Natural Sciences Tripos Part IB
MATERIALS SCIENCE Course D –Mechanics of Materials and Structures
Name .................................... College ......................... Dr. P.D. Bristowe Lent Term 2013-14
IB
Thomas Young 1773-1829
Leonhard Euler 1707-1783 Galileo Galilei 1564-1642
Woldemar Voigt 1850-1919
Otto Mohr 1835-1918 Simeon Poisson 1781-1840 Waloddi Weibull 1887-1979
Armand Considere 1841-1914 Charles Inglis 1875-1952
Alan Griffith 1893-1963
IB Course D
Mechanics of Materials and Structures
Lent 2014
DHi
Contents Introduction ..................................................................................................................................... 1 1 – Tensors...................................................................................................................................... 3 1.1 Introduction to Tensors ....................................................................................................... 4 1.2 Stress Components .............................................................................................................. 5 1.3 Transformation of Axes....................................................................................................... 7 1.4 Transformation (Rotation) of the Reference Axes of a Vector ........................................... 8 1.5 Einstein Summation Convention......................................................................................... 9 2 – Transformation of Stresses and Mohr’s Circle .................................................................. 10 2.1 Transformation of Second Rank Tensors .......................................................................... 10 2.2 Principal Stresses and the Secular Equation...................................................................... 11 2.3 Introduction to Mohr’s Circle for Plane Stress ................................................................. 13 3 – Mohr’s Circle in Practice...................................................................................................... 16 3.1 Construction of Mohr’s Circle .......................................................................................... 16 3.2 Simple Examples ............................................................................................................... 17 3.3 Example of a Thin-Walled Tube under Applied Torque and Axial Load......................... 18 3.4 Verification using the Secular Equation............................................................................ 21 4 – Representation of Strain ....................................................................................................... 23 4.1 Introduction ....................................................................................................................... 23 4.2 The Relative Displacement Tensor ................................................................................... 23 4.3 The Strain Tensor .............................................................................................................. 24 4.4 Some Simple Example Cases ............................................................................................ 26 4.5 Hydrostatic and Deviatoric Components of the Strain Tensor.......................................... 27 4.6 Field and Matter Tensors................................................................................................... 27 4.7 Stiffness and Compliance Tensors .................................................................................... 28 4.8 Relationship to Elastic Constants ...................................................................................... 28 4.9 Poisson Ratio ..................................................................................................................... 29 5 – Engineering Stresses and Strains and Plasticity ................................................................. 31 5.1 Engineering Shear Strains and the Shear Modulus ........................................................... 31 5.2 Bulk Modulus .................................................................................................................... 32 5.3 Relationships between Elastic Constants .......................................................................... 32 5.4 Large Uniaxial Tensile Strains and Plastic Deformation .................................................. 33 5.5 Plastic Instability and Considère’s Construction............................................................... 34 6 - Beam Bending ......................................................................................................................... 38 6.1 Introduction to Beam Bending .......................................................................................... 38 6.2 Beam Curvatures ............................................................................................................... 38 6.3 Bending Moments and Second Moments of Area............................................................. 39 6.4 Beam Stiffness................................................................................................................... 41 6.5 Maximising the Beam Stiffness ........................................................................................ 42
IB Course D
Mechanics of Materials and Structures
Lent 2014
DHii
7 – Examples of Beam Bending and Torsion .............................................................................. 44 7.1 Beam Deflections .............................................................................................................. 44 7.2 The Cantilever Beam......................................................................................................... 44 7.3 Symmetrical 3-Point Bending ........................................................................................... 45 7.4 Introduction to Torsion...................................................................................................... 46 7.5 Torsion of Thin-Walled Circular Tubes ............................................................................ 47 7.6 Torsion of Solid Cylindrical Bars ..................................................................................... 48 7.7 Torsion of Thick-Walled Circular Tubes .......................................................................... 50 8 – Failure in Compression: Elastic Buckling of Columns ........................................................ 52 8.1 Crushing ............................................................................................................................ 52 8.2 Elastic (Euler) Buckling .................................................................................................... 54 8.2 Freely-hinged columns ...................................................................................................... 54 9 – Failure in Compression: the Effect of Constraints and Column Shape ............................. 60 9.1 The effect of End-Constraint ............................................................................................. 60 9.2 The effect of Aspect Ratio................................................................................................. 63 9.3 The effect of Cross-Sectional Shape ................................................................................. 64 9.4 Local Buckling .................................................................................................................. 66 9.5 The effect of Internal Pressure .......................................................................................... 67 10 – Brittle Fracture in Tension ................................................................................................... 68 10.1 The Ideal Strength ............................................................................................................. 68 10.2 The Effect of Cracks – Stress Considerations ................................................................... 69 10.3 Energy Considerations ...................................................................................................... 71 11 – Stable Crack Growth............................................................................................................. 76 11.1 Fracture Toughness ........................................................................................................... 76 11.2 The Wedging Geometry .................................................................................................... 77 11.3 Energy Considerations ...................................................................................................... 77 11.4 Crack Healing.................................................................................................................... 80 11.5 The Nature of Cracking ..................................................................................................... 81 12 – Coping with a Scatter in Stength.......................................................................................... 83 12.1 Introduction ....................................................................................................................... 83 12.2 Weibull’s Approach .......................................................................................................... 84 12.3 Using Weibull’s Method ................................................................................................... 85 12.4 Practical Solutions ............................................................................................................. 87 12.5 Toughening........................................................................................................................ 88 Appendix – Summary of Materials Response in Compression and Tension ........................... 91 Nomenclature ................................................................................................................................. 94 Glossary .......................................................................................................................................... 95
IB Course D
Mechanics of Materials and Structures
Lent 2014
DH1
Course D: Materials and Structures Synopsis Any structure, man-made or biological, and whatever its primary function, must be able to withstand imposed loads. Building on ideas from the Part IA course which covered the mechanical behaviour of materials, this course aims to give a grounding in the interplay between the loads acting on a structure, its shape (“design”), relevant material properties and the resultant distributions of stress and strain, and associated deflections. This determines whether the structure is likely to fulfil its function, which normally requires avoiding excessive elastic deflections or plastic deformation, and also ensuring that fracture does not occur. In order to do this, it’s essential for the manipulation of stresses and strains to be well understood, so the course starts with an introduction to their representation as second rank tensors. Later parts of the course cover the behaviour of structures, particularly beams under bending, torsion and axial compression, and the ways in which they can fail due to buckling, crushing, fast fracture or fatigue.
Lecture Details (12 lectures) Tensors (Lecture 1) Introduction to Tensors. The Rank of a Tensor. Stresses as Second Rank Tensors. Transformation of Axes. Transformation (Rotation) of the Reference Axes of a Vector. The Einstein Summation Convention. Transformation of Stresses and Mohr’s Circle (Lecture 2) Transformation of Second Rank Tensors. Principal Stresses and the Secular Equation. Introduction to Mohr’s Circle for Plane Stress. Mohr’s Circle in Practice (Lecture 3) Construction of Mohr’s Circle. Examples of the use of Mohr’s Circle. Determination of the Principal Stresses and their Orientation for a Thin-Walled Tube under Torsion and Axial Tension. Representation of Strain (Lecture 4) Deformation of a Body. The Relative Displacement Tensor. Removal of the Rigid Body Rotation Component and Identification of the Strain Tensor. Hydrostatic and Deviatoric Components of the Strain Tensor. Stress-Strain Relationships and the Engineering Constants. The Stiffness and Compliance Tensors and their Relationship to the Elastic Constants. The Poisson Ratio. Engineering Stresses and Strains and Plasticity (Lecture 5) Engineering Shear Strains and the Shear Modulus. The Bulk Modulus and Inter-relationships between the Elastic Constants. Large Uniaxial Strains and Plastic Deformation. True Stress and Strain. Plastic Instability and Considère’s Construction. Beam Bending and Torsion (Lecture 6) Introduction. Beam Curvatures. Bending Moments and Second Moments of Area. Stress Distributions in Beams. Beam Stiffness. Maximising the Beam Stiffness. Examples of Beam Bending and Torsion (Lecture 7) Beam Deflections. The Cantilever Beam. Symmetrical 3-Point Bending. Torques and Resultant Twist Angles. Torsion of Thin-Walled Tubes, Thick-Walled Tubes and Solid Cylindrical Bars.
IB Course D
Mechanics of Materials and Structures
Lent 2014
DH2
Failure in Compression (Lecture 8) Crushing of Dense and Porous Bodies. Elastic (Euler) Buckling. Freely-Hinged Columns. Energy Balance Considerations. Failure in Compression: The Effect of Column Shape (Lecture 9) The Effect of End-Constraint, Column Slenderness and Cross-Sectional Shape. Local Buckling. The Effect of Internal Pressure on Buckling of a Hollow Cylinder. Brittle Fracture in Tension (Lecture 10) The Ideal Strength. The Effect of Cracks – Stress Considerations. Stress Concentrations at Atomic and Macroscopic Scales. Energy Considerations. Contributions to the Total Energy of the System. Changes in Mechanical and Surface Energies. Stable Crack Growth (Lecture 11) Fracture Toughness. The Wedging Geometry. Energy Considerations. Crack Healing. The Nature of Cracking. The Importance of Equilibrium. Catastrophic Failure. Coping with a Scatter in Strength (Lecture 12) Weibull’s Approach. Practical Solutions. Toughening. Crack Bridging. R-curves. Fatigue. S-N Curves.
Booklist Several textbooks are relevant to the course, although unfortunately there are sometimes differences in the conventions and symbolism which they use. The standard book on tensors is Physical Properties of Crystals – Their Representation by Tensors and Matrices, by JF Nye, although its coverage goes considerably beyond what is required here. The more structural aspects of the course, such as beam bending, torsion, buckling etc are well described in Mechanics of Materials by JM Gere and BJ Goodno. Mechanical Metallurgy, by GE Dieter is sound on basic mechanics and also deals with fracture and fatigue. Perhaps the most relevant book for background interest reading is Structures by JE Gordon. This omits mathematical treatments, but it does cover many of the ideas involved and it gives many entertaining examples of their application. It, and the earlier book of Gordon’s called The New Science of Strong Materials, should be read by all Materials Scientists.
Web-based Resources Most of the material associated with the course (handouts, question sheets, practical scripts etc) can be viewed on the web (www.msm.cam.ac.uk/teaching) and also downloaded. This includes model answers to question sheets, which are released after the work concerned should have been completed. In addition to this text-based material, DoITPoMS resources are also available (www.doitpoms.ac.uk). The following TLPs are directly relevant to this course: • Stress Analysis and Mohr’s Circle • Brittle Fracture
• Bending and Torsion of Beams • Tensors
Finally, there is a CamTools site for the course (https://camtools.cam.ac.uk/ - named MatSci Crs-D : NSTIB 13_14). Lecture summaries, and pointers to additional material, will be provided there.
IB Course D
Mechanics of Materials and Structures
Lent 2014
DH3
1 – Tensors Revision from Pt 1A Course D (Mechanical Behavior of Materials) Normal (σ ) and shear (τ ) stresses are defined as stresses induced by forces acting either normal or parallel to the sectional area of a body to which they are applied:
The response of the body to the applied stress is characterised by the strain. The normal strain (ε) is the relative change in length in the direction parallel to the original length. The shear strain (γ) is the angular deformation arising from a shear displacement in a particular plane:
IB Course D
Mechanics of Materials and Structures
Lent 2014
DH4
For elastic deformation, the normal stress and strain and the shear stress and strain are linearly proportional:
σ = Eε and τ = Gγ where E and G are the Young’s and shear moduli respectively.
There are two important points:
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1) In general, an applied set of forces generates both normal and shear stresses (and strains). These will, however, differ when different planes within the body are considered. It is therefore important to clearly specify the reference frame (set of axes) when stating stresses and strains. 2) Both stresses and strains are characterised by a magnitude and two directions. For stress, these are the direction of the force and the direction of the normal to the face on which it acts. For strain, these are the direction of the change in length and the direction of the reference length. Therefore they are NOT vectors and cannot be resolved into components. Characterisation using tensors is required.
1.1
Introduction to Tensors
The word “tensor” comes from the latin for “extension”†, and tensors are strongly associated with mechanics, although they are also widely used in other branches of science and engineering. The utility of tensors is mainly concerned with treating differences in the response or characteristics of a material in different directions within it. The usage is particularly valuable when treating anisotropic materials – and of course most (crystalline) materials are in practice anisotropic to at least some degree. However, even for isotropic materials, tensor analysis is necessary, or at least very helpful, when treating most types of mechanical loading. In the present course, only isotropic materials will be considered.
Woldemar Voigt (1850-1919) was the first to use the term tensor in its modern context to describe mechanical stress.
†
The term is also used in Robert Hooke’s famous expression “Ut tensio, sic vis – As the extension, so the force”. This is the origin of the simple form of Hooke’s law, relating the normal stress on a material to the resultant normal strain. Hooke’s experiments actually involved springs, which deform in a relatively complex manner, with the material being subjected to pure torsion – see lecture 7. Fortunately, there is still a linear relationship between load and extension.
IB Course D
Mechanics of Materials and Structures
Lent 2014
DH5
A tensor is an n-dimensional array of values, where n is the “rank” of the tensor. The simplest type is thus a tensor of zeroth rank, which is a scalar - ie a single numerical value. Properties like temperature and density are scalars. They are not associated with any particular direction in the material concerned, and the variable does not require any associated index (suffix§).
A first rank tensor is a vector. This is a 1-D array of values. There are normally 3 values in the array, each corresponding to one of 3 (orthogonal) directions. Each value has a single suffix, specifying the direction concerned. These suffices are commonly numerical {1, 2 & 3}, although sometimes other nomenclature, such as {x, y & z} or {r, θ & z}, may be used. Force and velocity are examples of vectors. The components of a vector can thus be written down in a form such as (1.1) in which each of the suffices {1, 2 & 3} refers to a specific direction, such as {x, y & z}.
1.2
Stress Components
There are other variables, such as stress, for which each component requires the specification of two directions, rather than one, so that two suffices are needed. These are called second rank tensors.
§
The term “suffix” is in common use for these indices, although they are employed as subscripts.
IB Course D
Mechanics of Materials and Structures
Lent 2014
DH6
In the case of stress, these two suffices specify, firstly, the direction in which a force is being applied and, secondly, the normal of the plane on which the force is acting†. Stress is thus an example of a second rank tensor and the components can be written down in the form of a 2-D array of values σ 11 σ 12 σ ij = σ 21 σ 22 σ 31 σ 32
σ 13 σ 23 σ 33
(1.2)
When the suffices i and j are the same, the force acts parallel to the plane normal, ie the component concerned is a normal stress. When they are different, it is a shear stress (and sometimes the symbol τ is used instead of σ for such components).
