Chapter 7
Boundary Value Problem: Formulation

7.1 Overview

In the previous chapters, we mastered the basic concepts of strain and stress along with the four basic equations, namely, the strain-displacement relation, compatibility condition, constitutive relation and the equilibrium equations. We are now in a position to find the stresses and displacement throughout the body - in the interior as well as in its boundary - given the displacement over some part of the boundary of the body and the traction on some other part of the boundary. Usually the part of the boundary where traction is specified, displacement would not be given and vice versa. In this course we shall formulate the boundary value problem for a body made up of isotropic material obeying Hooke’s law and deforming in such a manner that the relative displacement between its particles is small¹ . Then, we discuss three strategies to solve this boundary value problem.

Before we endeavor on the formulation of the boundary value problem and discuss strategies to solve it, we recollect and record the basic equations. If u represents the displacement that the particles undergo from a stress free reference configuration to the deformed current configuration with the deformation taking place due to application of the traction on the boundary, then we define the linearized strain as,

(7.1)

where, h = grad(u). The equation (7.1) is called as the strain displacement relation. The constitutive relation that relates the stress to the strain that is to be used in this course is the Hooke’s law. The various forms of the same Hooke’s law that would be used is recorded:

(7.5)

where ρ is the density, b is the body force per unit mass and a is the acceleration, for our purposes here reduces to:

(7.6)

since we have ignored the body forces and look at configurations that are in static equilibrium under the action of the applied static traction. That is the displacement is assumed not to depend on time. It is appropriate that we discuss these assumptions in some detail. By ignoring the body forces we are ignoring the stresses that arise in the body due to its own mass. Since, these stresses are expected to be much small in comparison to the stresses induced in the body due to traction acting on its boundary and these stresses practically do not vary over the surface of the earth, ignoring the body forces is justifiable. In the same spirit, all that we require is that the magnitude of acceleration be small, if not zero.

Finally, we document the compatibility conditions; the conditions that ensures the existence of a smooth displacement field given a strain field:

7.2 Formulation of boundary value problem

Formulation of the boundary value problem involves specification of the geometry of the body, the constitutive relation and the boundary conditions. To elaborate, we have to define the region of the Euclidean point space the body occupies in the reference configuration, which is denoted by . For example, the body might occupy a region that is a unit cube, then using Cartesian coordinates the body is defined as = {(X,Y,Z)|0 ≤ X ≤ 1, 0 ≤ Y ≤ 1, 0 ≤ Z ≤ 1}. Alternatively, the body might be the annular region between two annular cylinders of radius R_o and R_i and height H, then using cylindrical polar coordinates the body may be defined as = {(R, Θ,Z)|R_i ≤ R ≤ R_o, 0 ≤ Θ ≤ 2π, 0 ≤ Z ≤ H}. The boundary of the body, denoted as ∂ is a surface that encloses the body. For illustration, in the cube example, the surface is composite of 6 different planes, defined by X = 0, X = 1, Y = 0, Y = 1, Z = 0 and Z = 1 respectively and in the annular cylinder example the surface is composed of 4 different planes, defined by R = R_i, R = R_o, Z = 0, Z = H.

In this course, at least, the constitutive relation is known; it is Hooke’s law for isotropic materials. In real life problem, the weakest link in the formulation of the boundary value problem is the constitutive relation.

Then finally one needs to prescribe boundary conditions. These are specifications of the displacement or traction on the surface of the body. Depending on what is prescribed on the surface, there are four type of boundary conditions. They are

• Displacement boundary condition: Here the displacement is specified on the entire boundary of the body. This is also called as Dirichlet boundary condition
• Traction boundary condition: Here the traction is specified on the entire boundary of the body. This is also called as Neumann boundary condition
• Mixed boundary condition: Here the displacement is specified on part of the boundary and traction is specified on the remaining part of the boundary. Both traction as well as displacement are not specified over any part of boundary
• Robin boundary condition: Here both the displacement and the traction are specified on the same part of the boundary.

Once, the geometry of the body, constitutive relations and boundary conditions are prescribed then finding the Cauchy stress and displacement over the entire region of the body such that the displacement is continuous and differentiable over the entire region occupied by the body and the stress computed using this displacement field from the constitutive relation satisfies the equilibrium equations is called as solving the boundary value problem. If for a given geometry of the body, constitutive relations and boundary conditions, there exists only one displacement and stress field as a solution to the boundary value problem then the solution to the boundary value problem is said to be unique.

