Chapter 8 Axisymmetrically Loaded Members

8.1 Introduction

There are numerous practical situations in which the distribution of stress manifests symmetry about an axis. Examples include pressure vessels, compound cylinders, clad reactor elements, chemical reaction vessels, heat exchanger tubes, flywheels, solid or hollow spherical structures, turbine disks, and components of many other machines, from aerospace to household in application.

Consider a large thin plate having a small circular hole subjected to uniform pressure, as shown in Fig. 8.1. Note that axial loading is absent and therefore σ_z = 0. The stresses are clearly symmetrical about the z axis, and the deformations likewise display θ independence. The symmetry argument also dictates that the shearing stresses τ_rθ must be zero. Assuming z independence for this thin plate, the polar equations of equilibrium (3.28), reduce to

(8.1)

Figure 8.1. Large thin plate with a small circular hole.

Here σ_θ and σ_r denote the tangential (circumferential) and radial stresses acting normal to the sides of the element, and F_r represents the radial body force per unit volume, for example, the inertia force associated with rotation. In the absence of body forces, Eq. (8.1) reduces to

(8.2)

Consider now the radial and tangential displacements u and ν, respectively. There can be no tangential displacement in the symmetrical field; that is, ν = 0. A point represented by the shaded element abcd in the figure will thus move radially as a consequence of loading, but not tangentially. On the basis of displacements indicated, the strains given by Eqs. (3.30) become

(8.3)

Substituting u = rε_θ into the first expression in Eq. (8.3), a simple compatibility equation is obtained:

or

(8.4)

The equation of equilibrium [Eq. (8.1) or (8.2)], the strain-displacement or compatibility relations [Eqs. (8.3) or (8.4)], and Hooke’s law are sufficient to obtain a unique solution to any axisymmetrical problem withspecified boundary conditions.

8.2 Thick-Walled Cylinders

The circular cylinder, of special importance in engineering, is usually divided into thin-walled and thick-walled classifications. A thin-walled cylinder is defined as one in which the tangential stress may, within certain prescribed limits, be regarded as constant with thickness. The following familiar expression applies to the case of a thin-walled cylinder subject to internal pressure:

Here p is the internal pressure, r the mean radius (see Sec. 13.13), and t the thickness. If the wall thickness exceeds the inner radius by more than approximately 10%, the cylinder is generally classified as thick walled, and the variation of stress with radius can no longer be disregarded (see Prob. 8.1).

In the case of a thick-walled cylinder subject to uniform internal or external pressure, the deformation is symmetrical about the axial (z) axis. Therefore, the equilibrium and strain-displacement equations, Eqs. (8.2) and (8.3), apply to any point on a ring of unit length cut from the cylinder (Fig. 8.2). Assuming that the ends of the cylinder are open and unconstrained, σ_Z = 0, as shall be subsequently demonstrated. Thus, the cylinder is in a condition of plane stress and, according to Hooke’s law (3.31), the strains are given by

(8.5)

Figure 8.2. Thick-walled cylinder.

From these, σ_r and σ_θ are as follows:

(8.6)

Substituting this into Eq. (8.2) results in the following equidimensional equation in radial displacement:

(8.7)

having a solution

(a)

The radial and tangential stresses may now be written in terms of the constants of integration c₁ and c₂ by combining Eqs. (a) and (8.6):

(b)

(c)

The constants are determined from consideration of the conditions pertaining to the inner and outer surfaces.

Observe that the sum of the radial and tangential stresses is constant, regardless of radial position: σ_r + σ_θ = 2Ec₁/(1 – ν). Hence, the longitudinal strain is constant:

We conclude therefore that plane sections remain plane subsequent to loading. Then σ_z = Eε_z = constant = c. But if the ends of the cylinder are open and free,

or c = σ_z = 0, as already assumed previously.

For a cylinder subjected to internal and external pressures p_i and p_o, respectively, the boundary conditions are

(d)

where the negative sign connotes compressive stress. The constants are evaluated by substitution of Eqs. (d) into (b):

(e)

leading finally to

(8.8)

These expressions were first derived by French engineer G. LamÉ in 1833, for whom they are named. The maximum numerical value of σ_r is found at r = a to be p_i, provided that p_i exceeds p_o. If p_o > p_i, the maximum σ_r occurs at r = b and equals p_o. On the other hand, the maximum σ_θ occurs at either the inner or outer edge according to the pressure ratio, as discussed in Sec. 8.3.

Recall that the maximum shearing stress at any point equals one-half the algebraic difference between the maximum and minimum principal stresses. At any point in the cylinder, we may therefore state that

(8.9)

The largest value of τ_max is found at r = a, the inner surface. The effect of reducing p_o is clearly to increase τ_max. Consequently, the greatest τ_max corresponds to r = a and p_o = 0:

(8.10)

Because σ_r and σ_θ are principal stresses, τ_max occurs on planes making an angle of 45° with the plane on which σ_r and σ_θ act. This is quickly confirmed by a Mohr’s circle construction. The pressure p_yp that initiates yielding at the inner surface is obtained by setting τ_max = σ_yp/2 in Eq. (8.10):

(8.11)

Here σ_yp is the yield stress in uniaxial tension.

Special Cases

Internal Pressure Only. If only internal pressure acts, Eqs. (8.8) reduce to

(8.12)

(8.13)

(8.14)

Since b²/r² ≥ 1, σ_r is negative (compressive) for all r except r = b, in which case σ_r = 0. The maximum radial stress occurs at r = a. As for σ_θ, it is positive (tensile) for all radii and also has a maximum at r = a.

To illustrate the variation of stress and radial displacement for the case of zero external pressure, dimensionless stress and displacement are plotted against dimensionless radius in Fig. 8.3a for b/a = 4.

External Pressure Only. In this case, p_i = 0, and Eq. (8.8) becomes

(8.15

(8.16)

(8.17)

Figure 8.3. Distribution of stress and displacement in a thick-walled cylinder with b/a = 4: (a) under internal pressure; (b) under external pressure.

