PartIV ElasticityandThermodynamics ofReversibleProcessesweb.mit.edu/machen/Public/1.050 F09/Course...

Part IV

Elasticity and Thermodynamicsof Reversible Processes

175

Chapter 5

Elasticity: An EnergyApproach

Matter deforms when subjected to stresses. The English experimental physi-cist and inventor Robert Hooke (1635—1703), a contemporary of Isaac New-ton, �rst stated formally this observation as an empirical relation, ‘ut tensio,sic vis’, which translates into our language as ‘extension is directly propor-tional to force’. In Engineering Mechanics, this observation translates intothe theory of linear elasticity. Elasticity describes reversible material be-havior, and the notion of reversibility is a hallmark of another engineeringscience discipline: Thermodynamics. Thermodynamics is the study of en-ergy and energy transformations. Energy, which exists in many forms, suchas heat, light, chemical energy, and electrical energy, is the ability to bringabout change or to do work. Hence, an elastic, ie. reversible, behavior isreadily recognized to be one where the energy provided to a material orstructural system from the outside in form of work by forces and stresses,is stored as internal energy, which can be completely recovered at any time.Thermodynamics, therefore, provides a means to connect an observation toa fundamental physics law. The focus of this Chapter is to employ the lawsof thermodynamics as a backbone for the engineering investigation of linearelastic behavior of materials and structures.

176

5.1. 1-D ENERGY APPROACH 177

dF

x

SK

0

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0 0.1 0.2 0.3 0.4 0.5

dF

x

SK

1

)(x�

dF

x

SK

0

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0 0.1 0.2 0.3 0.4 0.5

dF

x

SK

1

)(x�

Figure 5.1: Elasticity: 1-D Thought Model of Elasticity (left), and force—displacement curve with free energy � (right).

5.1 1-D Energy Approach

The 1-D spring system subjected to a force � � on one side is the simplestrepresentation of an elastic behavior (Fig. 5.1). While equilibrium tells usthat the externally supplied force is equal to the force in the spring, that is,� � = ��, Hooke’s observational law links �� to the extension � at the pointof load application through the proportionality constant �� (of dimension[��] = �

�1 [� ] =��2):

� � ��= ��

��= �� (5.1)

Note that the elastic behavior of the spring may also be non-linear (Fig. 5.1).However, in all what follows we will restrict ourselves to linear elasticity.

Let us now take another route to the problem, by evoking the First Lawand the Second Law of Thermodynamics.

5.1.1 The First Law

The First Law of Thermodynamics states that energy can be changed fromone form to another, but that it cannot be created or destroyed. Energyis conserved. This is expressed by the internal energy U of a system (ofdimension [U ] = �2��2). Speci�cally the First Law informs us that the

178 CHAPTER 5. ELASTICITY: AN ENERGY APPROACH

change in internal energy �U of a system is due to work W and heat Qprovided from the outside. Formally we write:

�U�= W + Q (5.2)

where we employ the symbol ‘ (�)’ to specify that the rate is not necessarilythe time derivative of a function �. For instance, the work rate W isgenerally not the total time derivative of the work W, ie. W 6= �W. Forinstance, in the case of our 1-D spring system, the total work is W = �� ,and the time derivative is �W = �� + � �� . In contrast, the work rate Wis what the force � � realizes along the velocity �� = �0 at the point of loadapplication, that is, W = �� .

5.1.2 The Second Law

The Second Law of Thermodynamics speci�es the direction of spontaneouschange of energy, which is expressed by another physics quantity, the internalentropy S (of dimension [S] = �2��2��1). The entropy is a measure of thequality of the energy. Speci�cally, the Second Law informs us that the changeof the internal entropy is always greater or equal to the entropy supplied toa system in heat form; that is:

�S�� Q 0

(5.3)

where 0 � 0 stands for the absolute temperature, which is assumed to beconstant in what follows. The di�erence between the left and the right handside is recognized as the fraction of entropy which is produced internally ina spontaneous (ie. uncontrolled) fashion, and which manifests itself as aninternal heat source,

�D�= 0

�S�� Q � 0 (5.4)

This internal heat source is called dissipation. More speci�cally, if we substi-tute the First Law (5.2) into (5.4), we obtain:

�D�= W � �

�(U� 0S) � 0 (5.5)

In this form, the Second Law appears as a balance law between the externalwork rate W and some form of recoverable energy � (U� 0S). This energy


is called the free energy or Helmholtz Energy, which we denote by �, andwhich expresses the maximum internal capacity of a system to do work:

� = U� 0S (5.6)

By de�nition, an elastic behavior is one where the entire work rate suppliedfrom the outside is stored as recoverable energy; that is:

Elastic:�D�= W � ��

�

��= 0 (5.7)

Hence, since W is known, it su�ces to determine the expression of the freeenergy � of the spring. In the case of our 1-D spring system, the free energyis recognized to be the area below the � � � � curve (Fig. 5.1); thus in thecase of the linear behavior � = �(�) = 1

2��

2. Use in (5.7) yields:μ� � � ��

��

¶�� = 0 (5.8)

Since this relation must hold for any rate �� we come to the same result as(5.1):

� ��; � � =��

��= �� (5.9)

In contrast to (5.1) which was based on equilibrium and Hooke’s observationallaw, we here operate with energy states described by a state variable; here �.If we add now one additional equilibrium condition to (5.9) we have insteadof (5.1):

� � ��= ��

!=��

��= �� (5.10)

The partial derivative of � w.r.t. � is called state equation, and speci�es theconstitutive law of the spring force in equilibrium with the external force � �.

Finally, if the spring sti�ness �� is constant, we speak about linear elas-ticity. In return if �� (�) depends on �, we need to deal with a non linearelasticity problem. On the other hand, it does not need much imaginationto �gure out that the spring sti�ness �� cannot be negative, since it wouldmake not much sense that the spring extension �, in response to the load� �, takes o� in the opposite direction of the applied force. Behind this ob-servation, however, hides an important property of the free energy function


Pz

x

L 1 2 3

P

� �x�

0�PPz

x

L 1 2 3

P

� �x�

0�

Figure 5.2: 1-D Elasticity Example: Three-Truss System

� = �(�). In fact, the non-negativity of the spring constant is ensuredprovided that �(�) is a convex function of its arguments; that is:

�� =�2�

��2� 0 (5.11)

We will see later that this convexity property is of critical importance forsolving elasticity problems.

5.1.3 Example: Three-Truss System

By way of example, consider the truss system displayed in �gure 5.2. Thestructure is composed of three bars of same length � held on the top andconnected on the other side to a rigid beam device which is subjected to a load� at midspan between two trusses. In a �rst approach, such a truss systemcan be considered as a discrete system, in which each bar represents a springof sti�ness ��. We want to evaluate the forces in the trusses (ie. the springforces) and evaluate the displacement �0 at the point of load application.

Ad hoc Approach

To refresh our memory, let us start with equilibrium. Elementary forceequilibrium and moment equilibrium (around the point of load application)yields:

�1 +�2 +�3 = � (5.12a)

3�1 +�2 ��3 = 0 (5.12b)


We readily recognize that the structure is statically indeterminate of degree1; that is, there are only two equilibrium relations for the three unknowns,�1� �2 and �3. Let us now introduce a linear elastic constitutive law foreach truss of the form:

� = �� (5.13)

where stands for the elongation of bar � = 1� 3. We now need to ensure thekinematical compatibility between the bar elongations and the constraints onboth the top (where the bars are �xed) and the bottom where each bar is con-strained by the displacement � of the rigid bars at the nodes. Compatibilityrequires:

1 = �1 (5.14a)

2 = �2 = �1 +2

3(�0 � �1) (5.14b)

3 = �3 = �1 +4

3(�0 � �1) (5.14c)

While there are a priori three displacement unknowns in the problem, thegeometrical compatibility allows us to reduce these unknowns to two. Wehere choose �1 and �0, the �rst being the displacement of node 1 (Fig. 5.2),the second the displacement at the point of load application. Substitutionof the constitutive law (5.13) and the geometrical compatibility conditions(5.14) in the equilibrium relations (5.12) now yields a system of two equationsfor the two unknowns:

�� (�1 + 2�0) = � (5.15a)

��

μ11

3�1 �

2

3�0

¶= 0 (5.15b)

The linear elastic problem then is readily solved: �0 =1124�� and �1 =

112��. The truss forces are given by:

�1 =�

12; �2 =

�

3; �3 =

7�

12(5.16)

The problem nicely shows that the use of a constitutive law and geometricalcompatibility conditions allows one to obtain a unique solution for a stati-cally indeterminate structure. In this approach, we start with equilibrium,substitute for forces a constitutive law that links forces to displacements, and�nally ensure that the displacements are geometrically compatible.