Some of the stresses that could act on a body are depicted in Fig.1.1. Provided that the body is in static equilibrium, which is commonly assumed, then the normal forces acting on opposite faces so as to generate a normal stress (eg σ33 in Fig.1.1) must be equal in magnitude and anti-parallel in direction. (If this were not the case, then the body would translate.) For shear stresses, a further condition applies. Not only must the two forces generating the σ23 stress (see Fig.1.1) be equal and opposite, but the magnitude of the σ32 stress must be equal to that of the σ23 stress. (If this were not the case, then the body would rotate.) Shear stresses thus act in pairs. This applies to all shear stresses, so that σij = σji, and the tensor represented in Eqn.(1.2) must be symmetrical.
Fig.1.1
†
Illustration of the nomenclature of stresses acting on a body
This is the convention used by Nye. It’s also adopted in many other textbooks, and in the Part II courses on Tensors and Composite Materials. However, the reverse convention – ie that the first suffix gives the normal of the plane on which the force is acting and the second gives the direction of the force – can be employed and indeed is used by Dieter and by Gere and Goodno. Provided a convention is used consistently, problems are unlikely to arise.
IB Course D
1.3
Mechanics of Materials and Structures
Lent 2014
DH7
Transformation of Axes
It follows from this that there are just 6 independent components in a general stress state – 3 normal stresses and 3 shear stresses. The magnitude of these will, of course, depend on the directions of the axes that have been chosen to provide the frame of reference. However, it’s clear that the state of stress itself will be unaffected if we chose an alternative frame of reference. Any tensor can be transformed so as to refer to a new set of axes, provided the orientation of these with respect to the original set is specified. Axes transformations are important in the mechanics of materials for various reasons. For example we may know the stresses acting on 1-2 planes but really need to know the stresses acting on planes oriented at some angle φ to the 1-axis because these are slip planes in the material. Or perhaps we need to know the planes on which the normal stresses take their maximal value since these planes may fail by tensile cracking if the stresses exceed the tensile stress of the material. It turns out that any stress state can be expressed in terms of normal stresses only (ie all shear stresses are zero), provided that the appropriate set of axes is chosen. Since it’s often very helpful to be able to express a stress state in terms of this unique set of normal stresses, it’s important to be clear about the procedure for transforming tensors in this way. We will do this first for a vector (force) and then move on to treat stress in the second lecture.
IB Course D
1.4
Mechanics of Materials and Structures
Lent 2014
DH8
Transformation (Rotation) of the Reference Axes of a Vector
Consider a vector, ie a force, F (= [0, F2, F3]), with components which are referred to the axis set {1, 2, 3}. A specific reorientation of this set of axes is now introduced, namely a clockwise rotation by an angle φ about the 1-axis, to create a new axis set {1’, 2’, 3’} – see Fig.1.2.
Fig.1.2
Rotation, in the 2-3 plane, of the axes forming the reference frame for a vector F
In this case, the new 1’-axis coincides with the old 1-axis, but the 2’- and 3’-axes have been rotated with respect to the 2- and 3-axes. The values of F2’ and F3’ are found by resolving the components F2 and F3 onto the 2’ and 3’ axes and adding these resolved components together: F2 ' = F2 cos ( 2 '∧ 2 ) + F3 cos ( 2 '∧ 3) F3' = F2 cos ( 3'∧ 2 ) + F3 cos ( 3'∧ 3)
(1.3)
where the symbolism (x ^ y) represents the angle between the x and y axes. In terms of the angle φ, these two equations can be written as
F2 ' = F2 cos ( −φ ) + F3 cos ( −90 − φ ) = cF2 − sF3 F3' = F2 cos ( 90 − φ ) + F3 cos ( −φ ) = sF2 + cF3
(1.4)
in which c and s represent cos φ and sin φ respectively. Clearly these cosines of the angles between new and old axes are central to such transformations. They are commonly termed direction cosines and are represented by aij, which is conventionally the cosine of the angle between the new i direction (= i’) and the old j direction.
Of course, the rationale can be extended to cases in which all three axes have been reoriented, leading to the following set of equations F1' = a11F1 + a12 F2 + a13 F3 F2 ' = a21F1 + a22 F2 + a23 F3 F3' = a31F1 + a32 F2 + a33 F3
(1.5)
IB Course D
Mechanics of Materials and Structures
Lent 2014
DH9
It can be seen that the direction cosines form a matrix and this set of equations can be written more compactly in matrix form: F1' F1 F2' = [T ]F2 F3' F3
(1.6)
in which the transform matrix is given by a11 a12 a13 [T ] = a21 a22 a23 a31 a32 a33
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(1.7)
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1.5
Einstein Summation Convention
Sets of equations such as Eq.(1.6) can be written even more concisely by using the Einstein summation convention. This states that, when a suffix occurs twice in the same term, then this indicates that summation should be carried out with respect to that term. For example, in the equation
Fi ' = aij Fj
(1.8)
j is a dummy suffix, which is to be summed (from 1 to 3). The i suffix, on the other hand, is a free suffix, which can be given any chosen value. For example, Eqn.(1.8) could be used to create the equation
F1' = a11F1 + a12 F2 + a13 F3 and also the two other equations, corresponding to i being equal to 2 or 3.
(1.9)
IB Course D
Mechanics of Materials and Structures
Lent 2014
DH10
2 – Transformation of Stresses and Mohr’s Circle 2.1
Transformation of Second Rank Tensors
Extension of the treatment presented in the previous section to the case of second rank tensors, such as stress, follows quite logically. However, for a stress, we are concerned, not just with resolving three single components into new directions (eg F1, F2 and F3 in the case of a force), but rather we need to take account of the fact that both the force and the area on which it is acting will change when we refer them to different axes. In other words, the operation we carried out with respect to a single suffix in the case of force, Fi, has to be implemented with respect to two suffices for a stress, σij. Each stress component thus needs to be multiplied by two direction cosines, rather than one. Example of a single crystal under tensile load (see Pt IA course):
IB Course D
Mechanics of Materials and Structures
Lent 2014
DH11
For the case of a stress, the set of equations corresponding to Eqn.(1.8) can be written
σ 'ij = aik a jlσ kl
(2.1)
in which the prime is now applied to the symbol, rather than the individual suffices, to denote the transformed version. Eqn.(2.1) can be expanded into 9 equations, each giving one of the 9 components of the stress tensor when referred to the new set of axes (although only 6 are required, since the stress tensor is always symmetrical). Both k and l are dummy suffices in this equation (since they are repeated), while i and j are free suffices. The first equation of the set, corresponding to i=1 and j=1, can thus be expanded to
σ '11 = a11a11σ 11 + a11a12σ 12 + a11a13σ 13 +a12 a11σ 21 + a12 a12σ 22 + a12 a13σ 23 +a13a11σ 31 + a13a12σ 32 + a13a13σ 33
(2.2)
while that corresponding to i=3 and j=2 is given by
σ '32 = a31a21σ 11 + a31a22σ 12 + a31a23σ 13 +a32 a21σ 21 + a32 a22σ 22 + a32 a23σ 23 +a33a21σ 31 + a33a22σ 32 + a33a23σ 33
2.2
(2.3)
Principal Stresses and the Secular Equation
One of the main motivations for treating stresses within this rather cumbersome mathematical framework is that it facilitates identification of the Principal Stresses. These are the normal stresses acting on the Principal Planes, which are the planes on which there are no shear stresses. (For the mathematically-inclined, these principal stresses are the eigenvalues of the stress tensor.) Any general state of stress can thus be transformed such that it can be expressed in the form σ 1 0 σ ij = 0 σ 2 0 0
0 0 σ 3
with the single suffix being commonly used to denote a principal stress.
(2.4) (It’s important to
recognise, however, that these are still second rank tensors, and should thus, strictly speaking, always have two suffices.) Obtaining these principal stresses, for a general 3-D stress state, requires diagonalising the stress tensor. First a reminder about determinants:
IB Course D
Mechanics of Materials and Structures
Lent 2014
DH12
Second rank stress tensors are diagonalised by solving the determinant equation |σij - λI| = 0 where λ is a parameter and I is a unit matrix†. The solutions for the principal stresses are thus found from
σ 11 − λ σ 12 σ 13 σ 21 σ 22 − λ σ 23 = 0 σ 31 σ 32 σ 33 − λ
(2.5)
This is a cubic equation in λ, the roots of which (λ1, λ2, and λ3) are the principal stresses. It is often termed the secular equation.
Expanding the terms in Eqn.(2.5) leads to
λ 3 − I 1λ 2 + I 2 λ − I 3 = 0
(2.6)
in which the coefficients (termed the invariants, since they do not vary as the axes are changed) are given by I1 = σ 11 + σ 22 + σ 33 2 2 I 2 = σ 11σ 22 + σ 22σ 33 + σ 33σ 11 − σ 122 − σ 23 − σ 31
(2.7)
2 I 3 = σ 11σ 22σ 33 + 2σ 12σ 23σ 31 − σ 11σ 23 − σ 22σ 132 − σ 33σ 122
Solving (2.6) is straightforward if it factorises or if relations between the roots are known, otherwise numerical methods are needed. However the problem is simplified if one of the principal planes is known since the analysis becomes two dimensional as shown in the next section. A useful TLP on tensors elaborates on this further: http://www.doitpoms.ac.uk/tlplib/tensors/index.php
†
Proof of this is beyond the scope of the course, but in fact this is a fairly routine procedure in matrix algebra.
IB Course D
2.3
Mechanics of Materials and Structures
Lent 2014
DH13
Mohr’s Circle for plane stress
It’s fairly straightforward, if a little cumbersome, to find the principal stresses, and the orientation of the (normals of the) principal planes, for a 3-D stress state specified with respect to an arbitrary set of axes. It just requires a cubic equation to be solved. However, in practice it is common to know one principal plane (direction), but to be interested in finding the principal directions within that plane, or in establishing how the normal and shear stresses vary with direction in that plane. Provided the plane concerned is a principal one, this problem reduces to solving a quadratic equation, rather than a cubic. It can therefore be tackled via a geometrical construction, equivalent to using some simple trigonometry. This is the basis of Mohr’s circle, which was proposed in 1892 by Christian Otto Mohr, a German civil engineer who became a professor of mechanics in Dresden.
Christian Otto Mohr (1835-1918) developed the graphical method for analysing plane stress known as Mohr’s Circle.
Provided the 1-2 plane is a principal plane, the appropriate version of Eqn.(2.5) is
σ 11 − λ σ 12 0 σ 21 σ 22 − λ 0 =0 0 0 σ 33 − λ
(2.8)
and one principal stress is clearly σ33 (= σ3). Since σ12 = σ21, the secular equation reduces to
(σ 33 − λ)[(σ11 − λ)(σ 22 − λ) − σ122 ] = 0 ∴ λ2 − (σ11 + σ 22 ) λ + (σ11σ 22 − σ122 ) = 0
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which has the solution
(2.9)
IB Course D
Mechanics of Materials and Structures
Lent 2014
DH14
2
σ + σ 22 σ − σ 22 2 σ 1 , σ 2 = 11 ± 11 + σ 12 2 2
(2.10)
The form of the solution suggests a possible graphical interpretation involving a circle in which the first term is the circle’s centre and the second term is its radius. The circle is plotted on a graph with the normal stress along the abscissa and the shear stress along the ordinate.
To determine how the normal and shear stresses vary with direction in the 1-2 plane we start with the tensor in diagonalised form and then transform it to give σ'11, σ'22 and σ'12, for a specified orientation
σ '11 σ '12 0 σ1 0 0 σ '12 σ '22 0 = [T ]0 σ 2 0 0 0 0 σ 3 0 σ 3
(2.11)
Since we are rotating about a principal axis, the form of Eqn.(2.1) applicable in this case reduces to the following
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σ '11 = a11a11σ 1 + a12 a12σ 2 + a13a13σ 3 σ '22 = a21a21σ 1 + a22 a22σ 2 + a23a23σ 3 σ '12 = a11a21σ 1 + a12 a22σ 2 + a13a23σ 3
(2.12)
The set of direction cosines applicable here (assuming an anti-clockwise rotation) can be written as a11 a12 a13 cos φ [T ] = a21 a22 a23 = −sin φ a31 a32 a33 0
sinφ 0 cosφ 0 0 1
(2.13)
so it follows that these stresses are given by
σ11′ = cos2 φ σ1 + sin 2 φ σ 2
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σ ′22 = sin 2 φ σ1 + cos 2 φ σ 2 σ12′ = −sin φ cosφ σ1 + sin φ cosφ σ 2
€
(2.14)
IB Course D
Mechanics of Materials and Structures
Lent 2014
DH15
These equations can also be written in a form involving 2φ, rather than φ σ + σ 2 σ1 − σ 2 σ11′ = 1 + cos2φ 2 2 σ + σ 2 σ1 − σ 2 ′ = 1 σ 22 − cos2φ 2 2 σ − σ 2 σ12′ = − 1 sin2φ 2
(2.15)
These equations can be solved using the circle construction described above. The normal and shear stresses acting on a plane rotated by an angle φ from that on which σ1 acts are given by the € coordinates of a point rotated by 2φ around the circumference of a circle centred at the mean of the two principal stresses, and with a radius equal to half their difference. This is illustrated in Fig.2.1(b). This construction provides a convenient method of calculating the stresses acting on particular planes, establishing principal stresses and their orientations, finding the planes on which peak shear stresses† operate etc.
Fig.2.1
†
Mohr’s circle construction for a principal plane (1-2 plane), showing (a) principal stresses in the plane (σ1 & σ2, for a case in which σ1 is tensile and σ2 compressive) and (b) corresponding Mohr’s circle, giving, in addition to the principal stresses, the normal and shear stresses acting on a plane rotated about the 3-axis by an angle φ from that on which σ1 acts.
Care is needed with the signs of shear stresses, since there is more than one possible convention in use. However, this is not actually a point of any real concern, since the sign of a shear stress is solely an issue of convention, and has no physical significance (unlike the sign of a normal stress).