The maximum radial stress occurs at r = b and is compressive for all r. The maximum σ_θ is found at r = a and is likewise compressive.

For a cylinder with b/a = 4 and subjected to external pressure only, the stress and displacement variations over the wall thickness are shown in Fig. 8.3b.

Example 8.1

A thick-walled cylinder with 0.3-m and 0.4-m internal and external diameters is fabricated of a material whose elastic limit is 250 MPa. Let ν = 0.3. Determine (a) for p_o = 0, the maximum internal pressure to which the cylinder may be subjected without exceeding the elastic limit, (b) for p_i = 0, the maximum external pressure to which the cylinder can be subjected without exceeding the elastic limit, and (c) the radial displacement of a point on the inner surface for case (a).

Solution

a. From Eq. (8.13), with r = a,

(8.18)

or

b. From Eq. (8.16), with r = a,

(8.19)

Then

c. Using Eq. (8.14), we obtain

In the case of a closed-ended cylinder subjected to internal and external pressures, longitudinal or z-directed stresses exist in addition to the radial and tangential stresses. For a transverse section some distance from the ends, this stress may be assumed uniformly distributed over the wall thickness. The magnitude of σ_z is then determined by equating the net force acting on an end attributable to pressure loading to the internal z-directed force in the cylinder wall:

p_iπa² − p_oπb² = (πb² − πa²)σ_z

The resulting expression for longitudinal stress, applicable only away from the ends, is

(8.20)

Clearly, here it is again assumed that the ends of the cylinder are not constrained: ε_z ≠ 0 (see Prob. 8.13).

8.3 Maximum Tangential Stress

An examination of Figs. 8.3a and b shows that if either internal pressure or external pressure acts alone, the maximum tangential stress occurs at the innermost fibers, r = a. This conclusion is not always valid, however, if both internal and external pressures act simultaneously. There are situations, explored next, in which the maximum tangential stress occurs at r = b [Ref. 8.1].

Consider a thick-walled cylinder, as in Fig. 8.2, subject to p_i and p₀. Denote the ratio b/a by R, p_o/p_i by P, and the ratio of tangential stress at the inner and outer surfaces by S. The tangential stress, given by Eqs. (8.8), is written

(a)

Hence,

(8.21)

The variation of the tangential stress σ_θ over the wall thickness is shown in Fig. 8.4 for several values of S and P. Note that for pressure ratios P, indicated by dashed lines, the maximum magnitude of the circumferential stress occurs at the outer surface of the cylinder.

Figure 8.4. Maximum tangential stress distribution in a thick-walled cylinder under internal and external pressures.

8.4. Application of Failure Theories

Unless we are content to grossly overdesign, it is necessary to predict, as best possible, the most probable failure mechanism. Thus, while examination of Fig. 8.4 indicates that failure is likely to originate at the innermost or outermost fibers of the cylinder, it cannot predict at what pressure or stresses failure will occur. To do this, consideration must be given the stresses determined from LamÉ’s equations, the material strength, and an appropriate theory of failure consistent with the nature of the material.

Example 8.2

A steel cylinder is subjected to an internal pressure four times greater than the external pressure. The tensile elastic strength of the steel is σ_yp = 340 MPa, and the shearing elastic strength τ_yp = σ_yp/2 = 170 MPa. Calculate the allowable internal pressure according to the various yielding theories of failure. The dimensions are a = 0.1 m and b = 0.15 m. Let ν = 0.3.

Solution The maximum stresses occur at the innermost fibers. From Eqs. (8.8), for r = a and p_i = 4p_o, we have

(a)

The value of internal pressure at which yielding begins is predicted according to the various theories of failure, as follows:

a. Maximum shearing stress theory:

b. Energy of distortion theory [Eq. (4.5a)]:

c. Octahedral shearing stress theory: By use of Eqs. (4.6) and (1.38), we have

The results found in (b) and (c) are identical as expected (Sec. 4.8). The onset of inelastic action is governed by the maximum shearing stress: the allowable value of internal pressure is limited to 125.9 MPa, modified by an appropriate factor of safety.

8.5. Compound Cylinders

If properly designed, a system of multiple cylinders resists relatively large pressures more efficiently, that is, requires less material, than a single cylinder. To assure the integrity of the compound cylinder, one of several methods of prestressing is employed. For example, the inner radius of the outer member or jacket may be made smaller than the outer radius of the inner cylinder. The cylinders are assembled after the outer cylinder is heated, contact being effected upon cooling. The magnitude of the resulting contact pressure p, or interface pressure, between members may be calculated by use of the equations of Sec. 8.2. Examples of compound cylinders, carrying very high pressures, are seen in compressors, extrusion presses, and the like.

Referring to Fig. 8.5, assume the external radius of the inner cylinder to be larger, in its unstressed state, than the internal radius of the jacket, by an amount δ.

Figure 8.5. Compound cylinder.

This quantity is the shrinking allowance. Subsequent to assembly, the contact pressure, acting equally on both members, causes the sum of the increase in the inner radius of the jacket and decrease in the outer radius of the inner member to exactly equal δ. By using Eqs. (8.14) and (8.17), we obtain

(8.22)

From this expression, the interface pressure is then found as a function of the shrinking allowance:

(8.23)

The stresses in the jacket are then determined from Eqs. (8.12) and (8.13) by treating the contact pressure as p_i. Similarly, by regarding the contact pressure as p₀, the stresses in the inner cylinder are calculated from Eqs. (8.15) and (8.16). The case of a composite cylinder of different materials is treated by employing a procedure similar to that just described (see Prob. 8.18).

Example 8.3

A compound cylinder with a = 150 mm, b = 200 mm, c = 250 mm, E = 200 GPa, and δ = 0.1 mm is subjected to an internal pressure of 140 MPa. Determine the distribution of tangential stress throughout the composite wall.

