Chapter 4

 

Solutions to simple boundary and initial value problems

 

 

 

In this chapter, we derive exact solutions to several problems involving deformable solids.  The examples have been selected partly because they can easily be solved, partly because they illustrate clearly the role of the various governing equations and boundary conditions in controlling the solution, and partly because the solutions themselves are of some practical interest.

 

 

4.1 Axially and spherically symmetric solutions to quasi-static linear elastic problems

 

 

4.1.1 Summary of governing equations of linear elasticity in Cartesian components

 

At last, we have all the basic equations we need to solve problems involving elastic materials subjected to loading. 

 

Specifically, we are given the following information

1.       Geometry of the solid

2.       Constitutive law for the material (i.e. the linear elastic-stress-strain equations)

3.       Body force density  (per unit mass) (if any)

4.       Temperature distribution  (if any)

5.       Prescribed boundary tractions  and/or boundary displacements  

 

In addition, to simplify the problem, we make the following assumptions

1.       All displacements are small.  This means that we can use the infinitesimal strain tensor to characterize deformation; we do not need to distinguish between stress measures, and we do not need to distinguish between deformed and undeformed configurations of the solid when writing equilibrium equations and boundary conditions.

2.       The material is an isotropic, linear elastic solid, with Young’s modulus E and Poisson’s ratio , and mass density  

 

With these assumptions, we need to solve for the displacement field , the strain field  and the stress field  satisfying the following equations:

 

 Displacementstrain relation  

 Stressstrain relation  

 Equilibrium Equation  (static problems only  you need the acceleration terms for dynamic problems)

 Traction boundary conditions  on parts of the boundary where tractions are known.

 Displacement boundary conditions  on parts of the boundary where displacements are known.

 

 

4.1.2 Simplified equations for spherically symmetric linear elasticity problems

 

A representative spherically symmetric problem is illustrated in the picture.  We consider a hollow, spherical solid, which is subjected to spherically symmetric loading (i.e. internal body forces, as well as tractions or displacements applied to the surface, are independent of  and , and act in the radial direction only).  If the temperature of the sphere is non-uniform, it must also be spherically symmetric (a function of R only).

 

The solution is most conveniently expressed using a spherical-polar coordinate system, illustrated in the figure.  The general procedure for solving problems using spherical and cylindrical coordinates is complicated, and is discussed in detail in Appendix E.  In this section, we summarize the special form of these equations for spherically symmetric problems.

 

As usual, a point in the solid is identified by its spherical-polar co-ordinates . All vectors and tensors are expressed as components in the basis  shown in the figure.  For a spherically symmetric problem

 Position Vector        

 Displacement vector  

 Body force vector  

 

Here,  and  are scalar functions. The stress and strain tensors (written as components in  ) have the form

 

and furthermore must satisfy  . The tensor components have exactly the same physical interpretation as they did when we used a fixed  basis, except that the subscripts (1,2,3) have been replaced by .

 

For spherical symmetry, the governing equations of linear elasticity reduce to

 

 Strain Displacement Relations  

 StressStrain relations

 

 Equilibrium Equations

 

 Boundary Conditions

 

Prescribed Displacements  

Prescribed Tractions  

 

These results can either be derived as a special case of the general 3D equations of linear elasticity in spherical coordinates, or alternatively can be obtained directly from the formulas in Cartesian components.  Here, we briefly outline the the latter.

1.       Note that we can find the components of  in the  basis as follows. First, note that  is radial, and can be written in terms of the position vector as .  Next, note  and .  Using index notation, the components of the basis vectors  in  are therefore

 

where ,  is the Kronecker delta and  is the permutation symbol.

2.       The components of the (radial) displacement vector in the  basis are .

3.       To proceed with the algebra, it is helpful to remember that ,   and  

4.       The components of the strain tensor in the  basis therefore follow as

 

5.       The strain components  can then be found as ,  and .  Substituting for the basis vectors and simplifying gives the strain-displacement relations.  For example

 

where we have noted . The remaining components are left as an exercise.

