Mechanics of Materials Laboratory

Transcription

1 ME 354 Mechanics of Materials Laboratory University of Washington σ τ τ τ σ τ τ τ σ x xy xz xy y yz xz yz z τ σ P ε σ P Course website: Department of Mechanical Engineering University of Washington Seattle, Washington Michael G. Jenkins Associate Professor, Mechanical Engineering TEL: ; FAX: ; jenkinsm@u.washington.edu Rev.0 01 January 001

2 Mechanical Engineering 354 Mechanics of Materials Laboratory TABLE OF CONTENTS Topic Page 1. Introduction Laboratory Procedure Laboratory Report. Stress/Strain/Constitutive Relations...1 Stress Strain Constitutive Relations 3. Beams: Stress, Strain, Deflection (Lab Exercise) Strain Stress: Normal and Shear Deflections 4. Beams: Curved, Composite, Unsymmetrical Bending (Lab Exercise) 4.1 Curved Beams Unsymmetrical Bending Composite Beams 5. Mechanical Properties and Performance of Materials (Tension, torsion, hardness, impact tests) (Lab Exercise) Mechanical Testing Tension Hardness Torsion Impact Plasticity Relations 6. Stress Concentration and Stress Raisers (Lab Exercise) Stress Concentration Factors Effects of Stress Raisers Experimental Techniques 7. Fracture and Effects of Cracks (Lab Exercise) Time Dependent Behaviour: Creep Deformation (Lab Exercise) Time Dependent Behaviour: Cyclic Fatigue (Lab Exercise)...9.1

3 Topic Page 10. Compression and Buckling of Columns (Lab Exercise) Structures: Complex Stresses and Deflections (Lab Exercise) Failure Criteria Combined Stresses Types of Engineering Structures 1. Pressure Vessels: Combined Loading (Lab Exercise) Thin-Walled Pressure Vessels Thick-Walled Pressure Vessels 13. References... R.1 Appendix A: Lab Report Format... A.1 Appendix B: Lab Exercise Handouts... B.1

4 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 0 (formerly ENGR 0) 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 0 or their equivalents. 1.1

5 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 b e 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. 1.

6 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, becausefor, 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 b e 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 b e 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. 1.3

7 . 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 or move 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 M x = 0 M y = 0 M z = 0 (.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 F force to incremental area on which the force acts such that: lim is the stress which A 0 A can be further defined as the intensity of the internal force on a specific plane (area) passing through a point..1

8 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: F σ = lim n (.) A 0 A 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 = F lim t A 0 A (.3) The general state of stress is one which includes all the internal stresses acting on an incremental element as shown in Figure.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 positive coordinate direction. For example, σ xx (or just σ x ) acts normal to the positive x face in the 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 shown that τ xy = τ yx ;τ xz = τ zx ;τ yz = τ zy. z y τ yz τ zy σz σy τ yx τ xy τ τxz zx Figure.1 General and complete stress state shown on a three-dimensional incremental element. σ x x.

9 Therefore the complete state of stress contains six independent stress components (three normal stresses, σ x ;σ y ;σ z and three shear stresses, τ xy ;τ yz ;τ xz ) which uniquely describe the stress state for each particular orientation. This complete state of stress can be written either in vector form or in matrix form σ x σ y σ z τ xy τ xz τ yz σ x τ xy τ xz τ xy σ y τ yz τ xz τ yz σ z The units of stress are in general: Pa = N N or MPa = 106 m psi = lb f in or ksi = 103 lb f in Force Area = F L. (.4) (.5) In SI units, stress is m = N and in US Customary units, stress is mm = kip in. 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.., the applied force, P, can be written in terms of its normal, P N, and tangential, P T, components which are functions of the angle, θ, such that: P N = P cos θ P T = P sinθ (.6) The area, A θ, on which P N and P T act can also be written in terms of the area, A, normal to the applied load, P, and the angle, θ, such that A θ = A /cosθ (.7) The normal and shear stress relation acting on any area oriented at angle, θ, relative to the original applied force, P are: σ θ = P N A θ = P cos θ A / cosθ = P A cos θ = σ cos θ τ θ = P T = P sinθ A θ A /cosθ = P cosθ sinθ = σ cosθ sinθ A where σ is the applied unidirectional normal stress. (.8).3

10 A = A θ cos θ θ A θ P T P N θ Figure. 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 the surface (e.g., σ x ;σ y ; τ xy ). 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 by applying 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..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..4 By applying simple statics such that in the x'-direction, F x ' = 0 and σ x' = σ x cos θ + σ y sin θ + τ xy cosθ sinθ or P (.9) σ x' = σ x + σ y + σ x σ y cosθ + τ xy sinθ y' y Ax= A cos θ θ θ x' A Ay= A sin θ Figure.3 Rotated axes and functions for incremental area. X.4

11 τxy Ax y x' σx Ax y' θ θ σx' A θ θ τx'y' A τxy Ay θ X σy Ay θ Figure.4 Rotated coordinate axes and components of stresses/forces. Similarly, for the x'y'-direction, F y' = 0 and τ x'y' = (σ x σ y )cos θ sinθ + τ xy (cos θ + sin θ) or (.10) τ x' y' = σ x σ y Finally, for the y' direction, F y' = 0 and sinθ +τ xy cosθ σ y' = σ x sin θ + σ y cos θ τ xy cosθ sinθ or (.11) σ y' = σ x +σ y σ x σ y cosθ τ xy sinθ 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 there are always three principal normal stresses which are ordered such that σ 1 > σ > σ 3. Mathematically, the principal normal stresses are found by determining the angular direction, 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 dσ dθ = 0 before solving for the θ at which the principal stresses occur. Applying this idea to Eq..9, gives.5

12 dσ dθ = 0 = ( σ x σ y )sinθ +τ xy cosθ sinθ cosθ = tanθ = τ xy (.1) σ x σ y There are two solutions for the principal stress angle (i.e., for maximum and minimum) so that. θ N1 = 1 τ xy tan-1 σ x σ y θ N = 1 τ xy tan-1 + π = θ σ x σ N1 + π y Using trigonometry on the geometry shown in Fig..5 results in τ xy tanθ = ( σ x σ y ) / sinθ = cosθ = τ xy σ x σ y + τ xy ( σ x σ y ) / σ x σ y + τ xy (.1) (.14) Substituting the trigonometric relations of Eq..14 back into Eq..9 gives for the plane stress case: σ θ = σ 1, = σ x +σ y τ xy tanθ p = ( σ x σ y ) / ± σ x σ y +τ xy (.15) x σ σ ( y ) + τ xy τ xy θ (σ σ )/ Figure.5 Geometric representation of the principal direction relations x y.6

13 Note that for the plane stress case in the x-y plane, σ z = 0. Thus the 1 and subscripts in Eq..15 are only for the x-y plane and are not necessarily σ 1 and σ for the threedimensional general state of stress. Therefore, ordering of σ 1 and σ of Eq..15 is only preliminary, until they are compared to σ z and ordered according to convention σ 1 > σ > σ 3. For example, for a particular plane stress state σ 1 and σ found from Eq..15 are 100 and 0 MPa, then the principal stresses are σ 1 = 100, σ = 0, σ 3 = 0 MPa. However, if σ 1 and σ found from Eq..15 are 15 and -5 MPa, then the principal stresses are σ 1 = 15, σ = 0, σ 3 = 5 MPa. Finally, if σ 1 and σ found from Eq..14 are -5 and -85 MPa, then the principal stresses are σ 1 = 0, σ = -5, σ 3 = 85 MPa. Performing a similar substitution of the trigonometric relations of Eq..14 back into Eq..10 gives for the plane stress case: τ max = σ x σ y +τ xy ( ) (.16) σ ave = σ x +σ y and tanθ s = σ x σ y τ xy Note that the τ max of Eq..15 is only for the x-y plane. The maximum shear stress for the three-dimensional stress state can be found after the principal stresses are ordered σ 1 > σ > σ 3 such that: τ 1,3 = σ σ 1 3 Some general observations can be made about principal stresses. (.17) a) In a principal direction, when τ =0, then σ 's are maximum or minimum b) σ max and σ min ( τ max and τ min ) occur in directions 90 apart. c) τ max occurs in a direction midway between the directions of σ max and σ min +σ, τ y τ σ C= x + σy φ =θ σ +σ,+τ x R = ( σx - C) + τ ) tan φ = - τ σ( x - C) Figure.6 Mohr's circle representation of plane stress state.7

14 An interesting graphical relation occurs if the second equation in each of Eqs..9 and.10 are squared and added together: σ x' = σ θ = σ x +σ y + + σ x σ y cosθ + τ xy sinθ τ x'y ' =τ θ = σ x σ y sinθ +τ xy cos θ = σ θ σ x +σ y +τ θ = σ x σ y +τ xy ( x h) +y = r (.18) The result shown in Eq..18 is the equation for a circle (i.e., Mohr's circle) with radius, σ x σ y r= + τ σ x +σ y xy and displaced h= on the x=σ θ axis as illustrated in Fig..6. Examples of Mohr's circles are shown in Fig..7. A procedure for developing Mohr's circle for plane stress is shown in the following section. τ for x-y plane max τ max = σ 1 σ 3 σ τ σ 1 σ Mohr's circle for stresses in x-y plane σ 3 τ σ σ 1 σ Mohr's circle for stresses in x-y-z planes Figure.7 Examples of Mohr's circle for plane and three dimensional stress states..8

15 Graphical Description of State of Stress -D Mohr's Circle Y σ y τ xy σ x In this example all stresses acting in axial directions are positive as shown in Fig. M1. Fig. M1- Positive stresses acting on a physical element. +σ y X y-face +τ xy +σ x As shown in Figs. M and M3, plotting actual sign of the shear stress with x normal stress requires plotting of the opposite sign of the shear stress with the y normal stress on the Mohr's circle. Fig. M - Directionality of shear acting on x and y faces. τ +σ, τ y φ =θ σ +σ,+τ x σ C= x + σy R = ( σx - C) + τ ) tan φ = - τ σ( x - C) Fig. M3 - Plotting stress values on Mohr's circle. In this example σ x > σ y and τ xy is positive. By the convention of Figs. M and M3, φ = θ on the Mohr's circle is negative from the +σ axis. (Mathematical convention is that positive angle is counterclockwise). Note that by the simple geometry of Fig. M3, φ = θ appears to be negative while by the formula, tan θ = τ xy /(σ x -σ y ), the physical angle, θ, is actually positive. In-plane principal stresses are: σ 1 = C+R σ = C - R Maximum in-plane shear stress is: τ max =R=(σ 1 -σ )/.9

