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1Continuum Mechanics – HistoricalBackground

The fundamental problem that faces a structural engineer, civil, mechanical or aeronautical is to makeefficient use of the materials at their disposal to create shapes that will perform a certain function withminimum cost and high reliability whenever possible. There are two basic aspects of this process selectionof materials, and then selection of shape. Material scientists, on the basis of the demand generated byapplications, devote their efforts to creating the best possible materials for a given application. It is up tothe designer of the structure or mechanical component to make the best use of these materials by selectingshapes that will simultaneously provide the transfer of forces acting on the structure or component inan efficient, safe and economical fashion. Today, a designer has a variety of tools to achieve thesebasic goals.

These tools have evolved historically through a heritage that can be traced back to the great buildersof structures in 2700 BC Egypt, Greece and Rome, to the builders of cathedrals in the Middle Ages.Throughout the ancient and medieval period structural design was in the hands of master builders,helped by artisan masons and carpenters. During this period there is no evidence that structural theoriesexisted. The design process was based on empirical evidence, founded many times in trial and errorprocedures done at different scales. The Romans achieved great advances in structural engineering,building structures that are still standing today, like the Pantheon, a masonry semi-spherical vault witha bronze ring to take care of tension stresses in the right place. It took many centuries to arrive at thebeginning of a scientific approach to structures. It was the universal genius of the Renaissance LeonardoDa Vinci (1452–1519) one of the first designers that gives us evidence that scientific observations andrigorous analysis formed the basis of his designs. He was also an experimental mechanics pioneer andmany of his designs were based on extensive materials testing.

The text that follows will introduce the names of the most outstanding contributors to some of the basicideas of the mechanics of the continuum that we are going to review in this chapter. The next chapterprovides background on those who contributed further in the nineteenth century and early twentieth. Inthe twentieth century many of the basic ideas were reformulated in a more rigorous and comprehensivemathematical framework. At the same time basic principles were developed to formulate solid mechanicsproblems in terms of approximate solutions through numerical computation: Finite Element, BoundaryElement, Finite Differences.

The birth of the scientific approach to the design of structures can be traced back to Galileo Galilei.In 1638 Galileo published a manuscript entitled Dialogues Relating to Two New Sciences. This bookcan be considered as the precursor to the discipline Strength of Materials. It includes the first attempt

Experimental Mechanics of Solids, First Edition. Cesar A. Sciammarella and Federico M. Sciammarella.© 2012 John Wiley & Sons, Ltd. Published 2012 by John Wiley & Sons, Ltd.

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2 Experimental Mechanics of Solids

to develop the theory of beams by analyzing the behavior of a cantilever beam. A close successor ofGalileo was Robert Hook, curator of experiments at the Royal Society and professor of Geometry atGresham College, Oxford. In 1676, he introduced his famous Hooke’s law that provided the first scientificunderstanding of elasticity in materials.

At this point it is necessary to mention the contribution of Sir Isaac Newton, with the first systematicapproach to the science of Mechanics with the publication in 1687 of Philosophiae Naturalis PrincipiaMathematica. There is another important contribution of Newton and Gottfried Leibniz that helpedin the development of structural engineering; they established the basis of Calculus, a fundamentalmathematical tool in structural analysis.

From the eighteenth century, we must recall Leonard Euler, the mathematician who developed manyof the tools that are used today in structural analysis. He, together with Bernoulli, developed the funda-mental beam equation around 1750 by introducing the Euler-Bernoulli postulate of the plane sectionswhich remain plane after deformation. Another important contribution of Euler was his developmentsconcerning the phenomenon of buckling.

From the nineteenth century we recognize Thomas Young, English physicist and Foreign Secretary ofthe Royal Institute. Young introduced the concept of elastic modulus, the Young’s modulus, denoted asE, in 1807. The complete formulation of the basis of the theory of elasticity was done by Simon-DenisPoisson who introduced the concept of what is called today Poisson’s ratio.

Ausgustin-Louis Cauchy (1789–1857) the French mathematician, besides being an outstanding con-tribution to mathematics was one of the early creators of the field of what we call continuum mechanics,both through the introduction of the concept of stress tensor as well his extensive work on the theory ofdeformation of the continuum.

Claude-Louis Navier (1785–1836), a French engineer, professor of the Ecole de Ponts et Chaussees inParis, is considered to be the founder of structural analysis by developing many of the equations requiredfor the solution of structural problems and applying them to the construction of bridges.

Another contributor to the basic equations of the continuum is Gabriel Lame (1795–1870) Frenchmathematician, professor of physics at L’Ecole Polytechnique and professor of probability at the Sor-bonne and member of the French Academy. He made significant contributions to the elasticity theory(the Lame constants and Lame equations). He was one of the first authors to publish a book on the theoryof elasticity. In 1852 he published Lecons sur la theorie mathematique de l’elasticite des corps solides.Another outstanding contributor to the foundations of the mechanics of solids is the French engineerand mathematician Adhemar-Jean- Claude Barre de Saint Venant (1797–1880). His major contributionswere in the field of torsion and the bending of bars and the introduction of his principle that is keyto the formulation of the solutions in the continuum. The original statement was published in Frenchby Saint-Venant in 1852. The statement concerning his principle is to be found in Memoires sur latorsion des prismes. The Saint-Venant’s principle has made it possible to solve elasticity problems withcomplicated stress distributions, by transforming them into problems that are easier to solve.

