oversimplifications in engineering analysis.

Laboratory Procedure

Mechanics of Materials Laboratory, ME 354, is intended to give an experimental

understanding and verification of the coursework covered in Mechanics of Materials, CIVE 220

(formerly ENGR 220) and Introduction to Materials Science, MSE 170 (formerly ENGR 170).

No one should be enrolled in this course who has not taken or is not currently taking MSE 170

and CIVE 220 or their equivalents.

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1.2

Because of the nature of the laboratory experiments, it is necessary to conduct them as

a class activity with students either observing or directly participating in the exercises. In some

exercises, small by groups of students will conduct the experiment directly. In other exercises,

the instructor will take the lead in operating the equipment with some students participating as

assistants and others as observers and recorders. An instructor will always be available in the

laboratory to introduce the exercise, describe the operation of the equipment, discuss the

expected results and present the salient aspects of the analysis. Generally, students will be

expected to work as teams when required but must complete written reports independently.

Some laboratory exercises require formal written reports. Other exercises require the

completion of pre-formatted lab reports and their transmittal to the instructor in the form of a

short memo report Still other exercises require only the completion of pre-formatted write-ups

without any additional writing. All lab reports, regardless of type, must be turned in to receive a

passing course grade. Missing lab reports at the time of assignment of final grades will mean

the assignment of a final grade of X for one quarter following the course, regardless of the

quality of the rest of the coursework, until all reports are in. Failure to complete missing lab

reports or to make other arrangements after one quarter has passed following completion of the

course will result in the conversion of the X to a 0.0.

Laboratory reports

Reports are the primary basis for the course grade. Examination grades and discussion

participation are also considered in the final grade. Grades are important to the student for a

relatively short time; report writing will be important to the student's total career.

The laboratory reports provide an opportunity for the student to sharpen writing skills

and to increase the awareness of writing standards. Future employers will require standards

and will judge your professional or technical ability in part on your reporting capabilities.

Sherman (Sherman et al, 1975) has stated "It would be an overstatement, perhaps, to say that

a career in a technical profession will be impossible if you cannot write effectively. It is no

overstatement, however, to say that weakness in writing is a handicap that will weaken your

qualifications for many desirable positions, and that skill in writing is an asset that can make your

professional advancement faster and easier."

Technical writing involves style, neatness, grammar, usage of words, spelling, and

format. Of these attributes, first five are generally applicable, whereas the sixth (i.e., format) is

specific to the particular application. In the course, the format is non arbitrary and is detailed in

the appendix.

Neatness: All written communications should be neat in their final form. Reports should

be machine generated. Original data may be in pencil and is always included in the appendix

of the report.

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1.3

Grammar: Sentence structure, paragraph construction, and punctuation presumably

have been learned prior to taking ME 354. Errors in grammar will be noted by the grader so

that the student's writing skills will be improved.

Usage of Words: Misuse of words involves words and phrases that are problems for

many writers. A few examples of such "pairs" are: affect-effect, among-between, because-

for, fewer-less, like-as if, percent-portion, while-although, too-two, their-there.

Jargon is acceptable when properly used (i.e., not overused!). Specialized words are

acceptable to a particular profession but should not be used to impress an "outsider."

Debasing of the English language by the use of suffixes such as "ise" and "wise" is confusing

and unnecessary. Colloquialisms or contractions should not be used in formal technical writing.

Style: The style of writing is determined by the potential reader. A report may be

formal or informal, childish or mature, personal or impersonal, stilted or admirable, wordy or

succinct. The formal laboratory report may be read by a teaching assistant or a professor, but it

should be written for an engineering manager. Properly written laboratory reports may be

used for reference material; well-written reports will enhance this value.

Do not copy portions of these notes word for word in your report. Statements in "your

own words" will indicate understanding and descriptive conciseness ability. The use of future

tense or telling "what you are going to do" does not belong in a report of what you did.

Generally the tense of reports is such that anything in the report (e.g., tables, figures, section)

are referred to in the present tense. Anything done to produce the results of the reports is in

the past tense.

Traditionally, technical report writing has been conducted in the third person passive

voice (e.g., "The tests were conducted"). The use of "I" imparts a personal tone to the report

which is generally inappropriate. First person style emphasizes the writer's part in the

experiment or test rather than the material or equipment used. "I" and "we" are sometimes

used to reduce awkward or stilted language (such as using "one"). Such use should be kept to

a minimum, particularly in the Summary.

Spelling: Spelling words properly is a problem for many students. Incorrectly spelled

words, particularly simple words, indicate a juvenile approach to technical writing. To quote

Sherman (Sherman et al, 1975), "Even a weak speller, if he keeps a list of the words that he

misses, is usually surprised at its shortness.... He can often eliminate most of his errors b y

learning to spell no more than 40 or 50 words." Keep in mind that electronic spell checkers do

not have any bearing on word choice. The wrong word correctly spelled is still the wrong word.

Words frequently misspelled in this course include: yield, specimen, temperature,

Riehle, recommend, omission — to name just a few. The use of a word guide is strongly urged

for those students with a spelling problem.

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2. STRESS, STRAIN, AND CONSTITUTIVE RELATIONS

Mechanics of materials is a branch of mechanics that develops relationships

between the external loads applied to a deformable body and the intensity of internal

forces acting within the body as well as the deformations of the body. External forces can

be classified as two types: 1) surface forces produced by a) direct contact between two

bodies such as concentrated forces or distributed forces and/or b) body forces which

occur when no physical contact exists between two bodies (e.g., magnetic forces,

gravitational forces, etc.). Support reactions are external surface forces that develop at

the support or points of support between two bodies. Support reactions may include

normal forces and couple moments. Equations of equilibrium (i.e., statics) are

mathematical expressions of vector relations showing that for a body not to translate ormove along a path then F = 0∑ . For a body not to rotate, M = 0∑ . Alternatively, scalar

equations in three-dimensional space (i.e., x, y, z) are:

F x = 0∑ F y = 0 ∑ F z = 0∑Mx = 0∑ My = 0∑ Mz = 0∑

(2.1)

Internal forces are non external forces acting in a body to resist external loadings.

The distribution of these internal forces acting over a sectioned area of the body (i.e., force

divided by area, that is, stress) is a major focus of mechanics of materials. The response

of the body to stress in the form of deformation or normalized deformation, that is, strain is

also a focus of mechanics of materials. Equations that relate stress and strain are known

as constitutive relations and are essential, for example, for describing stress for a

measured strain.

Stress

If an internal sectioned area is subdivided into smaller and smaller areas, ∆A , two

important assumptions must be made regarding the material: it is continuous and it is

cohesive. Thus, as the subdivided area is reduced to infinitesimal size, the distribution of

internal forces acting over the entire sectioned area will consist of an infinite number of

forces each acting on an element, ∆A , as a very small force ∆F . The ratio of incremental

force to incremental area on which the force acts such that: lim∆A→0

∆F

∆A is the stress which

can be further defined as the intensity of the internal force on a specific plane (area)

passing through a point.

2.1

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Stress has two components, one acting perpendicular to the plane of the area and

the other acting parallel to the area. Mathematically, the former component is expressed

as a normal stress which is the intensity of the internal force acting normal to an

incremental area such that:

σ = lim∆A→0

∆Fn

∆A (2.2)

where +σ = tensile stress = "pulling" stress and -σ = compressive stress = "pushing"

stress. The latter component is expressed as a shear stress which is the intensity of the

internal force acting tangent to an incremental area such that:

τ = lim∆A→0

∆Ft

∆A (2.3)

The general state of stress is one which includes all the internal stresses acting on

an incremental element as shown in Figure 2.1. In particular, the most general state of

stress must include normal stresses in each of the three Cartesian axes, and six

corresponding shear stresses.

Note for the general state of stress that +σ acts normal to a positive face in the

positive coordinate direction and a +τ acts tangent to a positive face in a positivecoordinate direction. For example, σ xx (or just σ x ) acts normal to the positive x face inthe positive x direction and τ xy acts tangent to the positive x face in the positive y

direction.

Although in the general stress state, there are three normal stress component and

six shear stress components, by summing forces and summing moments it can be shownthat τ xy = τyx ;τxz = τzx ;τyz = τzy .

z

x

y

ττ

τ

τ

zy

zx

yxxy

τ

τ xz

yz

σ

σ

σ

y

z

x

Figure 2.1 General and complete stress state shown on a three-dimensional incrementalelement.

2.2

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Therefore the complete state of stress contains six independent stress components (threenormal stresses, σ x ;σy ;σ z and three shear stresses, τ xy;τyz;τxz ) which uniquely

describe the stress state for each particular orientation. This complete state of stress can

be written either in vector formσx

σy

σz

τ xy

τ xz

τ yz

(2.4)

or in matrix form

σ x τxy τxz

τxy σ y τyz

τxz τ yz σ z

(2.5)

The units of stress are in general: Force

Area=

F

L2 . In SI units, stress is

Pa =N

m2 or MPa = 106 N

m2 =N

mm2 and in US Customary units, stress is

psi =lbf

in 2 or ksi = 103 lbf

in 2 =kip

in 2 .

Often it is necessary to find the stresses in a particular direction rather than just

calculating them from the geometry of simple parts. For the one-dimensional case shown

in Fig. 2.2, the applied force, P, can be written in terms of its normal, PN, and tangential,PT, components which are functions of the angle, θ, such that:

PN = P cos θPT = P sinθ

(2.6)

The area, Aθ, on which PN and PT act can also be written in terms of the area, A, normal tothe applied load, P, and the angle, θ, such that

Aθ = A /cosθ (2.7)

The normal and shear stress relation acting on any area oriented at angle, θ, relative to

the original applied force, P are:

σθ =PN

Aθ=

P cos θA / cosθ

=P

Acos2 θ = σ cos2 θ

τθ = PT

Aθ= P sinθ

A /cosθ= P

Acosθ sinθ = σ cosθ sinθ

(2.8)

where σ is the applied unidirectional normal stress.

2.3

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θ

θP

P

P

A

A = A cos

θ

θ θ

N

T

Figure 2.2 Unidirectional stress with force and area as functions of angle, θ

For the two dimensional case (i.e., plane stress case such as the stress state at a

surface where no force is supported on the surface), stresses exist only in the plane of thesurface (e.g., σx ;σy ; τxy ). The plane stress state at a point is uniquely represented by

three components acting on a element that has a specific orientation (e.g., x, y) at the

point. The stress transformation relation for any other orientation (e.g., x', y') is found byapplying equilibrium equations ( F = 0 and∑ M = 0 )∑ keeping in mind that

F n = σA and F t = τA. The rotated axes and functions for incremental area are shown in

Fig. 2.3. The forces in the x and y directions due to F n = σA and F t = τA and acting on the

areas normal to the x and y directions are shown in Fig. 2.4By applying simple statics such that in the x'-direction, F x ' = 0∑ and

σx ' = σ x cos2θ + σy sin2 θ + 2τxy cosθ sinθ

or

σx ' =σ x + σy

2+

σx − σy

2cos2θ + τxy sin2θ

(2.9)

X

y

x'

y'θ

θ

∆A

∆Ay=∆A sin θ

∆Ax=∆A cos θ

Figure 2.3 Rotated axes and functions for incremental area.

2.4

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X

y x'

y' θ

θ

σy ∆Ay

σx ∆Ax

τxy ∆Ay

σx' ∆A

τx'y' ∆A

θ

θ

θ

θ

τxy∆Ax

Figure 2.4 Rotated coordinate axes and components of stresses/forces.

Similarly, for the x'y'-direction, Fy ' = 0∑ and

τx 'y ' = (σ x − σy )cos θ sinθ + τxy (cos2 θ + sin2θ)

or

τx ' y ' = −σx − σy

2sin2θ + τ xy cos2θ

(2.10)

Finally, for the y' direction, Fy ' = 0∑ and

σy ' = σ x sin2θ + σy cos2 θ − 2τ xy cosθ sinθ

or

σy ' =σ x +σ y

2−

σx − σy

2cos2θ − τxy sin2θ

(2.11)

If the stress in a body is a function of the angle of rotation relative to a given

direction, it is natural to look for the angle of rotation in which the normal stress is either

maximum or nonexistent. A principal normal stress is a maximum or minimum normal

stress acting in principal directions on principal planes on which no shear stresses act.

Because there are three orthogonal directions in a three-dimensional stress state thereare always three principal normal stresses which are ordered such that σ1 > σ2 > σ3 .

Mathematically, the principal normal stresses are found by determining the angulardirection, θ, in which the function, σ = f (θ ), is a maximum or minimum by differentiating

σ = f (θ ) with respect to θ and setting the resulting equation equal to zero such that dσdθ

= 0

before solving for the θ at which the principal stresses occur. Applying this idea to Eq. 2.9,

gives

2.5

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dσdθ

= 0 = σx −σ y( )sin2θ + 2τxy cos2θ ⇒ sin2θcos2θ

= tan2θ =2τxy

σx − σy(2.12)

There are two solutions for the principal stress angle (i.e., for maximum and minimum) so

that.

θN1 =12

tan-1 2τxy

σx − σy

θN2 =12

tan-1 2τxy

σx − σy+ π

= θN1 + π

2

(2.12)

Using trigonometry on the geometry shown in Fig. 2.5 results in

tan2θ =τ xy

σx −σ y( ) /2

sin2θ =τ xy

σ x − σy

2

2

+ τ xy2

cos2θ =σx − σy( ) /2

σx − σy

2

2

+ τ xy

2

(2.14)

Substituting the trigonometric relations of Eq. 2.14 back into Eq. 2.9 gives for the plane

stress case:

σθ = σ1,2 =σx +σ y

σ x − σy

2

2

+τ xy2

tan2θp =τxy

σx − σ y( ) /2

(2.15)

τ

(σ − σ )/2

√ x

y

xy

σ − σ ( )22 xy2+ τy

x

Figure 2.5 Geometric representation of the principal direction relations

2.6

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Note that for the plane stress case in the x-y plane, σz = 0 . Thus the 1 and 2 subscripts in

Eq. 2.15 are only for the x-y plane and are not necessarily σ1 and σ2 for the three-

dimensional general state of stress. Therefore, ordering of σ1 and σ2 of Eq. 2.15 is only

preliminary, until they are compared to σz and ordered according to conventionσ1 > σ2 > σ3 . For example, for a particular plane stress state σ1 and σ2 found from Eq.

2.15 are 100 and 20 MPa, then the principal stresses are σ1 = 100, σ2 = 20, σ3 = 0 MPa.

However, if σ1 and σ2 found from Eq. 2.15 are 125 and -5 MPa, then the principalstresses are σ1 = 125, σ2 = 0, σ3 = −5 MPa. Finally, if σ1 and σ2 found from Eq. 2.14 are

-25 and -85 MPa, then the principal stresses are σ1 = 0, σ 2 = -25, σ 3 = −85 MPa.

Performing a similar substitution of the trigonometric relations of Eq. 2.14 back into

Eq. 2.10 gives for the plane stress case:

τmax =σ x −σ y

2

2

+τ xy2

σave =σ x +σ y

2 and tan2θs =

− σx − σy( )2τxy

(2.16)

Note that the τ max of Eq. 2.15 is only for the x-y plane. The maximum shear stress for thethree-dimensional stress state can be found after the principal stresses are orderedσ1 > σ2 > σ3 such that:

τ1,3 =σ1 − σ3

2 (2.17)

Some general observations can be made about principal stresses.

a) In a principal direction, when τ =0, then σ 's are maximum or minimumb) σ max and σmin τmax and τmin( ) occur in directions 90° apart.

c) τ max occurs in a direction midway between the directions of σ max and σmin

σ

τ +σ ,+τ

+σ ,−τ

φ =2θ

x

y

C= 2

x + yσσR = ( x - C) + )

tan = -

σ τ 22

φ −τ σ ( x - C)

2

Figure 2.6 Mohr's circle representation of plane stress state

2.7

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An interesting graphical relation occurs if the second equation in each of Eqs. 2.9

and 2.10 are squared and added together:

σ x ' = σθ =σx +σ y

2+

σ x − σy

2cos2θ + τ xy sin2θ

2

+

τ x 'y '=τθ =−σx −σy

2sin2θ +τ xy cos 2θ

2

=

σθ −σ x +σ y

2

2

+τθ2 =

σx −σy

2

2

+τ xy2

x−h( )2 +y2 = r2

(2.18)

The result shown in Eq. 2.18 is the equation for a circle (i.e., Mohr's circle) with radius,

r=σ x −σy

2

2

+ τxy2 and displaced h=

σ x +σ y

2 on the x=σ θ axis as illustrated in Fig.

2.6. Examples of Mohr's circles are shown in Fig. 2.7. A procedure for developing Mohr's

circle for plane stress is shown in the following section.

σ

τ

1σσ

τ

2

maxfor x-y plane

Mohr's circle for stresses in x-y plane

σ

τ

1σσ

τ

3

max =

Mohr's circle for stresses in x-y-z planes

σ2

σ − σ2

1 3

Figure 2.7 Examples of Mohr's circle for plane and three dimensional stress states.

2.8

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Graphical Description of State ofStress

2-D Mohr's Circle

σ

σ

τxy

y

x

X

Y

In this example all stresses acting in axialdirections are positive as shown in Fig. M1.

Fig. M1- Positive stresses acting on a physical element.

+σx

y

xy

y-face As shown in Figs. M2 and M3, plottingactual sign of the shear stress with xnormal stress requires plotting of theopposite sign of the shear stress with the ynormal stress on the Mohr's circle.

Fig. M2 - Directionality of shear acting on x and y faces. In this example σx > σy and τxy is positive.

By the convention of Figs. M2 and M3, φ =2θ on the Mohr's circle is negative from the+σ axis. (Mathematical convention is thatpositive angle is counterclockwise).

σ

τ +σ ,+τ

+σ ,−τ

φ =2θ

x

y

C= 2

x + yσσR = ( x - C) + )

tan = -

σ τ 22

φ −τ σ ( x - C)

2

Note that by the simple geometry of Fig.M3, φ = 2θ appears to be negative while bythe formula, tan 2θ = 2τxy/(σx -σy), thephysical angle, θ, is actually positive.

In-plane principal stresses are: σ1= C+R σ2= C - R

Fig. M3 - Plotting stress values on Mohr's circle.

Maximum in-plane shear stress is:τmax=R=(σ1-σ2)/2

2.9

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θ

σσ

1

2Y

X

Directionof +θ The direction of physical angle, θ, is from

the x-y axes to the principal axes.

Fig. M4 - Orientation of physical element with only principal stresses acting on it.

θ

PrincipalAxis

Line of X-Y Stresses

Direction of

Fig. M5 - Direction of q from the line of x-y stresses to the principal stress

Note that the sense (direction) of thephysical angle, θ, is the same as on theMohr's circle from the line of the x-ystresses to the axes of the principalstresses.

axis.

Same relations apply for Mohr's circle for strain except interchange variables as

σ ⇔ ε and τ ⇔γ2

2.10

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Recall that all stress states are three-dimensional. Therefore, a more general

method is required to solve for the principal stresses. One such method is to solve for the

"eigenvalues" of the stress matrix where, σ is the principal stress:

σx −σ τ xy τxz

τxy σy −σ τ yz

τ xz τyz σz −σ

(2.19)

Finding the determinant for this matrix and grouping terms gives:

σ 3 − I1σ2 + I2σ − I3 = 0 (2.20)

where the stress invariants, I1 ,I2, I3( ), do not vary with stress direction:

I1 = σ x + σy + σz

I2 = σxσ y +σ yσz + σ zσx − τxy2 − τxz

2 −τ yz2

I3 = σ xσyσ z + 2τ xyτ xzτ yz − σxτ yz2 − σyτxz

2 −σ zτxy2

(2.21)

Note that if principal stresses are used in Eq. 2.21 for the σ terms then all terms

containing τ will be zero since by definition, principal stresses act in principal directions

on principal planes on which τ =0.Eq. 2.20 can then be solved for the three roots which when ordered σ1 > σ2 > σ3

are the principal stresses. Eq. 2.20 can be plotted as f σ( ) vs σ shown in Fig. 2.9 where

the principal stresses are the values of σ which occur when f σ( ) = 0.

σ

f( )σ

σσσ123

Figure 2.9 Solving for cubic routes for principal stresses

2.11

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Strain

Whenever a force is applied to a body, it will tend to change the body's shape and

size. There changes are referred to as deformation. Size changes are known as

dilatation (volumetric changes) and are due to normal stresses. Shape changes are

known as distortion and are due to shear stresses.

In order to describe the deformation in length of line segments and the changes in

angles between them, the concept of strain is used. Therefore, strain is defined as

normalized deformations within a body exclusive of rigid body displacements

There are two type of strain, one producing size changes by elongation or

contraction and the other producing shape changes by angular distortion.

Normal strain is the elongation or contraction of a line segment per unit length

resulting in a volume change such that

ε = limB →A along n

A' B'− AB

AB≡

L f − Lo

Lo

(2.22)

where + ε = tensile strain = elongation and -ε = compressive strain = contraction

Shear strain is the angle change between two line segments resulting in a shape

change such that:

γ = (θ =π2

) − θ ' ≈∆h

(for small angles ) (2.23)

where +γ occurs if π2

> θ ' and -γ occurs if π2

< θ '.

The general state is one which includes all the internal strains acting on an

incremental element as shown in Fig. 2.10. The complete state of strain has sixindependent strain components (three normal strains, εx; εy; εz and three engineering

shear strains, γ xy;γ yz;γ xz ) which uniquely describe the strain state for each particular

orientation. This complete state of strain can be written in vector formεx

εy

εz

γ xy

γ xz

γ yz

(2.24)

2.12

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A

ε xy ε

ε yx

ε

x

y

Engineering shear strain,

γ = ε + εxy xy yx

Figure 2.10 General and complete strain state shown on a three-dimensional incrementalelement.

Alternatively, the complete state of strain can be written in matrix form :

εx γ xy γ xz

γ xy εy γ yz

γ xz γ yz εz

(2.25)

The units of strain In general: Length

Length=

L

L, In SI units, strain is

m

m for ε and

m

m or radian for γ and in US Customary units, strain is

in

in for ε and

in

in or radian for γ

Just as in the stress case it is often necessary to find the stresses in a particular

direction rather than just calculating them from the geometry of simple parts. For the two-

dimensional, plane strain condition (e.g., strain at a surface where no deformation occursnormal to the surface), strains exist only in the plane of the surface (e.g., εx; εy;γ xy ). The

plane strain state at a point is uniquely represented by three components acting on a

element that has a specific orientation (e.g., x, y) at the point. The strain transformation

relation for any other orientation (e.g., x', y') is found by summing displacements in theappropriate directions keeping in mind that δ = ε Lo and ∆ = γ h as shown in Fig. 2.11

Simply adding components of displacements in the x' direction,displacements in x 'direction for Q to Q *∑ (see Fig. 2.12) gives

εx ' = εx cos2 θ + ε y sin2 θ + γ xy cosθ sinθ

or

εx ' =εx + εy

2+

εx − εy

2cos2θ +

γ xy

2sin2θ

(2.25)

2.13

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θ

x

y

}

}

}

y'

x'

Q

Q*

dy

δ = ε y y

δ = ε x x

∆ = γ dy

dx

ds

dy

dx

Figure 2.11 Rotated coordinate axes and displacements for x and y directions

Similarly, adding components of displacements in due to rotations of dx' and dy'rotation of dx ' and dy'∑ (see Fig. 2.12) gives

γ x 'y '

2= (εx − εy )cos θ sinθ +

γ xy

2(cos2 θ + sin2θ)

or

γ x 'y '

2= −

εx − εy

2sin2θ +

γ xy

2cos2θ

(2.26)

Finally, adding components of displacements in the y' direction,displacements in y 'direction for Q to Q *∑ (see Fig. 2.12) gives

εy ' = εx sin2θ + εy cos2 θ − γ xy cosθ sinθ

or

εy ' =εx + εy

2−

εx − εy

2cos2θ −

γ xy

2sin2θ

(2.27)

Q

Q*

x'y

xδ = ε x x

δ = ε y y

∆ = γ

θ

θ

θ

{δ = ε x' x'

dx

dy

ds

dy

cos

sin

θ =

θ =

ds

ds

dx

dy

Figure 2.12 Displacements in the x' direction for strains/displacements in the x and ydirections

2.14

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Just as for the stress in a body which is a function of the angle of rotation relative to

a given direction, it is natural to look for the angle of rotation in which the normal strain is

either maximum or minimum. A principal normal strain is a maximum or minimum normal

strain acting in principal directions on principal planes on which no shear strains act.

Because there are three orthogonal directions in a three-dimensional strain state thereare always three principal normal strains which are ordered such that ε1 > ε2 > ε3 .

Also as in the stress case, the principal strains for the plane strain case

ε1,2 =εx + εy

εx − εy

2

2

+γ xy

2

2

tan2θp =γ xy

εx − εy

(2.29)

Note that for the plane strain case in the x-y plane, εz = 0 . Thus, the 1 and 2 subscripts in

Eq. 2.29 are only for the x-y plane and are not necessarily ε1 and ε2 for the three-

dimensional general state of strain. Therefore, ordering of ε1 and ε2 of Eq. 2.29 is only

preliminary, until they are compared to εz and ordered according to conventionε1 > ε2 > ε3 .

For the shear strain case

γmax2

=εx − εy

2

2

+γ xy

2

τxy

2 ,

εave =εx + εy

2 and tan2θs =

− εx − εy( )γ xy

(2.30)

σ

γ/2

1εε

γ

2

max/2 for x-y plane

Mohr's circle for strains in x-y plane

σ

γ/2

1εε

γ

3

max =

Mohr's circle for strains in x-y-z planes

ε2

ε − ε1 3

Figure 2.13 Examples of Mohr's circle for strain.

2.15

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45°

60°

60°a

bc

a

bc

45° Rectangular 60° Deltax

y

x

Figure. 2.14 Rectangular and Delta rosettes.

Note that the γ max of Eq. 2.30 is only the x-y plane. The maximum shear strain for thethree-dimensional strain state can be found after the principal strains are orderedε1 > ε2 > ε3 such that:

γ 1,3 = ε1 − ε3 (2.31)

As with stress, the complete strain state can be represented graphically as a Mohr's

circle. Examples of Mohr's circles for strain are shown in Fig. 2.13. Note that the same

procedure for developing Mohr's circle for the plain strain case can be used as with the

plane stress case with by making the following substitutions: σ ⇔ ε and τ ⇔γ2

.