Energy Approach

Let us now take the alternative energy route to the problem. Instead ofequilibrium, we start out with the external work rate of the applied force � :

W = � ��0 (5.17)

According to the Thermodynamics of reversible processes expressed by (5.7),this external work rate is equal to the change in free energy of the system.Since energy is additive, the total free energy of the truss system is simplythe sum of the individual spring energies:

�(1� 2� 3) =1

2��

¡21 +

22 +

23

¢(5.18)

While we do not evoke any equilibrium, we need to ensure the geometriccompatibility conditions (5.14). This yields the free energy as a function of�0 and �1:

�(�0� �1) =1

2��

μ11

9�21 �

4

9�1�0 +

20

9�20

¶(5.19)

The change of the free energy is then given by:

��(�0� �1)

�=

��(�0� �1)

��0��0 +

��(�0� �1)

��1��1

=

�2

9�� (��1 + 10�0)

¸��0 +

�1

9�� (11�1 � 2�0)

¸��1(5.20)

Finally, substituting (5.17) and (5.20) in (5.7) yields:

�D�=

�� 2

9�� (��1 + 10�0)

¸��0 �

�1

9�� (11�1 � 2�0)

¸��1 = 0 (5.21)

Since the dissipation for any reversible process is zero irrespective of ��0 and��1, the energy approach yields the following set of two linear equations forthe two unknowns �0 and �1 of the problem:

�³��0� ��1

´;�D�= 0 �

�� 2

9�� (��1 + 10�0) = 0

19�� (11�1 � 2�0) = 0

(5.22)

We readily verify that the solution of this system of two equations yieldsthe same solution as (5.15). However, the way how we obtained it, did


not evoke any equilibrium, but just an energy balance and the displacementcompatibility.

Finally, let us note again that the free energy here is convex w.r.t. itsarguments �0 and �1. In fact, analogous to (5.11) it su�ces to determine thesti�ness matrix:

[�] =

��

�2�

��20

�2�

��0�1�2�

��1�0

�2�

��21

� =

��20

9�� 2

9��

�29��

11

9��

� (5.23)

From a mathematical point of view, the sti�ness matrix [�] derives from aquadratic form, that is, the free energy expression (5.18) which is quadratic indisplacements. For this reason, the matrix is positive de�nite, which impliesthat its determinant is positive, det [�] = 8

3�2� � 0, which is a generalization

of the convexity condition (5.11) of the 1 parameter system to a system withmultiple displacement degrees of freedoms. The sti�ness matrix, therefore,can be inverted, which allows solving the problem from the linear system ofequations:

[�] (�) = (�) (5.24)

where (�) = (�0� �1)� is the unknown nodal displacement vector and (�) =

(�� 0)� the nodal force vector.

5.2 3-D Energy Approach

5.2.1 Motivation

We want to extend the 1-D model to 3-D. We remind ourselves that theelementary system we are interested in is the r.e.v. �� (see Section ??, Fig.2.2). For this elementary system, at each material point ��, there are 15unknowns, namely six stresses �� = ��, six (linearized) strains �� = ��,and three displacements �. For these unknowns, we have three equilibriumrelations,

div� + �� = 0 (5.25)

and six (linearized) strain-displacement relations:

� =1

2

³grad�� + (grad��)�

´(5.26)


Thus, there is a total of 9 equations for 15 unknowns; 6 equations are missingfor the problem to be well posed. The missing equations need to be providedby relations linking stresses and strains. This is the focus of material laws.

5.2.2 External Work Rate and Strain Work Rate

The two type of forces that act upon a continuum system are surface forces�� (�� ) �� = � (�� ) · �� and volume forces �� . These forces realizean external work rate along a velocity �eld

�� . This velocity �eld must be

compatible with velocity boundary conditions, so to avoid that the forces dowork along a rigid body motion of the body. Such a velocity �eld is calledkinematically admissible (K.A.). The external work rate reads:

� =

Z �

�� · � · �� +

Z�

�� · �� (5.27)

Let us now apply the divergence theorem (2.77) to the �rst term in (5.27).In particular, letting f �� · � (�� ), application of (2.77) yields:Z

�

�� · � · �� =

Z�

div³�� · �

´��

=

Z�

³�� · div� + grad�� : �

´�� (5.28)

where the symbol ‘:’ stands for a double tensor contraction (a ‘double dotproduct’). Then, a substitution of (5.28) in (5.27) yields:

� =

Z�

grad�� : � ��+

Z�

�� · (div� + �� ) �� (5.29)

An interesting observation is that the second integral on the r.h.s. of (5.29)involves the local static momentum equation (5.25). Hence, if the stress �eld� is statically admissible —in the sense of the set of equations (3.21)— then thesecond integral on the r.h.s. vanishes, since div� + �� = 0. Furthermore,since the same conditions entail the symmetry of the stress tensor, �� = ��,we can rewrite (5.29) in the more convenient form:

� =

Z�

d : � �� =

Z�

tr (d · �) �� =Z�

�� (5.30)


where tr (a) stands for the trace of a, that is the sum of the diagonal terms,tr (a) = � (we also remind ourselves that repeated subscripts stands forsummation). Equation (5.30) is the strain work rate, that is, the work ratethe stresses �� realize along the strain rates ��, which —due to the aforemen-tioned symmetry reasons— represent the components of the symmetric partof the velocity gradient grad

�� :

d =1

2

μgrad

�� +

³grad

��´�¶

(5.31)

Eqn. (5.27) and (5.30) provide a link between the external work rateand the local strain work rate of the stresses. Because of its importance, wesummarize this link as the Strain Work Rate Theorem:

Theorem 4 (of Strain Work Rate (Continuum)) The external work raterealized by surface tractions and volume forces of a continuum system alonga velocity �eld

�� is equal to the strain work rate the local stresses �� realize

along the strain rates ��:

� =

Z �

�� · �� (�� ) ��+

Z�

�� · ��

Z�

�� (5.32)

Here, �� is statically admissible in the sense of (3.21) and the strain rates�� derive from a kinematically admissible velocity �eld that satisfy velocityboundary conditions; but �� and �� are a priori not dependent on each other;that is, they are not related by a material law.

Strictly speaking, the strain rate (5.31) is the rate of deformation of thematerial point in the deformed con�guration (see Section 4.2, Fig. 4.2). Inturn, within the limits of a linear deformation theory, in which we merge theinitial and deformed con�guration for spatial derivatives (see Section 4.3, Eq.(4.33)), the strain rate tensor is recognized to be the time derivative of thelinearized strain tensor (5.26):

Linear: d ' �� (5.33)

We will exclusively address this linear case in (almost) all what follows.


5.2.3 Thermodynamics of Reversible Processes

The Strain Rate Theorem (5.32) provides a �rst important quantity for con-structing the 3-D elasticity model, namely the external work rate and its linkto the local strain work rate of the r.e.v. This quantity is readily employedin both the First Law (5.2),

�U�=

Z�

�� + Q (5.34)

and the Second Law (5.7) written for reversible processes that take place inthe r.e.v.:

�D�=

Z�

μ��

�

¶��

��= 0 (5.35)

where � is the free energy volume density � (of dimension [�] = ��3 [�] =��1��2) of the r.e.v., so that the total free energy is � = ��. Since thedissipation must be zero for any material volume that undergoes elastic (ie.reversible) evolutions, equality (5.35) must equally hold for the r.e.v.; thatis:

Elastic: ��

�

��= 0 (5.36)

The Second Law of Thermodynamics of Reversible Processes, therefore, pro-vides a means to link the stresses to the free energy density contained in ther.e.v. Indeed, it su�ces to specify that � is a function of the strain tensor,ie. � = � (��). Use in (5.36) yields the extension of (5.8) and (5.9) for thecontinuum system; that is the stress equation of state:μ

��

��

¶��

��= 0

� (5.37)

�� =��

��

In other words, as soon as the expression of the free energy density is knownas a function of the strain, the link between the stresses and the strain isprovided by the state equation (5.37). Here lies all the strength of the energyapproach. Finally, the 1-D spring constant as de�ned by (5.11) extends to3-D as follows:

�� =��

=�2�

��(5.38)


The forth order tensor �� is called the elastic sti�ness tensor of the ma-terial. This sti�ness tensor has a priori 9× 9 = 81 elasticity constants (ninestresses derived w.r.t. to nine strain components). If we use the symmetryconditions of both stress tensor and strain tensor, these 81 constants reduceto 6× 6 = 36 (six stresses derived w.r.t. to six strain components):

��=��= ��

��=��= ��=��= �� (5.39)

Finally, from its energy de�nition (5.38), it is recognized that the order ofthe derivation of the free energy w.r.t. �� and �� does not play a role; whichimplies the following remarkable symmetry conditions:

�� =�2�

��

!=

�2�

��= �� (5.40)

These symmetry conditions reduce the 36 constants to 21. No doubt, sucha ‘monster’ tensor �� with 21 elasticity constants is mathematically muchmore involving; but let us keep in mind, that �� is, in a nutshell, just anextension of the 1-D spring constant to 3-D.

5.2.4 Isotropic Linear Elasticity

An isotropic material behavior is one that does not depend on a speci�corientation. As regards elasticity, this means that the free energy � dependsonly on the invariants of the strain tensor. What are such strain invariants?— As we have seen in Section 4.3, the linearized strain tensor � describes twomodes of deformation, length dilations ! (��) and half-distortions "

¡��¢.