Solution In the absence of applied internal pressure, the contact pressure is, from Eq. (8.23),

The tangential stresses in the outer cylinder associated with this pressure are found by using Eq. (8.13)

The stresses in the inner cylinder are, from Eq. (8.16),

These stresses are plotted in Fig. 8.6, indicated by the dashed lines kk and mm. The stresses owing to internal pressure alone, through the use of Eq. (8.13) with b = c, are found to be (σ_θ)_{r = 0.15} = 297.5 MPa, (σ_θ)_{r = 0.2} = 201.8 MPa, (σ_θ)_{r = 0.25} = 157.5 MPa, and are shown as the dashed line nn. The stress resultant is obtained by superposition of the two distributions, represented by the solid line. The use of a compound prestressed cylinder has thus reduced the maximum stress from 297.5 to 257.8 MPa. Based on the maximum principal stress theory of elastic failure, significant weight savings can apparently be effected through such configurations.

Figure 8.6. Example 8.3 Tangential stress distribution produced by a combination of shrink-fit and internal pressure in a thick-walled compound cylinder. Dashed lines nn represent stresses due to p_i alone.

It is interesting to note that additional jackets prove not as effective, in that regard, as the first one. Multilayered shrink fit cylinders, each of small wall thickness, are, however, considerably stronger than a single jacket of the same total thickness. These assemblies can, in fact, be designed so that prestressing owing to shrinking combines with stresses due to loading to produce a nearly uniform distribution of stress throughout.^* The closer this uniform stress is to the allowable stress for the given material, the more efficiently is the material utilized. A single cylinder cannot be uniformly stressed and consequently must be stressed considerably below its allowable value, contributing to inefficient use of material.

^*Refer, for example, to Ref. 8.2.

Example 8.4

Figure 8.7. shows a duplex hydraulic conduit composed of a thin steel cylindrical liner in a concrete pipe. For the steel and concrete the properties are E_s and E_c, v_c, respectively. The conduit is subjected to an internal pressure p_i. Determine the interface pressure p transmitted to the concrete shell.

Solution The radial displacement at the bore (r = a) of the concrete pipe is, from Eq. (8.14),

(a)

The steel sleeve experiences an internal pressure p_i and an external pressure p. Thus,

Figure 8.7. Example 8.4 Thick-walled concrete pipe with a steel cylindrical liner.

(b)

On the other hand, Hooke’s law together with Eq. (8.2) yields at r = a

(c)

Finally, evaluation of u from Eqs. (b) and (c) and substitution into Eq. (a) result in the interface pressure in the form:

(8.24)

Note that, for practical purposes, we can use E_s/E_c = 15, v_c = 0.2, and 2t/(2a − t) = t/a. Equation (8.24) then becomes

(8.25)

where R = b/a. From the foregoing, observe that, as the thickness t of the liner increases, the pressure p transmitted to the concrete decreases. However, for any given t/a ratio, the p increases as the diameter ratio (R) of the concrete increases.

8.6 Rotating Disks of Constant Thickness

The equation of equilibrium, Eq. (8.1), can be used to treat the case of a rotating disk, provided that the centrifugal “inertia force” is included as a body force. Again, stresses induced by rotation are distributed symmetrically about the axis of rotation and assumed independent of disk thickness. Thus, application of Eq. (8.1), with the body force per unit volume F_r equated to the centrifugal force ρω²r, yields

(8.26)

where ρ is the mass density and ω the constant angular speed of the disk in radians per second. Note that the gravitational body force ρg has been neglected. Substituting Eq. (8.6) into Eq. (8.26), we have

(a)

requiring a homogeneous and particular solution. The former is given by Eq. (a) of Sec. 8.2.

It is easily demonstrated that the particular solution is

The complete solution is therefore

(8.27a)

which, upon substitution into Eq. (8.6), provides the following expressions for radial and tangential stress:

(8.27b)

(8.27c)

The constants of integration may now be evaluated on the basis of the boundary conditions.

Annular Disk

In the case of an annular disk with zero pressure at the inner (r = a) and outer (r = b) boundaries (Fig. 8.8) the distribution of stress is due entirely to rotational effects. The boundary conditions are

(b)

Figure 8.8. Annular rotating disk of constant thickness.

These conditions, combined with Eq. (8.27b), yield two equations in the two unknown constants,

(c)

from which

(d)

The stresses and displacement are therefore

(8.28a-c)

Applying the condition dσ_r/dr = 0 to the first of these equations, it is readily verified that the maximum radial stress occurs at . Figure 8.9a is a dimensionless representation of stress and displacement as a function of radius for an annular disk described by b/a = 4.

Solid Disk

In this case, a = 0, and the boundary conditions are

(e)

To satisfy the condition on the displacement, it is clear from Eq. (8.27a) that c₂ must be zero. The remaining constant is now evaluated from the first expression of Eq. (d):

Combining these constants with Eqs. (8.27), the following results are obtained:

(8.29a-c)

The stress and displacement of a solid rotating disk are displayed in a dimensionless representation (Fig. 8.9b) as functions of radial location.

Figure 8.9. Distribution of stress and displacement in a rotating disk of constant thickness: (a) annular disk; (b) solid disk.

The constant thickness disks discussed in this section are generally employed when stresses or speeds are low and, as is clearly shown in Fig. 8.9b, do not make optimum use of material. Other types of rotating disks, offering many advantages over flat disks, are discussed in the sections to follow.

Clearly, the sharp rise of the tangential stress near r = a is observed in Fig. 8.9a. It can be shown that, by setting r = a in Eq. (8.28b) and letting a approach zero, the resulting σ_θ is twice that given by Eq. (8.29b). The foregoing conclusion applies the radial stress as well. Thus, a small hole doubles the stress over the case of no hole.

Example 8.5

A flat disk with t = 20 mm and a = 50 mm is shrunk onto a shaft, causing a contact pressure p = 25 MPa when not rotating (Fig. 8.10). The coefficient of static friction is μ = 0.15. Determine the torque T carried by the fit and power P transmitted to the disk when the assembly is rotating at n = 2400 rpm.

Figure 8.10. Example 8.5.

