Part IV Elasticity and Thermodynamics of Reversible Processes

Home , Elasticity (physics), Strain (chemistry)

175 Chapter 5

Elasticity: An Energy Approach

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 proportional to force’. In Engineering Mechanics, this observation translates into the theory of linear elasticity. Elasticity describes reversible material behavior, and the notion of reversibility is a hallmark of another engineering science discipline: Thermodynamics. Thermodynamics is the study of energy and energy transformations. Energy, which exists in many forms, such as heat, light, chemical energy, and electrical energy, is the ability to bring about change or to do work. Hence, an elastic, ie. reversible, behavior is readily recognized to be one where the energy provided to a material or structural 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 to a fundamental physics law. The focus of this Chapter is to employ the laws of thermodynamics as a backbone for the engineering investigation of linear elastic behavior of materials and structures.

176 5.1. 1-D ENERGY APPROACH 177

0.4 d F0.35

0.3

0.25

0.2 KS 0.15 (x) 0.1

0.05 K x S 0 1 0 0.1 0.2x 0.3 0.4 0.5 F d

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 ononesideisthesimplest representation of an elastic behavior (Fig. 5.1). While equilibrium tells us that the externally supplied force is equal to the force in the spring, that is, = , Hooke’s observational law links to the extension at the point of 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 Law and the Second Law of Thermodynamics.

5.1.1 The First Law The First Law of Thermodynamics states that energy can be changed from one form to another, but that it cannot be created or destroyed. Energy is conserved. This is expressed by the internal energy U of a system (of dimension [U]=22). Specically 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 Q provided from the outside. Formally we write: U W Q (5.2) = + where we employ the symbol ‘ ( )’ to specify that the rate is not necessarily the time derivative of a function . For instance, the work rate W is generally not the total time derivative of the work W, ie. W =6 W .For instance, 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 W is what the force realizes along the velocity = 0 at the point of load application, that is, W = .

5.1.2 The Second Law The Second Law of Thermodynamics species the direction of spontaneous change of energy, which is expressed by another physics quantity, the internal entropy S (of dimension [S]=221). The entropy is a measure of the quality of the energy. Specically, the Second Law informs us that the change of the internal entropy is always greater or equal to the entropy supplied to a system in heat form; that is: S Q (5.3) 0 where 0 0 stands for the absolute temperature, which is assumed to be constant in what follows. The dierence between the left and the right hand side is recognized as the fraction of entropy which is produced internally in aspontaneous(ie. uncontrolled) fashion, and which manifests itself as an internal heat source, D S Q = 0 0 (5.4) This internal heat source is called dissipation. More specically, if we substitute the First Law (5.2) into (5.4), we obtain: D W U S = ( 0 ) 0 (5.5) In this form, the Second Law appears as a balance law between the external work rate W and some form of recoverable energy (U 0S). This energy 5.1. 1-D ENERGY APPROACH 179 is called the free energy or Helmholtz Energy, which we denote by ,and which expresses the maximum internal capacity of a system to do work:

=U 0S (5.6)

By denition, an elastic behavior is one where the entire work rate supplied from the outside is stored as recoverable energy; that is:

D W (5.7) Elastic: = =0

Hence, since W is known, it suces to determine the expression of the free energy of the spring. In the case of our 1-D spring system, the free energy is recognized to be the area below the curve (Fig. 5.1); thus in the 1 2 case of the linear behavior =( )= 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 observational law, 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 instead of (5.1): ! = = = (5.10) The partial derivative of w.r.t. is called state equation, and species the constitutive law of the spring force in equilibrium with the external force . Finally, if the spring stiness is constant, we speak about linear elasticity. Inreturnif () depends on ,weneedtodealwithanon linear elasticity problem. On the other hand, it does not need much imagination to gure out that the spring stiness cannotbenegative,sinceitwould make not much sense that the spring extension ,inresponsetotheload , takes o in the opposite direction of the applied force. Behind this observation, however, hides an important property of the free energy function 180 CHAPTER 5. ELASTICITY: AN ENERGY APPROACH

L 123

z P 0 P

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

=(). In fact, the non-negativity of the spring constant is ensured provided 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 for solving elasticity problems.

5.1.3 Example: Three-Truss System By way of example, consider the truss system displayed in gure 5.2. The structure is composed of three bars of same length held on the top and connected 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 system can be considered as a discrete system, in which each bar represents a spring of stiness . We want to evaluate the forces in the trusses (ie. the spring forces) and evaluate the displacement 0 at the point of load application.

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

1 + 2 + 3 = (5.12a) 31 + 2 3 =0 (5.12b) 5.1. 1-D ENERGY APPROACH 181

We readily recognize that the structure is statically indeterminate of degree 1; that is, there are only two equilibrium relations for the three unknowns, 12 and 3. Let us now introduce a linear elastic constitutive law for each truss of the form: = (5.13) where stands for the elongation of bar =1 3. We now need to ensure the kinematical compatibility between the bar elongations and the constraints on both 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. Compatibility requires:

1 = 1 (5.14a) 2 2 = = + ( ) (5.14b) 2 1 3 0 1 4 3 = = + ( ) (5.14c) 3 1 3 0 1 While there are apriorithree displacement unknowns in the problem, the geometrical compatibility allows us to reduce these unknowns to two. We here 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. Substitution of the constitutive law (5.13) and the geometrical compatibility conditions (5.14) in the equilibrium relations (5.12) now yields a system of two equations for the two unknowns: μ ( 1 +2 0¶)= (5.15a) 11 2 =0 (5.15b) 3 1 3 0 11 The linear elastic problem then is readily solved: 0 = 24 and 1 = 1 12 . The truss forces are given by: 7 1 = ; 2 = ; 3 = (5.16) 12 3 12 The problem nicely shows that the use of a constitutive law and geometrical compatibility conditions allows one to obtain a unique solution for a statically 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. 182 CHAPTER 5. ELASTICITY: AN ENERGY APPROACH

Energy Approach Let us now take the alternative energy route to the problem. Instead of equilibrium, 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 simply the sum of the individual spring energies: ¡ ¢ 1 2 2 2 ( 1 2 3)= + + (5.18) 2 1 2 3 While we do not evoke any equilibrium, we need to ensure the geometric compatibility conditions (5.14). This yields the free energy as a function of 0 and 1: μ ¶ 1 11 2 4 20 2 ( )= + (5.19) 0 1 2 9 1 9 1 0 9 0 The change of the free energy is then given by: ( 0 1) ( 0 1) ( 0 1) = 0 + 1 0 ¸ 1 ¸ 2 1 = ( +10 ) + (11 2 ) (5.20) 9 1 0 0 9 1 0 1 Finally, substituting (5.17) and (5.20) in (5.7) yields: ¸ ¸ D 2 1 = ( +10 ) (11 2 ) =0 (5.21) 9 1 0 0 9 1 0 1