Both are de�ned w.r.t. speci�c directions, and cannot therefore be employedas invariants in an isotropic theory. A true invariant is recognized to be thevolume dilation (4.39), which combines the three length dilations:

�� = tr � = � : 1 (5.41)

where 1 = diag (1� 1� 1) is the second order unit tensor. By analogy withthe free energy of the linear spring system, we develop the energy associatedwith volume dilatation of a linear isotropic material in a quadratic form:

�� =1

2��2� (5.42)


where � is called the bulk modulus. Use of (5.42) in the stress state equation(5.37) readily yields the volume stress:

�� =��

= ��

= ��1 (5.43)

It is readily understood, that the limitation to volume dilations is not able tocapture shear stresses. Hence, we need to �nd a second invariant, one thatis able to capture the change in free energy due to distortions. A clever ideais to �rst subtract from the total strain � the average volume dilation; thatis:

e = �� 13��1 (5.44)

This generalized shear strain tensor is called strain deviator tensor, and has—by construction— a zero trace, ie. tr e = 0, and therefore is most suitableto describe distortions. Since tr e = 0, we need to consider the second ordernorm; for instance:

�� = 2

r1

2e : e =

s2

μ� : �� 1

3�2�

¶(5.45)

where the factor 2 before the root comes from the fact that the quantity�� represents a representative distortion (and not a half-distortion). Theassociated free energy of distortion is of the form:

�� =1

2#�2� (5.46)

where # is called the shear modulus. Proceeding as before, we substitute(5.46) into the stress state equation (5.37) to obtain the shear stress-straindeviator relation:

�� =��

= #��

= 2#e (5.47)

Moreover, since energies are additive, the total free energy density of a linearisotropic material is readily obtained by summing up the volume dilationcontribution (5.42) and the shear distortion contribution (5.46):

� = �� + �� =12(��2� +#�

2�) (5.48)

Moreover, if we assume that (��#) are constants, which constitutes thebasis for a linear isotropic elasticity model, the total stress equation of state


is obtained by summing up (5.43) and (5.47):

� = �� + �� = ��1+ 2#e

=

μ� � 2

3#

¶��1+ 2#� (5.49)

Finally, it always handy to have the inverse of this relation, relating strainsto stresses. This form is readily obtained from (5.43), (5.44), (5.47) and(5.49):

� =1

3��1+ e =

1

3

��

�+s

2#

=

μ1

3�� 1

2#

¶��1+

�

2#(5.50)

where we let:�� = ��1 =

1

3tr�; s = � � �� (5.51)

The stress �� is called mean stress or hydrostatic stress (positive in tension),and s is called stress deviator tensor (of zero trace tr s = 0). The strain-stressequation (5.51) could have been equally derived from the complimentary part�� (�) of the free energy:

�� = � (��) + �� (��) (5.52)

In fact, a substitution of (5.52) in the Second Law (5.36) readily yields:

�� + �� (��)�

��= 0 (5.53)

Whence the strain state equation:μ�� + �� (��)

��

¶��

��= 0

� (5.54)

�� =�� (��)��

Without di�culty we recognize that the complimentary energy �� (�) of anisotropic linear elastic material reads:

�� (��) = �� (��) + �� ($�) =

1

2

μ�2��+$2

#

¶(5.55)


(��#) � = 9��3�+�

% = 123��2�3�+�

! = � � 23# & = #

� = �3(1�2�) # = �

2(1+�)(�� %) ! = ��

(1+�)(1�2�) & = �2(1+�)

� = !+ 23& # = & � = &3�+2�

�+�% = �

2(�+�)(!� &)

Table 5.1: Sets of linear isotropic elasticity constants and conversions.

where $ can be seen as a generalized shear stress, called stress deviator in-variant:

$ =

r1

2s : s =

r1

2(� : � � 3�2�) (5.56)

In summary, the linear isotropic elasticity model is characterized by twomaterial sti�ness properties, � and #, which are sti�ness values associatedwith two particular deformation modes, ie. volume dilation and general-ized distortion, introducing a change of the free energy. More generally, thevery procedure applied here to deduce from energy considerations the linearisotropic elasticity model is of general interest. The key idea is to translatedeformation modes of an r.e.v. into a free energy expression, which —werecall— represents the maximum capacity of a system to do work. The ther-modynamics of reversible processes then provides a means to link stresses tostrains. This is what is often referred to as constitutive modeling.

Last, the choice of� and# to represent a linear isotropic elastic behavioris not mandatory. In fact, it is common practice to represent the linearisotropic behavior by other sets of two constants. The Box 5.1 provides asummary of di�erent sets of isotropic elasticity constants that are frequentlyemployed. This includes the Young’s modulus � and the Poisson’s ratio %and the so-called Lamé constants, ! and &.

5.3 Soil and Rock Mechanics Applications

5.3.1 Governing Equations

This section deals with the application of the linear isotropic elastic modelto solving some soil mechanics problems. To solve linear elastic problems,we need to ensure that the problem as a whole is linear in all its aspects,from the deformation theory to stress boundary conditions. As we have seenin Chapter 4, Section 4.3, the deformation theory is linear provided that the

5.3. SOIL AND ROCK MECHANICS APPLICATIONS 191

order of magnitude of the displacement gradient is in�nitessimal; hence:

H1:°°°Grad��°°°¿ 1� � = 1

2

μgrad�� +

³grad��

´�¶(5.57)

A second assumption we need for solving linear elastic problems is the smalldisplacement assumption:

H2:°°°��°°°¿ ('�(� )))) (5.58)

where '�(� ))) stand for characteristic length scales of the problem. Assump-tions H1 and H2 de�ne a true �rst-order strain—displacement theory. Theyare also referred to as small perturbation assumptions. On this basis, weare able to solve for stresses, strains and displacements in a linear elasticproblem. As a reminder, the solution stress �eld � is statically admissible inthe sense of (3.21), which we won’t get tired to recall:

(�) on ��

�� =

�� (�� )

�� · �� = 0

(*) on +�� (�� ) +�� (�� ) = 0

(,) in �

�� (�� ) = � · ��div� + �0

�� = 0�� = ��

(5.59)

We have added here an additional requirement for the problem to be linearand non-dissipative; that is, the contact must be frictionless,

�� · �� = 0.

Moreover, the solution displacement �eld �� is kinematically admissible; thatis, it satis�es the displacement boundary conditions where displacements areprescribed on the boundary. Furthermore, the contact must be perfect (no

unilateral boundary condition �� · �� = 0); hence:

(�) on ��

��= ��

�� · �� = 0

(*) in � � =1

2

μgrad�� +

³grad��

´�¶ (5.60)


Finally, the two solution �elds � and �, that satisfy separately (5.59) and(5.60), are linked through a linear material law; for instance in the case of alinear isotropic material by:

� =

μ� � 2

3#

¶��1+ 2#� (5.61)

� =

μ1

3�� 1

2#

¶��1+

1

2#� (5.62)

We are now ready to solve such linear elastic problems.

5.3.2 Elastic Soil Layer Consolidation

The �rst example we consider is the elastic consolidation of a soil layer ofheight ' on a rigid substrate, which is subjected to a surface pressure - (Fig.5.3). We start with formulating the boundary conditions of the problem. Oneboundary condition is the the pressure boundary condition at . = 0:

. = 0 :�� = -�/� � �� = �� = 0; �� = �- (5.63)

The second boundary condition is the zero displacement boundary conditionat . = ':

. = ' : ��= 0 (5.64)

To solve the problem, we start with a displacement �eld. Given the semi-in�nite nature of the problem, the displacement �elds has only one compo-nent in the .-direction, and depends only on .:

�� = �� (.)�/� (5.65)

The hypothesis of small perturbations (5.57) and (5.58) read here:¯̄̄¯��.

¯̄̄¯¿ 1;

¯̄̄¯��'¯̄̄¯¿ 1 (5.66)

The linearized strain tensor is readily recognized to have only one singlenon-zero component, namely:

�� =��.

(5.67)


H

z

p

g0�

zz�p

pgH 0�

z�

d��

dT�

H

z

p

g0�

zz�p

pgH 0�

z�

d��

dT�

Figure 5.3: Exercise: Elastic soil layer consolidation, with stress and dis-placement boundary.

Use of � = ��/� �/� in the linear isotropic elastic constitutive law (5.61)yields:

� =

μ� � 2

3#

¶��.1+ 2#

��.�/� �/� (5.68)

We now need to check that the so-obtained stress �eld is statically admissible.We start with the equilibrium condition. Since �� = �� (.), the stress �eldalso depends only on ., ie. � = � (.). Furthermore, the stress �eld (5.68)has no shear terms, the only equilibrium relation of interest is:

��.

+ �0� = 0�μ� +

4

3#

¶�2��.2

= ��0� (5.69)

We integrate twice to obtain the displacement:

�� (.) = 1 + 2. ��0�.

2

2¡� + 4

3#¢ (5.70)

where 1 and 2 are two integration constants, which need to be determinedby considering the boundary conditions (5.63) and (5.64); that is:

�� (. = 0) =

μ� +

4

3#

¶ 2 = �- (5.71a)

�� (. = ') = 1 + 2' ��0�'

2

2¡� + 4

3#¢ = 0 (5.71b)


It follows:

2 = � -¡� + 4

3#¢ (5.72a)

1 =-¡

� + 43#¢' + �0�'

2

2¡� + 4

3#¢ (5.72b)

The displacement solution reads:

�� (.) =-¡

� + 43#¢ (' � .)� �0� ('

2 � .2)2¡� + 4

3#¢ (5.73)

The stress solution reads:

�� (.) = � (-+ �0�.) (5.74a)

�� (.) = �� (.) =2#� 3�3� + 4#

(-+ �0�.) =%

1� %�� (.) (5.74b)

where % is the Poisson’s ratio (see Box 5.1). The stress solution shows thatthe zero-strain condition in the � and 0 direction generates stresses in thesedirections. In fact, as the material wants to dilate under the vertical pressureand own weight, this plane strain condition restrains the dilation, which leadsto ‘reaction-stresses’ in those directions.

