5.2.1 Surface Traction

Consider a solid subjected to external forces and in static equilibrium as shown in figure 5.1. We are interested in the state of stress at a point P in the interior of the solid. We cut the body into two halves by passing an imaginary plane through P. The unit vector normal to the plane is denoted by n [see figure 5.1(b)]. The portion of the body on the left side is in equilibrium because of the external forces f₁, f₂, and f₃ and also the internal forces acting on the cut surface. “Surface traction” is defined as the internal force per unit area or the force intensity acting on the cut plane. In order to measure the intensity or traction specifically at P, we consider the force ΔF acting over a small area ΔA that contains point P. Then the surface traction T⁽ⁿ⁾ acting at the point P is defined as

(5.1)

In eq. (5.1), the right superscript (n) is used to denote the fact that this surface traction is defined on a plane whose normal is n. It should be noted that at the same point P the traction vector T would be different on a different plane passing through P. It is clear from eq. (5.1) that the dimension of the traction vector is the same as that of pressure, or force per unit area.

Since T⁽ⁿ⁾ is a vector, one can resolve it into Cartesian components and write it as

(5.2)

and its magnitude can be computed from

(5.3)

5.2.2 Normal Stresses and Shear Stresses

The surface traction T⁽ⁿ⁾ defined by eq. (5.1) does not act in general in the direction of n, that is, T and n are not necessarily parallel to each other. Thus, we can decompose the surface traction into two components, one parallel to n and the other perpendicular to n, which will lie on the plane. The component normal to the plane or parallel to n is called the normal stress and is denoted by σ_n. The other component, parallel to the plane or perpendicular to n, is called the shear stress and is denoted by τ_n.

The normal stress can be obtained from the scalar product of T⁽ⁿ⁾ and n, as (see figure 5.3)

(5.4)

and the shear stress can be calculated from the relation

(5.5)

The angle between T⁽ⁿ⁾ and n can be obtained from the definition of scalar product given in eq. (A.18) in the appendix.

5.2.3 Rectangular or Cartesian Stress Components

The surface traction at a point varies depending on the direction of the normal to the plane, and therefore, one can obtain an infinite number of traction vectors T⁽ⁿ⁾ and corresponding normal and shear stresses for a given state of stress at a point. However, one might be interested in the maximum values of these stresses and the corresponding plane. Fortunately, the state of stress at a point can be completely characterized by defining the traction vectors on three mutually perpendicular planes passing through the point. That is, from the knowledge of T⁽ⁿ⁾ acting on three orthogonal planes, one can determine T⁽ⁿ⁾ on any arbitrary plane passing through the same point. For convenience, these planes are taken as the three planes that are normal to the x, y, and z axes.

Let us denote the traction vector on the yz–plane, which is normal to the x‐axis, as T^(x). Instead of decomposing the traction vector into its normal and shear components, we will use its components parallel to the coordinate directions and denote them as , and . That is,

(5.6)

It may be noted that in eq. (5.6) is the normal stress, and and are the shear stresses in the y‐ and z‐directions, respectively. In contemporary solid mechanics, the stress components in eq. (5.6) are denoted by σ_xx, τ_xy, and τ_xz, where σ_xx is the normal stress, and τ_xy and τ_xz are components of shear stress. In this notation, the first subscript denotes the plane on which the stress component acts—in this case, the plane normal to the x‐axis or simply the x‐plane—and the second subscript denotes the direction of the stress component. We can repeat this exercise by passing two more planes, normal to the y‐ and z‐axes, respectively, through point P. Thus the surface tractions acting on the plane normal to y‐plane will be τ_yx, σ_yy, and τ_yz. The stresses acting on the z‐plane can be written as τ_zx, τ_zy, and σ_zz.

The stress components acting on the three planes can be depicted using a cube as shown in figure 5.4. It must be noted that this cube is not a physical cube and hence has no dimensions. The six faces of the cube represent the three pairs of planes normal to the coordinate axes. The top face, for example, is the + z‐plane, and then the bottom face is the –z‐plane, whose normal is in the –z‐direction. Note that the three visible faces of the cube in figure 5.4 represent the three positive planes, that is, planes whose normal vectors are the positive x‐, y‐ and z‐axes. On these faces, all tractions are shown in the positive direction. For example, τ_yz is the traction on the y‐plane acting in the positive z‐direction. The description of all stress components is summarized in table 5.1.

Knowledge of the nine stress components is necessary in order to determine the components of the surface traction T⁽ⁿ⁾ acting on an arbitrary plane with normal n.

Stresses are second‐order tensors, and their sign convention is different from that of regular force vectors. Stress components, in addition to the direction of the force, contain information of the surface on which they are defined. A stress component is positive when both the surface normal and the stress component are either in the positive or in the negative coordinate direction. If the surface normal is in the positive direction and the stress component is in the negative direction, then the stress component has a negative sign.

Normal stress is positive when it is a tensile stress and negative when it is compressive. A shear stress acting on the positive face is positive when it is acting in the positive coordinate direction. The positive directions of all the stress components are shown in figure 5.4.