An important application of the strain transformation relation of Eq. 2.26 is to

experimentally determine the complete strain state since stress is an abstract engineering

quantity and strain is measurable/observable engineering quantity. Equation 2.26

requires that three strain gages be applied at arbitrary orientations at the same point on

the body to determine the principal strains and orientations. However, to simplify the data

reduction, the three strain gages are applied at fixed angles usually in the form of 45°

rectangular rosette or 60° Delta rosette as shown in Fig. 2.14. The resulting equations to

determine the local (for the strain gage rosettes shown in Fig. 2.14) x-y strains in

preparation for determining the principal strains:

45° Rectangular 60° Delta

Rosette Rosette

εx = εa εx = εa

εy = εc εy =13

(2εb + 2εc − εa )

γ xy = 2εb − (εa + εc ) γ xy =23

(εb − εc )

(2.32)

However, a limitation of measuring strains experimentally is that stresses are often

required to determine the engineering performance of the component. Thus, equations

which relate stress and strain are required.

2.16

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Constitutive Relations

If the deformation and strain are the response of the body to an applied force or

stress, then there must be some type of relations which allow the strain to be predicted

from stress or vice versa. The uniaxial stress-strain case is a useful example to begin to

understand this relation.

During uniaxial loading (see Fig. 2.15) by load, P, of a rod with uniform cross

sectional area, A, and length, L, the applied stress, σ ,is calculated simply as

σ = P

A(2.33)

The resulting normal strain, εL , in the longitudinal direction can be calculated from

the deformation response, ∆L , along the L direction:

εL = ∆L

L(2.34).

Another normal strain, εT , in the transverse direction can be calculated from the

deformation response, ∆D , along the D direction:

εT = ∆D

D(2.35).

A plot of σ vs. εL (see Fig. 2.16) shows a constant of proportionality between stress

and strain in the elastic region such that

y = mx + b ⇒ σ = EεL (2.36)

where E =∆σ∆εL

is known as the elastic modulus or Young's modulus and Eq. 2.36 is

known as unidirectional Hooke's law.

A

PDfD=Do

L=Lo

Lf

L=Lf - L

D=Df -D

P Deformed

Undeformed

Figure 2.15 Uniaxially-loaded rod undergoing longitudinal and transverse deformation

2.17

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Longitudinal strain, L Longitudinal strain, εε L

Ε−ν

Figure 2.16 Plots of applied stress vs. longitudinal strain and transverse strain versuslongitudinal strain

A plot of εT vs. εL (see Fig. 2.16) shows a constant of proportionality between

transverse and longitudinal strains in the elastic region such that

y = mx + b ⇒ εT = −νεL (2.37)

where ν = −ε

T

εL

is known as Poisson's ratio.

For the results for three different uniaxial stresses applied separately (see Fig.

2.17) in each of the orthogonal direction, x, y and z can give the general relations

between normal stress and normal strain know as generalized Hooke's law if the strain

components for each of the stress conditions are superposed:

εx = 1E

σx−ν σ

y+σ

z( )( )εy = 1

Eσ y −ν σ x +σ z( )( )

εz = 1E

σ z −ν σ x +σ y( )( )(2.38)

Since shear stresses are decoupled (i.e., unaffected) by stresses in other directions

the relations for shear stress-shear strain are:

γ xy = 1G

τxy

γ xz = 1G

τxz

γ yz = 1G

τyz

(2.39)

where the shear modulus is G =∆τ∆γ =

E

2(1+ ν).

2.18

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:

X

Y

Z

σ

ε

ε

εε

ε

ε

εε

ε

σ

σ

y

y y

y x

x

x

x

z

z

z

zStress εx εy εz

σ x Eσ x - νεx = -νEσx - νεx = -νEσ x

σ y -νεy = -νEσy Eσy -νεy = -νEσy

σz - νεz = -νEσ z -νεz = -νEσ z Eσz

Figure 2.17 Development of generalized Hooke's law.

Generalized Hooke's law is usually written in matrix form such that:ε{ } = S[ ] σ{ } (2.40)

where ε{ } is the strain vector, σ{ } is the stress vector and S[ ] is the compliance matrix.

Expanded, Eq. 2.40 is:

2.19

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εx

εy

εz

γ xy

γ xz

γ yz

=

1E

−νE

−νE

0 0 0

−νE

1E

−νE

0 0 0

−νE

−νE

1E

0 0 0

0 0 0 1G

0 0

0 0 0 0 1G

0

0 0 0 0 0 1G

σx

σy

σz

τxy

τ xz

τ yz

(2.41)

Although Eq. 2.41 is a very convenient form, it is not that useful since stress is

rarely known with strain being the unknown. Instead, strain is usually measured

experimentally and the associated stress needs to be determined. Therefore, the inverse

of the compliance matrix needs to be found such thatσ{ } = C[ ] ε{ } (2.42)

where C[ ] is the stiffness matrix such that C[ ] = S[ ]−1. Expanding Eq. 2.42 gives

σx

σy

σz

τ xy

τ xz

τ yz

=

E(1+ν)

1+ ν(1−2ν )

νE(1+ν )(1−2ν )

νE(1+ν )(1−2ν )

0 0 0

νE(1+ν )(1−2ν)

E(1+ν )

1+ ν(1−2ν )

νE(1+ν )(1−2ν )

0 0 0

νE(1+ν )(1−2ν)

νE(1+ν )(1−2ν )

E(1+ν )

1+ ν(1−2ν )

0 0 0

0 0 0 G 0 0

0 0 0 0 G 0

0 0 0 0 0 G

εx

εy

εz

γ xy

γ xz

γ yz

(2.43)

Generalized Hooke's can be simplified somewhat for the special case of planestress in the x-y plane since σz =0. Being orthogonal to the x-y plane, Since σz is a also

a principal stress by definition, all the shear stresses associated with the z-direction are

also zero. Thus, the stress-strain relations for plane stress in the x-y plane become

σx

σy

τxy

=

E

(1− ν2 )

νE

(1− ν2)0

νE

(1− ν2)

E

(1− ν2)0

0 0 G

εx

εy

γ xy

(2.44)

For the special case of plane stress, although σz =0, the strain in the z-direction is

not zero but instead can be determined such that

Plane stress : σ z = 0, εz ≠ 0 =−ν

1− ν(εx + εy) (2.45)

2.20

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Similarly for the special case of plane strain, although, εz =0, the stress in the z-

direction is not zero but instead can be determined such that

Plane strain : εz = 0, σ z ≠ 0 = ν(σx +σ y) (2.46)

For the special case of hydrostatic pressure, no distortion takes place, only size

changes. In this case the hydrostatic stress is

σH =σ x +σ y + σx( )

3(2.47)

and the dilatation (i.e., volumetric change) is

εV =∆VV

= εx + εy + εx( ) (2.48)

The ratio between the hydrostatic stress and the dilation is a special combination of

elastic constants called the bulk modulus.

k =σH

εV=

σ x + σy + σx( )3 εx + εy + εx( ) =

E

3(1− 2ν )(2.49)

It is useful to know that elastic constants are related to the atomic structure of the

material and thus are not affected by processing or component fabrication. For example,

E is related to the repulsion/attraction between two atoms. The force-displacement curve

for this interaction is shown in Fig. 2.18. Since this is the uniaxial case, recall that

σ =PA

and ε =∆L

L=

x

xe. Therefore, since E is defined as E =

∆σ∆ε =

dσdε with

dσ =dPA

and ε =dx

xe then E =

dσdε =

xe

A

dP

dx. Note from Fig. 2.18 that

dP

dx is the slope of

the force-displacement curve, thus making the elastic modulus E fixed by atomic

interaction. From a materials standpoint, covalent and ionic bonds such as those in

ceramics are stiff leading to high elastic moduli in those materials. Metallic bonds are

intermediate in stiffness leading to intermediate elastic moduli in metals. Secondary

bonds such as those found in polymers are least stiff leading to low elastic moduli in those

materials.

For Poisson's ratio, it is useful to consider the sphere model of atomic structure as

shown in Fig. 2.19. Before deformation the triangle between centers has a transverse

side length of 3R , a longitudinal side length of R and a hypotenuse of 2R. After

longitudinal deformation, the transverse side length is 3R -dx, the longitudinal side

2.21

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Figure 2.18 Repulsion-attraction force-displacement curve between two atoms.

length is R+dy and the hypotenuse is still 2R. Applying Pythagorean's theorem after

deformation gives

2R( )2 = 3R − dx( )2 + R + dy( )2 (2.50)

Expanding Eq. 2.50 and eliminating higher order terms leaves

2Rdy = 2 3Rdx ⇒dy

dx= 3 . (2.51)

Recalling that the longitudinal and transverse strains can be written in terms of the

deformations

εy =R + dy( ) − R

R=

dy

R⇒ dy = Rεy

εx =3R − dx( ) − 3R

3R= −dx

3R⇒ dx = − 3Rεx

(2.52)

Combining the two terms for dy and dx in Eq. 2.52 and equating them to Eq. 2.51 gives

dy

dx=

Rεy

− 3Rεx

= 3 ⇒ −εy

εx

≡ −1ν

= 3 (2.53)

which shows that from a simple sphere model and expected deformations, Poisson's ratio

is fundamentally linked to atomic interactions giving a value of 1/3 which is the range of

many dense materials.

2.22

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Figure 2.19 Undeformed and deformed sphere model for atomic structure anddetermination of Poisson's ratio

2.23

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3. BEAMS: STRAIN, STRESS, DEFLECTIONS

The beam, or flexural member, is frequently encountered in structures and

machines, and its elementary stress analysis constitutes one of the more interesting facets

of mechanics of materials. A beam is a member subjected to loads applied transverse to

the long dimension, causing the member to bend. For example, a simply-supported beam

loaded at its third-points will deform into the exaggerated bent shape shown in Fig. 3.1

Before proceeding with a more detailed discussion of the stress analysis of beams,

it is useful to classify some of the various types of beams and loadings encountered in

practice. Beams are frequently classified on the basis of supports or reactions. A beam

supported by pins, rollers, or smooth surfaces at the ends is called a simple beam. A

simple support will develop a reaction normal to the beam, but will not produce a moment

at the reaction. If either, or both ends of a beam projects beyond the supports, it is called

a simple beam with overhang. A beam with more than simple supports is a continuous

beam. Figures 3.2a, 3.2b, and 3.2c show respectively, a simple beam, a beam with

overhang, and a continuous beam. A cantilever beam is one in which one end is built into

a wall or other support so that the built-in end cannot move transversely or rotate. The

built-in end is said to be fixed if no rotation occurs and restrained if a limited amount of

rotation occurs. The supports shown in Fig. 3.2d, 3.2e and 3.2f represent a cantilever

beam, a beam fixed (or restrained) at the left end and simply supported near the other end

(which has an overhang) and a beam fixed (or restrained) at both ends, respectively.

Cantilever beams and simple beams have two reactions (two forces or one force

and a couple) and these reactions can be obtained from a free-body diagram of the beam

by applying the equations of equilibrium. Such beams are said to be statically

determinate since the reactions can be obtained from the equations of equilibrium.

Continuous and other beams with only transverse loads, with more than two reaction

components are called statically indeterminate since there are not enough equations of

equilibrium to determine the reactions.

Figure 3.1 Example of a bent beam (loaded at its third points)

3.1

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Figure 3.2 Various types of beams and their deflected shapes: a) simple beam, b) beam

with overhang, c) continuous beam, d) a cantilever beam, e) a beam fixed (or restrained)

at the left end and simply supported near the other end (which has an overhang), f) beam

fixed (or restrained) at both ends.

Examining the deflection shape of Fig. 3.2a, it is possible to observe that

longitudinal elements of the beam near the bottom are stretched and those near the top

are compressed, thus indicating the simultaneous existence of both tensile and

compressive stresses on transverse planes. These stresses are designated fibre or

flexural stresses. A free body diagram of the portion of the beam between the left end and

plane a-a is shown in Fig. 3.3. A study of this section diagram reveals that a transverse

force Vr and a couple Mr at the cut section and a force, R, (a reaction) at the left support

are needed to maintain equilibrium. The force Vr is the resultant of the shearing stresses

at the section (on plane a-a) and is called the resisting shear and the moment, Mr, is the

resultant of the normal stresses at the section and is called the resisting moment.

3.2

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Figure 3.3 Section of simply supported beam.

The magnitudes and senses of Vr and Mr may be obtained form the equations ofequilibrium Fy = 0∑ and MO = 0∑ where O is any axis perpendicular to plane xy (the

reaction R must be evaluated first from the free body of the entire beam). For the present

the shearing stresses will be ignored while the normal stresses are studied. The

magnitude of the normal stresses can be computed if Mr is known and also if the law of

variation of normal stresses on the plane a-a is known. Figure 3.4 shows an initially

straight beam deformed into a bent beam.

A segment of the bent beam in Fig. 3.3 is shown in Fig. 3.5 with the distortion highly

exaggerated. The following assumptions are now made

i) Plane sections before bending, remain plane after bending as shown in

Fig. 3.4 (Note that for this to be strictly true, it is necessary that the beam be

bent only with couples (i.e., no shear on transverse planes), that the beam

must be proportioned such that it will not buckle and that the applied loads

are such that no twisting occurs.

Figure 3.4 Initially straight beam and the deformed bent beam

3.3

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Figure 3.5 Distorted section of bent beam

ii) All longitudinal elements have the same length such the beam is initially

straight and has a constant cross section.

iii) A neutral surface is a curved surface formed by elements some distance,

c, from the outer fibre of the beam on which no change in length occurs. The

intersection of the neutral surface with the any cross section is the neutral

axis of the section.

Strain

Although strain is not usually required for engineering evaluations (for example,

failure theories), it is used in the development of bending relations. Referring to Fig. 3.5,

the following relation is observed:

δy

y =

δc

c(3.1)

where δy is the deformation at distance y from the neutral axis and δc is the deformation

at the outer fibre which is distance c from the neutral axis. From Eq. 3.1, the relation for

the deformation at distance y from the neutral axis is shown to be proportional to the

deformation at the outer fibre:

δy =δc

cy (3.2)

Since all elements have the same initial length, ∆x , the strain at any element can

be determined by dividing the deformation by the length of the element such that:

δy

∆x =

yc

δc

∆x⇒ ε = y

cεc (3.3)

3.4

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Figure 3.6 Undeformed and deformed elements

Note that ε is the in the strain in the x direction at distance y from the neutral axis and thatε =ε x . Note that Eq. 3.3 is valid for elastic and inelastic action so long as the beam does

not twist or buckle and the transverse shear stresses are relatively small.

An alternative method of developing Eq. 3.3 involves the definition of normal strain.

An incremental element of a beam is shown both undeformed and deformed in Fig. 3.6.

Note once again that any line segment ∆x located on the neutral surface does not

changes its length whereas any line segment ∆s located at the arbitrary distance y from

the neutral surface will elongate or contract and become ∆s' after deformation. Then by

definition, the normal strain along ∆s is determined as:

ε = lim∆s→0

∆s' −∆s

∆s(3.4)

Strain can be represented in terms of distance y from the neutral axis and radius ofcurvature ρ of the longitudinal axis of the element. Before deformation ∆s = ∆x but after

deformation ∆x has radius of curvature ρ with center of curvature at point O'. Since ∆θdefines the angle between the cross sectional sides of the incremental element,∆s = ∆x = ρ ∆θ . Similarly, the deformed length of ∆s becomes ∆s'= ρ − y( ) ∆θ .

Substituting these relations into Eq. 3.4 gives:

ε = lim∆θ→0

ρ − y( )∆θ − ρ ∆θρ ∆θ

(3.5)

3.5

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Eq. 3.5 can be arithmetically simplified as ε = −y / ρ . Since the maximum strainoccurs at the outer fibre which is distance c from the neutral surface, εmax = −c / ρ = εc ,

the ratio of strain at y to maximum strain is

εεmax

= −y / ρ−c / ρ (3.6)

which when simplified and rearranged gives the same result as Eq. 3.3:

ε =yc

εmax =

y

c

εc (3.7)

Note that an important result of the strain equations for ε = −y / ρ and εmax = −c / ρ = εc

indicate that the longitudinal normal strain of any element within the beam depends on its

location y on the cross section and the radius of curvature of the beam's longitudinal axis

at that point. In addition, a contraction (-ε ) will occur in fibres located "above" the neutral

axis (+y) whereas elongation (+ε ) will occur in fibres located "below" the neutral axis (-y).

Stress

The determination of stress distributions of beams in necessary for determining the

level of performance for the component. In particular, stress-based failure theories

require determination of the maximum combined stresses in which the complete stress

state must be either measured or calculated.Normal Stress: Having derived the proportionality relation for strain, εx , in the x-

direction, the variation of stress, σ x , in the x-direction can be found by substituting σ for

ε in Eqs. 3.3 or 3.7. In the elastic range and for most materials uniaxial tensile and

compressive stress-strain curves are identical. If there are differences in tension and

compression stress-strain response, then stress must be computed from the strain

distribution rather than by substitution of σ for ε in Eqs. 3.3 or 3.7.Note that for a beam in pure bending since no load is applied in the z-direction, σ z

is zero throughout the beam. However, because of loads applied in the y-direction toobtain the bending moment, σ y is not zero, but it is small enough compared to σ x to

neglect. In addition, σ x while varying linearly in the y direction is uniformly distributed in

the z-direction. Therefore, a beam under only a bending load will be in a uniaxial, albeit a

non uniform, stress state.

3.6

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Figure 3.7 Stress (force) distribution in a bent beam

Note that for static equilibrium, the resisting moment, Mr, must equal the appliedmoment, M, such that MO = 0∑ where (see Fig. 3.7):

Mr = dFy =A∫ σdAy

A∫ (3.8)

and since y is measured from the neutral surface, it is first necessary to locate this surfaceby means of the equilibrium equation Fx = 0∑ which gives σdA = 0

A∫ . For the case of

elastic action the relation between σ x and y can be obtained from generalized Hooke's

law σ x =E

1+ν( ) 1−2ν( ) 1−ν( )εx +ν εy +εz( )[ ] and the observation that εy = εz = −νεx .

The resulting stress-strain relation is for the uniaxial stress state such that σ x = Eεx

which when substituted into Eq. 3.3 or 3.7 gives

σ x =Eεc

c

y =

σc

cy (3.9)

Substituting Eq. 3.9 into Eq. 3.8 gives:

Mr = σdAyA∫ = σc

cy

2dA =

A∫

σ x

yy

2dA

A∫ (3.10)

Note that the integral is the second moment of the cross sectional area, also known as the

moment of inertia, I, such that

I = y2dA

A∫ (3.11)

3.7

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Figure 3.8 Action of shear stresses in unbonded and bonded boards

Substituting Eq. 3.11 into Eq. 3.10 and rearranging results in the elastic flexure

stress equation:

σ x = MyI

(3.12)

where σ x is the normal bending stress at a distance y from the neutral surface and acting

on a transverse plane and M is the resisting moment of the section. At any section of the

beam, the fibre stress will be maximum at the surface farthest from the neutral axis such

that.

σ max = McI

= MZ

(3.13)

where Z=I/c is called the section modulus of the beam. Although the section modulus

can be readily calculated for a given section, values of the modulus are often included in

tables to simplify calculations.

Shear Stress: Although normal bending stresses appear to be of greatest concern

for beams in bending, shear stresses do exist in beams when loads (i.e., transverse

loads) other than pure bending moments are applied. These shear stresses are of

particular concern when the longitudinal shear strength of materials is low compared to

the longitudinal tensile or compressive strength (an example of this is in wooden beams

with the grain running along the length of the beam). The effect of shear stresses can be

visualized if one considers a beam being made up of flat boards stacked on top of one

another without being fastened together and then loaded in a direction normal to the

surface of the boards. The resulting deformation will appear somewhat like a deck of

cards when it is bent (see Fig. 3.8a). The lack of such relative sliding and deformation in

an actual solid beam suggests the presence of resisting shear stresses on longitudinal

planes as if the boards in the example were bonded together as in Fib. 3.8b. The resulting

deformation will distort the beam such that some of the assumptions made to develop the

bending strain and stress relations (for example, plane sections remaining plane) are not

valid as shown in Fig. 3.9.

3.8

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Figure 3.9 Distortion in a bend beam due to shear

The development of a general shear stress relation for beams is again based onstatic equilibrium such that F = 0∑ . Referring to the free body diagram shown in Fig.

3.10, the differential force, dF1 is the normal force acting on a differential area dA and isequal to σ dA . The resultant of these differential forces is F1 (not shown). Thus,

F1 = σ dA∫ integrated over the shaded area of the cross section, where σ is the fibre

stress at a distance y from the neutral surface and is given by the expression σ =MyI

.

Figure 3.10 Free body diagram for development of shear stress relation

3.9

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When the two expressions are combined, the force, F1, becomes:

F1 = MI

y dA=MI

ty dyh

c∫∫ (3.14)

Similarly, the resultant force on the right side of the element is

F2 = M+∆M( )I

ty dyh

c

∫ (3.15)

The summation of forces in the horizontal direction on Fig. 3.10 gives

VH = F2 −F1 = ∆MI

ty dyh

c

∫ (3.16)

The average shear stress is VH divided by the area from which

τ = lim∆x →0

∆M∆x

1It

ty dy

h

c

∫ =dMdx

1It

ty dy

h

c

∫ (3.17)

Recall that V=dM/dx, which is the shear at the beam section where the stress is being

evaluated. Note that the integral, Q= ty dyh

c

∫ is the first moment of that portion of the cross

sectional area between the transverse line where the stress is being evaluated and the

extreme fiber of the beam. When Q and V are substituted into Eq. 3.17, the formula for the

horizontal / longitudinal shear stress is:

τ = VQIt

(3.18)

Note that the flexure formula used in this derivation is subject to the same

assumptions and limitations used to develop the flexure strain and stress relations. Also,

although the stress given in Eq. 3.18 is associated with a particular point in a beam, it is

averaged across the thickness, t, and hence it is accurate only if t is not too great. For

uniform cross sections, such as a rectangle, the shear stress of Eq. 3.18 takes on aparabolic distribution, with τ =0 at the outer fibre (where y=c and σ =σ max ) and

τ =τmax at the neutral surface (where y=0 and σ =0) as shown in Fig. 3.11.

3.10

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X

Y τ = 0

τ = 0

τ = τ Ν/Ασ = 0

σ = σ

σ = σ

max

max

min

Figure 3.11 Shear and normal stress distributions in a uniform cross section beam

Finally, the maximum shear stress for certain uniform cross section geometries can

be calculated and tabulated as shown in Fig. 3.12. Note that a first order approximation

for maximum shear stress might be made by dividing the shear force by the cross

sectional area of the beam to give an average shear stress such that τav ≈ V

A. However,

if the maximum shear stress is interpreted as the critical shear stress, than an error of 50%

would result for a beam with a rectangular cross section where τmax ≈ 3V2A

which is 1.5

times τav ≈ V

A.

τmax3V2A

=

τmax4V3A

=

τmax2V A

=

h

b

A=bh

A=( /4) dπ 2

A=( /4) (d - d )π 22o i

d

d

do

i

Figure 3.12 Maximum shear stresses for some common uniform cross sections

3.11

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Deflections

Often limits must be placed on the amount of deflection a beam or shaft may

undergo when it is subjected to a load. For example beams in many machines must

deflect just the right amount for gears or other parts to make proper contact. Deflections of

beams depend on the stiffness of the material and the dimensions of the beams as well as

the more obvious applied loads and supports. In order of decreasing usage four common

methods of calculating beam deflections are: 1) double integration method, 2)

superposition method, 3) energy (e.g., unit load) method, and 4) area-moment method.

The double integration method will be discussed in some detail here.

Deflections Due to Moments: When a straight beam is loaded and the action is

elastic, the longitudinal centroidal axis of the beam becomes a curve defined as "elastic

curve." In regions of constant bending moment, the elastic curve is an arc of a circle ofradius, ρ , as shown in Fig. 3.13 in which the portion AB of a beam is bent only with

bending moments. Therefore, the plane sections A and B remain plane and the

deformation (elongation and compression) of the fibres is proportional to the distance

from the neutral surface, which is unchanged in length. From Fig. 3.13:

θ = Lρ

=L + δρ + c

(3.19)

from whichcρ

= δL

= ε = σE

= McEI

(3.20)

and finally1ρ

= MEI

(3.21)

which relates the radius of curvature of the neutral surface of the beam to the bending

moment, M, the stiffness of the material, E, and the moment of inertia of the cross section,

I.

Figure 3.13 Bent element from which relation for elastic curve is obtained

3.12

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Equation 3.21 is useful only when the bending moment is constant for the interval

of the beam involved. For most beams the bending moment is a function of the position

along the beam and a more general expression is required.

The curvature equation from calculus is

= d2y / dx2

1+ dy / dx( )2[ ]3/2 (3.22)

which for actual beams can be simplified because the slope dy/dx is small and its square

is even smaller and can be neglected as a higher order term. Thus, with these

simplifications, Eq. 3.22 becomes

= d2y

dx 2 = y ' ' (3.23)

Substituting Eq. 3.23 into Eq. 3.21 and rearranging gives

EId2y

dx2 = Mx = EIy' ' (3.24)

which is the differential equation for the elastic curve of a beam.

An alternative method for obtaining Eq. 3.24 is to use the geometry of the bent

beam as shown in Fig. 3.14 where it is evident that dy/dx = tan θ ≈ θ for small angles and

that d2y /dx 2 = dθ /dx . From Fig. 3.14 it can be shown that

dθ =dLρ

=dxρ

(3.25)

for small angles and therefore.

y' ' =d2y

dx 2 = dθdx

= 1ρ

= M x

EI⇒ EI

d2y

dx2 = EIy' ' = Mx (3.26)

For the coordinate system shown in Fig. 3.15, the signs of the moment and second

derivative are as shown. It is also important to note the following physical quantities and

beam action.

Figure 3.14 Bent beam from which relation for elastic curve is obtained.