Finally, the solution strategy employed is based on a displacement �eld,which is followed through the set of equations to determine the linear elasticsolution

³��´, for which the stress is statically admissible, the displace-

ment kinematically admissible, and for which both are related by the linearelastic constitutive law. This solution strategy is called a displacement ap-proach. In contrast, a solution strategy based on a stress �eld as a startingpoint is referred to as stress approach, which we will use later on.

Last, given the linearity of the problem formulation, it is not surprisingthat the two load cases sum up linearly. Hence, we could have solved for thetwo load cases separately and superpose the solution �elds. This principleconstitutes the Theorem of Superposition, which can be formally written inthe form:

11

��

��

�0��

��(1)

+ 12

��

��

�0��

��(2)

7� 11

� �

��

��(1)

+ 12

� �

��

��(2)

(5.75)

where the left side represents the loading multiplied by load factors 1, whilethe right hand side stands for the linear elastic solution �elds.


5.3.3 Deep Tunneling

The second application of the continuum linear elastic model is taken fromrock mechanics: the deep tunneling problem. By deep tunneling we meanthe excavation of a cylindrical hole whose depth ' below ground is largecompared to the tunnel radius 2 (Fig. 5.4):

2¿ ' (5.76)

The length dimension in the tunnel direction is assumed to be in�nite.The deep tunneling problem is of great importance, not only for actual

tunnel engineering applications, but also for oil and gas drilling applicationsrecovering oil and gas through tunnel-shaped wells, bioengineering applica-tions, such as bone remodeling by cells that resorb bone by acidic activityalong resorption tunnels, and so on.

Dimensional Analysis

We are interested in the elastic displacement of the tunnel walls due to theexcavation. This displacement is readily recognized to depend on the speci�cweight �0� (where �0 is the initial mass density), the tunnel radius 2 andthe tunnel depth ', and the elastic properties (��#) of the excavated rock:

= 3 (�0��#�2�') (5.77)

The exponent matrix of dimension is readily established:

[] [�0�] [2] ['] [#] [�]

� 1 �2 1 1 �1 �1� 0 1 0 0 1 1� 0 �2 0 0 �2 �2

(5.78)

The rank of the matrix is 4 = 2, which corresponds to the number of dimen-sionally independent variables. Choosing # and 2 as dimensionally indepen-dent variables, application of the Pi-Theorem (1.14) yields a relation betweenthe dimensionless displacement and three dimensionless Pi-numbers:

�0 =

2= F

μ�01 =

�0�'

#; �2 =

'

2; �3 =

�

#

¶(5.79)


Photo Credit:Top: The Smithsonian Institution Bottom: P. Randy Trabold Collection, courtesy North Adams Transcript

Hoosac Railroad Tunnel, North Adams, MassachusettsBuilt: 1851–1873Length: 4.75 miles

r

�R

H

Photo Credit:Top: The Smithsonian Institution Bottom: P. Randy Trabold Collection, courtesy North Adams Transcript

Hoosac Railroad Tunnel, North Adams, MassachusettsBuilt: 1851–1873Length: 4.75 miles

r

�R

H

Figure 5.4: 3-D Elasticity exercise: Deep Tunneling


The �rst invariant1 �01 = �0�'�# is representative of the overload (persurface) generated at a characteristic depth '. The second invariant �2 ='�2 represents the depth-to-tunnel radius. In the deep tunneling situationthis invariant is known. In fact, according to the deep tunneling assumption(5.76), �2 goes to in�nity, respectively its inverse ��13 goes to zero. Hence,from the point of view of dimensionless analysis, �2 = '�2 does not a�ectthe solution of the problem, and can be neglected in the evaluation of thedisplacement. Finally, the third invariant �3 = ��#, ie. the elastic bulk-to-shear modulus ratio, is representative of the material’s elastic volume-to-shear deformation potential. Moreover, let us explore the linearity of theproblem. Provided that everything is linear in the problem, in the sense ofthe set of equations (5.57) to (5.62), the elastic displacement is proportionalto the applied load, which is represented here by the �rst invariant �01 =�0�'�#. This implies that the normalized displacement scales linearly with��'�#; thus:

lim��0

2� �!"=

�0�'

#Fμ�

#

¶� =

�0�2'

#Fμ�

#

¶(5.80)

Finally, let us remind us that we operate within the framework of the hy-pothesis of small displacements (5.58), which reads here || �2 ¿ 1. Thisassumption implies:

||2¿ 1�

¯̄̄¯�0�'# F

μ�

#

¶¯̄̄¯¿ 1 (5.81)

If we estimate function function F (��#) to be on the order of unity, whichwe will prove by the solution of the linear elastic boundary value problem,(5.81) provides a means to check whether the linear elastic model is relevantfor the speci�c tunneling problem, from �0�'�# ¿ 1. For instance, anexcavation of a ' = 1� 000 m tunnel in a soft rock (e.g. compacted clay)of density �0 2� 000 kg/m3 and shear modulus # 3 � 10 GPa, satis�esthe small displacement condition; while the same tunnel excavated in a verysoft clay with similar density but a much smaller sti�ness in the tens ofkilo-pascals, does not satisfy the small displacement assumption.

1Note that we let �01 = �1�2, where �1 = ��. We here make use of the possi-bility o�ered by D-analysis that we can always generate invariants as products and powerfunctions of previously identi�ed ones (see Section 1.1.3).


How does (Elastic) Tunneling Work? — Theorem of Superposition

How does tunneling work? From an engineering mechanics point of view,tunneling consists in driving, into a geologically prestressed continuum, atunnel of radius 2. In a �rst approach, the initial prestress can be assumedto be hydrostatic; that is:

�0 (.) = ��0�.1 (5.82)

where . � 0 is the coordinate below the surface at . = 0. In deep tunneling,the variation of the initial prestress due to deadweight variation over thetunnels diameter 22 can be neglected; since:

�0 (' �2) = ��0�'μ1� 2

'

¶1��¿1' �0 (') (5.83a)

�0 (' +2) = ��0�'μ1 +

2

'

¶1��¿1' �0 (') (5.83b)

where �0 (') = ��0�'1. By its very de�nition, this hydrostatic stressstate produces a stress vector whose magnitude is the same in all directions(see Section 2.4; Eq. (2.67)); that is,

�� (�� ) = �-0�� , where -0 = �0�'

and �� is the unit normal to any surface. For instance, it also holds forthe surface 5 = 2 oriented by �� = ��/" prior to the tunnel’s excavation.Furthermore, by de�nition, the natural prestressing is related to deformationat some geological time scales. It is therefore natural to choose this geologicalstate as reference state, and associate it with a zero displacement and zerostrain state. This initial load case and the corresponding system responsecan be summarized in the form:

�� 5 = 2 :�� = �-0��

5 �� : ��= 0

�0�� = 0

��(1)

7� � �0 = �-01

� = 0�� = 0

��(1)

(5.84)

The tunnel excavation then consists in removing the prescribed stressvector at 5 = 2. The �nal result is a stress-free boundary at the tunnel wall:

�� 5 = 2 :�� = 0

5 �� : ��= 0

�0�� = 0

�� 7�

� � =?

� =?�� =?

�� (5.85)


1� 00 p npTd

0

+ =1� 00 p npT

d

0

+ =

Figure 5.5: Theorem of Superposition applied to deep tunneling.

In the case of a linear elastic problem, for which the principle of superposition(5.75) applies, the di�erence between (5.84) and (5.85) is the change in stress,strain and displacement induced by tunneling excavation; that is:

�� 5 = 2 :�� = �-0��

5 �� : ��= 0

�0�� = 0

��(1)

+

�� 5 = 2 :

�� = +-0

��5 �� : ��

�= 0

�0�� = 0

��(2)

� (5.86) � �0 = �-01

� = 0�� = 0

��(1)

+

� � =?

� =?�� =?

��(2)

=

� � =?

� =?�� =?

��

where load case (2) represents the excavation load. This principle of super-position is displayed in �gure 5.5.

Linear Elastic Solution for the Excavation Load Case

The excavation load case consists of subjecting the tunnel walls to an inwardradial pressure (Fig. 5.5):

5 = 2 :�� = -0�� = �-0�/" (5.87)


The displacement �eld is zero at in�nity:

5�� : ��= 0 (5.88)

Given the overall radial symmetry and the in�nite extension of the prob-lem in the . -direction, the displacement �eld is recognized to be a radialdisplacement: ��

� = 6" (5) �/" (5.89)

We remind ourselves that we operate here in cylinder coordinates, for whichwe have already seen, in Section 4.4, that the non-zero strains in response toa radial displacement are a radial strain �"" and a hoop strain �##. In a moregeneral form, we have:

�"" =�6"�5; �## =

6"5

(5.90)

Using the displacement approach, we substitute (5.90) in the linear isotropicelastic constitutive law (5.61). The non-zero stresses are:

�"" =

μ� � 2

3#

¶μ�6"�5

+6"5

¶+ 2#

�6"�5

(5.91a)

�## =

μ� � 2

3#

¶μ�6"�5

+6"5

¶+ 2#

6"5

(5.91b)

�� =

μ� � 2

3#

¶μ�6"�5

+6"5

¶(5.91c)

We now need to ensure that these stresses are statically admissible. Westart with the equilibrium in cylinder coordinates, for which div� = 0 for adiagonal stress �elds reduces to (see Box 2.2):

��""�5

+1

5(�"" � �##) = 0 (5.92a)

��##�

= 0 (5.92b)

��.