Now it is appropriate to make a few comments regarding the choice of the independent variable for the boundary value problem. That is, when we say the region occupied by the body, we should have been more specific and said whether this is the region occupied by the body in the reference or current configuration. As described in section 3.4 of chapter 3, the displacement and the stresses can be given an Eulerian or Lagrangian description, i.e.,

(7.13)

In this course, to be specific, we follow the Eulerian description of these fields. That is we use the region occupied by the body in the current configuration as the spatial domain of the boundary value problem. However, this domain is not known a priori. Therefore, we approximate the region occupied by the body in the current configuration with the region occupied by the body in the reference configuration. This approximation is justified, since we are interested only in problems where the components of the gradient of the displacement field are small and the magnitude of the displacement is also small. Then, one might ponder as to why we cannot say we are following the Lagrangian description of these fields. In chapter 5 section 5.4.1, we showed that the conservation of linear momentum takes different forms depending on the description of the stress field. The equation used here (7.5) is derived assuming Eulerian or spatial description of the stress field. If we use (5.46) instead of (7.5) then we would have said we are following Lagrangian description of these fields. Following Lagrangian description also, we would obtain the same governing equations if we assume that the components of the gradient of the deformation field are small and the magnitude of the displacement is small; but we have to do a little more algebra than saying upfront that we are following an Eulerian or spatial description.

7.3 Techniques to solve boundary value problems

Depending on the boundary condition specified the solution can be found using one of the following two techniques. Outline of these methods is presented next.

7.3.1 Displacement method

Here we take the displacement field as the basic unknown that need to be determined. Then using this displacement field we find the strain using the strain displacement relation (7.1). The so computed strain is substituted in the constitutive relation written using Lamè constants (7.2) to obtain

(7.14)

where we have used the definition of divergence operator, (2.208) and the property of the trace operator (2.67). Substituting (7.14) in the reduced equilibrium equations, under the assumption that body forces can be ignored and the body is in static equilibrium, (7.5) we obtain

(7.15)

where we have used equation (3.31) to write the acceleration in terms of the displacement and E denotes the total time derivative. In addition, to obtain the equation (7.15), we have used the following identities:

• Since divergence is a linear operator,

  (7.16)

• It follows from equation (2.218) that
  
  where we have used the fact that div is a linear operator and 1m = m when m is any vector and 1 is the second order identity tensor.
• From the identity (2.222) it follows that,

  (7.19)

• Using definition of the Laplace operator (2.212)

  (7.20)

• Using the identity (2.211) we note that

  (7.21)

Thus, we obtain (7.15) by substituting successive substitution of equations (7.17) through (7.21) in (7.16).

In order to simplify equation (7.15), we make the following assumptions:

• The body is homogeneous. Hence λ and μ are constants
• Body forces can be ignored
• Body is in static equilibrium under the applied traction

In lieu of these assumptions, equation (7.15) reduces to

(7.22)

If body forces cannot be ignored but the other two assumptions hold, then (7.15) reduces to

(7.23)

In this course, we attempt to find the displacement field that satisfies (7.22) along with the prescribed boundary conditions. We compute the stress field corresponding to the determined displacement field, using equation (7.2) where the strain is related to the displacement field through equation (7.1). We illustrate this method in section 7.4.1.

7.3.2 Stress method

Here we use stress as the basic unknown that needs to be determined. This method is applicable only for cases when the inertial forces (ρa) can be neglected. Since, we have assumed stress as the basic unknown, we want to express the compatibility conditions (7.7) through (7.12) in terms of the stresses. For this we compute the strains in terms of the stresses using the constitutive relation (7.4) and substitute in the compatibility conditions to obtain the following 6 equations:

If the body forces could be obtained from a potential, β = (x,y,z) called as the load potential, as

(7.33)

then the Cartesian components of the Cauchy stress could be obtained from a potential, ϕ = (x,y,z) called as the Airy’s stress potential and the load potential, β as,

(7.34)

so that the equilibrium equations (7.30) through (7.32) is satisfied for any choice of ϕ. Substituting for the Cartesian components of the stress from equation (7.34) in the compatibility equations (7.24) through (7.29) and simplifying we obtain:

Plane stress formulation

Next, we specialize the above stress formulation for the plane stress case. Without loss of generality, let us assume that the Cartesian components of this plane stress state is

(7.41)

Further, let us assume that body forces are absent and that the Airy’s stress function depends on only x and y. Thus, β = 0 and ϕ = ϕ(x,y). Note that this assumption for the Airy’s stress function does not ensure σ_zz = 0, whenever + ≠ 0. Hence, plane stress formulation is not a specialization of the general 3D problem. Therefore, we have to derive the governing equations again following the same procedure.