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 for the two unknowns 0 and 1 of the problem: ³ ´ 2 D 9 ( 1 +10 0)=0 0 1 ; =0 (5.22) 1 9 (11 1 2 0)=0 We readily verify that the solution of this system of two equations yields the same solution as (5.15). However, the way how we obtained it, did 5.2. 3-D ENERGY APPROACH 183 not evoke any equilibrium, but just an energy balance and the displacement compatibility. Finally,letusnoteagainthatthefreeenergyhereisconvexw.r.t.its arguments 0 and 1. In fact, analogous to (5.11) it suces to determine the stiness matrix: 2 2 20 2 2 0 0 1 9 9 [ ]= 2 2 = (5.23) 2 11 2 10 1 9 9 From a mathematical point of view, the stiness matrix [] derives from a quadratic form, that is, the free energy expression (5.18) which is quadratic in displacements. For this reason, the matrix is positive denite, which implies 8 2 that its determinant is positive, det [ ]= 3 0, which is a generalization of the convexity condition (5.11) of the 1 parameter system to a system with multiple displacement degrees of freedoms. The stiness matrix, therefore, can be inverted, which allows solving the problem from the linear system of equations: []()=() (5.24) where ()=(01) 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 the elementary system we are interested in is the r.e.v. (see Section ??,Fig. 2.2). For this elementary system, at each material point ,thereare15 unknowns, namely six stresses = , six (linearized) strains = , and three displacements . For these unknowns, we have three equilibrium relations, div + =0 (5.25) and six (linearized) strain-displacement relations: ³ ´ 1 = grad +(grad) (5.26) 2 184 CHAPTER 5. ELASTICITY: AN ENERGY APPROACH

Thus, there is a total of 9 equations for 15 unknowns; 6 equations are missing for the problem to be well posed. The missing equations need to be provided by 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 realize an external work rate along a velocity eld . This velocity eld must be compatible with velocity boundary conditions, so to avoid that the forces do work along a rigid body motion of the body. Such a velocity eld is called kinematically 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 ³ ´ · · = div · Z ³ ´ = · div +grad : (5.28) where the symbol ‘:’ stands for a double tensor contraction (a ‘double dot product’). Then, a substitution of (5.28) in (5.27) yields: Z Z = grad : + · (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 the second 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 Z Z

= d : = tr (d · ) = (5.30) 5.2. 3-D ENERGY APPROACH 185 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 for summation). Equation (5.30) is the strain work rate, that is, the work rate the stresses realizealongthestrainrates , which —due to the aforemen- tioned symmetry reasons— represent the components of the symmetric part of the velocity gradient grad : μ ³ ´ ¶ 1 d = grad + grad (5.31) 2

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

Theorem 4 (of Strain Work Rate (Continuum)) Theexternalworkrate realized by surface tractions and volume forces of a continuum system along a 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 velocity boundary 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 the material point in the deformed conguration (see Section 4.2, Fig. 4.2). In turn, within the limits of a linear deformation theory, in which we merge the initial and deformed conguration for spatial derivatives (see Section 4.3, Eq. (4.33)), the strain rate tensor is recognized to be the time derivative of the linearized strain tensor (5.26):

Linear: d ' (5.33)

We will exclusively address this linear case in (almost) all what follows. 186 CHAPTER 5. ELASTICITY: AN ENERGY APPROACH

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 link to the local strain work rate of the r.e.v. This quantity is readily employed in both the First Law (5.2), Z U = + Q (5.34) and the Second Law (5.7) written for reversible processes that take place in the r.e.v.: Z μ ¶ D = =0 (5.35) where is the free energy volume density (of dimension []=3 [ ]= 12) of the r.e.v., so that the total free energy is =.Sincethe dissipation must be zero for any material volume that undergoes elastic (ie. reversible) evolutions, equality (5.35) must equally hold for the r.e.v.; that is: (5.36) Elastic: =0 The Second Law of Thermodynamics of Reversible Processes, therefore, provides a means to link the stresses to the free energy density contained in the r.e.v. Indeed, it suces 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 the continuum 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 known as a function of the strain, the link between the stresses and the strain is provided by the state equation (5.37). Here lies all the strength of the energy approach. Finally, the 1-D spring constant as dened by (5.11) extends to

3-D as follows: 2 = = (5.38) 5.2. 3-D ENERGY APPROACH 187

The forth order tensor is called the elastic stiness tensor of the material. This stiness tensor has a priori 9 × 9=81elasticity constants (nine stresses derived w.r.t. to nine strain components). If we use the symmetry conditions of both stress tensor and strain tensor, these 81 constants reduce to 6 × 6=36(six stresses derived w.r.t. to six strain components):

= = = = = = (5.39)

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

2 2 ! = = = (5.40)

These symmetry conditions reduce the 36 constants to 21. No doubt, such a ‘monster’ tensor with 21 elasticity constants is mathematically much more involving; but let us keep in mind, that is, in a nutshell, just an extension 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 specic orientation. As regards elasticity, this means that the free energy depends only on the invariants of the strain tensor. What are such strain invariants? —AswehaveseeninSection4.3,thelinearizedstraintensor describes¡ two¢ modes of deformation, length dilations ! ( ) and half-distortions " . Both are dened w.r.t. specic directions, and cannot therefore be employed as invariants in an isotropic theory. A true invariant is recognized to be the volume 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 with the free energy of the linear spring system, we develop the energy associated with volume dilatation of a linear isotropic material in a quadratic form: 1 = 2 (5.42) 2 188 CHAPTER 5. ELASTICITY: AN ENERGY APPROACH 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 to capture shear stresses. Hence, we need to nd a second invariant, one that is able to capture the change in free energy due to distortions. A clever idea is to rst subtract from the total strain the average volume dilation; that is: 1 e = 1 (5.44) 3 This generalized shear strain tensor is called strain deviator tensor,andhas —by construction— a zero trace, ie. tr e =0,andthereforeismostsuitable to describe distortions. Since tr e =0, we need to consider the second order norm; for instance: r s μ ¶ 1 1 2 =2 e : e = 2 : (5.45) 2 3 where the factor 2 beforetherootcomesfromthefactthatthequantity represents a representative distortion (and not a half-distortion). The associated free energy of distortion is of the form: 1 = #2 (5.46) 2 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-strain deviator relation: # #e = = =2 (5.47) Moreover, since energies are additive, the total free energy density of a linear isotropic material is readily obtained by summing up the volume dilation contribution (5.42) and the shear distortion contribution (5.46): 1 2 #2 = + = 2 ( + ) (5.48) Moreover, if we assume that ( #) are constants, which constitutes the basis for a linear isotropic elasticity model, the total stress equation of state 5.2. 3-D ENERGY APPROACH 189 is obtained by summing up (5.43) and (5.47): 1 #e = + = μ +2 ¶ 2 = # 1 +2# (5.49) 3 Finally, it always handy to have the inverse of this relation, relating strains to stresses. This form is readily obtained from (5.43), (5.44), (5.47) and (5.49):