= 0 (5.92c)

Since 6" = 6" (5), we readily recognize from (5.91b)—(5.92b) and (5.91c)—(5.92c) that the equilibrium in the - and . -direction is satis�ed. Fur-thermore, substitution of (5.91a) and (5.91b) in (5.92a) yields the following


di�erential equation for 6":

�26"�52

+1

5

�6"�5

� 6"52= 0 (5.93)

The solution of this di�erential equation is well known:

6 = 15 + 25

(5.94)

where 1 and 2 are two integration constants that need to be solved byboundary conditions. The displacement boundary condition at in�nity (5.88)readily yields 1 = 0. In return, the stress boundary condition (5.87) at5 = 2 provides a means to determine the second integration constant; from:

�"" = ��## = � 252� tr � = 0� �"" = �2# 2

52

� (5.95)

�"" (5 = 2) = �2# 222= -0 � 2 = �-02

2

2#

The linear isotropic elastic solution³��

��´(2)

for the excavation load case,

therefore, reads: �� 5 = 2 :

�� = �-0�/"

5 �� : ��= 0

�0�� = 0

��(2)

7�

�� = -0

¡�"

¢2(�/" �/" � �/# �/#)

� = $02�

¡�"

¢2(�/" �/" � �/# �/#)��

� = �$0�2

2�1"�/"

��(2)

(5.96)Finally, if we superpose, according to (5.86), the initial prestress response

(5.84) with the excavation load case response (5.96), we obtain the stress anddisplacement solution of the tunneling problem into a homogeneous linearisotropic elastic continuum:

� = �-0"1�

μ2

5

¶2(�/" �/" � �/# �/#)

#(5.97a)

�� = �-02

2

2#

1

5�/" (5.97b)

Without di�culties we verify that the displacement of the tunnel wall 5 = 2is of the form (5.80) with F (��#) = 1. The elastic tunneling turns out to


be a radial unloading process from an initially compressed state. Figure 5.6displays this unloading response in the Mohr-stress and Mohr-strain plane,which shows the generation of shear stresses and strains due to this unloading.The Mohr-stress representation allows us to evaluate the elastic limit state ofthe tunneling, which is reached at the tunnel wall 5 = 2, where the stressesare maximum; that is in terms of the principal stresses:

5 = 2 : �% = �"" = 0 � �%% = �� = �-0 � �%%% = �## = �2-0 (5.98)

Substitution of the linear elastic stress solution (5.98) in (3.44) yields thefollowing elastic limit Galileo number:

Elastic: N�!� =�0�'

�0 1

2(5.99)

where �0 is the uniaxial compressive strength, ie. �0 = 2, in the case ofa Tresca Material, and �0 = 2, cos7� (1� sin7) in the case of a Mohr-Coulomb material.

5.4 Linear Elastic Beam Section Constitu-tive Laws

We adopt here the two-scale approach of structural mechanics developed inChapter 2, Section 2.5, which we have already employed for the strengthof beam members in Chapter 3, Section 3.4. Following this approach, theapplication of the linear isotropic elasticity theory to structural members(trusses, beams, slabs, shells) is developed here in two stages. In this Sec-tion, we develop the linear elastic section constitutive laws of beam-typestructures, based on a combination of the beam stress model (Section 2.5.2),the linear beam deformation model (Section 4.6) and the continuum linearisotropic elastic theory (Section 5.2.4). In the next section, we will see howthese section constitutive laws are employed to determine the linear elasticforce-, moment- and de�ection �elds of beam-type structures.

5.4.1 Governing Equations

We consider a section + in the 0. plane of a beam oriented in the �-direction(Fig. 5.7).

5.4. LINEAR ELASTIC BEAM SECTION CONSTITUTIVE LAWS 203

0p�

0p�

0p�

0p�

0p�

0p�

11 1 2+ = 1

Rr � Rr

��r ��r

Rr �Rr

0

2pG�

0

2pG�

1 1+Rr � Rr

��r

0

2pG�

0

2pG�

= 1 1

Rr � Rr

��r

0

2pG�

0

2pG�

0p�

0p�

0p�

0p�

0p�

0p�

11 1 2+ = 1

Rr � Rr

��r ��r

Rr �Rr

0

2pG�

0

2pG�

1 1+Rr � Rr

��r

0

2pG�

0

2pG�

= 1 1

Rr � Rr

��r

0

2pG�

0

2pG�

Figure 5.6: Principle of superposition applied to deep tunneling in the Mohr-stress plane (top) and in the Mohr-strain plane. The �gure also illustratesthat the maximum shear stress 8max = 1

2(�% � �%%%) is related, in isotropic

linear elasticity, to the maximum half-distortion "max =12(�% � �%%%) by twice

the shear modulus 2#.

x

yz

dSnx

yz

dSn

Figure 5.7: Beam model.


We remind ourselves of the beam stress model introduced in Section 2.5:(1) the local stresses � (�� ) in the cross-section are of the form (2.85):

�� =

� �� sym�� 0�� 0 0

�� (5.100)

(2) the corresponding stress vector�� (�� = �/�) at any point �� = 0�/� + .�/�

in the cross-section is given by:�� (�� = ��/ �) = ��

��/ � + ��/ � + ��

��/ � (5.101)

and (3) the stress vector is linked to section forces and section moments bythe reduction formulas (2.86) and (2.88):

�� (�) =

Z�

�� (�� = �/�) �+ (5.102a)

��M� (�) =

Z�

�� ×�� (�� = �/�) �+ (5.102b)

Furthermore, on the deformation side, we restrict ourselves to the linearbeam deformation model introduced in Section 4.6: (1) the displacement �eldis de�ned by the Navier-Bernoulli assumption (4.82) and (4.90), for whichthe strains read:

Navier-Bernoulli:

� ��

�� =

� �0�� + 9

0�. � 90�0

�1290�.

1290�0

�� (5.103)

where �0�� stands for the axial strain, and 90�� 90� and 90� for the curvatures

(ie. the change of angles ��: (�) �� along �).The general aim of the linear elastic beam section constitutive law is to

link the section forces and moments to the section deformations quantities.We keep in mind that all quantities here de�ned depend only on the positionalong the beam’s axis �.

5.4.2 Stress Approach: Poisson’s E�ect and Young’sModulus

Following the two-scale approach for structural members (see Chapter 3,Section 3.4), we consider a beam element of length ; subjected to regular


moment and force boundary conditions at its end sides. We restrict ourselvesto bending moments and axial forces. As a consequence of the force andmoment equilibrium relations (2.91) and (2.93)2, the shear forces and thetorsion moments are zero, and the problem is entirely de�ned by:

�� &

= ��/� = �/�

Z�� + (5.104a)

M��/ � +M�

��/ �

��&=��M� = �/�

Z�

.��+| {z }M�

+ �/�

Z�

(�0��) �+| {z }M�

(5.104b)

We now need to link the stresses to strains. To this end, we substitute thebeam stress model (5.100) into the 3-D linear elastic constitutive law (5.61).Let us start by making use of the zero stresses in the 0 and . directions, whichas we have seen in Section 2.5.2, is a consequence of the beam’s slendernessratios. The conditions �� = �� = 0 yield �� and �� as a function of theaxial strain:

�� =

μ� +

4

3#

¶�� +

μ� � 2

3#

¶(�� + ��) = 0

�� =

μ� +

4

3#

¶�� +

μ� � 2

3#

¶(�� + ��) = 0

� (5.105)

�� = �� = �12

μ3� � 2#3� +#

¶�� = �%��

where % is the Poisson’s ratio (see Box 5.1). Hence, the axial deformation ��provokes a transversal length dilation of the beam section. Since this e�ectis quanti�ed, in the linear isotropic elastic model, by the Poisson’s ratio %,this transversal deformation is also often referred to as the Poisson’s e�ect.Furthermore, if we substitute (5.105) in the non-zero �� constitutive law,

2As a reminder, the vector of moment equilibrium (5.147) develops in the form (seeEq. (2.93)):

�M�

��= 0;

�M�

�� = 0;

�M�

��+� = 0

As M�M� and M� are constant over the beam length (regular boundary conditions),the shear forces � and � are zero.


we obtain the axial stress as a function of only the axial deformation; thatis:

�� =

μ� +

4

3#

¶�� +

μ� � 2

3#

¶(�� + ��) = �� (5.106)

where � = 9�#� (3� +#) is the Young’s modulus (see Box 5.1). It isthe uniaxial sti�ness of a material system when the transversal stresses arezero. Finally, if we note that �� = 2#�� = 0, we readily �nd that the onlynon-zero linear elastic beam stress in the problem is:

�� = �� = �¡�0�� + 9

0�. � 90�0

¢(5.107)

The stress constitutive equation (5.107) links the local stresses of the beammodel (5.100) to the local strains of the linear beam deformation model(5.103). The stresses are found to vary linearly over the cross-section. Wehave used this linear stress model in Section 3.4.1 (see Fig. 3.14) as onepossible statically admissible stress �eld for beam structures subjected tobending moment. The linear elastic constitutive law identi�es this stress�eld as the one associated with linear elastic bending.