6.       Finally, to derive the equilibrium equation, note that the stress tensor can be expressed as .  Substituting for the basis vectors from item(1) above gives

 

7.       Substitute the preceding result into the equilibrium equation

 

and work through a good deal of tedious algebra to see that

 

This result can also be obtained using the virtual work principle (see problems for Sect 2.4 for details)

 

 

 

4.1.3 General solution to the spherically symmetric linear elasticity problem

 

Our goal is to solve the equations given in Section 4.1.2 for the displacement, strain and stress in the sphere.  To do so,

1.       Substitute the strain-displacement relations into the stress-strain law to show that

 

2.       Substitute this expression for the stress into the equilibrium equation and rearrange the result to see that

 

 

Given the temperature distribution and body force this equation can easily be integrated to calculate the displacement u.  Two arbitrary constants of integration will appear when you do the integral  these must be determined from the boundary conditions at the inner and outer surface of the sphere.  Specifically, the constants must be selected so that either the displacement or the radial stress have prescribed values on the inner and outer surface of the sphere.

 

In the following sections, this procedure is used to derive solutions to various boundary value problems of practical interest.

 

 

4.1.4 Pressurized hollow sphere

 

Assume that

 No body forces act on the sphere

 The sphere has uniform temperature

 The inner surface R=a is subjected to pressure  

 The outer surface R=b is subjected to pressure  

 

 

 

The displacement, strain and stress fields in the sphere are

 

 

 

      

 

Derivation:  The solution can be found by applying the procedure outlined in Sect 4.1.3.

1.       Note that the governing equation for u (Sect 4.1.3) reduces to

 

2.       Integrating twice gives

 

where A and B are constants of integration to be determined.

3.       The radial stress follows by substituting into the stress-displacement formulas

 

4.       To satisfy the boundary conditions, A and B must be chosen so that  and  (the stress is negative because the pressure is compressive).  This gives two equations for A and B that are easily solved to find

 

5.       Finally, expressions for displacement, strain and stress follow by substituting for A and B in the formula for u in (2), and using the formulas for strain and stress in terms of u in Section 4.1.2.

 

 

4.1.5 Gravitating sphere

 

A planet under its own gravitational attraction may be idealized (rather crudely) as a solid sphere with radius a, with the following loading

 A body force  per unit mass, where g is the acceleration due to gravity at the surface of the sphere

 A uniform temperature distribution

 A traction free surface at R=a

 

The displacement, strain and stress in the sphere follow as

 

 

 

             

 

Derivation:

1.       Begin by writing the governing equation for u given in 4.1.3 as

 

2.       Integrating

 

where A and B are constants of integration that must be determined from boundary conditions.

3.       The radial stress follows from the formulas in 4.1.3 as

 

4.       Finally, the constants A and B can be determined as follows: (i) The stress must be finite at , which is only possible if .  (ii) The surface of the sphere is traction free, which requires  at R=a.  Substituting the latter condition into the formula for stress in (3) and solving for A gives

 

5.       The final formulas for stress and strain follow by substituting the result of (4) back into (2), and using the formulas in Section 4.1.2.

 

 

4.1.6 Sphere with steady state heat flow

 

The deformation and stress in a sphere that is heated on the inside (or outside), and has reached its steady state temperature distribution can be calculated as follows.  Assume that

 No body force acts on the sphere

 The temperature distribution in the sphere is

 

where  and  are the temperatures at the inner and outer surfaces.  The total rate of heat loss from the sphere is , where k is the thermal conductivity.

 The surfaces at R=a  and R=b are traction free.

The displacement, strain and stress fields in the sphere follow as

 

 

 

 

 

 

Derivation:

1.       The differential equation for u given in 4.1.3 reduces to

 

2.       Integrating

 

where A and B are constants of integration.