3.13

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Figure 3.15 Sign conventions used for deflection

deflection = y

slope =dydx

= y '

moment = Mx =EId2y

dx 2 = EIy' '

shear =dMdx

=EId3y

dx3= EIy' ' ' (for constant EI)

load =dVdx

=EId4ydx4

= EIyiv(for constant EI)

(3.27)

It is interesting to note that from Eqs. 3.24 and 3.26 can be written as

Mx = EIdθdx

(3.28)

from which

dθθA

θB

∫ = M x

EIdx

xA

xB

∫ (3.29)

Eqs. 3.28 and 3.29 show that except for the factor EI, the area under the moment

diagram between any two points along the beam gives the change in slope between the

same two points. Likewise, the area under the slope diagram between two points along a

beam gives the change in deflection between these points. These relations have been

used to construct the series of diagrams shown in Fig. 3.16 for a simply supported beam

with a concentrated load at the center of the span. The geometry of the beam was used to

locate the points of zero slope and deflection, required as the starting points for the

construction.

3.14

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Figure 3.16 Illustration of various elastic relations for a beam in three-point loading

It is important to remember that the calculation of deflections from elastic curve

relations is based on the following assumptions:

1) The square of the slope of the beam is assumed to be negligible

compared to unity

2) The beam deflection due to shear stresses is negligible (i.e., plane

sections remain plane)

3) The value of E and I remain constant for any interval along the beam.

The double integration method can be used to solve Eq. 3.24 for the deflection y as

a function of distance along the beam, x. The constants of integration are evaluated by

applying the applicable boundary conditions.

Boundary conditions are defined by a known set of values of x and y or x and dy/dx

at a specific point in the beam. One boundary condition can be used to determine one

and only one constant of integration. A roller or pin at any point in a beam (see Figs.

3.17a and 3.17b) represents a simple support which cannot deflect (y=0) but can rotate

(dy/dx≠0). At a fixed end (see Figs. 3.17c and 3.17d) the beam can neither deflect or

rotate (y=0 and dy/dx=0).

Matching conditions are defined as the equality of slope or deflection, as

determined at the junction of two intervals from the elastic curve equations for both

intervals.

3.15

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Figure 3.17 Types of boundary conditions

Calculating deflection of a beam by the double integration method involves four

definite steps and the following sequence for these steps is recommended.

1) Select the interval or intervals of the beam to be used; next, place a set of

coordinate axes on the beam with the origin at one end of an interval and then

indicate the range of values of x in each interval. For example, two adjacent

intervals might be: 0≤x≤L and L≤x≤3L

2) List the available boundary conditions and matching conditions (where two or

more adjacent intervals are used) for each interval selected. Remember that two

conditions are required to evaluate the two constants of integration for each interval

used.

3) Express the bending moment as a function of x for each interval selected, and

equate it to EI dy2/dx2 =EIy''.

4) Solve the differential equation or equations form item 3 and evaluate all

constants of integration. Check the resulting equations for dimensional

homogeneity. Calculate the deflection a specific points where required.

Deflections due to Shear: Generally deflections due to shear can be neglected as

small (<1%) compared to deflections due to moments. However, for short, heavily-loaded

beams, this deflection can be significant and an approximate method can be used to

evaluate it. The deflection of the neutral surface, dy due to shearing stresses in the

interval dx along the beam in Fig. 3.18 is

dy = γdx =τG

dx =VQ

GItdx (3.30)

from which the shear in Fig. 3.18 is negative such that

GIt

Q

dy

dx= −V ⇒

GIt

Qy' = −V (3.31)

3.16

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Since the vertical shearing stress varies from top to bottom of a beam the deflection

due to shear is not uniform. This non uniform distribution is reflected as slight warping of

a beam. Equation 3.31 gives values too high because the maximum shear stress (at the

neutral surface) is used and also because the rotation of the differential shear element is

ignored. Thus, an empirical relation is often used in which a shape factor, k, is employed

to account for this change of shear stress across the cross section such that

kAGy' = −V ⇒ y'=−V

kAG(3.32)

Often k is approximated as k≈1 but for box-like sections or webbed sections it is

estimated as:

1

k=

A total

Aweb(3.33)

A single integration method can be used to solve Eq. 3.32 for the deflection due to shear.

The constants of integration are then determined by employing the appropriate boundary

and matching conditions. The resulting equation provides a relation for the deflection due

to shear as a function of the distance x along the length of the beam. Note however that

unless the beam is very short or heavily loaded the deflection due to shear is generally

only about 1% of the total beam deflection.

Figure 3.18 Deflection due to shear stress

3.17

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An example of the use of integration methods is as follows for a simply supported

beam in three-point loading. The loading condition, free body, shear and moment

diagrams are shown in Fig. 3.19.

a b

a+b=L

P

R1 R2

P

Loading Diagram

Free Body Diagram

Shear Diagram

Moment Diagram

=Pb/L =Pa/L

Pb/L

Pa/L

Pbx/L

x

0 ≤ x ≤ a x=a a0 ≤ x ≤ L

(Pbx/L)-P(x-a)

Pba/L

Figure 3.19 Loading condition, free body, shear and moment diagrams

There are two boundary conditions: at x=0, y1=0 and at x=L, y2=0

There are two matching conditions: at x=a, y'1=y'2 and at x=a, y1=y2

3.18

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For 0 ≤ x≤ a (region 1) For a≤ x≤ L (region 2)

V= Pb/L V= Pa/LM=Pbx/L M=(Pbx/L)-P(x-a)

Double Integration Method Double Integration Method

EIy1' ' = −M = −Pbx

LEIy2 '' = −M = −

Pbx

L+ P(x − a)

EIy1' '∫ =-Pbx

L⇒ EIy1' =∫

-Pbx2

2L+ C1 EIy2 ''∫ =

−Pbx

L+ P(x − a) ⇒∫

EIy2 ' =−Pbx 2

2L+

P(x − a)2

2+ C2EIy1'∫ =

−Pbx2

2L+ C1 ⇒∫

EIy1 =−Pbx3

6L+ C1x + C3

EIy2 '∫ =−Pbx2

2L+

P(x -a)2

2+C2 ⇒∫

EIy2 =−Pbx3

6L+

P(x -a)3

6+ C2x + C4

Applying the matching conditons at x=a, y'1=y'2

y1'=1

EI

-Pba2

2L+ C1

=1

EI

-Pba2

2L+

P a-a( )2

2+C2

= y2 '

so that C1 = C2

and at x=a, y1=y2

y1 =1

EI

−Pba3

6L+ C1a + C3

=1

EI

−Pba3

6L+

P(a -a)3

6+ C2a + C4

= y2

Since C1 = C2 ,then C3 = C4

3.19

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Applying the boundary conditons at x=0, y1=0

y1 = 0 =1

EI

−Pb 03

6L+ C10 + C3

So C3 = 0

and at x=L, y2=0

y2 = 0 = 1EI

−PbL3

6L+ P(L-a) 3

6+ C2L + C4

=

Since (L-a) = b and C1 = C2 ,and C3 = C4 = 0

then C2 =PbL

6−

Pb3

6L

=Pb

6LL2 − b2[ ]

Finally, the equations for deflection due to the bending moment are:

For 0 ≤ x≤ a (region 1) For a≤ x≤ L (region 2)

y1 =−Pbx

6EILL2 − b2 − x2[ ] y2 =

−Pbx

6EILL2 − b2 − x2[ ] −

P x -a( )3

6EI

The deflection due to the shear component is:

For 0 ≤ x≤ a (region 1) For a≤ x≤ L (region 2)

V=Pb/L V=Pa/L

kAG y'1= −V ⇒ y'1=−V

kAG= −

Pb

L

1

kAG kAG y'1= −V ⇒ y'1=

−V

kAG= −

Pa

L

1

kAG

y'1∫ =−Pb

L

1

kAG∫ ⇒ y1 =−Pb

kLAGx + C1 y'1∫ =

−Pa

L

1

kAG∫ ⇒ y1 =−Pa

kLAGx + C2

3.20

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Applying the boundary condition at x=0, y1=0,

y1 = 0 =−Pb

kLAG0 + C1

So C1 = 0 and y1 = −PbkLAG

x for 0 ≤ x ≤ a

Applying the boundary condition at x=L, y2=0,

y2 = 0 =−Pa

kLAGL + C2

So C2 = PakAG

and y2 = PakAG

− xL

+1

Now the total deflection relation for both bending and shear is:

For 0 ≤ x≤ a (region 1) For a≤ x≤ L (region 2)

V=Pa/L

y2 = −Pbx6EIL

L2 − b2 − x2[ ] − P x-a( )3

6EI

−Pa

kLAG

−x

L+1

y1 = −Pbx

6EILL2 − b2 − x2[ ]

−Pb

kLAGx

3.21

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4.1

4. BEAMS: CURVED, COMPOSITE, UNSYMMETRICAL

Discussions of beams in bending are usually limited to beams with at least one

longitudinal plane of symmetry with the load applied in the plane of symmetry or to

symmetrical beams composed of longitudinal elements of similar material or to initially straight

beams with constant cross section and longitudinal elements of the same length. If any of these

assumptions are violated, the simple equations which describe the beam bending stress and

strain are no longer applicable. The following sections discuss curved beams, composite

beams and unsymmetrical beams.

Curved Beams

One of the assumptions of the development of the beam bending relations is that all

longitudinal elements of the bean have the same length, thus restricting the theory to initially

straight beams of constant cross section. Although considerable deviations from this restriction

can be tolerated in real problems, when the initial curvature of the beams becomes significant,

the linear variations of strain over the cross section is no longer valid, even though the

assumption of plane cross sections remaining plane is valid. A theory for a beam subjected to

pure bending having a constant cross section and a constant or slowly varying initial radius of

curvature in the plane of bending is developed as follows. Typical examples of curved

beams include hooks and chain links. In these cases the members are not slender but rather

have a sharp curve and their cross sectional dimensions are large compared with their radius of

curvature.

Fig 4.1 Curved beam element with applied moment, M

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4.2

Fig 4.1 is the cross section of part of an initially curved beam. The x-y plane is the plane of

bending and a plane of symmetry. Assumptions for the analysis are: cross sectional area is

constant; an axis of symmetry is perpendicular to the applied moment; M, the material is

homogeneous, isotropic and linear elastic; plane sections remain plane, and any distortions of

the cross section within its own plane are neglected. Since a plane section before bending

remains a plane after bending, the longitudinal deformation of any element will be proportional

to the distance of the element from the neutral surface.

In developing the analysis, three radii, extending from the center of curvature, O’, of the

member are shown in Fig 4.1. The radii are: r that references the location of the centroid of the

cross sectional area; R that references the location of the neutral axis; and r references some

arbitrary point of area element dA on the cross section. Note that the neutral axis lies within the

cross section since the moment M creates compression in the beams top fibers and tension in

its bottom fibers. By definition, the neutral axis is al line of zero stress and strain.

If a differential segment is isolated in the beam (see Fig 4.2). The stress deforms the

material in such a way that the cross section rotates through an angle of δθ /2. The normal strain

in an arbitrary strip at location r can be determined from the resulting deformation. This strip has

an original length of Lo=r dθ . The strip’s total change in length, ∆L R r= −2 2( ) /δθ . The normal

strain can be written as:

. εδθ

θ= =

−∆L

L

R r

o

( )r d (4.1)

To simplify the relation, a constant k is defined as k d= δθ θ/ such that the normal strain can be

rewritten as:

ε =−

kR r

r(4.2)

Fig 4.2 Isolated differential element is a curved beam

Note that Eq 4.2 shows that the normal strain is a nonlinear function of r varying in a

hyperbolic fashion. This is in contrast to the linear variation of strain in the case of the straight

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4.3

beam (i.e., ερ

=−y

). The nonlinear strain distribution for the curved beam occurs even though

the cross section of the beam remains plane after deformation. The moment, M, causes the

material to deform elastically and therefore Hooke’s law applies resulting the following relation

for stress:

σ ε= =−

E EkR r

r(4.3)

Because of the linear relation between stress and strain, the stress relation is also

hyperbolic. However, with the relation for stress determined, it is possible to determine the

location of the neutral axis and thereby relate the applied moment, M to this resulting stress.

First a relation for the unknown radius of the neutral axis from the center of curvature, R, is

determined. Then the relation between the stress, σ , and the applied moment, M is

determined.

Force equilibrium equations can applied to obtain the location of R (radius of the neutral

axis). Specifically, the internal forces caused by the stress distribution acting over the cross

section must be balanced such that the resultant internal force is zero:

F

F dA

x∑

∑ ∫

=

= =

=

0

0

Now since =dFdA

then dF = dA and

dA =

or EkR r

rdA 0

σ σ

σ σ

(4.4)

Because Ek and R are constants Eq 4.4 can be rearranged such that:

EkRr

dA -rr

dAdAr

- dA

= ⇒ =∫∫ ∫∫0 0R (4.5)

Solving Equation 4.5 for R results in:

R =dAdAr

=AdAr

∫∫ ∫ (4.6)

where R is the location of the neutral axis referenced from the center of curvature, O’, of the

member, A is the cross sectional area of the member and r is the arbitrary position of

Table 4.1 Areas and Integrals for Various Cross Sections

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4.4

the area element dA on the cross section and is referenced from the center of curvature, O’, of

the member. Equation 4.6 can be solved for various cross sections with examples of common

cross sections listed in Table 4.1

Moment equilibrium equations can be applied to relate the applied moment, M, to the

resulting stress, σ . Specifically, the internal moments caused by the stress distribution acting

over the cross section about the neutral must be balanced such that the resultant internal

moment balances the applied moment:

Recall that since =dFdA

then dF = dA

if y is the distance from the neutral axis such that y = R - r

then dM = y dF or dM = y dA

Applying moment equilibrium such that M 0 gives

y dA 0 or M y dA

Substituting the derived relations for y and gives

M = y dA

σ σ

σ

σ σ

σ

σ

( )=

− ( )= = ( )

( )= −−

∑∑ ∑

∫∑

M

R r EkR r

rdA( ) (4.7)

Again realizing that Ek and R are constants, Eq 4.7 can be expanded and grouped such that:

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4.5

M Ek RdA

rR dA rdA= − +

⇓ ⇓ ⇓

∫∫∫2 2 such that

R AR

2R A rA which gives

M = Ek R AR

2R A + rA = EkA(r -R)

2

2(4.8)

Note that the third integral term in Eq 4.8, comes from the geometric determination of the

centroid such that r = r dA/A∫ . Equation 4.8 can now be solved for Ek such that Ek =M

A(r -R).

Substituting this relation for Ek into Eq 4.3 gives:

σ ε= =−

=−

=−

E EkR r

r

R r

r

R rMA(r -R)

M( )Ar(r -R)

(4.9)

Substituting the relations involving y into Eqs 4.9 (y=R-r and r=R-y) along with an term for

“eccentricity” e=r -R gives:

σ =−

=M( )Ar(r -R)

MyAe(R - y)

R r(4.10)

which is the so-called curved-beam flexure formula where σ is the normal stress, M is the

applied moment, y is the distance from the neutral axis (y=R-r), A is the area of the cross

section, e is the “eccentricity” (e=r -R) and R is the radius of the neutral axis (R =AdAr

∫).

Note that Eq. 4.10 gives the normal stress in a curved member that is in the direction of

the circumference (a.k.a. circumferential stress) and is nonlinearly distributed across the cross

section (see Fig 4.3). It is worth noting that due to the curvature of the beam a compressive

radial stress (acting in the direction of r) will also be developed. Typically the radial stress is

small compared to the circumferential stress and can be neglected, especially if the cross

section of the member is a solid section. Sometimes, such as the case of thin plates or thin

cross sections (e.g., I-beam), this radial stress can become large relative to the circumferential

stress. If the beam is loaded with forces (instead of pure moments) then additional stresses will

occur on radial planes. Because the action is elastic, the principle of superposition applies and

the additional normal stresses can be added to the flexural stresses obtained in Eq. 4.10.

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4.6

Fig 4.3 Nonlinear stress distribution across cross section is a curved beam

The curved beam flexure formula is usually used when curvature of the member is

pronounced as in the cases of hooks and rings. A rule of thumb, for rectangular cross sections

for which the ratio of radius of curvature to depth ( r /h) is >5, shows that the curved beam

flexure formula agrees well with experimental, elasticity, and numerical results. If the flexure

formula used, a difference of 7% from the maximum stress determined from the curved beam

flexure formula can result at r /h=5. As this ratio increases (i.e., at the beam becomes less

curved and more straight), the difference of the maximum stress calculated from the flexure

formula for the straight beam and the curved beam flexure formula becomes much less.

Fig. 4.4 Crane hook with rectangular cross section

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4.7

A common machine element problem involving curved beams is the crane hook shown

in Fig. 4.4. In this problem, the load, F is 22,240 N, the cross sectional thickness, t=b=19.05

mm and the cross sectional width, h=W= is 101.6 mm. Since A=bh, dA=bdr and from Eq. 4.6:

RAdA

r

bh

b

rdr

h

r

rr

ro

ii

o= = =

∫ ∫ ln

(4.11)

From Figs 4.4a and 4.4b, ri=50.8 mm, ro=152.4 mm, and A=5161.3 mm2.

Substituting the appropriate values into Eq. 4.11 gives R=92.5 mm. The eccentricity,

e=(101.6-92.5)=9.1 mm. Since the moment, M is positive such that M = F r ( r = radialdistance to the centroid where r =(ro+ri)/2).

In the case, the axial force, F, superposes an axial stress on the bending stress such

that

σ = +−

= +−F

A

My

Ae R y

r

r( ),

.( , * . )( . )

. ( . )22 2405161 3

22 240 101 6 92 55161 3 9 1

(4.12)

Substituting values of r from ri=50.8 mm to ro=152.4 mm results in the stress distribution

shown in Fig. 4.5 (in psi). The stresses on the inner and radii are 116.5 MPa and -38.6 MPa,

respectively. Note that if the flexure stress relations for an initially straight beam are used such

that:

σ = + = +−

( )F

A

My

I

r22 2405161 3

22 240 101 6 101 6

19 05 101 6

12

3

,.

( , * . )( . )

. .(4.13)

The maximum and minimum stresses are ±80.4 MPa. A straight beam assumption thus

underpredicts the maximum tensile stress and overpredicts the maximum compressive stress.

Fig 4.5 Stress distribution across the cross section of a crane hook

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4.8

Unsymmetrical Bending

Another of the limitations of the usual development of beam bending equations is that

beams are assumed to have at least one longitudinal plane of symmetry and that the load is

applied in the plane of symmetry. The beam bending equations can be extended to cover

pure bending (i.e., bent with bending moments only and no transverse forces) of 1) beams

with a plane of symmetry but with the load (couple) applied not in or parallel to the plane of

symmetry or 2) beams with no plane of symmetry.

Fig 4.6 depicts a beam of unsymmetrical cross section loaded with a couple, M, in a

plane making an angle, α , with the xy plane, where the origin of coordinates is at the centroid of

the cross section. The neutral axis, which passes through the centroid for linearly elastic actionmakes an unknown angle, β , with the z axis. The beam is straight and of uniform cress section

and a plane cross section is assumed to remain plane after bending. Note that the following

development is restricted to elastic action.

Since the orientation of the neutral axis is unknown, the usual flexural stress distributionfunction (i.e., σ ε σ= =E c y c yc c( / ) ( / ) ) cannot be expressed in terms of one variable.

Fig 4.6 Beam undergoing unsymmetrical bending

However, since the plane section remains plane, the stress variation can be written as:

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4.9

σ = +k y k z1 2 (4.14)

The resisting moments with respect to the z and y axes can be written as

M dAy k y dA k zydA k k

M dAz k yzdA k z dA k k

rzA AA

ryA AA

= = + = +

= = + = +

∫ ∫∫

∫ ∫∫

σ

σ

12

2 1 2

1 22

1 2

I I

I I

z yz

yz y

(4.15)

where I Iy zand are the moment of inertia of the cross sectional area with respect to the z and y

axes, respectively, and Iyz is the product of inertia with respect to these two axes. It will be

convenient to let the y and z axes be principal axes, Y and Z; then Iyz is zero. Equating the

applied moment to the resisting moment and solving for k and k1 2 gives:

M k M kM

M k M kM

rZ

rY

= =

= =

1 1

2 2

I =I

I =I

ZZ

YY

coscos

sinsin

αα

αα

(4.16)

Substituting the expressions for k given in Eq. 4.16 into Eq. 4.14 gives the elastic flexure

formula for unsymmetrical bending.

σα α

= Y + Z M Mcos sin

I IZ Y

(4.17)

Since σ is zero at the neutral axis, the orientation of the neutral axis is found by setting

Eq. 4.17 equal to zero, for which

cos sinα αI IZ Y

Y = - Z (4.18)

or

Y = -tan Z αII

Z

Y

(4.19)

where Y is the equation of the neutral axis in the YZ plane. The slope of the line is the dY/dZand since dY/dZ= tanβ , the orientation of the neutral axis is given by the expression

tan = - tan β αII

Z

Y

(4.20)

The negative sign indicates that the angles, α and β are in adjacent quadrants.

Note that the neutral axis is not perpendicular to the plane of loading unless 1) the angle,

α , is zero, in which case the plane of loading is (or is parallel to) a principal angle, or 2) the two

principal moments of inertia are equal. This reduces to the special kind of symmetry where all

centroidal moments of inertia are equal (e.g., square, rectangle, etc.)

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4.10

Composite Beams

The method of "fibre" stress calculation for basic beam bending is sufficiently general to

cover symmetrical beams composed of longitudinal elements (layers) of different materials.

However, for many real beams of two materials (often referred to as reinforced beams) a

method can be developed to allow the use of the elastic flexure formula, thus reducing the

computational labor involved. Of course, the method is applicable to the elastic region only.

The assumption of plane sections remaining plane is still valid, provided that the

different materials are securely bonded together so as to give the necessary resistance to

longitudinal shearing stresses. Therefore, the usual linear transverse distribution of longitudinal

strains is valid.

The beam shown in Fig. 4.7 is composed of a central portion of material A and two

outer layers of material B. The beam will serve as the model for the development of the stress

distribution. The section is assumed to be symmetrical with respect to the xy and xz planes

and the moment is applied in the xy plane. As long as neither material is subjected to stresses

greater than the proportional limit stress, then Hooke's law applies and the strain relation is:

ε εb a= ba

(4.21)

which, in terms of stress, becomes

σ σb

B

a

AE E

= ba

( ) (4.22)

After rearranging, Eq. 4.22 becomes the relation for the stress distribution:

σ σb aB

A

E

E=

ba

(4.23)

From this relation, it is evident that the junction between the two materials where

distances a and b are equal, there is an abrupt change in the stress determined by the ratio,

nEE

B

A=

, of the two elastic moduli (see Fig. 4.7). Using Eq. 4.23, the normal force on a

differential end area of element B is given by the expression:

dF n aB b a= ≤ ≤

σ σ σdA =

ba

dA =ba

(nt)dy for c y c A B (4.24)

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4.11

where t is the thickness (width) of the beam at a distance b from the neutral surface. The first

term in parentheses represents the linear stress distribution in a homogeneous material A. The

second term in the parentheses may be interpreted as the extended width of the beam fromy=cA to y=cB if material B were replaced by material A, thus resulting in an equivalent or

transformed cross section for a beam of homogeneous material. The transformed section is

obtained by replacing either material by an equivalent amount of the other material as

determined by the ratio, n, of their elastic moduli. The method is not limited to two materials:

however the use of more than two materials in a beam might be unusual.

Fig 4.7 Composite beam with two different materials

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5. MECHANICAL PROPERTIES AND PERFORMANCE OF MATERIALS

Samples of engineering materials are subjected to a wide variety of mechanical

tests to measure their strength, elastic constants, and other material properties as well as

their performance under a variety of actual use conditions and environments. The results

of such tests are used for two primary purposes: 1) engineering design (for example,

failure theories based on strength, or deflections based on elastic constants and

component geometry) and 2) quality control either by the materials producer to verify the

process or by the end user to confirm the material specifications.

Because of the need to compare measured properties and performance on a

common basis, users and producers of materials use standardized test methods such as

those developed by the American Society for Testing and Materials (ASTM) and the

International Organization for Standardization (ISO). ASTM and ISO are but two of many

standards-writing professional organization in the world. These standards prescribe the

method by which the test specimen will be prepared and tested, as well as how the test

results will be analyzed and reported. Standards also exist which define terminology and

nomenclature as well as classification and specification schemes.

The following sections contain information about mechanical tests in general as

well as tension, hardness, torsion, and impact tests in particular.

Mechanical Testing

Mechanical tests (as opposed to physical, electrical, or other types of tests) often

involves the deformation or breakage of samples of material (called test specimens or test

pieces). Some common forms of test specimens and loading situations are shown in Fig

5.1. Note that test specimens are nothing more than specialized engineering components

in which a known stress or strain state is applied and the material properties are inferred

from the resulting mechanical response. For example, a strength is nothing more than a

stress "at which something happens" be it the onset of nonlinearity in the stress-strain

response for yield strength, the maximum applied stress for ultimate tensile strength, or

the stress at which specimen actually breaks for the fracture strength.

Design of a test specimen is not a trivial matter. However, the simplest test

specimens are smooth and unnotched. More complex geometries can be used to

produce conditions resembling those in actual engineering components. Notches (such

as holes, grooves or slots) that have a definite radius may be machined in specimens.

Sharp notches that produce behaviour similar to cracks can also be used, in addition to

actual cracks that are introduced in the specimen prior to testing.

5.1

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Figure 5.1 Geometry and loading scenarios commonly employed in mechanical testing ofmaterials. a) tension, b) compression, c) indentation hardness, d) cantilever flexure, e)

three-point flexure, f) four-point flexure and g) torsion

Equipment used for mechanical testing range from simple, hand-actuated devices

to complex, servo-hydraulic systems controlled through computer interfaces. Common

configurations (for example, as shown in Fig. 5.2) involve the use of a general purpose

device called a universal testing machine. Modern test machines fall into two broad

categories: electro (or servo) mechanical (often employing power screws) and servo-

hydraulic (high-pressure hydraulic fluid in hydraulic cylinders). Digital, closed loop

Figure 4.2 Example of a modern, closed-loop servo-hydraulic universal test machine.