5.4.3 Linear Elastic Section Properties: AreaMomentsof Inertia

We are now ready to relate these local stresses to section forces and momentsby means of the reduction formulas (5.104). In the case of a homogeneoussection (same elastic properties throughout the section), a substitution of(5.107) in (5.104) yields: � ��

M�

M�

�� = �

��

R��+

R�. �+ � R

�0 �+R

�. �+

R�.2 �+ � R

�0.�+

� R�0 �+ � R

�0. �+

R�02�+

� � �0��

90�90�

�� (5.108)

Eq. (5.108) de�nes the section constitutive laws that relate the section forceand section moments to section deformation properties by means of sectionrigidity quantities, which are —for homogenous sections— the product of theelastic sti�ness � and an area moment of di�erent order: the section + =R��+ is a zero-order area moment, the integrals +� =

R�. �+ and +� =R

�0 �+ are �rst-order area moments (of dimension [+�] = [+�] = �3), and

the integrals <�� =R�02�+� <�� =

R�.2�+ and <�� =

R�0.�+ are second-

order area moments (of dimension [<�] = �4). From a geometrical point of


view, the �rst-order area moments de�ne the centroid (0'� .') of the cross-section; that is:

+� =

Z�

0 �+ = 0'+; +� =

Z�

. �+ = .'+ (5.109)

Hence, if we take as reference axis the beam’s centroid, the �rst-order areamoments are zero, and the axial force — axial deformation constitutive lawof the section reduces to:

Centroid: �� = �+�0�� (5.110)

On the other hand, the second-order area moments are called area mo-ments of inertia, and need to be evaluated for any speci�c section underconsideration. For instance, for a rectangular section, those moment of iner-tia are readily derived:

<�� =

Z�

02�+ =

Z �=(�2

�=�(�2

Z �=)�2

�=�)�202�0�. =

*3=

12(5.111a)

<�� =

Z�

.2�+ =

Z �=(�2

�=�(�2

Z �=)�2

�=�)�2.2�0�. =

*=3

12(5.111b)

<�� =

Z�

0.�+ =

Z �=(�2

�=�(�2

Z �=)�2

�=�)�20. �0�. = 0 (5.111c)

The fact that the mixed second-order area moment is zero is due to the factthat <�� was evaluated in the centroid of the section, and that the rectangularsection posses double symmetry w.r.t. the 0 and . axis. More generally,<�� = 0 for any section that possesses at least one symmetry around one ofthe section’s axis. In this case, the bending moment relations reduce to thefollowing moment—curvature relations:

Section Symmetry: M� = �<��90�; M� = �<��9

0� (5.112)

Finally, in this symmetric case with the beam’s reference axis placed in thesection’s centroid, relations (5.110) and (5.112) provide an easy way to de-termine the section deformation quantities, �0�� 9

0� and 9

0�, which are readily

employed in (5.107) to determine the axial stress distribution in the cross-section:

�� (0� .) =��

++

.

<��M� � 0

<��M� (5.113)


In the more general case, that is, when the section does not possess symmetry,a similar approach can be carried out that starts with the inversion of thesection state equations (5.108), which yields:

� �0��90�90�

�� =

1

�

��

1�

0 0

0 %��%��%��%2��

%��%��%��%2��

0 %��%��%��%2��

%��%��%��%2��

� � ��

M�

M�

�� (5.114)

Whence the axial stress distribution:

�� (0� .) =��

++.<�� 0<��<��<�� <2��

M� � 0<�� .<��<��<�� <2��

M� (5.115)

Eqs. (5.113) and (5.115), which are valid only for homogeneous cross-sections,provide a convenient way to check the order of magnitude of the linear elas-tic stresses in a beam’s section and to identify where the maximum elasticstress due to combined axial force—bending loading occurs. Given the lin-earity of the deformation theory and of the elastic constitutive law, it is notsurprising that the e�ects of axial force and bending moments on the lo-cal stress �� (0� .) simply sum up. Last, without di�culties we verify that(5.113) and (5.115) is statically admissible in the sense of the set of equa-tions (3.21). Indeed, subjected to the regular force and moment boundaryconditions (5.104), ��M� andM� are constant over the beam length, as is�� (0� .); thus:

div� =��

�/� = 0 (5.116)

5.4.4 Torsional Shear: Saint-Venant Torsion Model

So far we have not considered torsion and the torsional rigidity. Proceeding asbefore, we subject a beam to a regular torsion moment boundary condition:

��M� =M��/ �

��&= �/�

Z�

(0�� .��) �+| {z }M�

(5.117)

where �� and �� are shear stresses. All other stresses are zero. In order to bestatically admissible, the shear stresses need to satisfy the local equilibriumrelations:

div� =

μ��0

+��.

¶�/� +

��

�/� +��

�/� = 0 (5.118)


and the zero-stress boundary condition at the boundary �+ of the cross-section oriented by the unit normal �� = ��/� + ��/�:

on �+ : �� (��) = 0� �� + �� = 0 (5.119)

Let us now consider the beam strain �eld (5.103) in the 3-D linear elasticconstitutive law (5.61):

�� = 2#�� = �#90�. (5.120a)

�� = 2#�� = #90�0 (5.120b)

where 90� = 90� (�) is the torsion curvature or twist. We readily verify that the

stress �eld (5.120) satis�es the local equilibrium conditions (5.118), providedthat (homogeneous section):

div� = 0� �90��

= 0 (5.121)

On the other hand, the zero-stress boundary condition (5.119) becomes:

on �+ : �.�� + 0�� = 0 (5.122)

This boundary condition is strictly satis�ed for a circular cross-section, forwhich in cylinder coordinates (5� ):

0 = 5 cos ; �� = cos . = 5 sin ; �� = sin

(5.123)

In this case, use of (5.120) in (5.117) yields:

M� = #> 90� (5.124)

where #> is the torsional rigidity and > is the area moment of torsion inertia(or polar section moment):

> =

Z�

¡02 + .2

¢�+ =

Z�=*�2

52�+ =

Z 2*

0

Z �

0

53�5� =1

2?24 (5.125)

where 2 is the radius of the cross-section. For non-circular sections, it isintuitively understood that the boundary condition (5.122) leads to staticallyadmissible shear stress �elds that are less regular than the one prevailing in acircular cross-section. As a consequence, a non-circular section (for instancea rectangular section) cannot achieve a similar high torsional rigidity. Thisobservation is known as Saint-Venant’s Theorem:


Among all simply connected cross-sections of a given area,only the circular cross section achieves the largest torsional rigid-ity.

In other words, for general cross-section shapes, Eq. (5.121) may welloverestimate the area moment of torsion inertia. For such cross-sections, analternative approach is in order, one that automatically satis�es both theequilibrium relations (5.118) and the zero-stress boundary condition (5.119).This approach, which is also referred to as the Saint-Venant torsion the-ory, consists in introducing a stress function or stress potential @ (0� .) =#90�7 (0� .), from which the linear elastic shear stresses �� and �� derive:

�� = ��@�.= �#90�

�7

�.; �� =

�@

�0= #90�

�7

�0(5.126)

From a substitution of (5.126) in (5.118), we readily see that this stresspotential approach is compatible with div� = 0, for which it was indeeddevised. Furthermore, the boundary condition (5.119) becomes:

on �+ :�@

�.�� =

�@

�0�� (5.127)

Noting that ||��|| =p�2� + �2� = 1, the only way how the boundary condition(5.127) can be satis�ed, is by letting @ have a constant value on �+. Let usreturn now, for a minute, to the circular section of radius 2, for which thestress function is @ = 1

2#90� (0

2 + .2 �22); hence 7 = 12(02 + .2 �22). The

stress function has a zero-value at the boundary 5 = 2, and has a maximumin the section’s center. Generalizing this shape for arbitrary connected cross-sections, we let:

on �+ : @ (0� .) = 0 (5.128)

and choose @ (0� .) in a form to satisfy (5.120); that is:

�@ =��0

� ��.

=�2@

�02+�2@

�.2= 2#90�; �7 =

�27

�02+�27

�.2= 2 (5.129)

where � stands for the 2-D expression of the Laplacian. With the help ofsome mathematical operations (don’t worry, just some integration by parts),the area moment of torsional inertia is recognized to be:

> =

Z�

μ0�7

�0+ .

�7

�.

¶�+ = �2

Z�

7 (0� .) �+ (5.130)


Hence, what it takes to determine the torsional rigidity of non-circular sec-tions is the determination of the stress function @ (0� .), which can be viewedas a soap bubble that covers the cross-section. Its volume is close to a mul-tiplying constant the area moment of torsional inertia.

By way of application, let us consider an elliptical section of half-axis� and *. A stress function @ (0� .) which satis�es the boundary conditions(5.128) and the Laplacian relations (5.129) reads:

@ (0� .) = #90��2*2

�2 + *2

μ³0�

´2+³.*

´2� 1¶

(5.131)

In fact, since (0��)2 + (.�*)2 = 1 represents the ellipse, this stress function,by design, automatically satis�es the zero-stress function condition on �+.Furthermore, the stresses (5.126) read:

�� = ��@�.= �#90�

2�2

�2 + *2.; �� =

�@

�0= #90�

2*2

�2 + *20 (5.132)

The Laplacian of (5.131) is recognized to satisfy (5.129):

�@ =��0

� ��.

= #90�

μ2*2

�2 + *2+

2�2

�2 + *2

¶= 2#90� (5.133)

Therefore, we can evaluate the area moment of torsion inertia from (5.131),for 7 (0� .) = @ (0� .) �#90�:

> = �2 �2*2

�2 + *2

Z�

μ³0�

´2+³.*

´2� 1¶�+ = ?