3.       The radial stress follows from the formulas in 4.1.3 as

 

4.       The boundary conditions require that  at r=a and r=b.  Substituting these conditions into the result of step (3) gives two equations for A and B which can be solved to see that

 

 

 

 

4.1.7 Simplified equations for axially symmetric linear elasticity problems

 

 

 

 

Two examples of axially symmetric problems are illustrated in the picture.  In both cases the solid is a circular cylinder, which is subjected to axially symmetric loading (i.e. internal body forces, as well as tractions or displacements applied to the surface, are independent of  and , and act in the radial direction only).  If the temperature of the sphere is non-uniform, it must also be axially symmetric (a function of r only).  Finally, the solid can spin with steady angular velocity about the  axis.

 

The two solids have different shapes.  In the first case, the length of the cylinder is substantially greater than any cross-sectional dimension.  In the second case, the length of the cylinder is much less than its outer radius. 

 

 

The state of stress and strain in the solid depends on the loads applied to the ends of the cylinder. Specifically

 If the cylinder is completely prevented from stretching in the  direction a state of plane strain exists in the solid.  This is an exact solution to the 3D equations of elasticity, is valid for a cylinder with any length, and is accurate everywhere in the cylinder.

 If the top and bottom surface of the short plate-like cylinder are free of traction, a state of plane stress exists in the solid.  This is an approximate solution to the 3D equations of elasticity, and is accurate only if the cylinder’s length is much less than its diameter.   

 If the top and bottom  ends of the long cylinder are subjected to a prescribed force (or the ends are free of force) a state of generalized plane strain exists in the cylinder.  This is an approximate solution, which is accurate only away from the ends of a long cylinder.  As a rule of thumb, the solution is applicable approximately three cylinder radii away from the ends.

 

The solution is most conveniently expressed using a spherical-polar coordinate system, illustrated in the figure.  A point in the solid is identified by its spherical-polar co-ordinates . All vectors and tensors are expressed as components in the basis  shown in the figure.  For an axially symmetric problem

  Position Vector        

  Displacement vector  

  Body force vector  

  Acceleration vector  

Here,  and  are scalar functions.

 

The stress and strain tensors (written as components in  ) have the form

 

 

For axial symmetry, the governing equations of linear elasticity reduce to

 

 Strain Displacement Relations  

 StressStrain relations (plane strain and generalized plane strain)

 

    where  for plane strain, and constant for generalized plane strain.

 StressStrain relations (plane stress)

 

 

 Equation of motion

 

 Boundary Conditions

 

Prescribed Displacements  

Prescribed Tractions  

Plane strain solution  

Generalized plane strain solution, with axial force  applied to cylinder:

 

 

These results can either be derived as a special case of the general 3D equations of linear elasticity in spherical coordinates, or alternatively can be obtained directly from the formulas in Cartesian components.  Here, we briefly outline the the latter.

1.       Note that we can find the components of  in the  basis as follows. First, note that  is radial  a radial unit vector can be written in terms of the position vector as .  Next, note  and .  Using index notation, the components of the basis vectors  in  are therefore

 

where , and we use the convention that Greek subscripts range from 1 to 2.

2.       The components of the (radial) displacement vector in the  basis are .

3.       To proceed with the algebra, it is helpful to remember that ,   and  

4.       The components of the strain tensor in the  basis therefore follow as

 

5.       The strain components  can then be found as ,  and .  Substituting for the basis vectors and simplifying gives the strain-displacement relations.  For example

 

where we have noted . The remaining components are left as an exercise.

6.       Finally, to derive the equilibrium equation, note that the stress tensor can be expressed as .  Substituting for the basis vectors from item(1) above gives

 

7.       Substitute the preceding result into the equilibrium equation

 

and crank through a good deal of tedious algebra to see that

 

This result can also be obtained using the virtual work principle (see problems for Sect 2.4 for details)

 

 

4.1.8 General solution to the axisymmetric boundary value problem

 

Our goal is to solve the equations given in Section 4.1.2 for the displacement, strain and stress in the sphere.  To do so,

1.       Substitute the strain-displacement relations into the stress-strain law to show that, for generalized plane strain

 

where  is constant.  The equivalent expression for plane stress is

 

2.       Substitute these expressions for the stress into the equilibrium equation and rearrange the result to see that, for generalized plane strain

 

while for plane stress

 

 

Given the temperature distribution and body force these equations can be integrated to calculate the displacement u.  Two arbitrary constants of integration will appear when you do the integral  these must be determined from the boundary conditions at the inner and outer surface of the sphere.  Specifically, the constants must be selected so that either the displacement or the radial stress have prescribed values on the inner and outer surface of the cylinder.  Finally, for the generalized plane strain solution, the axial strain  must be determined, using the equation for the axial force acting on the ends of the cylinder.