Since, we assume that there are no body forces and the Airy’s stress function depends only on x and y, the Cartesian components of the stress are related to the Airy’s stress function as,

(7.42)

Substituting for stress from equation (7.42) in the constitutive relation (7.4) we obtain,

(7.43)

Substituting for strain from equation (7.43) in the compatibility condition (7.7) through (7.12), the non-trivial equations are

(7.48)

where α_i’s are constants. Differentiating equation (7.48) with respect to x twice and adding to this the result of differentiation of equation (7.48) with respect to y twice, we obtain equation (7.44). Thus, for equations (7.44) through (7.47) to hold it suffices that ϕ satisfy equation (7.48) along with the prescribed boundary conditions. Comparing the expression for ϵ_zz in equation (7.43) and the requirement (7.48) arising from compatibility equations (7.8), (7.9) and (7.11) is a restriction on how the out of plane normal strain can vary, i.e., ϵ_zz = α₁x + α₂y + α₃, where α_i’s are some constants. Hence, this requirement that ϕ satisfy equation (7.48) does not lead to solution of a variety of boundary value problems. Due to Poisson’s effect plane stress does not lead to plane strain and vice versa, resulting in the present difficulty. To overcome this difficulty, it has been suggested that for plane problems one should use the 2 dimensional constitutive relations, instead of 3 dimensional constitutive relations that we have been using till now.

Since, the constitutive relation is 2 dimensional, plane stress implies plane strain and the three independent Cartesian components of the plane stress and strain are related as,

(7.49)

Inverting the above equations we obtain

(7.50)

By virtue of using (7.49) to compute the strain for the plane state of stress given in equation (7.42), the only non-trivial restriction from compatibility condition is (7.7) which requires that

(7.51)

Equation (7.51) is called as the bi-harmonic equation. Thus, for two dimensional problems formulated using stress, one has to find the Airy’s stress potential that satisfies the boundary conditions and the bi-harmonic equation (7.51). Then using this stress potential, the stresses are computed using (7.42). Having estimated the stress, the strain are found from the two dimensional constitutive relation (7.49). Finally, using this estimated strain, the strain displacement relation (7.1) is integrated to obtain the smooth displacement field. We study bending problems in chapter 8 using this approach.

Recognize that equation (7.51) is nothing but Δ(Δ(ϕ)) = 0, where Δ(⋅) is the Laplacian operator.

Next, we would like to formulate the plane stress problem in cylindrical polar coordinates. Let us assume that the cylindrical polar components of this plane stress state is

(7.52)

Further, let us assume that body forces are absent and that the Airy’s stress function depends on only r and θ, i.e. ϕ = (r,θ). For this case, the Cauchy stress cylindrical polar components are assumed to be

(7.53)

so that it satisfies the static equilibrium equations in the absence of body forces, equation (7.6). Then, using a 2 dimensional constitutive relation, the cylindrical polar components of the strain are related to the cylindrical polar components of the stress through,

(7.54)

As shown before, the only non-trivial restriction from compatibility condition in 2 dimensions is (7.7) and this in cylindrical polar coordinates takes the form,

(7.55)

Substituting equation (7.54) and (7.53) in (7.55) and simplifying we obtain,

(7.56)

A general periodic solution to the bi-harmonic equation in cylindrical polar coordinates, (7.56) is

(7.57)

7.4 Illustrative example

Having formulated the boundary value problem and seen at techniques to solve it, in this section we illustrate the same by solving some standard boundary value problems.

7.4.1 Inflation of an annular cylinder

In the first boundary value problem that is of interest, the body is in the form of an annular cylinder as shown in the figure 7.1. We use cylindrical polar coordinates to study this problem. Consequently, the body in the reference configuration is assumed to occupy a region in the Euclidean point space defined by, = {(r,θ,z)|r_i ≤ r ≤ r_o, 0 ≤ θ ≤ 2π, 0 ≤ z ≤ L} that is the region between two coaxial right circular cylinders of radius r_i and r_o respectively and of length L.

The boundary conditions that is of interest are also illustrated in figure 7.1. Thus, the cylinder is held fixed at constant length. Consequently, there is no axial displacement of the planes defined by z = 0 and z = L of the cylinder,

(7.58)

where u_z represents the axial component of the displacement field. Then, the outer surface defined by r = r_o, of the cylinder is traction free, i.e.,

(7.59)

On the remaining surface, the inner surface defined by r = r_i, of the cylinder only radial stress acts as shown in figure 7.1. Therefore,

(7.60)

where p_i is some positive constant. By virtue of the boundary conditions being independent of time and the constitutive relation is that of an elastic² response, we are justified in assuming that the body is in static equilibrium under the action of boundary traction and hence a = o. Strictly, the traction boundary condition has to be applied on the deformed surface and not on the original surface. However, as discussed in section 7.2, we approximate the deformed surface with the original surface since they are close, in lieu of our assumptions that the magnitude of the displacement is small and that the magnitude of the components of the displacement gradient is also small.