IB Course D
Mechanics of Materials and Structures
Lent 2014
DH16
3 – Mohr’s Circle in Practice 3.1
Construction of Mohr’s Circle
Mohr’s Circle is a useful graphical representation of the plane stress transformations given in the previous section (eqn. 2.15). It allows visualisation and determination of principal stresses, maximum shear stresses and the variation of normal and shear stresses on arbitrary planes. Given values of σ11, σ22 and σ12, for example, the principal stresses σ1 and σ2 in the 1-2 principal plane may be found using the following procedure: (a) Construct a graph with normal stress along the abscissa and shear stress along the ordinate, using the same scale on both axes. Use the convention that positive shear stress points downwards. Along the abscissa normal tensile stresses are positive and point to the right. Normal compressive stresses are negative and point to the left.
(b) Plot the points (σ11, -σ12) and (σ22, σ12) and connect them with a straight line. Where the straight line crosses the abscissa locates the centre of the circle. Label the angle 2φ between the straight line (the circle diameter) and the abscissa. (c) Draw the circle centred at ((σ11+ σ22)/2, 0). (d) The principal stresses are located at the points where the circle intercepts the abscissa. The maximum shear stresses are located at the top and bottom of the circle. (e) The angle of rotation of the diameter (2φ) is twice the angle of rotation of the axes in real space. (f) The procedure works for any symmetrical second rank tensor and therefore also for strain. (g) The procedure can be used to represent a three-dimensional stress state transformed into its principal stresses. The Mohr’s circles for the three principal planes are superimposed on the same graph.
IB Course D
3.2
Mechanics of Materials and Structures
Simple Examples
(a) Uniaxial tension σ11 = σ1; σ22 = σ12 = 0
σ1 0 0 σ ij = 0 0 0 0 0 0
€
(b) Biaxial stress σ11 = σ1; σ22 = -σ1; σ12 = 0
σ1 0 0 σ ij = 0 - σ1 0 0 0 0
€
Representation in the 3 principal planes
Lent 2014
DH17
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Mechanics of Materials and Structures
Lent 2014
DH18
(c) Triaxial tension σ11 = σ1; σ22 = σ2; σ33 = σ3 (σ1 > σ2 > σ3 > 0)
σ1 0 0 σ ij = 0 σ 2 0 0 0 σ 3
€
3.3
Example of a Thin-Walled Tube under Applied Torque and Axial Load
Consider the following problem. “A torque of 100 N m is applied to a thin-walled metal tube (t = 2 mm), of diameter 30 mm. An axial tensile force of 6 kN is then also applied. Find the principal stresses, and their orientation, before and after the axial force has been applied.”
Firstly, it is convenient to use cylindrical polar coordinates. The appropriate stress tensor is given by σ zz σ zθ σ zr σ ij = σθz σθθ σθr σ rz σ rθ σ rr
where σzz, σθθ and σrr are the axial, hoop and radial stresses respectively. €
The stress tensor can be simplified considerably. Since the wall is thin, and the surfaces are free, there are no through-thickness stresses – ie σrr = σrz = σrθ = 0. The plane of the surface of the tube is thus a principal plane and the Mohr’s circle construction can be applied to it. The physical situation and Mohr’s circle for the case where no axial force is applied are shown in Fig.3.1(a).
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Mechanics of Materials and Structures
Lent 2014
DH19
Fig.3.1(a) Physical situation and corresponding Mohr’s circle when only torque is applied to a thin-walled tube. There are no normal stresses acting in either axial (z) or hoop (θ) directions. The Mohr’s circle is centred on the origin - this is sometimes termed a state of pure shear.
There are, however, normal stresses acting on all planes other than those with their normals in axial or hoop directions. Those with the largest and the smallest (most tensile and most compressive) normal stresses (ie the principal stresses) are oriented at ±45˚ to these (since they are 90˚ from the axial and hoop directions in Mohr space) and have values of ±σθz.
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Mechanics of Materials and Structures
Lent 2014
DH20
The value of σθz is found by recognising that a torque results from a tangential force given by the product of σθz and the sectional area of the tube: T = Ftan r ∴ σθz =
€
€
Ftan = σθz (2πrt )
T = 35.4 MPa 2πr 2 t
When the σzz stress (= +31.8 MPa) is added, the Mohr’s circle becomes larger and its centre moves horizontally in the tensile direction - see Fig.3.1(b). The principal stresses also change in both magnitude and direction. The shear stresses on the θ and z planes, however, remain unaffected (σθz = σzθ = 35.4 MPa) and there is no normal stress on the θ plane.
Fig.3.1(b) Physical situation and corresponding Mohr’s circle when both a torque and an axial force are applied to a thin-walled tube.
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Mechanics of Materials and Structures
Lent 2014
DH21
The angle φ between σ1 stress and z-axis is given by tan 2φ =
€
σ zθ 35.4 = σ zz /2 15.9
∴ φ = 32.9°
(3.1)
The principal stresses are also readily found, since the radius, R, of the Mohr’s circle is given by cos 2φ =
σ zz /2 R
∴R =
15.9 = 38.8 MPa cos (65.8°)
(3.2)
The centre of the circle has been displaced from the origin by σzz/2, ie by 15.9 MPa, so the principal stresses are 15.9±38.8, ie +54.7 MPa and -22.9 MPa. €
3.4
Verification using the Secular Equation
It may be noted that this problem, and indeed all such problems tackled using Mohr’s circle, could alternatively have been solved by manipulating the secular equation (Eqn.(2.5)). For the torque only situation, this can be written
(3.3) which can be expanded to
λ2 − σθ2z = 0 ∴ σ1 = +σθz = +35.4 MPa and σ 2 = −σθz = −35.4 MPa
€
(3.4)
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Mechanics of Materials and Structures
Lent 2014
DH22
For the torque plus axial load case, the corresponding equation is
(3.5) which expands to
(3.6)
The orientation of the principal stresses, ie the value of the angle φ, can be obtained from Eqn.(3.1) in the same way as was done using Mohr’s circle. (This equation is simply derived from those presented as Eqn.(2.15).) However, it should also be noted that, while use of Mohr’s circle is never essential, it can nevertheless be very helpful and convenient, particularly in terms of visualising stress states. A useful TLP is available which provides an interactive tool for plotting a Mohr’s circle according to a user’s specified set of stresses: http://www.doitpoms.ac.uk/tlplib/metal-forming-1/index.php
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Mechanics of Materials and Structures
Lent 2014
DH23
4 – Representation of Strain 4.1
Introduction
Stresses tend to deform a body. This deformation could be elastic or plastic, but we are concerned here, at least in the first half of this course, with elastic (reversible) strains only. (For most materials, the elastic limit is less than 1%, so elastic strains are normally small.) Care is needed in defining a strain tensor since it could contain an unwanted rigid body rotation as well as the genuine shape change of the body. The true strain tensor should reflect only the shape change. It will become evident that the simple definitions of strain given in the first lecture (particularly shear strain) do not satisfy this requirement.
4.2
The Relative Displacement Tensor
The application of a set of stresses causes all points in the body to be displaced, relative to their initial locations, so that their coordinates, in a given frame of reference, will change. Provided that the body is macroscopically homogeneous, these displacements can be related to the stress state, via a description of the elastic properties of the material, which may, of course, be anisotropic. (However, the focus here is on the motion of points in the body, and there is no reference at this stage to the stress state that produced this motion.) The relative displacement tensor (sometimes, rather confusingly, called the deformation tensor), indicates how any point in the body becomes displaced (extended and/or sheared). It is a second rank tensor, with the first suffix representing the direction in which the displacement has occurred and the second one the reference direction. This is illustrated in Fig.4.1. All of the terms are (dimensionless) ratios of two distances. The shear terms (eij, with i≠j) can also be considered as angles - since they are small, they are approximately equal to their tangents. This is how strain was defined in the first lecture although not using tensor terminology.
Fig.4.1
Illustration, in the y-z (2-3) plane, of how the terms of the relative displacement tensor are defined, showing (a) a normal term, e33, and (b) a shear term, e23.
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Mechanics of Materials and Structures
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DH24
The sign of a shear component is taken as positive when the positive axis is rotated towards the positive direction of the other axis. For example, in Fig.4.1(b) the e23 component would be positive. It is clear, however, that for a general two-dimensional strain the relative displacement tensor will not be symmetric, e.g. e23 ≠ e32. This is illustrated on the left side of Fig.4.2, for shear deformation in the y-z (2-3) plane. Nevertheless it is possible to decompose the relative displacement tensor into two parts, one symmetrical (the true strain) and the other anti-symmetrical (the rigid body rotation). This is seen on the right side of Fig. 4.2.
Fig.4.2
4.3
Illustration of how, in the 2-3 plane, the relative displacements e23 and e32 can be represented as the sum of a strain, ε23 (=ε32) and a rigid body rotation, ω23 (=-ω32).
The Strain Tensor
The separation illustrated in Fig.4.2 can be expressed more generally. Any second rank tensor can be expressed as the sum of a symmetrical tensor and an anti-symmetrical tensor. For the relative displacement tensor, this can be written as (4.1) Since εij is a symmetrical tensor (4.2)
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Mechanics of Materials and Structures
Lent 2014
DH25
whereas for ωij, the rotation tensor, which is anti-symmetrical (4.3) It’s quite easy to see that such an anti-symmetrical tensor represents only a rotation. For example, in the 2-3 plane, the components of such a tensor can be written
0 e23 ω ij = −e23 0
€
(4.4)
and it can be seen in Fig.4.3 that this represents solely rigid body rotation. The residual component of the relative displacement tensor, ie the εij tensor, represents the strain.
Fig.4.3
Illustration of how an anti-symmetrical relative displacement tensor, in the 2-3 plane, represents only a rotation, with the body not being subject to any strain.
As in the case of a stress tensor, a strain tensor can be diagonalised to give the principal strains. These are the normal strains in the principal directions. As with stress, the principal directions are a unique set of (orthogonal) plane normals. For these directions, there are no shear strains. A strain tensor can be manipulated in a very similar way to a stress tensor. For example, Mohr’s circle can be used to find the principal strains or to facilitate other calculations or visualisations concerning the strain.
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Mechanics of Materials and Structures
Lent 2014
DH26
In full the strain tensor is given by
(e12 + e21 ) /2 (e13 + e31 ) /2 ε11 ε12 ε13 e11 (e23 + e32 ) /2 ε12 ε22 ε23 = (e21 + e12 ) /2 e 22 ε13 ε23 ε33 (e31 + e13 ) /2 (e32 + e23 ) /2 e 33 Referred to principal axes it is given by €
ε1 0 0 εij = 0 ε2 0 0 0 ε3
€
4.4
Some Simple Example Cases
We can now consider some familiar examples of states of strain. Fig.4.4 depicts states of pure shear, pure rotation and simple shear. The latter involves both rotation and strain, whereas the first involves no rotation and the second involves no strain.
Fig.4.4
Pictorial representation, in 2-D, of combinations of the shear components of the relative displacement tensor, corresponding to three simple limiting cases.
It will be seen in the next lecture that tensorial shear strains actually differ from the shear strains that are commonly used in engineering practice. However, for the moment we should just note that the state of strain in a body is fully and rigorously represented by the strain tensor.
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4.5
Mechanics of Materials and Structures
Lent 2014
DH27
Hydrostatic and Deviatoric Components of the Strain Tensor
A point worth noting at this stage is that the strain tensor can itself be divided into two components – one representing the change in volume of the body and the other the change in its shape. These are commonly termed the hydrostatic and deviatoric components. Consider a cube subjected to principal strains ε1, ε2 and ε3. The volumetric strain, often called the dilatation, can be written as (see 1A Course D): (4.5)
In fact, we can write Δ = εii for any strain tensor, since the sum of the diagonal terms of a second rank tensor is an invariant, and hence independent of the reference axes – see Eqn.(2.7). The mean hydrostatic strain tensor has one third of the dilation for all three of the normal terms, and no shear terms. The residual part is the deviatoric component
ε11 ε12 ε13 Δ /3 0 0 ε11 − Δ /3 ε12 ε13 ε22 − Δ /3 ε23 ε12 ε22 ε23 = 0 Δ /3 0 + ε12 ε13 ε23 ε33 0 0 Δ/3 ε13 ε23 ε33 − Δ /3
(4.6)
€
4.6
Field and Matter Tensors
The tensors we have been treating so far have all been field tensors. These are imposed on bodies, or regions of space, in some way. Relationships between field tensors, for example between stress and resultant strain, depend on properties on the material concerned. These are characterised by matter tensors. Matter tensors reflect the symmetry exhibited by the material. In this course, we will be treating only isotropic materials, so the arrays of values in the matter tensors we will be handling will tend to exhibit a high degree of symmetry. However, it’s helpful to appreciate how tensors can be used to analyse more complex cases, involving anisotropic materials.
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4.7
Mechanics of Materials and Structures
Lent 2014
DH28
Stiffness and Compliance Tensors
The relationship between a stress (tensor) and the strain (tensor) generated by it can be written (4.7) in which Cijkl is the stiffness tensor. This is a fourth-rank tensor, with 34 (=81) components. (The rank of a tensor is equal to the number of its suffices, although there are some situations in which a convention may be used such this is not the case - an example is provided by principal stresses and strains being given a single suffix.) Eqn.(4.7) is a generalised expression of Hooke’s law. It represents 9 equations, generated according to the Einstein summation convention. For example, the first of these is
(4.8)
The stress-strain relationship can also be expressed in the inverse sense
ε ij = Sijklσ kl
(4.9)
in which Sijkl is the compliance tensor. (Note that the symbols conventionally used for stiffness and compliance are the reverse of the initial letters of these words.)
4.8
Relationship to Elastic Constants
This looks a little cumbersome and daunting, but in practice the treatment can be simplified. The symmetry of stress and strain tensors when the body is in static equilibrium means that
Cijkl = Cijlk = C jikl = C jilk
(4.10)
reducing the number of independent components from 81 to 36. Furthermore, the symmetry exhibited by the material commonly results in further reductions in this number. In fact, for a cubic material it reduces to 3. For an isotropic material (eg a polycrystal) where the stiffnesses and
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Mechanics of Materials and Structures
Lent 2014
DH29
compliances are independent of the choice of reference axes, it is reduced to just 2, so the elastic behaviour of isotropic materials is fully specified by the values of 2 constants. In terms of engineering constants, the relationship between (normal) stress and strain can be obtained by considering the application of a single stress σ1, generating a strain ε1, so that (4.11) in which E is the Young’s modulus. One might be tempted to deduce that E is given by C1111. However, this is incorrect, since the application of a stress σ1 generates, not only a (direct) strain ε1, but also Poisson strains ε2 and ε3 (see below), which would appear in the full (tensorial) equation for σ1.