5.2.4 Traction on an Arbitrary Plane through a Point

If the components of stress at a point, say P, are known, it is possible to determine the surface traction acting on any plane passing through that point. Let n be the unit normal to the plane on which we want to determine the surface traction. The normal vector can be represented as,

(5.7)

For convenience, we choose P as the origin of the coordinate system, as shown in figure 5.5, and consider a plane parallel to the intended plane but passing at an infinitesimally small distance h away from P. Note that the normal to the face ABC is also n. We will calculate the tractions on this plane first and then take the limit as h tends to zero. We will consider the equilibrium of the tetrahedron PABC. If A is the area of the triangle ABC, then the areas of triangles PAB, PBC, and PAC are given by An_z, An_x, and An_y, respectively. Let be the surface traction acting on the face ABC.

From the definition of surface traction in eq. (5.1), the force on the surface can be calculated by multiplying the stresses by the surface area. Since the tetrahedron should be in equilibrium, the sum of the forces acting on its surfaces should be equal to zero. Force balance in the x‐direction yields

In the above equation, we have assumed that the stresses acting on a surface are uniform. This will not be true if the size of the tetrahedron is not small. However, the tetrahedron is infinitesimally small, which is the case as h approaches zero. Dividing the above equation by A, we obtain the following relation:

(5.8)

Similarly, force balance in the y‐ and z‐directions yield

(5.9)

From eqs. (5.8) and (5.9) it is clear that the surface traction acting on the surface whose normal is n can be determined if the nine stress components are available. Using matrix notation, we can write these equations as

(5.10)

where

(5.11)

[σ] is called the stress matrix and it completely characterizes the state of stress at a given point.

5.2.5 Symmetry of Stress Matrix and Vector Notation

The nine components of the stress matrix can be reduced to six using the symmetry property of the stress matrix. Figure 5.6 shows an infinitesimal portion (Δl  ×  Δl) of a solid with shear stresses acting on its surface. The dimension in the z‐direction is taken as unity. The direction of the shear stress τ_xy on the surface BD is in the positive y‐direction, while on the surface AC, it is in the negative y‐direction. As the body is in static equilibrium, the sum of the moments about the z‐axis must be equal to zero, which implies that the shear stresses τ_xy and τ_yx must be equal to each other as shown below.

Similarly, we can derive the following relations:

(5.12)

Thus, we need only six components to fully represent the stress at a point. In some occasions, stress at a point is written as a 6 × 1 pseudovector as shown below:

(5.13)

Other textbooks often use a single subscript for normal stresses when the stress is written in a vector form.

5.2.6 Principal Stresses

As shown in the previous section, the normal and shear stresses acting on a plane passing through a given point in a solid change as the orientation of the plane is changed. Then a natural question is: Is there a plane on which the normal stress is the maximum? Similarly, we would also like to find the plane on which the shear stress attains a maximum. These questions are not only academic but also have significance in predicting the failure of the material at that point. In the following, we will provide some answers to the above questions without furnishing the proofs. The interested reader is referred to books on continuum mechanics, for instance, L. E. Malvern³ or advanced solid mechanics,⁴ for a more detailed treatment of the subject.

It can be shown that at every point in a solid there are at least three mutually perpendicular planes on which the normal stress attains extremum (maximum or minimum) values. On all these planes, the shear stresses vanish. Thus the traction vector T⁽ⁿ⁾ will be parallel to the normal vector n, that is, T⁽ⁿ⁾ = σ_n n, on these planes. Of these three planes, one plane corresponds to the global maximum value of the normal stress and the other to the global minimum. The third plane will carry the intermediate normal stress. These special normal stresses are called the principal stresses at that point, the planes on which they act are the principal stress planes, and the corresponding normal vectors are the principal stress directions. The principal stresses are denoted by σ₁, σ₂, and σ₃ such that σ₁ ≥ σ₂ ≥ σ₃.

Based on the above observations, the principal stresses can be calculated as follows. When the normal direction to a plane is the principal direction, the surface normal and the surface traction are in the same direction (T⁽ⁿ⁾ || n). Thus, the surface traction on a plane can be represented by the product of the normal stress σ_n and the normal vector n, as

(5.14)

Combining eq. (5.14) with eq. (5.10) for the surface traction, we obtain

(5.15)

Equation (5.15) represents the eigenvalue problem, where σ_n is the eigenvalue and n is the corresponding eigenvector (see section A.4 and eq. A.58). Equation (5.15) can be rearranged as

(5.16)

where [I] is a 3 × 3 identity matrix. In the component form, the above equation can be written as

(5.17)

Note that n = 0 satisfies the above equation, which is not only a trivial solution but also physically not possible as ||n|| must be equal to unity. The above set of linear simultaneous equations will have a nontrivial, physically meaningful solution if and only if the determinant of the coefficient matrix is zero, that is,

(5.18)

Expanding this determinant, we obtain the following cubic equation in σ_n:

(5.19)

where

(5.20)

In the above equation, I₁, I₂, and I₃ are the three invariants of the stress matrix [σ], which can be shown to be independent of the coordinate system. The three roots of the cubic equation (5.19) correspond to the three principal stresses. We will denote them by σ₁, σ₂, and σ₃ in the order of σ₁ ≥ σ₂ ≥ σ₃. A method to solve the cubic equation is described in the appendix.