G. B. Airy (1801–1892) mathematician and professor of Astronomy at Cambridge, introduced in 1862the concept of stress function. The idea of stress function was applied by Lame in his work on thickwalled vessels, by Boussineq in his work of contact stresses and by Charles Edward English, professorat the Department of Engineering at Cambridge University who applied the idea of stress functionsto the solution of problems of stress concentration (1913). August Edward Hough Love (1863–1940),English Mathematician Professor of Natural Philosophy at Oxford author of many papers on the field ofElasticity, author of, A treatise in the Mathematical Theory of Elasticity, first published in 1892.

Tulio Levi-Civita (1871–1941), professor of Rational Mechanics at the University of Padova. He wasone of the outstanding mathematicians of the 19th century. He introduced the idea of tensors and tensorcalculus that played a fundamental role in the field of mechanics of solids and in the Theory of Relativity.The contributors to the mechanics of solids includes the names of many outstanding mathematiciansand physicists of the nineteenth century: James Clerk Maxwell, H. Herzt, Eugenio Beltrami, John HenriMitchell, Carlo Alberto Castigliano, Luigi Federico Menabrea.


Continuum Mechanics – Historical Background 3

Let us start with a basic approach to see how these different schools of thought are utilized. Hereis the scenario: Given a certain body subjected to given loads and given form of support what arethe stresses? In strength of materials (i.e., buckling of columns, late eighteenth century) assumptionsare made on how body deformations occur and from that stress distributions are obtained. Forthis approach intuition and experimental measurements are necessary in order to provide an educatedguess of how the body deforms. From deformations strains are obtained and then, by using elastic law,stresses are obtained.

Theory of elasticity, a mathematical model of the behavior of materials subjected to deformations(formalized in the late nineteenth and early twentieth century) has a different approach. In theory ofelasticity there is no need to make any assumptions in the way the body deforms. All that is needed tosolve the problem is:

1. Certain differential equations; and2. The postulated boundary conditions for the body.

If the solution meets all the conditions of the theory it is possible to say that an exact solution wasachieved. At this stage the following question may be asked: What value does this solution have? Ifexperiments are performed using (experimental mechanics) the solution that was obtained using thetheory of elasticity will be in agreement with the experiment within a certain number of significantfigures. It should be noted that using the theory of elasticity is more complicated than using the strengthof materials approach, but it is worth understanding.

The main reason why the theory of elasticity is worth using is because it yields solutions that wouldnot be possible to get using strength of materials. A very simple example of this concept is the case ofbending a beam. Strength of material gives the strain and the stress distribution of a section of a beambut these distributions are the correct answers under special conditions: pure bending and away from theapplied load. If we have a beam with a concentrated load the stress distribution in the section where theload is applied will be quite different from that given in strength of materials. In many cases the solutionof theory of elasticity agrees with strength of materials solutions, but the understanding that comes fromtheory of elasticity allows us to have a good grasp of the validity of the solutions. In particular it ispossible to know when the solutions can be applied to a particular problem.

Today, numerical techniques (i.e., Finite Element Analysis “FEA”) are used in almost all applications.A FEA practically provides the solution for any possible problem of the theory of elasticity. One maygo so far as to say that FEA is all that is necessary to solve problems. However, it should be mentionedat this point that the ability of numerical analysis to provide the solutions is due to the understandinggained through theory of elasticity and continuum mechanics. Another very important distinction shouldbe made between the solution obtained by theory of elasticity and one that is obtained by a numericalmethod. The theory of elasticity solution provides the answer for all possible solutions of a given problem.The numerical solution provides the answer for specific dimensions and loads. For example, if one wantsto analyze what influence a given variable has on a given problem, this can be done in FE but it willrequire continual computations for all the range of values of interest of the variable. If one knows thetheory of elasticity solution the effect of a variable can be deduced directly from this solution. At thisstage of our knowledge the possibility of obtaining solutions directly from the theory of elasticity islimited and hence numerical techniques such as FE allow us to solve numerically any possible problemof the theory of elasticity if we have correct information concerning the boundary conditions and theinitial conditions in time if we have dynamic problems.

What follows is a review of the basic concepts upon which the theory of the continuum is built.Continuum mechanics is a branch of classical mechanics. It deals with the analysis of the kinematicsand the mechanical behavior of materials modeled as a continuous rather than as an aggregate of discreteparticles such as atoms. The French mathematician Augustin Louis Cauchy was the first to formulatethis model in the early nineteenth century. The continuum model is not only utilized in mechanics, but



also in many branches of physics. It is a very powerful concept that helps in the mathematical modelingof complex problems. A continuum can be continually sub-divided into infinitesimal elements whoseproperties are those of the bulk material. The continuum hypothesis has at its basis the concepts of arepresentative volume element. What is a representative volume element? It is an actual volume, withgiven dimensions. To this volume we can apply continuum mechanics and get results that can be verifiedby experimental mechanics. It is a concept that depends on scales, for example, when we consider alarge structure like a dam, the representative volume may be in the order of centimeters, if we considera metal the representative volume will be of the order of 10 microns or less. What we measure inexperimental mechanics is a certain statistical average of what occurs at the level of the microstructure.This characteristic of the continuum model leads us to ambiguities in language, for example, when wetalk of properties at a point of the continuum we are in reality referring to the representative volume thathas a definite size.

1.1 Definition of the Concept of StressThe concept of stress is one of the building blocks of continuum mechanics. The stress vector at a pointis defined as a force per unit area as in

�� = lim �F�A

�A → 0(1.1)

where �F denotes the force acting on �A, this vector depends on the orientation of the surface definedby its normal. This vector is not necessarily normal to the surface.