Expressed in terms of compliance, however, the equation we need (from Eqn.(4.9)) is (4.12) so it follows that E=
4.9 €
1 S1111
(4.13)
Poisson Ratio
The second elastic constant specified for an isotropic material is commonly the Poisson ratio, ν. This gives the transverse contraction strain which accompanies an axial extension strain. Again considering a single (normal tensile) stress σ1, generating principal strains ε1, ε2 and ε3, the Poisson ratio is defined by (4.14) The tensorial expression for ε2, with only σ1 applied, is (4.15) so that (4.16)
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Mechanics of Materials and Structures
Lent 2014
DH30
Since Poisson strains simply superimpose during multi-axial loading, we can write the following expressions for the (principal) strains arising from a set of (principal) stresses
(4.17)
For an isotropic material, the 36 components of the compliance tensor can be represented by a 6x6 matrix leading to a pseudovector-matrix form of Eqn (4.9):
ε1 1 - ν - ν ε2 -ν 1 - ν ε3 1 -ν - ν 1 = 0 0 2ε23 E 0 2ε31 0 0 0 0 0 2ε12 0
€
0
0
0
σ1 0 0 0 σ 2 0 0 0 σ 3 2(1 + ν ) 0 0 σ 23 0 2(1 + ν ) 0 σ 31 0 0 2(1 + ν )σ12
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Mechanics of Materials and Structures
Lent 2014
DH31
5 – Engineering Stresses and Strains and Plasticity 5.1
Engineering Shear Strains and the Shear Modulus
A minor complication arises when considering shear strains and the shear modulus. One might imagine that the shear modulus, G, would be defined as shear stress divided by shear strain, eg σ12/ε12. Unfortunately, G ≠ σ12/ε12. The problem is associated with the difference between a tensorial shear strain, such as ε12, and the corresponding engineering shear strain, γ12. The shear modulus is conventionally (see first lecture) defined by (5.1) in which the value of γ12 is simply the measured strain, with no account being taken of the fact that it really represents the sum of two shear strains, one being generated by each of the pair of shear stresses which is being applied. This is illustrated in Fig.5.1.
Fig.5.1
Illustration of how a pure shear deformation, in the 1-2 plane, when rotated about the 3-axis, can be viewed as a simple shear, and used to define the shear modulus, G, in terms of the engineering shear strain, γ12. The presence of the other shear stress acting in this plane, σ21, is neglected in this definition.
The array of engineering strains (ε ii and γ ij) thus do not form a tensor and cannot be transformed using the standard procedures for second rank tensors. However, we can still define the shear modulus in terms of components of the compliance tensor. With only σ12 and σ21 acting, Eqn(4.9) gives
ε12 = S1212σ12 + S1221σ 21 = 2S1212σ12 ∴G =
€ €
σ12 σ12 1 = = γ12 2ε12 4S1212
(5.2)
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Mechanics of Materials and Structures
Lent 2014
DH32
In fact, similar procedures can be used to establish all of the components of the compliance and stiffness tensors, in terms of (measured) engineering elastic constants. Of course, this procedure is more complex for anisotropic materials, for which there are more than 2 independent elastic constants, but it can still be carried out.
5.2
Bulk Modulus
It may finally be noted that the bulk modulus, K, is defined by (see 1A Course D): K=
σ H (σ1 + σ 2 + σ 3 ) /3 σ ii = = Δ 3ε jj (ε1 + ε2 + ε3 )
(5.3)
where σH is the hydrostatic component of the stress state. The bulk modulus is a measure of the resistance of the material to volume change. A general stress tensor, referred to the principal axes, €
can be written as the sum of a hydrostatic component (pressure) and a deviatoric stress, in a way similar to that discussed for strain in section 4.5.
σ1 0 0 σ ii /3 0 0 σ1 - σ ii /3 0 0 σ 2 - σ ii /3 0 0 σ 2 0 = 0 σ ii /3 0 + 0 0 0 σ 3 0 0 σ ii /3 0 0 σ 3 − σ ii /3
€
5.3
Relationships between Elastic Constants
Since only two elastic constants are required to fully define the behaviour of an (isotropic) material, it follows that there must be inter-relationships between the 4 that have been presented here. For example, adding up the 3 equations in Eqn.(4.17) gives
(5.4)
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Mechanics of Materials and Structures
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DH33
Several other relationships are commonly quoted, and can be useful. For example, the shear modulus is given by (5.5)
5.4
Large Uniaxial Tensile Strains and Plastic Deformation
Application of a uniaxial tensile stress σ1 generates contraction strains given by
ε2 = ε3 = −νε1 The volumetric strain is thus
€
Δ = ε1 + ε2 + ε3 = ε(1− 2ν )
(5.6)
For a ductile material (e.g. a metal) ν ≅ 1/3 and a small elastic strain of 0.5% would produce a € volume change of about 0.2%. Eqn. (5.6) indicates that the elastic volume change decreases as ν increases. For a plastically deforming metal subject to large strains the volume change is zero and ν is effectively 1/2 (although ν is only properly defined for elastic deformation). Since volume is conserved during plastic deformation then, for the sample shown below, A0l0 = Aili
Fig. 5.2
Change in sample dimensions under uniaxial tensile load which results in permanent deformation.
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DH34
Under constant load the cross-sectional area of the sample decreases and hence the instantaneous stress increases. This instantaneous or “true” stress is defined as σT = F/Ai. It is clearly larger than the nominal (or engineering) stress σN = F/A0. The difference between the two quantities increases with increasing plastic deformation. A true strain can also be defined based on the instantaneous sample length. It is determined by summing the incremental strains:
εT =
δl δl δl δl + + + ......... = ∑ l0 l1 l2 i li
where l1 = l0 + δl, l2 = l1 + δl etc. €
In differential form (1A Course D): li
εT =
∫
l0
l dl = ln i l l0
The true and nominal stresses and strains can be related as follows: €
li − l0 l0 A l F F A0 σT = = = σN 0 = σN i Ai A0 Ai Ai l0
εT = ln(1+ εN )
where εN =
thus σ T = σ N (1+ εN )
For small stresses and strains (i.e. before yielding) σT ≅ σN and εT ≅ εN and there is no distinction € between the quantities.
5.5
Plastic Instability and Considère’s Construction
The figure below compares a typical nominal stress-strain curve for a polycrystalline metal with the corresponding true stress-strain curve.
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Mechanics of Materials and Structures
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DH35
It is seen that the nominal curve passes through a maximum (point a) beyond which continued straining requires an ever-decreasing applied force. The maximum, sometimes called the ultimate tensile strength, signals the onset of a plastic or geometric instability. It is geometric in the sense that it does not reflect changes in material properties. The instability manifests itself in the form of a neck as shown below:
Fig. 5.3
Illustration of neck formation under large uniaxial tensile deformation.
Necking occurs when the increase in yield stress due to work hardening cannot compensate for the larger increase in stress due to the reduction in cross-sectional area. Once necking begins the strain is restricted to the necked region. Voids form where the local stress is large and eventually the metal fractures at the neck (point b). The true stress-strain curve shows that σT > σN beyond the elastic limit (point c) as expected. It is also concave down but does not exhibit a turning point before fracture (point b’). Nevertheless the onset of necking can still be predicted from a true stress-strain curve using Considère’s criterion (1885). Necking occurs at maximum load, i.e. dF = 0. Therefore necking begins when:
dF = Ai dσ T + σ T dAi = 0 From conservation of volume we have
€
dV = Ai dli + li dAi = 0 Hence by combining these equations we find
€
dσ T dA dl = − i = i = dεT σT Ai li and so at the onset of necking it follows that
€
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Mechanics of Materials and Structures
Lent 2014
DH36
dσ T = σT dεT This is Considère’s criterion for necking. When the slope of the true stress-strain curve € equals the true stress necking begins. This is represented graphically by a tangent line which intersects the true strain axis 1 unit away from the relevant strain.
Fig. 5.4
Schematic true stress-strain curve showing Considère’s tangent construction.
In practice the tangent construction is usually made by plotting the true stress against the nominal strain (or extension ratio, λ = εN + 1) which also does not exhibit a turning point. From the relationships above it is easily shown that
dσ T σ = T dεN εN + 1 so that in this case the tangent line intersects the nominal strain axis at -1.
€
Fig. 5.5
Schematic true stress versus nominal strain curve showing Considère’s tangent construction.
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Mechanics of Materials and Structures
Lent 2014
DH37
Nominal stress-strain curves are useful in determining strength and ductility data for purposes of engineering design. True stress-strain curves are useful for understanding the work hardening process in metals and also for predicting the drawability of different materials. Brittle materials, for example, exhibit concave upwards curves where no tangent can be constructed and polymers exhibit sigmoidal curves where two tangents can be constructed. This is discussed further in the Pt II Materials Science course.
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Mechanics of Materials and Structures
Lent 2014
DH38
6 - Beam Bending 6.1
Introduction to Beam Bending
While mechanics treatments, and mechanical testing procedures, are often based on the uniform application of a normal stress to a body, such a loading configuration is in fact relatively unusual in engineering practice. It’s much more likely that a component or structure will be subjected to bending moments. In fact, virtually all structures, including bridges, buildings and many natural structures (trees, bones etc) are commonly subjected to significant applied moments.
Beam stiffness is an important concept for many types of structure, particularly those with slender shapes (large span-to-height ratio). Inadequate beam stiffness can lead to large deflections, and may also cause high, localised, stresses and a danger of failure in that region. Slightly less common, but still of considerable importance, are situations in which bodies are subjected to twisting, or torsional moments (torques). This lecture is focussed on beam bending and the next lecture covers torsion.
6.2
Beam Curvatures (Pt 1A revision)
The concept of the curvature of a beam, κ, is central to beam bending. Fig.6.1 shows that the axial strain in the beam, εxx, is given by the ratio y/R, where y is the distance from the neutral axis (neutral plane in 3-D) and R is the radius of curvature. Equivalently, 1/R (the “curvature”, κ) is equal to the through-thickness gradient of axial strain.
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Fig.6.1
6.3
Mechanics of Materials and Structures
Lent 2014
DH39
Relation between the radius of curvature, R, the beam curvature, κ, and the strains within a beam subjected to a bending moment
Bending Moments, Stresses and Second Moments of Area
Bending moments are produced by transverse loads applied to beams. The simplest case is the cantilever beam, widely encountered in balconies, aircraft wings, diving boards etc - see Fig.6.2. One end of the beam is fixed while the other is subjected to a concentrated load and is free to bend. Loads distributed along the beam’s length will have the same effect. In Fig. 6.2 the load is downwards and the beam is hogging (i.e tension in the top surface, compression in the lower surface). The mass of the beam is neglected in determining the loading action. In addition to the bending moment there is a shearing force acting vertically along the length of the beam and in this case is constant and equal to F.
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Fig.6.2
Mechanics of Materials and Structures
Lent 2014
DH40
Balancing the external and internal moments during the bending of a cantilever beam
The bending moment acting on a section of the beam, due to an applied transverse force, is given by the product of the applied force and its distance from that section. For the cantilever, this gives (6.1) Moments thus have units of N m. Note that it is necessary to adopt a sign convention for M. Here M is taken as positive when the upper surface is in tension, i.e. the beam shape is concave downwards and hogging. However, this convention is somewhat arbitrary and could be completely reversed as seen in many mechanics textbooks. It is thus seen that in the case of the end-loaded cantilever (Fig. 6.2) the bending moment has a maximum value of +FL at the fixed end and decreases linearly to zero at the free end. The moment, as expressed in Eqn.(6.1), may be regarded as being externally imposed, and tending to rotate the beam. It is balanced by the internal moment arising from the axial stresses generated in the beam. This moment is given by a summation of all of the internal moments acting on individual elements within the section. These are equal to the product of the force acting on the element (stress times area of element) and its distance from the neutral axis, y. The internal moment can therefore be written (6.2) where dA is an element of the section at a uniform distance, y, from the neutral axis. This can be presented more compactly by defining I (the second moment of area, or “moment of inertia”) as I=
2
∫ y dA A
€
(6.3)
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Mechanics of Materials and Structures
Lent 2014
DH41
The units of I are m4. The value of I is dependent solely on the beam sectional shape. The moment can now be written as (6.4) Since the product EI is a constant for a given beam shape, the beam curvature κ varies in the same way as M along the length of the beam, which in the case of the cantilever is a linearly decreasing function of x as seen in Fig. 6.2.
6.4
Beam Stiffness
The product EI is termed the “beam stiffness”, or sometimes the “flexural rigidity”. It is a measure of how strongly the beam resists deflection. The beam curvature, κ, is thus given by the ratio of applied moment to beam stiffness, in an analogous way to axial strain being equal to the ratio of applied stress to Young’s modulus, under uniaxial loading.
Fig.6.3
Illustration of how the integration is carried out in order to find the expression for I in the cases of (a) rectangular section and (b) circular section beams
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Mechanics of Materials and Structures
Lent 2014
DH42
The expression for I, for a given sectional shape, is established by carrying out the integration of Eqn.(6.3). For example, for a rectangular beam of width w and thickness h (see Fig.6.3(a)), it is given by (6.5) For a circular section (see Fig.6.3(b)), the integration is more complex, although the final expression is again a simple one
(6.6)
Note that these expressions for I are obtained with respect to the neutral (centroidal) axis. Other expressions would be obtained using a different axis. In general I increases as the reference axis is moved parallel to itself further from the centroid (centre of mass for a beam of uniform density). This is indicated by the parallel axis theorem which states I parallel = Ineutral + d 2 A
where d is the distance between the neutral and parallel axes and A is the area of the section. €
€
i.e. I parallel =
6.5
2
2
2
∫ ( y + d ) dA = ∫ y dA + 2d ∫ ydA + d ∫ dA
Maximising the Beam Stiffness
Expressions such as Eqns.(6.5) and (6.6) are clearly useful in beam bending calculations. However, even without such equations, some simple guidelines can be identified for maximising the value of I, for a given sectional area (ie for a given beam mass). It’s clear that I is raised by ensuring that much of the section is at large distances from the neutral axis. The classical Ibeam section, commonly used in many constructions (see Fig.6.4) is an example of the application of this principle.
Fig.6.4
Photo showing I-beams being used in a civil engineering construction
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Another commonly-employed approach is to use hollow sections, which also tends to ensure that the sectional area close to the neutral axis is reduced. Calculation of I for hollow beams is very straightforward, since it is obtained by simply subtracting the I of the missing section from that of the overall section. For example, that for a cylindrical tube is given by (6.7) where D is the outer diameter of the tube and d is the inner diameter.
