executes stress tensor transformations between curvilinear and Cartesian coordinates

⎣σ11 σ12 σ13

σ12 σ22 σ23

σ13 σ23 σ33

⎤

⎦. (1.90)

Hence, the components of the stress tensor in the new system,σ= {σi j}, are obtained by the standard tensor transformation

σi j=σabTiaTjb. (1.91)

This operation may be expressed using matrix notation as well, as

σ= T ·σ · T^T. (1.92)

To simplify the relations between the stress tensor components before and after transforma-tion, we shall look at the above formula using the vectorsσ andσ, which contain the stress components before and after transformation, namely,

σ= [σxx,σyy,σzz,σyz,σxz,σxy]^T, σ= [σxx,σyy,σzz,σyz,σxz,σxy]^T. (1.93) These vectors may be related asσ= Mσ·σ, where Mσ is a (non-symmetric) 6× 6 matrix, which is clearly a function of the transformation (Euler’s) rotation anglesψ,θandφ. Example terms are

M_σ(3,2) = (cosφsinθsinψ− sinφcosψ)², M_σ(2,3) = sin²φcos²θ. (1.94) The reader may activate P.1.9 to generate all terms of Mσsymbolically. Figure 1.4 presents an example of axis rotation and the associated Mσmatrix. Such a view for different sets of rotation angles may be obtained by activating P.1.10.

To transform a stress tensor given in orthogonal curvilinear coordinates to Cartesian coor-dinates, we select the transformation tensor of (1.216). The result is a function of the spe-cific point under discussion, and may be written as a function of the curvilinear coordinates α1,α2,α3. The corresponding transformation angles may be determined by (1.208).

P.1.11 executes stress tensor transformations between curvilinear and Cartesian coordinates.

x_new

We shall now seek expressions for the principal stresses at a point. In essence, we are looking for a set of rotation angles that will define a new system orientation, in which only the normal stress componentsσ_α(α= x,y,z) are nonzero at a point. Clearly, these values of rotation angles will ensure that the vector

σL= [σyz,σxz,σxy]^T (1.95) will vanish, or in other words, the state whereσ is diagonal, and its elements are called the principal stresses at a point. One way to carry out this task is to set the following set of three equations:

σL = Lσ·σ= 0, (1.96)

where Lσ is a 3× 6 matrix (essentially the last three lines of the matrix Mσ). For a given stress tensor, these are three equations in the three unknown orientation anglesψ,θ, andφ. The solution of such a system is not trivial, and it is much more convenient to adopt a more standard way. In this new course, we carry out an eigenvalue analysis forσ and obtain its eigenvalues {σ^P_i}i=1,2,3and eigenvectors{vi}i=1,2,3so that

σ · vi=σ^P· vi. (1.97)

Hence, in the principal stress state,σii=σ^P_i andσi j= 0 (i = j).

Implementation of the above eigenvalue analysis requires that the system matrix determinant will vanish, namely, 

which leads to the cubic polynomial equation

σ³−Θ1σ²+Θ2σ−Θ3= 0. (1.99) Θiare usually referred to as the stress invariants and are given by the matrix minors as

Θ1=trσ=σ11+σ22+σ33, (1.100a)

SinceΘiare invariants, the resulting eigenvalues are invariants as well. Therefore, the above invariants may be expressed in a more compact form using the principal stresses by ignoring the underlined terms (1.100a–c) and replacingσiiwithσ^P_i. From this point on, we shall assume that the three eigenvalues obtained by solving (1.99) are put in a decreasing order,σ^P₁≥σ^P₂≥σ^P₃, and their eigenvectors vi= {vi(1),vi(2),vi(3)} are put in the same order.

To determine the above stress eigenvalues, it is convenient to define the stress deviator tensor σ^D_{i j}=σi j−¹₃Θ1δi j, which is a second-order tensor as well, and its eigenvalues will be denoted

In addition, the invariants of the stress deviator tensor may be expressed as functions of the stress tensor invariants, namely, Hence, to determineσ^DP_i , one needs to solve the cubic equation

(σ^D)³+Θ^D₂σ^D−Θ^D₃ = 0, (1.103) which is simpler than (1.99). Onceσ^DP_i are known,σ^P_i are directly obtained viaσ^P_i =σ^DP_i +

1

3Θ1, see also Remark 1.4.