Solution The force (axial or tangential) required for the assembly equals F = 2πaοpt. The torque capacity of the fit is therefore

(f)

Substituting the given numerical values,

T = 2π(50²)(0.15)(25)(0.02) = 1.178 kN·m

Power transmitted is expressed by p = Tω. Here the angular velocity of the disk ω = 2400(2π)/60 = 251.3 rad/s. We thus have P = (1.178) (251.3) = 296 kW.

Example 8.6

A flat 0.5-m outer diameter, 0.1-m inner diameter, and 0.075-m-thick steel disk is shrunk onto a steel shaft (Fig. 8.11). If the assembly is to run at speeds up to 6900 rpm, determine (a) the shrinking allowance, (b) the maximum stress when not rotating, and (c) the maximum stress when rotating. The material properties are ρ = 7.8 kN.s²/m⁴, E = 200 GPa, and v = 0.3.

Solution

a. The radial displacements of the disk (u_d) and shaft (u_s) are, from Eqs. (8.28) and (8.29),

We observe that u_s may be neglected as it is less than 1% of u_d of the disk at the common radius. The exact allowance is

b. Applying Eq. (8.23), we have

Therefore, from Eq. (8.18),

c. From Eq. (8.28), for r = 0.05,

Figure 8.11. Example 8.6. Tangential stresses produced by the combination of a shrink-fit and rotation in a disk of constant thickness.

A plot of the variation of stress in the rotating disk is shown in Fig. 8.11.

8.7. Rotating Disks of Variable Thickness

In Sec. 8.6, the maximum stress in a flat rotating disk was observed to occur at the innermost fibers. This explains the general shape of many disks: thick near the hub, tapering down in thickness toward the periphery, as in a steam turbine. This not only has the effect of reducing weight, but also results in lower rotational inertia.

The approach employed in the analysis of flat disks can be extended to variable thickness disks. Let the profile of a radial section be represented by the general hyperbola (Fig. 8.12),

(8.30)

Figure 8.12. Disk of hyperbolic profile.

where t₁ represents a constant and s a positive number. The shape of the curve depends on the value selected for s; for example, for s = 1, the profile is that of an equilateral hyperbola. The constant t₁ is simply the thickness at radius equal to unity. If the thickness at r = a is t_i and that at r = b is t_o, as shown in the figure, the hyperbolic curve is fitted by forming the ratio

and solving for s. Clearly, Eq. (8.30) does not apply to solid disks, as all values of s except zero yield infinite thickness at r = 0.

In a turbine application, the actual configuration may have a thickened outer rim to which blades are affixed and a hub for attachment to a shaft. The hyperbolic relationship cannot describe such a situation exactly, but sometimes serves as an adequate approximation. If greater accuracy is required, the hub and outer ring may be approximated as flat disks, with the elements of the assembly related by the appropriate boundary conditions.

The differential equation of equilibrium (Eq. 8.26) must now include t(r) and takes the form

(8.31)

Equation (8.31) is satisfied by a stress function of the form

(a)

Then the compatibility equation (8.4), using Eqs. (a) and Hooke’s law, becomes

(b)

Introducing Eq. (8.30), we have

(c)

This is an equidimensional equation, which the transformation r = e^α reduces to a linear differential equation with constant coefficients:

(d)

The auxiliary equation corresponding to Eq. (d) is given by

m² + sm − (1 + vs) = 0

and has the roots

(8.32)

The general solution of Eq. (d) is then

(e)

The stress components for a disk of variable thickness are therefore, from Eqs. (a),

(8.33)

Note that, for a flat disk, t = constant; consequently, s = 0 in Eq. (8.30) and m = 1 in Eq. (8.32). Thus, Eqs. (8.33) reduce to Eqs. (8.29), as expected. The constants c₁ and c₂ are determined from the boundary conditions

(f)

The evaluation of the constants is illustrated in the next example.

Example 8.7

The cross section of the disk in the assembly given in Example 8.6 is hyperbolic with t_i = 0.075 m and t_o = 0.015 m; a = 0.05 m, b = 0.25 m, and δ = 0.05 mm. The rotational speed is 6900 rpm. Determine (a) the maximum stress owing to rotation and (b) the maximum radial displacement at the bore of the disk.

Solution

a. The value of the positive number s is obtained by the use of Eq. (8.30):

Substituting t_i/t_o = 5 and b/a = 5, we obtain s = 1. The profile will thus be given by t = t_i/r. From Eq. (8.32), we have

, m₁ = 0.745, m₂ = –1.745

Hence the radial stresses, using Eqs. (8.33) and (f) for r = 0.05 and 0.25, are

from which

The stress components in the disk, substituting these values into Eqs. (8.33), are therefore

(g)

The maximum stress occurs at the bore of the disk and from Eqs. (g) is equal to (σ_θ)_{r = 0.05} = 0.0294 ρ ω². Note that it was 0.052 ρ ω² in Example 8.6. For the same speed, we conclude that the maximum stress is reduced considerably by tapering the disk.

b. The radial displacement is obtained from the second equation of (8.5), which together with Eq. (g) gives u_r=0.05 = (rσ_θ/E)_r=0.05 = 0.00147ρ ω²/E. Again, this is advantageous relative to the value of 0.0026 ρ ω²/E found in Example 8.6.

8.8 Rotating Disks of Uniform Stress

If every element of a rotating disk is stressed to a prescribed allowable value, presumed constant throughout, the disk material will clearly be used in the most efficient manner. For a given material, such a design is of minimum mass, offering the distinct advantage of reduced inertia loading, as well as lower weight. What is sought, then, is a thickness variation t(r) such that σ_r = σ_θ = σ = constant everywhere in the body. Under such a condition of stress, the strains, according to Hooke’s law, are ε_r = ε_θ = ε = constant, and the compatibility equation (8.4) is satisfied.

The equation of equilibrium (8.31) may, under the conditions outlined, be written

or

(8.34)

which is easily integrated to yield

(8.35)

This variation assures that σ_θ = σ_r = σ = constant throughout the disk. To obtain the value of the constant in Eq. (8.35), the boundary condition t = t₁ at r = 0 is applied, resulting in c₁ = t₁ (Fig. 8.13).