A similar result is obtained for a hollow box section. The procedure can be particularly useful in obtaining I for asymmetrical shapes like a T-section. For these shapes the location of the neutral axis (NA) also has to be determined. This can be done by calculating the coordinates (z’, y’) of the centroid of the shape. These are given by the first moment of inertia divided by the area i.e. z'=
€
∫ zdA ∫ dA
y'=
∫ ydA ∫ dA
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7 - Examples of Beam Bending and Torsion 7.1
Beam Deflections
Eqn.(6.4) allows the curvature distribution along the length of a beam (ie its shape), and the stress distribution within it, to be calculated for any given set of applied forces (ie any variation of the applied moment along the length). The deflection at any point along the length can therefore be calculated. In this section we illustrate this by deriving expressions for the beam deflection along an end-loaded cantilever and along a beam subjected to symmetrical 3-point bending. The beam support and loading configuration is different in each case. The cantilever is fixed at one end and therefore experiences a reaction moment at that end. In 3-point bending the ends are “simply-supported” which means they rest on knife-edges or frictionless pins perpendicular to the plane of bending and do not experience reaction moments. The cantilever hogs (positive M) whereas in 3-point bending the beam sags (negative M). In both cases the deflection is downwards which is taken to be the positive y-direction and the x-y origin is at the left end of the beam. The orientation of the axes and location of the origin is somewhat arbitrary and the analysis could equally be performed with other choices.
7.2 Cantilever beam The beam curvature, κ, is approximately equal to the curvature of the line traced by the neutral axis, d2y/dx2, as shown in Fig. 7.1a.
Fig.7.1a
Approximation involved in equating beam curvature to the curvature of the neutral axis
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Hence d2y M =κ = 2 dx EI
Therefore using Eqn (6.1) for the bending moment of an end-loaded cantilever: €
EI
d2y = M = F(L − x) dx 2
The deflection is found by integrating this expression, using the boundary conditions to establish the integration constants € EI
dy Fx 2 = FLx − + C1 dx 2
at x = 0, €
EIy = €
€
€
€
dy = 0, ∴C1 = 0 dx
FLx 2 Fx 3 − + C2 2 6
at x = 0, y = 0, ∴C2 = 0 The defections along the length of the beam, and specifically at the loaded end, are thus given by y=
Fx 2 ( 3L − x ) 6EI
FL3 δ= 3EI
€ 7.3 Symmetrical 3-point bending
Fig.7.1b
Symmetrical 3-point bending
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The bending moment on the left hand side of the beam (0 ≤ x ≤ L/2) is given by (7.1) From Eqn.(6.4), and writing the curvature, κ, as d2y/dx2, (7.2) Integrating with respect to x, and applying boundary conditions to find the integration constants
(7.3)
(7.4) so the equation for the deflection is (0 ≤ x ≤ L/2)
(7.5)
and the deflection of the centre of the beam (x = L/2) is given by (7.6)
A useful TLP is available which provides an interactive tool for simulating beam bending according to a user’s specified loading geometry and beam stiffness: http://www.doitpoms.ac.uk/tlplib/beam_bending/index.php
7.4
Introduction to Torsion
Torsion is the twisting of a beam under the action of a torque (twisting moment). It is systematically applied to screws, nuts, axles, drive shafts etc, and is also generated more randomly under service conditions in car bodies, boat hulls, aircraft fuselages, bridges, springs and many other structures and components. A torque, T, has the same units (N m) as a bending moment, M. Both are the product of a force and a distance. In the case of a torque, the force is tangential and the distance is the radial distance between this tangent and the axis of rotation. While the treatment of torsion follows logically from that beam bending, there are some differences that need to be noted.
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Fig.7.2
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Torsion of Thin-Walled Circular Tubes
Application of a torque, T, to a thin-walled tube, showing the stress state experienced by a small volume element of the tube wall
In section 3.3 we considered the shear stresses acting on a thin-walled tube under torsional loading. When a torque is applied, the wall is subjected to a state of pure shear, as illustrated in Fig.7.2. The relation between the torque and the resultant shear stress in the plane of the wall, σθz (= σzθ), was found to be (7.7) Because the wall is thin (t << r) it is assumed that this stress does not vary significantly across the thickness of the wall. This is not true, however, for thick-walled tubes or solid cylinders as shown below.
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Torsion of Solid Cylindrical Bars
In most cases, the cylindrical bodies to which torques are applied are not thin-walled tubes, but are solid. The analysis is more complex in such cases, since the shear stresses induced will vary with radial location (distance from the axis of the torque). Furthermore, in order to analyse the situation properly, we need to consider the response of the cylinder to the torque – see Fig.7.3.
Fig.7.3
Torsion of a Cylindrical Bar
In contrast to the case of beam bending, the shape of the cylinder† does not change when a torque is applied. However, there is a rotation about the torque axis, which would be apparent if there were reference lines marked along the axis of the cylinder, as in Fig.7.3. It can be seen that the (engineering) shear strain in an element of the bar is given by
(7.8)
†
This is, of course, only true for a cylinder. It would not, for example, apply to a square section bar, or other prismatic bodies with various sectional shapes. Furthermore, torques are commonly applied to bodies which are not prismatic, but have sections which change along their length – such as bolts. However, while the analysis is clearly more complex in such cases, the principles described here are still applicable to them.
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This equation applies both at the surface of the bar, as shown, and also for any other radial location, using the appropriate value of r. Clearly, the shear strain varies linearly with r, from zero at the centre of the bar to a maximum value at the free surface (r =R) γ θmax = z
Rdφ dL
The quantity dφ/dL is called the rate of twist. In the special case of pure torsion, the rate of twist equals the total angle of twist divided by the length of the bar:
€
γθmax = z
€
Rφ L
The shear stress, σθz, at any radial location, is related to the shear strain by
σ θ z = Gγ θ z
(7.9)
where G is the shear modulus (see Eqn.(5.1). Substituting from Eqn.(7.8), it follows that
σ θ z = Gr
dφ dL
(7.10)
The torque, T, can therefore be written as T = ∫ dT = ∫ σ θ z r dA = ∫ G r 2 A
A
A
dφ dA dL
(7.11)
As with beam bending, the geometrical integral is represented as a (polar) second moment of area
I P = ∫ r 2 dA
(7.12)
A
Note the relationship between the polar second moment of area and the (rectangular) second moments of area defined earlier for beam bending: 2
∫ r dA = ∫ (z
2
+ y 2 dA =
i.e. IP = Iz + Iy €
)
2
2
∫ z dA + ∫ y dA
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The torque is thus given by T = G I P
dφ dL
(7.13)
Comparing this equation with the corresponding one for beam bending (Eqn.(6.4))
M = E I κ
(7.14)
it can be seen that the torsional analogue for the curvature of a bent beam is the rate of twist along the length of the bar, dφ/dL. For a solid cylinder of diameter D, IP can be written as D /2
D /2
πr 4 IP = ∫ r dA = ∫ r 2πrdr = 2 0 A 0 2
€
2
πD4 = 32
(7.15)
By combining Eqns (7.10) and (7.13) it is seen that the maximum shear stress, which is on the surface of the cylinder, is given by
σθmax = z
€ 7.7
TR 16T = IP πD 3
(7.16)
Torsion of Thick-Walled Circular Tubes
Most of the torsion equations derived for solid cylinders apply to thick-walled circular tubes with suitable modification for the range of allowable radii (e.g. inner r1 to outer r2). For example the shear stress is given by
σθz =
€
Tr π π where IP = ( D24 − D14 ) = ( r24 − r14 ) IP 32 2
(7.17)
If r is defined as the average radius of the tube equal to (r1 + r2)/2 then the expression for IP € becomes IP =
πrt 4r 2 + t 2 ) ( 2
(7.18)
where t = r2 – r1, the tube thickness. €
If the thickness of the tube is small (t << r) then Eqn (7.18) reduces to IP ≈ 2πr 3 t
€
which is effectively the same as that used in Eqn (7.7).
Thick-walled circular tubes can be more efficient at resisting torsional loads than solid bars if weight is taken into consideration. For example, for a fixed mass per unit length, a tube will have a larger outer radius than the solid bar and a larger polar second moment of area resulting in a
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higher torsional stiffness and a lower maximum shear stress for a given applied torque. As a consequence large drive shafts, propeller shafts and generator shafts often have tubular crosssections.
It is emphasised that the torsion equations for circular cylinders and tubes do not apply to bars with other shapes. For example bars with rectangular cross-sections do not remain plane under torsion and their maximum stresses are not located at the furthest point from their centroid.
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8 – Failure in Compression: Elastic Buckling of Columns 8.1 Crushing 8.1.1 Porous bodies If a body is porous then failure can occur by material being squashed into the pores. This gives rise to a crease of compacted material that spreads out perpendicular to the compressive stress (i.e. in the example below, horizontally).
8.1.2 Non-porous bodies However, if the body is solid it is not at all obvious how it will break when the atoms are pushed together. There do not seem to be any forces pulling the atoms apart. However, consider the stress state of uniaxial compression represented with Mohr’s circle:
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This tells us that there are shear stresses within the material, and that these are highest on planes which lie at 45° to the principal axes. In ductile materials, failure could occur by plastic flow on these planes. In brittle materials, failure occurs by the growth of cracks, causing the material to break into pieces on the 45° planes.
Often this occurs by the formation of small cracks that lie parallel to the compression axis. Failure occurs by the linking of these cracks along the 45° plane. This failure mechanism is known as crushing and is commonly seen in concrete in compression.
From http://www.tfhrc.gov/safety/pubs/05063/chapt3.htm
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8.2 Elastic (Euler) Buckling 8.2.1 Freely-hinged columns We will consider now another way in which a structure might fail in compression under a load less than that required to cause crushing failure of the material.
A number of important assumptions must be made for a straightforward analysis of elastic buckling: 1. The column has a uniform prismatic cross section 2. The material is homogeneous, defect-free, isotropic and linear-elastic 3. The forces due to the weight of the column itself are negligible 4. The force is axial – with no eccentricity Also, to begin with, we will assume that the ends are free to rotate The top pin remains vertically above the bottom pin
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Let us consider the situation if the column were to deflect. The resulting scenario is that of a bent beam If the force F moves downwards by an increment δx, there is a change in the mechanical energy of the system.
However that will cause an increase in elastic energy in the beam.
If |δUE| > |δUF| then the buckling leads to an overall increase in the energy of the system (δUE + δUF would be positive) and so it is more favourable for the column to straighten itself out. If |δUF| > |δUE| then the buckling leads to an overall decrease in the energy of the system (δUE + δUF would be negative) and so it is favourable for the column to buckle.
We can now consider the total energy stored in the beam as a function of the sideways displacement, y.
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We will consider the case of neutral equilibrium – this will allow us to calculate the critical force (when the energies driving and resisting buckling are equal). Consider some point along the column, which we will label Q.
Summing moments about point Q, we have the moment due to the applied external force, (8.1) which must be balanced by the bending moment arising from the stresses in the beam, which is equal to the curvature, κ multiplied by EI (see section 6.3). (8.2) and therefore just by balancing these moments, (8.3)
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Eqn (8.3) is similar to eqn (7.2) for symmetrical three-point bending. The important difference, however, is that for buckling the bending moment is a function of the deflection y. This is not the case for three-point bending. For buckling the equilibrium equations take account of the geometry of the deformed structure. Rearranging (8.3) gives (8.4)
where
. This is an ordinary second order linear homogeneous differential equation with
constant coefficients and has the general solution (8.5) where A and B are determined from the boundary conditions. The boundary conditions are that the end-points of the column do not leave the x-axis during loading, i.e.
Using eqn. 8.6a, (8.7) hence B = 0.
Using eqn. 8.6b, (8.8) One solution to this is A = 0, but as B = 0, that means that y = 0 and so the column has not deflected.
If
, then the column will have deflected. This is only possible if (8.9) , where n = 0, ±1, ±2, ±3.....
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i.e. k2L2 = n2π2 and k2 = F/E.I, giving the force required for elastic buckling, FEB; (8.10) Hence we can sketch possible solutions to the configurations which the column could, in theory achieve:
n=0
n=1
n=2
Hence for low forces the column will not buckle (for any slight sideways perturbation, the column will restore itself to reduce the energy). This is a stable equilibrium. However at a critical force, when F=FEB, the n = 1 solution is achieved and the column is in neutral equilibrium. If the force exceeds FEB the column is unstable and any slight sideways perturbation will cause the column to buckle catastrophically. The force at which a column in compression will buckle is therefore: (8.11)
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Note the critical force depends on the flexural rigidity EI of the column and not on the strength of the material itself as represented, for example, by the yield stress.
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9 – Failure in Compression: the effect of Constraints and Column Shape 9.1 The Effect of End-Constraint So far, we have considered both ends to be hinged. However in many situations this may not be the case. First we will consider the situation where one end is freely hinged (i.e. free to rotate and to move laterally) but the other end is fixed (a).
It can be seen that the force which would cause this column to buckle is equal to the force which would cause a completely freely-jointed column of twice the length (b) to buckle, i.e. (9.1) that is to say that the column with one end fixed and one end freely jointed buckles under one quarter of the load which would cause a freely jointed column of the same length to buckle. (9.2)
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Another scenario likely to occur is that in which both ends are clamped (a):
It can be seen that the force which would cause this column to buckle is equal to the force which would cause a completely freely-jointed column of half the length (b) to buckle, i.e. (9.3)
that is to say that the column with two fixed ends buckles under four times the load which would cause a freely jointed column of the same length to buckle. (9.4)
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Hence we can write a general expression for elastic buckling as:
(9.5)
c=
1/4
1
4
DH62
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9.2 The effect of aspect ratio (or slenderness) We wish to consider the effects of shape rather than size, so we must think in terms of the stress,
σEB, rather than the force for elastic buckling. The compressive stress under which our column will buckle is given by (9.6) If the column is a solid cylindrical rod, then we can substitute for I and A;
(9.7) In other words the buckling stress depends on the ratio (L/R)2. So as we know from everyday experience the longer and more slender a rod becomes, the lower is the stress required to make it buckle. The important shape parameter is L/R, which we call the “slenderness ratio”. The only materials property which determines the buckling stress is E, the Young’s Modulus. Hence we can make a simple plot of the buckling stress for columns as a function of the slenderness.
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9.3 The effect of cross-sectional shape As the column is bending when it buckles then we would expect the shape of the cross-section to be important. We know that if we are designing a beam, an ‘I’ section can be a good choice. There is one important difference. In bending the direction of the applied force (and hence the resulting deformation) is usually known, whereas in buckling it can occur in any direction perpendicular to the longitudinal axis of the column. It will therefore buckle in the direction in which EI is smallest, which normally means the direction in which I is smallest.