To proceed further one has to decide whether to use displacement or stress as the basic unknown. Here we use displacement as the basic unknown. In general, one needs to solve the three partial differential equations involving the components of the displacement field, (7.22). Note by using (7.22) instead of (7.15), we have ignored the body forces and as per the discussion above we have assumed the body to be in static equilibrium.

Next, in order to avoid solving partial differential equations, appropriate assumptions are made for the displacement field so that the governing equation (7.22) reduces to a ordinary differential equation. This is possible because, for mixed boundary condition boundary value problems for bodies in static equilibrium and in the absence of body forces, it is shown in section 7.5.1 that there exist a unique solution for a body made up of a material that obeys isotropic Hooke’s law. In light of the boundary condition (7.58) we assume u_z = 0. Further, we expect the cross section of the cylinder to deform as shown in the figure 7.2. That is we expect the circular cross section to remain circular but with a different radius and initially straight radial lines on the cross section to remain straight after deformation. Further we assume that there is no axial variation in the displacement field, that is any section along the axis of the cylinder deforms in the same fashion. These suggest that there is no circumferential component of the displacement field, i.e., u_θ = 0 and that the radial component of the displacement field vary radially only. Thus,

(7.61)

Substituting equation (7.61) for displacement field in (7.22) and using equations (2.259), (2.260) and (2.261) we obtain

(7.62)

Since, (λ + 2μ) ≠ 0, for equation (7.62) to hold we require

(7.63)

Solving the ordinary differential equation (7.63) we get

(7.64)

where C₁ and C₂ are integration constants to be found from the boundary conditions (7.59) and (7.60). Having found the unknown function in the displacement field (7.61), using (2.259) the cylindrical polar coordinate components of linearized strain can be computed as,

(7.65)

Substituting the above strain in the constitutive relation (7.2) the cylindrical polar coordinate components of the Cauchy stress is obtained as

(7.66)

Recognizing that for the above state of stress, t_(er) = [(λ + μ)C₁ - μC₂∕r²]e _r, boundary conditions (7.59) and (7.60) yield

(7.67)

Solving the above equations we obtain:

(7.68)

Substituting equation (7.68) in (7.66),

(7.69)

where we have used (6.79) to deduce that 2ν = λ∕(λ + μ). Substituting equation (7.68) in (7.64),

(7.70)

Figure 7.3 plots the variation of the radial (σ_rr) and hoop (σ_θθ) stresses with respect to r as given in equation (7.69) when r_i = r_o∕2. It is also clear from equation (7.69) that the axial stresses do not vary across the cross section. When r_i = r_o∕2, the constant axial stress value is 2p_iν∕3. Thus, we find that due to inflation of the annular cylinder, tensile hoop stresses develop and the radial stresses are always compressive. The magnitude of the hoop stress is greater than radial stress. The magnitude of the axial stress is nearly one-tenth of the maximum hoop stress and it is tensile when ν > 0 and compressive when ν < 0 and is zero when ν = 0. Next, we physically explain this variation of the axial stresses with the Poisson’s ratio. The hoop stresses by virtue of being tensile in nature will cause a reduction in the axial length due to Poisson’s effect³ if the Poisson’s ratio, ν > 0 and increase in length if ν < 0 and no change if ν = 0. Similarly, the radial stress being compressive in nature will cause an increase in the axial length when ν > 0, decrease in axial length when ν < 0 and no change in axial length when ν = 0. However, the actual change in axial length would be a sum of both the change in length due to hoop and radial stress. Since, the hoop stress is more than the radial stress, if no axial force is applied, the axial length would reduce when ν > 0, will not change when ν = 0 and will increase when ν < 0. Thus, if the length is to remain unchanged, a tensile axial force has to be applied to counter the reduction in length when ν > 0 and a compressive axial force when ν < 0 and no axial force is required when ν = 0. It can be seen from the expression for axial stress in equation (7.69) that the expression for the axial stresses is consistent with the expectation that it be positive when ν > 0, negative when ν < 0 and zero when ν = 0.