5.2

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control (e.g., force, displacement, strain, etc.) along with computer interfaces and user-

friendly software are common. Various types of sensors are used to monitor or control

force (e.g., strain gage-based "load" cells), displacement (e.g., linear variable differential

transformers ( LVDT's) for stroke of the test machine), strain (e.g., clip-on strain-gaged

based extensometers). In addition, controlled environments can also be applied through

self-contained furnaces, vacuum chambers, or cryogenic apparati. Depending on the

information required, the universal test machine can be configured to provide the control,

feedback, and test conditions unique to that application.

Tension Test

The tension test is the commonly used test for determining "static" (actually quasi-

static) properties of materials. Results of tension tests are tabulated in handbooks and,

through the use of failure theories, these data can be used to predict failure of parts

subjected to more generalized stress states. Theoretically, this is a good test because of

the apparent simplicity with which it can be performed and because the uniaxial loading

condition results in a uniform stress distribution across the cross section of the test

specimen. In actuality, a direct tensile load is difficult to achieve (because of

misalignment of the specimen grips) and some bending usually results. This is not

serious when testing ductile materials like copper in which local yielding can redistribute

the stress so uniformity exists; however, in brittle materials local yielding is not possible

and the resulting non uniform stress distribution will cause failure of the specimen at a

load considerably different from that expected if a uniformly distributed load were applied.

The typical stress-strain curve normally observed in textbooks with some of the common

nomenclature is shown in Fig. 5.3. This is for a typical low-carbon steel specimen. Note

that there are a number of definitions of the transition from elastic to plastic behavior. A

few of these definitions are shown in Fig. 5.3. Oftentimes the yield point is not so well

defined as for this typical steel specimen. Another technique for defining the beginning of

plastic behavior is to use an offset yield strength defined as the stress resulting from the

intersection of a line drawn parallel to the original straight portion of the stress strain

curve, but offset from the origin of this curve by some defined amount of strain, usually 0.1

percent ( ε = 0.001) or 0.2 percent ( ε = 0. 002) and the stress-strain curve itself. The total

strain at any point along the curve in Fig. 5.3 is partly plastic after yielding begins. The

amount of elastic strain can be determined by unloading the specimen at some

deformation, as at point A. When the load is removed, the specimen shortens by an

amount equal to the stress divided by elastic modulus (a.k.a., Young's modulus). This is,

in fact, the definition of Young's modulus E =∆σ∆ε

in the elastic region.

5.3

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Figure 5.3 Engineering stress-strain diagram for hot-rolled carbon steel showingimportant properties (Note, Units of stress are psi for US Customary andMPa for S.I. Units of strain are in/in for US Customary and m/m for S.I.

Other materials exhibit stress-strain curves considerably different from carbon-steel

although still highly nonlinear. In addition, the stress-strain curve for more brittle

materials, such as cast iron, fully hardened high-carbon steel, or fully work-hardened

copper show more linearity and much less nonlinearity of the ductile materials. Little

ductility is exhibited with these materials, and they fracture soon after reaching the elastic

limit. Because of this property, greater care must be used in designing with brittle

materials. The effects of stress concentration are more important, and there is no large

amount of plastic deformation to assist in distributing the loads.

5.4

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As shown in Fig. 5.3, often basic stress-strain relations are plotted using

engineering stress, σ , and engineering strain, ε . These are quantities based on the

original dimensions of the specimen, defined as

σ =

Load

Original Area=

P

Ao(5.1)

ε =

Deformed length - Original length

Original length=

L − Lo

Lo(5.2)

The Modulus of Resilience is the amount of energy stored in stressing the material

to the elastic limit as given by the area under the elastic portion of the σ - ε diagram and

can be defined as

Ur = σ dε ≈σoεo

20

ε o

∫ (5.3)

where σo is the proportional limit stress and εo is the strain at the proportional limit stress.

Ur is important in selecting materials for energy storage such as springs. Typical values

for this quantity are given in Table 5.1.

The Modulus of Toughness is the total energy absorption capabilities of the

material to failure and is given by the total area under the σ - ε curve such that

Ut = σ dε ≈(σo + Su)

20

ε f

∫ εf (5.4)

where Su is the ultimate tensile strength, σo is the proportional limit stress and ε f is the

strain at fracture. Ut is important in selecting materials for applications where high

overloads are likely to occur and large amounts of energy must be absorbed. This

modulus has also been shown to be an important parameter in ranking materials for

resistance to abrasion or cavitation. Both these wear operations involve tearing pieces of

metal from a parent structure and hence are related to the "toughness" of the material.

Table 5.1 Energy properties of materials in tension

MaterialYield

Strength(MPa)

UltimateStrength

(MPa)

Modulus ofResilience,

(kJ/m3)

Modulus ofToughness,

(kJ/m3)SAE 1020 annealed 276 414 186 128755SAE 1020 heat treated 427 621 428 91047Type 304 stainless 207 586 103 195199Cast iron 172 586Ductile cast iron 400 503 462 50352Alcoa 2017 276 428 552 62712Red brass 414 517 828 13795

5.5

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The ductility of a material is its ability to deform under load and can be measured

by either a length change or an area change. The percent elongation, which is the

percent strain to fracture is given by:

%EL = 100εf = 100L f − Lo

Lo

= 100

Lf

Lo

−1

(5.5)

where Lf is the length between gage marks at fracture. We should note that this quantity

depends on the gage length used in measuring L, as non uniform deformation occurs in a

certain region of the specimen during necking just prior to fracture, hence, the gage length

should always be specified. The percent reduction in area is a cross-sectional area

measurement of ductility defined as

%RA = 100Ao − Af

Ao

= 100 1−

Af

Ao

(5.6)

where Af is the cross-sectional area at fracture. Note %RA is not sensitive to gage length

and is somewhat easier to obtain, only a micrometer is required. It should be realized that

the stress-strain curves just discussed, using engineering quantities, are fictitious in the

sense that the σ and ε are based on areas and lengths that no longer exist at the time of

measurement. To correct this situation true stress (σT ) and true strain (εT ) quantities are

used. The quantities are defined as:

σT =P

Ai

(5.7)

where Ai is the instantaneous area at the time P is measured. Also

εT =dL

LLo

L

∫ = lnL

Lo

(5.8)

or

εT = −dA

AAo

A

∫ = lnAo

A(5.9)

where L is the instantaneous length between gage mark at the time P is measured.

These two definitions of true strain are equivalent in the plastic range where the

material volume can be considered constant during deformation as shown below.

Since

AL = AoLo (5.10)

then

L Lo = Ao A (5.11)

The constant volume condition simply says the stressed volume AL is equal to theoriginal volume AoLo. (Note this is only true in the plastic range of deformation, in the

5.6

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elastic range the change in volume ∆V per unit volume is given by the bulk modulus B

(where B =E

3(1− υ) and υ

is Poisson's ratio).

Prior to necking, engineering values can be related to true values by noting that

εT = lnLi

Lo

= lnLo + ∆L

Lo

(5.12)

thenεT = ln(1+ ε) (5.13)

and since

Ao

A=

L

Lo=

Lo +∆ L

Lo (5.14)

so

A =

Ao

1 + ε(5.15)

and since

σT =P

A (5.16)

then

σT =P

Ao

1+ ε( ) = σ(1+ ε ) (5.17)

where σ and ε are the engineering stress and strain values at a particular load.

True stress and true strain values are particularly necessary when one is working

with large plastic deformations such as large deformation of structures or in metal forming

processes. In the elastic region the relation between stress and strain is simply the linear

equation

σ = Eε (5.18)

and also

σT = EεT (5.19)

In the plastic region, a commonly used relation to define the relation between

stress and strain is given byσT = K (εT )n = H(εT )m (5.20)

where strength coefficient, K or H, is the stress when εT =1 and m or n is an exponent

often called the strain hardening coefficient. Typically values for K or H and m or n are as

given in Table 5.2.

5.7

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Table 4.2 Material constants m or n and K or H for different sheet materials

Material Treatment n or m K or H(MPa)

SpecimenThickness

(mm)0.05% carbon rimmed steel Annealed 0.261 532 240.05/0.07% phosphorus low-carbon steel Annealed 0.156

644 24

SAE 4130 steel Annealed 0.118 1168 24Type 430 stainless steel (17% Cr) Annealed 0.229

986 32

Alcoa 24-S aluminum Annealed 0.211 386 258

The last two equations can be written in the form

log σT = log E + log εT (elastic) (5.21)

and

log σT = log K+ m log εT (plastic) (5.22)

by taking logarithms of both sides of the equations.

In this form we see that when plotted on log-log graph paper the following are true.

1. The elastic part of the deformation plots as a 45° line on true stress and

true strain coordinates. The extrapolated elastic (45° line) at a value of

strain of one corresponds to a stress value equal to the elastic modulus

(a.k.a., Young's modulus).

2. the plastic part of the deformation is a straight line of slope m. The

strength coefficient, K or H, is that stress existing when the strain is one. This

type of graph is shown for a particular aluminum alloy in Fig. 5.4. This type

of plot clearly shows the difference in elastic and plastic behavior of ductile

materials and the distinct transition between ductile and brittle behavior.

Test specimens used for tensile experiments may be either cut from flat sheet stock

or turned from round stock. The round specimens have the advantage of being usable for

many types of materials. The specimens typically have a 0.505 in (51 mm) diameter

reduced section (giving a cross sectional area of 0.2 in2.) and may have either a button

head or threaded ends for mounting in the machine. The button heads are used more

commonly for brittle materials because of less chance of failure in the heads as can occur

with threaded specimens.

The load in the specimen is read directly from the testing machine, while the

elongation is measured with some type of extensometer. In the elastic region the strains

are so small that some type of magnification of the deformations are necessary. There are

many ways to achieve this magnification.

5.8

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Figure 5.4 Logarithmic true stress-logarithmic (true) strain data plotted on

logarithmic coordinates

In the plastic region, the strains become sufficiently large that a finely graduated

scale used in conjunction with a pair of dividers to measure linear strain, or a micrometer

to measure lateral strain can be used.

In the U.S., generally a 2.0 in (50.8 mm) gage length is used to measure

deformations. The 2.0 in (50.8 mm) interval is often marked off with a special tool that

marks the interval with punch marks. These punch marks should be very light or fracture

will occur at this point. Alternatively an indelible marker can be used to avoid damaging

the surface of the test specimen.

5.9

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Hardness

In the field of engineering, hardness is often defined as the resistance of a material

to penetration. Methods to characterize hardness can be divided into three primary

categories:

1) Scratch Tests

2) Rebound Tests

3) Indentation Tests

Scratch tests commonly involve comparatively scratching progressively harder

materials. In mineralogy, a Mohs hardness scale is used as shown in Fig. 5.5. Diamond,

the hardest material, is assigned a value of 10. Decreasing values are assigned to other

minerals, down to 1 for the soft mineral, talc. Decimal fractions, such as 9.7 for tungsten

carbide, are used for materials intermediate between the standard ones. Where a

material lies on the Mohs scale is determined by a simple manual scratch test. If two

materials are compared, the harder one is capable of scratching the softer one, but not

vice versa. This allows materials to be ranked as to hardness, and decimal values

between the standard ones are assigned as a matter of judgment.

Rebound tests may employ techniques to assess the resilience of material by

measuring changes in potential energy. For example, the Sceleroscope hardness test

employs a hammer with a rounded diamond tip. This hammer is dropped from a fixed

height onto the surface of the material being tested. The hardness number is proportional

to the height of rebound of the hammer with the scale for metals being set so that fully

hardened tool steel has a value of 100. A modified version is also used for polymers.

Indentation tests actually produce a permanent impression in the surface of the

material. The force and size of the impression can be related to a quantity (hardness)

which can be objectively related to the resistance of the material to permanent

penetration. Because the hardness is a function of the force and size of the impression,

the pressure (and hence stress) used to create the impression can be related to both the

yield and ultimate strengths of materials. Several different types of hardness tests have

evolved over the years. These include macro hardness test such as Brinell, Vickers, and

Rockwell and micro hardness tests such as Knoop and Tukon.

Brinell Hardness Test In this test, a relatively large steel ball, specifically 10 mm in

diameter is used with a relatively large force. The force is usually obtained with either

3000 kg for relatively hard materials such as steels and cast irons or 500 kg for softens

materials such as copper and aluminum alloys. For very hard materials, the standard

steel ball will deform excessively and a tungsten carbide ball is used. The Brinell

5.10

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hardness dates from the late 1800's and is probably the most common hardness test in

the world.

The Brinell hardness number is obtained by dividing the applied force, P, in kg, by

the actual surface area of the indentation which is a segment of a sphere as illustrated in

Fig. 5.6 such that:

BHN = HB =P

πDt=

2P

πD D − D 2 −d 2( )[ ] (5.23)

where D is the diameter of the ball in mm, t is the indentation depth from the surface in

mm, and d is the diameter of the indentation at the surface in mm.

Brinell hardness is good for averaging heterogeneities over a relatively large area,

thus lessening the influence of scratches or surface roughness. However the large ball

size precludes the use of Brinell hardness for small objects or critical components where

large indentations may promote failure. Another limitation of the Brinell hardness test is

that because of the spherical shape of the indenter ball, the BHN for the same material

will not be the same for different loads if the same size ball is used. Thus, geometric

similitude must be imposed by maintaining the ratio of the indentation load and indenter

diameter such that:P1

D1

=P2

D2

=P3

D3

= etc. (5.24)

The Meyers hardness test is a variation on the Brinell hardness test and addresses

this lack of geometric similarity by using the projected area of the indentation such that:

MHN = HM =P

πd 2 /4(5.25)

Although the Meyers hardness is less sensitive to applied load and is a more fundamental

measure of hardness, it is rarely used.

Vickers Hardness Test (a.k.a.. diamond pyramid hardness) In this test, the same

general principles of the Brinell test are applied. However, a four-sided diamond pyramid

is implied as an indenter rather than a ball to promote geometric similarity of indentation

regardless of indentation load (see Fig. 5.7). The included angle between the faces of the

pyramid is 136° which corresponds to a desired d/D ratio for the Brinell test of 0.25. The

resulting Vickers indentation has a depth, h equal to 1/7 of the indentation size, L,

measured on the diagonal. The Vickers hardness is obtained by dividing the applied

force by the surface area of the paramedical impression such that:

VHN = HV =2P

L2 sinα2

(5.26)

where P is the indentation load which typically ranges from 0.1 to 1 kg but may be as high

as 120 kg, L is the diagonal of the indentation in mm and α is the included angle of 136°.

5.11

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Figure 5.5 Approximate relative hardnesses of metals and ceramics for Mohs scale and

indention scales.

The primary advantage of the Vickers hardness is that the result is independent of

load. However, disadvantages are that it is somewhat slow since careful surface

preparation is required. In addition, the result may be prone to personal error in

measuring the diagonal length along with interpretation of anomalies such as "pin

cushioning" for soft materials and "barreling/ridging" for hard materials.

5.12

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P=3000 kg or 500 kg

D=10 mm

dt

Steel or tungstencarbideball

Side view

0 1 2

d

Figure 5.6 Brinell hardness test.

Rockwell Hardness Test The Rockwell test is the most widely accepted hardness

test in the United States. In this test, penetration depth is measured, with the hardness

reported as the inverse of the penetration depth. A two step procedure is used as

illustrated in Fig. 5.8. The first step "sets" the indenter in the material and the second step

is the actual indentation test. The conical diamond or spherical indenter tips produce

indentation depths, the inverse of which are used to display hardness on the test machine

directly. The reported hardness is in arbitrary units, but the Rockwell scale which

identifies the indentation load and indenter tip must be reported with the hardness

number (otherwise the number is useless). Rockwell scales include those in Table 5.3.

Figure 5.7 Vickers hardness indenter

5.13

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Table 5.3 Representative Rockwell indenter specifics

Rockwell

Scale

Indenter Major Load

(kg)

A Brale 60

B 1/16" Ball 100

C Brale 150

D Brale 100

E 1/8" Ball 100

F 1/16" Ball 60

M 1/4" Ball 100

* Brale is a conical diamond indenter

Some important points concerning Rockwell hardness testing include the following

1) Indenter and anvil should be clean and well seated.

2) Surface should be clean, dry, smooth, and free from oxide

3) Surface should be flat and normal

A primary advantage of the Rockwell hardness test is that it is automatic and self-

contained thereby given and instantaneous readout of hardness which lends itself to

automation and rapid through put.

Elastic/Plastic Correlations and Conversions

The deformations caused by a hardness indenter can be correlated to those

produced at the yield and ultimate tensile strengths in a tensile test. However, an

important difference is that the material cannot freely flow outward, sot that a complex

triaxial state of stress exists under the indenter (see Fig. 5.9). Nonetheless, various

correlations have been established between hardness and tensile properties.

For example, the elastic constraint under the hardness indenture reaches a limiting

value of 3 such that the yield strength can be related to the pressure exerted by the

indenter tip:Sy = BHN x 9.816 / 3 (5.27)

where Sy is the yield strength of the material in MPa and BHN is the Brinell hardness in

kg/mm2.

5.14

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Figure 5.8 Rockwell hardness indentation for a minor load and for a major load.

Empirical relations have also been developed to correlate different hardness

number as well as hardness and ultimate tensile strength. For example, for low- and

medium carbon and alloy steels,Su = 3.45 x BHN (5.28)

where Su is the ultimate tensile strength of the material in MPa and BHN is the Brinell

hardness in kg/mm2.

Figure 5.9 Plastic deformation under a Brinell hardness indenter.

5.15

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Note that for both these relations, there is considerable scatter in actual data, so

that these relationships should be considered to provide rough estimates only. For other

classes of material, the empirical constant will differ, and the relationships may even

become nonlinear. Similarly, the relationships will change for different types of hardness

tests. Rockwell hardness correlates well with ultimate tensile strength and with other

types of hardness tests, although the relationships can be nonlinear. This situation results

from the unique indentation-depth basis of this test. For carbon and alloy steels,

conversion charts for estimating various types of hardness from one another as well as

ultimate tensile strengths are contained in an ASTM standard, ASM handbooks and

information supplied by manufacturers of hardness testing equipment

Torsion

The torsion test is another fundamental technique for obtaining the stress-strain

relationship for a metal. Because the shear stress and shear strain are obtained directly

in the torsion test, rather than tensile stress and tensile strain as in the tension test, many

investigators actually prefer this test to the tension test. Since all deformation of ductile

materials is by shear, the torsion test would seem to be the more fundamental of the two.

The torsion test is accomplished by simply clamping each end of a suitable

specimen in a twisting machine that is able to measure the torque, T, applied to the

specimen. Care must be used in gripping the specimen to avoid any bending. A device

called a troptometer is used to measure angular deformation. This device consists of two

collars which are clamped to the specimen at the desired gage length. One collar is

equipped with a pointer the other with a graduated scale, so the relative twist between the

gage marks can be determined. The troptometer is useful for measuring strains up to and

slightly past the elastic limit. For larger plastic strains, complete revolutions of the collars

are counted.

The test, then, consists of measuring the angle of twist, θ (radians) at selected

increments of torque T (N-m). Expressing the twist as θ '= θ /L, the angular deflection per

unit gage length, one is able to plot a T - θ ' diagram that is analogous to the load-

deflection diagram obtained in the tension test. To be useful for engineering purposes, itis necessary to convert this T - θ 'diagram to a shear stress (τ ) - shear strain (γ ) diagram

similar to the previous normal stress ( σ ) - normal strain (ε ) diagram. Of course, one canalso convert the τ - γ diagram to a σ - ε diagram as will be shown later.

5.16

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Figure 5.10 Torsion of cylindrical test bar

Two possible approaches are used: 1) a mechanistic approach which requires no

a priori knowledge of the properties of the particular material, only the form of the resulting

stress-strain relations, and 2) a materials approach which requires a priori knowledge of

the properties of the particular material along with the form of the resulting stress-strain

relations.

For the mechanistic approach, consider first a circular, thin-walled specimen as

shown in Figure 5.10The shear strain γ is the relative rotation of one circular cross-section with respect

to a section one unit length away or:

γ =

rθL

(5.29)

where θ is in radians. This relation is true in either the elastic or plastic range.

The shear stress τ is simply the average applied force at the tube cross section

(T/r) divided by the cross-sectional area. This is so because the stress can be assumed

uniformly distributed across the thickness of the tube, t. This gives:

τ =

T

2πr2t(5.30)

Using the Eqs. 5.29 and 5.30, the complete τ - γ diagram in the elastic and plastic range

can be obtained.

5.17

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Figure 5.11 Elastic Shear Stress Distribution

The τ - γ diagram can also be obtained from T - θ information obtained using a

solid circular test specimen. This specimen has the advantage of being somewhat easier

to grip in the testing machine and has no tendency to collapse during twisting. This is the

specimen type to be used in this laboratory. For the solid specimen, the shear strain

relation remains the same as for the tubular specimen, i.e. γ =

rθL

.

The shear stress distribution is somewhat more difficult to obtain because we can

no longer assume the stress distribution to be uniform across the section. The derivation

for the equation giving τ from T-θ data is as follows:

In the elastic deformation range the stress is distributed uniformly across the

section as shown in Fig. 5.11

Considering the very thin circumferential ring shown above, the torque resisted by

this ring is given by

dT = (shear stress) × (area) × (lever arm)

dT= τ × τr ×

a

r= 2πa2τda (5.31)

since

dA = 2π ada. (5.32)Since the stress distribution is linear, at any radius, a, the shear stress, γ , is related to themaximum shear stress, τr , existing at r by

τ = τr ×

a

r(5.33)

so substituting in the equation for dT, the torque on a small area becomes:

dT= 2πτr

ra3da (5.34)

and integrating over the entire cross-sectional area, the total external torque is equal to

5.18

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T=2πτr

r or a3da

= 2πτ r

rr4

4= πτ rr

3

2(5.35)

and the shear stress at the outermost fibers is

τr =

2T

πr 3 (5.36)

Note that Eqs. 5.33 to 5.34 applies only in the elastic region. When the metal starts to

deform plastically, the shear stress distribution is no longer linear, but is as shown in Fig.

5.12.

The relation between T and τ is no longer the same. To evaluate this relation we

begin as before, noting that the torque at a very thin ring of radius a is again given by

dT = 2πτa2da (5.37)

So the total external torque resisted across the section is then

T = 2π τ

o

a∫ a2da (5.38)

The shear strain relation γ =

aθL

at any radius a is still valid, however, and substituting this

in Eq. 5.38, we obtain

T = 2π

τγ 2L2

θ2or∫

L

θdγ (5.39)

The shear stress T at any radius a is also a function of γ only, i.e.

τ = f γ( ) (5.40)

so the expression for torque T can be written in terms of γ only as

Tθ3 = 2πL3 f γ( )

o

γr∫ γ 2dγ (5.41)

Fig. 5.12 Elastic-Plastic Shear Stress Distribution

5.19

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Differentiating both sides of this equation with respect to 9, one obtains

d

dθ= Tθ3( ) = 2πL3f γ r( )γ

r2 dγr

dθ` (5.42)

since γ r =

rθL

then

dγ r

dθ=

r

L(5.44)

and substituting these quantities in the equation for d dθ Tθ3( ) and working out the

derivative, one obtains

3Tθ2 +

dt

dθθ3 = 2πτr 3θ2 (5.45)

and

3T +θ

dt

dθ= 2πτr3 (5.44)

Solving for the shear stress, the result is

τ =

1

2πr3

θ

dt

dθ+ 3T

(5.45)

This was rather a lengthy derivation, but the application is easy. Refer to the typical

T - θ diagram as obtained from a torsion test shown in Fig. 5.13.

Fig. 5.13 Example of torque-twist curve used for data

5.20

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At the typical point P at which it is desired to obtain the shear stress, observe that

θ = BC and that dTdθ

=PCBC

such that T = AP. Substituting these quantities, the result is

τ =

1

2πr3 BCPC

BC+ 3AP

or

τ =

PC + 3AP

2πr3 .

With this last relation it is then a simple matter to obtain values of τ at various θ

positions of the plastic part of the T - θ curve. Remember that γ =

rθL

, the complete τ - γ

curve can be obtained.

For the materials approach, it is possible to again make the valid assumption that

γ =

rθL

. However, τ is determined as one of two functions of γ depending on whether the

internal stress state is in the elastic or plastic range. However, calculating this internal

stress state requires a priori knowledge of material properties usually determined from a

tensile test. In particular, E is required to calculate G =E

2(1+ υ), σo is required to calculate

τo =σ o

3, K and n are required to calculate Kτ =

K3( n +1)/2 and n for shear equals n for

tension. Once these relations are established, then it is possible to calculate the shear

stress from the shear strain for the elastic or plastic condition as follows.

τ = Gγ for τ ≤ τo and/or γ ≤ γ o (5.46)

orτ = Kτ γ n for τ > τo and/or γ > γ o (5.47)

where G is the shear modulus, Kτ =K

3( n +1)/2 in which K and n are the strength coefficient

and strain hardening exponent from the tension test, respectively. The shear strain at

yield can be determined from an effective stress-strain relation from plasticity such that

γ o =τo

G(5.48)

where τo =σ o

3 in which σo is the "yield" strength from a tension test.

For any given T-θ combination, it is possible to calculate the shear strain at the

surface of the specimen (that is, r=R) as γ =

rθL

. Comparing this shear strain to that

calculated in Eq. 5.48, allows the choice of either Eq. 5.46 or 5.47.

Note that when the shear stress at r=R is plastic, the total torque, T, required toproduce the deformation, θ , will have two components: an elastic torque, Te and a plastic

torque, Tp since the shear stress across the cross section of the specimen will have both a

5.21

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plastic part and an elastic part as shown in Fig. 5.14. The relation for T can then be

written as:Ttotal = Telastic + T plastic (5.50)

where

Telastic = τ dA r =0

τo

∫ Gγ dA r0

γ o

∫ (5.51)

Tplastic = τ dA r =τ o

τR

∫ Kτγ n dA rγ R

γ R

∫ (5.52)

For convenience it is possible to rewrite the integration variables in Eqs. 5.51 and 5.52 in

terms for the specimen radius, r, only such that

Telastic =τ y

ry

2πr 3dr0

ry

∫ (5.53)

Tplastic =K

3rθ '

3

n

2πr 2drr y

R

∫ (5.54)

where θ ' =θL

and ry =θL

γ o =θL

τo

G= θ '

τo

G.