So there is not much advantage to an I-beam if its job is to resist buckling. Because the buckling failure will always occur in whichever direction has the lowest EI, we obtain the most efficient use of material if that value is the same in all directions. Therefore cylindrical columns and those with a section that is a regular polygon (e.g. equilateral triangle, square etc) will, for a given height and volume, be more resistant to buckling than rectangular-section columns.
There is a way in which we can increase the second moment of area of a column whilst retaining uniformity in all directions. That is to have a hollow cross section (moving material away from the neutral axis). Hence for the same amount of material we can increase the resistance to buckling by
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using a hollow cylinder (which can therefore have a larger external diameter), increasing the value of I.
Hence a drinking straw has more resistance to elastic buckling than a solid cylinder of the same material with the same length and volume.
Does elastic buckling cause failure? We have described buckling as an elastic phenomenon, which in principle should therefore be recoverable. However it often leads to other types of failure such as brittle fracture or local buckling (which we will deal with shortly). Let us consider a column being tested under “displacement control”. That is we apply a force to some end-plates (as much as is needed) and control the distance between them. At first the displacement can be achieved by uniaxial elastic compression, but quickly the buckling force will be reached. However the column will not instantaneously collapse at this point. It will buckle in a stable manner and if the test is stopped the material will be able to recover (as long as the buckling is not taken so far as to cause some other failure mechanism).
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Another way we can test a column is under “load control”. One way we could achieve this is to apply a constantly increasing force to the top of a column (e.g. by stacking weights on top). Recall that we derived the force for elastic buckling by saying that it will happen when the energy driving buckling exceeds the energy resisting it. When this occurs, collapse of the column will be sudden and instantaneous. However it is possible that the material will remain elastic and be able to recover once the load is removed.
9.4 Local Buckling Our analysis of hollow cylinders suggests that we can prevent buckling, even in a very long tube simply by having a large external diameter. However, assuming we have a fixed amount of material, the larger we make the external diameter, the thinner the wall will become and we know in practice that there is a limit to how far we can reduce the wall-thickness. The failure mechanism for thin-walled tubes is called local buckling. A paper straw kinking is an example. And just like the straw (and unlike elastic buckling) this is irreversible. The force required for local buckling to occur is given by: (9.8) where k~0.5 but depends on any surface imperfections. A drinks can is a good example of a thin walled structure which we know fails by local buckling and it is straightforward to calculate the load which it can support.
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9.5 The Effect of Internal Pressure In Part IA Course D you looked at the effect that pressure has on fast fracture of cylinders. Now we will see how it affects buckling. Consider a pressurised cylinder with some internal pressure which is greater than the external pressure. We will call this pressure difference ΔP and in this case it will have a positive value.
We know that a positive internal pressure has the effect of creating a longitudinal tensile stress in the wall and so the total stress in the wall will be reduced:
Hence this positive internal pressure helps to prevent buckling by reducing the overall compressive stress in the wall. A negative internal pressure would lead to buckling under a lower applied load.
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10 – Brittle Fracture in Tension 10.1 The Ideal Strength We can start by considering when we would expect the atoms within a material to separate. If we look at a typical force vs. displacement curve between neighbouring atoms we see that bonds usually break at about 10-20% strain, and hence we should expect a chain of atoms to have a tensile strength between E/10 and E/5.
Hence for a material with a Young’s modulus of 100 GPa, which is typical for a metal, we might predict that its tensile strength would be around 10 GPa. In reality however, most practical materials fail at much lower stresses than this. Having said that, experiments carried out by Griffith measuring glass fibres showed that the tensile strength was a function of fibre thickness with very thin samples having strengths approaching the theoretical strength.
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This is not surprising – we should expect that in the limit of a sample which is a linear chain of atoms it will have the theoretical strength. The question is not why thin samples are so strong, but why larger samples are so weak. The answer is due to the presence of flaws. There are two ways of thinking about the effect of the presence of a crack in a material. Inglis considered how the concentration of stress near the tip of a crack would enable the crack to extend whilst Griffith considered whether it would energetically favourable for a crack to grow.
10.2 The Effect of Cracks – Stress Considerations 10.2.1 Stress concentrations at an atomic scale
From this picture of stresses in a crystal it is clear that where a crack is present, an applied stress will have to be taken up by other bonds. This stress is not redistributed equally but is concentrated near to the crack tip. It is apparent that the extent of the stress concentration is dependent not only on the crack length but also on whether (on a macroscopic scale) the crack is sharp or blunt.
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10.2.2 Stress concentrations at macroscopic scale Inglis was able to calculate the stress concentration factor in a continuum material near to a through-thickness elliptical crack. If the far-field stress is σ0, the stress at the elliptical opening was calculated to be:
σ0
(10.1)
where c is the half-length of the crack and R is its radius of curvature. Hence for a circular opening, the stress locally is raised by a factor of 3.
If we apply this argument to a sharp rather than smooth hole, and assume that the diameter of the crack tip is of the order of an atomic spacing, this gives us R = 0.1 nm. A surface crack one micron in depth would then produce a stress concentration of 200 which leads to a more reasonable value for the strength of glass. However, windows can often contain cracks which are on a cm scale (producing a concentration factor of 20,000) and large structures such as ships and bridges can have cracks and openings on even larger length scales. Hence if the stress concentration were the whole story, materials would end up being weaker than they actually are. The fact is that the stress concentration is only a mechanism to break a material apart. To work out whether a material will fracture, we also need to consider whether it is energetically favourable for it to do so.
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10.3 Energy Considerations 10.3.1 Contributions to the total energy of the system When a material is subject to a stress, there is some mechanical energy available which could potentially cause fracture. If the extension of a crack decreases the mechanical energy of the system, that is clearly a driving force for crack growth and subsequent fracture. However, growth of a crack produces new surfaces, which have an associated energy. This provides some energetic resistance to crack growth and fracture. By considering these energies, we can work out whether it will be favourable for the crack to extend.
10.3.2 The change in mechanical energy of the system, UM The mechanical energy of the system can have two parts – there is an elastic strain energy stored in a body under stress, but if the force which is applied is able to do work, we must consider that also. There are two ways to calculate the value of the change in mechanical energy (driving force for crack extension). It is possible to put a material in tension with a fixed extension (a simpler analysis) or apply a fixed load (a more common scenario). These are equivalent in terms of the total change in mechanical energy (though not in terms of the change in elastic strain energy stored) when a crack is introduced.
i) Strained material under constant extension
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ii) Strained material under constant load
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Hence we simply need to work out the change in elastic strain energy in a material of Young’s modulus, E, subjected to a far-field stress σ if it contains an interior crack of length 2c. This is derived from Inglis’ analysis and is: (10.2)
This term is negative and always results in a decrease in the total energy of the system.
We can see qualitatively why the term scales with the square of the crack length by thinking about the volume of the region in which the stress is relieved.
10.3.3 The energy required to make new, cracked surface, US This is more straightforward – the work required to create the 2 new surfaces for an internal crack of length 2c is: (10.3)
where Γ is the amount of energy needed to create 2 new surfaces, each of unit area. This term is positive (creates new surfaces) and always results in an increase in the total energy of the system.
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10.3.4 The total energy change as a function of crack length Combining the mechanical and surface contributions, the change in energy (per unit thickness of sample) due to the presence of a through thickness crack of length 2c is: (10.4) We can see that extending very small cracks actually causes the overall energy of the system to increase because the creation of new surfaces outweighs the benefit of relieving some elastic strain. There is then an unstable equilibrium – a value of crack length for which the energy used in creating new surfaces is exactly balanced by the effect of strain energy relief. Any crack which is larger than the equilibrium length is favoured to grow and will grow catastrophically as the energy taken to create the new surfaces is more than made up for by the release of strain energy.
It is then straightforward to work out the equilibrium crack length by differentiating: for c = ce. ce =
€
EΓ πσ 2
(10.5) (10.6)
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We can plot the gradient (energy per unit crack length) as a function of c:
2Γ
ce
c
Clearly we could plot both G and Γ as positive, in which case the equilibrium crack length is found where the lines cross. Recall that Γ is the energy (per unit area of crack) which is resisting fracture. So we can define G as the energy (per unit area of crack) driving fracture such that there is a critical value Gc which corresponds to Gc=Γ. G is often called the “strain energy release rate”. It is the amount by which the mechanical energy driving the system varies per unit area of crack (not per unit time).
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Lent 2014
DH76
11 – Stable Crack Growth 11.1 Fracture toughness We see from the previous section that there are two materials parameters which determine the stress at which fracture occurs for a given crack size (or equally which determine the critical crack size which will propagate for a given tensile stress). So we can write the relationship between the stress and the crack size as:
(11.1)
K c = σ πc e
(MPa m )
(11.2)
So we have now defined a materials property, Kc as a value which depends on the stiffness and € about the critical relationship between crack length and stress. In fracture energy and which tells us
fact, more strictly we have only really looked at one particular loading geometry (known as mode I loading) and so we call this value KIc. This mode, is illustrated below, along with other possible crack opening modes.
K is often described as the “stress intensity factor”, such that Kc tells us when the stress reaches a critical value. Whilst this is evidently true, it is fundamentally based on calculating the energy of the system subjected to a given stress. Typically GIIc >> GIc (and hence KIIc >> KIc) since there is normally a lot of frictional sliding between the crack flanks during mode II loading (i.e. the crack tip is shielded from much of the applied load). The compressive failure stress of a brittle material is therefore much higher than the tensile failure stress.
IB Course D
Mechanics of Materials and Structures
Lent 2014
DH77
11.2 The Wedging Geometry In 1930, Obreimoff conducted an experiment with mica, where plates of the material were cleaved apart by inserting a wedge. This clearly creates new surface by propagating cracks, but the geometry is significantly different from cracking in tension.
Consider a wedge of thickness h inserted beneath a thin layer of material (of thickness d) attached to a block of unit width causing the thin layer to peel off.
11.3 Energy Considerations 11.3.1 The change in mechanical energy of the system, UM One contribution to consider is the work done by the force (equal to F multiplied by the distance moved in the direction of its operation). F operates at the point where the wedge meets the beam and does not therefore move in the direction of its operation (it moves sideways) so this contribution is zero.
IB Course D
Mechanics of Materials and Structures
Lent 2014
DH78
The only contribution to the mechanical energy is therefore elastic strain energy in the peeling layer from C to the point where the wedge touches the layer (O): (11.3)
IB Course D
Mechanics of Materials and Structures
Lent 2014
DH79
11.3.2 The change in surface energy of the system, US Just as with fracture in tension, breaking the adhesive bond between the mica plates causes an increase in energy of the surfaces (11.4)
11.3.3 The change in total energy of the system as a function of crack length (11.5) The variation of the individual terms and that of the sum are shown in the figure below.
This equilibrium point (the turning point) can be found by differentiating the overall energy, U(c), with respect to c and finding the value of c where this differential is equal to zero. That is where (11.6)
(11.7)
(11.8) which gives the equilibrium crack length, ce, as (11.9)
IB Course D
Mechanics of Materials and Structures
Lent 2014
DH80
If E, d and h are known, ce can be measured to give a value for Γ.
3Ed 3 h 2 3× 2 ×1011 ×10−9 × 8 2 ×10−6 Γ = = 4 8 ×10−4 8c e
€
11.4 Crack Healing If our system is truly thermodynamically reversible the crack tip should move backwards (that is the crack should heal) if we pull the wedge out. To investigate this, Obreimoff cleaved crystals of mica. On pushing the wedge in he found that the crack grew at a constant distance ahead of the crack tip, as predicted, and estimated a value of Γ for mica close to that measured elsewhere. If he tested the sample in vacuum he found that the crack would heal as he withdrew the wedge. However when the sample was tested in air this did not occur suggesting that chemical groups attach themselves to the surface of the mica, lowering, if not eliminating, the ability of the crack to heal if the wedge is withdrawn. Obreimoff's experiment clearly demonstrated the reversible, thermodynamic nature of cracking and that the principle outlined by Griffith was correct. All modern theories of fracture come from this basic idea.
IB Course D
Mechanics of Materials and Structures
Lent 2014
DH81
11.5 The Nature of Cracking 11.5.1 The importance of equilibrium The total energy of a cracked body under load, U(c), is (11.10) where UM is the change in mechanical energy of the system (a combination of the work done by the applied force and the change in elastic energy on cracking) and US is the work required to create two new surfaces. Equilibrium will occur when (11.11) or
(11.12)
From our expression for cracking in tension (11.13) and (11.14)
Hence (11.15)
So equilibrium will occur when (11.16) We could perform a similar operation for the expression we derived for wedging.
IB Course D
Mechanics of Materials and Structures
Lent 2014
DH82
11.5.2 Will failure be catastrophic? It is incorrect to say that failure will necessarily occur when
There will be some cracking but remember that complete failure also requires that (11.17) or (11.18)
In other words failure will be catastrophic when the rate of increase of the driving force with crack growth is greater than the rate of change in Γ with crack growth. So far we have taken Γ as a constant, so catastrophic when
has been zero and any crack propagation would be
>0 (which is always the case when cracks propagate in tension).
But cracking would be stable (even in tension) if (11.19) i.e. (11.20) So if we can make Γ increase with crack length, we should be able to get stable crack propagation, even in tension as long as
IB Course D
Mechanics of Materials and Structures
Lent 2014
DH83
12 – Coping with a Scatter in Strength 12.1 Introduction Virtually all materials (even metals) fail by cracking. Normally this occurs due to tension or bending loads, where the equilibrium is unstable and the failure catastrophic. Such cracks normally develop from pre-existing flaws. However the Griffith expression shows that the stress required for failure in tension is dependent on the size of the largest flaw, c, according to
.
Consider materials subjected to stresses of ~100 MPa Metals:
Ceramics:
How do we cope with this? A way around this problem is to determine how likely failure will be under a given applied stress by testing samples from a given batch of material. To design the component, one must decide what is an acceptable probability of failure and then set the stress to the associated value.
Having set the tolerable failure level, one then estimates the stress which will produce this probability of failure and use this as the design strength. You have to accept that a given number are likely fail. This is very different from designing with a metal, where one has confidence that the strength will have a given value. The scatter in strength is described using Weibull statistics.
IB Course D
Mechanics of Materials and Structures
Lent 2014
DH84
12.2 Weibull’s Approach The Weibull approach considers a chain of N links. So the survival of the chain under load requires that ALL the links survive.