To determine the transformation matrix, TE, from a given state to the principal directions along with the corresponding transformation angles ψ, θand φ, we use the above derived eigenvectors vias

To preserve orientation (the transformation matrix determinant must be equal to a unit), one should change the sign of one of the eigenvectors if required. Also, by simple eigenvalue analysis argumentation we obtain

Example 1.1 A Generic Stress State.

To demonstrate the above derivation, we shall examine a generic stress state as shown in Fig. 1.5(a). This state serves as an example only (while units will not be indicated). The above procedure shows that the eigenvalues areσ^P₁= .0823,σ^P₂= −.0187,σ^P₃= −.0937 and that The corresponding principal stress state is shown in Fig. 1.5(b). The principal stress directions at a point may be derived by P.1.12.

1

Figure 1.5:Example of a stress tensor and its principal state.

Remark 1.4 Two additional stress invariants, Θ4 andΘ5 (which are identical for both the stress and the stress deviator tensors) may be defined. The first one is

Θ4= 1

−Θ^D₂. The second invariant is defined by

tan(Θ5) = 1

|Θ5| ≤π/6. The above definitions enable us to write σ^DP₁ ,σ^DP₃ =2Θ√4

3 sin(Θ5±2π

3 ), σ^DP₂ =2Θ√4

3 sin(Θ5), (1.109) which allows graphical interpretation ofΘ4andΘ5as presented in Fig. 1.6(a).

1.3.3.2 Visualizing the State of Stress at a Point

Many visualization methods of the state of stress at a point have been discussed extensively in the literature. In view of the powerful modern visualization tools, the classical methods seem less attractive and important. We will describe the main ideas in this area briefly.

A good starting point is the examination of the stresses over a face of an infinitesimal cube having general orientation so that the normal to the face under discussion is oriented at the x direction as shown in Fig. 1.6(b). Also, the normal stress and the resultant shear stress over this face are denoted as

σN=σ11, (1.110a)

σT=

σ²₁₂+σ²₁₃. (1.110b)

We shall now assume that the original x-, y-, z-axes are the directions of the principal stress state, and write the above two equations using (1.91), in addition to (1.204) (with i= 1). This

(a)Geometrical relations between theΘ4,Θ5 in-variants and the principalσ^DP_i eigenvalue.

(b)An infinitesimal cube at an ar-bitrary orientation.

Figure 1.6:Stress invariants and transformation notation.

procedure constitutes the following set of three equations:

σN=σ^P₁T₁₁²+σ^P₂T₁₂²+σ^P₃T₁₃², (1.111a) σ²_T= (σ^P₁)²T₁₁²+ (σ^P₂)²T₁₂²+ (σ^P₃)²T₁₃²−σ²_N, (1.111b)

1= T₁₁²+ T₁₂²+ T₁₃², (1.111c)

which may be solved for T₁₁², T₁₂² and T₁₃² as

T₁₁² =σ²_T+

σN−σ^P₂

σN−σ^P₃

σ^P₂−σ^P₁

σ^P₃−σ^P₁ , (1.112a)

T₁₂² =σ²_T+

σN−σ^P₃

σN−σ^P₁

σ^P₃−σ^P₂

σ^P₁−σ^P₂ , (1.112b)

T₁₃² =σ²_T+

σN−σ^P₁

σN−σ^P₂

σ^P₁−σ^P₃

σ^P₂−σ^P₃ . (1.112c)

In view of the eigenvalues’ order (σ^P₁ ≥σ^P₂ ≥σ^P₃) and the fact that the l.h.s. of the above equations are non-negative, the following inequalities are obtained:

σ²_T+

σN−σ^P₂

σN−σ^P₃

≥ 0, (1.113a)