1 1 s = 1 + e = + 3 μ3 2# ¶ 1 1 = 1 + (5.50) 3 2# 2# wherewelet: 1 = 1 = tr ; s = (5.51) 3 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-stress equation (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 diculty we recognize that the complimentary energy () of an isotropic linear elastic material reads: μ ¶ 2 2 1 ( )= ( )+ ( )= + (5.55) 2 # 190 CHAPTER 5. ELASTICITY: AN ENERGY APPROACH

# 9 % 1 32 ! 2 # & # ( ) = 3+ = 2 3+ = 3 = # % ! & = 3(12) = 2(1+) ( ) = (1+)(12) = 2(1+) ! 2 & # & & 3+2 % ! & = + 3 = = + = 2(+) ( ) Table 5.1: Sets of linear isotropic elasticity constants and conversions. where can be seen as a generalized shear stress, called stress deviator invariant: r r 1 1 = s : s = ( : 32 ) (5.56) 2 2 In summary, the linear isotropic elasticity model is characterized by two material stiness properties, and #, which are stiness values associated with two particular deformation modes, ie. volume dilation and generalized distortion, introducing a change of the free energy. More generally, the very procedure applied here to deduce from energy considerations the linear isotropic elasticity model is of general interest. The key idea is to translate deformation modes of an r.e.v. into a free energy expression, which —we recall— represents the maximum capacity of a system to do work. The thermodynamics of reversible processes then provides a means to link stresses to strains. This is what is often referred to as constitutive modeling. Last, the choice of and # to represent a linear isotropic elastic behavior is not mandatory. In fact, it is common practice to represent the linear isotropic behavior by other sets of two constants. The Box 5.1 provides a summary of dierent sets of isotropic elasticity constants that are frequently employed. 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 model to 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 seen in 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 innitessimal; hence: ° ° μ ³ ´ ¶ ° ° ¿ 1 H1: °Grad ° 1 = 2 grad + grad (5.57)

A second assumption we need for solving linear elastic problems is the small displacement assumption: ° ° ° ° H2: °° ¿ (' ( )))) (5.58) where ' ( ))) stand for characteristic length scales of the problem. Assump- tions H1 and H2 dene a true rst-order strain—displacement theory.They are also referred to as small perturbation assumptions. On this basis, we are able to solve for stresses, strains and displacements in a linear elastic problem. As a reminder, the solution stress eld is statically admissible in the sense of (3.21), which we won’t get tired to recall: = ( ) () on · =0 () on + ( )+ ( )=0 (5.59) ( )= · ( ) in div + 0 =0 =

We have added here an additional requirement for the problem to be linear and non-dissipative; that is, the contact must be frictionless, · =0. Moreover, the solution displacement eld is kinematically admissible; that is, it satises the displacement boundary conditions where displacements are prescribed on the boundary. Furthermore, the contact must be perfect (no unilateral boundary condition · =0); hence: =

( ) on · (5.60) μ =0 ³ ´ ¶ 1 () in = grad + grad 2 192 CHAPTER 5. ELASTICITY: AN ENERGY APPROACH

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 a linear isotropic material by: μ ¶ 2 = # 1 +2# (5.61) μ 3 ¶ 1 1 1 = 1 + (5.62) 3 2# 2#

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 of height ' on a rigid substrate, which is subjected to a surface pressure (Fig. 5.3). We start with formulating the boundary conditions of the problem. One boundary condition is the the pressure boundary condition at =0:

=0: = = =0; = (5.63)

The second boundary condition is the zero displacement boundary condition ' at = : = ' : =0 (5.64) To solve the problem, we start with a displacement eld. Given the semi- innite nature of the problem, the displacement elds has only one component 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 single non-zero component, namely: = (5.67) 5.3. SOIL AND ROCK MECHANICS APPLICATIONS 193

p p d T zz z z

H 0 g

d 0gH p

Figure 5.3: Exercise: Elastic soil layer consolidation, with stress and displacement boundary.

Use of = in the linear isotropic elastic constitutive law (5.61) yields: μ ¶ 2 = # 1 +2# (5.68) 3 We now need to check that the so-obtained stress eld is statically admissible. We start with the equilibrium condition. Since = (), the stress eld also depends only on , ie. = (). Furthermore, the stress eld (5.68) has no shear terms, the only equilibrium relation of interest is: μ ¶ 2 4 + =0 + # = (5.69) 0 3 2 0 We integrate twice to obtain the displacement: 2 0 ()= 1 + 2 ¡ ¢ (5.70) 4 # 2 + 3 where 1 and 2 are two integration constants, which need to be determined by considering the boundary conditions (5.63) and (5.64); that is: μ ¶ 4 ( =0)= + # 2 = (5.71a) 3 2 0' ( = ')= 1 + 2' ¡ ¢ =0 (5.71b) 4 # 2 + 3 194 CHAPTER 5. ELASTICITY: AN ENERGY APPROACH

It follows: 2 = ¡ ¢ (5.72a) 4 # + 3 2 0' 1 = ¡ ¢' + ¡ ¢ (5.72b) 4 # 4 # + 3 2 + 3 Thedisplacementsolutionreads: ('2 2) ()=¡ ¢ (' ) 0 ¡ ¢ (5.73) 4 # 4 # + 3 2 + 3 The stress solution reads:

()= ( + 0) (5.74a) 2# 3 % ()= ()= ( + )= () (5.74b) 3 +4# 0 1 % where % is the Poisson’s ratio (see Box 5.1). The stress solution shows that the zero-strain condition in the and direction generates stresses in these directions. In fact, as the material wants to dilate under the vertical pressure and own weight, this plane strain condition restrains the dilation, which leads to ‘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 elastic ³ ´ solution , for which the stress is statically admissible, the displacement kinematically admissible, and for which both are related by the linear elastic constitutive law. This solution strategy is called a displacement approach. In contrast, a solution strategy based on a stress eld as a starting point is referred to as stress approach, which we will use later on. Last, given the linearity of the problem formulation, it is not surprising that the two load cases sum up linearly. Hence, we could have solved for the two load cases separately and superpose the solution elds. This principle constitutes the Theorem of Superposition, which can be formally written in the form: 1 1 7 1 1 1 + 2 1 + 2 (5.75) 0 (1) 0 (2) (1) (2) where the left side represents the loading multiplied by load factors 1, while the right hand side stands for the linear elastic solution elds. 5.3. SOIL AND ROCK MECHANICS APPLICATIONS 195

5.3.3 Deep Tunneling The second application of the continuum linear elastic model is taken from rock mechanics: the deep tunneling problem. By deep tunneling we mean the excavation of a cylindrical hole whose depth ' below ground is large compared to the tunnel radius 2 (Fig. 5.4):

2 ¿ ' (5.76)

The length dimension in the tunnel direction is assumed to be innite. The deep tunneling problem is of great importance, not only for actual tunnel engineering applications, but also for oil and gas drilling applications recovering oil and gas through tunnel-shaped wells, bioengineering applications, such as bone remodeling by cells that resorb bone by acidic activity along resorption tunnels, and so on.