The stress vector does not characterize the state of stress at a given point of the space in the continuum.The state of stress is characterized by a more complex quantity know as the stress tensor �ij. The stresstensor has nine components, of which only six are independent. The stress components are representedin a Cartesian system of coordinates by the stress Cartesian tensor that was originally introducedby Cauchy.

[�] =∣∣∣∣∣∣

�x �xy �xz�yx �y �yz�zx �zy �z

∣∣∣∣∣∣

(1.2)

The cube shown in Figure 1.1 represents the stress tensor at a point with its nine components(

�ij = �ji)

.This definition has the ambiguity in language we have pointed out before. Figure 1.1 represents a cube

in the continuum, but as we said before ideally it represents a system of three mutually perpendicularplanes that go through a point. Each of these planes are defined by their normals, in this case the basevector of an orthogonal Cartesian system x, y, z. At each face of the cube there is a resultant stress vector

that we have represented by→�

(ei )with i = x, y, z. As can be seen these vectors are not perpendicular

to the faces of the cube. The components of the stress tensor are the projections of the stress vectorsin the direction of the coordinate axis. Mathematically, the tensor is a point function that, according tocontinuum mechanics, is continuous and has continuous derivatives up to the third order. However, whenwe want to measure it we need to make the measurement in a finite volume. If the finite volume is toosmall compared to the representative volume, what we measure will appear to us as a random quantity.The fact that we have talked about measuring a stress tensor is again an ambiguity in language. There isno way to measure stresses directly, we will be able to measure deformations and changes of geometryfrom which we will compute the values of stresses.



Figure 1.1 (a) Elementary cube with stress vectors for the faces of the cube. (b) Components of thestress vectors of the faces. (c) Loaded body showing the elementary cube inside the volume.

1.2 Transformation of CoordinatesAll our measurement procedures will require us to define a coordinate system that we need to specify. Butto handle this information in posterior manipulations it may be necessary to switch coordinate systems.A tensor is an entity that mathematically is defined by the way it transforms. In the following derivationswe are going to go in an inverse way, define the components and then find out how they transform. It isa classical way through which historically the stress tensor was defined. We consider the equilibrium ofa tetrahedron, as in Figure 1.2.

Figure 1.2 Equilibrium of a tetrahedron at point P of a continuum. (a) Component of the stress vectorsacting on the different elementary areas. (b) Angular orientation of the rotated axis.



Introducing an arbitrary oblique plane, where it intersects the three mutually perpendicular referenceplanes creates a tetrahedron. A tetrahedral element about a point P is defined. The axis x′ of the rotatedCartesian coordinates system is perpendicular to oblique plane whereas y′ and z′ are tangent to the planeorientation of the axis x′ and can be established by the angles shown in Figure 1.2 (b). Areas for thetriangular elements formed by the coordinates axis and by the intersection of the oblique plane with thecoordinates planes are given by,

Ax = Aonx′x, Ay = Aonx′ y, Az = Aonx′z (1.3)

Where nx′i are the direction cosines of the normal→n with respect to the coordinate axis. The projection

equations of static equilibrium can be applied to get the components shown in Figure 1.2. To utilize theprojection equations, the first step is to obtain the summation of forces in the x′ direction. Recall thatthe force corresponding to each stress is: � × A� . Next it is important to obtain the component of theforce in x′ direction. Force due to �x is �x Ax = �x Aonx′x. The component of force in x′ is given as(�x Aonx′x) nx′x.

The same procedure is utilized for the other components and the summation of forces in x′ direc-tion gives,

�x′ = �xn2x′x + �yn2

x′ y + �zn2x′z + 2�xynx′xnx′ y + 2�yznx′ ynx′z + 2�zxnx′znx′x (1.4)

For a complete transformation of the stress components with respect to the arbitrary oblique surface, theshear stresses �x′ y′ and �x′z′ must be computed. Directional cosines for y′ and z′ as in x′ are defined as,

�x′ y′ = �xnx′xny′x + �ynx′ yny′ y + �znx′zny′z + �xy(nx′xny′ y + nx′ yny′x) + �yz(nx′ yny′z + nx′zny′ y)

+ �zx(nx′xny′z + nx′zny′x) (1.5)

�z′x′ = �xnx′xnz′x + �ynx′ ynz′ y + �znx′znz′z + �xy(nx′xnz′ y + nx′ ynz′x) + �yz(nx′ ynz′z + nx′znz′ y)

+ �zx(nx′xnz′z + nx′znz′x) (1.6)

These equations are sufficient for the determination of the stress components on any internal surface inwhich an arbitrarily selected tangential set of coordinates is used (y′z′). For a complete transformation ofthe stress tensor shown earlier to that of a rectangular element oriented by the x′y′z′ coordinate system,the six stresses on the two surfaces with normals in the y′ and z′ must also be determined. The component�y′ , �z′ , �y′z′ are:

�y′ = �xn2y′x + �yn2

y′ y + �zn2y′z + 2�xyny′xny′ y + 2�yzny′ yny′z + 2�zxny′zny′x (1.7)

�y′z′ = �xny′xnz′x + �yny′ ynz′ y + �zny′znz′z + �xy(ny′xnz′ y + ny′ ynz′x) + �yz(ny′ ynz′z + ny′znz′ y)

+ �zx(ny′xnz′z + ny′znz′x) (1.8)

�z′ = �xn2z′x + �yn2

z′ y + �zn2z′z + 2�xynz′xnz′ y + 2�yznz′ ynz′z + 2�zxnz′znz′x (1.9)

The above equations give all the components of the stress tensor when the Cartesian axis orientation ischanged. Although these equations have been derived using a finite tetrahedron the postulation is thatthese relationships continue to be valid in the limit when the tetrahedron dimensions go to zero and thetetrahedron merges with the point P.