16 Y σ σ 1 θ Direction of +θ X 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. Principal Axis Direction of Line of X-Y Stresses Fig. M5 - Direction of q from the line of x-y stresses to the principal stress axis. Note that the sense (direction) of the physical angle, θ, is the same as on the Mohr's circle from the line of the x-y stresses to the axes of the principal stresses. θ Same relations apply for Mohr's circle for strain except interchange variables as σ ε and τ γ.10

17 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: σ x σ τ xy τ xz τ xy σ y σ τ yz τ xz τ yz σ z σ (.19) Finding the determinant for this matrix and grouping terms gives: σ 3 I 1 σ + I σ I 3 = 0 (.0) where the stress invariants, ( I 1,I, I 3 ), do not vary with stress direction: I 1 = σ x + σ y + σ z I = σ x σ y +σ y σ z + σ z σ x τ xy τxz τ yz I 3 = σ x σ y σ z + τ xy τ xz τ yz σ x τ yz σy τ xz σ z τ xy (.1) Note that if principal stresses are used in Eq..1 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..0 can then be solved for the three roots which when ordered σ 1 > σ > σ 3 are the principal stresses. Eq..0 can be plotted as f σ the principal stresses are the values of σ which occur when f σ ( ) vs σ shown in Fig..9 where ( ) = 0. f( σ ) σ3 σ σ1 σ Figure.9 Solving for cubic routes for principal stresses.11

18 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 = A' B' AB lim L f L o (.) B A along n AB L o 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 = π ) θ ' (for small angles ) (.3) h where +γ occurs if π > θ ' and -γ occurs if π < θ '. The general state is one which includes all the internal strains acting on an incremental element as shown in Fig..10. The complete state of strain has six independent strain components (three normal strains, ε x ; ε y ; ε z 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 form ε x ε y ε z γ xy γ xz γ yz (.4).1

19 ε y ε yx A γ = ε + ε xy xy yx Engineering shear strain, Figure.10 General and complete strain state shown on a three-dimensional incremental element. Alternatively, the complete state of strain can be written in matrix form : ε x γ xy γ xz γ xy ε y γ yz γ xz γ yz ε z ε xy ε x (.5) The units of strain In general: Length Length = L, L In SI units, strain is m m for ε and m or radian for γ m and in US Customary units, strain is in in for ε and or radian for γ in in 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 twodimensional, plane strain condition (e.g., strain at a surface where no deformation occurs normal to the surface), strains exist only in the plane of the surface (e.g., ε x ; ε y ;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 the appropriate directions keeping in mind that δ = ε L o and = γ h as shown in Fig..11 Simply adding components of displacements in the x' direction, displacements in x 'direction for Q to Q * (see Fig..1) gives ε x' = ε x cos θ + ε y sin θ + γ xy cosθ sinθ or (.5) ε x' = ε x + ε y + ε x ε y cosθ + γ xy sinθ.13

20 y' y } dy = γ dy } ds Q x' Q* δ = ε y y dy dx x Figure.11 Rotated coordinate axes and displacements for x and y directions θ } δ x = ε x Similarly, adding components of displacements in due to rotations of dx' and dy' rotation of dx ' and dy' (see Fig..1) gives dx γ x'y' or = (ε x ε y )cos θ sinθ + γ xy (cos θ + sin θ) (.6) γ x'y' = ε x ε y sinθ + γ xy cosθ Finally, adding components of displacements in the y' direction, displacements in y 'direction for Q to Q * (see Fig..1) gives ε y' = ε x sin θ + ε y cos θ γ xy cosθ sinθ or (.7) ε y' = ε x + ε y ε x ε y cosθ γ xy sinθ y x' δ = ε x' Q { θ x' ds θ θ Q* = γ dy δ y = ε y dy cos θ = sin θ = dx ds dy ds δ x = ε x dx x Figure.1 Displacements in the x' direction for strains/displacements in the x and y directions.14

21 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 there are always three principal normal strains which are ordered such that ε 1 > ε > ε 3. Also as in the stress case, the principal strains for the plane strain case ε 1, = ε x + ε y ± tanθ p = γ xy ε x ε y ε x ε y + γ xy (.9) Note that for the plane strain case in the x-y plane, ε z = 0. Thus, the 1 and subscripts in Eq..9 are only for the x-y plane and are not necessarily ε 1 and ε for the threedimensional general state of strain. Therefore, ordering of ε 1 and ε of Eq..9 is only preliminary, until they are compared to ε z and ordered according to convention ε 1 > ε > ε 3. For the shear strain case γ max = ε ave = ε x + ε y ε x ε y + γ xy τ xy, ( ) and tanθ s = ε x ε y γ xy (.30) γ / for x-y plane max γ max = ε ε 1 3 ε γ/ ε 1 σ γ/ Mohr's circle for strains in x-y plane Mohr's circle for strains in x-y-z planes Figure.13 Examples of Mohr's circle for strain. ε 3 ε ε 1 σ.15

22 y c b a c b 60 a x x 45 Rectangular 60 Delta Figure..14 Rectangular and Delta rosettes. Note that the γ max of Eq..30 is only the x-y plane. The maximum shear strain for the three-dimensional strain state can be found after the principal strains are ordered ε 1 > ε > ε 3 such that: γ 1,3 = ε 1 ε 3 (.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..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 γ. An important application of the strain transformation relation of Eq..6 is to experimentally determine the complete strain state since stress is an abstract engineering quantity and strain is measurable/observable engineering quantity. Equation.6 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..14. The resulting equations to determine the local (for the strain gage rosettes shown in Fig..14) x-y strains in preparation for determining the principal strains: 45 Rectangular 60 Delta Rosette Rosette ε x = ε a ε x = ε a ε y = ε c ε y = 1 3 (ε b + ε c ε a ) (.3) γ xy = ε b (ε a + ε c ) γ xy = 3 (ε b ε c ) 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..16

23 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..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 (.33) The resulting normal strain, ε L, in the longitudinal direction can be calculated from the deformation response, L, along the L direction: ε L = L L (.34). Another normal strain, ε T, in the transverse direction can be calculated from the deformation response, D, along the D direction: ε T = D D (.35). A plot of σ vs. ε L (see Fig..16) shows a constant of proportionality between stress and strain in the elastic region such that where E = σ ε L y =mx + b σ = Eε L (.36) is known as the elastic modulus or Young's modulus and Eq..36 is known as unidirectional Hooke's law. P D=Do Df Undeformed Deformed A P D=Df -D L=Lo Lf L=Lf - L Figure.15 Uniaxially-loaded rod undergoing longitudinal and transverse deformation.17

24 Ε ν Longitudinal strain, εl Longitudinal strain, εl Figure.16 Plots of applied stress vs. longitudinal strain and transverse strain versus longitudinal strain A plot of ε T vs. ε L (see Fig..16) shows a constant of proportionality between transverse and longitudinal strains in the elastic region such that where ν = ε T ε L y =mx + b ε T = νε L (.37) is known as Poisson's ratio. For the results for three different uniaxial stresses applied separately (see Fig..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: ( ( )) ε x = 1 E σ x ν σ y +σ z ε y = 1 ( E σ y ν( σ x +σ z )) ε z = 1 ( E σ z ν( σ x +σ y )) (.38) Since shear stresses are decoupled (i.e., unaffected) by stresses in other directions the relations for shear stress-shear strain are: γ xy = 1 G τ xy γ xz = 1 G τ xz (.39) where the shear modulus is G = τ γ = γ yz = 1 G τ yz E (1+ ν)..18

25 ε y ε z ε σ x x Stress ε x ε y ε z σ x Eσ x - νε x = -νeσ x - νε x = -νeσ x ε z σ y ε y ε x σ y -νε y = -νeσ y Eσ y -νε y = -νeσ y ε z σ z εx ε Y y σ z - νε z = -νeσ z -νε z = -νeσ z Eσ z X Z : Figure.17 Development of generalized Hooke's law. Generalized Hooke's law is usually written in matrix form such that: { ε} = [ S ] σ where { ε} is the strain vector, { σ }is the stress vector and S Expanded, Eq..40 is: { } (.40) [ ] is the compliance matrix..19

26 ε x 1 ν ν σ x ε y E E E ν 1 ν σ y ε E E E z ν ν σ γ = E E E z xy G γ 0 0 τ xy xz G 0 τ xz γ yz τ G yz (.41) Although Eq..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 that σ { } = C [ ] ε { } (.4) where [ C]is the stiffness matrix such that [ C] = [ S] 1. Expanding Eq..4 gives σ x E σ y (1+ν) 1+ ν νe νe ε x (1 ν ) (1+ν )(1 ν) (1+ν)(1 ν) ε σ νe E z (1+ν)(1 ν) (1+ν ) 1+ y ν νe (1 ν ) (1+ν )(1 ν) ε z τ = νe νe E xy (1+ν)(1 ν) (1+ν)(1 ν) (1+ν) 1+ ν (1 ν) γ xy G 0 0 τ xz G 0 γ xz τ yz G γ yz (.43) Generalized Hooke's can be simplified somewhat for the special case of plane stress in the x-y plane since σ z =0. Being orthogonal to the x-y plane, Since σ z 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 E νe σ x (1 ν ) (1 ν 0 ) ε νe E x σ y = (1 ν τ xy ) (1 ν 0 ε y ) 0 0 G γ xy (.44) For the special case of plane stress, although σ z =0, the strain in the z-direction is not zero but instead can be determined such that Plane stress : σ z = 0, ε z 0 = ν 1 ν (ε x + ε y ) (.45).0

27 Similarly for the special case of plane strain, although, ε z =0, the stress in the z- direction is not zero but instead can be determined such that Plane strain : ε z = 0, σ z 0 = ν(σ x +σ y ) (.46) For the special case of hydrostatic pressure, no distortion takes place, only size changes. In this case the hydrostatic stress is ( ) σ H = σ x +σ y + σ x 3 and the dilatation (i.e., volumetric change) is (.47) ε V = V V = ( ε x + ε y + ε x ) (.48) The ratio between the hydrostatic stress and the dilation is a special combination of elastic constants called the bulk modulus. ( ) ( ) = E k = σ H = σ x + σ y + σ x ε V 3 ε x + ε y + ε x 3(1 ν ) (.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..18. Since this is the uniaxial case, recall that σ = P L and ε = A L = x. Therefore, since E is defined as E = σ x e ε = dσ with dε dσ = dp A dx and ε = x e then E = dσ dε = x e A dp dx. Note from Fig..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..19. Before deformation the triangle between centers has a transverse side length of 3R, a longitudinal side length of R and a hypotenuse of R. After longitudinal deformation, the transverse side length is 3R -dx, the longitudinal side.1

28 Figure.18 Repulsion-attraction force-displacement curve between two atoms. length is R+dy and the hypotenuse is still R. Applying Pythagorean's theorem after deformation gives ( R ) = ( 3R dx ) + ( R + dy) (.50) Expanding Eq..50 and eliminating higher order terms leaves Rdy = 3Rdx dy dx = 3. (.51) Recalling that the longitudinal and transverse strains can be written in terms of the deformations ε ( y = R + dy ) R R ( 3R dx ) 3R ε x = 3R = dy R dy = Rε y = dx 3R dx = 3Rε x (.5) Combining the two terms for dy and dx in Eq..5 and equating them to Eq..51 gives dy dx = Rε y 3Rε x = 3 ε y ε x 1 ν = 3 (.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..