�3*3

�2 + *2(5.134)

Hence, as soon as an appropriate stress function is available for a speci�csection, the torsion area moment of homogeneous sections is determined usingthe Saint-Venant Torsion Theory.3

5.4.5 Return to Energy Approach

We cannot conclude this Section on Beam constitutive laws without brie�yevoking the energy approach. We start with the strain work rate theorem

3Expressions for � for sections commonly employed in the engineering practice can befound in Engineering Design Books.


(5.32), in which we substitute the beam stress model (5.100) and the Navier-Bernoulli strain rates (5.103):

� =

Z+

Z�

[�� + 2 (�� + �� )] �+ ��

=

Z+

Z�

h��

³��0�� +

�90

�. � �90

�0´+ �9

0

� (�.�� + 0��)i�+ ��

=

Z+

��

0�� + �M� ·

��9

¸�� (5.135)

where we made use of the reduction formulas (5.102) and where��9 = �9

0

� �/�+�90

� �/�+�90

� �/� is the vector of curvature rates. Next, we express the free energyof a beam per unit beam length:

�� =

Z�

��+ =

Z�

1

2

¡��2�� + 4#

¡�2�� + �

2��

¢¢�+ (5.136)

Use of the Navier-Bernoulli strains (5.103) in (5.136) yields:

�� =

Z�

1

2

³�¡�0�� + 9

0�. � 90�0

¢2+#

¡90�¢2 ¡

02 + .2¢´�+ (5.137)

Following thermodynamics of reversible processes, the external work rate(5.135) is equal to the change in free energy; that is:

�� 0�� + �M� ·

��92 � ,!-=

��

=��0��

��0�� +��

��9·��9 (5.138)

It follows from (5.138):

�� =��0��

=

Z�

�¡�0�� + 9

0�. � 90�0

¢�+

�M� =��

��9:

��

M� =��90�

=

Z�

#¡02 + .2

¢90� �+

M� =��90�

=

Z�

�¡.�0�� + .

290� � 0.90�¢�+

M� =��90�

=

Z�

�¡�0�0�� 90�0. + 0290�¢ �+

(5.139)

For homogeneous sections, we won’t have any di�culty to identify the derivedset of state equations (5.139) with the state equations (5.108) and (5.124).


5.4.6 What’s the Matter with (Transverse) Shear inBeams?

One (very) last problem we want to address is transverse shear in beams.Due to the restrictions to regular moment and axial force conditions, we con-sciously excluded the e�ects of a bending moment gradient. For illustration,we consider the 2-D beam situation, for which a combination of Eqn. (2.87)and (2.93) yields:

A� =

Z�

�� (�� .) �+ =�M�

��(5.140)

where �� is a shear stress. The simultaneous presence of bending and shearrequires the shear stress to satisfy the local equilibrium condition:

��

+��.

= 0 (5.141)

where �� (�� .) = .�<��M� (�) is the linear elastic axial stress according to(5.115). Use of this relation in (5.141) yields (for a constant section along�):

��

=.

<��

�M�

��=

.

<��A� � ��

�.(5.142)

and after integration over .:

�� (�� .) = � .2

2<��A� (�) + (�) (5.143)

where (�) = �� (�� 0) is recognized to be the shear stress in the section’scentroid, where it takes a maximum value. In fact, use of (5.143) in the �rstpart of Eq. (5.140) provides a means to determine it from:

A� (�) = �R�.2�+

2<��A� (�)+ (�)+ � (�) = �� (�� 0) =

3

2

A� (�)

+(5.144)

where we made use of <�� =R�.2�+. The shear stress distribution over the

beam section’s height thus reads:

�� (�� .) =A� (�)

+

μ3

2� .2+

2<��

¶(5.145)

The shear stress distribution is found to have a parabolic shape over thecross-section, with a maximum value in the centroid. To complete the shear


model, we still need to check that the shear stress at the beam’s surfacesis zero. For that matter, let us consider a rectangular section + = *= and<�� = *=3�12, for which �� (�� .) =

.�(�)�

¡32� 6 (.�=)2¢. In this case, the

shear stress is readily found to satisfy the boundary condition at . = ±=�2.While statically admissible, one question that immediately comes to mind

is how the found shear stress distribution is compatible with the Navier-Bernoulli beam deformation assumption (see Section 4.6) that plane sectionsremain plane throughout the beam’s deformation and perpendicular to thebeam’s reference axis, according to which —in the absence of torsion— �0�� = 0.To make a long story short, it is not compatible, since:

�� (�� 0) = 2#�0�� 6= 0 (5.146)

This incompatibility is due to the fact that we have not foreseen in the free en-ergy expression (5.137) a shear energy contribution other than due to torsion.In other words, in the absence of torsion, the linear elastic Navier-Bernoullibeam theory does not provide a shear stress or shear force state equation4.More advanced shear beam theories try to circumvent this problem, but theygo beyond the scope of our presentation. As a consolidation, let us remindus that the beam theory is strictly valid only for large slenderness ratios,;�=À 1, for which any possible shear deformation associated with the shearstresses is negligible compared to the beam de�ection quantities.

5.5 Structural Mechanics Application

5.5.1 Linear Elastic Beam De�ection: Di�erential Ap-proach

Let us now move upwards in scales, from the section constitutive equationsto beam structures. We recall the section force and moment equilibrium

4This is an analogous to the incompressibilty assumption frequently made in �uidmechanics, for which thermodynamics due to the assumed incompressibility of the �uid,doe not provide a �uid pressure — volume change state equation. In this case, the pressureis solved from equilibrium relations, very much in the same vain as we have done here forthe shear stress.

5.5. STRUCTURAL MECHANICS APPLICATION 215

relations (2.91) and (2.93):

�� [0� ;] :

��

��+��3 ��/ = 0

��M�

��+��/ � ×�� = 0

(5.147)

which can be combined reading:

�2��M�

��2��/ � ×��3 ��/ = 0 :

��

�2M�

��2= 0

�2M�

��2+ 3� = 0

�2M�

��2� 3� = 0

(5.148)

A substitution of the bending section constitutive equations (5.112) for sym-metric sections in (5.148) yields:

Section Symmetry:

��2¡�<��9

0�

¢��2

+ 3� = 0

�2¡�<��9

0�

¢��2

� 3� = 0(5.149)

Finally, we remind ourselves that the curvatures 90� and 90� are related, inthe Navier-Bernoulli beam theory (see Section ??, Eq. (4.90)) to the sectionrotation, :� and :�, and the displacement components �0� and �0� by:

90� =�:��

= ��2�0��2

; 90� =�:��

=�2�0��2

(5.150)

Then, a substitution of (5.150) in (5.149) yields, for constant �<�� and �<��along �, the following forth-order di�erential equations of linear elastic beambending:

Section Symmetry(�<�� <��) = const

:

��

(�) ��<�� 4�0��4

+ 3� = 0

(*) �<��4�0��4

� 3� = 0(5.151)


The integration of those di�erential equations with four appropriate beamforces, moment, rotation and curvature conditions yields the linear elasticsolution of the beam problem.

5.5.2 Examples of Statically Determined and Indeter-minate Beam Structures

By way of example, we reconsider the 2-D beam examples of Section 2.5.4,subjected to a line load:

��3 ��/ = �B�/� � 3� = �B (5.152)

Integration of the di�erential equation of beam de�ection (5.151(�)) yields:

A� (�) = ��<�� 3�0��3

= A� (0)�Z3� �� = A� (0) + B� (5.153a)

�� (�) = ��<�� 2�0��2

=�� (0) +

ZA� (�) �� (5.153b)

= �� (0) +A� (0)�+1

2B�2

:� (�) = ��0�

��= :� (0) +

1

�<��

Z�� (�) �� (5.153c)

= :� (0) +1

�<��

μ�� (0)�+

1

2A� (0)�

2 +1

6B�3¶

�0� (�) = �0� (0)�Z:� (�) �� (5.153d)

= �0� (0)� :� (0)��1

�<��

μ1

2�� (0) �

2 +1

6A� (0)�

3 +1

24B�4¶

where the integration constants A� (0) �� (0) � :� (0) � �0� (0) are force, mo-

ment, rotation and displacement boundary conditions at � = 0. Those needto be solved for the particular beam boundary conditions, as detailed below.


Cantilever Beam

The boundary conditions of the cantilever beam are zero displacement androtation at � = 0 and zero force and moments at � = ;; that is:

� = 0 (�� )

��0� (0) = 0

:� (0) = ��0�

��= 0

� = ; (�� )

½A� (;) = 0�� (;) = 0

(5.154)

Use of these boundary conditions in (5.153) yieldsA� (0) = �B; and�� (0) =12B;2; whence (Fig. 5.8):

A� (�) = ��<�� 3�0��3

= B (�� ;) (5.155a)

�� (�) = ��<�� 2�0��2

=B

2(�� ;)2 (5.155b)

:� (�) = ��0�

��=

B;3

6�<��

¡1 + (��;� 1)3¢ (5.155c)

�0� (�) = � B;4

24�<��

μ4�

;+³�;� 1´4� 1¶

(5.155d)

The maximum displacement occurs at � = ; where:

�<��B;4

�0� (� = ;) = �0)125 (5.156)

Simply Supported Beam

The displacement, force and moment boundary conditions of the simply sup-ported beam are:

(�� )

½�0� (� = 0) = 0�0� (� = ;) = 0

(�� )

½�� (� = 0) = 0�� (� = ;) = 0

(5.157)


4

04

dxdEIqf z

zzz�

x

yz

0

0.1

0.2

0.3

0.4

0.5

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

-1-0.9-0.8-0.7-0.6-0.5-0.4-0.3-0.2-0.1

00 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

0

0.05

0.1

0.15

0.2

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

0

0.02

0.04

0.06

0.08

0.1

0.12

0.14

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

� � 3

03

dxdEIxQ z

zzz�

� � 2

02

dxdEIxM z

zzy�

� �dxdx z

y

0��

� �xz�

Figure 5.8: Cantilever beam: Linear elastic solution �elds.