1.3 Stress Tensor RepresentationThe nine components of �ij, with i, j = x, y, z of the stress vectors are the components of a second-orderCartesian tensor called the Cauchy stress tensor, which completely defines the state of stresses at a given



point, with the notation→�

∧(ei ) = T

∧(ei ), i = x, y, z and is defined as,

� =⎡

⎣

T(ex)

T(ey)

T(ez)

⎤

⎦ =∣∣∣∣∣∣

�xx �xy �zx�xy �yy �yz�zx �yz �zz

∣∣∣∣∣∣

=∣∣∣∣∣∣

�x �xy �zx�xy �y �yz�zx �yz �z

∣∣∣∣∣∣

(1.10)

The first index i indicates that the stress acts on a plane normal to the xi axis, and the second indexj denotes the direction in which the stress acts. A stress component is positive if it acts in the positivedirection of the coordinate axes, and if the plane where it acts has an outward normal vector pointing inthe positive coordinate direction. The above notation is a standard notation in continuum mechanics andsometimes the coordinate axis are represented by xi with i = 1, 2, 3. In such a case the components ofthe stress tensor become �ij with i, j = 1, 2, 3. We have derived the expressions of how the stress tensortransforms under a change of the coordinate system; from an xi system to a x′

i system. The components�ij in the initial system are transformed into the components �′

ij in the new system according to the tensortransformation rule that utilizing matrix notation can be represented by,

�′ = R� RT (1.11)

In (1.11) R is the rotation matrix and the symbol T indicates the transpose matrix∣∣∣∣∣∣∣∣

�′xx �′

xy �′zx

�′xy �′

yy �′yz

�′zx �′

yz �′zz

∣∣∣∣∣∣∣∣

=∣∣∣∣∣∣


∣∣∣∣∣∣

∣∣∣∣∣∣


∣∣∣∣∣∣

∣∣∣∣∣∣


∣∣∣∣∣∣

T

(1.12)

The above operation can be accomplished by using MATLAB R© matrix routines. In MATLAB R© matricescan be entered manually, or by using some pre-defined MATLAB R© functions.

1.3.1 Two Dimensional Case

Figure 1.3 represents the stress tensor transformation in 3D. This figure can be simplified if one has a2D state of stresses. The cube of the 3D space becomes a square in two dimensions and the tetrahedron

Figure 1.3 Transformation of the stress tensor.



Figure 1.4 Rotation of the stress tensor in 2D. The normal indicates the plane where the componentsare computed, the angle � defines the rotation. (a), (b), (c) components resulting from each one of thecomponents of the stress tensor.

becomes a triangle. Let us say that the stress tensor is such that: �z = � zx = � zy = 0 the stresstensor becomes,

� =[

�x �xy�xy �y

]

(1.13)

Figure 1.4 illustrates the rotation of the stress tensor in two dimensions. The normal defines the cor-responding plane where the components of the stress tensor need to be computed. The normal is theoutwards normal and the positive rotation is counterclockwise.

The components are given by equations (1.14) to (1.16).

�x′ = �x cos2 � + �y sin2 � + 2�yz cos � sin � (1.14)

�y′ = �x sin2 � + �y cos2 � − 2�xy sin � cos � (1.15)

�x′ y′ = −(�x − �y) sin � cos � + �xy(cos2 � − sin2 �) (1.16)

1.4 Principal StressesIn a 3D state of stress there are three mutually orthogonal planes such that the corresponding stressvectors are normal to the corresponding planes. This means these planes have no shear components. Theorientations of the planes are called principal directions (also known as Eigen values of the tensor). Thevalues of the stress vectors are called principal stresses (�1, �2, �3).

The principal stresses can be ordered in a way such that �1〉�2〉�3. In the algebraic sense �1 is thelargest value. It is important when dealing with principal stresses to include the corresponding sign. Forexample, if the stresses are positive �1 is the largest in absolute value and �3 is the smallest. If negativethere are two basic cases: If the signs are different for two of the three components the algebraic definitionshould be upheld (example: +�1 − �2 − �3) where �3 would be the largest negative value.

The components �ij of the stress tensor depend in particular on the coordinate system at the pointunder consideration. However, the stress tensor is a physical quantity and hence it is independent of thecoordinate system chosen to represent it. That is it has a fix position in the 3D space. Therefore thereare invariant quantities associated with a stress tensor. The word invariants implies that these quantitiesare independent of the coordinate system; or saying it in a different way, they have the same values nomatter what system of coordinates we select. A stress tensor has three independent invariant quantities



associated with it. One set of invariants is the values of the principal stresses of the stress tensor. Inmathematics the values of the principal stresses are called Eigen values. The directions of the principalstresses in the space are the second set of invariants. Their direction vectors are the principal directionsor eigenvectors. Since the basic property of the principal stresses is the direction of the normal to theface of the plane, we can write,

�T (n) = ��n, T(n)i = �i n with i = 1, 2, 3. (1.17)

where � is a constant of proportionality, and in this particular case corresponds to the magnitude �i ofthe normal stress vector or principal stress.