Principal Directions:

Once the principal stresses have been computed, we can substitute them one at a time into eq. (5.17) to obtain the linear system of equations that have three unknowns, n_x, n_y, and n_z. Corresponding to each principal value, we will get a principal direction that will be denoted as n¹, n², and n³. For example, if σ_n is replaced by the first principal stress σ₁, then we obtain

(5.21)

In eq. (5.21), are components of n¹. The above equations are three linear simultaneous equations in three unknowns. However, since the determinant of the matrix is zero (i.e., the matrix is singular), they are not independent. Thus, an infinite number of solutions exist. We need one more equation to find a unique value for the principal directions nⁱ. Note that n is a unit vector, and hence its components must satisfy the following relation:

(5.22)

It can be shown that the planes on which the principal stresses act are mutually perpendicular. Let us consider any two principal directions nⁱ and n^j, with i ≠ j. If σ_i and σ_j are the corresponding principal stresses, then they satisfy the following equations:

(5.23)

Multiplying the first equation by n^j and the second equation by nⁱ, we obtain

(5.24)

Considering the symmetry of [σ], that is, [σ] = [σ]^T, and the rule for the transpose of matrix products (eq. (A.35) in the appendix), one can show that . Then subtracting the first equation from the second in eq. (5.24), we obtain

(5.25)

This implies that if the principal stresses are distinct, that is, σ_i ≠ σ_j, then

(5.26)

which means that nⁱ and n^j are orthogonal. The three planes on which the principal stresses act are mutually perpendicular.

There are three different possibilities for principal stresses and directions:

1. σ₁, σ₂, and σ₃ are distinct ⇒ principal stress directions are three unique mutually orthogonal unit vectors.
2. σ₁ = σ₂ ≠ σ₃ ⇒ n³ is a unique principal stress direction, and any two orthogonal directions on the plane that is perpendicular to n³ are the other principal directions.
3. σ₁ = σ₂ = σ₃ ⇒ any three orthogonal directions are principal stress directions. This state of stress is called hydrostatic or isotropic state of stress.

5.2.7 Transformation of Stress

Let the state of stress at a point be given by the stress matrix [σ]_xyz, where the components are expressed with reference to the xyz coordinates as shown in figure 5.7. The question is what the components will look like in a different coordinate system, say x′y′z′, or how to determine [σ]_x′y′z′. It will be shown that the stress matrix [σ]_x′y′z′ can be obtained by rotating the stress matrix [σ]_xyz using a transformation matrix of direction cosines.

Let us define the x′y′z′ coordinate system whose unit vectors b¹, b², and b³, respectively, are given in the xyz coordinate system as

For example, b¹ = {1, 0, 0}^T in the x′y′z′ coordinate system, while in the xyz coordinates. Using these vectors, let us define a transformation matrix as

(5.27)

Matrix [N] transforms a vector in the x′y′z′ coordinates into a vector in the xyz coordinates, while its transpose [N]^T transforms a vector in the xyz coordinates into a vector in the x′y′z′ coordinates. For example, the unit vector in x′y′z′ coordinates can be represented in the xyz coordinates as

(5.28)

The transformation of a stress matrix is more complicated than the transformation of a vector. We will perform the transformation in two steps. First, we will determine the traction vectors on the three planes normal to the x'‐, y'‐ and z'‐axes. This can be accomplished by multiplying the stress matrix and the corresponding direction cosine vector as described in eq. (5.10). The three sets of traction vectors are written as columns of a square matrix as shown below:

(5.29)

Equation (5.29) represents the three surface traction vectors on planes perpendicular to b¹, b², and b³ in the xyz coordinates system. In the next step, we would like to transform the traction vectors to the x'y'z' coordinate system. This transformation can be accomplished by using eq. (5.28). It may be recognized that the traction vectors , and represented in the x'y'z' coordinate system will be the transformed stress matrix. Thus, the stress matrix in the new coordinate system can be obtained by

(5.30)

5.2.8 Maximum Shear Stress

Maximum shear stress plays an important role in the failure of ductile materials. Let σ₁, σ₂, and σ₃ be the three principal stresses at a point such that σ₁ ≥ σ₂ ≥ σ₃. The stress state at this point can be described using three Mohr’s circles as shown in figure 5.9.