Figure 8.13. Profile of disk of uniform strength.

Example 8.8

A steel disk of the same outer radius, b = 0.25 m, and rotational speed, 6900 rpm, as the disk of Example 8.6 is to be designed for uniform stress. The thicknesses are t₁ = 0.075 m at the center and t₂ = 0.015 m at the periphery. Determine the stress and disk profile.

Solution From Eq. (8.35),

or

Thus,

Recall that the maximum stress in the hyperbolic disk of Example 8.7 was 0.0294 ρω². The uniformly stressed disk is thus about 34% stronger than a hyperbolic disk with a small hole at its center.

In actual practice, fabrication and design constraints make it impractical to produce a section of exactly constant stress in a solid disk. On the other hand, in an annular disk if the boundary condition is applied that the radial stress is zero at the inner radius, constancy of stress dictates that σ_θ and σ_r be zero everywhere. This is clearly not a useful result for the situation as described. For these reasons, the hyperbolic variation in thickness is often used.

8.9 Thermal Stresses in Thin Disks

In this section, our concern is with the stresses associated with a radial temperature field T(r) that is independent of the axial dimension. The practical applications are numerous and include annular fins and turbine disks. Because the temperature field is symmetrical with respect to a central axis, it is valid to assume that the stresses and displacements are distributed in the same way as those of Sec. 8.1, and therefore the equations of that section apply here as well.

In this case of plane stress, the applicable equations of stress and strain are obtained from Eq. (3.31) with reference to Eq.(3.23):

(8.36)

The equation of equilibrium, Eq. (8.2), is now

(a)

Introduction of Eq. (8.3) into expression (a) yields the following differential equation in radial displacement:

(b)

This is rewritten

(c)

to render it easily integrable. The solution is

(8.37)

where a, the inner radius of an annular disk, is taken as zero for a solid disk, and v and α have been treated as constants.

Annular Disk

The radial and tangential stresses in the annular disk of inner radius a and an outer radius b may be found by substituting Eq. (8.37) into Eq. (8.3) and the results into Eq. (8.36):

(d)

(e)

The constants c₁ and c_s are determined on the basis of the boundary conditions (σ_r)_r=a = (σ_r)_r=b = 0. Equation (d) thus gives

The stresses are therefore

(8.38)

Solid Disk

In the case of a solid disk of radius b, the displacement must vanish at r = 0 in order to preserve the continuity of material. The value of c₂ in Eq. (8.37) must therefore be zero. To evaluate c₁, the boundary condition (σ_r)_r=b = 0 is employed, and Eq. (d) now gives

Substituting c₁ and c₂ into Eqs. (d) and (e), the stresses in a solid disk are found to be

(8.39)

Given a temperature distribution T(r), the stresses in a solid or annular disk can thus be determined from Eqs. (8.38) or (8.39). Note that T(r) need not be limited to those functions that can be analytically integrated. A numerical integration can easily be carried out for σ_r, σ_θ to provide results of acceptable accuracy.

8.10 Thermal Stress in Long Circular Cylinders

Consider a long cylinder with ends assumed restrained so that ω = 0. This is another example of plane strain, for which ε_z = 0. The stress–strain relations are, from Hooke’s law,

(8.40)

For ε_z = 0, the final expression yields

(a)

Substitution of Eq. (a) into the first two of Eqs. (8.40) leads to the following forms in which z stress does not appear:

(8.41)

Inasmuch as Eqs. (8.2) and (8.3) are valid for the case under discussion, the solutions for u, σ_r, and σ_θ proceed as in Sec. 8.9, resulting in

(b)

(c)

(d)

Finally, from Eq. (a), we obtain

(e)

Solid Cylinder

For the radial displacement of a solid cylinder of radius b to vanish at r = 0, the constant c₂ in Eq. (b) must clearly be zero. Applying the boundary condition (σ_r)_r=b = 0, Eq. (c) may be solved for the remaining constant of integration,

(f)

and the stress distributions determined from Eqs. (c), (d), and (e):

(8.42)

(8.43a)

To derive an expression for the radial displacement, c₂ = 0 and Eq. (f) are introduced into Eq. (b).

The longitudinal stress given by Eq. (8.43a) is valid only for the case of a fixed-ended cylinder. In the event the ends are free, a uniform axial stress σ_z = s₀ may be superimposed to cause the force resultant at each end to vanish:

This expression together with Eq. (e) yields

(g)

The longitudinal stress for a free-ended cylinder is now obtained by adding s₀ to the stress given by Eq. (e):

(8.43b)

Stress components σ_r and σ_θ remain as before. The axial displacement is obtained by adding u₀ = –νs₀r/E, a displacement due to uniform axial stress, s₀, to the right side of Eq. (b).

Cylinder with Central Circular Hole

When the inner (r = a) and outer (r = b) surfaces of a hollow cylinder are free of applied load, the boundary conditions (σ_r)_r=a = (σ_r)_r=b = 0 apply. Introducing these into Eq. (c), the constants of integration are

(h)

Equations (c), (d), and (e) thus provide

(8.44)

(8.45a)

If the ends are free, proceeding as in the case of a solid cylinder, the longitudinal stress is described by

(8.45b)

Implementation of the foregoing analyses depends on a knowledge of the radial distribution of temperature T(r).

Example 8.9

Determine the stress distribution in a hollow free-ended cylinder, subject to constant temperatures T_a and T_b at the inner and outer surfaces, respectively.

Solution The radial steady-state heat flow through an arbitrary internal cylindrical surface is given by Fourier’s law of conduction:

(i)

Here Q is the heat flow per unit axial length and K the thermal conductivity. Assuming Q and K to be constant,

This is easily integrated upon separation:

(j)

Here c₁ and c₂, the constants of integration, are determined by applying the temperature boundary conditions (T)_r=a = T_a and (T)_r=b = T_b. By so doing, Eq. (j) may be written in the form

(8.46)

When this is substituted into Eqs. (8.44) and (8.45a), the following stresses are obtained:

(8.47)

We note that, for v = 0, Eqs. (8.47) provide a solution for a thin hollow disk. In the event the heat flow is outward, that is, T_a > T_b, examination of Eqs. (8.47) indicates that the stresses σ_θ and σ_z are compressive (negative) on the inner surface and tensile (positive) on the outer surface. They exhibit their maximum values at the inner and outer surfaces. On the other hand, the radial stress is compressive at all points and becomes zero at the inner and outer surfaces of the cylinder.