Use of (5.157) in (5.153) yields A� (0) = �12B; and :� (0) = B;3� (24�<��);

whence (Fig. 5.9):

A� (�) = ��<�� 3�0��3

= B

μ�� 1

2;

¶(5.158a)

�� (�) = ��<�� 2�0��2

=B

2

¡�2 � ;�¢ (5.158b)

:� (�) = ��0�

��=

B

24�<��

¡;3 � 6�2;+ 4�3¢ (5.158c)

�0� (�) = � B

24�<��

¡;3�� 2�3;+ �4¢ (5.158d)

The maximum de�ection along � occurs where the rotation is zero, which ishere in mid-span (Fig. 5.9):

:� (� = ;�2) = 0� �<��B;4

�0� (;�2) = �1) 302 1× 10�2 (5.159)

Statically Indeterminate Beam

The displacement, rotation, force andmoment boundary conditions of a beamclamped at � = 0 and simply supported at � = ; are:

(�� )

��0� (� = 0) = 0

:� (� = 0) = ��0�

��= 0

�0� (� = ;) = 0

(�� ) �� (� = ;) = 0

(5.160)

Note that although the beam is statically indeterminate, the displacementboundary conditions plus a linking material law permit the determination offorces and moments in the structure. Indeed, the boundary conditions (5.160)provide a means to solve for the unknowns in (5.153), and in particular for


4

04

dxdEIqf z

zzz�

x

yz

- 0 . 15

-0 . 1

- 0 . 0 5

0

0 0 . 1 0 . 2 0 . 3 0 . 4 0 . 5 0 . 6 0 . 7 0 . 8 0 . 9 1

-0.5

-0.3

-0.1

0.1

0.3

0.5

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

-0 . 0 6

-0 . 0 4

-0 . 0 2

0

0 . 0 2

0 . 0 4

0 . 0 6

0 0 . 1 0 . 2 0 . 3 0 .4 0 . 5 0 . 6 0 . 7 0 . 8 0 . 9 1

-0 . 015

-0 . 0 1

- 0 . 00 5

0

0 0 . 1 0 . 2 0 . 3 0 . 4 0 . 5 0 . 6 0 . 7 0 . 8 0 . 9 1

� � 3

03

dxdEIxQ z

zzz�

� � 2

02

dxdEIxM z

zzy�

� �dxdx z

y

0��

� �xz�

� �8

)0(max2�qQxM zy

� �)0(max yyy Mx��

� �)0(max yzz x ��

4

04

dxdEIqf z

zzz�

x

yz

- 0 . 15

-0 . 1

- 0 . 0 5

0

0 0 . 1 0 . 2 0 . 3 0 . 4 0 . 5 0 . 6 0 . 7 0 . 8 0 . 9 1

-0.5

-0.3

-0.1

0.1

0.3

0.5

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

-0 . 0 6

-0 . 0 4

-0 . 0 2

0

0 . 0 2

0 . 0 4

0 . 0 6

0 0 . 1 0 . 2 0 . 3 0 .4 0 . 5 0 . 6 0 . 7 0 . 8 0 . 9 1

-0 . 015

-0 . 0 1

- 0 . 00 5

0

0 0 . 1 0 . 2 0 . 3 0 . 4 0 . 5 0 . 6 0 . 7 0 . 8 0 . 9 1

� � 3

03

dxdEIxQ z

zzz�

� � 2

02

dxdEIxM z

zzy�

� �dxdx z

y

0��

� �xz�

� �8

)0(max2�qQxM zy


� �)0(max yzz x ��

Figure 5.9: Simply supported beam: linear elastic solution �elds.

5.6. CHAPTER SUMMARY: SOLVINGLINEARELASTIC PROBLEMS221

�� (0) and A� (0) from:

�0� (� = ;) = �1

�<��

μ1

2�� (0) ;

2 +1

6A� (0) ;

3 +1

24B;4¶= 0

�� (� = ;) =�� (0) +A� (0) ;+1

2B;2 = 0 (5.161)

�� (0) =

1

8B;2; A� (0) = �5

8B;

The linear-elastic beam solution �elds read (Fig. 5.10):

A� (�) = ��<�� 3�0��3

= B

μ�� 5

8;

¶(5.162a)

�� (�) = ��<�� 2�0��2

= B

μ1

8;2 +

1

2�2 � 5

8;�

¶(5.162b)

:� (�) = ��0�

��=

B

�<��

μ1

8;2�+

1

6�3 � 5

16;�2¶

(5.162c)

�0� (�) = � B

�<��

μ1

16;2�2 +

1

24�4 � 5

48;�3¶

(5.162d)

Let us note that the minimum and maximum bending moments occur at� = 0, as well as at ��; = 5�8 = 0)625 where the shear force is zero, andwhere� (� = 5�8;) = � (9�128) B;2. Similarly, the minimum and maximumdisplacement occurs where the rotation is zero, that is, at � = 0 and at��; =

¡15��33¢ �16 = 0)578, where (Fig. 5.10):

�<��B;4

�0� (� = 0)578;) = �5) 4161× 10�3 (5.163)

5.6 Chapter Summary: Solving Linear Elas-tic Problems

This �rst chapter on elasticity was dedicated to the linear elastic stress, forceand moment equations of state, and to solution schemes that can be employedfor solving linear elastic continuum and structural mechanics problems. Thesalient feature of elasticity is its reversibility: the work rate provided from


4

04

dxdEIqf z

zzz�

x

yz

-0.1

-0.05

0

0.05

0.1

0.15

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

-1

-0.5

0

0.5

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

-0.025-0.02

-0 .015-0.01

-0.0050

0.0050.01

0 .0150.02

0 0.1 0.2 0 .3 0.4 0.5 0.6 0.7 0 .8 0.9 1

-0. 006

-0 . 005

-0 . 004

-0 . 003

-0 . 002

-0 . 001

0

0 0 . 1 0 . 2 0 . 3 0 . 4 0 . 5 0 . 6 0 . 7 0 . 8 0 . 9 1

� � 3

03

dxdEIxQ z

zzz�

� � 2

02

dxdEIxM z

zzy�

� �dxdx z

y

0��

� �xz�

� �)0(max zyy QxMM


� �)0(max yzz x ��

4

04

dxdEIqf z

zzz�

x

yz

-0.1

-0.05

0

0.05

0.1

0.15

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

-1

-0.5

0

0.5

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

-0.025-0.02

-0 .015-0.01

-0.0050

0.0050.01

0 .0150.02

0 0.1 0.2 0 .3 0.4 0.5 0.6 0.7 0 .8 0.9 1

-0. 006

-0 . 005

-0 . 004

-0 . 003

-0 . 002

-0 . 001

0

0 0 . 1 0 . 2 0 . 3 0 . 4 0 . 5 0 . 6 0 . 7 0 . 8 0 . 9 1

� � 3

03

dxdEIxQ z

zzz�

� � 2

02

dxdEIxM z

zzy�

� �dxdx z

y

0��

� �xz�

� �)0(max zyy QxMM


� �)0(max yzz x ��

Figure 5.10: Statically indeterminate beam: linear elastic solution �elds.

5.6. CHAPTER SUMMARY: SOLVINGLINEARELASTIC PROBLEMS223

the outside to the system is entirely stored in free energy. This reversibilityprovides a direct link, the state equations, between stresses and strains, forcesand strains, moments and curvatures, and so on. The solution of a linearelastic problem then involves the following three elements:

1. Statically admissible stresses, forces and moments: The linear elasticstress solution �eld �� (��) and the section forces and moments of beam

elements³�� (�) �

��M� (�)´are statically admissible. They satisfy the

stress, force and moment boundary conditions (on ��), and the mo-mentum conservation (in �). We have discussed these conditions indetail in Chapter 2.

2. Kinematically admissible displacements: The linear elastic displace-ment solution is kinematically admissible; that is, the displacement�eld satis�es the displacement boundary conditions (on ��), and,in the case of beam elements, rotational boundary conditions. Localstrains and curvatures (in �) derive from such a kinematically admis-sible displacement �eld. We have discussed deformation and strains indetail in Chapter 4.

3. Equations of State: The two elements of the solution, ie. the staticallyadmissible stress �eld and the kinematically admissible displacement�eld, are linked by the linear elastic constitutive laws, ie. for continuumsystems by the state equation of stress (5.49):

Continuum: � =μ� � 2

3#

¶��1+ 2#� (5.164)

and for beam-type structures by the section force and section momentsstate equations (5.108) and (5.124), which can be summarized in thesymbolic form:

Structures: �A� = K�� · �B� (5.165)

where �A� is the vector that assembles section forces and section mo-ments, �B� is the vector of section deformation which assembles axialdeformation, twist and curvatures; while the matrix K�� stands forthe section rigidity properties.

Direct solving methods of elasticity combine these three elements to ob-tain the linear elastic stress—displacement solution �eld,

³��

´.

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