29 Figure.19 Undeformed and deformed sphere model for atomic structure and determination of Poisson's ratio.3

30 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.a, 3.b, and 3.c 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.d, 3.e and 3.f 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

31 Figure 3. 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.a, 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 A study of this section diagram reveals that a transverse force V r and a couple M r at the cut section and a force, R, (a reaction) at the left support are needed to maintain equilibrium. The force V r is the resultant of the shearing stresses at the section (on plane a-a) and is called the resisting shear and the moment, M r, is the resultant of the normal stresses at the section and is called the resisting moment. 3.

32 Figure 3.3 Section of simply supported beam. The magnitudes and senses of V r and M r may be obtained form the equations of equilibrium F y = 0 and M O = 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 M r 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

33 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: δ y y = δ c c (3.1) where δ y is the deformation at distance y from the neutral axis and δ c 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: δ y = δ c c y (3.) 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: δ y x = y c δ c x ε = y c ε c (3.3) 3.4

34 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 that ε =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 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: s' s ε = lim s 0 s (3.4) Strain can be represented in terms of distance y from the neutral axis and radius of curvature 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 θ defines 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: ( ρ y) θ ρ θ ε = lim θ 0 ρ θ (3.5) 3.5

35 Eq. 3.5 can be arithmetically simplified as ε = y / ρ. Since the maximum strain occurs at the outer fibre which is distance c from the neutral surface, ε max = c / ρ = ε c, the ratio of strain at y to maximum strain is ε = y / ρ ε max c / ρ (3.6) which when simplified and rearranged gives the same result as Eq. 3.3: ε = y c ε max = y c ε c (3.7) Note that an important result of the strain equations for ε = y / ρ and ε max = c / ρ = ε c 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, ε x, 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 to obtain 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

36 Figure 3.7 Stress (force) distribution in a bent beam Note that for static equilibrium, the resisting moment, M r, must equal the applied moment, M, such that M O = 0 where (see Fig. 3.7): M r = dfy = σday (3.8) A and since y is measured from the neutral surface, it is first necessary to locate this surface by means of the equilibrium equation F x = 0 which gives σda = 0. For the case of A elastic action the relation between σ x and y can be obtained from generalized Hooke's E law σ x = 1 ν 1+ν [ ( )ε x +ν ( ε y +ε )] z and the observation that ε y = ε z = νε x. ( )( 1 ν ) The resulting stress-strain relation is for the uniaxial stress state such that σ x = Eε x which when substituted into Eq. 3.3 or 3.7 gives σ x =E ε c c Substituting Eq. 3.9 into Eq. 3.8 gives: A y = σ c c y (3.9) M r = σday A = σ c c y σ da = x A y y da (3.10) A Note that the integral is the second moment of the cross sectional area, also known as the moment of inertia, I, such that I = y da (3.11) A 3.7

37 Figure 3.8 Action of shear stresses in unbonded and bonded boards Substituting Eq into Eq and rearranging results in the elastic flexure stress equation: σ x = My I (3.1) 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 = Mc I = M Z (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

38 Figure 3.9 Distortion in a bend beam due to shear The development of a general shear stress relation for beams is again based on static equilibrium such that F = 0. Referring to the free body diagram shown in Fig. 3.10, the differential force, df 1 is the normal force acting on a differential area da and is equal to σ da. The resultant of these differential forces is F 1 (not shown). Thus, F 1 = 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 = My I. Figure 3.10 Free body diagram for development of shear stress relation 3.9

39 When the two expressions are combined, the force, F 1, becomes: F 1 = M I y da= M c ty dy (3.14) I h Similarly, the resultant force on the right side of the element is F = ( M+ M) c ty dy (3.15) I h The summation of forces in the horizontal direction on Fig gives V H = F F 1 = M I The average shear stress is V H divided by the area from which τ = lim x 0 M 1 c ty dy x It h c ty dy (3.16) h = dm dx 1 c ty dy (3.17) It Recall that V=dM/dx, which is the shear at the beam section where the stress is being c evaluated. Note that the integral, Q= ty dy is the first moment of that portion of the cross h 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 = VQ (3.18) It 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 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 takes on a parabolic distribution, with τ =0 at the outer fibre (where y=c and σ =σ max ) and τ =τ max at the neutral surface (where y=0 and σ =0) as shown in Fig h 3.10

40 Y τ = 0 σ = σ min X τ = τ max σ = 0 Ν/Α τ = 0 σ = σ max 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 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 τ av 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 τ max 3V which is 1.5 A times τ av V A. h τ 3V max = A A=bh b d τ 4V max = 3A A=( π/4) d di do τ V max = A A=( π/4) (d o - d i ) Figure 3.1 Maximum shear stresses for some common uniform cross sections 3.11

41 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, ) 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 of radius, ρ, as shown in Fig 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 = L ρ = L + δ ρ + c from which and finally c ρ = δ L = ε = σ E = Mc EI 1 ρ = M EI (3.19) (3.0) (3.1) 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 3.1

42 Equation 3.1 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 1 ρ = d y / dx 1+ dy / dx [ ( ) ] 3/ (3.) 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. becomes 1 ρ = d y = y'' (3.3) dx Substituting Eq. 3.3 into Eq. 3.1 and rearranging gives EI d y dx = M x = EIy'' (3.4) which is the differential equation for the elastic curve of a beam. An alternative method for obtaining Eq. 3.4 is to use the geometry of the bent beam as shown in Fig where it is evident that dy/dx = tan θ θ for small angles and that d y /dx = dθ /dx. From Fig it can be shown that for small angles and therefore. dθ = dl ρ = dx ρ (3.5) y'' = d y dx = dθ dx = 1 ρ = M x EI EI d y dx = EIy'' =M x (3.6) 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

43 Figure 3.15 Sign conventions used for deflection deflection = y slope = dy dx = y ' moment = M x =EI d y dx = EIy'' shear = dm dx =EI d3 y = EIy''' (for constant EI) 3 dx load = dv dx =EI d4 y dx 4 = EIyiv (for constant EI) (3.7) It is interesting to note that from Eqs. 3.4 and 3.6 can be written as M x = EI dθ (3.8) dx from which θ B x B M dθ = x EI dx (3.9) θ A x A Eqs. 3.8 and 3.9 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 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

44 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 ) 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. 3.4 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

45 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 ) 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 dy /dx =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 is from which the shear in Fig is negative such that dy = γdx = τ VQ dx = dx (3.30) G GIt GIt Q dy dx = V GIt Q y' = V (3.31) 3.16

46 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 estimated as: kagy' = V y'= V kag (3.3) Often k is approximated as k 1 but for box-like sections or webbed sections it is 1 k = A total A web (3.33) A single integration method can be used to solve Eq. 3.3 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

47 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 a P b a+b=l Loading Diagram P Free Body Diagram R1 =Pb/L R =Pa/L Pb/L Pa/L Shear Diagram Pbx/L Pba/L (Pbx/L)-P(x-a) 0 x a x=a a0 x L Moment Diagram Figure 3.19 Loading condition, free body, shear and moment diagrams x There are two boundary conditions: at x=0, y 1 =0 and at x=l, y =0 There are two matching conditions: at x=a, y' 1 =y' and at x=a, y 1 =y 3.18

48 For 0 x a (region 1) For a x L (region ) V= Pb/L V= Pa/L M=Pbx/L M=(Pbx/L)-P(x-a) Double Integration Method EIy 1 ''= M = Pbx L EIy 1 '' = -Pbx L EIy 1 ' = -Pbx L + C 1 EIy 1 ' = Pbx L + C 1 EIy 1 = Pbx3 6L + C 1 x + C 3 Double Integration Method EIy '' = M = Pbx L EIy '' = Pbx + P(x a) L EIy ' = Pbx L EIy ' = Pbx EIy = Pbx3 6L L + P(x a) + P(x a) P(x -a) + +C + C P(x -a)3 + +C 6 x + C 4 Applying the matching conditons at x=a, y' 1 =y' y 1 '= 1 -Pba EI L + C 1 = 1 -Pba EI L + P( a-a) +C = y ' so that C 1 = C and at x=a, y 1 =y y 1 = 1 Pba 3 EI 6L + C 1 a + C 3 = 1 Pba 3 EI 6L P(a -a) C a +C 4 = y Since C 1 = C,then C 3 = C

49 Applying the boundary conditons at x=0, y 1 =0 y 1 = 0 = 1 Pb 0 3 EI 6L + C C 3 and at x=l, y =0 So C 3 = 0 y = 0 = 1 PbL 3 EI 6L + P(L-a)3 6 +C L +C 4 = Since (L-a) = b and C 1 = C,and C 3 = C 4 = 0 [ ] then C = PbL 6 Pb3 6L = Pb 6L L b Finally, the equations for deflection due to the bending moment are: For 0 x a (region 1) For a x L (region ) [ ] y = Pbx [ ] P x -a y 1 = Pbx 6EIL L b x 6EIL L b x ( )3 6EI The deflection due to the shear component is: For 0 x a (region 1) For a x L (region ) V=Pb/L kag y' 1 = V y' 1 = V kag = Pb 1 L kag V=Pa/L kag y' 1 = V y' 1 = V kag = Pa 1 L kag y' 1 = Pb 1 y L kag 1 = Pb klag x + C 1 y' 1 = Pa 1 y L kag 1 = Pa klag x + C 3.0

50 Applying the boundary condition at x=0, y 1 =0, y 1 = 0 = Pb klag 0 + C 1 So C 1 = 0 and y 1 = Applying the boundary condition at x=l, y =0, y = 0 = Pa klag L + C Pb klag x for 0 x a So C = Pa kag and y = Pa kag x L +1 Now the total deflection relation for both bending and shear is: For 0 x a (region 1) For a x L (region ) V=Pa/L [ ] y 1 = Pbx 6EIL L b x Pb klag x [ ] P x-a y = Pbx 6EIL L b x ( )3 6EI Pa x klag L

51 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 4.1

52 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.). The stress deforms the material in such a way that the cross section rotates through an angle of δθ /. The normal strain in an arbitrary strip at location r can be determined from the resulting deformation. This strip has an original length of L o =r dθ. The strip s total change in length, L= ( R r) δθ /. The normal strain can be written as: L ( R r) δθ. ε = = (4.1) L r dθ o To simplify the relation, a constant k is defined as k = δθ / dθ such that the normal strain can b e rewritten as: ε = k R r r (4.) Fig 4. Isolated differential element is a curved beam Note that Eq 4. 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 4.