In practice, a pressure loading is often superimposed on the thermal stresses, as in the case of chemical reaction vessels. An approach is to follow similarly to that already discussed, with boundary conditions modified to reflect the pressure; for example, (σ_r)_r=a = –p_i, (σ_r)_r=b = 0. In this instance, the internal pressure results in a circumferential tensile stress (Fig. 8.3a), causing a partial cancellation of compressive stress owing to temperature.

Note that, when a rotating disk is subjected to inertia, loading combined with an axisymmetrical distribution of temperature T(r), the stresses and displacements may be determined by superposition of the two cases.

8.11 Finite Element Solution

In the previous sections, the cases of axisymmetry discussed were ones in which along the axis of revolution (z) there was uniformity of structural geometry and loading. In this section, the finite element approach of Chapter 7 is applied for computation of displacement, strain, and stress in a general axisymmetric structural system, formed as a solid of revolution having material properties, support conditions, and loading, all of which are symmetrical about the z axis, but that may vary along this axis. A simple illustration of this situation is a sphere uniformly loaded by gravity forces.

"Elements" of the body of revolution (rings or, more generally, tori) are used to discretize the axisymmetric structure. We shall here employ an element of triangular cross section, as shown in Fig. 8.14. Note that a node is now in fact a circle; for example, node i is the circle with r_i as radius. Thus, the elemental volume dV appearing in the expressions of Sec. 7.8 is the volume of the ring element (2πr dr dz).

Figure 8.14. Triangular cross section of axisymmetric solid finite element.

The element clearly lies in three-dimensional space. Any randomly selected vertical cross section of the element, however, is a plane triangle. As already discussed in Sec. 8.1, no tangential displacement can exist in the symmetrical system; that is, ν = 0. Inasmuch as only the radial displacement u and the axial displacement w in a plane are involved (rz plane), the expressions for displacement established for plane stress and plane strain may readily be extended to the axisymmetric analysis [Ref. 8.3].

8.12 Formulation of Axisymmetric Element

The theoretical development will follow essentially the procedure given in Chapter 7, with the exception that, in the present case, cylindrical coordinates will be employed (r, θ, z), as shown in Fig. 8.14.

Strain, Stress, and Elasticity Matrices

The strain matrix, from Eqs. (2.3), (3.30a), and (3.30b), may be defined as follows:

(8.48)

The initial strain owing to a temperature change is expressed in the form

{ε₀}_e = {αT, αT, αT, 0}

It is observed from Eq. (8.48) that the tangential strain ε_θ becomes infinite for a zero value of r. Thus, if the structural geometry is continuous at the z axis, as in the case of a solid sphere, r is generally assigned a small value (for example, 0.1 mm) for the node located at this axis.

It can be demonstrated that the state of stress throughout the element {σ}_e is expressible as follows:

(8.49)

A comparison of Eqs. (8.49) and (7.35) yields the elasticity matrix

(8.50)

Displacement Function

The nodal displacements of the element are written in terms of submatrices δ_u and δ_ν:

(a)

The displacement function {f}_e, which describes the behavior of the element, is given by

(8.51a)

or

(8.51b)

Here the α’s are the constants, which can be evaluated as follows. First, we express the nodal displacements {δ}_e:

u_i = α₁ + α₂r_i + α₃z_i

u_j = α₁ + α₂r_j + α₃z_j

u_m = α₁ + α₂r_m + α₃z_m

w_i = α₄ + α₅r_i + α₆z_i

w_j = α₄ + α₅r_j + α₆z_j

w_m = α₄ + α₅r_m + α₆z_m

Then, by the inversion of these linear equations,

(8.52)

where A is defined by Eq. (7.45) and

(8.53)

Finally, upon substitution of Eqs. (8.52) into (8.51), the displacement function is represented in the following convenient form:

(8.54)

or

{f}_e = [N]{δ}_e

with

(8.55)

The element strain matrix is found by introducing Eq. (8.54) together with (8.55) into (8.48):

(8.56)

where

(8.57)

with

It is observed that the matrix [B] includes the coordinates r and z. Thus, the strains are not constant, as is the case with plane stress and plane strain.

The Stiffness Matrix

The element stiffness matrix, from Eq. (7.38), is given by

(8.58a)

and must be integrated along the circumferential or ring boundary. This may thus be rewritten

(8.58b)

where the matrices [D] and [B] are defined by Eqs. (8.50) and (8.57), respectively. It is observed that integration is not easily performed as in the case of plane stress problems, because [B] is also a function of r and z. Although tedious, the integration can be carried out explicitly. Alternatively, approximate numerical approaches may be used. In a simple approximate procedure, [B] is evaluated for a centroidal point of the element. To accomplish this, we substitute fixed centroidal coordinates of the element

(8.59)

into Eq. (8.57) in place of r and z to obtain . Then, by letting , from Eqs. (8.58), the element stiffness is found to be

(8.60)

This simple procedure leads to results of acceptable accuracy.

External Nodal Forces

In the axisymmetric case, "concentrated" or "nodal" forces are actually loads axisymmetrically located around the body. Let q_r and q_z represent the radial and axial components of force per unit length, respectively, of the circumferential boundary of a node or a radius r. The total nodal force in the radial direction is

(8.61a)

Similarly, the total nodal force in the axial direction is

(8.61b)

Other external load components can be treated analogously. When the approximation leading to Eq. (8.60) is used, we can, from Eq. (7.35), readily obtain expressions for nodal forces owing to the initial strains, body forces, and any surface tractions (see Prob. 8.39).