In the diagram in figure 5.9, any point (σ, τ) located in the shaded area represents the normal and shear stresses, σ_n and τ_n, on a plane through the point. One can note that the maximum shear stress is given by the radius of the largest Mohr’s circle as

(5.31)

The plane on which the shear stress attains maximum bisects the first and third principal stress planes. The normal stress on this plane is given by

(5.32)

5.3.1 Definition of Strains

Figure 5.10 shows a body before and after deformation. Let the points P, Q, and R in the undeformed body moves to P', Q', and R', respectively, after deformation. The displacement of P can be represented by three displacement components, u, v, and w in the x‐, y‐, and z‐directions. Thus the coordinates of P' are (x + u, y + v, z + w). The functions u(x,y,z), v(x,y,z), and w(x,y,z) are components of a vector field that is referred to as the deformation field or the displacement field. The displacements of point Q will be slightly different from that of P. They can be written as

(5.33)

Similarly, displacements of point R are

(5.34)

The coordinates of P, Q, and R before and after deformation are as follows:

(5.35)

Length of the line segment P'Q' can be calculated as

(5.36)

Substituting for the coordinates of P' and Q' from eq. (5.35) we obtain

(5.37)

It may be noted that we have used a two‐term binomial expansion in deriving an approximate expression for the change in length. In this book, we will consider only small deformations such that all deformation gradients are very small compared to unity, that is, , . Then we can neglect the higher‐order terms in eq. (5.37) to obtain

(5.38)

Now we invoke the definition of normal strain as the ratio of change in length to original length to derive the expression for strain as

(5.39)

Thus, the normal strain ε_xx at a point can be defined as the change in length per unit length of an infinitesimally long line segment originally parallel to the x‐axis. Similarly, we can derive normal strains in the y‐ and z‐directions as

(5.40)

The engineering shear strain, say γ_xy, is defined as the change in the angle between a pair of infinitesimal line segments that were originally parallel to the x‐ and y‐axes. From figure 5.10, the angle between PQ and P'Q' can be derived as

(5.41)

Similarly, the angle between PR and P'R' is

(5.42)

Using the aforementioned definition of shear strain,

(5.43)

Similarly, we can derive shear strains in the yz‐ and zx‐planes as

(5.44)

The shear strains, γ_xy, γ_yz, and γ_zx, are called engineering shear strains. We define tensorial shear strains as

(5.45)

It may be noted that the tensorial shear strains are one half of the corresponding engineering shear strains. It can be shown that the normal strains and the tensorial shear strains transform from one coordinate system to another following tensor transformation rules.

In the general three‐dimensional case, the strain matrix is defined as

(5.46)

As is clear from the definition in eq. (5.45), the strain matrix is symmetric. Like the stress vector, the symmetric strain matrix can be represented as a pseudovector

(5.47)

where γ_yz, γ_zx, and γ_xy are used instead of ε_yz, ε_zx, and ε_xy. The six components of strain completely define the deformation at a point. The normal strain in any arbitrary direction at that point and also the shear strain in any arbitrary plane passing through the point can be calculated using the above strain components, as explained in the following section. Strain is also a tensor and therefore it has properties similar to a stress tensor. For example, the transformation of strain, principal strains, and corresponding principal strain directions can be determined using the procedures we described for stresses.

5.3.2 Transformation of Strain

Let the state of strain at a point be given by the strain matrix [ε]_xyz, where the components are expressed in the xyz coordinates shown in figure 5.7. As for the case of stresses, the question again is: What are the components of strain at the same point in a different coordinate system, x′y′z′ (i.e., [ε]_x′y′z′)?

Let b¹, b², and b³ be unit vectors along the axes x′, y′, and z′, respectively, represented in xyz coordinates. The strain matrix expressed with respect to the x′y′z′ coordinates system is

(5.48)

which is the same transformation used for stresses in eq. (5.30). It should be emphasized that tensorial shear strains must be used when using the transformation in eq. (5.48).

Let us consider the term in eq. (5.48). It can be written as

(5.49)

We note that the normal strain in the x'‐direction depends only on the strain tensor and the direction cosines of the x'‐axis. In fact, this relation can be generalized to any arbitrary direction n and written as

(5.50)

Similarly, one can derive the shear strain in a plane containing two mutually perpendicular vectors m and n as

(5.51)

5.3.3 Principal Strains

Strains, like stresses, are second‐order tensors, and therefore one can compute eigenvalues and eigenvectors for a given strain matrix. The eigenvalues are the three principal strains, and the eigenvectors are the corresponding principal strain directions. The principal strain directions represent the directions in which the normal strain ε_nn in eq. (5.50) takes an extremum value. The maximum and minimum of the three principal strains are the global maximum and minimum, respectively. The three principal planes intersect along the principal strain directions. The shear strain vanishes on the principal strain planes. That is, a pair of infinitesimally small line segments parallel to any two principal directions remain perpendicular to each other after deformation, or, equivalently, the angle between them does not change after deformation. In other words, the shear strain on this plane is equal to zero.

The physical meaning of strains can be explained as follows. Consider an infinitesimal cube of size Δa × Δa × Δa whose sides are parallel to the coordinate axes before deformation. After deformation, the rectangular parallelepiped will become an oblique parallelepiped of size Δa(1+ε_xx) × Δa (1+ε_yy) × Δa(1+ε_zz). The angles between the edges of the parallelepiped will reduce by γ_xy, γ_yz, and γ_zx (see figure 5.11). If the same parallelepiped is oriented such that its edges are parallel to the principal strain directions, then after deformation the parallelepiped will remain as a rectangular parallelepiped, and its dimensions will be Δa(1+ε₁) × Δa(1+ε₂) × Δa(1+ε₃). Since the shear strains in the principal strain planes are zero, the angle between the edges will remain as 90°.