53 beam (i.e., ε = y ). The nonlinear strain distribution for the curved beam occurs even though ρ 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 = E ε= Ek R 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 x = 0 Now since σ = df then df = σ da and da F = σ da = σda= 0 or Ek R r da = 0 r (4.4) Because Ek and R are constants Eq 4.4 can be rearranged such that: R Ek da - r Solving Equation 4.5 for R results in: r r da da - da = 0 R r = 0 (4.5) da R= = A da da (4.6) r r 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 4.3

54 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 σ = df then df = σ da da if y is the distance from the neutral axis such that y = R - r then dm = y df or dm = y ( σ da) Applying moment equilibrium such that M = 0 gives M y ( σ da)= 0 or M = y ( σ da) Substituting the derived relations for y and σ gives M = y ( σ da)= ( R r) Ek R r da (4.7) r Again realizing that Ek and R are constants, Eq 4.7 can be expanded and grouped such that: 4.4

55 da M = Ek R R da + rda such that r R A R R A ra which gives M=Ek R A R A + ra = EkA(r - R) R (4.8) Note that the third integral term in Eq 4.8, comes from the geometric determination of the M centroid such that r = r da/a. Equation 4.8 can now be solved for Ek such that Ek = A(r - R). Substituting this relation for Ek into Eq 4.3 gives: σ = ε= = = E Ek R r M R r M( R r) r A(r - R) r 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: M( R r) My σ = = (4.10) Ar(r - R) Ae(R - y) 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= A ). da r Note that Eq 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

56 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 4.6

57 A common machine element problem involving curved beams is the crane hook shown in Fig In this problem, the load, F is,40 N, the cross sectional thickness, t=b=19.05 mm and the cross sectional width, h=w= is mm. Since A=bh, da=bdr and from Eq. 4.6: A bh h R = = r = (4.11) da o b r r dr r o ln r ri From Figs 4.4a and 4.4b, ri=50.8 mm, ro=15.4 mm, and A= mm. Substituting the appropriate values into Eq gives R=9.5 mm. The eccentricity, e=( )=9.1 mm. Since the moment, M is positive such that M = F r (r = radial distance to the centroid where r =(ro+ri)/). In the case, the axial force, F, superposes an axial stress on the bending stress such that F My, 40 (, 40 * )( 9. 5 r) σ = + = + A Ae( R y) (.) r i (4.1) Substituting values of r from ri=50.8 mm to ro=15.4 mm results in the stress distribution shown in Fig. 4.5 (in psi). The stresses on the inner and radii are MPa and MPa, respectively. Note that if the flexure stress relations for an initially straight beam are used such that: F My, 40 (, 40 * )( r) σ = + = + (4.13) 3 A I ( ) 1 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 4.7

58 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 ) 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 action makes 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 distribution function (i.e., σ = E( ε / c) y = ( σ / c) y) cannot be expressed in terms of one variable. c c Fig 4.6 Beam undergoing unsymmetrical bending However, since the plane section remains plane, the stress variation can be written as: 4.8

59 σ = ky 1 + kz (4.14) The resisting moments with respect to the z and y axes can be written as M = σday = k y da + k zyda = k I + k I rz A M = σdaz = k yzda + k z da = k I + k I ry A A A 1 A 1 A 1 z yz 1 yz y (4.15) where I y and Iz are the moment of inertia of the cross sectional area with respect to the z and y axes, respectively, and I yz is the product of inertia with respect to these two axes. It will b e convenient to let the y and z axes be principal axes, Y and Z; then I yz is zero. Equating the applied moment to the resisting moment and solving for k and k 1 gives: M cosα MrZ = k1i Z = Mcosα k1 = I M sinα MrY = ki Y = Msinα k = I Z Y (4.16) Substituting the expressions for k given in Eq into Eq gives the elastic flexure formula for unsymmetrical bending. Mcosα Msinα σ = Y + Z I I Z Y (4.17) Since σ is zero at the neutral axis, the orientation of the neutral axis is found by setting Eq equal to zero, for which or cosα sinα Y = - Z (4.18) I I Z Y Y = -tanα I Z IZ Y (4.19) where Y is the equation of the neutral axis in the YZ plane. The slope of the line is the dy/dz and since dy/dz= tan β, the orientation of the neutral axis is given by the expression IZ tan β = - tan α I Y (4.0) 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 ) 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.) 4.9

60 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: εb= ε b a a (4.1) which, in terms of stress, becomes σ b σ a = b a E E B A () (4.) After rearranging, Eq. 4. becomes the relation for the stress distribution: E B σb= b σ a E a A (4.3) 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, E n B =, of the two elastic moduli (see Fig. 4.7). Using Eq. 4.3, the normal force on a EA differential end area of element B is given by the expression: df = n b σbda = σa a B da = b σ (nt)dy for c y c a a A B (4.4) 4.10

61 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 from y=c A to y=c B 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 4.11

62 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 ) 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

63 Figure 5.1 Geometry and loading scenarios commonly employed in mechanical testing of materials. 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.) 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 servohydraulic (high-pressure hydraulic fluid in hydraulic cylinders). Digital, closed loop Figure 4. Example of a modern, closed-loop servo-hydraulic universal test machine. 5.

64 control (e.g., force, displacement, strain, etc.) along with computer interfaces and userfriendly 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 quasistatic) 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 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 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. percent ( ε = 0. 00) 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 in the elastic region. ε 5.3

65 Figure 5.3 Engineering stress-strain diagram for hot-rolled carbon steel showing important properties (Note, Units of stress are psi for US Customary and MPa 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

66 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 A o (5.1) ε = Deformed length - Original length Original length = L L o L o (5.) 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 ε o U r = σ dε σ ε o o (5.3) 0 where σ o is the proportional limit stress and ε o 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 ε f U t = σ dε (σ + S ) o u ε f (5.4) 0 where Su is the ultimate tensile strength, σ o 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. Material Table 5.1 Energy properties of materials in tension Yield Ultimate Modulus of Strength Strength Resilience, (MPa) (MPa) (kj/m 3 ) Modulus of Toughness, (kj/m 3 ) SAE 100 annealed SAE 100 heat treated Type 304 stainless Cast iron Ductile cast iron Alcoa Red brass

67 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 = 100ε f = 100 L f L o = 100 L f 1 (5.5) L o L o 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 = 100 A A o f = A f (5.6) A o A o 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 (σ T ) and true strain (ε T ) quantities are used. The quantities are defined as: σ T = P A i (5.7) where Ai is the instantaneous area at the time P is measured. Also or L ε T = dl = ln L (5.8) L L o L o ε T = A A o da A = ln A o A where L is the instantaneous length between gage mark at the time P is measured. (5.9) 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. then Since AL = A o L o (5.10) L L o = A o A (5.11) The constant volume condition simply says the stressed volume AL is equal to the original volume AoLo. (Note this is only true in the plastic range of deformation, in the 5.6

68 elastic range the change in volume V per unit volume is given by the bulk modulus B E (where B = 3(1 υ) and υ is Poisson's ratio). then Prior to necking, engineering values can be related to true values by noting that ε T = ln L i = ln L + L o (5.1) L o L o ε T = ln(1+ ε) (5.13) and since so and since then A o A = L L o = L o + L L o (5.14) A = A o 1 + ε σ T = P A σ T = P A o (5.15) (5.16) ( 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 and also σ = Eε (5.18) σ T = Eε T (5.19) In the plastic region, a commonly used relation to define the relation between stress and strain is given by σ T = K (ε T ) n = H(ε T ) m (5.0) where strength coefficient, K or H, is the stress when ε T =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

69 Table 4. Material constants m or n and K or H for different sheet materials Specimen Material Treatment n or m K or H (MPa) Thickness (mm) 0.05% carbon rimmed steel Annealed /0.07% phosphorus low-carbon steel Annealed SAE 4130 steel Annealed Type 430 stainless steel (17% Cr) Annealed 0.9 Alcoa 4-S aluminum Annealed The last two equations can be written in the form log σ T = log E + log ε T (elastic) (5.1) and log σ T = log K+ m log ε T (plastic) (5.) 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).. 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 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 in (51 mm) diameter reduced section (giving a cross sectional area of 0. in.) 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

70 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.0 in (50.8 mm) gage length is used to measure deformations. The.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

71 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 ) 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 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

72 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 = P [ ( )] πd D D d (5.3) 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: P 1 D 1 = P D = P 3 D 3 = etc. (5.4) 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: P MHN = HM = (5.5) πd /4 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 0.5. 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 = P L sin α (5.6) where P is the indentation load which typically ranges from 0.1 to 1 kg but may be as high as 10 kg, L is the diagonal of the indentation in mm and α is the included angle of

73 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. 5.1

74 Steel or tungsten carbide ball P=3000 kg or 500 kg D=10 mm d t Side view 0 1 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 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

75 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. ) 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 selfcontained 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: S y = BHN x / 3 (5.7) where Sy is the yield strength of the material in MPa and BHN is the Brinell hardness in kg/mm. 5.14