Now, we study the changes that occur to the thickness of the annular cylinder. Towards this, the ratio of change in thickness of the cylinder, Δt to its original thickness is obtained as

(7.71)

where we have used (6.80) to write Lamè constants in terms of the Poisson’s ratio, ν and Young’s modulus E. Figure 7.4 plots the variation in the ratio of change in thickness of the cylinder to its original thickness as a function of the thickness of the cylinder for various possible values of Poisson’s ratio for a given radial component of the normal stress at the inner surface, p_i. It can be seen from the figure that while the thickness decreases when ν ≥ 0, thickness increases when ν < 0 for thickness less than a critical value. By virtue of the radial component of the normal stress being compressive in nature one would expect a reduction in the thickness of the cylinder. However, the hoop stress being in tension, due to Poisson’s effect there would be reduction in thickness when ν > 0 and an increase when ν < 0. The actual change in the thickness is the sum of both the reduction due to radial stress and alteration due to hoop stress. Hence, the thickness decreases when ν ≥ 0 and increases when ν < 0 for thickness less than a critical value.

Finally, we would like to show that due to inflation the inner as well as outer radius of the cylinder increases. Noting that the deformed inner radius of the cylinder is r_i + u_r(r_i) and that of the deformed outer radius is r_o + u_r(r_o), for these radius to increase we need to show that u_r(r_i) > 0 and u_r(r_o) > 0. Towards this, using equation (7.70) and (6.80) we obtain,

(7.72)

Recollecting from table 6.1, that the physically possible values for the Poisson’s ratio is: -1 < ν ≤ 0.5, it is straightforward to see that (1 - 2ν)∕(1 - ν) ≥ 0 for these possible range of values for ν. Also, it can be seen from table 6.1 that μ > 0. Further, by definition r_o > r_i and p_i > 0. Therefore, u_r(r) > 0 for any r, since each term in the expression for u_r is positive. Hence, in particular u_r(r_i) > 0 and u_r(r_o) > 0 and hence the radius of the cylinder increases due to inflation, as one would expect.

Pressurized hole in an infinite medium

As a limiting case of the above solution, we would like to study the problem of a hole subjected to uniform radial component of the normal stress in an infinite medium as shown in figure 7.5. We continue to use cylindrical polar coordinates. Therefore the body in the reference configuration is assumed to occupy the region of the Euclidean point space defined by = {(r,θ,z)|r_i ≤ r < ∞, 0 ≤ θ ≤ 2π,-∞ < z < ∞}. The boundary conditions for this problem is essentially same as that in the inflation of an annular cylinder, except that now r_o tends to ∞. For completeness the boundary conditions for the present problem is:

(7.76)

(7.77)

Thus, we find that both the stress and the displacement tend to zero as r tends to ∞. This means that the effect of pressurized hole is not felt at a distance far away from the hole.

Having studied the problem of pressurized hole, in the following section we shall study the influence of the hole to uniaxial tensile load applied to the infinite medium with the hole.

7.4.2 Uniaxial tensile loading of a plate with a hole

In the second boundary value problem that we study, the body is an infinite medium with a circular stress free hole subjected to a far field tension along the x-direction as shown in the figure 7.6. We envisage to solve this boundary value problem using the stress formulation. In particular we assume plane stress state and the body to be two dimensional and use cylindrical polar coordinates to study this problem. Thus, the body in the reference configuration is assumed to occupy a region in the Euclidean point space defined by = {(r,θ)|a ≤ r < ∞, 0 ≤ θ ≤ 2π}, where a is a constant characterizing the size of the hole in the infinite medium. The boundary conditions that is of interest are

(7.83)

where σ_ij^* represents the value of the stress at (∞,θ). Of the four equations only 3 are independent. Picking any three equations and solving for the cylindrical polar components of the stress, we obtain

Based on the observation that the far field stresses are a function of cos(2θ), we infer that the Airy’s stress function, ϕ also should contain cos(2θ) term. Then, since the solution at θ equal to 0 and 2π should be the same, the Airy’s stress function can depend on θ only through the trigonometric functions present in the general solution (7.57). We assume Airy’s stress function with only five constants from the general solution (7.57) as,

(7.87)

and examine whether the prescribed boundary conditions can be met. Consideration that the stress at ∞ be finite required dropping of the terms r² ln(r) and r⁴. The constant a ₀₁ does not enter the expressions for stress and hence indeterminate. So we assumed it to be zero. For the assumption of Airy’s stress function (7.87), the cylindrical polar components of the stress is computed using (7.53) to be

(7.91)

where we have equated the coefficients of cos(2θ), sin(2θ) and the constant to the values of the right hand side of these equations. It can be seen that the equations (7.85) and (7.86) yield the same last two equations in (7.91). Solving the linear equations (7.91) for a_ij’s, we obtain

(7.92)

Since, we are able to meet the required boundary conditions with the assumed form of the Airy’s stress function, this is the required stress function.