Dimensional Analysis We are interested in the elastic displacement of the tunnel walls due to the excavation. This displacement is readily recognized to depend on the specic weight 0 (where 0 is the initial mass density), the tunnel radius 2 and the tunnel depth ', and the elastic properties ( #) of the excavated rock:

= (0 #2 ') (5.77)

The exponent matrix of dimension is readily established:

[ ][0][2]['][#][] 1 211 1 1 (5.78) 01 0 0 1 1 0 200 2 2

The rank of the matrix is =2, which corresponds to the number of dimensionally independent variables. Choosing # and 2 as dimensionally independent variables, application of the Pi-Theorem (1.14) yields a relation between the dimensionless displacement and three dimensionless Pi-numbers: μ ¶ ' ' F 0 0 0 = 2 = 1 = # ;2 = 2 ;3 = # (5.79) 196 CHAPTER 5. ELASTICITY: AN ENERGY APPROACH

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

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

Figure 5.4: 3-D Elasticity exercise: Deep Tunneling 5.3. SOIL AND ROCK MECHANICS APPLICATIONS 197

1 0 The rst invariant 1 = 0'# is representative of the overload (per surface) generated at a characteristic depth '. The second invariant 2 = '2 represents the depth-to-tunnel radius. In the deep tunneling situation this invariant is known. In fact, according to the deep tunneling assumption 1 (5.76), 2 goes to innity, respectively its inverse 3 goes to zero. Hence, from the point of view of dimensionless analysis, 2 = '2 does not aect the solution of the problem, and can be neglected in the evaluation of the displacement. 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 the problem. Provided that everything is linear in the problem, in the sense of the set of equations (5.57) to (5.62), the elastic displacement is proportional 0 to the applied load, which is represented here by the rst invariant 1 = 0'#. This implies that the normalized displacement scales linearly with '#;thus: μ ¶ μ ¶ ! ' 2' lim = 0 F = 0 F (5.80) 0 2 # # # #

Finally, let us remind us that we operate within the framework of the hypothesis of small displacements (5.58), which reads here | | 2 ¿ 1.This assumption implies: ¯ μ ¶¯ | | ¯ ' ¯ ¿ ¯ 0 F ¯ ¿ 2 1 ¯ # # ¯ 1 (5.81)

If we estimate function function F (#) to be on the order of unity, which we 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 relevant for the specic tunneling problem, from 0'# ¿ 1. For instance, an excavation of a ' =1 000 m tunnel in a soft rock (e.g. compacted clay) 3 of density 0 2 000 kg/m and shear modulus # 3 10 GPa, satises the small displacement condition; while the same tunnel excavated in a very soft clay with similar density but a much smaller stiness in the tens of kilo-pascals, does not satisfy the small displacement assumption.

1 0 Note that we let 1 =12,where1 = . We here make use of the possi- bility oered by D-analysis that we can always generate invariants as products and power functions of previously identied ones (see Section 1.1.3). 198 CHAPTER 5. ELASTICITY: AN ENERGY APPROACH

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, a tunnel of radius 2. In a rst approach, the initial prestress can be assumed to be hydrostatic; that is:

0 ()=01 (5.82) where 0 isthecoordinatebelowthesurfaceat =0.Indeeptunneling, the variation of the initial prestress due to deadweight variation over the tunnels diameter 22 can be neglected; since: μ ¶ 2 ¿1 ' 2 ' 1 ' ' 0 ( )= 0 1 ' 0 ( ) (5.83a) μ ¶ 2 ¿1 ' 2 ' 1 ' ' 0 ( + )= 0 1+' 0 ( ) (5.83b) where 0 (')=0'1. By its very denition, this hydrostatic stress state produces a stress vector whose magnitude is the same in all directions (see Section 2.4; Eq. (2.67)); that is, ( )=0 ,where0 = ' 0 and istheunitnormaltoany surface. For instance, it also holds for the surface = 2 oriented by = prior to the tunnel’s excavation. Furthermore, by denition, the natural prestressing is related to deformation at some geological time scales. It is therefore natural to choose this geological state as reference state, and associate it with a zero displacement and zero strain state. This initial load case and the corresponding system response canbesummarizedintheform: = 2 : = 0 0 = 01 7 =0 : =0 (5.84) =0 0 =0 (1) (1)

The tunnel excavation then consists in removing the prescribed stress vector at = 2. The nal result is a stress-free boundary at the tunnel wall: = 2 : =0 =? 7 =? : =0 (5.85) 0 =0 =? 5.3. SOIL AND ROCK MECHANICS APPLICATIONS 199

1 d 0 p0 T p0 n

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 dierence between (5.84) and (5.85) is the change in stress, strain and displacement induced by tunneling excavation; that is: = 2 : = 0 = 2 : =+0 : =0 + : =0 0 =0 (1) 0 =0 (2) (5.86) 0 = 01 =? =? =0 =? =? + = =0 (1) =? (2) =? where load case (2) represents the excavation load. This principle of superposition 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 inward radial pressure (Fig. 5.5):

= 2 : = 0 = 0 (5.87) 200 CHAPTER 5. ELASTICITY: AN ENERGY APPROACH

The displacement eld is zero at innity:

: =0 (5.88)

Given the overall radial symmetry and the innite extension of the problem in the -direction, the displacement eld is recognized to be a radial displacement: = () (5.89) We remind ourselves that we operate here in cylinder coordinates, for which we have already seen, in Section 4.4, that the non-zero strains in response to a radial displacement are a radial strain and a hoop strain ##.Inamore general form, we have: = ; ## = (5.90) Using the displacement approach, we substitute (5.90) in the linear isotropic elastic constitutive law (5.61). The non-zero stresses are: μ ¶μ ¶ 2# # = + +2 (5.91a) μ 3 ¶μ ¶ 2# # ## = + +2 (5.91b) μ 3 ¶μ ¶ 2 = # + (5.91c) 3

We now need to ensure that these stresses are statically admissible. We start with the equilibrium in cylinder coordinates, for which div =0for a diagonal stress elds reduces to (see Box 2.2): 1 + ( ##)=0 (5.92a) ## =0 (5.92b) =0 (5.92c)