(12.1) How does the probability of survival/failure of a single link depend on applied stress?
(12.2)
(12.3)
(12.4)
(12.5) 1
σ + constant σ0
hence lnln = m ln S
€
(12.6)
IB Course D
Mechanics of Materials and Structures
Lent 2014
DH85
Clearly when σ = σ0, SL = 1/e. Hence σ0 is the value of σ that 37% of the links can be subjected to without failure.
12.3 Using Weibull’s Method To determine the survival probability associated with each stress we start by testing some samples, in this case some silicon carbide bars. The strengths are (in MPa) 949, 985, 921, 771, 779, 653, 945, 732, 858 This gives a mean strength of 843 MPa. σ0 = 921 MPa. First organise the data into descending order with the highest strength being 1, the next highest 2 and the nth highest n. A probability of survival, Sn, is then assigned to each value so that (12.7) where nT is the total number of samples. If a load is applied corresponding to the highest observed strength then there would be a relatively low probability of the material surviving, in this case a 1 in 10 chance. If the load corresponds to the lowest strength observed then there is a much higher chance – 9 out of 10 – of the component surviving. Using eqn (12.6) a plot of
against ln(σ /σ0) should
give a straight line with a slope m. Using this relationship it is now possible to work out the value of the stress corresponding to the survival (or failure) probabilities that are appropriate for the design we have in mind. The strength data arranged to estimate the Weibull modulus n
σ (MPa)
S=n/nT+1
ln (σ/σ0)
ln ln (1/S)
1 2 3 4 5 6 7 8 9
985 949 945 921 858 779 771 732 653
0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9
0.067 0.030 0.026 0 -0.071 -0.167 -0.178 -0.230 -0.344
0.83403 0.47588 0.18563 –0.08742 –0.36651 –0.67173 –1.0309 –1.4999 –2.2504
IB Course D
Mechanics of Materials and Structures
Lent 2014
And we can extrapolate this data back to an acceptable value of S.
For S = 0.99999 (F = 10-5): lnln(1/S) = -11.5
This usable strength (so that only 10 per million fail) is much less than the average.
DH86
IB Course D
Mechanics of Materials and Structures
Lent 2014
DH87
12.4 Practical Solutions If a component is being manufactured to support a given load, there are a number of possible ways in which to ensure that fewer components fail in service. 1. Redesign This is rather obvious, but by increasing the cross-sectional area a given load can be supported under a lower stress. Drawbacks include:
2. Change material It may be possible to use a material with a higher average failure strength or a larger Weibull modulus, but this may involve increased cost and there may be other materials properties which cannot be met. Alternatively it may be possible to toughen the material (increasing its fracture energy) by producing a composite.
3. Improve processing If the component is a ceramic then the processing route will play a large part in determining the flaw size. By changing the processing method (i.e. hot isostatic pressing at larger pressures, higher temperatures or for longer times) it may be possible to remove the larger flaws.
4. Throw some components away It may seem odd but the most cost-effective method is simply to throw away (or better, recycle) a number of the components. If all the components are proof tested up to the stress which they will be subjected to in service, a small number of them will fail, but these will be the ones which had the largest flaws and the remainder will have been shown not to contain flaws which propagate under the service load.
IB Course D
Mechanics of Materials and Structures
Lent 2014
DH88
12.5 Toughening 12.5.1 Crack bridging It is commonly observed that cracks can grow stably in a structure over a period of time. This happens even though the body is being loaded in an unstable manner (i.e. in tension), so that we would expect any cracking to occur catastrophically. Oddly this phenomenon tends to occur in tougher materials. This suggests that toughening does not simply increase the magnitude of the fracture energy but changes the way in which a crack grows. Consider a material toughened by crack bridging in which intact ligaments across the crack faces are left behind as the crack grows. Toughening occurs because separating the crack faces then requires that extra work is done in order to either stretch the ligament or pull it out of the matrix in which it is embedded. The fibres are normally discontinuous.
IB Course D
Mechanics of Materials and Structures
Lent 2014
DH89
12.5.2 R-curves Bridging leads to a change in Γ with c. This type of behaviour is known as R-curve behaviour. We can see that as we load the material containing a small flaw, it will begin to grow (under an increasing applied crack driving force) until the process zone is fully developed. Once the zone has developed fully then the whole crack will move forward with the process zone size remaining a constant size. As process zones exist in all toughened materials (that is where Γ > 2γ), we might expect that they would all show R-curves and this is the case as shown below.
IB Course D
Mechanics of Materials and Structures
Lent 2014
DH90
Γ
c
Γ
c
By controlling the microstructure and properties of the material to give R-curve behaviour then we can (over certain limits of flaw size) ensure that cracks grow in a stable manner even though the loading state would normally have caused catastrophic failure.
IB Course D
Mechanics of Materials and Structures
Lent 2014
DH91
Appendix Summary of Materials Response in Compression and Tension A.1 In Compression A.1.1 Structures If compression elements fail by crushing then the behaviour really just depends on the crosssectional area (and the compressive strength), but if it occurs by buckling then the failure stress depends on (i) the slenderness ratio and (ii) the elastic modulus of the beam compressive failure by crushing or plastic yielding
Elastic buckling
L/R Hence if we decide to change the: Cross sectional shape: doesn’t affect crushing subdividing is bad for buckling hollow tubes help prevent elastic buckling
Length: Making longer makes buckling more likely
(but worry about local buckling)
IB Course D
Mechanics of Materials and Structures
Lent 2014
DH92
A.1.2 Materials If the structure is short and stocky, we need to design against crushing/yielding. Compressive strengths of ceramics are very high and so bricks, stone and concrete are a good (cheap!) option as long as weight is not a factor.
If the structure has a large aspect ratio we need to be more careful. Difficult to improve material’s resistance to buckling (just depends on E). Use stiff material, such as ceramic (e.g. stone pillars in buildings) -make it wide enough to have a large safety margin If applied load is intermittent, may be OK if material buckles and recovers. (e.g. grass)
A.2 In Tension A.2.1 Structures If we decide to change the: Cross sectional shape: Subdivide – no adverse effect on tensile failure – can use wire instead of rod In brittle materials, subdividing is good to prevent possibility of large flaws
IB Course D
Mechanics of Materials and Structures
Lent 2014
DH93
Length: Longer structures are more likely to contain significant cracks (Weibull analysis- more links)
Tensile strengths are usually much larger than those which would cause buckling, hence it is usually most efficient to use primarily tensile elements.
A.2.2 Materials Metals large Γ, less susceptible to fast fracture than ceramics. Composites Addition of fibres improves toughness.
A.3 Conclusions Tension elements tend to be long and sub-divided. Compression elements tend to be shorter and made from stiffer materials. It is what we see in structures around us. Think of the difference between a compression structure such as a stone bridge and a structure with more tensile elements such as the Viaduc de Millau.
IB Course D
Mechanics of Materials and Structures
Lent 2014
Nomenclature Symbols 2
δ
deflection (m)
A
cross-sectional area (m )
ε
(normal) strain (-)
c
crack length (m) constant in elastic buckling expression which depends on end-constraints (-)
εij
strain tensor (-)
εN
nominal strain (-)
d
(inner) diameter or linear dimension (m)
εT
true strain (-)
D
(outer) diameter or linear dimension (m)
θ
angle (rad. or ˚)
eij
relative displacement tensor (-)
κ
curvature (m–1)
E
Young’s modulus (Pa ≡ N m–2)
ν
Poisson ratio (-)
F
force (N) probability of failure (-)
ρ
density (kg m–3)
σ
normal stress (Pa ≡ N m–2)
σij
stress tensor (Pa ≡ N m–2)
–2
G
strain energy release rate (J m ) shear modulus (Pa ≡ N m–2)
h
height of beam (m)
I
second moment of area (m4)
k
constant (various)
K
stress intensity factor (Pa m1/2) bulk modulus (Pa ≡ N m–2)
σN nominal stress (Pa ≡ N m–2) σT
true stress (Pa ≡ N m–2)
Σ
beam stiffness or flexural rigidity (N m2)
τ
shear stress (Pa ≡ N m–2)
φ
angle (rad. or ˚)
L
distance or length (m)
m
Weibull modulus (-)
Subscripts
M
bending moment (N m)
0
N
number of chain links (-) number of cycles to failure (-)
base level average
1
first, in direction 1 etc
P
pressure (Pa)
I, II
loading mode
r
radial distance or radius (m)
c
critical
R
radius of curvature (m)
e
equilibrium
radius (m)
EB
Elastic buckling
s
aspect ratio (-)
E
stored elastically
S
probability of survival (-) stress amplitude (Pa)
F
due to movement of a force failure
t
thickness (m) time (s)
i
in the i direction
j
in the j direction
T
temperature (K or ˚C) torque (N m)
L
link
U
energy (J)
LB
Local buckling
v
velocity (m s-1)
M
mechanical
V
volume (m3)
P
polar
w
width (m)
r
in radial direction
W
energy (J)
S
surface
x
distance (along a principal axis) (m)
T
total
y
distance (along a principal axis) (m)
Y
yield
z
distance (along a principal axis) (m)
x
in x direction in y direction
γ
(engineering) shear strain (-) surface energy (J m–2)
y z
in z direction
Γ
energy resisting crack growth (J m-2)
θ
in hoop direction
DH94
IB Course D
Mechanics of Materials and Structures
Lent 2014
DH95
Glossary Words in italics provide cross-references to other entries in the glossary. Anti-symmetrical (2nd Rank Tensor). Tensor for which the values of the off-axis components (Sij, with i≠j) are reflected across the diagonal, and their signs changed, so that Sij = -Sji. Beam Stiffness, EI. Product of Young’s modulus and second moment of area, characterising the resistance to deflection when subjected to a bending moment. Sometimes termed the flexural rigidity. Bending Moment, M. Turning moment generated in a beam by a set of applied forces. The bending moment is balanced at each point along the beam by the moment of the internal stresses. Brittle Fracture. Fracture not involving gross plastic flow (although some local plastic deformation may occur at the tip of the crack). Bulk Modulus, K. Constant of proportionality between hydrostatic stress and hydrostatic strain, for an isotropic body. Compliance Tensor, Sijkl. Tensor of 4th rank reflecting the elastic properties of a material. The product with the stress tensor is the strain tensor. Crushing. A compressive failure mode for a brittle material, arising from propagation of cracks due to shear stresses. Curvature, κ. Reciprocal of the radius of curvature adopted by a beam subject to a bending moment. Also equal to the through-thickness gradient of strain in the beam. Deviatoric Component. Component of a second rank tensor representing the shape change (in the case of strain) or the driving force for shape change (in the case of stress). Diagonalising (a 2nd Rank Tensor). Transforming a tensor so that it is referred to the principal planes, and hence has non-zero terms only along the diagonal. Dilation, Δ. Volume change associated with a strain, given by the sum of the diagonal components of the strain tensor. Driving Force for Crack Extension. Any contribution which would decrease the total energy of a system if a crack were to propagate, e.g. mechanical energy (a negative value if the energy decreases). Ductile Fracture. Fracture involving gross plastic flow. Einstein Summation Convention. Convention relating to abbreviated formulation of sets of equations, stating that, when a suffix occurs twice in the same term, then this indicates that summation (from 1 to 3) should be carried out with respect to that term. Elastic Buckling. See Euler Buckling.
IB Course D
Mechanics of Materials and Structures
Lent 2014
DH96
Engineering Shear Strain, γ ij. Distortional deformation (angle) of a body arising from a shear stress (actually a pair of shear stresses). Equilibrium Crack Length. The crack length at which (under a given loading scenario) the energy of the system is a maximum (for an unstable cracking scenario) or a minimum (for a stable cracking scenario). Euler Buckling. An elastic failure mode, whereby a thin column in compression becomes unstable against lateral perturbations. Fatigue. A common failure mode in materials subjected to stresses which vary with time. A crack may propagate gradually over millions of cycles before the reduced cross sectional area of the component leads to sudden fracture. Flexural Rigidity. See beam stiffness. Fracture Energy, Gc. Energy per unit area required to extend a crack. Also termed the critical strain energy release rate. See surface energy. Fracture Toughness Kc. Critical value of the stress intensity factor, characterising the resistance of a material to crack propagation. Griffith Criterion. Energy-based condition for fracture. Hooke’s Law. Stress is proportional to strain during elastic deformation. Hydrostatic Component. Component of a second rank tensor representing the volume change (in the case of strain) or the driving force for volume change (in the case of stress). Invariants (of the Secular Equation). Coefficients of the secular equation, which are independent of the choice of axes. These combinations of the components of the tensor therefore remain unchanged during transformation of axes. Isotropic Material. A material in which the properties of interest are the same in all directions within it. Local Buckling. A failure mode for hollow cylinders in compression. If the wall thickness is small, the walls may crumple under a lower stress than is required for either elastic buckling or compressive failure. An empty soft-drinks can is a good example. Mechanical Energy (of a Loaded Body). The total mechanical energy of the system arising from the stored elastic energy within the material plus the energy contribution to the system if the load moves and does work . Mohr’s Circle. Geometrical construction for solving the secular equation, applicable only in a principal plane. Neutral Axis (of a beam). Axis (strictly, a plane) parallel to the length of a beam, along which there is no change in axial length on bending. Nominal (or engineering) strain. The strain based on the original length of the sample.
IB Course D
Mechanics of Materials and Structures
Lent 2014
DH97
Nominal (or engineering) stress. The stress based on the original cross-sectional area of the sample. Normal Strain, ε ii. Deformation in which change in length is parallel to original length. Normal Stress, σ ii. Stress induced by a force acting normal to the sectional area to which it is applied. Poisson Ratio, ν . Ratio of the transverse contraction to the axial extension induced in an isotropic body by an applied normal stress. Polar Second Moment of Area, IP. Parameter dependent on sectional shape, characterising the resistance a body offers to twisting under an applied torsional moment, or torque. Principal Directions. Normals to the principal planes. Principal Planes. A unique set of orthogonal planes, on which there are no shear stresses acting (or on which there are no shear strains). Principal Strains, ε i. A unique set of normal strains, acting on the principal planes, which represents the strain state of a body. Principal Stresses, σ i. A unique set of normal stresses, acting on the principal planes, which represents the stress state of a body. R-curve. A plot of the fracture energy vs. crack length in systems where the fracture energy increases (towards some upper limit) as the crack grows. Relative Displacement Tensor, eij. Second rank tensor giving the displacement of points within a body. Resistance to Crack Extension. Any contribution which would increase the total energy of a system if a crack were to propagate, e.g. surface energy (a positive value if the energy increases). Rotation Tensor, ω ij. Anti-symmetrical second rank tensor representing the rigid body rotation component of the relative displacement tensor. Second Moment of Area, I. Parameter dependent on sectional shape, characterising the resistance a beam offers to deflection under an applied bending moment. Also called the moment of inertia. Secular Equation (of a 2nd Rank Tensor). A cubic equation, the roots of which are the principal values of the tensor. Shear Modulus, G. Constant of proportionality between shear stress and (engineering) shear strain, when an isotropic body is subjected to a single shear stress (actually a pair of shear stresses, although this is not taken into account in the definition). Shear Stress, σ ij. Stress induced by a force acting parallel to the sectional area to which it is applied.