In summary, the solution of an axisymmetric problem can be obtained, having generated the total stiffness matrix [K], from Eq. (8.58) or (8.60), and the load matrices {Q}. Then Eq. (7.37) provides the numerical values of nodal displacements {δ} = [K]^-1{Q}. The expression (8.56) together with Eq. (8.57) yields values of the element strains. Finally, Eq. (8.49), upon substitution of Eq. (8.56), is used to determine the element stresses.

Problems

Secs. 8.1 through 8.5

8.1. A cylinder of internal radius a and external radius b = 1.10a is subjected to (a) internal pressure p_i only and (b) external pressure p_o only. Determine for each case the ratio of maximum to minimum tangential stress.

8.2. A thick-walled cylinder with closed ends is subjected to internal pressure p_i only. Knowing that a = 0.6 m, b = 1 m, σ_all = 140 MPa, and τ_all = 80 MPa, determine the allowable value of p_i.

8.3. A cylinder of inner radius a and outer radius na, where n is an integer, has been designed to resist a specific internal pressure, but reboring becomes necessary. (a) Find the new inner radius r_x required so that the maximum tangential stress does not exceed the previous value by more than Δσ_θ, while the internal pressure is the same as before. (b) If a = 25 mm and n = 2 and after reboring the tangential stress is increased by 10%, determine the new diameter.

8.4. A steel tank having an internal diameter of 1.2 m is subjected to an internal pressure of 7 MPa. The tensile and compressive elastic strengths of the material are 280 MPa. Assuming a factor of safety of 2, determine the wall thickness.

8.5. Two thick-walled, closed-ended cylinders of the same dimensions are subjected to internal and external pressure, respectively. The outer diameter of each is twice the inner diameter. What is the ratio of the pressures for the following cases? (a) The maximum tangential stress has the same absolute value in each cylinder. (b) The maximum tangential strain has the same absolute value in each cylinder. Take

8.6. Determine the radial displacement of a point on the inner surface of the tank described in Prob. 8.4. Assume that outer diameter 2b = 1.2616 m, E = 200 GPa, and ν = 0.3.

8.7. Derive expressions for the maximum circumferential strains of a thick-walled cylinder subject to internal pressure p_i only, for two cases: (a) an open-ended cylinder and (b) a closed-ended cylinder. Then, assuming that the allowable strain is limited to 1000 μ, p_i = 60 MPa, and b = 2 m, determine the required wall thickness. Use E = 200 GPa and

8.8. An aluminum cylinder (E = 72 GPa, v = 0.3) with closed and free ends has a 500-mm external diameter and a 100-mm internal diameter. Determine maximum tangential, radial, shearing, and longitudinal stresses and the change in the internal diameter at a section away from the ends for an internal pressure of p_i = 60 MPa.

8.9. Rework Prob. 8.8 with p_i = 0 and p_o = 60 MPa.

8.10. A steel cylinder of 0.3-m radius is shrunk over a solid steel shaft of 0.1-m radius. The shrinking allowance is 0.001 m/m. Determine the external pressure p_o on the outside of the cylinder required to reduce to zero the circumferential tension at the inside of the cylinder. Use E_s = 200 GPa.

8.11. A steel cylinder is subjected to an internal pressure only. (a) Obtain the ratio of the wall thickness to the inner diameter if the internal pressure is three-quarters of the maximum allowable tangential stress. (b) Determine the increase in inner diameter of such a cylinder, 0.15 m in internal diameter, for an internal pressure of 6.3 MPa. Take E = 210 GPa and

8.12. Verify the results shown in Fig. 8.4 using Eqs. (a) and (8.21) of Sec. 8.3.

8.13. A thick-walled cylinder is subjected to internal pressure p_i and external pressure p_o. Find (a) the longitudinal stress σ_z if the longitudinal strain is zero and (b) the longitudinal strain if σ_z is zero.

8.14. A cylinder, subjected to internal pressure only, is constructed of aluminum having a tensile strength σ_yp. The internal radius of the cylinder is a, and the outer radius is 2a. Based on the maximum energy of distortion and maximum shear stress theories of failure, predict the limiting values of internal pressure.

8.15. A cylinder, subjected to internal pressure p_i only, is made of cast iron having ultimate strengths in tension and compression of σ_u = 350 MPa and MPa, respectively. The inner and outer radii are a and 3a. Determine the allowable value of p_i using (a) the maximum principal stress theory and (b) the Coulomb-Mohr theory.

8.16. A flywheel of 0.5-m outer diameter and 0.1-m inner diameter is pressed onto a solid shaft. The maximum tangential stress induced in the flywheel is 35 MPa. The length of the flywheel parallel to the shaft axis is 0.05 m. Assuming a coefficient of static friction of 0.2 at the common surface, find the maximum torque that may be transmitted by the flywheel without slippage.

8.17. A solid steel shaft of 0.1-m diameter is pressed onto a steel cylinder, inducing a contact pressure p₁ and a maximum tangential stress 2p₁ in the cylinder. If an axial tensile load of P_L = 45 kN is applied to the shaft, what change in contact pressure occurs? Let

8.18. A brass cylinder of outer radius c and inner radius b is to be press-fitted over a steel cylinder of outer radius b + δ and inner radius a (Fig. 8.5). Demonstrate that the interface pressure is given by

Calculate the maximum stresses in both materials for δ = 0.02 mm, a = 40 mm, b = 80 mm, and c = 140 mm. Let E_b = 110 GPa, v_b = 0.32, E_s = 200 GPa, and v_s = 0.28.

8.19. An aluminum alloy cylinder (E_a = 72 GPa, v_a = 0.33) of outer and inner diameters of 300 and 200 mm is to be press-fitted over a solid steel shaft (E_s = 200 GPa, v_s = 0.29) of diameter 200.5 mm. Calculate (a) the interface pressure p and (b) the change in the outer diameter of the aluminum cylinder.

8.20. A thin circular disk of inner radius a and outer radius b is shrunk onto a rigid plug of radius a + δ₀ (Fig. P8.20). Determine (a) the interface pressure; (b) the radial and tangential stresses.