Since = (), we readily recognize from (5.91b)—(5.92b) and (5.91c)— (5.92c) that the equilibrium in the -and -direction is satised. Fur- thermore, substitution of (5.91a) and (5.91b) in (5.92a) yields the following 5.3. SOIL AND ROCK MECHANICS APPLICATIONS 201 dierential equation for : 2 1 2 + 2 =0 (5.93) The solution of this dierential equation is well known:

2 = 1 + (5.94) where 1 and 2 are two integration constants that need to be solved by boundary conditions. The displacement boundary condition at innity (5.88) readily yields 1 =0. In return, the stress boundary condition (5.87) at = 2 provides a means to determine the second integration constant; from:

2 # 2 = ## = 2 tr =0 = 2 2 (5.95) 2 2 02 ( = 2)=2# = 0 2 = 22 2# ³ ´ The linear isotropic elastic solution for the excavation load case, (2) therefore, reads: ¡ ¢ 2 = 2 : = 0 = 0 ( # #) ¡ ¢2 7 0 : =0 = 2 ( # #) 2 0 1 0 =0 (2) = 2 (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 and displacement solution of the tunneling problem into a homogeneous linear isotropic elastic continuum: " μ ¶ # 2 2 1 = 0 ( # #) (5.97a) 2 02 1 = (5.97b) 2# Without diculties we verify that the displacement of the tunnel wall = 2 is of the form (5.80) with F (#)=1.Theelastic tunneling turns out to 202 CHAPTER 5. ELASTICITY: AN ENERGY APPROACH be a radial unloading process from an initially compressed state. Figure 5.6 displays 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 of the tunneling, which is reached at the tunnel wall = 2, where the stresses are maximum; that is in terms of the principal stresses:

= 2 : % = =0%% = = 0 %%% = ## = 20 (5.98)

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

0' 1 Elastic: N! = (5.99) 0 2 where 0 is the uniaxial compressive strength, ie. 0 =2 in the case of a Tresca Material, and 0 =2 cos 7 (1 sin 7) 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 in Chapter 2, Section 2.5, which we have already employed for the strength of beam members in Chapter 3, Section 3.4. Following this approach, the application 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-type structures, based on a combination of the beam stress model (Section 2.5.2), the linear beam deformation model (Section 4.6) and the continuum linear isotropic elastic theory (Section 5.2.4). In the next section, we will see how these section constitutive laws are employed to determine the linear elastic force-, moment- and deection elds of beam-type structures.

5.4.1 Governing Equations We consider a section + in the plane of a beam oriented in the -direction (Fig. 5.7). 5.4. LINEAR ELASTIC BEAM SECTION CONSTITUTIVE LAWS 203

p0 p0 p0

r r

1 12 p 1 p0 +=p0 1 0

r R r R r R r R

2 2 2 G G G p0 p0 p0

r r

2 2 2 G G G 1 1 p 1 1 p p0 + 0 = 0

r R r R r R r R

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 illustrates 8 1 that the maximum shear stress max = 2 ( % %%%) is related, in isotropic " 1 linear elasticity, to the maximum half-distortion max = 2 ( % %%%) by twice the shear modulus 2#.

z y

x n dS

Figure 5.7: Beam model. 204 CHAPTER 5. ELASTICITY: AN ENERGY APPROACH

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 (5.100) 00 (2) the corresponding stress vector ( = ) at any point = + in the cross-section is given by: ( = )= + + (5.101) and (3) the stress vector is linked to section forces and section moments by the reduction formulas (2.86) and (2.88): Z ()= ( = ) + (5.102a) Z M ()= × ( = ) + (5.102b) Furthermore, on the deformation side, we restrict ourselves to the linear beam deformation model introduced in Section 4.6: (1) the displacement eld is dened by the Navier-Bernoulli assumption (4.82) and (4.90), for which the strains read: 0 0 0 + 9 9 1 90 Navier-Bernoulli: = 2 (5.103) 1 90 2 0 0 0 0 where stands for the axial strain, and 99 and 9 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 dened depend only on the position along the beam’s axis .

5.4.2 Stress Approach: Poisson’s Eect and Young’s Modulus Following the two-scale approach for structural members (see Chapter 3, Section 3.4), we consider a beam element of length ; subjected to regular 5.4. LINEAR ELASTIC BEAM SECTION CONSTITUTIVE LAWS 205 moment and force boundary conditions at its end sides. We restrict ourselves to bending moments and axial forces. As a consequence of the force and moment equilibrium relations (2.91) and (2.93)2, the shear forces and the torsion moments are zero, and the problem is entirely dened by: Z & = = + (5.104a) Z Z & M + M = M = + + () + (5.104b) | {z } | {z } M M

We now need to link the stresses to strains. To this end, we substitute the beam 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 and directions, which as we have seen in Section 2.5.2, is a consequence of the beam’s slenderness ratios. The conditions = =0yield and as a function of the axial strain: μ ¶ μ ¶ 4 2 = + # + # ( + )=0 μ 3 ¶ μ 3 ¶ 4 2 = + # + # ( + )=0 3 3 (5.105) μ ¶ 1 3 2# = = = % 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 eect is quantied, in the linear isotropic elastic model, by the Poisson’s ratio %, this transversal deformation is also often referred to as the Poisson’s eect. Furthermore, if we substitute (5.105) in the non-zero constitutive law,

2 As a reminder, the vector of moment equilibrium (5.147) develops in the form (see Eq. (2.93)): M M M =0; =0; + =0

As M M and M are constant over the beam length (regular boundary conditions), the shear forces and are zero. 206 CHAPTER 5. ELASTICITY: AN ENERGY APPROACH we obtain the axial stress as a function of only the axial deformation; that is: μ ¶ μ ¶ 4 2 = + # + # ( + )= (5.106) 3 3 where =9#(3 + #) is the Young’s modulus (see Box 5.1). It is the uniaxial stiness of a material system when the transversal stresses are zero. Finally, if we note that =2# =0, we readily nd that the only non-zero linear elastic beam stress in the problem is: ¡ ¢ 0 0 0 = = + 9 9 (5.107) The stress constitutive equation (5.107) links the local stresses of the beam model (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. We have used this linear stress model in Section 3.4.1 (see Fig. 3.14) as one possible statically admissible stress eld for beam structures subjected to bending moment. The linear elastic constitutive law identies this stress eld as the one associated with linear elastic bending.