IB Course D
Mechanics of Materials and Structures
Lent 2014
DH98
Slenderness ratio. The ratio L/R for a column of length L and radius R. Stiffness Tensor, Cijkl. Tensor of 4th rank reflecting the elastic properties of a material. The product with the strain tensor is the stress tensor. Strain Energy Release Rate, G. Elastic strain energy released per unit of created crack area. Strain Tensor, ε ij. Symmetrical second rank tensor representing the strain component of the relative displacement tensor. Stress Intensity Factor, K. Parameter characterising the crack driving force, in terms of applied stress level and crack length. Surface Energy, γ. The energy, γ, associated with one square metre of surface. Note that the fracture energy, Γ, of an ideally brittle material is the energy required to form two new surfaces, each of unit area. Hence Γ =2γ. Symmetrical (2nd Rank Tensor). Tensor for which the values of the off-axis components (Sij, with i≠j) are reflected across the diagonal, so that Sij = Sji. Tensor (of nth Rank). Array of values (with n dimensions) representing either an imposed field (eg velocity, thermal gradient, stress etc) or properties of a specified type of matter (eg stiffness, conductivity etc). Torque, T. Twisting moment generated in a body by a set of applied forces, which tend to rotate the body about an axis lying within it. Torsional Rigidity. Product of shear modulus and polar second moment of area, characterising the resistance to twisting when subjected to a torque. Toughening. Some mechanism which increases the resistance to crack extension. Examples are crack bridging and plastic deformation. Transform Matrix. Array of terms which, when concatenated with a tensor, effects a transformation of axes operation on the tensor. Transformation of Axes (of a Tensor). Re-evaluation of the components of a tensor, so that they relate to a different set of reference axes. True Stress. The stress based on the instantaneous cross-sectional area of the sample. True Strain. The strain based on the instantaneous length of the sample. Twisting Moment. See torque. Wedging. A loading geometry where a wedge exerts a force in a direction normal to the direction in which it moves. Weibull Modulus. A parameter which quantifies the variability in failure strengths of components.
IB Course D
Mechanics of Materials and Structures
Lent 2014
DH99
Weibull Statistics. A method of predicting the failure strength distribution of brittle components. As there will always be some range of failure strengths, Weibull statistics allow us to predict the probability of failure at a given stress. Young’s Modulus, E. Constant of proportionality between normal stress and normal strain, when an isotropic body is subjected to a single principal stress.
DQ1
IB Materials Science - Course D: Mechanics of Materials and Structures
DQ1
Question Sheet DQ1 1.
At a particular location on the earth’s surface, magnetic north is 10˚ E of true north. Write down a 3×3 transform matrix for converting a velocity measured using a compass to establish its direction to one which is specified relative to the true geographical directions.
2.
A metallic beer can, of radius r and wall thickness t (<
0 σ1 2 0
0 0 0
where σ1 is given by Pr/t. Sketch Mohr’s circle for each of the 3 principal planes. Hence derive an expression for the peak shear stress experienced by the material of the can and, on a sketch of a volume element, indicate the planes on which it acts. If t = 0.3 mm, r = 30 mm and the shear yield stress of the material is 15 MPa, what pressure would be needed to cause yielding? On a hot day, and after being agitated, the pressure in the beer can reaches 2 bar (about two thirds of that needed to cause yielding). It is then subjected to a torque of 15 N m, applied about the axis of the can. Calculate the new value of the peak shear stress in the can wall and decide whether the material will yield under these new conditions. 3.
(a) Separate the following tensors into symmetrical and antisymmetrical components:
If these tensors were describing general strains, what would be the significance of the symmetrical and antisymmetrical components? (b) The stress tensor is symmetrical. Separate the following stress tensors into hydrostatic and deviatoric components:
(c) Diagonalise the second stress tensor given in part (b). Separate the diagonalised tensor into hydrostatic and deviatoric components. Show that the deviatoric stress can be expressed as the sum of three pure shear stresses. -1-
PDB: LT14
DQ1
4.
IB Materials Science - Course D: Mechanics of Materials and Structures
DQ1
(a) Describe what is meant by a principal stress. If σ3 is a principal stress write down a secular equation which can be used to determine the two other principal stresses, σ1 and σ2. Solve this equation to show that
where σii and σij are normal and shear stresses respectively. (b) Describe a possible graphical interpretation of the above solution and how it can be used to characterize a state of plane stress. (c) An elastic material, in a two-dimensional stress state, has normal stresses of 85 MPa (tensile) and 35 MPa (compressive) acting on two mutually perpendicular planes. A shear stress also acts on these planes. Of the two principal stresses present the largest one is limited to 110 MPa (tensile). Use Mohr’s circle to determine (i) the shear stress (ii) the second principal stress (iii) the normal stress on the plane of maximum shear (iv) the maximum shear stress and (v) the orientation of the principal planes. Confirm the results analytically. [modified Tripos 2012]
-2-
PDB: LT14
DQ2
IB Materials Science - Course D: Mechanics of Materials and Structures
DQ2
Question Sheet DQ2 1. (a) Show how the following stress states can be represented using Mohr’s circles and write down the corresponding stress tensors: (i) uniaxial compression (ii) pure shear and (iii) hydrostatic pressure. (b) Use the relative displacement tensor and suitable sketches to distinguish between states of pure shear and simple shear and show how this difference affects the definition of the shear modulus. (c) Express a general stress tensor, referred to principal axes, in terms of its hydrostatic and deviatoric components. Show that the deviatoric stress can be written as the sum of three pure shear stresses. (d) Determine the pure shear components of the following stress tensor and show that the hydrostatic component is zero.
" 1 2 0 % $ ' $ 2 1 0 ' $ 0 0 !2 ' # & [modified Tripos 2013]
2.
(a) Define the terms principal stress and principal strain. Describe how these quantities are computed from a general stress or strain state. Determine the principal stresses of the following stress state: σ σ σ σij = σ σ σ σ σ σ (b) Consider an isotropic material subjected to a set of principal tensile stresses σ1, σ2 and σ3. € the resulting principal strains are given by Show that 1 σ − ν (σ 2 + σ 3 ) E 1 1 ε 2 = σ 2 − ν (σ1 + σ 3 ) E 1 ε 3 = σ 3 − ν (σ1 + σ 2 ) E
ε1 =
(
)
(
)
(
)
where E is Young’s modulus and ν is Poisson’s ratio. € 1
PDB: LT14
DQ2
IB Materials Science - Course D: Mechanics of Materials and Structures
DQ2
(c) An isotropic material is subjected to a compressive load in the 1-direction but restrained from extending in the 2 and 3-directions. Show that the Young’s modulus of the material
2ν 2 effectively changes from E to E 1 − (1 − ν ) (d) A uniform bar of isotropic material 10 cm long and 2 cm square is loaded in compression € bar is laterally restrained in the plane normal to the load, determine the along its length. If the change in length of the bar resulting from a load of 40kN. (For the material E = 100 GPa, ν = 0.3). [modified Tripos 2011]
3.
Determine an expression for the second moment of area of the semicircular section shown below with respect to the neutral axis (NA) parallel to its base.
4.
(a) When a uniform beam carries a transverse load it experiences a bending moment and deflects. Derive a relationship between the bending moment, the beam curvature and the beam stiffness assuming the defection is small.
(b) A uniform cantilever of length L and stiffness EI (E = Young’s modulus, I = moment of inertia), is loaded at its free end as shown in the schematic diagram below. Show that the deflection along the beam is given by
y=
F x 3 L2 x L3 + − EI 6 2 3
€
2
PDB: LT14
DQ2
IB Materials Science - Course D: Mechanics of Materials and Structures
DQ2
(c) Where along its length does the beam exhibit its greatest slope ( dy dx ) and its greatest curvature ( d 2 y dx 2 ) and state their values?
€ below, derive an expression for the (d) If the cantilever is now loaded at its mid-point as shown € deflection at the free end. Note that the relationship derived in (a) now only applies to the left hand side of the beam.
[Tripos 2011]
3
PDB: LT14
DQ3
IB Materials Science - Course D: Mechanics of Materials and Structures
DQ3
Question Sheet DQ3 1. (a) Determine an expression for the second moment of area of the equilateral triangle shown below with respect to its base (the x-axis).
(b) Consider three freely hinged, unconstrained, columns of the same material and length but with the following cross-sections (i) a circle (ii) a square and (iii) an equilateral triangle. If the volume of each column is the same show that the critical compressive load for elastic buckling of the square and triangular columns is about 5% and 21% greater respectively than that of the circular column. (c) Three identical columns, which have cross-sections that are equilateral triangles, are placed together without bonding as shown below and loaded axially in compression. Assuming freely hinged, unconstrained, conditions show that the critical load for elastic buckling increases by a factor of five if the columns are bonded along their lengths.
[Tripos 2012]
1
PDB: LT14
DQ3
IB Materials Science - Course D: Mechanics of Materials and Structures
DQ3
2. (a) Define the terms bending moment and beam stiffness. Show that the deflection y(x) along the axis of a uniform beam subject to a bending moment M is given by y(x) =
M
∫∫ EI dx dx + Ax + B
where E is Young’s modulus, I is the second moment of area and A and B are constants. State the conditions under which this expression is valid and describe how the constants € are determined. (b) A simply supported uniform beam of length L carries a central concentrated load F as shown below.
Using the centre of the beam as the origin show that the deflection along its length is given by 3 Fx Lx 2 L3 y(x) = − + EI 12 8 48 Sketch the variation of bending moment and curvature along the length of the beam. (c) If the€loading configuration in (b) is changed to that shown below, write down expressions for the bending moments along the regions of the beam defined by 0 ≤ x ≤ L/6 and L/6 ≤ x ≤ L/2.
Hence sketch the new variation of bending moment along the beam. How will this variation change as each load is successively divided in half and configured symmetrically along the beam? [Tripos 2012]
2
PDB: LT14
DQ3
IB Materials Science - Course D: Mechanics of Materials and Structures
DQ3
3. A solid cylindrical rod of radius 3 mm is tested in uniaxial compression with its ends free to rotate. (a) Describe the failure mechanism of the rod for (i) very short lengths (several millimetres) and (ii) very long lengths (several metres). (b) Stating your assumptions, calculate how the failure load depends on sample length, and plot this on graph paper. (c) The solid rod is replaced with a hollow rod which has an external diameter of 10 mm, but the same mass per unit length as the solid rod. Plot the failure load vs length for this hollow rod on the same graph. [Tripos 2011] 4. Consider a simply supported beam of length L which is subject to symmetrical 4point bending as shown below.
Show that the maximum deflection is given by y max =
Fa 3L2 − 4a 2 24 EI
(
)
€
3
PDB: LT14
DQ4
IB Materials Science - Course D: Mechanics of Materials and Structures
DQ4
Question Sheet DQ4 1. (a) Explain, with an example, how sample size affects the average failure strength of a brittle material in tension. Using Weibull statistics show that the probability of failure increases exponentially with sample volume. (b) If a rod of brittle material is subject to a transverse (bending) load rather than a tensile load explain whether the average failure strength increases, decreases or remains the same. (c) Given below are the tensile strengths (in MPa) of a batch of graphite/epoxy composite samples. Determine the Weibull modulus and comment on its value. Estimate the usable strength of the composite if the tolerable failure rate in practice must be 1 in 104. 509, 452, 571, 555, 536, 522, 498, 468, 477 (d) Estimate how the mean strength of the samples would change if their volume doubled or halved. [Tripos 2013] 2. (a) Explain why two samples fabricated from the same brittle material using an identical process can fail at different tensile loads. (b) Fifty samples were tested to failure in uniaxial tension by loading them in increments of 100 g. The accumulated mass at which each sample failed is listed below. Calculate the mean failure load. Accumulated mass / g 700 800 900 1000 1100 1200 1300 1400 1500 (c)
No. of samples which failed 0 4 5 7 10 11 8 4 1
Make a Weibull plot of the data and calculate the Weibull modulus.
1
PDB: LT14
DQ4
IB Materials Science - Course D: Mechanics of Materials and Structures
DQ4
(d)
Estimate the load for which the probability of failure is a) 0.1 and b) 0.001. Comment of the relative confidence in these estimates.
(e)
A new batch of samples is fabricated using the same processing method. They have the same cross-sectional area, but are three times longer. Explain whether the average tensile strength is likely to increase, decrease or remain the same. Estimate the median tensile failure load of these longer samples. [Tripos 2009]
3. (a) Explain why, under certain conditions, a crack grows so that the body fails catastrophically, whereas under other conditions crack growth may start and then stop. (b) A wedge 500 µm thick is used to cleave a surface coating, with a Young’s modulus of 200 GPa, off a thick, square substrate with a side length of 40 mm. Assuming that the fracture energy of the interface between the substrate and the coating is 1 Jm-2, show how increasing the coating thickness from 200 µm to 1 mm might influence the way the crack is seen to grow. [Tripos 2004] 4.
(a) Explain how crack bridging gives rise to toughening and comment on the microstructural features that influence the magnitude of the toughening. (b) A material contains elongated grains 10 µm in length that act as bridging ligaments and has a fracture energy (Γ) of 250 Jm-2 and a Young’s modulus (E) of 380 GPa. Stating your assumptions, estimate the length of the zone where the ligaments bridge the crack. [In a homogeneous material, the crack opening (2u) is 1/ 2 4K r related to the distance from the crack tip (r ) by the expression u = where E 2π
K is the fracture toughness]. €
2
[modified Tripos 2003]
PDB: LT14
IB Course D
Mechanics of Materials and Structures
Lent 2014
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IB Course D
Mechanics of Materials and Structures
Lent 2014
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IB Course D
Mechanics of Materials and Structures
Lent 2014
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IB Course D
Mechanics of Materials and Structures
Lent 2014
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