Equations 5.53 and 5.54 can be solved either closed form or numerically for any

combination of T and θ .

Radial distance, r

τγ

r=Rr=0 ry

=K n

=f( )γ

τ=Gγ τ γτ

=r /Lθ

Elastic Plastic

Figure 5.14 Shear stress and shear strain as functions of radial distance

5.22

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Once the shear stress-strain curve is obtained, engineering properties are easily

calculated. A few of the more important quantities will be discussed. As in the tension test,

yield strengths for shearing stress can be defined, such as a proportional limit or an offsetyield strength. The Modulus of Rupture is the total area under the τ - γ curve determined

at r=R and represents the total energy absorption abilities of the material in shear.

Figure 5.15 Mohr's circles for the tensile test and torsion test

5.23

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As in the tension test, the Modulus of Resilience is the area under the elasticportion of the τ - γ curve such that

Ur = τ do

γ o

∫ γ (5.55)

Similarly, the Modulus of Toughness is the area under the total τ - γ curve such

that

Ut = τ do

γ f

∫ γ (5.56)

The Modulus of Rigidity (or Shear Modulus), G, is the slope of the τ - γ curve in the

elastic region and is comparable to Young's Modulus, E, found in tension. Recall that the

relation between E and G is G =∆τ∆γ

=E

2(1+υ ).

The true shear stress-strain curve can be compared to the tensile true stress-strain

curve by converting the normal values to shear values. The conversion is as follows:

Elastic range: τ equivalent =

σ2

; γ equivalent = 1.25ε (5.57)

Plastic range: τ equivalent =

σ2

; γ equivalent = 1.5ε (5.58)

That these values are correct can be seen from Mohr's circle of stress and of strain

for the elastic and plastic ranges (Fig. 5.15). Knowledge of Poisson's ratio, υ , is needed

for Mohr's circles of strain for the tensile test. For mild steel in the elastic range, υ = .0.30;

in the plastic range, υ =0.5 as a result of the constant volume assumption.

Impact

The static properties of materials and their attendant mechanical behavior are very

much functions of factors such as the heat treatment the material may have received as

well as design factors such as stress concentrations.

The behavior of a material is also dependent on the rate at which the load is

applied. Polymeric materials and metals which show delayed yielding are most sensitive

to load application rate. Low-carbon steel, for example, shows a considerable increase in

yield strength with increasing rate of strain. In addition, increased work hardening occurs

at high-strain rates. This results in reduced local necking, hence, a greater overall

material ductility occurs. A practical application of these effects is apparent in the

fabrication of parts by high-strain rate methods such as explosive forming. This method

5.24

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results in larger amounts of plastic deformation than conventional forming methods and,

at the same time, imparts increased strength and dimensional stability to the part.

In design applications, impact situations are frequently encountered, such as

cylinder head bolts, in which it is necessary for the part to absorb a certain amount of

energy without failure. In the static test, this energy absorption ability is called

"toughness" and is indicated by the modulus of rupture. A similar "toughness"

measurement is required for dynamic loadings; this measurement is made with a

standard ASTM impact test known as the Izod or Charpy test. When using one of these

impact tests, a small notched specimen is broken in flexure by a single blow from a

swinging pendulum. With the Charpy test, the specimen is supported as a simple beam,

while in the Izod it is held as a cantilever. Figure 5.16 shows standard configurations for

Izod (cantilever) and Charpy (three-point) impact tests.

A standard Charpy impact machine is used. This machine consists essentially of a

rigid specimen holder and a swinging pendulum hammer for striking the impact blow (see

Fig. 5.17). Impact energy is simply the difference in potential energies of the pendulum

before and after striking the specimen. The machine is calibrated to read the fracture

energy in N-m or J directly from a pointer which indicates the angular rotation of the

pendulum after the specimen has been fractured.

Figure 5.16 Charpy and Izod impact specimens and test configurations

5.25

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h1

h2

mass, m

IMPACT ENERGY=mg(h1-h2)

Figure 5.17 Charpy and Izod impact specimens and test configurations

The Charpy test does not simulate any particular design situation and data

obtained from this test are not directly applicable to design work as are data such as yield

strength. The test is useful, however, in comparing variations in the metallurgical structure

of the metal and in determining environmental effects such as temperature. It is often

used in acceptance specifications for materials used in impact situations, such as gears,

shafts, or bolts. It can have useful applications to design when a correlation can be found

between Charpy values and impact failures of actual parts.

Curves as shown in Fig. 5.18 showing the energy to fracture as measured by a

Charpy test indicate a transition temperature, at which the ability of the material to absorb

energy changes drastically. The transition temperature is that temperature at which,

under impact conditions, the material's behavior changes from ductile to brittle. This

change in the behavior is effected by many variables. Metals that have a face-centered

cubic crystalline structure such as aluminum and copper have many slip systems and are

the most resistant to low-energy fracture at low-temperature. Most metals with body-

centered cubic structures (like steel) and some hexagonal crystal structures show a sharp

transition temperature and are brittle at low temperatures.

Considering steel; coarse grain size, strain hardening, and certain minor impurities

can raise the transition temperature whereas fine grain size and certain alloying elements

will increase the low temperature toughness. Figure 5.18 shows the effect of heat

treatment on alloy steel 3140 and 2340. Note that a transition temperature as high as

about 25°C is shown. This material, then should not be in service below temperature of

25°C when impact conditions are likely to exist.

5.26

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Figure 5.18 Variation in transition-temperature range for steel in the Charpy test

In defining notch "toughness", a number of criteria have been proposed to define

the transition temperature. These include:

a. some critical energy level

b. a measure of ductility such as lateral contraction of the specimen after fracture

c. fracture surface appearance - the brittle fracture surface has a crystallineappearance, while the portion of the specimen which fracture in a ductilemanner will have a so-called fibrous appearance.

Any of these criteria are usable. Perhaps the most direct criteria for a particular

metal is to define the transition temperature as that temperature at which some minimum

amount of energy is required to fracture. During World War II, allied Victory ships literally

broke in two in conditions as mild as standing at the dock because of the use of steel with

a high-transition temperature, coupled with high-stress concentrations. It was found that

specimens cut from plates of these ships averaged only 9 J. Charpy energy absorption at

the service temperature. Ship plates were resistant to failure if the energy absorption

value was raised to 20 J at the service temperature by proper control of impurities.

5.27

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Plasticity Relations

Plasticity can be defined as non recoverable deformation beyond the point of

yielding where Hooke's law (proportionality of stress and strain) no longer applies. Flow

curves are the true stress vs. true strain curves which describe the plastic deformation. As

shown in Fig. 5.19, are several simple approximations made to represent mathematically

represent actual plastic deformation.

σο

εT

Rigid-Perfectly Plastic

σο

εTElastic-Perfectly Plastic

σο

εTElastic-Linear HardeningElastic-Power Hardening

PowerLinear

EE

Figure 5.19 Mathematical approximations of plot curves

The hardening- flow curve is the most generally applicable type of flow curve. This

type of plastic deformation behaviour has been modeled two different ways: Simple

Power Law and Ramberg-Osgood.

In the Simple Power Law model, the stress strain curve is divide into two discreteregion, separate at σ = σo such that:

Elastic : σ = Eε (σ ≤ σo) (5.59)

Plastic : σ = Hεn (σ ≥ σo) (5.60)

In the Ramberg-Osgood relationship the stress-strain curve is modeled as a

continuous function such that the total strain is sum of elastic and plastic parts:

ε = εe + εp =σE

+ εp (5.61)

and

σ = H εpn ⇒ ε =

σE

+σH

1n

(5.62)

5.28

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For the Ramberg-Osgood relation, σo is not distinct "break" in the stress-strain curve, but

is instead calculated from the elasticity and plasticity relations such that

σo = EH

E

11− n (5.63)

General stress-strain relations can be developed for deformation plasticity theory

such that the effective stress is

σ eff =12

(σ1 − σ2)2 + (σ2 −σ3)2 +(σ3 − σ1)2 (5.64)

and the effective total strain is

ε =σE

+ ε p (5.65)

where the effective plastic strain is

ε p =23

(εp 1 − εp 2)2 + (εp2 − εp 3)2 + (εp 3 − εp1)2 (5.66)

The resulting effective stress-effective strain curve is independent of the state of stress

and is used to estimate the stress-strain curves for other states of stress. Of particular

importance for are equations which allow correlation of plasticity relations for tension and

torsion such that:τ = Kτ γ n (5.67)

where Kτ =K

3( n +1)/2 in which K and n are the strength coefficient and strain hardening

exponent from the tension test, respectively. In addition,

τo =σ o

3(5.68)

where σo is the yield strength from a tension test.

Strain

Strain Hardening

σο

Ε

εεp e

Figure 5.20 Elastic-plastic stress-strain curve

5.29

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6. STRESS CONCENTRATION AND STRESS RAISERS

It is very important for the engineer to be aware of the effects of stress raisers such

as notches, holes or sharp corners in his/her design work. Stress concentration effects in

machine parts and structures can arise from internal holes or voids created in the casting

or forging process, from excessively sharp corners or fillets at the shoulders of stepped

shafts, or even from punch or stamp marks left during layout work or during inspection of

parts.

Stress Concentration Factors

Such discontinuities in a part can cause a large rise in stress above the nominal

P/A value that might be expected for example in a uniaxially loaded member such as a

tensile specimen. A discontinuity such as a circular, circumferential groove is a stress

raiser. The effects of stress raisers are usually given in terms of a stress concentration

factor, K, which is the factor by which the stress at the considered discontinuity is raised

over the nominal stress in the area of the discontinuity. Figures 6.1 and 6.2 show design

data for stress concentration factors, K=σlocal / σremote or K= σw/ notch / σw/o notch, for a

stepped flat tensile bar and a grooved cylindrical tensile bar, respectively. The nominal

stress at the reduced area is computed as shown on the graph and the actual stress

existing in the immediate vicinity of the notch is found by multiplying this nominal stress

value by the factor K.

Curves as shown in Figs. 6.1 and 6.2 can be computed theoretically for simple

shapes using advanced techniques such as elasticity, but are more often determined

using either various techniques of experimental stress analysis or via numerical methods

such as finite element analysis (FEA). Compendiums of stress concentrations factors

such as (Stress Concentration Factors, R.E. Peterson, John Wiley and Sons, Inc., 1974)

are excellent sources of information when "common" stress raisers are encountered.

Effects of Stress Raisers

The stress raising effects of a circular groove in a tensile bar are shown in Fig. 6.2,

where a stress concentration, K, of 2.0 might be expected, then since the stress in the

area of the groove is twice the nominal stress in a region removed from the groove, the

specimen would fail at one-half the load required for an unnotched specimen. Such is not

often not the case since stress concentrating factors are valid only while the material

behaves elastically. Beyond the elastic limit, plastic flow action can cause a stress

redistribution such that the high peak stress caused by the groove is redistributed to an

6.1

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almost uniform stress across the cross section, as if the groove didn't exist at all. This

plastic flow action is the reason why notches and holes in ductile materials may not lower

the ultimate strength when the specimen is tested statically, and is why stress

concentrations are sometimes ignored when designing with ductile materials. If the

groove is sufficiently deep, the large amount of material adjacent to the groove may

prevent any plastic flow action from occurring, and the specimen will fail at a stress higher

than an ungrooved specimen, stress being based on the reduced section area as shown

in Fig. 6.2. This is an instance when a stress concentration can be dangerous in a ductile

material.

Very little of the energy-absorbing plastic flow will occur with such a severe notch,

and such a member may fail in a brittle manner with a small shock load. In addition it

should be remembered that any grooves at all are dangerous in ductile materials if the

load fluctuates in magnitude, since fatigue crack initiation is a surface phenomenon and

the resulting fatigue strength is strongly influenced by surface finish.

The effects of a discontinuity in a brittle material are very much different than in a

ductile material. With these materials, no stress relieving plastic flow action is possible

and the full value of the stress concentration is valid right up to the fracture strength. For

these materials, then, we expect the fracture strength to be reduced from the unnotched

fracture strength by the value of K. In fact, one method for determining K is to use brittle

plaster test specimens with notches of various severity.

Design with brittle materials must be done with a great deal of care to avoid

undesirable failures. Generous fillets are used, holes eliminated, and attachments

carefully worked out. Considerable care must be taken to avoid even surface scratches

during fabrication.

Experimental Techniques

Elastomer Models: Geometric models can be used when the concern is with the

elastic behavior of materials. Metal specimens are not particularly good for demonstrating

elastic behavior and stress concentration effects because they are so stiff. It is much

easier to visualize elastic behavior if an elastomer specimen is used. There is, however,

one important difference between the behavior of most elastomers, such as rubber, and

that of metal. The stress-strain relation is linear, elastic (to yielding) for metal and is

nonlinear, elastic for rubber. This difference is offset by the large, easily measured

strains, which occur in rubber.

Usually a square grid of lines is printed on the surface of the specimen. A loading

frame can be used on which the specimen is stretched to approximately twice its original

length. The shape of the grid network is then carefully observed while the specimen is in

6.2

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the frame. At the junction of large and small portions of the specimen, it can be observed

that the strains are significantly greater than those removed from the junction. In fact, the

exact region of maximum strain can be seen on the deformed grid and the stress (strain)

concentration factor can be calculated.

Brittle Coating: One of the most straight-forward methods of experimental stress

analysis involves the use of a brittle coating. During testing, brittle materials fracture with

a clean, square break that is always oriented so the fracture surface is normal to the

direction of the largest principal stress. The brittle coating technique utilizes this property

to gage the magnitude and direction of stresses in a loaded member.

The use of brittle coatings in stress analysis has a long history, but its real

beginning was in the observation that hot-rolled steel with a mill-scale coating would

behave in a most unusual manner when stressed. In tension tests, for example, the mill

scale would crack in a geometric pattern indicating principal stress direction. In a tensile

test the cracks appear normal to the direction of load, while in a torsion test the cracks

appear in a 45° helix pattern.

Figure 6.1 Stress concentration factors for a stepped, flat tensile specimen.

6.3

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Figure 6.2 Stress concentration factors for a circumferential grove in a tensile specimen.

The usefulness of the brittle oxide mill scale is limited by the fact that the yield point

of the material must be exceeded before cracking occurs. Today a much more sensitive

brittle coating known as Stresscoat™ is available. This material is a patented mixture

which can be sprayed on the structure to be analyzed and after drying will crack at strainlevels as low as 400 µ m/m.

Stresscoat™ possesses many of the important characteristics of a brittle material,

however, it also has several limitations which must be allowed for during usage. One of

these is that the Stresscoat™ must dry for several hours after application before it can be

used. In addition, although Stresscoat™ is now available in aerosol cans, the grade to be

used depends on the temperature of the test room only. The material is simply sprayed on

to an average film thickness of about 0.01 mm or, with practice, until the correct uniform

yellow shade is obtained. At the same time the model is sprayed a number of calibration

specimens are also sprayed and all are allowed to dry in the test environment.

At the time of testing the Stresscoat™ is first "calibrated" by loading the calibration

specimens as a cantilever beams in a special loading fixture. A series of fine cracks

normal to the long axis of the beam will be evident and the last crack nearest the loading

6.4

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cam is marked with a soft pencil; the strain level at this crack, as indicated when the bar is

held in a fixture, is the material sensitivity. This is true because the strain level in a

cantilever beam is a maximum at the rigid end and decreases uniformly to zero at the

loaded end. Somewhere, then, along the length of the bar the strain will decrease to a

level that is insufficient to crack the coating. The last crack appearing nearest the loading

end is the critical level of strain.

A typical Stresscoat™ test is as follows: An estimate of the maximum load to be

applied is made and load increments to reach this load decided upon. Because the

Stresscoat™ is sensitive to the duration of load application, a loading interval is used.

The specimen is loaded to the level of the first interval, inspected for cracks and then

unloaded within the time interval allotted. The specimen is then allowed to remain

unloaded for about five min before loading to a load increased by the desired increment.

Each loading inspection and unloading cycle must be done within the same time interval,

probably 100 s per interval is reasonable. As the crack pattern progresses with

increasing loads, the locus of points of crack tips is marked with a grease pencil. These

marked lines are points of known strain value as found from the calibration bar. The most

critical cracks in this experiment are the initial cracks that form at the reduced section.

These will be the first cracks that form and considerable care should be exercised in

obtaining the load at which they initiate.

The calibration bars are loaded in one-second intervals, and the sensitivity thus

obtained is corrected to the actual sensitivity caused by the longer loading cycle in the

model by using a creep correction chart supplied by the instructor.

Photoelastic Technique: The photoelastic technique is one of the most powerful of

experimental stress analysis techniques. The photoelastic technique is valuable because

it gives an overall picture of the stress field, quickly showing regions of stress

intensification. In addition, the direction of principal stresses is also easily determined.

Like all experimental techniques, photoelasticity requires some practice to yield accurateresults, in particular, the determination of the principal stresses σ1 and σ2 on the interior

of the model requires considerable effort. Often, one is interested only in determining the

stress on the boundary of the model where one of the principal stresses is zero.

In the photoelastic method a model of the shape to be investigated is made from a

suitable transparent material. The model is then loaded in a manner similar to the actual

part and an accurate description of the stress magnitude and direction is obtained by

measuring the change in optical properties of the transparent model. These changes in

properties are measured by viewing the model in a special equipment called a

polariscope, so named because polarized light or light vibrating in a single plane only, is

used.

6.5

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The property of the model material that makes it suitable for stress field studies is

termed birefringence. The effects of this property are as follows:

1. A polarized light beam passing through a birefringent material becomes split

into two components, parallel to each direction of the principal stress axes.

2. These split polarized beams are out of phase by an amount that is dependent ofthe difference of the principal stresses, i. e. to ( σ1 −σ 2 ) at a point on the loaded

model.

The theoretical background of photoelasticity is beyond the scope of these

laboratory notes, although numerous references are also available on this experimental

technique. Simply note how the engineer can quickly use photoelasticity to determine

stress concentrations Typically, the polariscope is used in what is termed a circularily

polarized light configuration. In this configuration the model is located between the

polarizing elements as sketched in Fig. 6.3.

Note in the Fig. 6.3 that special filters called polarizers are used, one at each end of

the polariscope. Inside these filters is another set of polarizing filters called quarter-wave,

λ 4( ) , plates. These elements can be arranged so the background light is either light,

called light field, or completely extinguished, called dark field.

Figure 6.3 Circular polariscope

6.6

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When the loaded model is viewed in this type polariscope, a fringe pattern termed

an isochromatic pattern is apparent. These patterns are the loci of constant principal

stress difference. That is, if we know the calibration constant f of the photoelastic material

then

f =t

N(σ1 − σ2 ) (6.1)

where f is the stress-optical coefficient, N is the fringe order, t is the model thickness, andσ1 and σ2 are the plane-stress principal stresses.

Each dark band (See Fig. 6.4) for a dark field arrangement corresponds to an

integral (0, 1, 2, 3, etc.) fringe order. In this experiment, we are simply interested in the

maximum fringe order at the radii. The fringe order can be determined in at least two

ways. One method is to count the fringe order to the point of interest by beginning at apoint of zero fringe order such as a free unloaded corner. At such a corner σ1 = σ2 = 0 .

hence σ1 = σ2 = 0 and N must be zero. The second method is to observe to increase in

fringe order at the point of interest as the model is slowly loaded from zero load.

Once the maximum fringe order has been determined at the edge of the notch,

including estimates of fractional fringes orders, the stress can be calculated such that:

(σ1 − σ2 ) = fN

t(6.2)

where f is the stress-optical coefficient determined previously, N is the fringe order, t is themodel thickness, and σ1 and σ2 are the plane-stress principal stresses in which one of the

plane-stress principal stresses is equal to zero at the free surface of the notch edge.

6.7

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Figure 6.4 Photoelastic model as viewed in polariscope. Fringe value is 0 at externalsharp corners and 3 in narrow leg. Note that the model is a uniaxially loadedtensile specimen.

6.8

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7. FRACTURE

Fracture can be defined as the process of separation (or fragmentation) of a solid

into two or more parts under the action of a stress. So-defined, fracture can certainly be

identified as one type of engineering failure in which a design can no longer perform its

intended function. Ductile fracture is accompanied by gross deformation and shear

fracture due to slip (sometimes described as "graceful failure" due to the resulting

nonlinear load-displacement curve). Brittle fracture occurs with no gross deformation and

often is accompanied by rapid crack propagation and catastrophic failure. Brittle fracture

is not confined to materials normally thought of as brittle but can occur in ductile materials

at high strain rates (impact), at low temperatures, at the root of notches, and in the

presence of cracks.

The presence of cracks may weaken the material such that fracture occurs at

stresses much less than the yield or ultimate strengths (see Fig. 7.1). Fracture mechanics

is the methodology used to aid in selecting materials and designing components to

minimize the possibility of fracture from cracks.

In understanding the effect of brittle fracture in performance of engineering designs,

it is useful to first examine the strength potential of engineering materials. Consider the

force (or stress)-displacement relationship between two atoms as shown in Fig. 7.2. If this

relation is modeled as a sine wave then the equation can be written as:

σ = σmax sinπx

λ / 2

(7.1)

where σmax is the maximum stress on the curve, x is the displacement and λ /2 is the halfwave length of the sine wave. Applying uniaxial Hooke's law (σ = Eε ) where

ε =x − xe

xe

and small angle approximations (sinθ = θ ) of Eq. 7.1 gives

σ = Eε = Ex − xe

xe

= σmax

πx

λ /2

(7.2)

σALLOWABLESTRESS,

CRACK LENGTH, a

Low KHigh KIc

Ic

σALLOWABLESTRESS,

CRACK LENGTH, a

σοa t = transition crack length

between yield and fracture

Figure 7.1 Cracks lower the material's tolerance (allowable stress) to fracture.

7.1

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σ

Separation distance, x

Separation stress,

Sine Wave Approximation

Equilibrium Spacing, xe

Half Wave Length, /2

Repulsion Attractionσmax

λ

Figure 7.2 Attractive/repulsive force-distance curve between two atoms

where E is the elastic modulus and xe is the equilibrium distance (a.k.a., lattice spacing)

between the atoms. Assuming λ /2 =xe and solving Eq. 7.2 for σmax gives.

σmax =E

π

(7.3)

where σmax is the maximum cohesive strength (i.e., greatest strength potential) of a

material.

Another approach is to evaluate the work done in pulling the atoms apart:

W = F (x )dx =o

x

∫ σ(x )Adx =o

x

∫ σmax sinπx

λ /2Adx

o

λ /2

∫ (7.4)

Solving the integral for the work per unit area (W/A) and setting this equal to the

resisting energy to create new fracture surfaces gives:W

A= σmax

λπ

= 2γ s (7.5)

where γs is the fracture surface energy. Solving Eq. 7.5 for λ results in λ =2γ

σmax

.

Substituting this expression for λ into Eq. 7.2 and solving for σmax

σmax =Eγ

sxe

(7.6)

To check how well expressions in Eq. 7.3 and 7.6 compare to actual strengths of

materials, steel is used as an example where E = 200 GPa, γs = 1 J/m2, and

xe=ao=2.5x10-10 m. In this case, Eq. 7.3 gives σmax = 63.7 GPa and Eq. 7.6 gives σmax =

28.3 GPa. Since most steels have ultimate strengths in the 0.5 to 1 GPa range, something

7.2

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in real materials must be preventing them from reaching the strength potential predicted

by the maximum cohesive strength.

In engineering design, stress raisers such as holes can raise the local stress to

levels much greater than the remote or applied stress. For example, for a plate with a

hole in it subjected to uniaxial, uniform tensile stress (see Fig. 7.3) the elasticity solutions

in polar coordinates with variables r and θ are:

σrr = σ2

1− R2

r2

1+ 3R2

r 2 −1

cos2θ

σθθ =σ2

1 +R2

r2

+ 1 +

3R2

r2

cos2θ

(7.7)

At the edge of the hole for θ=0°, Eqs. 7.7 can be rewritten in Cartesian coordinates as:

σ xx = σrr = σ2

1 − R2

r2

3R2

r2

σ yy = σθθ =σ2

2 +R2

r2 +3R2

r2

(7.8)

Finally at the edge of the hole, where r=R, the stresses are:σ xx = σrr = 0

σyy = σθθ = 3σ (7.9)

The stress concentration factor can be defined as kt =σ local

σremote. For the stress in the y-

direction, kt is 3. Thus, the local stress at the edge of the hole is three times the remotely

applied stress regardless of the size of the hole as long as the hole diameter is small

relative to the width of the place.

In the case of an elliptical hole in uniaxial, uniform tension (see Fig. 7.3), the stress

in the y-direction can be related to the major and minor axes such that:

σ yy = σ 1 + 2c

a

= σ 1+ 2

c

ρ

(7.10)

where the radius of the ellipse tip is ρ =a2

c. The stress concentration factor for the

elliptical hole is no longer constant as it is for the circular hole but is a function of the

ellipse geometry:

kt = 1 + 2c

ρ

≈ 2

c

ρ

(7.11).

Note in this case that as the minor axis 2a shrinks to zero (e.g., as in a crack), ρ goes to

zero and the stress concentration factor in Eq. 7.11 goes to infinity.

7.3

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σ

σ

R

ry

σ

σ

y

x

ρ=a_c

2

2c

2a

Figure 7.3 Plates with circular and elliptical holes, subjected to uniform tensile stresses

Thus, elliptically-shaped features can have local stresses which are much greater

than remote stresses. If these local stresses approach the maximum cohesive strength on

a microscopic scale, then failure can initiate at such features ultimately leading to failure

of the material.

However, stress analyses such as these do not lend themselves to application in

engineering because infinite stress concentration factors are not usable in design.

Therefore, E. Orowan proposed the following approach:

a) assume cracks exist

b) assume these cracks are elliptically shaped with the appropriate stress solution

c) assume at fracture that the theoretical cohesive strength, σmax, is exceeded by

the stress, σyy, at the tip of the ellipse.