σ²_T+

σN−σ^P₃

σN−σ^P₁

≤ 0, (1.113b)

σ²_T+

σN−σ^P₁

σN−σ^P₂

≥ 0. (1.113c)

We may now plot the above conditions in theσN−σT plane, in order to find a region where all inequalities are satisfied, which will yield the valid combinations of normal and shear stress at the point under discussion. This graphical analysis is shown in Fig. 1.7(a) and is traditionally referred to as Mohr’s diagram. The first inequality shows a region outside a circle, the diameter of which is located over theσN-axis betweenσN =σ^P₂ andσN=σ^P₃. The second inequality shows a region inside a circle, the diameter of which is located over the σN-axis between

σN =σ^P₃ andσN=σ^P₁. The third inequality shows a region outside a circle, the diameter of which is located over theσN-axis betweenσN=σ^P₁ andσN=σ^P₂. Note that only the upper part of theσN-σT plane is of interest in this case since by definition σT ≥ 0, see (1.110b).

An immediate and clear result of Mohr’s diagram is that the maximal shear stress is given by σ^m_T =^σP1−σ₂ ^P3, which occurs atσ^m_N =^σP1+σ₂ ^P3. For the state of stress described in Example 1.1, P.1.12 has produced Mohr’s diagram shown in Fig. 1.7(b).

(a)General notation.

0 0.05

0.082 –0.019

–0.095 –0.058

(b)The values of Example 1.1.

Figure 1.7:Mohr’s diagram.

σ

T

σ

N

For each combination ofσN andσT in Mohr’s diagram corresponds a pair of angles (ψ,θ).

Note that the value ofφhas no importance in this case, since it represents a rotation about the x-direction (as it will change neitherσN norσT). To determine the rotation angles between the principal directions and the coordinate systems of the point under discussion for a given set of σN andσT, we evaluate T11, T12by (1.112a,b) and use (1.208:a,b). The inverse problem is of course much easier to solve since for each set of two rotation anglesψandθ, one can calculate σN andσT directly from (1.111a,b), (1.207), see also Remark 1.5.

Suppose now that we wish to calculate the anglesψandθthat are required to reach a given set ofσN andσT from a state of stress, which is not principal. To do that, we substituteσN

andσT in the l.h.s. of (1.110a,b) while in the r.h.s. we substitute the expressions obtained from (1.92) withφ= 0, namely

σ11= cos²θcos²ψσ11+ 2 cos²θcosψsinψσ12− 2 cosθcosψsinθσ13

+ cos²θsin²ψσ22− 2 cosθsinψsinθσ23+ sin²θσ33, (1.114a) σ12= −cosθcosψsinψσ11+ cosθ(cos²ψ− sin²ψ)σ12

+ sinθsinψσ13+ cosθsinψcosψσ22− sinθcosψσ23, (1.114b) σ13= sinθcos²ψcosθσ11+ 2 sinθcosψcosθsinψσ12+ (cos²θ− sin²θ)cosψσ13

+ sinθsin²ψcosθσ22+ (cos²θ− sin²θ)sinψσ23− cosθsinθσ33. (1.114c) Substitutions of the above in (1.110a,b) converts these equations into a system of two equations that should be simultaneously solved forψandθ,

Since in the general case, the expressions are not simple enough for analytic solution, a graphical representation of these equations is given in Fig. 1.8 for two cases (produced by P.1.12). In Fig. 1.8(a), the angles are measured from the non-principal stress state of Exam-ple 1.1, while in Fig. 1.8(b), the angles are measured from the corresponding principal stress state. In these figures, the thick line shows solutions for (1.110a), while the thin line shows so-lutions for (1.110b) (angles are given in radians). The desired soso-lutions are therefore the points where the above lines coincide.