5.4.3 Linear Elastic Section Properties: Area Moments of Inertia We are now ready to relate these local stresses to section forces and moments by means of the reduction formulas (5.104). In the case of a homogeneous section (same elastic properties throughout the section), a substitution of (5.107) in (5.104) yields: R R R + + + 0 R R R M + 2 + + 90 = R R R (5.108) M + + 2+ 0 9 Eq. (5.108) denes the section constitutive laws that relate the section force and section moments to section deformation properties by means of section rigidity quantities, which are —for homogenous sections— the product of the elasticR stiness and an area moment of dierent order:R the section + = + + + + R is a zero-order area moment, the integrals = and = + + + 3 are rst-orderR area momentsR (of dimension [ R]=[ ]= ), and < 2+ < 2+ < + the integrals = = and = are second- 4 order area moments (of dimension [< ]= ). From a geometrical point of 5.4. LINEAR ELASTIC BEAM SECTION CONSTITUTIVE LAWS 207 view, the rst-order area moments dene the centroid () of the cross- section; that is: Z Z

+ = + = +; + = += + (5.109) Hence, if we take as reference axis the beam’s centroid, the rst-order area moments are zero, and the axial force — axial deformation constitutive law of the section reduces to:

0 Centroid: = + (5.110)

On the other hand, the second-order area moments are called area moments of inertia, and need to be evaluated for any specic section under consideration. For instance, for a rectangular section, those moment of inertia are readily derived:

Z Z = 2 Z = 2 3 2 2 < = + = = (5.111a) = 2 = 2 12 Z Z = 2 Z = 2 3 2 2 < = + = = (5.111b) = 2 = 2 12 Z Z = 2 Z = 2 < = + = =0 (5.111c) = 2 = 2 The fact that the mixed second-order area moment is zero is due to the fact that < was evaluated in the centroid of the section, and that the rectangular section posses double symmetry w.r.t. the and axis. More generally, < =0for any section that possesses at least one symmetry around one of the section’s axis. In this case, the bending moment relations reduce to the following moment—curvature relations:

0 0 Section Symmetry: M = <9; M = <9 (5.112)

Finally, in this symmetric case with the beam’s reference axis placed in the section’s centroid, relations (5.110) and (5.112) provide an easy way to de- 0 0 0 termine the section deformation quantities, 9 and 9, which are readily employed in (5.107) to determine the axial stress distribution in the cross- section: ()= + M M (5.113) + < < 208 CHAPTER 5. ELASTICITY: AN ENERGY APPROACH

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 the section state equations (5.108), which yields: 1 0 00 0 1 % % 9 0 2 2 M = %%% %%% (5.114) 0 % 9 % M 0 2 2 %%% %%% Whence the axial stress distribution: < < < < M M ( )= + 2 2 (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 elastic stresses in a beam’s section and to identify where the maximum elastic stress due to combined axial force—bending loading occurs. Given the linearity of the deformation theory and of the elastic constitutive law, it is not surprising that the eects of axial force and bending moments on the local stress ( ) simply sum up. Last, without diculties we verify that (5.113) and (5.115) is statically admissible in the sense of the set of equations (3.21). Indeed, subjected to the regular force and moment boundary conditions (5.104), M and M are constant over the beam length, as is ( );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 as before, we subject a beam to a regular torsion moment boundary condition: Z & M = M = ( ) + (5.117) | {z } M where and are shear stresses. All other stresses are zero. In order to be statically admissible, the shear stresses need to satisfy the local equilibrium relations: μ ¶ div = + + + =0 (5.118) 5.4. LINEAR ELASTIC BEAM SECTION CONSTITUTIVE LAWS 209 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 elastic constitutive law (5.61): 0 =2# = #9 (5.120a) 0 =2# = #9 (5.120b) 0 0 where 9 = 9 () is the torsion curvature or twist.Wereadilyverifythatthe stress eld (5.120) satises the local equilibrium conditions (5.118), provided that (homogeneous section): 90 div =0 =0 (5.121) On the other hand, the zero-stress boundary condition (5.119) becomes:

on + : + =0 (5.122) This boundary condition is strictly satised for a circular cross-section, for which in cylinder coordinates ( ):

= cos ; =cos (5.123) = sin ; =sin In this case, use of (5.120) in (5.117) yields: 0 M = #> 9 (5.124) where #> is the torsional rigidity and > is the area moment of torsion inertia (or polar section moment): Z Z Z Z ¡ ¢ 2* 1 > = 2 + 2 + = 2+ = 3 = ?24 (5.125) =*2 0 0 2 where 2 is the radius of the cross-section. For non-circular sections, it is intuitively understood that the boundary condition (5.122) leads to statically admissible shear stress elds that are less regular than the one prevailing in a circular cross-section. As a consequence, a non-circular section (for instance a rectangular section) cannot achieve a similar high torsional rigidity. This observation is known as Saint-Venant’s Theorem: 210 CHAPTER 5. ELASTICITY: AN ENERGY APPROACH

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

In other words, for general cross-section shapes, Eq. (5.121) may well overestimate the area moment of torsion inertia. For such cross-sections, an alternative approach is in order, one that automatically satises both the equilibrium relations (5.118) and the zero-stress boundary condition (5.119). This approach, which is also referred to as the Saint-Venant torsion theory, consists in introducing a stress function or stress potential @ ( )= 0 #97 ( ), from which the linear elastic shear stresses and derive: @ 7 @ 7 #90 #90 = = ; = = (5.126) From a substitution of (5.126) in (5.118), we readily see that this stress potential approach is compatible with div =0, for which it was indeed devised. Furthermore, the boundary condition (5.119) becomes: @ @ + on : = (5.127) p 2 2 Noting that || || = + =1, the only way how the boundary condition (5.127) can be satised, is by letting @ have a constant value on +.Letus return now, for a minute, to the circular section of radius 2,forwhichthe @ 1 #90 2 2 22 7 1 2 2 22 stress function is = 2 ( + ); hence = 2 ( + ).The stress function has a zero-value at the boundary = 2,andhasamaximum in the section’s center. Generalizing this shape for arbitrary connected cross- sections, we let: on + : @ ( )=0 (5.128) and choose @ ( ) in a form to satisfy (5.120); that is: 2@ 2@ 27 27 @ #90 7 = = 2 + 2 =2 ; = 2 + 2 =2 (5.129) where stands for the 2-D expression of the Laplacian. With the help of some mathematical operations (don’t worry, just some integration by parts), the area moment of torsional inertia is recognized to be: Z μ ¶ Z 7 7 > = + + = 2 7 ( ) + (5.130) 5.4. LINEAR ELASTIC BEAM SECTION CONSTITUTIVE LAWS 211

Hence, what it takes to determine the torsional rigidity of non-circular sections is the determination of the stress function @ (),whichcanbeviewed as 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 @ () which satises the boundary conditions (5.128) and the Laplacian relations (5.129) reads: μ ¶ 22 ³ ´2 ³ ´2 @ ( )=#90 + 1 (5.131) 2 + 2

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

2 2 @ 0 2 @ 0 2 = = #9 ; = = #9 (5.132) 2 + 2 2 + 2

The Laplacian of (5.131) is recognized to satisfy (5.129): μ ¶ 2 2 2 2 @ = = #90 + =2#90 (5.133) 2 + 2 2 + 2