1.4.1 How to Calculate Principal Stresses after Making the TransformationLooking back to the transformation coordinates carried out in Section 1.2 our new axes are defined. It isnow necessary and very important to relate this new axis to satisfy the equilibrium condition. Applyingthe equilibrium conditions, means that for the new axes x′y′ and z′ we must satisfy the conditions,

Fx + Fx′ = 0Fy + Fy′ = 0Fz + Fz′ = 0

whereFx′ = �iAnxFy′ = �iAnyFz′ = �iAnz

(1.18)

Calling �i the principal stresses and summing up the forces in the x′, y′ and z′ directions the followingequilibrium conditions are obtained:

(�x − �i ) nx + �xyny + �zxnz = 0�xynx + (�y − �i ) ny + �zxnz = 0�xynx + �yzny + (�z − �i ) nz = 0

(1.19)

Recalling that

n2x + n2

y + n2z = 1 (1.20)

Since the n’s are the directional cosines, there is a homogeneous system that has three equations withthree unknowns (�i and two of the directional cosines). A theorem of algebra tells us that in order tohave a solution different from the trivial solution zero, the determinant of the coefficients must be equalto zero.

∣∣∣∣∣∣

�x − �i �xy �zx�xy �y − �i �yz�zx �yz �z − �i

∣∣∣∣∣∣

= 0 (1.21a)

Expanding the determinant gives us the so called characteristic equation of the tensor, a cubic equation:

�3i − (�x + �y + �z)�2

i +(

�x�y + �y�z + �z�x − �2yz − �2

zx − �2xy

)

+ �i +(

�x�y�z + 2�yz�zx�xy − �x�2yz − �y�2

zx − �z�2xy

)

= 0(1.21b)

The above equation can be written,

− �3i + I1�2

i − I2�i + I3 = 0 (1.22)

Where

I1 = �x + �y + �z (1.23)

I2 = �x�y + �y�z + �z�x − �2xy − �2

yz − �2xz (1.24)

I3 = �x�y�y + 2�xy �yz�xz − �x�2yz − �y�2

xz − �z�2xy = det �ij (1.25)



As said before, the principal stresses are unique for a given stress tensor. Hence, it follows from thecharacteristic equation that I1, I2 and I3, called the first, second, and third stress invariants, are invariantsregardless of the particular system of coordinates selected. Since equation (1.22) is a cubic equation, itdoes not have a closed form solution. The literature presents a number of approaches to the solution ofthe cubic equation. MATLAB R© has routines that can be utilized to compute the solution of the cubicequation. Once principal stresses are determined one can go back to the equations below and solve forthe directional cosines.

(�x − �i ) nx + �xyny + �zxnz = 0�xynx + (�y − �i ) ny + �zxnz = 0�xynx + �yzny + (�z − �i ) nz = 0

(1.26)

1.4.2 Maximum and Minimum Shear Stresses

The maximum shear stress is equal to one-half the difference between the largest and smallest principalstresses, and acts on the plane that bisects the angle between the directions of the largest and smallestprincipal stresses, that is, the plane of the maximum shear stress is oriented 45o from the principal stressplanes. The maximum shear stress is expressed as

�max = 1

2(�max − �min) (1.27)

If, �1 ≥ �2 ≥ �3 then,

�max = 1

2(�1 − �3) (1.28)

The normal stress component acting on the plane of the maximum shear stress

�n = 1

2(�1 + �3) (1.29)

1.5 Principal Stresses in Two DimensionsThe equations derived above become simplified when dealing with a state of stresses in two dimensions.In fact (1.11) becomes,

[

�x − �i �xy�xy �y − �i

]

= 0 (1.30)

This equation gives the second degree equation,

�2i − �i (�x + �y) + �x�y − �xy = 0 (1.31)

The solution of this equation is,

�i = �x + �y

2±

√

�x − �y

2+ �2

xy (1.32)

The direction of the principal stresses can be found directly by making the shear stress given by (1.16)equal to zero,

tan 2� = − 2�xy

�x − �y(1.33)



The above equation gives two solutions that represent the two orthogonal principal stresses. Since (1.33)provides two solutions, to know without ambiguity the direction of �1 it is necessary to compute anadditional trigonometric function,

sin 2� = − �xy√

�x− �y

2 + �2xy

(1.34)

Knowing both tangent and sine it is possible to establish without ambiguity the direction of �1 becausethe quadrant of the angle 2� is defined. The maximum shear is defined by

�xy =√

�x − �y

2+ �2

xy (1.35)

The angle is given by,

tan 2�� max = �x − �y

2�xy(1.36)

Again to determine the angle without ambiguity,

sin 2�� max = �x − �y

2√

�x− �y

2 + �2xy

(1.37)

1.6 The Equations of EquilibriumIn the previous developments the concept of a stress tensor and the associated transformations are consid-ered. Those concepts correspond to properties that are defined at a point. Now the emphasis shifts to whathappens between two neighboring points. This way the equations of equilibrium can be derived. Theseequations are partial differential equations that involve the components of the stress tensor. These equa-tions are required to have a stress function that satisfies the equilibrium of the continuum, see Figure 1.5.

The cube represents neighbor planes in the continuum. One set of planes has the components ofthe stress tensor. The other plane contains stress tensor components of a neighboring point. This is amathematical model that defines the behavior of the continuum. This model is in agreement with allthe experimental determinations. Towards the middle of the last century other definitions have beenintroduced but their applicability is reduced to some very special media. To analyze the equilibrium wemust introduce forces per unit of volume, F (for example weight, or centrifugal force).