76 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, S u = 3.45 x BHN (5.8) where Su is the ultimate tensile strength of the material in MPa and BHN is the Brinell hardness in kg/mm. Figure 5.9 Plastic deformation under a Brinell hardness indenter. 5.15

77 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 loaddeflection diagram obtained in the tension test. To be useful for engineering purposes, it is 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 can also convert the τ - γ diagram to a σ - ε diagram as will be shown later. 5.16

78 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 ) 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 5.10 The shear strain γ is the relative rotation of one circular cross-section with respect to a section one unit length away or: γ = rθ L where θ is in radians. This relation is true in either the elastic or plastic range. (5.9) 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: T τ = πr (5.30) t Using the Eqs. 5.9 and 5.30, the complete τ - γ diagram in the elastic and plastic range can be obtained. 5.17

79 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 = rθ L. 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 Considering the very thin circumferential ring shown above, the torque resisted by this ring is given by since dt = (shear stress) (area) (lever arm) dt= τ τ r a r = πa τda (5.31) da = π ada. (5.3) Since the stress distribution is linear, at any radius, a, the shear stress, γ, is related to the maximum shear stress, τ r, existing at r by τ = τ r a r (5.33) so substituting in the equation for dt, the torque on a small area becomes: dt= π τ r r a3 da (5.34) and integrating over the entire cross-sectional area, the total external torque is equal to 5.18

80 T= πτ r r a 3 da = πτ r r o r r 4 4 = πτ r r3 (5.35) and the shear stress at the outermost fibers is τ r = T πr 3 (5.36) Note that Eqs 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 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 = πτa da (5.37) So the total external torque resisted across the section is then a T = π τa da (5.38) o The shear strain relation γ = aθ L in Eq. 5.38, we obtain at any radius a is still valid, however, and substituting this r T = π τγ L o θ L θ dγ (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 γ r Tθ 3 = πl 3 f( γ) γ dγ (5.41) o Fig. 5.1 Elastic-Plastic Shear Stress Distribution 5.19

81 Differentiating both sides of this equation with respect to 9, one obtains d dθ = ( ) Tθ3 = πl 3 dγ f( γ r )γ r r dθ ` (5.4) since γ r = rθ L then dγ r dθ = r L and substituting these quantities in the equation for d dθ Tθ 3 derivative, one obtains and ( ) and working out the (5.44) 3Tθ + dt dθ θ3 = πτr 3 θ (5.45) 3T +θ dt dθ = πτr3 (5.44) Solving for the shear stress, the result is 1 τ = πr 3 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 Fig Example of torque-twist curve used for data 5.0

82 At the typical point P at which it is desired to obtain the shear stress, observe that θ = BC and that dt dθ = PC such that T = AP. Substituting these quantities, the result is BC τ = 1 PC πr 3 BC BC + 3AP PC + 3AP or τ = πr 3. 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 = rθ, the complete τ - γ L curve can be obtained. For the materials approach, it is possible to again make the valid assumption that γ = rθ. However, τ is determined as one of two functions of γ depending on whether the L 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 E tensile test. In particular, E is required to calculate G = (1+ υ), σ o is required to calculate τ o = σ o 3, K and n are required to calculate K τ = K ( n +1)/ and n for shear equals n for 3 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. or τ = Gγ for τ τ o and/or γ γ o (5.46) τ = K τ γ n for τ > τ o and/or γ > γ o (5.47) K where G is the shear modulus, K τ = ( n +1)/ in which K and n are the strength coefficient 3 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 where τ o = σ o 3 in which σ o γ o = τ o G is the "yield" strength from a tension test. (5.48) 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 = rθ. Comparing this shear strain to that L calculated in Eq. 5.48, allows the choice of either Eq or Note that when the shear stress at r=r is plastic, the total torque, T, required to produce 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 5.1

83 plastic part and an elastic part as shown in Fig The relation for T can then be written as: Ttotal = Telastic + T plastic (5.50) where τ o T elastic = τ da r = Gγ da r (5.51) 0 γ o 0 γ R T plastic = τ da r = K τ γ n da r (5.5) τ R τ o γ R For convenience it is possible to rewrite the integration variables in Eqs and 5.5 in terms for the specimen radius, r, only such that r y T elastic = τ y πr 3 dr r 0 y (5.53) R n K rθ' T plastic = πr dr 3 3 (5.54) r y where θ' = θ L and r y = θ L γ o = θ L τ o G = θ' τ o G. Equations 5.53 and 5.54 can be solved either closed form or numerically for any combination of T and θ. γ τ θ γ =r /L =f( ) Elastic Plastic n τ=gγ τ =K τ γ r=0 r y r=r Radial distance, r Figure 5.14 Shear stress and shear strain as functions of radial distance 5.

84 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 offset yield 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 5.3

85 As in the tension test, the Modulus of Resilience is the area under the elastic portion of the τ - γ curve such that that γ o U r = τ d γ (5.55) o Similarly, the Modulus of Toughness is the area under the total τ - γ curve such γ f U t = τ d γ (5.56) o 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 (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 = σ ; γ equivalent = 1.5ε (5.57) Plastic range: τ equivalent = σ ; 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 5.4

86 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 5.5

87 mass, m h1 h IMPACT ENERGY=mg(h1-h) 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 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 bodycentered 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 340. Note that a transition temperature as high as about 5 C is shown. This material, then should not be in service below temperature of 5 C when impact conditions are likely to exist. 5.6

88 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 crystalline appearance, while the portion of the specimen which fracture in a ductile manner 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 0 J at the service temperature by proper control of impurities. 5.7

89 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. σ ο ε T Rigid-Perfectly Plastic E σ ο ε T Elastic-Perfectly Plastic σ ο E Power Linear ε T Elastic-Linear Hardening Elastic-Power Hardening 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 discrete region, separate at σ = σ o such that: Elastic : σ = Eε (σ σ o ) (5.59) Plastic : σ = Hε n (σ σ o ) (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 = ε e + ε p = σ E + ε p (5.61) and n σ σ = H ε p ε = E + σ H 1 n (5.6) 5.8

90 For the Ramberg-Osgood relation, σ o is not distinct "break" in the stress-strain curve, but is instead calculated from the elasticity and plasticity relations such that 1 σ o = E H 1 n (5.63) E General stress-strain relations can be developed for deformation plasticity theory such that the effective stress is and the effective total strain is σ eff = 1 (σ 1 σ ) + (σ σ 3 ) +(σ 3 σ 1 ) (5.64) where the effective plastic strain is ε = σ E + ε p (5.65) ε p = 3 (ε p 1 ε p ) + (ε p ε p 3 ) + (ε p 3 ε p1 ) (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 = K τ γ n (5.67) K where K τ = ( n +1)/ in which K and n are the strength coefficient and strain hardening 3 exponent from the tension test, respectively. In addition, τ o = σ o (5.68) 3 where σ o is the yield strength from a tension test. σ ο Strain Hardening Ε ε p ε e Strain Figure 5.0 Elastic-plastic stress-strain curve 5.9

91 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. show design data for stress concentration factors, K=σ local / σ remote or K= σ w/ notch / σ w/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. 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., where a stress concentration, K, of.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

92 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.. 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.

93 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

94 Figure 6. 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 strain levels 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

95 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 accurate results, in particular, the determination of the principal stresses σ 1 and σ 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

96 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.. These split polarized beams are out of phase by an amount that is dependent of the difference of the principal stresses, i. e. to (σ 1 σ ) 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 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

97 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 (σ 1 σ ) (6.1) where f is the stress-optical coefficient, N is the fringe order, t is the model thickness, and σ 1 and σ are the plane-stress principal stresses. Each dark band (See Fig. 6.4) for a dark field arrangement corresponds to an integral (0, 1,, 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 a point of zero fringe order such as a free unloaded corner. At such a corner σ 1 = σ = 0. hence σ 1 = σ = 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: (σ 1 σ ) = f N t (6.) where f is the stress-optical coefficient determined previously, N is the fringe order, t is the model thickness, and σ 1 and σ 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

98 Figure 6.4 Photoelastic model as viewed in polariscope. Fringe value is 0 at external sharp corners and 3 in narrow leg. Note that the model is a uniaxially loaded tensile specimen. 6.8

99 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.. If this relation is modeled as a sine wave then the equation can be written as: x σ = σ max sinπ (7.1) λ / where σ max is the maximum stress on the curve, x is the displacement and λ / is the half wave length of the sine wave. Applying uniaxial Hooke's law (σ = Eε ) where ε = x x e x e and small angle approximations (sinθ = θ ) of Eq. 7.1 gives σ = Eε = E x x e πx = σ max λ / x e (7.) ALLOWABLE STRESS, σ CRACK LENGTH, a High K Ic Low K Ic ALLOWABLE STRESS, σ a t σ ο = transition crack length between yield and fracture CRACK LENGTH, a Figure 7.1 Cracks lower the material's tolerance (allowable stress) to fracture. 7.1

100 Repulsion σ max Attraction Separation stress, σ Sine Wave Approximation Equilibrium Spacing, x Half Wave Length, / e λ Separation distance, x Figure 7. Attractive/repulsive force-distance curve between two atoms where E is the elastic modulus and x e is the equilibrium distance (a.k.a., lattice spacing) between the atoms. Assuming λ /=x e and solving Eq. 7. for σ max gives. σ max = E π (7.3) where σ max 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: x x λ / W = F (x )dx = σ(x )Adx = σ max sin πx λ / Adx (7.4) o o o 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 = σ λ max π = γ s (7.5) where γ s is the fracture surface energy. Solving Eq. 7.5 for λ results in λ = γ s π σ max. Substituting this expression for λ into Eγ Eq. 7. and solving for σ max σ max = s (7.6) x e 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 = 00 GPa, γ s = 1 J/m, and xe=ao=.5x10-10 m. In this case, Eq. 7.3 gives σ max = 63.7 GPa and Eq. 7.6 gives σ max = 8.3 GPa. Since most steels have ultimate strengths in the 0.5 to 1 GPa range, something 7.