The principal strain is the extensional strain on a plane, where there is no shear strain. Principal strains and their directions are the eigenvalues and eigenvectors of the following eigenvalue problem:

(5.52)

where n is the eigenvector of matrix . The above eigenvalue problem has three solutions , and ε₃, which are the principal strains.

If the principal strains ε₁, ε₂, and ε₃ are known, then the maximum engineering shear strain γ_max can be computed as

(5.53)

where ε₁ and ε₃ are the maximum and minimum principal strains, respectively.

5.3.4 Stress vs. Strain

Stresses and strains defined in the previous two sections are second‐order tensors and hence share some common properties as shown in table 5.2. When the material is isotropic, which means the material properties are the same in all directions or independent of the coordinates system, the principal stress directions and principal strain directions coincide. If the material is anisotropic, the principal stress and strain directions are in general different. In this textbook, we will focus on isotropic materials only.

[σ] is a symmetric 3 × 3 matrix	[ε] is a symmetric 3 × 3 matrix
Normal stress in the direction n is	Normal strain in the direction n is
Shear stress in a plane containing two mutually perpendicular unit vectors m and n is	Shear strain in a plane containing two mutually perpendicular unit vectors m and n is
Transformation of stresses	Transformation of strains
Three mutually perpendicular principal directions and principal stresses can be computed as eigenvectors and eigenvalues of the stress matrix:	Three mutually perpendicular principal directions and principal strains can be computed as eigenvectors and eigenvalues of the strain matrix:

5.4.1 Linear Elastic Relationship (Generalized Hooke’s Law)

The one‐dimensional stress–strain relationship described in the previous section can be extended to the three‐dimensional state of stress. When the stress–strain relationship is linear, it can be written as

(5.54)

where

Matrix [C] is called the stress–strain matrix, or elasticity matrix. It can be shown that [C] must be a symmetric matrix and hence the number of independent coefficients or elastic constants for an anisotropic material is only 21. Many composite materials, naturally occurring composites such as wood or bone, and man‐made materials such as fiber‐reinforced composites can be modeled as an orthotropic material with nine independent elastic constants. Some composites are transversely isotropic and require only five independent elastic constants. Most materials are isotropic, and for such materials, the 21 constants in the symmetric matrix [C] can be expressed in terms of two independent constants called engineering elastic constants.

For isotropic materials, the relation between stress and strain can be written as:

(5.55)

where E, Young’s modulus, andν, Poisson’s ratio, are the two independent elastic constants, and G is the shear modulus defined by

(5.56)

Note that there are only two independent constants for isotropic materials.

Alternately, stresses can be written as a function of strains by inverting the relations in eq. (5.55), as

(5.57)

The elasticity matrix [C] in eq. (5.54) can be written as

(5.58)

5.4.2 Simplified Laws for Two‐Dimensional Analysis

The general three‐dimensional stress–strain relationship in eq. (5.54) can be simplified for certain special situations that often occur in practice. The two‐dimensional stress–strain relationship can be categorized into three cases: plane stress, plane strain, and axisymmetric.

Most practical structures consist of thin plate‐like components in order to be efficient. Assume a thin plate that is parallel to the xy‐plane. If we assume that the top and bottom surfaces of the plate are not subjected to any significant forces, that is, the plate is subjected to forces in its plane only, in the x‐ and y‐directions, then the transverse stresses (stresses with a z subscript) vanish on the top and bottom surfaces, that is, on the top and bottom surfaces. If the thickness is much smaller compared to the lateral dimensions of the plate, then we can assume that the aforementioned transverse stresses are approximately zero through the entire thickness. Then the plate is said to be in a state of plane stress parallel to the xy‐plane or normal to the z‐axis.

In order to derive the stress–strain relations for the state of plane stress we start with eq. (5.55). We set on the right‐hand side of the equations to obtain

(5.59)

Inverting the above relations, we obtain

(5.60)

Similar to plane stress, one can define a state of plane strain in which strains with a z subscript are all equal to zero. This situation corresponds to a structure whose deformation in the z‐direction is constrained (i.e., w = 0), so that the following relation holds:

(5.61)

Plane strain can also be used if the structure is infinitely long in the z‐direction. The stress–strain relations for the case of plane strain can be derived by starting with eq. (5.54) with the stress–strain matrix [C] in eq. (5.58) and setting the strains with a z subscript, ε_zz, γ_xz, and γ_yz, equal to zero to obtain

(5.62)

The inverse relation is given by

(5.63)

Note that the normal stress σ_zz is not zero in the plane strain problem but can be calculated from ε_xx and ε_yy:

(5.64)