In one set of planes (for x-dir) we have the components �x, �xy and �xz. The next plane has theincrements of these components. By definition the increments are given as

�x + ∂�x

∂x�x; �y + ∂�y

∂y�y; �z + ∂�z

∂z�z

Summing the components in the x direction gives (where Fx is body force − weight).(

�x + ∂�x

∂x�x

)

�y�z +(

�xy + ∂�xy

∂y�y

)

�z �x +(

�zx + ∂�zx

∂z�z

)

�x�y

− �x�y�z − �xy�z�x − �zx�x�y + F x�x�y�z = 0 (1.38)

Simplifying previous equation results in

∂�x

∂x+ ∂�xy

∂y+ ∂�zx

∂z+ Fx = 0 (1.39)



Figure 1.5 Equilibrium of the elementary cube at a point of the continuum.

This is also true for y and z directions

∂�xy

∂x+ ∂�y

∂y+ ∂�yz

∂z+ F y = 0

∂�zx

∂x+ ∂�yz

∂y+ ∂�z

∂z+ F z = 0

(1.40)

Recall that F is a force per unit volume. In particular, what if I have a solution that provides thecomponents of the stress tensor? These components must satisfy the previous equations because if they

Figure 1.6 Moment with respect to O is equal to zero.



do not it means the solution is incorrect. The forces projection equations of statics are satisfied. Whatabout equilibrium of the moments? The fact that the stress tensor is symmetric (for example �xy = �yxensures the validity of the moment equilibrium equations.)

The condition of equilibrium with respect to the centroid of cube requires that, �xy = �yx, Figure 1.6.Similar relationships can be derived for all the other shear components. Then in general

�ij = �ji (1.41)

1.7 Strain TensorIn the previous derivations we defined the stress tensor. Parallel to the forces in continuum mechanicswe need to develop geometry of deformations that correspond to those forces. In continuum mechanics,there is a general theory of deformation. This theory has to satisfy a metric of the Cartesian space thatrequires that the distances between points are given by the sum of the squares of the components. Thisleads to non linear strain tensors. These tensors complicate the solution of problems of the mechanics ofthe continuum because they transform the system of equations into non linear systems. The developersof continuum mechanics quickly realized this difficulty and proceeded to create the small deformationtheory, or also small displacement theory. Of course the basis of the adoption of this theory was the factthat deformations of structural materials are small quantities compared to 1. Consequently, this theorydeals with infinitesimal deformations of the continuum. By an infinitesimal deformation it is meant thatthe displacements ‖u‖ << 1 and the displacement gradients are small compared to unity, ‖�u‖ << 1making feasible the linearization of the Lagrangian finite strain tensor, that is, all the second order termsare removed. The resulting linearized tensor violates one of the basic requirements of the definition ofstrain, it is not invariant upon rigid body motions. Hence, even if the deformations are small, but rotationsare important, the tensor will give non zero strain components. This aspect is very important in the caseof methods that measure displacements such us moire, and speckle methods. With these basic conceptsunderlying the infinitesimal deformation theory, strain and a point is defined,

ε = lim��

�L

�L → 0 (1.42)

This equation deals with the change of segment length.Besides the change of length the geometry of deformation includes the change of the angle made by

two segments, see Figure 1.7.

Figure 1.7 The equation should be mentioned after change of the angle made by two segments.



Figure 1.8 Analysis of the deformation of the continuum at the neighborhood of a point.

The derivation of the linearized strain tensor can be explained with the help of an infinitesimal geometrydrawing Figure 1.8. We consider an infinitesimal rectangular material element with dimensions dx, dy(Figure 1.8), which after deformation, takes the form of a rhombus.

An element of area is represented by a square with size �x, �y. A deformation is applied to themedium causing the element to change position and shape. The displacements of the points in themedium are represented by two continuous functions of x and y. They are written as

u = u (x, y) v = v (x, y) (1.43)

It is possible to define the displacement in the neighborhood of a point, as the strain change at which apoint was defined.

u (x, �x) = u + ∂u

∂xdx v (x,�x) = v + ∂v

∂xdx (1.44)

A similar expression can be applied for the y direction

u (y, �y) = u + ∂u

∂ydy v (y,�y) = v + ∂v

∂ydy (1.45)

The previous equations are utilized to define the deformation experienced by the segments �x and �yas the medium is deformed. Utilizing our definition of strain given initially

εx = Q′ D′ − Q D

Q D(1.46)

As explained before this definition leads to a non linearity in the strain tensor due to the fact that Q′D′has to be computed as the sum of the square of the two components given in the figure. In view ofthe difficulties that arise when this occurs the earlier developers of continuum mechanics simplifiedthe relationship by replacing Q′D′ by its projection on the x axis. Practically it means that the anglealpha shown in the figure has to be a small angle such that cos � ∼ 1. This means that the equationsthat we are going to derive are limited to small deformations and small rotations. Therefore, if we



want to compute the deformations of a steel component that experiences small deformations the aboveequations can be used. For the deformation of car tires these equations will not work! The strain tensorcomponents resulting from these simplifications as stated before are called the linearized strain tensors.The simplification introduced produces the following equation

εx = [�x + (∂u/∂x) �x] − �x

�x= ∂u

∂xlim x → 0

(1.47)

Similar derivation for the y direction yields

εy = ∂v

∂y(1.48)

The shear component can be computed utilizing a similar simplification

tan � = (∂v/∂x) �x

�x= ∂v

∂xand tan � = (∂u/∂y) �y

�y= ∂u

∂y(1.49)

Since the angles should be small the tangent is equal to the arc (tan � ≈ �)

xy = ∂v

∂x+ ∂u

∂y(1.50)

Defining w as the displacement in the z direction we can generalize the above quantities

εz = ∂w

∂z

yz = ∂w

∂y+ ∂v

∂z

zx = ∂u

∂z+ ∂w

∂x

(1.51)

Utilizing the derivations just made we can now define the strain tensor at a given point of a continuummedium

[ε] =∣∣∣∣∣∣

εxεxyεzx

εxyεyεyz

εzxεyzεz

∣∣∣∣∣∣

(1.52)

Where εxy = xy

2 , εzx = zx2 , εyz = yz

2The tensor defined here is linearized and has the same transformation equations as the stress tensor.