 

In the following sections, this procedure is used to derive solutions to various boundary value problems of practical interest.

 

 

4.1.9 Long (generalized plane strain) cylinder subjected to internal and external pressure.

 

We consider a long hollow cylinder with internal radius a and external radius b as shown in the figure. 

Assume that

 No body forces act on the cylinder

 The cylinder has zero angular velocity

 The sphere has uniform temperature

 The inner surface r=a is subjected to pressure  

 The outer surface r=b is subjected to pressure  

 For the plane strain solution, the cylinder does not stretch parallel to its axis.  For the generalized plane strain solution, the ends of the cylinder are subjected to an axial force  as shown.  In particular, for a closed ended cylinder the axial force exerted by the pressure inside the cylinder acting on the closed ends is  

 

The displacement, strain and stress fields in the cylinder are

 

 

 

where  for plane strain, while

 

for generalized plane strain.

 

Derivation: These results can be derived as follows.  The governing equation reduces to

 

The equation can be integrated to see that

 

The radial stress follows as

 

The boundary conditions are  (the stresses are negative because the pressure is compressive).  This yields two equations for A and B that area easily solved to see that

 

The remaining results follow by elementary algebraic manipulations.

 

 

 

4.1.10 Spinning circular plate

 

We consider a thin solid plate with radius a that spins with angular speed  about its axis. Assume that

 No body forces act on the disk

 The disk has constant angular velocity

 The disk has uniform temperature

 The outer surface r=a and the top and bottom faces of the disk are free of traction.

 The disk is sufficiently thin to ensure a state of plane stress in the disk.

 

            

 

 

 

 

Derivation: To derive these results, recall that the governing equation is

 

The equation can be integrated to see that

 

The radial stress follows as

 

The radial stress must be bounded at r=0, which is only possible if B=0.  In addition, the radial stress must be zero at r=a, which requires that

 

The remaining results follow by straightforward algebra.

 

 

4.1.11 Stresses induced by an interference fit between two cylinders

 

Interference fits are often used to secure a bushing or a bearing housing to a shaft.  In this problem we calculate the stress induced by such an interference fit.

 

Consider a hollow cylindrical bushing, with outer radius b and inner radius a.  Suppose that a solid shaft with radius , with  is inserted into the cylinder as shown.  (In practice, this is done by heating the cylinder or cooling the shaft until they fit, and then letting the system return to thermal equilibrium)

 No body forces act on the solids

 The angular velocity is zero

 The cylinders have uniform temperature

 The shaft slides freely inside the bushing

 The ends of the cylinder are free of force.

 Both the shaft and cylinder have the same Young’s modulus E and Poisson’s ratio  

 The cylinder and shaft are sufficiently long to ensure that a state of generalized plane strain can be developed in each solid.

 

The displacements, strains and stresses in the solid shaft  (r<a) are

 

 

 

In the hollow cylinder, they are

 

 

 

 

Derivation: These results can be derived using the solution to a pressurized cylinder given in Section 4.1.9. After the shaft is inserted into the tube, a pressure p acts to compress the shaft, and the same pressure pushes outwards to expand the cylinder.  Suppose that this pressure induces a radial displacement  in the solid cylinder, and a radial displacement  in the hollow tube.  To accommodate the interference, the displacements must satisfy

 

Evaluating the relevant displacements using the formulas in 4.1.9 gives

 

Here, we have assumed that the axial force acting on both the shaft and the tube must vanish separately, since they slide freely relative to one another.  Solving these two equations for p shows that

 

This pressure can then be substituted back into the formulas in 4.1.9 to evaluate the stresses.

 

 

 

 

(c) A.F. Bower, 2008
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