101 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: σ rr = σ 1 R r 1+ 3R r 1 cosθ σ θθ = σ 1 + R r R r cosθ (7.7) At the edge of the hole for θ=0, Eqs. 7.7 can be rewritten in Cartesian coordinates as: σ xx = σ rr = σ 1 R r 3R r (7.8) σ yy = σ θθ = σ + R r + 3R r Finally at the edge of the hole, where r=r, the stresses are: σ xx = σ rr = 0 σ yy = σ θθ = 3σ (7.9) The stress concentration factor can be defined as k t = σ local σ remote. 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 + c a = σ 1+ c (7.10) ρ where the radius of the ellipse tip is ρ = a. The stress concentration factor for the c elliptical hole is no longer constant as it is for the circular hole but is a function of the ellipse geometry: k t = 1 + c ρ c (7.11). ρ Note in this case that as the minor axis a shrinks to zero (e.g., as in a crack), ρ goes to zero and the stress concentration factor in Eq goes to infinity. 7.3

102 σ σ y R θ r x a y ρ= a _ c x c σ σ 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, σ max, is exceeded by the stress, σ yy, at the tip of the ellipse. Combining Eqs and 7.6 (simplifying per Eq. 7.11) gives: σ max = σ yy Eγ s = σ c (7.1) a o ρ If fracture occurs at an atomistically-sharp ellipse tip, then let ρ a o and the remote stress required to cause brittle fracture is: σ f = Eγ s 4c (7.13) 7.4

103 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, c, the stored elastic strain energy is: U E = 1 σ σεv = ( E πc t) σ ( E πc t) (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, γ s, and the surface area of the crack, A= (ct), such that: U s = γ s A = γ s ((ct)) = γ s (4ct) (7.15) The change in stored energy from the uncracked panel to the cracked panel is: U = U s U E = γ s A = γ s (4 ct) σ E πc t ( ) (17.16) σ >>>W t c Figure 7.4 Uniformly stress plate of thickness, t, and width, W, into which a crack of length, c, will be introduced. σ 7.5

104 The three energy terms, U s, U E and U, are plotted as functions of half crack length, c, in Fig It is apparent that the critical condition for unstable, brittle fracture is reached when c=c c and d U =0 at the peak of the U curve. Differentiating Eq. 17.6, dc setting it equal to zero, and solving for the stress at which brittle fracture occurs results in the following: Plane stress (σ z = 0) σ f = Plane strain (ε z = 0) σ f = γ s E πc γ s E (1 ν )πc (17.17) It is interesting to compare Eqs and and note that they only differ by factors of 1/4 and /π under the radical, respectively. However, from a physical standpoint Eq is more "satisfying" because it is based purely on an energy balance, making no assumptions about crack shapes or stress solutions. Although Eq 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). Us U cc Crack Length, c Figure 7.5 Graphical representation of the Griffith energy balance Ue 7.6

105 Irwin used the Griffith approach as the basis for his subsequent development, first denoting a new engineering term, G, as the strain energy release rate and crack driving force: G = du da du (7.18) tda 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 γ s such that the Griffith relation for fracture stress is written in LEFM as: σ f = GE (7.19) πa 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 = 1 P dc da Fixed displacement G = - 1 P dc da for t = 1 for t = 1 (7.0) where dc da is the compliance change (compliance is C = δ ) with crack extension, da, P is P 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 critical condition when the plate fractures, P=Pc and the dimensionless compliance is EBC c = EB δ c P c geometry, dc c da c (see Fig. 7.6) From the master compliance curve for the particular is determined and the critical G at fracture is: G c = 1 P c dc c (7.1) da c Although Eq. 7.1 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

106 P, δ P P c W B=t δ c δ Test Results a EBC EBC c c EBW dc da c a /W c a/w P, δ Master Compliance Curve Figure 7.6 Illustration of method for determining critical strain energy release rate σ xx = σ a r cos θ 1-sin θ sin 3θ σ yy = σ a r cosθ 1+sin θ sin 3θ σ xy = τ xy = σ a r cos θ sin θ cos 3θ +... σ zz = 0 and τ yz = τ zx = 0 for plane stress or σ zz = ν(σ xx + σ yy ) for plane strain (7.) Note in Eqs. 7. 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 (7.3) The units of K are a bit unusual as MPa m. 7.8

107 σ y σ r yy θ σ xy σ xx x a σ 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 = But, since K = σ GE πa then σ = GE πa and G = σ πa E πa then G = K E (7.4) Now Eq. 7. can be written in terms of K: σ xx = σ K πr cos θ 1-sin θ sin 3θ σ yy = σ K πr cos θ 1+sin θ sin 3θ σ xy = τ xy = σ K πr cosθ sin θ cos 3θ +... σ zz = 0 and τ yz = τ zx = 0 for plane stress or σ zz = ν(σ xx + σ yy ) for plane strain (7.5) Again, note that although Eqs. 7.5 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= K c K (7.6) 7.9

108 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 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 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. 7.3 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. 7.3 can be generalized as K = Yσ πa = ασ πa = Fσ πa (7.7) 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 ~1.1. In other words, the stress intensity factor is about 1% 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 I OPENING MODE MODE II SLIDING MODE MODE III TEARING MODE Figure 7.8 Fracture modes 7.10

109 Figure 7.9 Common and practical stress intensity factors 7.11

110 Plane Stress Fracture Toughness, K c Bs Plane Strain Fracture Toughness, K Ic Material Thickness, B 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 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.5 K Ic S ys (7.8) 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: P max P Q 1.1 (7.9) where P max is the maximum load recorded during the fracture test and P Q 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) 7.1

111 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. 7.1). 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 K Ic = K I K Ic = Yσ πa in which K Ic is a material property Y is a function of the geometry σ is a design stress a is an allowable flaw (crack) (7.30) 0-00 MPa m MPa m 0-10 MPa m 1-5 MPa m Ceramics Glasses Polymers Metals Composites Figure 7.1 Relative comparisons of fracture toughness for various materials 7.13

112 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

113 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 = D O exp Q RT (8.1) where D is the diffusion rate, D o 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 ) ) is know as the homologous temperate. Creep is a relative T mp (abs ) phenomenon which may occur at temperatures not normally considered "high." Several examples illustrate this point. a) Ice melts at 0 C=73 K and is known to creep at -50 C=3 K. The homologous temperature is 3 = 0.8 which is greater than 0.5 so this is consistent with the 73 definition of creep. b) Lead/tin solder melts at ~00 C=473 K and solder joints are known to creep at room temperature of 0 C=93 K. The homologous temperature is = 0.6 which is greater than 0.5 so this is consistent with the definition of creep. 8.1

114 Creep Stress Rupture Strain T/Tmp >0.5 Constant Load Displacement - Low Loads - High Loads - Precision Strain - Gross Strain Measurement ( ε f<0.5%) Measurement ( ε f up to 50%) - Long term (000-10,000 h) - Short term (<1000 h) - Expensive equipment - Less expensive equipment Emphasis on minimum strain rate at stress and temperature Emphasis on time to failure at at stress and temperature 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 = 0.50 which 1773 is equal to 0.5 so this is consistent with the definition of creep. d) Silicon nitride melts/dissociates at ~1850 C=13 K and is known to creep in advanced heat engine applications of 1300 C=1573 K. The homologous temperature is 1573 = 0.74 which is greater than 0.5 so this is consistent with the 13 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.

115 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. shows a typical creep testing setup. The basic results of a creep test are the strain versus time curve shown schematically in Fig 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 L o. The primary region (I) is characterized by transient creep with decreasing creep strain rate ( dε dt = 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. Typical creep test set-up 8.3

116 Constant Load Constant Stress Primary I dε dt Secondary II Time, t Tertiary III t f 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 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 An empirical relation which describes the strain-time relation is: ε = ε i ( 1+ βt 1/3 ) exp(kt) (8.) 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 + ε t ( 1 exp(rt) ) + t ε ss (8.3) where r is a constant, ε t 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: where B, m, D, a and b are empirical constants. ε = ε i +Bσ m t + Dσ α ( 1 exp(βt ) (8.4) 8.4

117 σ > σ >σ 3 1 σ > σ 1 Τ > Τ >Τ 3 1 Τ > Τ 1 σ 1 Τ 1 Time, t Time, t Iso thermal Tests Iso stress Tests Figure 8.4 Effect of stress and temperature on strain time creep curves In this relation, if t >t transient then ε = ε i +Bσ m t + Dσ α (8.5) and the strain rate is the steady-state or minimum strain rate: dε dt = 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 - 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 - 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 = Aσ n 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 = Bσ n (8.8) Equation 8.8 can be linearized by taking logarithms of both sides such that logε ss = logε min = log B + n log σ (8.9) 8.5

118 = + + ε i ε i Time, t Time, t Time, t Time, t 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). in For iso stress tests, the power dependence of stress becomes a constant resulting ε ss = ε min = C exp Q RT Equation 8.10 can be linearized by taking natural logarithms of both sides such that lnε ss = lnε min = ln C Q R 1 T (8.10) (8.11) Log-linear plots of ε min = ε ss versus 1 T (see Fig. 8.7) results in a linear relation in which the slope, Q R, is related to the activation energy, Q, for creep. n>1 (power law creep). n=1 (diffusion creep) log σ Figure 8.6 Log-log plot of minimum creep strain rate versus applied stress showing diffusion creep and power law creep. 8.6

119 . -Q/R 1/T Figure 8.7 Log-linear plot of minimum creep strain rate versus reciprocal of temperature showing 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 ) 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 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 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 t f = C ε f (8.1) 8.7

120 Stress, or log σ σ Τ1 Τ >Τ1 Τ 3 > Τ > Τ 1 Change in failure mechanism Time to failure, t f or log t f Figure 8.8 Stress rupture plots for various temperatures Equation 8.1, known as the Monkman-Grant relation, should give a slope of -1 on a 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: P SD = logθ = logt f - log e R Q T (8.13) where P SD 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 P SD. The allowable stress for an combination of time to failure and temperature (i.e., P SD ) can then be determined from the curve. In the Larson-Miller method, θ, is the temperature compensated time such that: P LM = log e R Q =T ( logt f +(logθ =C) ) (8.14) 8.8

121 log t f σ 1 < σ < σ 3 1/T Experimental Results Q log e R Allowable σ P = log t - log e Q SD f R T "Universal" Sherby Dorn Relation Figure 8.9 Summary of Sherby-Dorn relation where P LM is the Larson-Miller parameter, Q is assumed to a function of stress only, and C is a constant of ~0 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 P LM. The allowable stress for an combination of time to failure and temperature (i.e., P LM ) 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-p SD relation is found to be σ = f ( P SD ) = 11.3P SD 14 (5 σ 85 MPa) (8.15) The design problem is to determine the allowable stress to give 000 h life at 00 C. For this alloy, the activation energy, Q, is kj/mole. Using Q=150,500 J/mole, R=8.314 J/mole K, t f =000 h, and T=473 K, P SD is calculated as Substituting this value of P SD into Eq gives an allowable stress of 5 MPa. log t f σ 1< σ < σ 3 Allowable σ -C Q log e R 1/T Experimental Results P = T (log t - C) LM f Figure 8.10 Summary of Larson-Miller relation "Universal" Larson-Miller Relation 8.9

122 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, N f, in the range of <10 to 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 N f >10 5 (sometimes >10 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 quasistatic 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

123 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.