The tensor will have three mutually orthogonal directions that define the principal directions and alongthese directions act the principal strains, ε1, ε2, ε3. All the derivations of tensorial properties that wehave tried concerning the stress tensor apply also to the strain tensor.

1.8 Stress – Strain RelationsAn experimental relationship between stresses and strains is incorporated into a logical framework ofmechanics to produce formulas for the analysis and design of structural members. If a material modeldoes not fit experimental data well, we get errors in theoretical predictions. Yet a model that can fit theexperimental data very well may be so complex that no analytical model can be built. The choice ofmaterial model is dictated both by the experimental data and by accuracy needs of analysis.

The stress and strain tensors are related to each other. The constitutive equations provide the relation-ship between the two. While the mechanics of continuum mathematically models the physical reality,



constitutive equations need to be based on experiments that allow the corresponding properties of realmaterials to be derived.

Each strain is dependent on each stress

εx = C11�x + C12�y + C13�z + C14�xy + C15�yz + C16�zx + C17�xz + C18�zy + C19�yx (1.53)

This must be repeated for the other eight terms (81 – constants). Since shear normally is symmetricthis reduces to six terms each (36 – constants). This is known as the compliance matrix – that at least36 material constants are required to describe the most general linear relationship between stressesand strains. Recalling that C12 = C21, C13 = C31 (shown that W1 = W2) where the w’s are energies,reducing independent constants to 21.

There are important simplifications to constitutive equations. Homogeneous – elastic properties donot change from point to point. Isotropy, regardless of direction in space shows the properties are thesame. Many engineering materials are not isotropic so for these the 21 independent material constantsmust be used. Isotropic materials only require two independent material constants to describe its linear�, ε relationship. How to classify depends on: (1) material properties with orientation, (2) scale we arelooking at and (3) type of information that is desired.

1.8.1 Homogeneous or Not?

At the atomic, crystalline or grain size level materials are non homogeneous. Depending on what typeof information is required crystalline bodies can be grouped into classes for the purpose of defining theindependent material constants needed in the linear stress-strain relationship. A few classifications ofmaterial groups will be analyzed. If a material has one plane of symmetry (xy), there can be no interactionbetween the out of plane shear stresses and remaining strains.

⎧

⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎩

εxεyεzxyyzzx

⎫

⎪⎪⎪⎪⎪⎪⎬

⎪⎪⎪⎪⎪⎪⎭

=

⎡

⎢⎢⎢⎢⎢⎢⎣

C11 C12 C13 C14 0 0C21 C22 C23 C24 0 0C31 C32 C33 C34 0 0C41 C42 C43 C44 0 0

0 0 0 0 C55 C560 0 0 0 C65 C66

⎤

⎥⎥⎥⎥⎥⎥⎦

⎧

⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎩

�x�y�z�xy�yz�zx

⎫

⎪⎪⎪⎪⎪⎪⎬

⎪⎪⎪⎪⎪⎪⎭

(1.54)

Orthotropic materials have two orthogonal planes of symmetry. If we rotate a sample by 90◦ about the xor y axis we will obtain the same � ε relation

⎧

⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎩

εxεyεzxyyzzx

⎫

⎪⎪⎪⎪⎪⎪⎬

⎪⎪⎪⎪⎪⎪⎭

=

⎡

⎢⎢⎢⎢⎢⎢⎣

C11 C12 C13 0 0 0C21 C22 C23 0 0 0C31 C32 C33 0 0 0

0 0 0 C44 0 00 0 0 0 C55 00 0 0 0 0 C66

⎤

⎥⎥⎥⎥⎥⎥⎦

⎧

⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎩


⎫

⎪⎪⎪⎪⎪⎪⎬

⎪⎪⎪⎪⎪⎪⎭

(1.55)

For orthotropic materials the normal strains are not affected by the shear stresses, and the shear strainsare not affected by the normal stresses. This is not true for general anisotropic materials.

1.8.2 Material Coordinate System

If the coordinate system is transformed from the xyz coordinate system (1.54) and (1.55) will change.These equations are valid for a specific coordinate system. This system is known as the material coordinatesystem. Looking at composite materials as in Figure 1.9, that are made out of a matrix reinforced withcontinuous fibers, the material properties will change with the orientation of the fibers.



Figure 1.9 Looking at composite materials, that is a material made out of a matrix reinforced withcontinuous fibers, the material properties will change with the orientation of the fibers.

Fibers are inherently stiffer and stronger than the bulk material, due to the reduction of defects andalignment of crystals along the fiber axis. It is clear that the mechanical properties will be different inthe direction of the fiber and perpendicular to the fiber. If the properties of fiber and epoxy are averagedeach lamina can be regarded as an orthotropic material and the directions parallel and perpendicular tothe fiber are the material axis directions. Stacking the laminate with different fiber orientations createsa composite laminate. Overall the properties can be controlled by orientation of fibers and stackingsequence – for certain stacking sequences laminate will respond like an orthotropic material. If theproperties of the orthotropic material are identical in all three directions the material is said to have acubic structure (FCC, BCC). Below are definitions that provide the relationship between the laminates.