Another class of problems that can be treated as two‐dimensional is the axisymmetric problems where the geometry, the applied loads, and the boundary conditions are all symmetric about an axis. Examples of axisymmetric geometry are a cylinder, a cone, or any geometry that has rotational symmetry about an axis. If the loads and boundary conditions are symmetric about the same axis, then the deformation will be in the radial and axial direction only. That is, there will not be any twisting or rotation about that axis. Therefore, such problems can be treated as two‐dimensional with the displacement vector having only two components. It is convenient to use a cylindrical coordinate system for such problems so that the displacement components, (u_r, u_z), are the radial and axial components. A two‐dimensional section of this geometry is the domain of analysis. Even though there is no rotation or twisting, there will be strain and stress in the circumferential direction. This component of stress and strain are called hoop stress and hoop strain. The hoop strain occurs whenever there is radial expansion on contraction because any radial displacement would cause the circumference of the object to increase or decrease. In a cylindrical coordinate system, the stress and strain relation can be written as:

(5.65)

This relation can be obtained by dropping the equations for the two out‐of‐plane shear components τ_rθ and τ_zθ from the 3D stress‐strain relation in eq. (5.58).

5.5.1 Equilibrium Equations

As we discussed earlier, the state of stress at a point is defined by the six stress components, three normal and three shear stresses. These components, in general, vary within the solid. In static problems the stresses can be represented by six functions of the spatial coordinates x, y, and z: σ_xx(x,y,z), σ_yy(x,y,z), and so forth. These six functions cannot be arbitrary and must satisfy certain relations called stress equilibrium equations. The functions are also called the stress field in the solid.

Consider the equilibrium of a differential element represented in two dimensions by a square (see figure 5.14) whose center is (x, y) and its sides are dx and dy, respectively. The dimension in the z‐direction is taken as unity. The stresses are assumed to be independent of z, and the solid is in a state of plane stress.

Equilibrium in the x‐direction yields the following equation:

(5.66)

If the first‐order Taylor series expansion is used to represent stresses on the surfaces of the rectangle in terms of stresses at the center, the first two terms in eq. (5.66) can be approximated by

Similarly, the last two terms can be approximated by

By substituting these two equations into eq. (5.66), we obtain an equilibrium equation in the x‐direction as

(5.67)

Similarly, equilibrium in the y‐direction yields the following equation:

(5.68)

Equations (5.67) and (5.68) are the equilibrium equations for a solid subjected to a two‐dimensional state of stress. We can similarly derive the equations for a three‐dimensional state of stress, by considering the equilibrium of a three‐dimensional differential element to obtain,

(5.69)

Equation (5.69) is obtained by considering force equilibrium in the x‐, y‐, and z‐directions. As has been shown in eq. (5.12), moment equilibrium yields symmetry of the stress matrix.

5.5.2 Traction or Stress Boundary Conditions

While the stress field must satisfy the differential equations of equilibrium in eq. (5.69), there are other conditions the stress field must satisfy on the boundaries of a solid. These are called traction or stress boundary conditions. Consider the surface of a solid subjected to distributed forces such that the tractions in the x‐, y‐ and z‐directions are, t_x, t_y, and t_z, respectively (see figure 5.15). Let the direction cosines of the normal to the surface be n_x, n_y, and n_z. Then the state of stress at a point on the surface of the body must satisfy the boundary conditions shown below:

(5.70)

The derivation of the above boundary conditions is similar to those of eq. (5.10).

5.5.3 Boundary Value Problems

Consider an arbitrary body subjected to external forces and displacement constraints as shown in figure 5.16. Our goal is to determine the displacement field u(x,y,z). Once the displacements are determined, strains can be found using the strain displacement relations, that is in eqs. (5.39), (5.40), and (5.45), and then stress field can be determined using the constitutive relations, that is in eq. (5.54). This problem, typical of solid and structural mechanics, is called the elastostatic boundary value problem. Techniques for solving the boundary value problems in two and three dimensions are considered in advanced courses in solid mechanics and theory of elasticity.⁶ The elasticity problems can be simplified by making certain approximations in the displacement field, for instance, Euler‐Bernoulli beam theory or Kirchhoff plate theory. The finite element method is a numerical method that can solve complex two‐ and three‐dimensional elasticity problems and will be discussed in later chapters.

In general, solving an elasticity problem involves solving the three equilibrium equations in eq. (5.69) in conjunction with the strain–displacement relations in eqs. (5.39), (5.40), and (5.45) and constitutive relations in eq. (5.54). The boundary conditions also need to be provided in terms of given displacements or prescribed traction forces.