124 Figure 9. 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.. This machine is called a rotating-beam type, and from Fig. 9. it can be seen that a constant moment of magnitude equal to onehalf 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 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 = σ min σ max, of -1. Other nomenclature and symbols for fatigue are shown in Fig σ σ min σ max σ σa t σ m Figure 9.3 Examples of stress cycles for completely-reversed and tension-tension loading 9.3

125 σ max = Maximum stress σ min = Minimum stress σ m = Mean stress = σ max + σ min σ = Stress range = σ max σ min σ a = Stress amplitude = σ = (σ max σ m ) = (σ m σ min ) Note: tension = +σ and compression = σ. Completely reversed R= 1, σ m = 0. R = Stress ratio = σ min σ max A = Amplitude ratio = σ a σ m = 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 in which the data can sometimes be represented by two straight lines. S Ferrous and Ti-based alloys 6 σ e = fatigue limit or endurance limit ( σ cycles) Non-ferrrous materials (e.g Al or Cu alloys) ( σ 8 10 cycles) 10 6 log N 8 f 10 Figure 9.5 Schematic representation of S-N curves for ferrous and non ferrous materials. 9.4

126 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 10 3 to In the range of 10 5 to 10 6 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 the material and is often designated as Se or σ e. 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 materials under completely reversed bending. 9.5

127 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 S max of the specimen (σ max in a simple tension test).. On a diagram of S vs. log N, plot fatigue strength values of a. 0.9 S max at 10 3 cycles b. 0.5 S max at 10 6 cycles. 3. Join these points together to form a S-N diagram similar to that shown in Figs 9.5 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: k t = σ LOCAL σ REMOTE (9.1) where σ LOCAL is the maximum local stress at the stress raiser and σ REMOTE 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: UN NOTCHED k f = σ e NOTCHED (9.) σ e where σ e UN NOTCHED and σ e NOTCHED are the endurance limits for un-notched and notched fatigue specimens as illustrated in Fig S σe σ NOTCHED e UN-NOTCHED log N f Figure 9.7 Illustration of endurance limits for notched and un-notched fatigue tests. 9.6

128 A notch sensitivity factor can be defined which relates the material behavior, k f, and the component parameter, k t such that q = k f 1 k 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 S UTS increases Generally, k f << k t for ductile materials and sharp notches but k f k t 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 K I 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, S e =σ e, is also lowered by factors such as surface finish (m a ), type of loading (m t ), size of specimen (m d ), miscellaneous effects (m o ) such that: σ e ' = m a m t m d m o σ e (9.4) Note that σ e can be estimated from the ultimate tensile strength of the material such that: σ e m e S UTS where m e = 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 σ m =0, then the allowed amplitude stress, σ a is the endurance limit measured for completely reversed fatigue test loading with R=-1. However, if σ a =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 e σ a Goodman Soderberg σ m S ys S uts Figure 9.8 Illustration of effect of mean stress on allowable amplitude stress. 9.7

129 The mathematical expression for this effect is the equation of the line such that: σ a = σ e 1 σ m (9.5) which is known as the Goodman line. If S UTS is replaced with S ys, then Eq. 9.5 is known as the Soderberg relation. S UTS If a factor of safety, FS, and / or fatigue factors, k f, are used then the following expressions results for either brittle or ductile materials. Brittle σ a = σ e FS k f σ 1 m (S UTS /(k f k t )FS) (9.6a) Ductile σ a = σ e FS k f σ 1 m (S UTS / 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: N 1 N f1 + N N f + N 3 N f3 = N j = 1 (9.7) N fj where N 1 and N f1, etc. are the actual number of cycles and number of S-N fatigue cycles at stress, σ a1 etc. (see Fig. 9.10) respectively. σ a3 σa1 σ a σ t N1 N N 3 Figure 9.9 Variable amplitude loading with a constant mean stress 9.8

130 σa σa3 σ a σa1 N N f N 3 f f1 N f 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, ) 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 9.1. 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 (d g ), 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

131 Slip Band Crack Propagation Slip Band Crack Initiation Tensile Stress Crack Propagation 1- d g 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 9.1 Woods model for fatigue 9.10

132 Saturation Cracking Number of Cycles N Figure 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 rearranged such that: N f N i dn = a f a i a da f C( K) = da (9.9) m C(F σ πa) m a i Assuming that F can be approximated as nearly constant over the range of crack growth and assuming that m and C are constant, then: (1 (m a /)) (1 (m /)) N f = f a i C F( σ) π [ ] m 1 (m/) [ ] (9.10) where a i is the initial crack length which is either assumed (~ d g ) or determined by non destructive evaluation and a f = 1 K Ic. π Fσ max I II III log da/dn m K th log K Figure 9.14 Crack growth rate versus stress intensity factor range. 9.11

133 a da/dn Figure 9.15 Crack length vs. number cycles relations from which crack propagation rate is found. If F is a function of crack length, i.e. F(a,W, etc.), then numerical integration must be used such that:. N f a f a da f dn = C( K) = da (9.11) m C F(a,W,etc) σ πa N i a i N a i [ ] m 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 da/dn For >(+R) crack remains in tension (open) longer and can tolerate less K for the same da/dn +R -R log K Figure 9.16 Effect of R-ratio on crack growth rates. 9.1

134 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, P cr. 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 Two moments are generated about the pinned base, O. The first moment is the restoring moment: FL = K ( x )L = K (Lθ)L = KL θ (10.1) where F is the restoring force, L is the length of the column, K is the spring constant, and θ is the angle. The second moment is the overturning moment: where P is the axial force. P( x) = P(Lθ)L (10.) P P L F=K x θ O Figure 10.1 Model of a slender column under axial load with a restoring force/moment 10.1

135 P Unstable Pcr=KL Stable θ 0 +θ Figure 10. Illustration of stable and unstable conditions along with critical load The following conditions apply as illustrated in Fig. 10.: If PLθ < KθL then P < KL and a stable condition results If PLθ > KθL then P > KL and an unstable condition results (10.3) The critical load separating stability and instability is P cr = KL (10.4) This situation can also be thought of as a ball on a surface as shown in Fig 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''= d v dx, 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.

136 P P P L x v P M Ideal Column Internal Force and Moment Figure 10.4 Ideal column and internal force/moment Substituting this relation for M into Eq gives EIv '' = Pv v '' + P EI v = 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 = C 1 sin P EI x +C cos P EI x (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 C =0. In addition, at x=l, v=0. In this case the trivial solution is that C 1 =0 which means that the column is always straight. The other possibly P is that sin EI L P =0 which is satisfied if L =nπ. This solution can be rearranged such EI that P = n π EI L (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: P cr = π EI L (10.8) 10.3

137 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. P cr = π EI L P cr = π EI 4L P cr = 4π EI L P cr =.046π EI L Pinned Pinned Free Fixed Fixed Fixed Pinned Fixed (10.9) The equations in Eq are a bit cumbersome because four different relations need to be remembered. So instead an effective length is defined such that L e = KL. Now a single equation can be used to replace the four equations of Eq 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 and the single equation is: P cr = π EI L e = π EI (KL) (10.10) For purposes of design, it is more useful to express the critical loading condition in terms of a stress such that: σ cr = P cr A = π EI AL e σ cr = π E (L e / k) (10.11) where k= I / A is the smallest radius of gyration determined from the least moment of inertia, I, for the cross section. The term, L e /k is known as the slenderness ratio and contains information about the length and the cross section. 10.4

138 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, σ cr, and the generalized yield strength, σ o. At this point there is a transition between yield and buckling: σ = σ o = σ cr σ o = π E (L e / k ) (10.1) If Eq is solved for L e /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: L e k min = π E σ o (10.13) This transition can be illustrated by plotting the relation between stress, σ, and slenderness ratio, L e /k, as shown in Fig Note that in reality there is no sharply divided transition between yielding and buckling. Instead the σ vs. L e /k curve can be divided into three regions as shown in Fig Region I is the short-column region in which general yielding occurs when σ =σ o. 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

139 σ o σ σ= π E ( Le/k ) Le/k min Le/k Figure 10.6 Stress vs. slenderness ratio relation Regions I and III are straight forward to determine (either apply σ =σ o. in Region I or σ cr = π E in Region III). Region II is material/ geometry/loading dependent and (L e /k) empirical relations are often used. One approach is to fit a parabola to the σ vs. L e /k curve from σ =σ o. to σ =σ o / such that. σ = σ o 1 and σ = ( L e / k ) ( L e / k min ) for 0 L e k L e k min π E (L e / k) for L e k > L e k min (10.14) σ o Parabolic Approximation σ Region I Region II Region III σ= π E ( Le/k ) Le/k min Le/k Figure 10.7 Stress vs. slenderness ratio relation with parabolic fit in Region II 10.6

140 In summary, note that buckling loads and stresses are sensitive to the following 1) stiffness of material ) 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

141 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 σ 1, σ, σ 3 [ ] S UTS (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

142 σ σ σ σ =-Suts =Suts 1 1 =Suts "safe" σ σ =-Suts Figure 11.1 Failure envelope for maximum normal stress criterion A factor of safety can be defined based on Eq such that: S Fractures if FS 1 where FS= UTS MAX σ 1, σ, σ 3 [ ] (11.) This fracture criterion can be represented in a plane stress state (σ z =0) where σ is the ordinate and σ 1 is the abscissa. As shown in Fig. 11.1, any combination of σ 1 and σ that falls within the square box (i.e., FS=1 for Eq. 11. where ±S UTS =σ 1 or ±S UTS =σ ) 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 τ 1 = σ 1 σ ) (, τ 13 = σ 1 σ 3) (, τ 3 = σ σ 3) τ o = σ o (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 such that: Yields if FS 1 τ where FS= o =σ o / MAX τ ( 1 = σ 1 σ ) (, τ 13 = σ 1 σ 3) (, τ 3 = σ σ 3) (11.4) 11.