C11 = C22 = C33 = 1

E

C12 = C13 = C23 = C21 = C31 = C32 = −

E

C44 = C55 = C66 = 1

G

⎧

⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎩

εxεyεzxyyzzx

⎫

⎪⎪⎪⎪⎪⎪⎬

⎪⎪⎪⎪⎪⎪⎭

= 1

E

⎡

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

1 − − 0 0 0− 1 − 0 0 0− − 1 0 0 0

0 0 0E

G0 0

0 0 0 0E

G0

0 0 0 0 0E

G

⎤

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

⎧

⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎩


⎫

⎪⎪⎪⎪⎪⎪⎬

⎪⎪⎪⎪⎪⎪⎭

(1.56)

If isotropic you only need constants E and G or E and where G = E/2(1 + )

⎧

⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎩

εxεyεzxyyzzx

⎫

⎪⎪⎪⎪⎪⎪⎬

⎪⎪⎪⎪⎪⎪⎭

= 1

E

⎡

⎢⎢⎢⎢⎢⎢⎣

1 − − 0 0 0− 1 − 0 0 0− − 1 0 0 00 0 0 2 (1 + ) 0 00 0 0 0 2 (1 + ) 00 0 0 0 0 2 (1 + )

⎤

⎥⎥⎥⎥⎥⎥⎦

⎧

⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎩


⎫

⎪⎪⎪⎪⎪⎪⎬

⎪⎪⎪⎪⎪⎪⎭

(1.57)



1.8.3 Linear, Elastic, Isotropic Materials. Lame Constants

The utilization of Lame constants is an alternative way to express the stress-strain relationship, thefollowing equation is an example expressed in indicial notation where x, y and z are replaced by 1, 2and 3. In the equation that follows when an index variable appears twice in a single term it implies thatwe are summing over all of its possible values.

�ij = 2�εij + ��ijεkk (1.58)

Where i, j = 1, 2, 3Where:

�ij = 1 when i = j�ij = 0 when i = j

(1.59)

and

εi i = ε11 + ε22 + ε33 (1.60)

The above expression is an invariant of the strain tensor, the sum of the normal strains. The followingequations are the relation between the Lame constants and the Young’s modulus and Poisson’s ratio.

� = E

2 (1 + )(1.61)

� = E

(1 + ) (1 − 2)(1.62)

1.9 Equations of CompatibilityIn continuum mechanics, a compatible strain tensor field in a deformed body is a field that is producedby a continuous, single-valued, displacement field. One can prove that such a field is unique. Theequations of compatibility are mathematical expressions that provide the conditions under which thestrain field satisfy the conditions of continuity and uniqueness. These equations were first derived forlinear elasticity by Barre de Saint-Venant in 1864. Later Beltrami in 1886 provided a more generalexpression. An intuitive understanding of the equations can be gained by imagining the continuummade of infinitesimal volumes. Each volume is deformed but in doing so it must be connected to allthe neighbor elements without gaps or overlaps. The equations of compatibility are the mathematicalconditions that must be satisfied to guarantee that in the process of deformation gaps or overlaps are notintroduced. The adjective compatible applied to a deformation indicates that the basic postulate of thecontinuum is satisfied. The mathematical expression of compatibility conditions depends on the utilizedtensor field. If the tensor is non-linear the mathematical conditions of compatibility are quite involvedand are called the Mainardi-Codazzi conditions of compatibility. If the tensor is linearized the expressionof the equations become much more simple.

When working through a problem it is possible to define either displacements or stresses as unknowns,or a combination of the two. A problem arises if stresses are chosen as unknowns. The stresses may givedisplacements that violate the basic assumption of continuum mechanics. The solution may result in thepresence of discontinuities inside the continuum, for example voids or overlaps, as previously explained.To avoid this problem, the strains computed after the stress-strain relationship are applied must satisfy asystem of partial equations of the second order. These equations are called equations of compatibility,also called the Beltrami equations. Using the indicial notation, and indicating the second derivatives withtwo indices following a semicolon

εij, km + εkm, ij − εik, jm − εjk, ik = 0 (1.63)



Expanding (1.63) one obtains,

∂2εx

∂y2+ ∂2εy

∂x2= ∂xy

∂x∂y

∂2εy

∂z2+ ∂2εz

∂y2= ∂yz

∂y∂z

∂2εz

∂x2 + ∂2εx

∂z2 = ∂zx

∂z∂x

(1.64)

If the solution is done using displacements as unknowns the compatibility equations will automaticallybe satisfied. If we choose solutions that involve linear stresses the compatibility equations will also beautomatically satisfied because the compatibility is second order equations.

ReferencesThe following publications are provided for more information on the topics presented in this chapter.

1. Nair, S. (2000) Introduction to Continuum Mechanics, Cambridge.2. Mase, G.E. (1969) Theory and Problems of Continuum Mechanics, Schaum’s Outline Series, McGraw Hill.3. Boresi, A.P. and Chong, K.P. (2000) Elasticity in Engineering Mechanics, 2nd edn, John Wiley & Sons, Inc., NY.4. Liu, I.-S. (2002) Continuum Mechanics, Springer.5. Timoshenko, S. History of Strength of Materials, Dover Paper Back, February 1, 1983.6. Todhunter, I. (1960) History of the Theory of Elasticity and of the Strength of Materials from Galilei to Lord

Kelvin Two Volumes in Three Parts, Dover.

continuum mechanics – historical background …€¦ · continuum mechanics ... efﬁcient use of...

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