Therefore, we can evaluate the area moment of torsion inertia from (5.131), 0 for 7 ( )=@ ( ) #9: Z μ ¶ 22 ³ ´2 ³ ´2 33 > + ? = 2 2 2 + 1 = 2 2 (5.134) + +

Hence, as soon as an appropriate stress function is available for a specic section, the torsion area moment of homogeneous sections is determined using the Saint-Venant Torsion Theory.3

5.4.5 Return to Energy Approach We cannot conclude this Section on Beam constitutive laws without briey evoking the energy approach. We start with the strain work rate theorem

3 Expressions for for sections commonly employed in the engineering practice can be found in Engineering Design Books. 212 CHAPTER 5. ELASTICITY: AN ENERGY APPROACH

(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 ´ i 0 0 0 0 = + 9 9 + 9 ( + ) + + Z ¸ 0 = + M · 9 (5.135) + 0 where we made use of the reduction formulas (5.102) and where 9 = 9 + 0 0 9 +9 is the vector of curvature rates. Next, we express the free energy of a beam per unit beam length: Z Z ¡ ¡ ¢¢ 1 2 2 2 = + = +4# + + (5.136) 2 Use of the Navier-Bernoulli strains (5.103) in (5.136) yields: Z ³ ¡ ¢ ¡ ¢ ¡ ¢´ 1 0 0 0 2 0 2 2 2 = + 9 9 + # 9 + + (5.137) 2 Following thermodynamics of reversible processes, the external work rate (5.135) is equal to the change in free energy; that is: 0 M · 9 2 ,! 0 · 9 + = = 0 + (5.138) 9 It follows from (5.138): Z ¡ ¢ 0 90 90 + = 0 = + Z ¡ ¢ M # 2 2 90 + = 0 = + 9 Z ¡ ¢ 0 2 0 0 M = : M = 0 = + 9 9 + (5.139) 9 9 Z ¡ ¢ 0 0 2 0 M = 0 = 9 + 9 + 9 For homogeneous sections, we won’t have any diculty to identify the derived set of state equations (5.139) with the state equations (5.108) and (5.124). 5.4. LINEAR ELASTIC BEAM SECTION CONSTITUTIVE LAWS 213

5.4.6 What’s the Matter with (Transverse) Shear in Beams? 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 eects 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: Z M A = ( ) + = (5.140) where is a shear stress. The simultaneous presence of bending and shear requires 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 ( )= A ()+ () (5.143) 2< where ()= ( 0) is recognized to be the shear stress in the section’s centroid, where it takes a maximum value. In fact, use of (5.143) in the rst part of Eq. (5.140) provides a means to determine it from: R 2+ 3 A () A ()= A ()+ () + ()= ( 0) = (5.144) 2< 2 + R < 2+ wherewemadeuseof = . The shear stress distribution over the beam section’s height thus reads: μ ¶ 2 A () 3 + ( )= (5.145) + 2 2< The shear stress distribution is found to have a parabolic shape over the cross-section, with a maximum value in the centroid. To complete the shear 214 CHAPTER 5. ELASTICITY: AN ENERGY APPROACH model, we still need to check that the shear stress at the beam’s surfaces is zero. For that matter, let us consider a¡ rectangular¢ section + = and < 3 .() 3 2 = 12,forwhich ( )= 2 6( ) .Inthiscase,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 sections remain plane throughout the beam’s deformation and perpendicular to the 0 beam’s reference axis, according to which —in the absence of torsion— =0. To make a long story short, it is not compatible, since:

0 ( 0) = 2# =06 (5.146)

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

5.5 Structural Mechanics Application

5.5.1 Linear Elastic Beam Deection: Dierential Ap- proach

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

4 This is an analogous to the incompressibilty assumption frequently made in uid mechanics, 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 pressure is solved from equilibrium relations, very much in the same vain as we have done here for the shear stress. 5.5. STRUCTURAL MECHANICS APPLICATION 215 relations (2.91) and (2.93): + =0 [0;]: (5.147) M × + =0 which can be combined reading: 2M 2 =0 2 2 M M × =0: (5.148) 2 2 + =0 2 M =0 2 A substitution of the bending section constitutive equations (5.112) for symmetric sections in (5.148) yields: ¡ ¢ 2 0 <9 2 + =0 Section Symmetry: ¡ ¢ (5.149) 2 < 90 2 =0 0 0 Finally, we remind ourselves that the curvatures 9 and 9 are related, in the Navier-Bernoulli beam theory (see Section ??, Eq. (4.90)) to the section 0 0 rotation, : and :, and the displacement components and by: : 20 : 20 90 90 = = 2 ; = = 2 (5.150)

Then, a substitution of (5.150) in (5.149) yields, for constant < and < along , the following forth-order dierential equations of linear elastic beam bending: 40 < ( ) 4 + =0 Section Symmetry < < : (5.151) ( )=const 40 < ( ) 4 =0 216 CHAPTER 5. ELASTICITY: AN ENERGY APPROACH

The integration of those dierential equations with four appropriate beam forces, moment, rotation and curvature conditions yields the linear elastic solution 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:

= = (5.152)

Integration of the dierential equation of beam deection (5.151())yields:

Z 30 A < A A ( )= 3 = (0) = (0) + (5.153a) Z 20 < A ( )= 2 = (0) + ( ) (5.153b)

1 2 = (0) + A (0) + 2 Z 0 : : 1 ( )= = (0) + < ( ) (5.153c) μ ¶ : 1 1A 2 13 = (0) + < (0) + (0) + Z 2 6 0 0 ()= (0) : () (5.153d) μ ¶ 0 1 1 2 1 3 1 4 = (0) : (0) (0) + A (0) + < 2 6 24

0 where the integration constants A (0) (0) : (0) (0) are force, moment, rotation and displacement boundary conditions at =0. Those need to be solved for the particular beam boundary conditions, as detailed below. 5.5. STRUCTURAL MECHANICS APPLICATION 217

Cantilever Beam The boundary conditions of the cantilever beam are zero displacement and rotation at =0and zero force and moments at = ;;thatis: 0 (0) = 0 0 =0( ) : (0) = =0 (5.154) ½ A (;)=0 ; = ( ) (;)=0

Use of these boundary conditions in (5.153) yields A (0) = ; and (0) = 1 ;2 2 ; whence (Fig. 5.8):

30 A < ; ( )= 3 = ( ) (5.155a) 2 0 2 ()= < = ( ;) (5.155b) 2 2 0 ;3 ¡ ¢ : ; 3 ( )= = < 1+( 1) (5.155c) 6μ ¶ 4 ³ ´4 0 ; ()= 4 + 1 1 (5.155d) 24 < ; ;

The maximum displacement occurs at = ; where:

< 0 ( = ;)=0)125 ;4 (5.156)