Combining Eqs. 7.10 and 7.6 (simplifying per Eq. 7.11) gives:

σmax = σ yy ⇒Eγ s

ao= σ 2

c

ρ

(7.12)

If fracture occurs at an atomistically-sharp ellipse tip, then let ρ ⇒ ao and the

remote stress required to cause brittle fracture is:

σ f =Eγ s

4c(7.13)

7.4

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A. A. Griffith proposed a different solution by assuming neither a crack shape or a

stress solution. Instead, Griffith stated that the criterion for brittle fracture is:

"A crack will propagate when the decrease in elastic strain energy is as least equal

to the energy required to create new crack surfaces."

The starting point for Griffith's solution is to evaluate the strain energy in the volume

surrounding the area in a plate (see Fig. 7.4) of thickness, t, and width, >>>W≈∞ , under

uniform uniaxial tension, σ, into which a crack could be introduced. For a crack of length,

2c, the stored elastic strain energy is:

UE =1

2σεV =

σ2

2Eπc2t( ) ≈

σ 2

Eπc2t( ) (7.14)

The resistance to fracture is related to the energy required to create new fracture

surfaces, Us, which is a function of the fracture surface energy, γs, and the surface area of

the crack, A=2 (2ct), such that:

Us = γ sA = γ s(2(2ct)) = γ s(4ct) (7.15)

The change in stored energy from the uncracked panel to the cracked panel is:

∆U = Us − UE = γ sA = γ s(4 ct) − σ 2

Eπc2

t( ) (17.16)

σ

σ

2c

t>>>W

Figure 7.4 Uniformly stress plate of thickness, t, and width, W, into which a crack of length,2c, will be introduced.

7.5

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The three energy terms, Us, UE and ∆U, are plotted as functions of half crack

length, c, in Fig. 7.5. It is apparent that the critical condition for unstable, brittle fracture is

reached when c=cc and d∆U

dc=0 at the peak of the ∆U curve. Differentiating Eq. 17.6,

setting it equal to zero, and solving for the stress at which brittle fracture occurs results in

the following:

Plane stress (σz = 0) σ f =2γ s E

πc

Plane strain (εz = 0) σ f =2γ s E

(1− ν2 )πc

(17.17)

It is interesting to compare Eqs. 17.13 and 17.17 and note that they only differ by

factors of 1/4 and 2/π under the radical, respectively. However, from a physical standpoint

Eq 17.17 is more "satisfying" because it is based purely on an energy balance, making no

assumptions about crack shapes or stress solutions.

Although Eq. 17.17 gives a physical relation for the critical fracture stress, it has

little engineering application because it is specific to the case of uniform stress and infinite

relative dimensions. G. Irwin attempted to place an engineering sense on the

understanding of crack/structure/material interactions. The resulting discipline is known

as Fracture Mechanics, and when dealing with non plastic situations, Linear Elastic

Fracture Mechanics (LEFM).

c

Ue

Us

U∆

Crack Length, cc

Figure 7.5 Graphical representation of the Griffith energy balance

7.6

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Irwin used the Griffith approach as the basis for his subsequent development, firstdenoting a new engineering term, G , as the strain energy release rate and crack driving

force:

G =dU

dA≈

dU

tda(7.18)

where U is the potential energy of the cracked panel, a is now defined as the half crack

length (note, a=c), and t is the plate thickness. In this context, G ≈2γs such that the Griffith

relation for fracture stress is written in LEFM as:

σ f =

GEπa

(7.19)

The utility of G for engineering purposes can be illustrated by an example of a

center cracked panel subjected to a uniform, uniaxial tensile stress (see Fig. 7.6). In this

case, it can be shown that:

Fixed load G =12

P2 dCda

for t = 1

Fixed displacement G = -1

2P2 dC

da

for t = 1

(7.20)

where dC

da is the compliance change (compliance is C =

δP

) with crack extension, da, P is

the applied load, and δ is the displacement. For a known geometry, the dimensionless

compliance, EBC (E is elastic modulus and B=t is the specimen thickness) is plotted

versus dimensionless crack length, a/W (W is the specimen width). At the criticalcondition when the plate fractures, P=Pc and the dimensionless compliance is

EBCc = EBδc

Pc (see Fig. 7.6) From the master compliance curve for the particular

geometry, dCc

dac is determined and the critical G at fracture is:

G c =

12

Pc2 dCc

dac

(7.21)

Although Eq. 7.21 gives a readily accessible engineering quantity for critical

fracture, it still lacks utility for general applicability to engineering design. An alternative

approach is to look at the stress at the crack tip. The near field stresses for a through

crack in an infinite plate, subjected to a uniform, uniaxial tensile stress are (see Fig. 7.7):

7.7

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2a

B=tW

P, δ

P, δ a/W

EBC

EBCc

a /Wc

EBWdC

da_ _c

c

P

δ

P c

δc

Test Results

Master Compliance Curve

Figure 7.6 Illustration of method for determining critical strain energy release rate

σxx = σa

2rcos

θ2

1-sinθ2

sin3θ2

+.....

σyy = σa

2rcos

θ2

1+sinθ2

sin3θ2

+.....

σxy = τ xy = σa

2rcos

θ2

sinθ2

cos3θ2

+.....

σzz = 0 and τyz = τ zx = 0 for plane stress

or

σzz = ν(σ xx + σ yy ) for plane strain

(7.22)

Note in Eqs. 7.22 that as r ⇒ 0 , the stresses predicted by the equations, all become

infinite, thus rendering the equations unusable for calculating the critical stress at the

crack tip for engineering design purposes. Irwin, however, took a different approach and

defined a stress intensity factor, K, which could uniquely define the stress state at the

crack tip, without the need to determine the actual stress such that:

K = σ πa (7.23)

The units of K are a bit unusual as MPa m .

7.8

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x

y

2a

σ

σ

σ

σσ

xx

yyxy

Figure 7.7 Near field stress state for a crack in an infinite plate subjected to uniform,

uniaxial tension.

Irwind did not chose the epression for K arbitrarily. Note that K and G are related:

If σ f =GEπa

then σ 2 =GEπa

and G =σ 2πa

E

But, since K = σ πa then G =K2

E

(7.24)

Now Eq. 7.22 can be written in terms of K:

σxx = σK

2πrcos

θ2

1-sinθ2

sin3θ2

+.....

σyy = σK

2πrcos

θ2

1+sinθ2

sin3θ2

+.....

σxy = τ xy = σK

2πrcos

θ2

sinθ2

cos3θ2

+.....

σzz = 0 and τyz = τ zx = 0 for plane stress

or

σzz = ν(σ xx + σ yy ) for plane strain

(7.25)

Again, note that although Eqs. 7.25 still predict infinite stresses at the crack tip, the

stress intensity factor can now be used to define the stress state at the crack tip. From an

engineering standpoint, the stress intensity factor can be used like stress to predict the

critical condition at fracture:

Fracture OCCURS if FS ≤ 1 where FS=Kc

K(7.26)

7.9

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where Kc is a material property known as fracture toughness and K is the stress intensity

factor at a particular combination of crack length and stress in the component. Several

things are important in Eq. 7.26.

The first is related to the "mode" of the fracture or loading. There are three modes

designated via Roman numerals as: I (opening), II (sliding) and III (tearing) as illustrated

in Fig. 7.8. Note that the most critical mode is Mode I because the crack tip carries all the

stress whereas in Modes II and III some of the stress is carried by interaction of the

opposing crack faces.

A second point is that the stress intensity factor defined in Eq. 7.23 is for the special

case of idealized crack in an infinite plate. Real cracks are affected by the geometry of the

component, the applied stress field, and other factors. Thus, Eq. 7.23 can be generalized

as

K = Yσ πa = ασ πa = Fσ πa (7.27)

where Y, F and a are geometrical correction factors which maintain the uniqueness of the

stress intensity factor by accounting for the particular geometry. For example, a center

through crack in a plate of finite width W, has a Y of ~1.12. In other words, the stress

intensity factor is about 12% greater in a finite width plate than for an infinitely wide plate.

Geometrical correction factors are found from closed form solutions, finite element

analyses, and experimental methods such as photoelastcity. Many handbooks of stress

intensity factors for common and not so common geometries have been complied. Some

common geometries are shown in Fig. 7.9

MODE IOPENINGMODE MODE II

SLIDINGMODE

MODE IIITEARINGMODE

Figure 7.8 Fracture modes

7.10

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Figure 7.9 Common and practical stress intensity factors

7.11

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Material Thickness, B

Plane StressFracture Toughness,K

Plane StrainFracture Toughness,K

Bsc Ic

Figure 7.10 Fracture toughness as a function of material thickness.

The third point is that the most critical fracture toughness of the material must be

measured so as to be geometry independent. Specifically, the plane stress fracture

toughness is a function of material thickness and is an important consideration when

determining fracture toughness in sheet material. Plane strain fracture toughness is

independent of material thickness and is less than the plane stress fracture toughness as

illustrated in Fig. 7.10.

When conducting fracture toughness tests, especially of metals, three critical

aspects must be addressed if a valid test is to result. The first is a valid stress intensity

factor for the particular geometry being tested. Geometries often used include center

cracked panel, three- or four-point flexure and compact tension (see Fig. 7.9). The

second is that the dimensions are in proper proportion to assure a valid stress intensity

factor and a thick enough specimen for plane strain conditions. ASTM recommends a

thickness such that:

B ≥ 2.5K Ic

2

Sys

2

(7.28)

where Sys is the yield strength of the material and KIc is the plane strain fracture

toughness. Finally, a plasticity requirement must be met where for a valid test:Pmax

PQ≤ 1.1 (7.29)

where Pmax is the maximum load recorded during the fracture test and PQ is the load at

which the load-displacement curve deviates from linearity.

Various trends in fracture toughness are related to material type. For example, as

yield point increases, fracture toughness generally decreases since yield points close to

the ultimate tensile strength of the material indicate low ductility and tendency to brittle

fracture (see Fig. 7.11)

7.12

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Yield Strength, Syp

Figure 7.11 General relation between fracture toughness and yield strength

The type of atomic bonding can indicate tendency to brittle fracture (see Fig. 7.12).

In general, covalent and ionic bonds as are found in ceramics, glasses and polymers

usually leas to lower fracture toughness. Metallic bonds usually lead to higher fracture

toughness. Some materials, such as composites, may have relatively high apparent

fracture toughness values although the applicability of LEFM to these materials is

debatable.

Design philosophies can take several forms, all based on the basic relation that

fracture will occur when the stress intensity factor in the component is equal to the fracture

toughness of the material such that:

Fractures OCCURS if KIc = KI ⇒ KIc = Yσ πa

in which KIc is a material property

Y is a function of the geometry

σ is a design stress

a is an allowable flaw (crack)

(7.30)

CeramicsGlasses

Polymers Metals Composites

0-10MPa√m 1-5

MPa√m

20-200MPa√m

10-100MPa√m

Figure 7.12 Relative comparisons of fracture toughness for various materials

7.13

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For example, in the design of a nuclear pressure vessel:

a) Material is chosen for certain properties (corrosion resistance, etc.). This fixes KIc.

b) In the component, allow for the presence of large flaw since these are detectable and correctable. This fixes Y and ac.

c) Design for the allowable stress, σ, to accommodate KIc, Y and ac.

In a different example, an aerospace application:

a) Material is chosen for certain properties (high yield strength, low density, etc). This fixes KIc.

b) In the component, fix the design stress, σ, for high performance or high

payload to weight ratio, etc. This fixes σ.c) Use nondestructive testing to find a before it reaches critical Y and ac.

Another example is the clever use of cracks to provide a warning before

catastrophic fracture can occur. In the leak before break philosophy, the thickness of a

pressure vessel is chosen so that a crack will penetrate the wall before it can reach a

critical size to cause brittle fracture. In this way, the presence of the crack can be easily

spotted without expensive (and often unreliable) non destructive test methods.

7.14

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8. TIME DEPENDENT BEHAVIOUR: CREEP

In general, the mechanical properties and performance of materials change with

increasing temperatures. Some properties and performance, such as elastic modulus

and strength decrease with increasing temperature. Others, such as ductility, increase

with increasing temperature.

It is important to note that atomic mobility is related to diffusion which can be

described using Ficks Law:

D = DO exp −Q

RT

(8.1)

where D is the diffusion rate, Do is a constant, Q is the activation energy for atomic motion,

R is the universal gas constant (8.314J/mole K) and T is the absolute temperature. Thus,

diffusion-controlled mechanisms will have significant effect on high temperature

mechanical properties and performances. For example, dislocation climb, concentration

of vacancies, new slip systems, and grain boundary sliding all are diffusion-controlled and

will affect the behaviour of materials at high temperatures. In addition, corrosion or

oxidation mechanisms, which are diffusion-rate dependent, will have an effect on the life

time of materials at high temperatures.

Creep is a performance-based behaviour since it is not an intrinsic materials

response. Furthermore, creepis highly dependent on environment including temperature

and ambient conditions. Creep can be defined as time-dependent deformation at

absolute temperatures greater than one half the absolute melting. This relative

temperature (T (abs )

Tmp (abs )) is know as the homologous temperate. Creep is a relative

phenomenon which may occur at temperatures not normally considered "high." Several

examples illustrate this point.

a) Ice melts at 0°C=273 K and is known to creep at -50°C=223 K. The homologous

temperature is 223273

= 0.82 which is greater than 0.5 so this is consistent with the

definition of creep.

b) Lead/tin solder melts at ~200°C=473 K and solder joints are known to creep at

room temperature of 20°C=293 K. The homologous temperature is 293473

= 0.62

which is greater than 0.5 so this is consistent with the definition of creep.

8.1

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Creep Stress Rupture

T/Tmp >0.5Strain

Constant Load

- Low Loads - High Loads- Precision Strain - Gross Strain Measurement ( f<0.5%) Measurement ( f up to 50%)- Long term (2000-10,000 h) - Short term (<1000 h)- Expensive equipment - Less expensive equipment

Emphasis on minimum Emphasis on time to failure atstrain rate at stress and at stress and temperaturetemperature

εε

Displacement

Figure 8.1 Comparison of creep and stress rupture tests

c) Steel melts at ~1500°C=1773 K and is known to creep in steam plant

applications of 600°C=873 K. The homologous temperature is 873

1773= 0.50 which

is equal to 0.5 so this is consistent with the definition of creep.

d) Silicon nitride melts/dissociates at ~1850°C=2123 K and is known to creep in

advanced heat engine applications of 1300°C=1573 K. The homologous

temperature is 15732123

= 0.74 which is greater than 0.5 so this is consistent with the

definition of creep.

Conceptually a creep test is rather simple: Apply a force to a test specimen and

measure its dimensional change over time with exposure to a relatively high temperature.

If a creep test is carried to its conclusion (that is, fracture of the test specimen), often

without precise measurement of its dimensional change, then this is called a stress

rupture test (see Fig 8.1). Although conceptually quite simple, creep tests in practice are

more complicated. Temperature control is critical (fluctuation must be kept to <0.1 to

0.5°C). Resolution and stability of the extensometer is an important concern (for low

8.2

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creeping materials, displacement resolution must be on the order of 0.5 µm).

Environmental effects can complicate creep tests by causing premature failures unrelated

to elongation and thus must either mimic the actual use conditions or be controlled to

isolate the failures to creep mechanisms. Uniformity of the applied stress is critical if the

creep tests are to interpreted. Figure 8.2 shows a typical creep testing setup.

The basic results of a creep test are the strain versus time curve shown

schematically in Fig. 8.3. The initial strain, ε i =σ i

E, is simply the elastic response to the

applied load (stress). The strain itself is usually calculated as the engineering strain,

ε =∆L

Lo. The primary region (I) is characterized by transient creep with decreasing creep

strain rate (dεdt

= ˙ ε ) due to the creep resistance of the material increasing by virtue of

material deformation. The secondary region (II) is characterized by steady state creep(creep strain rate, ˙ ε min = ˙ ε ss , is constant) in which competing mechanisms of strain

hardening and recovery may be present. The tertiary region (III) is characterized by

increasing creep strain rate in which necking under constant load or consolidation of

failure mechanism occur prior to failure of the test piece. Sometimes quaternary regions

are included in the anlaysis of the strain-time curve as well, although these regions are

very specific and of very short duration.

Figure 8.2 Typical creep test set-up

8.3

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ddtε

Time, t

Primary I

Secondary II

Tertiary III

Constant Load

Constant Stress

tf

Figure 8.3 Strain time curve for a creep test

In principle, the creep deformation should be linked to an applied stress. Thus, as

the specimen elongates the cross sectional area decreases and the load needs to be

decreased to maintain a constant stress. In practice, it simpler to maintain a constant

load. When reporting creep test results the initial applied stress is used. The effect of

constant load and constant stress is shown in Fig. 8.3. Note that in general this effect

(dashed line for constant stress) only really manifests itself in the tertiary region, which is

beyond the region of interest in the secondary region. The effects of increasing

temperature or increasing stress are to raise the levels and shapes of the strain time

curves as shown in Figure 8.4. Note that for isothermal tests, the shapes of the curves for

increasing stress may change from dominant steady state to sigmoidal with little steady

state to dominant primary. Similar trends are seen for iso stress tests and increasing

temperature (see Fig. 8.4).

Creep mechanisms can be visualized by using superposition of various strain-time

curves as shown in Fig. 8.5. An empirical relation which describes the strain-time relation

is:ε = ε i 1+ βt 1/3( ) exp(kt) (8.2)

where β is a constant for transient creep and k is related to the constant strain rate. A

"better" fit is obtained by:

ε = ε i + εt 1− exp(rt)( ) + t˙ ε ss (8.3)

where r is a constant, εt is the strain at the transition from primary to secondary creep and˙ ε ss is the steady-state strain rate. Although no generally-accepted forms of nonlinear

strain-time relations have been developed, one such relations is:

ε = ε i +Bσ mt + Dσα 1− exp(βt )( ) (8.4)

where B, m, D, a and b are empirical constants.

8.4

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Time, t

σ

σ > σσ > σ >σ

1

112

23

Iso thermal Tests

Time, t

Τ

Τ > ΤΤ > Τ >Τ

1

112

23

Iso stress Tests

Figure 8.4 Effect of stress and temperature on strain time creep curves

In this relation, if t >ttransient then

ε = ε i +Bσ mt + Dσα (8.5)

and the strain rate is the steady-state or minimum strain rate:dεdt

= Bσ m = ˙ ε ss (8.6)

The steady state or minimum strain rate is often used as a design tool. For

example, what is the stress needed to produce a minimum strain rate of 10-6 m/m / h ( or

10-2 m/m in 10,000 h) or what is the stress needed to produce a minimum strain rate of 10-

7 m/m / h ( or 10-2 m/m in 100,000 h). An Arrhenius-type rate model is used to include the

effect of temperature in the model of Eq. 8.6 such that:

˙ ε ss = ˙ ε min = Aσn exp−Q

RT

(8.7)

where n is the stress exponent, Q is the activation energy for creep, R is the universal gas

constant and T is the absolute temperature.

To determine the various constants in Eq. 8.7 a series of isothermal and iso stress

tests are required. For isothermal tests, the exponential function of Eq. 8.7 becomes a

constant resulting in

˙ ε ss = ˙ ε min = Bσn (8.8)

Equation 8.8 can be linearized by taking logarithms of both sides such that

log ˙ ε ss = log ˙ ε min = log B + n log σ (8.9)

8.5

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Time, t Time, t Time, t Time, t

= + +ε i ε i

Total Creep Curve Sudden Strain Transient Creep Viscous Creep

Figure 8.5 Superposition of various phenomenological aspects of creep

Log-log plots of ˙ ε min = ˙ ε ss versus σ (see Fig. 8.6) often results in a bilinear

relation in which the slope, n, at low stresses is equal to one indicating pure diffusion

creep and n at higher stresses is greater than one indicating power law creep with

mechanisms other than pure diffusion (e.g., grain boundary sliding).

For iso stress tests, the power dependence of stress becomes a constant resulting

in

˙ ε ss = ˙ ε min = C exp−QRT

(8.10)

Equation 8.10 can be linearized by taking natural logarithms of both sides such that

ln ˙ ε ss = ln ˙ ε min = ln C −QR

1T

(8.11)

Log-linear plots of ˙ ε min = ˙ ε ss versus 1T

(see Fig. 8.7) results in a linear relation in which

the slope, −Q

R, is related to the activation energy, Q, for creep.

log

n=1 (diffusion creep)

n>1 (power law creep)

σ

.

Figure 8.6 Log-log plot of minimum creep strain rate versus applied stress showingdiffusion creep and power law creep.

8.6

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.

1/T

-Q/R

Figure 8.7 Log-linear plot of minimum creep strain rate versus reciprocal of temperatureshowing determination of activation energy.

The goal in engineering design for creep is to predict the behaviour over the long

term. To this end there are three key methods: stress-rupture, minimum strain rate vs. time

to failure, and temperature compensated time. No matter which method is used, two

important rules of thumb must be borne in mind: 1) test time must be at least 10% of

design time and 2) creep and/or failure mechanism must not change with time,

temperature or stress.

Stress-rupture This is the "brute force method" is which a large number of tests are

run at various stresses and temperatures to develop plots of applied stress vs. time to

failure as shown in Fig. 8.8. While it is relatively easy to use these plots to provide

estimates of stress rupture life within the range of stresses and lives covered by the test

data, extrapolation of the data can be problematic when the failure mechanism changes

as a function of time or stress as shown by the "knee" in Fig. 8.8.

Minimum strain rate vs. time to failure This type of relation is based on the

observation that strain is the macroscopic manifestation of the cumulative creep damage.

As such, it is implied that failure will occur when the damage in the material in form of

creep cavities and cracks resulting from coalesced creep cavities reaches a critical level.

This critical level of damage is manifested as the failure which can be predicted from the

minimum strain rate and the time to failure such that.

˙ ε min tf = C ≈ εf (8.12)

8.7

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Stress, or log

Time to failure, tor log t

ff

σσ

Τ

Τ >Τ

Τ > Τ > ΤChange in failure mechanism

1

123

2 1

Figure 8.8 Stress rupture plots for various temperatures

Equation 8.12, known as the Monkman-Grant relation, should give a slope of -1 ona log-log plot of ˙ ε min versus t f regardless of temperature or applied stress for a particular

material It then becomes a simple matter to predict a time to failure either by measuring

the minimum strain at a given stress and temperature or predicting the minimum strain

rate from Eq. 8.7 for the given temperature and stress once the A and Q are determined.

Having found the minimum strain rate, the time to failure can be found from the Monkman-

Grant plot for the particular material.

Temperature-compensated time In these methods, a higher temperature is used at

the same stress so as to cause a shorter time to failure such that temperature is traded for

time. In this form of accelerated testing it is assumed that the failure mechanism does not

change and hence is not a function of temperature or time. In addition, assumptions can

be made that Q is stress and temperature independent. Two of the more well-known

relations are Sherby-Dorn and Larson Miller.In the Sherby-Dorn method, θ is the temperature compensated time such that:

PSD = logθ = logt f -log e

RQT

(8.13)

where PSD is the Sherby-Dorn parameter and Q is assumed independent of temperature

and stress. In this method, a number of tests are run at various temperatures and stresses

to determine the times to failure and activation energy. A "universal" plot (see Fig. 8.9) is

then made of the stress as a function of PSD. The allowable stress for an combination of

time to failure and temperature (i.e., PSD) can then be determined from the curve.In the Larson-Miller method, θ, is the temperature compensated time such that:

PLM = log eR

Q =T logtf +(logθ = C)( ) (8.14)

8.8

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1/T

log t fQ log e

R

σ < σ < σ1 2 3

Allowable

P = log t -

σ

SD flog e Q

TRExperimental Results "Universal" Sherby Dorn Relation

Figure 8.9 Summary of Sherby-Dorn relation

where PLM is the Larson-Miller parameter, Q is assumed to a function of stress only, and

C is a constant of ~20 for most materials. In this method, a number of tests are run at

various temperatures and stresses to determine the times to failure and activation energy.

A "universal" plot (see Fig. 8.10) is then made of the stress as a function of PLM. The

allowable stress for an combination of time to failure and temperature (i.e., PLM) can then

from the curve.

An example of the application of the Sherby-Dorn relation is as follows. For a

certain aluminum-magnesium alloy the stress-PSD relation is found to be

σ = f PSD( ) = −11.3PSD −124 (25 ≤ σ ≤ 85 MPa) (8.15)

The design problem is to determine the allowable stress to give 2000 h life at 200°C. For

this alloy, the activation energy, Q, is 150.5 kJ/mole. Using Q=150,500 J/mole, R=8.314

J/mole K, tf=2000 h, and T=473 K, PSD is calculated as -13.21. Substituting this value of

PSD into Eq. 8.15 gives an allowable stress of 25 MPa.

1/T

log t f

Q log e

R

σ < σ < σ1 2 3

Allowable

P = T (log t - C)

σ

LM

Experimental Results "Universal" Larson-Miller Relation

-C

f

Figure 8.10 Summary of Larson-Miller relation

8.9

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9. TIME DEPENDENT BEHAVIOUR: CYCLIC FATIGUE

A machine part or structure will, if improperly designed and subjected to a repeated

reversal or removal of an applied load, fail at a stress much lower than the ultimate

strength of the material. This type of time-dependent failure is referred to as a cyclic

fatigue failure. The failure is due primarily to repeated cyclic stress from a maximum to a

minimum caused by a dynamic load. A familiar example of a fatigue failure is the final

fracture of a piece of wire that is bent in one direction then the other for a number of

cycles. This type of behavior is termed low-cycle fatigue and is associated with large

stresses causing considerable plastic deformation with failure cycles, Nf, in the range of

<102 to 104. The other basic type of fatigue failure is termed high-cycle fatigue and is

characterized by loading which causes stress within the elastic range of the material and

many thousands of cycles of stress reversals before failure occurs often with Nf>105

(sometimes >102 to 104).

Fatigue has been a major concern in engineering for over 100 years, and there is a

very large amount of literature available on the fatigue problem. The importance of a

knowledge of fatigue in engineering design is emphasized by one estimation that 90

percent of all service failures of machines are caused by fatigue and 90 percent of these

fatigue failures result from improper design.

Fatigue failures of normally ductile materials in structural and machine members

are very much different in appearance than failure under a static loading. Under quasi-

static loading of the tensile test, considerable plastic flow of the metal precedes fracture

and the fracture surface has a characteristic fibrous appearance. This fibrous appearance

can also be noted in the ductile part of the fracture surface of Charpy impact specimens.