Figure P8.20.

8.21. When a steel sleeve of external diameter 3b is shrunk onto a solid steel shaft of diameter 2b, the internal diameter of the sleeve is increased by an amount δ₀. What reduction occurs in the diameter of the shaft? Let

8.22. A cylinder of inner diameter b is shrunk onto a solid shaft. Find (a) the initial difference in diameters if the contact pressure is p and the maximum tangential stress is 2p in the cylinder and (b) the axial compressive load that should be applied to the shaft to increase the contact pressure from p to p₁. Let

8.23. A brass solid cylinder is a firm fit within a steel tube of inner diameter 2b and outer diameter 4b at a temperature T₁°C. If now the temperature of both elements is increased to T₂°C, find the maximum tangential stresses in the cylinder and in the tube. Take α_s = 11.7 × 10^−6/°C, α_b = 19.5 × 10^−6/°C, and , and neglect longitudinal friction forces at the interface.

8.24. A gear of inner and outer radii 0.1 and 0.15 m, respectively, is shrunk onto a hollow shaft of inner radius 0.05 m. The maximum tangential stress induced in the gear wheel is 0.21 MPa. The length of the gear wheel parallel to shaft axis is 0.1 m. Assuming a coefficient of static friction of 0.2 at the common surface, what maximum torque may be transmitted by the gear without slip?

Secs. 8.6 through 8.12

8.25. Show that for an annular rotating disk the ratio of the maximum tangential stress to the maximum radial stress is given by

FigureP8.25.

8.26. Determine the allowable speed ω_all in rpm of a flat solid disk employing the maximum energy of distortion criterion. The disk is constructed of an aluminum alloy with σ_yp = 260 MPa, , ρ = 2.7 kN · s²/m⁴, and b = 125 mm.

8.27. A rotating flat disk has 60-mm inner diameter and 200-mm outer diameter. If the maximum shearing stress is not to exceed 90 MPa, calculate the allowable speed in rpm. Let v = 1/3 and ρ = 7.8 kN · s²/m⁴.

8.28. A flat disk of outer radius b = 125 mm and inner radius a = 25 mm is shrink-fitted onto a shaft of radius 25.05 mm. Both members are made of steel with E = 200 GPa, ν = 0.3, and ρ = 7.8 kN · s²/m⁴. Determine (a) the speed in rpm at which the interface pressure becomes zero and (b) the maximum stress at this speed.

8.29. A flat annular steel disk (ρ = 7.8 kN · s²/m⁴, ν = 0.3) of 4c outer diameter and c inner diameter rotates at 5000 rpm. If the maximum radial stress in the disk is not to exceed 50 MPa, determine (a) the radial wall thickness and (b) the corresponding maximum tangential stress.

8.30. A flat annular steel disk of 0.8-m outer diameter and 0.15-m inner diameter is to be shrunk around a solid steel shaft. The shrinking allowance is 1 part per 1000. For v = 0.3, E = 210 GPa, and ρ = 7.8 kN s²/m⁴, determine, neglecting shaft expansion, (a) the maximum stress in the system at standstill and (b) the rpm at which the shrink fit will loosen as a result of rotation.

8.31. Show that in a solid disk of diameter 2b, rotating with a tangential velocity ν, the maximum stress is . Take .

8.32. A steel disk of 500-mm outer diameter and 50-mm inner diameter is shrunk onto a solid shaft of 50.04-mm diameter. Calculate the speed (in rpm) at which the disk will become loose on the shaft. Take ν = 0.3, E = 210 GPa, and ρ = 7.8 kN · s²/m⁴.

8.33. Consider a steel rotating disk of hyperbolic cross section (Fig. 8.12) with a = 0.125 m, b = 0.625 m, t_i = 0.125 m, and t_o = 0.0625 m. Determine the maximum tangential force that can occur at the outer surface in newtons per meter of circumference if the maximum stress at the bore is not to exceed 140 MPa. Assume that outer and inner edges are free of pressure.

8.34. A steel turbine disk with b = 0.5 m, a = 0.0625 m, and t_o = 0.05 m rotates at 5000 rpm carrying blades weighing a total of 540 N. The center of gravity of each blade lies on a circle of 0.575-m radius. Assuming zero pressure at the bore, determine (a) the maximum stress for a disk of constant thickness and (b) the maximum stress for a disk of hyperbolic cross section. The thickness at the hub and tip are t_i = 0.4 m and t_o = 0.05 m, respectively. (c) For a thickness at the axis t_i = 0.02425 m, determine the thickness at the outer edge, t_o, for a disk under uniform stress, 84 MPa. Take ρ = 7.8 kN · s²/m⁴ and g = 9.81 m/s².

8.35. A solid thin flat disk is restrained against displacement at its outer edge and heated uniformly to temperature T. Determine the radial and tangential stresses in the disk.

8.36. Show that for a hollow disk or cylinder, when subjected to a temperature distribution given by T = (T_a – T_b) In (b/r)/ In (b/a), the maximum radial stress occurs at

8.37. Calculate the maximum thermal stress in a gray cast-iron cylinder for which the inner temperature is T_a = –8°C and the outer temperature is zero. Let a = 10 mm, b = 15 mm, E = 90 GPa, α = 10.4 × 10.4 × 10⁻⁶ per °C, and ν = 0.3.

8.38. Verify that the distribution of stress in a solid disk in which the temperature varies linearly with the radial dimension, T(r) = T₀(b-r)/b, is given by

FigureP8-38.

Here T₀ represents the temperature rise at r = 0.

8.39. Redo Example 7.7, with the element shown in Fig. 7.15 representing a segment adjacent to the boundary of a sphere subjected to external pressure p = 14 MPa.

8.40. A cylinder of hydraulic device having inner radius a and outer radius 4a is subjected to an internal pressure p_i. Using a finite element program with CST (or LST) elements, determine the distribution of the tangential and radial stresses. Compare the results with the exact solution shown in Fig. 8.3a.