5.5.4 Compatibility Conditions

In the previous section, we talked about the displacement field u(x,y,z). Usually, solutions to the boundary value problems are obtained by trial and error or inverse methods. We first assume a physically suitable displacement field and adjust the terms in the solution to satisfy the equilibrium equations and boundary conditions. More often one needs to start with a physically possible displacement field. In fact, a set of any three nonsingular functions will represent a possible displacement field. Of course, the forces required to produce such a displacement field may be complex and difficult in practice, but still physically possible. However, the same thing cannot be said about the strain field or stress field. That is, one cannot choose six arbitrary functions in spatial coordinates and claim the set to be a possible strain field. The particular strain field may not be physically possible. For example, a valid strain field should not create a discontinuous deformation or overlapping of certain points. The six strain functions must satisfy three relations called compatibility equations. The derivation of the three‐dimensional compatibility equation is beyond the scope of this book. The interested reader is referred to more advanced elasticity books, such as Timoshenko.⁷ For two‐dimensional (plane) problems, the compatibility equation is given as

(5.71)

5.6.1 Strain Energy in a Plane Solid

Consider a plane elastic solid as illustrated in figure 5.18. The strain energy is a form of energy that is stored in the solid due to the elastic deformation. Formally, it can be defined as

(5.72)

where h is the thickness of the plane solid (h = 1 for plane strain) and [C] = [C_σ] for plane stress and [C] = [C_ε] for plane strain. Since stress and strain are constant throughout the thickness, the volume integral is converted into the area integral by multiplying by the thickness in the second relation in eq. (5.72). The linear elastic relation in eq. (5.54) has been used in the last relation.

5.6.2 Potential Energy of Applied Loads

When a force acting on a body moves through a small distance, it loses its potential to do additional work, and hence its potential energy is given by the negative of the product of the force and corresponding displacement. For example, when concentrated forces are applied to the solid, the potential energy becomes

(5.73)

where F_i is the force in the i‐th DOF, q_i is the displacement in the direction of the force, and ND is the total number of concentrated forces acting on the body. The negative sign indicates that the potential energy decreases as the force has expended some energy performing the work given by the product of force and corresponding displacement.

When distributed forces, such as a pressure load, act on the edge of a body, the summation sign in the above expression is replaced by integration over the edge of the body as shown below:

(5.74)

where T_x and T_y are the components of applied surface forces in the x‐ and y‐direction, respectively. In the finite element discretization, the vector of displacements, {u}, is approximated by shape functions and nodal DOFs {q}. Therefore, the potential energy of the applied loads is a function of nodal DOFs {q} after discretization by finite elements.

If body forces (forces distributed over the volume) are present, work done by these forces can be computed in a similar manner. The gravitational force is an example of body force. In this case, integration should be performed over the volume. We will discuss this further when we derive the finite element equations.

5.6.3 Principle of Minimum Potential Energy

As with the beam problem, the potential energy is defined as the sum of the strain energy and the potential energy of applied loads:

(5.75)

where U is the strain energy and V is the potential energy of applied loads. The principle of minimum total potential energy states that of all possible displacement configurations of a solid/structure, the equilibrium configuration corresponds to the minimum total potential energy. That is, at equilibrium, we have

(5.76)

where q₁, q₂, …, q_N are nodal DOFs (i.e., displacements) that define the deformed configuration of the body. In finite element analysis, the deformation of the body is defined in terms of the displacements of the nodes. In the next chapter, we will use the principle of minimum total potential energy to derive finite element equations for different types of elements.

5.7.1 Strain Energy

When a force is applied to a solid, it deforms. Then, we can say that work is done on the solid, which is proportional to the force and deformation. The work done by the applied force is stored in the solid as potential energy, which is called the strain energy. The strain energy in the solid may not be distributed uniformly throughout the solid. We introduce the concept of strain energy density, which is strain energy per unit volume, and we denote it by U₀. Then the strain energy in the body can be obtained by integration as follows:

(5.77)

where the integration is performed over the volume V of the solid. In the case of uniaxial stress state, the strain energy density is equal to the area under the stress–strain curve (see figure 5.20). Thus, it can be written as

(5.78)

For the general 3‐D case the strain energy density is expressed as

(5.79)

If the material is elastic, then the strain energy can be completely recovered by unloading the body.

The strain energy density in eq. (5.79) can be further simplified. Consider a coordinate system that is parallel to the principal stress directions. In this coordinate system, no shear components exist. Extending eq. (5.79) to this stress states yields

(5.80)

From section 5.3, we know that stresses and strains are related through the linear elastic relationship. For example, in case of principal stresses and strains,

(5.81)

Substituting from eq. (5.81) into eq. (5.80), we can write the strain energy density in terms of principal stresses as

(5.82)

The strain energy density can be thought of as consisting of two components: one due to dilation or change in volume and the other due to distortion or change in shape. The former is called dilatational strain energy and the latter, distortional energy. Many experiments have shown that ductile materials can be hydrostatically stressed to levels beyond their ultimate strength in compression without failure. This is because the hydrostatic state of stress reduces the volume of the specimen without changing its shape.