A fatigue crack, however, appears entirely different. The crack begins at a surface, often

at the point of high stress concentration. Once the crack begins, the crack itself forms an

area of even higher stress concentration (also stress intensify factor), and it proceeds to

propagate progressively with each application of load until the remaining stressed area

finally becomes so small that it cannot support the load statically and a sudden fracture

results. In fatigue failures, then, a characteristic appearance is always evident. The

fatigue portion begins at the point of high-stress concentration and spreads outward

showing concentric rings (known as beach marks) as it advances with repeated load. The

final fracture surface has the same appearance as that of a ductile tensile specimen with a

deep groove. The fracture is brittle due to constraint of the material surrounding the

groove and has a crystalline appearance. The failure was not because the material

crystallized as is sometimes supposed, it always was of crystalline structure.

9.1

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Fatigue cracks, then, begin at a point of high-stress concentration. The severity of

those stress concentrations will vary, even with carefully prepared laboratory specimens.

In addition, the rapidity with which the crack propagates will vary. The cracks are irregular

and will follow various paths around regions of stronger metal. A consequence of this

manner of crack propagation is wide variation in time to failure of a number of seemingly

identical test specimens loaded with the same load. For this reason a number of

statistical procedures have been developed for interpreting fatigue data.

The basic mechanism of a high-cycle fatigue failure is that of a slowly spreading

crack that extends with each cycle of applied stress. In order for a crack to propagate, the

stress across it must be tension; a compression stress will simply close the crack and

cause no damage. One way, then, of preventing fatigue or at least extending the fatigue

life of parts is to reduce or eliminate the tensile stresses that occur during loading by

creating a constant compressive surface stress, called a residual stress, in the outer

layers of the specimen. We can picture how it is possible to induce a constant

compressive stress of this type by a simple analogy. Consider a bar with a small slot cut

in the surface. If we force a wedge tightly into this slot, the wedge and the material in the

bar itself adjacent to the slot will be in a compressive stress state. If a tensile stress is now

applied to the bar, no tensile stress can exist in the region of the slot until the compressive

stress caused by the wedge is overcome. In other words, the tensile stress in the region

of the slot will always be less than in a region removed from the slot by an amount equal

to the magnitude of the residual compressive stress.

Favorable compressive stresses can be induced in parts by more practical

methods than cutting slots. One of the most common methods is that of shot peening.

Shot peening is a process in which the surface of the part is impacted by many small steel

balls moving at high velocity. This process plastically deforms the surface of the part and

actually tends to make it somewhat bigger than it was. The effect is the same as the

wedge. A now larger surface is forced to exist on a smaller sublayer of material with the

net result that a compressive stress is induced in the outer layer making it more resistant

to fatigue failure.

Figure 9.1 Typical fatigue test specimen

9.2

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Figure 9.2 One type of rotating beam reversed stress testing machine (a.k.a. R.R. Moore

rotating bending fatigue test machine)

A large portion of fatigue testing is done under conditions of sinusoidal loading in

pure bending. The test specimen is shown in Fig. 9.1 and a schematic of the test machine

(proposed by R. R. Moore) is shown in Fig. 9.2. This machine is called a rotating-beam

type, and from Fig. 9.2 it can be seen that a constant moment of magnitude equal to one-

half the applied load W multiplied by the distance d between the two bearings.

Since the specimen rotates while the constant moment is applied to the beam, the

stress at any point in the beam makes a complete cycle from, say tension at a point on the

bottom of the specimen to compression as the specimen rotates so the point comes to the

top then back to compression as the rotation cycle to the bottom is completed. Thus, for

each complete revolution of the specimen a point on the specimen experiences a

complete stress cycle of tension and compression as shown on the left side of Fig. 9.3.

The maximum and minimum values of this stress are equal and opposite at the

specimen's surface, occurring at the peaks and valleys of the stress cycle, while the mean

value of the stress is zero. This stress cycle is known as completely reversed loading and

has a stress ratio, R =σmin

σmax

, of -1. Other nomenclature and symbols for fatigue are shown

in Fig. 9.4.

σt

σ

σσ

∆σ

σ

min

maxm

a

Figure 9.3 Examples of stress cycles for completely-reversed and tension-tension loading

9.3

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σmax = Maximum stress

σmin = Minimum stress

σm = Mean stress =σmax + σmin

2∆σ = Stress range = σmax − σmin

σa = Stress amplitude = ∆σ2

= (σmax − σm ) = (σm − σmin )

Note: tension = +σ and compression = −σ. Completely reversed R= −1, σm = 0.

R = Stress ratio =σmin

σmax

A = Amplitude ratio = σa

σm

= 1− R

1+ R

Figure 9.4 Nomenclature and symbols for fatigue

Some test machines turn at 10,000 RPM. and are equipped with a counter that

counts once for each 1000 revolutions. The machines are also equipped with an

automatic cutout switch which immediately stops the motor when the specimen fails.

Fatigue test data are obtained in a number of ways. The most common is to test a

number of different specimens by determining the time to failure for a specimen when

stressed at a certain stress level. Figure 9.5 shows schematic representation of these test

results on a type of graph called an S-N curve which shows the relationship between the

number of cycles N for fracture and the maximum (or mean or amplitude or range) value

of the applied cyclic stress. Generally, the abscissa is the logarithm of N, the number of

cycles, while the vertical axis may be either the stress S or the logarithm of S. Typical test

data are shown plotted in Fig. 9.6. in which the data can sometimes be represented by

two straight lines.

10 10

σS

log N f6 8

e

Ferrous and Ti-based alloys

Non-ferrrous materials (e.g Al or Cu alloys) ( @ 10 cycles)

= fatigue limit or endurance limit ( @10 cycles)σe6

8σe

Figure 9.5 Schematic representation of S-N curves for ferrous and non ferrous materials.

9.4

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The fatigue strength is then the stress required to fracture for a set number of

cycles. The usual trend is for the fatigue strength to decrease rapidly in the cycle range of

103 to 106. In the range of 105 to 106 the curve flattens out and for some materials such

as steel actually becomes flat, indicating this material will last "forever" in fatigue loading,

if the applied stress is kept below a certain value. The stress corresponding to failure at

an infinite number of cycles is called the endurance limit (or endurance limit) of thematerial and is often designated as Se or σe. It should be recalled, in light of the previous

discussion, that a considerable amount of scatter will exist and values such as endurance

limit will be only approximate when a small number of test specimens are used.

Figure 9.6 Typical S-N curves for determining endurance limits of selected materialsunder completely reversed bending.

9.5

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Many times, in preliminary design work, it is necessary to approximate the S-N

curve without actually running a fatigue test. For steel it has been found that a good

approximation of the S-N curve can be drawn if the following rules are used.

1. Obtain the ultimate tensile strength Smax of the specimen ( σmax in a simpletension test).

2. On a diagram of S vs. log N, plot fatigue strength values of

a. 0.9 Smax at 103 cyclesb. 0.5 Smax at 106 cycles.

3. Join these points together to form a S-N diagram similar to that shown in Figs9.5 or 9.6

Many factors affect fatigue failures. Not only do fatigue failures initiate at surfaces,

but stress raisers create stresses which are greatest at surfaces. Fatigue factors

Recall the stress concentration factor:

kt =σ LOCAL

σ REMOTE

(9.1)

where σLOCAL is the maximum local stress at the stress raiser and σREMOTE is the remote or

net stress. The effect of the stress raiser on fatigue strength can be evaluated by first

defining a fatigue strength reduction factor:

kf =σ e

UN − NOTCHED

σeNOTCHED (9.2)

where σ eUN − NOTCHED and σ e

NOTCHED are the endurance limits for un-notched and notched

fatigue specimens as illustrated in Fig. 9.7.

σS

log N f

eσe

NOTCHED

UN-NOTCHED

Figure 9.7 Illustration of endurance limits for notched and un-notched fatigue tests.

9.6

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A notch sensitivity factor can be defined which relates the material behavior, kf, and

the component parameter, kt such that

q =k f − 1k t − 1

(9.3)

where q=0 for no notch sensitivity, q=1 for full sensitivity. Note that q increases as notch

radius, r, increases and q increases as SUTS increases

Generally, kf<< kt for ductile materials and sharp notches but kf≈kt for brittle

materials and blunt notches. This is due to i) steeper dσ/dx for a sharp notch so that the

average stress in the fatigue process zone is greater for the blunt notch, ii) volume effect

of fatigue which is tied to average stress over a larger volume for blunt notch, iii) crack

cannot propagate far from a sharp notch because steep stress gradient lowers KI quickly.

In design, avoid some types of notches, rough surfaces, and certain types of loading. As

mentioned, compressive residual stresses at surfaces (from shot peening, surface rolling,

etc.) can increase fatigue lives.

Endurance limit, Se=σe, is also lowered by factors such as surface finish (ma), type

of loading (mt), size of specimen (md), miscellaneous effects (mo) such that:

σe' = mamtmd moσe (9.4)

Note that σe can be estimated from the ultimate tensile strength of the material such that:σe ≈ meSUTS where me=0.4-0.6 for ferrous materials.

Note also that most fatigue data is generated for R=-1 and a mean stress of zero.

Non zero mean stresses can also play a large part in resulting fatigue data. This is

illustrated in Fig. 9.8 where the lines on the plot represent lines of infinite life. Note that

when mean stress is σm=0, then the allowed amplitude stress, σa is the endurance limit

measured for completely reversed fatigue test loading with R=-1. However, if σa=0 then

the allowable mean stress is either the yield or ultimate strength from a monotonic test

since the stress is no fluctuating when the stress amplitude is zero.

σ

σa

m

SSys uts

Goodman

Soderberg

Figure 9.8 Illustration of effect of mean stress on allowable amplitude stress.

9.7

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The mathematical expression for this effect is the equation of the line such that:

σa = σ e 1 −σ m

SUTS

(9.5)

which is known as the Goodman line. If SUTS is replaced with Sys, then Eq. 9.5 is known

as the Soderberg relation.

If a factor of safety, FS, and / or fatigue factors, kf, are used then the following

expressions results for either brittle or ductile materials.

Brittle σa =σe

FS •k f1 −

σ m

(SUTS /(k f ≈ k t)FS)

(9.6a)

Ductile σa =σe

FS •k f1 −

σm

(SUTS / FS)

(9.6b)

The previous understanding of fatigue behavior assumed constant amplitude or

constant mean stress conditions and infinite life. Sometimes it is necessary to be able to

understand the effect of variable amplitude about a constant mean stress as illustrated in

Fig. 9.9 for non infinite life. Often a linear damage model (Palmgren-Miner rule) is

applied such that:

N1

Nf1

+N2

Nf2

+N3

Nf3

=Nj

Nfj∑ = 1 (9.7)

where N1 and Nf1, etc. are the actual number of cycles and number of S-N fatigue cycles

at stress, σa1 etc. (see Fig. 9.10) respectively.

σ t

σ σ

σa3

a2a1

N NN1 2 3

Figure 9.9 Variable amplitude loading with a constant mean stress

9.8

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Nf

σ σσ

N NN 123

a1a2

a3a σ

fff

Figure 9.10 Illustration of cycles to failure on S-N curve for different stress amplitudes

S-N curves are generally used to design for infinite life in fatigue. That is, the

fatigue data are used to chose stresses such that fatigue cracks will never develop.

However, the fatigue mechanism is such that is possible to analyze the growth of fatigue

cracks using linear elastic fracture mechanics. This is useful in extending the service life

of components when a fatigue crack is noticed, rather than discarding a part which might

still have useful life.

The fatigue process (see Fig. 9.11) consists of 1) crack initiation, 2) slip band crack

growth (stage I crack propagation) 3) crack growth on planes of high tensile stress (stage

II crack propagation) and 4) ultimate failure. Fatigue cracks initiate at free surfaces

(external or internal) and initially consist of slip band extrusions and intrusions as

illustrated in Figure 9.12. When the number of slip bands reaches a critical level

(saturation) cracking occurs (see Figure 9.13). These slip band cracks will grow along

directions of maximum shear for only one to two grain diameters (dg), when the crack

begins growing along a direction normal to maximum tensile stress. Although, fatigue

striations (beach marks) on fracture surfaces represent successive crack extensions

normal to tensile stresses where one mark is approximately equal to one cycle (i.e. 1

mark≈1N), beach marks only represent fatigue cycles during crack propagation and do

not represent the number of cycles required to initiate the crack. Thus, summing the

beach marks does not represent the total number of fatigue cycles.

9.9

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1-2 dg

Slip BandCrack Initiation

Slip BandCrack Propagation

Tensile StressCrack Propagation

Figure 9.11 Illustration of fatigue process

During fatigue crack propagation (stage II may dominate as shown in Fig 9.14)

such that crack growth analysis can be applied to design: a) cracks are inevitable, b)

minimum detectable crack length can be used to predict total allowable cycles, c) periodic

inspections can be scheduled to monitor and repair growing cracks, d) damage tolerant

design can be applied to allow structural survival in presence of cracks.

The most important advance in understanding fatigue crack propagation was

realizing the dependence of crack propagation on the stress intensity factor. The

mathematical description of this is known as the Paris power law relation:

da

dN= C(∆K )m (9.8)

where da/dN is the crack propagation rate (see Fig. 9.15), ∆K = F(∆σ ) πa in which F is

the geometry correction factor, a is the crack length, and C and n are material constants

found in stage II of the da/dN vs ∆K curve (see Fig. 9.14)

Slip bands on first loading

Slip band Intrusions and Extrusions under cyclic loading

Figure 9.12 Woods model for fatigue

9.10

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Saturation

Cracking

Number of Cycles NFigure 9.13 Slip band saturation and the onset of fatigue cracking

To predict the crack propagation life of a component, Eq. 9.8 can be rearrangedsuch that:

dN =da

C(∆K)m =a i

a f

∫N i

N f

∫da

C(F∆σ πa)m

ai

af

∫ (9.9)

Assuming that F can be approximated as nearly constant over the range of crack growth

and assuming that m and C are constant, then:

N f =a f

(1− (m /2)) − ai(1 − (m /2) )

C F(∆σ) π[ ]m1− (m /2)[ ]

(9.10)

where ai is the initial crack length which is either assumed (~2 dg) or determined by non

destructive evaluation and af =

1

πKIc

Fσmax

2

.

log K

logda/dN m

Kth

I II III

∆∆

Figure 9.14 Crack growth rate versus stress intensity factor range.

9.11

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N

ada/dN

Figure 9.15 Crack length vs. number cycles relations from which crack propagation rate isfound.

If F is a function of crack length, i.e. F(a,W, etc.), then numerical integration must beused such that:.

dN =da

C(∆K)m =a i

a f

∫N i

N f

∫da

C F(a,W,etc)∆σ πa[ ]m

ai

af

∫ (9.11)

Crack propagation rates are also highly sensitive to R ratios primarily because

crack propagation only occurs during tensile loading. Thus, the longer the crack stays

open (i.e. R ≥0) the more time out of each cycle that crack propagation occurs. Thus, the

more negative R, the more tolerant the crack is of ∆K and vice versa (see Fig. 9.16)

log K

logda/dN

+R

-R

For >(+R) crack remains in tension (open) longer and can tolerate less K for the same da/dN

Figure 9.16 Effect of R-ratio on crack growth rates.

9.12

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10. COMPRESSION AND BUCKLING

Whenever a structural member is designed, it is necessary that it satisfies specific

strength, deflection and stability requirements. Typically strength (or in some cases

fracture toughness) is used to determine failure, while assuming that the member will

always be in static equilibrium. However, when certain structural members are subjected

to compressive loads, they may either fail due to the compressive stress exceeding the

yield strength (i.e., yielding) or they may fail due to lateral deflection (i.e., buckling).

The maximum axial load that a structural component (a.k.a., column) can support

when it is on the verge of buckling is called the critical load, Pcr. Any additional load

greater than P will cause the column to buckle and therefore deflect laterally. Buckling is

a geometric instability and is related to material stiffness, column length, and column

cross sectional dimensions. Strength does not play a role in buckling but does play a role

in yielding.

Recall a slender column under load with a restoring force related to lateral

deflection as shown in Fig. 10.1. Two moments are generated about the pinned base, O.

The first moment is the restoring moment:

FL = K (∆x )L = K (Lθ)L = KL2θ (10.1)

where F is the restoring force, L is the length of the column, K is the spring constant, and θis the angle. The second moment is the overturning moment:

P(∆x) = P(Lθ)L (10.2)

where P is the axial force.

F=K x

PP

L

θ

O

Figure 10.1 Model of a slender column under axial load with a restoring force/moment

10.1

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P

Pcr=KL

Unstable

Stable

0 +θ−θ

Figure 10.2 Illustration of stable and unstable conditions along with critical load

The following conditions apply as illustrated in Fig. 10.2:

If PLθ < KθL2 then P < KL and a stable condition results

If PLθ > KθL2 then P > KL and an unstable condition results(10.3)

The critical load separating stability and instability is

Pcr = KL (10.4)

This situation can also be thought of as a ball on a surface as shown in Fig. 10.3.

The ideal column (pinned at both ends as shown in Fig. 10.4) can be modeled

mathematically using the equation of the elastic curve. Note that assumptions for the

ideal column include: i) column is initially perfectly straight, ii) load is applied through the

center of the cross section, iii) material is homogeneous and linear elastic, and iv) column

buckles and bends in a single plane. The elastic curve equation is:

EIv '' = −M (10.5)

where E is the elastic modulus, I is the moment of inertia, v''=d2v

dx2 , and M is the internal

moment introduced by defection, v, and the applied load, P, such that M=Pv.

Stable Unstable Neutral

Figure 10.3 Illustration of stable, unstable and neutral conditions

10.2

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PP

IdealColumn

Internal Forceand Moment

P

P

Mx

v

L

Figure 10.4 Ideal column and internal force/moment

Substituting this relation for M into Eq. 10.5 gives

EIv '' = −Pv ⇒ v '' +P

EIv = 0 (10.6)

Equation 10.6 is a homogeneous, second-order, linear differential equation with

constant coefficients. The general solution for this equation is:

v = C1sinP

EIx

+C 2 cos

P

EIx

(10.6)

The constants of integration are found from the boundary conditions at the ends of

the column. Since v=0 at x=0, then C2=0. In addition, at x=L, v=0. In this case the trivial

solution is that C1=0 which means that the column is always straight. The other possibly

is that sinP

EIL

=0 which is satisfied if

P

EIL =nπ. This solution can be rearranged such

that

P =n2π 2EI

L2 (10.7)

The smallest value of P (in other words, the value of P reached first when loading from

P=0 to some critical load) is obtained when n=1, so that the critical load for the column

(known as the Euler load) is:

Pcr =π2EI

L2 (10.8)

10.3

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Note that this critical load is independent of the strength of the material, but rather

depends on the column dimensions (actual length, L, and smallest moment of inertia, I)

and material stiffness, E. Note also that load-carrying ability increases as the moment of

inertia increases, as elastic modulus increases but at length decreases. Thus, short, fat

columns made of stiff material will have very low tendency to buckle. A key to

understanding buckling is to create a design whereby the required (i.e., applied) design

load is always much less than the theoretical buckling load. The reasons for this are that

the theoretical buckling load is determined based on an ideal column with many inherent

assumptions. Thus, a large safety margin must be placed between the design load and

the calculated critical buckling load.

Equation 10.8 is the solution for an ideal column with pinned/pinned end

conditions. If other boundary conditions are applied then the following critical load

equations result for the indicated end conditions.

Pcr =π2EI

L2 Pinned −Pinned

Pcr = π2EI

4L2 Free −F ixed

Pcr =4π 2EI

L2 Fixed −Fixed

Pcr =2.046π 2EI

L2 Pinned − Fixed

(10.9)

The equations in Eq. 10.9 are a bit cumbersome because four different relationsneed to be remembered. So instead an effective length is defined such that Le = KL .

Now a single equation can be used to replace the four equations of Eq. 10.9 and effective

length constants, K, for each end condition can be used to define the effective length. The

effective length constants are as shown in Fig. 10.5 and the single equation is:

Pcr =π2EI

Le

2 =π 2EI

(KL)2 (10.10)

For purposes of design, it is more useful to express the critical loading condition in

terms of a stress such that:

σcr =Pcr

A=

π 2EI

ALe

2 ⇒ σcr =π2E

(Le / k)2 (10.11)

where k= I / A is the smallest radius of gyration determined from the least moment ofinertia, I, for the cross section. The term, Le /k is known as the slenderness ratio and

contains information about the length and the cross section.

10.4

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Figure 10.5 Effective length constants for various end conditions

It is interesting to investigate the case where the applied stress, σ , is equal to the

critical buckling stress, σcr , and the generalized yield strength, σo . At this point there is a

transition between yield and buckling:

σ = σo = σcr ⇒ σo =π2E

(Le / k )2(10.12)

If Eq. 10.12 is solved for Le /k , the resulting relation marks the combination of

length and cross section at which the compressive behaviour transitions from yielding to

buckling. This relation is known as the minimum slenderness ratio:

Le

k min=

π 2E

σo(10.13)

This transition can be illustrated by plotting the relation between stress, σ , and

slenderness ratio, Le /k , as shown in Fig. 10.6. Note that in reality there is no sharplydivided transition between yielding and buckling. Instead the σ vs. Le /k curve can be

divided into three regions as shown in Fig. 10.7. Region I is the short-column region inwhich general yielding occurs when σ =σo . Region II is an intermediate-column region

in which i) the column may yield or may buckle and ii) empirical relations are used to

approximate the resulting curve. Region III is for long columns and buckling will occur.

10.5

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σLe/k

Le/k | min

σo

σ= π E2

2( )

Le/k

Figure 10.6 Stress vs. slenderness ratio relation

Regions I and III are straight forward to determine (either apply σ =σo . in Region

I or σcr =π2E

(Le /k)2 in Region III). Region II is material/ geometry/loading dependent and

empirical relations are often used. One approach is to fit a parabola to the σ vs. Le /k

curve from σ =σo . to σ =σo / 2 such that.

σ = σo 1−Le / k( )2

2Le / k min( )2

for 0 ≤ Le

k≤ Le

k min

and

σ =π2E

(Le / k)2 for Lek

>Lek min

(10.14)

σ

Le/kLe/k | min

σo

σ= π E2

2( )

Le/k

Region

I

Region

III

Region

II

Parabolic Approximation

Figure 10.7 Stress vs. slenderness ratio relation with parabolic fit in Region II

10.6

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In summary, note that buckling loads and stresses are sensitive to the following

1) stiffness of material

2) length of column

3) cross section dimensions

4) cross sectional shape

5) end conditions

6) initial eccentricity

7) eccentric loading

8) end conditions

Thus, the moral of the story is to always keep design loads well below the calculated

critical buckling loads.

10.7

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11. Structures: Complex Stresses and Deflections

Engineering structures may take many forms, from the simple shapes of square

cross section beams to the complex and intricate shapes of trusses. Regardless of the

shape another important aspect of structures is that often the stresses, strains, and

deflections of the structure do not lend themselves to simple and straight-forward

analyses of simple components such as those used for materials testing (e.g., the

uniaxially-loaded and uniformly-stressed tensile specimen). Further complicating the

analyses of engineering structures is the need to apply failure criteria to evaluate the

probable success (or non success) of the design.

Failure Criteria

Engineering failure can be broadly defined as the "inability to perform the intended

function." An obvious failure is a broken part (unless of course the intended function is to

fail as in the case of shear pins or explosive bolts!) which is known as fracture. However,

excessive elastic or plastic deformation without fracture can also constitute a failure. In

addition, a component with too much or not enough "give" such as with too compliant or

too stiff of a spring-like component can be a failure. A cracked component such as in a

pressure vessel would constitute a failure if a leak occurred. Thus, failure criteria can be

based on stress, strain, deflection, crack length, time or cycles, or any other engineering

parameter we choose to apply.

The most common failure criteria are stress-based. The basic premise is that

failure will occur in the component or structure when the combined stress state is equal

that which caused failure in the same material subjected to a uniaxial tensile test. Two

primary types of stress-based failure criteria are used: yield (for ductile materials with

%el>5) and fracture (for brittle materials with %el<5). In both cases, one can consider

failure to occur at the onset of non linearity in the tensile stress strain curve (the yield point

in ductile materials and ultimate tensile strength in brittle materials).

Fracture Criterion: The simplest failure criterion is that failure is expected when the

greatest principal stress reaches the uniaxial tensile strength of the material. Thus, the

Maximum Normal Stress fracture criterion (a.k.a., Rankine) can be specified as:

Fractures if MAX σ1, σ2 , σ3[ ] ≥ SUTS (11.1)

where the function MAX indicate the greater of the absolute values of the principal normal

stresses. Note that it is assumed that the ultimate strength of the material is the same in

compression or tension.

11.1

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σ

σ

σ

σ

σσ

=-Suts

=-Suts=Suts

=Suts1 1

2

2"safe"

Figure 11.1 Failure envelope for maximum normal stress criterion

A factor of safety can be defined based on Eq. 11.1 such that:

Fractures if FS ≤ 1 where FS=SUTS

MAX σ1, σ2 , σ3[ ] (11.2)

This fracture criterion can be represented in a plane stress state (σ z =0) where σ 2 is the

ordinate and σ1 is the abscissa. As shown in Fig. 11.1, any combination of σ1 and σ 2

that falls within the square box (i.e., FS=1 for Eq. 11.2 where ±SUTS=σ1 or ±SUTS=σ 2) is

"safe" and the perimeter is fracture.

Yield criteria: There are two relatively well-accepted yield criteria: Maximum Shear

Stress criterion (a.k.a., Tresca) and Octahedral Shear Stress criterion (a.k.a., Distortional

Energy or Von Mises). Each is discussed as follows.

The simplest yield criterion is that yield failure is expected when the greatest shear

stress reaches the shear strength of the material. Thus, the maximum shear stress yield

criterion can be specified as:

Yields if MAX τ12 =σ1 − σ2( )

2, τ13 =

σ1 − σ3( )2

, τ23 =σ2 −σ 3( )

2

≥ τo = σ o

2(11.3)

where the function MAX indicates the greater of the absolute values of the principal shear

stresses.