Substituting the constants (7.92) in the equations (7.88) through (7.90) we obtain

(7.99)

(7.100)

where g(r) is some function of r. Substituting equations (7.99) and (7.100) in equation (7.98) and simplifying we obtain

(7.101)

Since g is a function of only r and f only of θ, we require

(7.102)

where C₀ is a constant. Solving the linear ordinary differential equations (7.102) we obtain

(7.103)

where C_i’s are constants. To ensure that the tangential displacement of the ray θ = 0 and θ = 2π to be the same and there be no rigid body displacement we require that,

(7.104)

The condition (7.104) translates into requiring C₁ = C₂ = C₃ = 0. Thus, the displacement field is computed to be,

First we begin with hoop stress, σ_θθ given in equation (7.94). Recollecting that the extremum value of a function could occur either at the boundary points or at points where the derivative of the function goes to zero, the extremum hoop stress could occur at a radial location, r = a, when cos(2θ) > 0 or at the boundary points, r = a and r tending to ∞. In figure 7.7 we plot the hoop stress as a function of r for various values of θ. As expected, the function is monotonically decreasing function of r when cos(2θ) < 0 and therefore the maximum value occurs at r = a for this case. When cos(2θ) > 0, the minimum occurs at r = a, then the maximum occurs when r = a and then the stress decreases and asymptotically reaches the value of σ_a[1 - cos(2θ)]∕2. Therefore, at r = a an extremum of the hoop stress occurs. The circumferential variation of the hoop stress at r = a is given by σ_a[1 - 2 cos(2θ)]. Hence, the maximum hoop stress occurs at r = a and θ = π∕2 (or θ = 3π∕2) and the value of the maximum hoop stress is 3σ_a. Thus, the stress concentration factor, defined as the ratio of the maximum stress in the structure to the far field stress is 3 (= σ_θθ^max∕σ _a).

Next, we study the circumferential shearing stress, σ_rθ given in equation (7.95). The extremum value of this stress occurs at a radial location r = a and the value of the circumferential shearing stress at this radial location is -2σ_a sin(2θ)∕3. At the boundary points the circumferential shearing stress takes values 0 at r = a and -σ_a sin(2θ)∕2. The variation of σ_rθ with the radial location, r and the orientation of the ray, θ is shown in figure 7.8. Thus, it can be seen that the absolute maximum value of the stress, σ_rθ occurs at r = a, θ = mπ∕4, where m takes one of the values from the set {1, 3, 5, 7} and this maximum value is 2σ_a∕3.

Finally, we examine the radial stress, σ_rr given in equation (7.93). An extremum value of this stress occurs at

(7.107)

when⁴ 0 ≤ θ ≤ π∕6 or 2π∕6 ≤ θ ≤ 4π∕6 or 5π∕6 ≤ θ ≤ 7π∕6 or 8π∕6 ≤ θ ≤ 10π∕6 or 11π∕6 ≤ θ ≤ 2π. This extremum value of the radial stress is -σ_a{11 cos(2θ)∕9 + 1∕9 + 5∕[36 cos(2θ)]}∕2. At the boundary the radial stresses takes the value 0 at r = a and σ_a[1 + cos(2θ)]∕2 as r tends to ∞. Figure 7.9 portrays the variation of the radial stress with radial location, r for various orientations of the ray, θ. It can be seen from the figure that for certain orientations (say, θ = 90 degrees) the maximum value of the radial stress occurs at r = a and for some other orientations (say, θ = 15 degrees) the maximum value occurs when r tends to ∞. To understand this, in figure 7.10 we plot the two extremum radial stresses - one at r = r_crit and the other value that occurs when r tends to ∞. It can be seen from the figure that when π∕3 ≤ θ ≤ 2π∕3 and 8π∕3 ≤ θ ≤ 10π∕3 the maximum radial stress occurs at r_crit and for all other values of θ it occurs as r tends to ∞. In any case the maximum value is always less than σ_a.

This concludes our illustration of the two techniques to solve boundary value problems in this chapter. However, in the remaining chapters we shall see employment of these techniques to solve more boundary value problems of interest in engineering.

7.5 General results

In this section, we record two results that are useful while solving boundary value problems. One result tells when one can expect an unique solution to a given boundary value problem. The other allows us to construct solutions to complex boundary conditions from simple cases.

7.5.1 Uniqueness of solution

Now, we are interested in showing that there could at most be one solution that could satisfy the prescribed displacement or mixed boundary conditions, in a given body made of a material that obeys Hooke’s law and in static equilibrium with no body forces acting on it. If traction boundary condition is specified, we shall see that the stress is uniquely determined but the displacement is not for the bodies made of a material that obeys Hooke’s law and in static equilibrium with no body forces acting on it.