143 σ σ = σ 1 o σ σ1 =σ ο σ =σ ο σ =σ ο σ 1 =σ ο "safe" σ 1 σ σ =σ 1 o Figure 11. Failure envelope for maximum stress criterion This yield criterion can be represented in a plane stress state (σ z =0) where σ is the ordinate and σ 1 is the abscissa. As shown in Fig. 11., any combination of σ 1 and σ that plots within the parallelogram (i.e., FS=1 for Eq where ±σ o =(σ 1 -σ ), ±σ o =σ 1 or ±σ o =σ ) is "safe" and the perimeter is yielding. A more complicated yield criterion is that yield failure is expected when the octahedral shear stress,τ h, reaches the octahedral shear stress at yield of the material, τ ho. Thus, the octahedral shear stress yield criterion can be specified as: and Yields if τ h τ ho where τ h = 1 3 ( σ 1 σ ) + ( σ σ 3 ) + ( σ 3 σ 1 ) (11.5) τ ho = 3 σ o (11.6) when the stress state of a uniaxial tensile test at yielding (σ 1 =σ o, σ =σ 3 =0) are substituted into the relation for τ h given in Eq If Eq and 11.6 are set equal to each other the yield criterion can be expressed in terms of normal stresses: Yields if τ h τ ho 1 3 ( σ 1 σ ) + ( σ σ 3 ) + ( σ 3 σ 1 ) 3 σ o (11.7) 11.3

144 σ σo σ ο=σ 1 σ 1σ +σ σ ο "safe" σο σ 1 σ ο Figure 11.3 Failure envelope for maximum stress criterion A factor of safety can be defined based on Eq such that: Yields if FS 1 where FS= 1 σ o ( σ 1 σ ) + ( σ σ 3 ) + ( σ 3 σ 1 ) = σ o σ H (11.8) where σ H 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 is the ordinate and σ 1 is the abscissa. As shown in Fig. 11.3, any combination of σ 1 and σ that plots within the ellipse (i.e., FS=1 for σ o = σ1 σ1 σ + σ is "safe" and the perimeter is yielding. The usefulness of the three failure criteria presented here is shown in Fig 11.4 for the failure envelopes for the plane stress case where σ 1 and σ are normalized to S UTS 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

145 Figure 11.4 Failure criteria and failure points plotted on normalized plane stress Combined Stresses coordinates The previous discussion of failure criteria was based on two premises: 1) the uniaxial tensile behavior of the material was known (i.e., S UTS or σ o.) and ) 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-dimensional stress σ z acting on it On the other hand, in a realistic situation, the engineer is usually faced with a twoor 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 Figure 11.5 Illustration of the uniaxial tensile test 11.5

146 Figure 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 be acted upon by normal stresses σ x, σ y and σ z acting in the x, y and z directions, as shown. Often, the stresses in the z directional are zero, or much smaller than σ x or σ y. 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, the principal stresses σ 1 and σ and the maximum shear stresses τ max. 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/t. 11.6

147 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 Then an application of Mohr's circle of stress will give the principal stresses σ 1 and σ and the maximum shear stress τ max. 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 a two-dimensional object, the unknown strains are the normal or elongation strains ε x and ε y 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 a point to determine the three strains ε x, ε y, and γ xy at a point. This follows directly from the equations that give strain in a direction "a" (ε a ), oriented at an angle θ from the x axis, as shown in Figure 11.8, when strains along the x and y axes, ε x and ε y, and the shear strain γ xy are known. Thus: ε a = 1 ( ε x + ε y ) + 1 ( ε x ε y )cosθ + 1 γ xy sinθ (11.9) 11.7

148 Figure 11.8 Orientation of "a" direction and the x-y axes If we measure three strains, ε a1, ε a, ε a3, at three different angles θ 1, θ, θ 3, we can substitute the values in the above equations and obtain three equations to solve for ε x, ε y, and γ xy. Knowing ε x, ε y, and γ xy, Mohr's circle of strain can then be used to find ε 1, ε, 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 added advantage of directly producing the Mohr's circle of strain, i.e., ε 1, ε 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 for a rectangular rosette. 1. Assume we are using a rosette with strain gages oriented along lines a, b, and c oriented at angles α and β apart, in this case, α = β = 45. It is desired to find the maximum strains ε 1 and ε.. Along an arbitrary horizontal axis x - x, lay off three vertical lines a, b, and c corresponding to the three measured strains ε a. ε b, ε c measured from the y axis. The y - y axis will be the γ shear strain axis. 11.8

149 Figure 11.9 Mohr's strain circle as determined graphically from strain valves ε a. ε b, and ε c. 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 for the three gages. 8. Values of principal strains are determined by the intersection of the circle and the ε axis. The angular orientation of ε 1 from gage A is shown as θ. 11.9

150 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: ε 1 = σ 1 E νσ E ε = σ E νσ 1 E (11.10) (11.11) or more conveniently, the inverse of these ( σ 1 = ε 1 +νε )E 1 ν ( ) ( σ = ε +νε 1)E 1 ν ( ) (11.1) (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 stressstrain 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 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 ) 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

151 Figure 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 Example of a Simple Truss 11.11

152 Y Z X Figure 11.1 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. 11.1) 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 the first step is to calculate the reactions at joints C and D. In this case, F =0 and M =0 such that M C = 0 = PL R D L R D = P (11.15) and F = 0 F x F y = 0 = P + R xc R xc = P = 0 = P P + R yc R yc = 3P (11.16) The resulting free body diagram is shown in Fig

153 P P A L B L L L y C L D x Figure Example of a simple truss Using the method of joints, BD F =0 at joint D such that CD P D F = 0 F x F y = 0 = F CD F CD = 0 = 0 = P +F BD F BD = P (11.17) and since F BD pulls on the joint, then the joint must pull back on the member so member BD is in tension. P P A L B L L L C P 3P L P D Figure Free body diagram for simple truss 11.13

154 Using the method of joints, F =0 at joint C such that AC 1 P 3P C 1 CD F = 0 F x = 0 = P 1 F BC F BC = P F y = 0 = 3P 1 (11.18) F BC F AC F AC = P Since F AC pushes on the joint, then the joint must push back on the member so member AC is in compression. Furthermore, since F CB 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, P F =0 at joint A such that P A AB AC=P F = 0 F x = 0 = P + F AB F AB = P F y = 0 = P + P checks (11.19) Since F AB 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 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 P (compression) BC P (compression) BD P (tension) CD

155 For axially loaded members, the displacement is: N = N U N L dx (11.0) EA where N U is the axial force in the member due to a unit load applied at the point and direction of interest, N L 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 = M U M L dx (11.1) EI where M U is the bending moment in the member due to a unit load applied at the point and direction of interest, M L 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 = T U T L dx (11.) GJ where T U is the torque in the member due to a unit load applied at the point and direction of interest, T L 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 = V U V L dx (11.3) GA where V U is the transverse shear in the member due to a unit load applied at the point and direction of interest, V L 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

156 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 = N U N L dx + M U M L EA dx + V U V L EI dx + T U T L GA dx (11.4) GJ An example of the unit load method applied to the simple truss example is shown in Fig 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

157 Figure Example of application of unit load method to find a deflection in a simple truss 11.17

158 1. 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 ) 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 1.1). 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 1.1, two types of normal stresses are generated: hoop, σ h, and axial, σ a, that both exhibit tension of the material. 1.1

159 Figure 1.1 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 1.. A free body diagram of a half segment along with the pressurized working fluid is shown in Fig. 1.3 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, A p. For equilibrium in the x-direction we sum forces on the incremental segment of width dy to be equal to zero such that: Fx = 0 σha pap 0 σh t dy or solving for σ [ ] = = [ ] h p r dy (1.1) pr σ h = t where dy = incremental length, t = wall thickness, r = inner radius, p = gauge pressure, and σ h is the hoop stress. Figure 1. Cylindrical Thin-Walled Pressure Vessel Showing Coordinate Axes and Cutting Planes (a, b, and c) 1.

160 Figure 1.3 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 1.. A free body diagram of a half segment along with the pressurized working fluid is shown in Fig. 1.4 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: Fy = 0 σaa pae = 0 = σaπ( ro r ) p πr or solving for σ a p πr σ a = π( ro r ) substituting r o = r + t gives (1.) p πr r r a = ([ ] r ) = p π σ ( r + rt+ t r ) = p π r+t π ( rt + t ) since this is a thin wall with a small t,t is smaller and can be neglected such that after simplification p r σ a = t where r o = inner radius andσ a is the axial stress. 1.3

161 Figure 1.4 Free-Body Diagram of End Section of Cylindrical Thin-Walled Pressure Vessel Showing Pressure and Internal Axial Stresses Note that in Equations 1.1 and 1., 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 1-5, the axial stress results from the action of the pressure acting on the projected area of the sphere such that Fy = 0 σaa pae = 0 = σaπ( ro r ) p πr or solving for σ a p πr σ a = π( ro r ) substituting r o = r + t gives (1.3) p πr r r a = ([ ] r ) = p π σ ( r + rt+ t r ) = p π r+t π ( rt + t ) since this is a thin wall with a small t,t is smaller and can be neglected such that after simplification p r σa = = σh t 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. 1.4

162 Figure 1.5 Free-Body Diagram of End Section of Spherical Thin-Walled Pressure Vessel Showing Pressure and Internal Hoop and Axial Stresses The analyses of Equations 1.1 to 1.3 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 r which 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 the limiting 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 1.1 to 1.3 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: 1.5

163 σ σ σ h a r = = = r p i ( ro ri ) ri p ( ro ri ) r p i ( ro ri ) r o 1 + r r o 1 r (1.4) where r o =outer radius, r i =inner radius, and r is the radial variable. Equations 1.4 apply for any wall thickness and are not restricted to a particular r/t ratio as are the Equations 1.1 and 1.. 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 1.6 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 1.4 can be generalized for the case of internal and external pressures such that ri p- i ropo ri ro ( po p i) / r σ h = r r σ σ h h = = ( o i ) ri p- i ropo ( ro ri ) ri p- i ropo + ri ro ( po p i) / r ( ro ri ) where p o =is the outer gauge pressure and, p i =inner gage pressure. (1.5) a) hoop stress b) radial stress Figure 1.6 Stress distributions of hoop and radial stresses 1.6