Simply Supported Beam The displacement, force and moment boundary conditions of the simply supported beam are: ½ 0 ( =0)=0 ( ) 0 ( = ;)=0 ½ (5.157) ( =0)=0 ( ) ( = ;)=0 218 CHAPTER 5. ELASTICITY: AN ENERGY APPROACH

z y

d 4 0 f q EI z z zz dx4 -1 -0.9 -0.8 -0.7 3 0 -0.6 d -0.5 z -0.4 Q x EI 3 -0.3 z zz -0.2 dx -0.1 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 0

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 0 2 0 0.1 d 0.2 M x EI z y zz dx2 0.3 0.4

0.5

0. 2

0. 15 0 d 0. 1 x z 0. 05 y 0 dx 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

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 z x 0.1 0.12

0.14

Figure 5.8: Cantilever beam: Linear elastic solution elds. 5.5. STRUCTURAL MECHANICS APPLICATION 219

A 1 ; : ;3 < Use of (5.157) in (5.153) yields (0) = 2 and (0) = (24 ); whence (Fig. 5.9):

μ ¶ 3 0 1 A ()= < = ; (5.158a) 3 2 2 0 ¡ ¢ 2 ()= < = ; (5.158b) 2 2 0 ¡ ¢ 3 2 3 : ()= = ; 6 ; +4 (5.158c) 24 < ¡ ¢ 0 3 3 4 ()= ; 2 ; + (5.158d) 24 <

The maximum deection along occurs where the rotation is zero, which is here in mid-span (Fig. 5.9):

< 0 2 : ( = ;2) = 0 (;2) = 1) 302 1 × 10 ;4 (5.159)

Statically Indeterminate Beam

The displacement, rotation, force and moment boundary conditions of a beam clamped at =0and simply supported at = ; are: 0 ( =0)=0 0 ( ) : ( =0)= =0 0 (5.160) ( = ;)=0

; ( ) ( = )=0

Note that although the beam is statically indeterminate, the displacement boundary conditions plus a linking material law permit the determination of forces 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 220 CHAPTER 5. ELASTICITY: AN ENERGY APPROACH

z y

d 4 0 f q EI z z zz dx4

-0.5

-0.3 3 0 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 -0.1 d z 0.1 Qz x EI zz 3 0. 3 dx 0. 5

q2 max M x(Q 0) y z 8

2 0 -0.15 -0. 1 d -0.05 z 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 M x EI 0 y zz dx2 max ( 0) y y x M y 0.06

0.04 0 0.02 d z 0 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 y x -0.02 dx -0.04 -0.06 max ( 0) z z x y

0 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 -0.005 z x -0.01 -0. 015

Figure 5.9: Simply supported beam: linear elastic solution elds. 5.6. CHAPTER SUMMARY: SOLVING LINEAR ELASTIC PROBLEMS221

(0) and A (0) from: μ ¶ 0 1 1 2 1 3 1 4 ( = ;)= (0) ; + A (0) ; + ; =0 < 2 6 24 1 2 ( = ;)= (0) + A (0) ; + ; =0 (5.161) 2

1 2 5 (0) = ; ; A (0) = ; 8 8 The linear-elastic beam solution elds read (Fig. 5.10): μ ¶ 3 0 5 A ()= < = ; (5.162a) 3 8 μ ¶ 2 0 1 2 1 2 5 ()= < = ; + ; (5.162b) 2 8 2 8 μ ¶ 0 : 1;2 13 5 ;2 ( )= = < + (5.162c) μ 8 6 16¶ 0 1 2 2 1 4 5 3 ()= ; + ; (5.162d) < 16 24 48

Let us note that the minimum and maximum bending moments occur at =0,aswellasat; =58=0)625 where the shear force is zero, and where ( =58;)= (9128) ;2. Similarly, the minimum and maximum displacement¡ occurs¢ where the rotation is zero, that is, at =0and at ; = 15 33 16 = 0)578, where (Fig. 5.10): < 0 ) ; ) × 3 ;4 ( =0578 )= 5 4161 10 (5.163)

5.6 Chapter Summary: Solving Linear Elas- tic Problems

This rst chapter on elasticity was dedicated to the linear elastic stress, force and moment equations of state, and to solution schemes that can be employed for solving linear elastic continuum and structural mechanics problems. The salient feature of elasticity is its reversibility: the work rate provided from 222 CHAPTER 5. ELASTICITY: AN ENERGY APPROACH

z y

d 4 0 f q EI z z zz dx4

-1

3 0 -0.5 d 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 z 0 Qz x EI zz 3 dx 0. 5 max ( 0) M y M y x Qz -0.1

-0.05 00.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 2 0 0 0.05 d z M y x EI zz 2 0.1 dx 0.15 max ( 0) y y x M y

0.02 0.015 0 0.01 0.005 d z 0 x -0.005 0 0 .1 0.2 0 .3 0.4 0 .5 0.6 0.7 0 .8 0.9 1 y -0.01 dx -0.015 -0.02 -0.025 max ( 0) z z x y

-0.001 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

-0.002

-0.003 -0.004 z x -0.005 -0.006

Figure 5.10: Statically indeterminate beam: linear elastic solution elds. 5.6. CHAPTER SUMMARY: SOLVING LINEAR ELASTIC PROBLEMS223 the outside to the system is entirely stored in free energy. This reversibility provides a direct link, the state equations, between stresses and strains, forces and strains, moments and curvatures, and so on. The solution of a linear elastic problem then involves the following three elements: 1. Statically admissible stresses, forces and moments: The linear elastic stress solution³ eld ()´and the section forces and moments of beam elements () M () are statically admissible. They satisfy the

stress, force and moment boundary conditions (on ), and the momentum conservation (in ). We have discussed these conditions in detail in Chapter 2. 2. Kinematically admissible displacements: The linear elastic displacement solution is kinematically admissible; that is, the displacement eld satises the displacement boundary conditions (on ), and, in the case of beam elements, rotational boundary conditions. Local strains and curvatures (in ) derive from such a kinematically admissible displacement eld. We have discussed deformation and strains in detail in Chapter 4. 3. Equations of State: The two elements of the solution, ie. the statically admissible stress eld and the kinematically admissible displacement eld, are linked by the linear elastic constitutive laws, ie. for continuum systems by the state equation of stress (5.49): μ ¶ 2 Continuum: = # 1 +2# (5.164) 3 and for beam-type structures by the section force and section moments state equations (5.108) and (5.124), which can be summarized in the symbolic form: Structures: A = K · (5.165) where A is the vector that assembles section forces and section moments, is the vector of section deformation which assembles axial deformation, twist and curvatures; while the matrix K stands for the section rigidity properties. Direct solving methods of elasticity combine these three³ elements´ to obtain the linear elastic stress—displacement solution eld, .