Towards this, we consider a general setting with being some region occupied by the body and ∂ the boundary of the body. Let us also assume that displacement is prescribed over some part of the boundary ∂_u and traction specified on the remaining part of the boundary, ∂_σ. Let (u₁, σ₁) and (u₂, σ₂) be the two distinct solutions to a given boundary value problem. Let us define

(7.108)

By virtue of (u₁,σ₁) and (u₂, σ₂) satisfying the specified boundary conditions,

First, we examine the term,

(7.111)

here to obtain the last equality we have used (2.104) and the fact that the Cauchy stress σ is a symmetric tensor. Since, we are interested in materials that obey isotropic Hooke’s law, we substitute for the stress from⁵

(7.112)

in (7.111) to obtain

(7.113)

As recorded in table 6.1 G > 0 and K > 0. Further recognizing that irrespective of the sign of the Cartesian components of the strain,

(7.114)

(7.115)

as long as ϵ ≠ 0.

Next, we want to express

(7.116)

in terms of the boundary conditions alone. Towards this, using the result in equation (2.219) we write

(7.117)

Since the body is in static equilibrium and has no body forces acting on it, from equation (7.6) div(σ) = o. Using div(σ) = o in equation (7.117) and substituting the result in (7.116) we get

(7.118)

where to obtain the last equality we have used the result (2.267). Substituting the conditions (7.109) and (7.110) on the displacement and stress, in (7.118) that

(7.119)

Since, from equation (7.115) the integrand in the equation (7.119) is positive and W = 0 only if ϵ = 0, it follows that ϵ = 0 everywhere in the body. Then, it is straight forward to see that σ = 0, everywhere in the body. It can then be shown that, we do this next for a special case, the conditions ϵ = 0 everywhere in the body and (7.109) imply u = o everywhere in the body.

Integrating the first order differential equations on the Cartesian components of the displacement field when ϵ = 0, we obtain

(7.120)

where K_i’s are constants. Knowing the displacement of 3 points, say (x₁,y₁,z₁), (x₂,y₂,z₂) and (x₃,y₃,z₃), that are not collinear to be zero, the constants K_i’s could be uniquely determined by solving the following system of equations:

(7.121)

Hence, we conclude that the solution to the boundary value problem involving a body in static equilibrium, under the absence of body forces and made of a material that obeys isotropic Hooke’s law is unique except in cases where only traction boundary condition is specified. As a consequence of this theorem, if a solution has been found for a given boundary conditions it is the solution for a body in static equilibrium, under the absence of body forces and made of a material that obeys isotropic Hooke’s law.

7.5.2 Principle of superposition

This principle states that For a given body made up of a material that obeys isotropic Hooke’s law, in static equilibrium and whose magnitude of displacement is small, if {u⁽¹⁾,σ⁽¹⁾} is a solution to the prescribed body forces, b⁽¹⁾ and boundary conditions, {u_b⁽¹⁾,t _(n)⁽¹⁾} and {u⁽²⁾,σ⁽²⁾} is a solution to the prescribed body forces, b⁽²⁾ and boundary conditions, {u _b⁽²⁾,t _(n)⁽²⁾} then {u⁽¹⁾ + u⁽²⁾,σ⁽¹⁾ + σ⁽²⁾} will be a solution to the problem with body forces, b⁽¹⁾ + b⁽²⁾ and boundary conditions, {u _b⁽¹⁾ + u _b⁽²⁾,t _(n)⁽¹⁾ + t _(n)⁽²⁾}

The proof of the above statement follows immediately from the fact that equation (7.23) is linear. That is,

This is one of the most often used principles to solve problems in engineering. We shall illustrate the use of this principle in chapter 11.

7.6 Summary

In this chapter, we formulated the boundary value problem for an isotropic material undergoing small elastic deformations obeying Hooke’s law. Two techniques were outlined to solve this problem. These two techniques are summarized in figure 7.11. Thus, in the displacement approach, one starts with the displacement use the strain displacement relation to compute the strain and then the constitutive relation to find the stress and finally use the equilibrium equations to get the governing equation that the displacement field has to satisfy along with the prescribed boundary conditions. In the stress approach, stress field is assumed in terms of a potential, such that equilibrium equations hold, then the strain is computed using the constitutive relation. In order to be able to obtain a smooth displacement field using this strain field, it is required that the stress potential satisfy the compatibility conditions along with the prescribed boundary conditions. We then illustrated these techniques by solving two boundary value problems, that of inflation of an annular cylinder and that of a plate with a hole subjected to uniaxial tension far away from the hole.