0

As shown, there are a number of solutions in the region under discussion, and the picture has a period ofπinθand 2πinψ. To visualize the above solution, one may also plot a graph

0.2 0

Figure 1.9:σNas function ofψandθmeasured from the (principal) state of stress, see Example 1.1.

of σN=σ11 over theψ-θplane, see Fig. 1.9(a), and similarly, by substituting (1.114b,c) in (1.110b),σT as described by the surface in Fig. 1.10(a) is obtained.

One may also create a three-dimensional (spherical) surface ofσN =σ11(θ,ψ), andσT =

σ²₁₂(θ,ψ) +σ²₁₃(θ,ψ). To do that we use the results of Remark 1.12 and replaceψ byθ^s

andθbyφ^s−^π₂, respectively, whereθ^s andφ^s are spherical angles, see Fig. 1.20(b). These spherical plots are shown in Figs. 1.9(b), 1.10(b) where each point on the surface represents an orientation of the x-axis of the transformed system (by connecting the origin with it). Thus,σN

andσT are directly proportional to the distance that is measured along the x-axis between the origin and the corresponding surfaces.

0.2 0 Figure 1.10:σTas a function ofψandθmeasured from the (principal) state of stress, see Example 1.1.

Further on, we may also plot graphs ofψandθas functions ofσN andσT over the valid region of Mohr’s diagram. From (1.112a–c), for each (σN,σT) point we calculate T11, T12and T13, and then use (1.208) to determineψandθ, respectively. The resulting diagrams are shown in Fig. 1.11 produced by P.1.12. Note that each surface is plotted over the valid area of Mohr’s diagram only. See also Remark 1.6.

–0.08

Figure 1.11:The anglesψandθthat transform a principal stress state for eachσN,σTpoint.

Remark 1.5 One may define the mean shear stress,σ⁰_T, by integratingτT over the surface,ω, of an infinitesimally small sphere, the center of which is located at the point under discussion.

This operation may be expressed as

σ⁰_T = lim

The above relation connects the mean shear stress with the principal stresses. According to (1.109), the maximal shear stress may be written as σ^m_T = ¹₂(σ^P₁−σ^P₃) = ¹₂(σ^DP₁ −σ^DP₃ ) = 2Θ4cos(Θ5), and thus,σ⁰_T/σ^m_T = √

cos(Θ2/5₅) which, by taking into account the valid range ofΘ5

in (1.109), yields  8 15≥σ⁰_T

σ^m_T ≥

2

5. (1.116)

Remark 1.6 An additional, rather classical visualization tool is based on the well-known Stress Quadric of Cauchy. To derive the related equations, we start from the principal stress directions (namely, assume that the x,y,z system origin is located at the point under discussion while the coordinate lines coincide with the principal directions at that point). As a first step, we look at a point located at a distance
A and arbitrary direction, which may be expressed by ψandθ. The coordinates of this point are, see Remark 1.12,

x=
A cosθcosψ, y=
A cosθsinψ, z= −
A sinθ. (1.117) As a second step, we defineσN =σ11as the normal stress obtained in the x-direction of the new system by rotating the coordinate system with the above angles. This stress component may be easily obtained from (1.114a) by settingσii=σ^P_i andσi j= 0(i = j), as

σ11= cos²θcos²ψσ^P₁+ cos²θsin²ψσ^P₂+ sin²θσ^P₃, (1.118) and by using (1.117)

A
²σ11=σ^P₁x²+σ^P₂y²+σ^P₃z². (1.119) We shall now call forσ11to be proportional to the inverse of
A, namely,σ11= c/
A²where c is a normalization constant that may take both positive and negative values. Such a requirement yields a relatively simple quadratic surface, which is given by

σ^P₁x²+σ^P₂y²+σ^P₃z²= c. (1.120) The above surface may be classified by the four types: (1) ellipsoid, (2) unparted and biparted hyperboloids, (3) cylinder over ellipse, (4) hyperbola and parallel planes. Figure 1.12 (pro-duced by P.1.12) presents an illustrative case where the value c= 1 yields the unparted

hyper-–1 –1.5 0 –0.5

1 0.5

1.5 x

–0.5 0 0.5 y

–1 0 1 z

Figure 1.12:Three-dimensional plot of (1.120) for the state of stress described in Example 1.1.

boloid (i.e. the outer surface), and the value c= −1 yields the biparted hyperboloid (i.e. the inner surface). Hence, when applying stress transformation from a given system to a “new”

system (denoted by a bar), the interpretation of the surface in Fig. 1.12 is as follows: the value ofσN is inversely proportional to the square of the distance that is measured along the x-axis between the origin and the surface.