5.7.2 Decomposition of Strain Energy

The strain energy density at a point in a solid can be divided into two parts: dilatational strain energy density, U_h, that is due to change in volume, and distortional strain energy density, U_d, that is responsible for the change in shape. In order to compute these components, we divide the stress matrix also into similar components, dilatational stress matrix, σ_h, and deviatoric stress matrix, σ_d. For convenience, we will use the stress components with respect to the principal stress coordinates. Then the aforementioned stress components can be derived as follows:

(5.83)

The dilatational component σ_h is defined as

(5.84)

which is also called the volumetric stress. Note that 3σ_h is the first invariant I₁ of the stress matrix in eq. (5.20). Thus, it is independent of the coordinate system. Note that σ_h is a state of hydrostatic stress and hence the subscript h is used to denote the dilatational stress component as well as dilatational energy density

The dilatational energy density can be obtained by substituting the stress components of the hydrostatic stress state in eq. (5.84) into the expression for strain energy density in eq. (5.82),

(5.85)

and using the relation in eq. (5.84),

(5.86)

The distortion part of the strain energy is now found by subtracting eq. (5.86) from eq. (5.82), as

(5.87)

It is customary to write U_d in terms of an equivalent stress called von Mises stress, σ_VM, as

(5.88)

The von Mises stress is defined in terms of principal stresses as

(5.89)

5.7.3 Distortion Energy Theory (von Mises)

According to von Mises’s theory, a ductile solid will yield when the distortion energy density reaches a critical value for that material. Since this should be true for the uniaxial stress state also, the critical value of the distortional energy can be estimated from the uniaxial test. At the instance of yielding in a uniaxial tensile test, the state of stress in terms of principal stress is given by σ₁ = σ_Y (yield stress) and σ₂ = σ₃ = 0. The distortion energy density associated with yielding is

(5.90)

Thus, the energy density given in eq. (5.90) is the critical value of the distortional energy density for the material. Then according to von Mises’s failure criterion, the material under multiaxial loading will yield when the distortional energy is equal to or greater than the critical value for the material:

(5.91)

Thus, the distortion energy theory states that material yields when the von Mises stress exceeds the yield stress obtained in a uniaxial tensile test.

The von Mises stress in eq. (5.87) can be rewritten in terms of stress components as

(5.92)

For a two‐dimensional plane stress state, σ₃ = 0, the von Mises stress can be defined in terms of principal stresses as

(5.93)

and in terms of general stress components as

(5.94)

The two‐dimensional distortion energy equation in eq. (5.94) becomes an ellipse when plotted on the σ₁‐σ₂ plane as shown in figure 5.21. The interior of this ellipse defines the region of combined biaxial stress where the material is safe against yielding under static loading.

Consider a situation in which only a shear stress exists, such that σ_x = σ_y = 0, and τ_xy = τ. For this stress state, the principal stresses are σ₁ = –σ₃ = τ and σ₂ = 0. On the σ₁‐σ₃ plane, this pure shear state is represented as a straight line through the origin at –45° as shown in figure 5.21. The line intersects the von Mises failure envelope at two points, A and B. The magnitude of σ₁ and σ₂ at these points can be found from eq. (5.93) as

(5.95)

Thus, in a pure shear stress state, the material yields when the shear stress reaches 0.577σ_Y. This value will be compared to the maximum shear stress theory described below.

5.7.4 Maximum Shear Stress Theory (Tresca)

According to the maximum shear stress theory, the material yields when the maximum shear stress at a point equals the critical shear stress value for that material. Since this should be true for a uniaxial stress state, we can use the results from uniaxial tension test to determine the maximum allowable shear stress. The stress state in a tensile specimen at the point of yielding is given by σ₁ = σ_Y, σ₂ = σ₃ = 0. The maximum shear stress is calculated as

(5.96)

This value of maximum shear stress is also called the yield shear stress of the material and is denoted by τ_Y. Note that τ_Y = σ_Y/2. Thus, Tresca’s yield criterion is that yielding will occur in a material when the maximum shear stress equals the yield shear strength, τ_Y, of the material.

The hexagon in figure 5.22 represents the two‐dimensional failure envelope according to maximum shear stress theory. The ellipse corresponding to von Mises’s theory is also shown in the same figure. The hexagon is inscribed within the ellipse and contacts it at six vertices. Combinations of principal stresses σ₁ and σ₃ that lie within this hexagon are considered safe based on the maximum shear stress theory, and failure is considered to occur when the combined stress state reaches the hexagonal boundary. This is obviously more conservative failure theory than distortion energy theory as it is contained within the latter. In the pure shear stress state, the shear stress at points C and D correspond to 0.5σ_Y, which is smaller than 0.577σ_Y, the failure point according to the distortion energy theory.

5.7.5 Maximum Principal Stress Theory (Rankine)

According to the maximum principal stress theory, a brittle material ruptures when the maximum principal stress in the specimen reaches some limiting value for the material. Again, this critical value can be inferred as the tensile strength measured using a uniaxial tension test. In practice, this theory is simple but can only be used for brittle materials. Some practitioners have modified this theory for ductile materials as

(5.97)

where σ₁ is the maximum principal stress, and σ_U is the ultimate strength described in table 5.3. Figure 5.23 shows the failure envelope based on the maximum principal stress theory. Note that the failure envelopes in the first and third quadrants are coincident with those of the maximum shear stress theory and contained within the distortion energy theory. However, the envelopes in the second and fourth quadrants are well outside of the other two theories. Hence, the maximum principal stress theory is not considered suitable for ductile materials. However, it can be used to predict failure in brittle materials.