7.7 Self-Evaluation

• A body in the form of an annular cylinder is pressurized from outside as shown in figure 7.12. Obtain the stress and displacement field for this boundary value problem and show that the developed hoop stresses are compressive in nature and that the thickness of the annular cylinder increases due to the applied external pressure. Assume that the body is homogeneous and is made of a material that obeys isotropic Hooke’s law.
• Using the solution obtained in problem 1, study the problem of a stress free hole in an infinite medium under equal biaxial loading at infinity, as shown in figure 7.13. Recognize that this solution can be obtained by tending r_o to ∞ in the solutions to problem 1. Then show that the hoop stresses developed at the boundary of the hole is twice the far field equal biaxial stress applied.

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  Figure 7.14: An annular cylinder pressurized from inside and outside. (Figure for Problems 3, 5 and 6)

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  Figure 7.15: An annular cylinder pressurized from outside with its inner surface prevented from radially displacing. (Figure for Problem 4)

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• The annular cylinder shown in figure 7.14 is pressurized from both inside and outside. Assuming the cylinder to be homogeneous and made of a material that obeys isotropic Hooke’s law with Young’s modulus, E = 200 GPa and Poisson’s ratio, ν = 0.3,
  • Obtain the location and magnitude of the maximum and minimum hoop stress and radial stress.
  • Obtain the maximum principal stress, its location and magnitude.
  • Use thin cylinder approximation to obtain the hoop and radial stress. Use this to obtain the maximum principal stress.
  • Plot the ratio of the maximum principal stress obtained in part b to that in part c as a function of (1-R_i∕R_o)-1 for R_i∕R_o = 0.05 to 0.99. Plot the same for (p_i,p_o) = {(1, 0), (0, 1), (1, 0.5), (0.5, 1)}. Using the above plots verify the reasonableness of the following classification: d∕t < 20 thick cylinder, d∕t > 40 thin cylinder, where d is the mean diameter of the cylinder and t is its thickness.
• Assuming the annular cylinder shown in figure 7.15, to be homogeneous and made of a material that obeys isotropic Hooke’s law with Young’s modulus, E = 200 GPa and Poisson’s ratio, ν = 0.3 repeat parts (a) and (b) in problem 3. Here the inner boundary is fixed, i.e., u_r(r_i,θ,z) = 0. Find the hoop stress and radial stress as R_i∕R_o tends to 0 and comment on the significance of this limit.
• Now assume the annular cylinder shown in figure 7.14 is inhomogeneous and is composed of two annular cylinders made of aluminium and steel respectively. Solve parts (a) to (d) in problem 3 assuming
  • Aluminium cylinder occupies the region = {(r,θ,z)|R_i ≤ r ≤ (R_i + R_o)∕2, 0 ≤ θ ≤ 2π, 0 ≤ z ≤ H} and steel cylinder occupies the region = {(r,θ,z)|(R_i + R_o)∕2 ≤ r ≤ R_o, 0 ≤ θ ≤ 2π, 0 ≤ z ≤ H}.
  • Steel cylinder occupies the region = {(r,θ,z)|R_i ≤ r ≤ (R_i + R_o)∕2, 0 ≤ θ ≤ 2π, 0 ≤ z ≤ H} and aluminium cylinder occupies the region = {(r,θ,z)|(R_i + R_o)∕2 ≤ r ≤ R_o, 0 ≤ θ ≤ 2π, 0 ≤ z ≤ H}.

  Take Young’s modulus of steel and aluminium to be E_steel = 200 GPa and E_aluminium = 70 GPa respectively and Poisson’s ratio of these materials to be ν_steel = 0.3 and ν_aluminium = 0.27.

• Assume that an aluminium annular cylinder occupying the region = {(r,θ,z)|R_i ≤ r ≤ (δ + (R_i + R_o)∕2), 0 ≤ θ ≤ 2π, 0 ≤ z ≤ H} in the stress free state is shrunk and fitted into a steel annular cylinder occupying the region = {(r,θ,z)|(R_i + R_o)∕2 ≤ r ≤ R_o, 0 ≤ θ ≤ 2π, 0 ≤ z ≤ H} in the stress free state. Then this composite shrink fitted cylinder is pressurized as shown in figure 7.14. For this case solve parts (a) to (d) in problem 3. Assume δ = 0.0001R_o. Take Young’s modulus of steel and aluminium to be E_steel = 200 GPa and E_aluminium = 70 GPa respectively and Poisson’s ratio of these materials to be ν_steel = 0.3 and ν_aluminium = 0.27.