1.3.4 Strain Tensor Transformation due to Coordinate System Rotation

Since both the strain and the stress at a point are described as symmetric second-order tensors, their transformation, invariants and principal axes are derived by the same rules.

Hence, similar to the discussion of S.1.3.3, one may transform the strain components at a point by applying the tensor transformation or by using the matrix notation

εab=εabTiaTjb, ⇔ ε = T ·ε· T^T. (1.121) Compared with S.1.3.3, the discussion regarding the strain tensor transformation will be brief in view of the above similarities between the stress and the strain tensor transformation, the associated eigenvalue analysis, the principal axis determination, etc., Hence, the discussion regarding the stress tensor is applicable in a direct manner by replacingσi jwithεi j, etc. Sub-sequently, a matrix Mε= Mσmay be defined, and the strain tensor invariantsΞ1,Ξ2,Ξ3, the eigenvaluesε^P_i and the principal axes may be calculated analogously.

One may also draw a Mohr’s diagram in theεN-εTplane (whereεN=ε11,εT=

ε²₁₂+ε²₁₃) using the same technique that has been described for the stress description in theσN-σT plane (see S.1.3.3), and similarly, all other visualization methods are applicable in this case as well, see also discussion in Remark 2.3.

The reader may activate P.1.9, P.1.10 to transform strain coefficients by coordinate system rotation and P.1.12 for visualization of a state of strain at a point. P.1.11 executes transforma-tions of a strain tensor between curvilinear and Cartesian coordinates.

1.4 Energy Theorems

In this section we shall review some work- and energy-based measures that are encountered in the theory of elasticity. These measures may be further exploited to create functionals that may be combined with variational analysis to derive the associated Euler’s equations, see S.1.5, and other solutions in this area.

In general, two main measures, the “Potential Energy” and the “Complementary Energy”, should be addressed. These measures serve as the basis to the well known Theorem of Mini-mum Potential Energy, the Theorem of MiniMini-mum Complementary Energy and the Theorem of Reciprocity.

1.4.1 The Theorem of Minimum Potential Energy

Suppose that a body of volume B and surface S= SL+ SD, has reached equilibrium under the action of distributed load Fs over the SL part of its surface, and body force, Fb, that applies at each material point. The deformation u= (u1, u2, u3) over the remaining SD part of the surface is known as well. We introduce a variation of them,δu (i.e. “virtual displacements”), that vanish over SD. Since the body and the surface loads are fixed during the application of the virtual displacements, andδu= 0 on SD, one can write the “virtual work” as

δU=^

S_LFs·δu+^

BFb·δu=δ(^

S_LFs· u +^

BFb· u) (1.122) where “· ” is a scalar product of vectors. Further on, the strain energy is given by

U≡^

BW, (1.123)

where W is the volume density of the strain energy, see S.2.10. Equating the variation of the external load potential and the strain energy, may be written asδV= 0, where

V=^

BW−^

S_LFs· u −^

BFb· u. (1.124)

This enables us to conclude that the potential energy, V , has a stationary value among all admissible variations of the displacements uifrom the equilibrium state.

Since V(u +δu) ≥ 0 for any variationδu, the Theorem of Minimum Potential Energy may be expressed as:

Among all displacements that satisfy the boundary conditions, those that satisfy the equilib-rium equations as well make the potential energy an absolute minimum.

The converse theorem is true as well, see (Sokolnikoff, 1983).

Remark 1.7 In the time-dependent case, the above statement may be extended to the

In document About the Authors Omri Rand Vladimir Rovenski (Page 37-47)