A factor of safety can be defined based on Eq. 11.3 such that:

Yields if FS ≤ 1

where FS=τ o =σo /2

MAX τ12 =σ1− σ2( )

2, τ13 =

σ1 −σ 3( )2

, τ 23 =σ2 −σ 3( )

2

(11.4)

11.2

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σ

σ

σ

σ

σ

σ1

1

2

2

"safe"1

2

=σο

=σοο=σ

ο=σσ − σ =−σ1 2 o

σ − σ =σ1 2 o

Figure 11.2 Failure envelope for maximum stress criterion

This yield criterion can be represented in a plane stress state (σ z =0) where σ 2 is

the ordinate and σ1 is the abscissa. As shown in Fig. 11.2, any combination of σ1 and σ 2that plots within the parallelogram (i.e., FS=1 for Eq. 11.4 where ±σ o=(σ1-σ 2), ±σ o=σ1

or ±σ o=σ 2) is "safe" and the perimeter is yielding.

A more complicated yield criterion is that yield failure is expected when theoctahedral shear stress,τh , reaches the octahedral shear stress at yield of the material,

τho . Thus, the octahedral shear stress yield criterion can be specified as:

Yields if τh ≥ τho

where τh = 13

σ1−σ 2( )2+ σ 2 −σ 3( )2

+ σ3 −σ 1( )2 (11.5)

and

τho = 23

σo (11.6)

when the stress state of a uniaxial tensile test at yielding (σ1=σ o , σ 2=σ 3=0) are

substituted into the relation for τh given in Eq. 11.5. If Eq 11. 5 and 11.6 are set equal to

each other the yield criterion can be expressed in terms of normal stresses:

Yields if τh ≥ τho ⇒ 13

σ1−σ 2( )2+ σ 2 −σ 3( )2

+ σ3 −σ 1( )2≥

23

σo (11.7)

11.3

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σ

σ

σσ

σ 11 22

"safe"

2

ο

σο

ο

οσ =σ − σ 2σ +σ

1

2o

2

Figure 11.3 Failure envelope for maximum stress criterion

A factor of safety can be defined based on Eq. 11.7 such that:

Yields if FS ≤ 1

where FS=σ o

12

σ 1−σ 2( )2+ σ 2 −σ 3( )2

+ σ3 −σ1( )2= σo

σH

(11.8)

where σH

is the effective stress based on the octahedral shear stress criterion.

This yield criterion can be represented in a plane stress state (σ z =0) where σ 2 is

the ordinate and σ1 is the abscissa. As shown in Fig. 11.3, any combination of σ1 and σ 2

that plots within the ellipse (i.e., FS=1 for σo2 = σ1

2 −σ1σ 2 + σ22 is "safe" and the

perimeter is yielding.

The usefulness of the three failure criteria presented here is shown in Fig 11.4 forthe failure envelopes for the plane stress case where σ1 and σ 2 are normalized to SUTS

or σ o . Note for the brittle material (cast iron) that the actual failure points follow the

maximum normal stress criterion envelope (i.e., FS=1) and for the ductile materials (steels

and aluminums) that the actual failure points fall between the maximum shear stress and

octahedral shear stress criteria envelopes (i.e., FS=1). Since the maximum difference

between the two yield criteria is about 15%, it is often advisable to err on the side of

conservatism and use the simpler maximum shear stress criterion for ductile materials.

11.4

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Figure 11.4 Failure criteria and failure points plotted on normalized plane stress

coordinates

Combined Stresses

The previous discussion of failure criteria was based on two premises: 1) theuniaxial tensile behavior of the material was known (i.e., SUTS or σ o .) and 2) the principal

stresses based on all the coordinate stresses was known.

When determining the mechanical properties and performance of a material, such

as its yield strength, it is desirable to choose a fundamental test that will give the required

property in the most direct manner. Thus, for yield strength, a simple one-dimensional

tensile test specimen is usually used. Figure 11.5 illustrates the fundamental tensile test

and shows a free body diagram of an infinitesimal element with the one-dimensionalstress σz acting on it

On the other hand, in a realistic situation, the engineer is usually faced with a two-

or three-dimension load condition in which, at any point, P, the loaded member may be

subject to a combination of tension, compression and shear stresses as idealized in Fig.

11.6.

Figure 11.5 Illustration of the uniaxial tensile test

11.5

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Figure 11.6. General three-dimensional stress state

Figure 11.6 shows an arbitrary body loaded with forces P and moments M. At a

point O, shown enlarged at the right, an infinitesimal three-dimensional element can beacted upon by normal stresses σx ,

σy and σz acting in the x, y and z directions, as

shown. Often, the stresses in the z directional are zero, or much smaller than σx or σy .

This is a condition called plane stress, and the analysis is simplified.

Thus, in order to determine the margin of safety of a loaded structure, it is

necessary to relate the two- or three-dimensional stress state that usually occurs, with the

fundamental strength, like tensile yield strength, that is obtained in the laboratory. This

relationship is defined through the previously-discussed failure criteria of which there are

a number in addition to those already discussed.

Prerequisite to applying a failure criterion, is to deduce from the general two- or

three-dimensional element in which both shear and normal stresses are present, theprincipal stresses σ1 and σ2 and the maximum shear stresses τmax . These stresses can

be determined either analytically or experimentally.

The analytical calculation of principal stress and maximum shear stress involves

the superposition of normal and shear stress to determine the total stress acting at a

critical point. Thus, normal stresses are computed from P/A for simple tension and Mc/I for

bending. Shear stresses are computed from Tr/J for torsion and VQ/It for direct shear. For

thin-walled, pressurized cylindrical pressure vessels the hoop stress is given by pr/t and

the axial stress by pr/2t.

11.6

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Figure 11.7 A general two-dimensional stress state and stresses resulting from

element rotation.

Since the final stress state is independent of the order in which the loads are

applied, the total stress existing at a point with a combined loading consisting, perhaps, of

tension, torsion, pressure and shear can be found by simply adding normal stresses

together and shear stresses together taking proper account of their sign. Finally, after

considering the stresses caused by each load, a two-dimensional element at the point

considered may appear as shown in Figure 11.7.Then an application of Mohr's circle of stress will give the principal stresses σ1 and

σ2 and the maximum shear stress τmax . Stress is the quantity that causes failure, yet we

realize from earlier work, that one cannot measure stress directly. This is because stress

is related to the force in a part, which, except in very simple cases, is not easily measured.

On the other hand, strain is easily measured and these values can easily be converted to

stress values.

Calculation of principal strains can be performed as follows. If we consider only atwo-dimensional object, the unknown strains are the normal or elongation strains εx and

εy and the shear strain

γ xy . Since electrical resistance strain gages can measure only

normal strains (ε ), not shear strains (γ ), we need to measure three normal strains at apoint to determine the three strains εx ,

εy , and

γ xy at a point. This follows directly from

the equations that give strain in a direction "a" ( εa ), oriented at an angle θ from the x axis,as shown in Figure 11.8, when strains along the x and y axes, εx and

εy , and the shear

strain γ xy are known. Thus:

εa =12

εx + εy( ) +12

εx − εy( )cos2θ +12γ xy sin2θ (11.9)

11.7

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Figure 11.8 Orientation of "a" direction and the x-y axes

If we measure three strains, εa1, εa2 , εa3 , at three different angles θ1, θ2 , θ3 , we can

substitute the values in the above equations and obtain three equations to solve for εx ,

εy , and

γ xy . Knowing εx ,

εy , and

γ xy , Mohr's circle of strain can then be used to find

ε1, ε2 , and γ max .

A special type of electrical resistance strain gage called a rosette is available for

measuring the three normal strains at a point. These rosettes are simply three strain

gages, mounted one directly on top of the other, or near each other, and oriented at

precise angular relationship with respect to each other. Several types of rosettes are

available, the most common being a rectangular rosette with 45° between gages and a

delta rosette with 60° between gages.

Although the analytical method for calculating the principal strains has already

been described, a graphical method for accomplishing the same result also exists. The

graphical method has the inherent advantages of graphical techniques with the addedadvantage of directly producing the Mohr's circle of strain, i.e., ε1, ε2 and γ max are given

directly. This method is known as Murphy's method. This method is described as follows

and is illustrated in Fig. 11.9 for a rectangular rosette.

1. Assume we are using a rosette with strain gages oriented along lines a, b, andc oriented at angles α and β apart, in this case, α = β = 45°. It is desired tofind the maximum strains ε1 and ε2 .

2. Along an arbitrary horizontal axis x - x, lay off three vertical lines a, b, and ccorresponding to the three measured strains εa . εb , εc measured from the y

axis. The y - y axis will be the γ 2 shear strain axis.

11.8

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Figure 11.9 Mohr's strain circle as determined graphically from strain valves

εa . εb , and εc .

3. Let the b gage direction lie between the a and c directions as shown. Then

from a convenient point F on line b, lay off lines whose directions correspond

to the direction of the gauging lines of the rosette, maintaining the same

directional sense as in the rosette. These lines intersect the lines a and c at

points A and C.

4. Erect perpendicular bisectors to line FA and FC, to intersect at O.

5. Draw a circle with center at O and passing through points A, F, and C.

6. From points C, B and A, draw radii to O. Draw the strain axis ε horizontal

through O. These radii will have the same angular orientation sense as the

corresponding gauging lines of the rosette; the angle between the radii will be

twice the actual angle between the gages.

7. The point A, B, and C on the circle give the values of ε and γ 2 for the three

gages.

8. Values of principal strains are determined by the intersection of the circle andthe ε axis. The angular orientation of ε1 from gage A is shown as 2θ .

11.9

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Thus this simple graphical technique results in a Mohr's circle of strain. Strain

values at any angular orientation can be found. Once the principal strains are found the

principal stresses follow directly from the Hooke's relations, considering Poisson's effect:

ε1 =σ1

E−

νσ2

E(11.10)

ε2 =σ 2

E−

νσ1

E(11.11)

or more conveniently, the inverse of these

σ1 =ε1+νε2( )E

1 −ν2( ) (11.12)

σ 2 =ε2 +νε1( )E

1−ν 2( ) (11.13)

where E is the elastic modulus and ν is Poisson's ratio.

The shear stress-strain relation is completely independent of the normal stress-

strain relation and is given by

τ = Gγ (11.14)

where G is the shear modulus of the material.

Types of Engineering Structures

One type of engineering structure is one which is composed of a few simple

elements but subjected to a complex loading condition as shown in Fig. 11.10. In this

figure the loading condition involves a torque and bending moment and possibly an

internal pressure. The stresses due to these loading conditions can be calculated and

appropriately superposed before performing the transformations to determine the

principal stresses.

Another type of engineering structure is one which is composed of many similarly

loaded elements subjected to either a relatively simple or slightly more complex loading

condition. Trusses (see Fig. 11.11) are an example of one of the major types of

engineering structures, providing practical and economical solutions to many engineering

situations. Trusses consist of straight members connected at joints (for example, see

Figure 1). Note that truss members are connected at their extremities only: thus no truss

members are continuous through a joint.

11.10

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Figure 11.10 Relatively simple engineering component subjected to a complex loading

condition

In general, truss members are slender and can support little lateral load.

Therefore, major loads must be applied to the various joints and not the members

themselves. Often the weights of truss members are assumed to be applied only at the

joints (half the weight at each joint). In addition, even though the joints are actually rivets

or welds, it is customary to assume that the truss members are pinned together (i.e., the

force acting at the end of each truss member is a single force with no couple). Each truss

member may then be treated as a two force member and the entire truss is treated as a

group of pins and two-force members.

Truss Members

Joints

Figure 11.11 Example of a Simple Truss

11.11

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X

Y

Z

Figure 11.12 Illustration of a bicycle frame as a truss-like structure

A bicycle frame, on first inspection, appears to be an example of a truss (see Fig.

11.12) Each tube (truss member) is connected to the other at a joint, the principal loads

are applied at joints (e.g., seat, steering head, and bottom bracket), and the reaction loads

are carried at joints as well (e.g., front and rear axles). Although the joints are not pinned,

a reasonable first approximation for analyzing forces, deflections, and stresses in the

various tubes of the bicycle frame might be made using a simple truss analysis.

Forces in various truss members can be found using such analysis techniques as

the method of joints or the method of sections. Deflections at any given joint may be

found by using such analysis techniques as the unit load method of virtual work.

An example of the use of the method of joint to solve for the axial loads in each

truss member is as follows. For the simple truss shown in Figure 11.13 the first step is tocalculate the reactions at joints C and D. In this case, F∑ =0 and M∑ =0 such that

MC∑ = 0 = PL − RDL ⇒ RD = P ↓ (11.15)

andF∑ = 0 ⇒ Fx∑ = 0 = −P + RxC ⇒ RxC = P →

Fy∑ = 0 = −2P − P + RyC ⇒ RyC = 3P ↑(11.16)

The resulting free body diagram is shown in Fig. 11.14

11.12

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AB

C D

2P

P

L

L

L

L

√2 L y

x

Figure 11.13 Example of a simple truss

Using the method of joints, F∑ =0 at joint D such that

D

BD

CDP

F∑ = 0 ⇒ Fx∑ = 0 = −FCD ⇒ FCD = 0 →

Fy∑ = 0 = P +FBD ⇒ FBD = P ↑(11.17)

and since FBD pulls on the joint, then the joint must pull back on the member so member

BD is in tension.

A B

CD

2PP

L

L

L

L

√2 L

3P

PP

Figure 11.13 Free body diagram for simple truss

11.13

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Using the method of joints, F∑ =0 at joint C such that

C

AC

CD

3P

1

1

√2

PF∑ = 0 ⇒ Fx∑ = 0 = P −

1

2FBC ⇒ FBC = 2P ←

Fy∑ = 0 = 3P −1

2FBC − FAC ⇒ FAC = P ↓

(11.18)

Since FAC pushes on the joint, then the joint must push back on the member so member

AC is in compression. Furthermore, since FCB pushes on the joint, then the joint must

push back on the member so member CB is in compression.

Finally, using the method of joints, F∑ =0 at joint A such that

A

AC=2P

AB

2P

P

F∑ = 0 ⇒ Fx∑ = 0 = −P + FAB ⇒ FAB = P → Fy∑ = 0 = −2P + 2P ⇒ checks

(11.19)

Since FAB pulls on the joint, then the joint must pull back on the member so member AB is

in tension.

The summary of the member forces is shown in Table 11.1.

Although finding deflections in complex structures is more involved than finding

deflections in simple components, it is not difficult. A useful technique is the unit load

method in which the displacements can be found from simple deflection equations at

joints which do not have forces acting on them. The unit load method works for linearly

elastic materials and superposition applies.

Table 11.1 Summary of Truss Member Forces

Member Force

AB P (tension)

AC 2P (compression)

BC √2P (compression)

BD P (tension)

CD 0

11.14

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For axially loaded members, the displacement is:

∆ N =NUNL

EA∫ dx (11.20)

where NU is the axial force in the member due to a unit load applied at the point and

direction of interest, NL is the actual force in the member due to the actual applied load on

the structure, E and A are the elastic modulus and cross sectional area of the individual

member. The integral sign signifies that the calculated quantities for each member are

summed via integration to give the final total deflection at the point and direction of

interest.

For members subjected to bending moments, the displacement is:

∆ M =MUML

EI∫ dx (11.21)

where MU is the bending moment in the member due to a unit load applied at the point

and direction of interest, ML is the actual bending moment in the member due to the actual

applied load on the structure, E and I are the elastic modulus and cross sectional moment

of inertia of the individual member. The integral sign signifies that the calculated

quantities for each member are summed via integration to give the final total deflection at

the point and direction of interest.

For members subjected to torsion, the displacement is:

∆T =TU TL

GJ∫ dx (11.22)

where TU is the torque in the member due to a unit load applied at the point and direction

of interest, TL is the actual torque in the member due to the actual applied load on the

structure, G and J are the shear modulus and polar moment of inertia of the individual

member. The integral sign signifies that the calculated quantities for each member are

summed via integration to give the final total deflection at the point and direction of

interest.

For members subjected to transverse shear, the displacement is:

∆ v =VUVL

GA∫ dx (11.23)

where VU is the transverse shear in the member due to a unit load applied at the point

and direction of interest, VL is the actual transverse shear in the member due to the actual

applied load on the structure, G and A are the shear modulus and cross sectional area of

the individual member. The integral sign signifies that the calculated quantities for each

11.15

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member are summed via integration to give the final total deflection at the point and

direction of interest.

The total deflection due to each of these contributions can then be found by adding

the individual contribution such that

∆ t =NU NL

EA∫ dx +MU ML

EI∫ dx +VUVL

GA∫ dx +TUTL

GJ∫ dx (11.24)

An example of the unit load method applied to the simple truss example is shown

in Fig. 11.14 in which only the axial loading contributions are required since truss

members are pinned and no bending moments, transverse shear, or torque can be

carried in the members.

11.16

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Figure 11.14 Example of application of unit load method to find a deflection ina simple truss

11.17

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12.1

12. Pressure Vessels: Combined Stresses

Cylindrical or spherical pressure vessels (e.g., hydraulic cylinders, gun barrels,

pipes, boilers and tanks) are commonly used in industry to carry both liquid s and gases

under pressure. When the pressure vessel is exposed to this pressure, the material

comprising the vessel is subjected to pressure loading, and hence stresses, from all

directions. The normal stresses resulting from this pressure are functions of the radius of

the element under consideration, the shape of the pressure vessel (i.e., open ended

cylinder, closed end cylinder, or sphere) as well as the applied pressure.

Two types of analysis are commonly applied to pressure vessels. The most

common method is based on a simple mechanics approach and is applicable to “thin

wall” pressure vessels which by definition have a ratio of inner radius, r, to wall thickness,

t, of r/t≥10. The second method is based on elasticity solution and is always applicable

regardless of the r/t ratio and can be referred to as the solution for “thick wall” pressure

vessels. Both types of analysis are discussed here, although for most engineering

applications, the thin wall pressure vessel can be used.

Thin-Walled Pressure Vessels

Several assumptions are made in this method.

1) Plane sections remain plane

2) r/t ≥ 10 with t being uniform and constant

3) The applied pressure, p, is the gage pressure (note that p is the

difference between the absolute pressure and the atmospheric pressure)

4) Material is linear-elastic, isotropic and homogeneous.

5) Stress distributions throughout the wall thickness will not vary

6) Element of interest is remote from the end of the cylinder and other

geometric discontinuities.

7) Working fluid has negligible weight

Cylindrical Vessels: A cylindrical pressure with wall thickness, t, and inner radius,

r, is considered, (see Figure 12.1). A gauge pressure , p, exists within the vessel by the

working fluid (gas or liquid). For an element sufficiently removed from the ends of the

cylinder and oriented as shown in Figure 12.1, two types of normal stresses aregenerated: hoop, σ h, and axial, σ a, that both exhibit tension of the material.

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12.2

Figure 12.1 Cylindrical Thin-Walled Pressure Vessel

For the hoop stress, consider the pressure vessel section by planes sectioned by planes

a, b, and c for Figure 12.2. A free body diagram of a half segment along with the

pressurized working fluid is shown in Fig. 12.3 Note that only the loading in the x-

direction is shown and that the internal reactions in the material are due to hoop stress

acting on incremental areas, A, produced by the pressure acting on projected area, Ap.

For equilibrium in the x-direction we sum forces on the incremental segment of width dy to

be equal to zero such that:

F

A pA

pr

t

x

h p h

h

h

∑ =

[ ] − = = [ ] −

=

0

2 0 2σ σ

σ

σ

t dy p 2r dy

or solving for (12.1)

where dy = incremental length, t = wall thickness, r = inner radius, p = gauge pressure,and σ h is the hoop stress.

Figure 12.2 Cylindrical Thin-Walled Pressure Vessel Showing Coordinate Axes and

Cutting Planes (a, b, and c)

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12.3

Figure 12.3 Free-Body Diagram of Segment of Cylindrical Thin-Walled Pressure Vessel

Showing Pressure and Internal Hoop Stresses

For the axial stress, consider the left portion of section b of the cylindrical pressure

vessel shown in Figure 12.2. A free body diagram of a half segment along with the

pressurized working fluid is shown in Fig. 12.4 Note that the axial stress acts uniformly

throughout the wall and the pressure acts on the endcap of the cylinder. For equilibrium

in the y-direction we sum forces such that:F

A pA r r r

r

r r

r

r

r

r rt t r

r

rt t

y

a e a o

a

a

o

a

∑ =

− = = −( ) −

=−( )

=[ ] −( )

=+ + −( )

=+( )

0

0

2 2

2 2 2

2

2 2

2

2 2

2

2 2 2

2

2

σ σ π π

σ

σπ

π

σπ

π

π

π

p

or solving for

p

substituting r = r + t gives

p

r + t

p p

since this is a thin wall with a small t,t is smaller and can

o

2

be neglected such that after simplificationbe neglected such that after simplification

p σ a

r

t=

2

(12.2)

where ro = inner radius andσ a is the axial stress.

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12.4

Figure 12.4 Free-Body Diagram of End Section of Cylindrical Thin-Walled Pressure

Vessel Showing Pressure and Internal Axial Stresses

Note that in Equations 12.1 and 12.2, the hoop stress is twice as large as the axial

stress. Consequently, when fabricating cylindrical pressure vessels from rolled-formed

plates, the longitudinal joints must be designed to carry twice as much stress as the

circumferential joints.

Spherical Vessels: A spherical pressure vessel can be analyzed in a similar

manner as for the cylindrical pressure vessel. As shown in Figure 12-5, the “axial” stress

results from the action of the pressure acting on the projected area of the sphere such thatF

A pA r r r

r

r r

r

r

r

r rt t r

r

rt t

y

a e a o

a

a

o

a

∑ =

− = = −( ) −

=−( )

=[ ] −( )

=+ + −( )

=+( )

0

0

2 2

2 2 2

2

2 2

2

2 2

2

2 2 2

2

2

σ σ π π

σ

σπ

π

σπ

π

π

π

p

or solving for

p

substituting r = r + t gives

p

r + t

p p

since this is a thin wall with a small t,t is smaller and can

o

2

be neglected such that after simplificationbe neglected such that after simplification

p σ σa h

r

t= =

2

(12.3)

Note that for the spherical pressure vessel, the hoop and axial stresses are equal

and are one half of the hoop stress in the cylindrical pressure vessel. This makes the

spherical pressure vessel a more “efficient” pressure vessel geometry.

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12.5

Figure 12.5 Free-Body Diagram of End Section of Spherical Thin-Walled Pressure Vessel

Showing Pressure and Internal Hoop and Axial Stresses

The analyses of Equations 12.1 to 12.3 indicate that an element in either a

cylindrical or a spherical pressure vessel is subjected to biaxial stress (i.e., a normal

stress existing in only two directions). In reality, the element is subjected to a radial stress,σ rwhich acts along a radial line. The stress has a compressive value equal to the

pressure, p, at the inner wall, and decreases through the wall to zero at the outer wall

(plane stress condition) since the gage pressure there is zero. For thin walled pressure

vessels, the radial component is assumed to equal zero throughout the wall since thelimiting assumption of r/t=10 results in σ h being 10 times greater than σ r=p and σ a being

5 time greater than σ r=p. Note also that the three normal stresses are principal stresses

and can be used directly to determine failure criteria.

Note that the relations of Equation 12.1 to 12.3 are for internal gauge pressures

only. If the pressure vessel is subjected to an external pressure, it may cause the

pressure vessel to become unstable and collapse may occur by buckling of the wall.

Thick-Walled Pressure Vessels

Closed-form, analytical solutions of stress states can be derived using methods

developed in a special branch of engineering mechanics called elasticity. Elasticity

methods are beyond the scope of the course although elasticity solutions are

mathematically exact for the specified boundary conditions are particular problems. For

cylindrical pressure vessels subjected to an internal gage pressure only the following

relations result:

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12.6

σ

σ

σ

ho i

o

ao i

ro i

o

r

r r

r

r

r

r r

r

r r

r

r

=−( )

+

=−( )

=−( )

i

i

i

p

p

p

2

2 2

2

2

2

2 2

2

2 2

2

2

1

1

(12.4)

where ro=outer radius, ri=inner radius, and r is the radial variable. Equations 12.4 apply

for any wall thickness and are not restricted to a particular r/t ratio as are the Equations12.1 and 12.2. Note that the hoop and radial stresses(σ h and σ r) are functions of r (i.e.

vary through the wall thickness) and that the axial stress, σ a, is independent of r (i.e., is

constant through the wall thickness. Figure 12.6 shows the stress distributions through

the wall thickness for the hoop and radial stresses. Note that for the radial stress

distributions, the maximum and minimum values occur, respectively, at the outer wall(σ r=0) and at the (σ r=-p) as noted already for the thin walled pressure vessel.

Equations 12.4 can be generalized for the case of internal and external pressures

such that

σ

σ

σ

ho i

ho i

ho i

r r r r r

r r

r r

r r

r r r r r

r r

=− −( )

−( )

=−( )

=+ −( )

−( )

i i o o i o o i

i i o o

i i o o i o o i

p - p p p /

p - p

p - p p p /

2 2 2 2 2

2 2

2 2

2 2

2 2 2 2 2

2 2

(12.5)

where po=is the outer gauge pressure and, pi=inner gage pressure.

a) hoop stress b) radial stress

Figure 12.6 Stress distributions of hoop and radial stresses

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12.7

Combined Loading

Typical formulae for stresses in mechanics of materials are developed for specific

conditions. For example

Axial Loading,

Beam Bending, and

Direct Shear,

Torsional Shear,

Pressure Vessels, Shear, p

, p

,

σ

σ τ

τ

τ

σ σ σ

=

= =

=

=

= = =

P

A

My

I

VQ

ItP

A

Tr

Jr

t

r

t

N

N

T

T

h a r20

(12.6)

Often, the cross section of a member is subjected to several types of loadings

simultaneously and as a result the method of superposition can be applied to determine

the resultant stress distribution caused by the loads. In superposition, the stress

distribution due to each loading is first determined, and then these distributions are

superimposed to determine the resultant stress distributions. Note that only stresses of

the same type and in the same direction can be superimposed. The principle of

superposition can be used for the purpose provided that a linear relationship exists

between the stress and the loads. In addition, the geometry of the member should not

undergo significant change when the loads are applied. This is necessary in order to

endure that the stress produced by one load is not related to the stress produced by any

other loads. The following procedure is taken from (Hibbeler, 1997)

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12.8