46

§2.13 Reduction for Ring In the event that the shell is very short, then the solution can be simplified. When the loading is self-equilibrating, the behavior is that of a beam (without foundation). When a resultant force and/or moment per unit length of the circumference exists, the behavior is that of a ring. A derivation can be found in many texts; here it is shown that the shell equations do yield the correct result. The approximation is that the ring cross section remains undistorted and the only significant stress is the component in the circumferential direction. For simplicity, a short flat plate element of length L, with L << r, is considered. The constitutive relations (2.8.15-16) give

N θ = E t ε θ (2.13.1) while the equilibrium equation (2.4.10) gives N θ = r p H (2.13.2)

The quantities can be integrated over the element. When E and p H are constant, the

result is

f = LpH = ELt h

r2 = EA h

r2 (2.13.3)

where f is the total radial force per unit length in the circumferential direction, and A is the cross-sectional area in the meridional plane of the element. A moment loading of the flat-plate element is provided by Q. From (2.8.17-18) is obtained

M θ =

E t3

12 κ θ (2.13.4) while (2.4.12) gives M θ = – r Q (2.13.5) Together these give

m = – L Q = E t3L

12 r2χ = E I

r2χ

(2.13.6) in which m is the externally applied moment per unit length of the circumference. For the general cross section of the ring, I is the area moment of inertia about the axis passing through the centroid and parallel to the r - axis. The results may be placed into the form of a stiffness matrix:

47

r m

r f =

E Ir 0

0 E Ar

• χh

(2.13.7)

48

Chapter 3

Cylindrical Shell

The cylinder is a very fundamental shape which is extensively utilized for pipes, pressure vessels, liquid storage tanks, and as the basic structural element in tubular frame structures, including bicycles, off-shore drilling platforms, and the long bones of the skeleton in vertebrates. §3.1 Equations and Analogy with Beam on an Elastic Foundation The cylinder is relatively easy to fabricate and analyze, since the geometry is rather simple:

ϕ = π / 2r = constant (3.1.1-2)

The Novozhilov equation (2.11.11) for the homogeneous, isotropic shell with constant thickness reduces to an equation with constant coefficients:

– i c r d2H

d s2 + H = i c d

d s r2 pH – ν r V

(3.1.3) while the coefficient matrix of the equation (2.11.22) reduces to:

A =

0 1 0 0

0 0 0 Ε t r

1r Ε t c2

0 0 0

0 0 – 1 0

(3.1.4) while the load vector is:

P =

0 r pH - ν V

00

(3.1.5) These equations have exactly the same form as the equations for a beam on an elastic foundation (Winkler foundation ) with the following replacements:

49

Cylindrical Beam on Elastic Shell Foundation E t c2 E I Beam bending stiffness per

unit width E t / r2 k Foundation stiffness modulus pH– ν V / r pH Pressure load (3.1.6-8)

The meridional bending stiffness is the same in beam and shell, except for the dependence on Poisson's ratio for the shell. The load term differs only by the Poisson contraction due to the axial resultant in the shell. The significant feature is the correspondence of the elastic foundation stiffness and the term in the shell equation which can be identified as the circumferential ring stiffness. For the beam on a foundation, much of the load is carried by the foundation; for the shell, much of the load is carried by the important circumferential stress. The resulting behavior can be quite unlike that of a beam supported only at a few points. By straight-forward elimination, the matrix equation can be reduced to the scalar equation with the normal displacement as the dependent variable, which is the customary form for the beam equation:

d2d s2

E I d2h

d s2 + k h = pH

(3.1.9) or, for the cylindrical shell,

d2d s2

E t c2 d2h

d s2 + E t

r2 h = pH – ν V / r

(3.1.10) The solution of these equations can be obtained by standard techniques. First, the equations are linear and nonhomogeneous, so the principle of superposition can be used, by which the total solution can be divided into a particular solution and a complementary solution. The particular solution is any solution of the entire equation ( i.e., with the right-hand-side load term). The complementary solution is a solution of the homogeneous equation ( i.e., without the load term), with a sufficient number of independent, arbitrary constants with which the boundary conditions can be satisfied. Closed-form solutions for some special cases are treated in the next sections. §3.2 Particular Solution When the properties and the right-hand-side terms are constant, the exact particular solution is just:

50

h = r2

E t p H– ν V / r (3.2.1)

This includes the two important special cases of a pure axial load and a pressure load. §3.2.1 Axial Force For only an axial force P acting on the shell, the results for all the quantities are:

V = N s = P / 2 π rN θ = 0

h = – ν P

2 π E t

v = P2 π r E t

s – s0 (3.2.2-5)

and the remaining quantities are all zero:

χ , Ms , Mθ (3.2.6) This is simple axial stress in the shell. The only difference between this and the elementary solution for a rod in tension or compression is the shell equations yield the radial displacement due to the Poisson's ratio (3.2.4). §3.2.2 Pressure The next important case is uniform pressure loading, described by: p H= p = constant p s = 0 (3.2.7) There are situations in which no axial force is present. However, the more typical case is indicated in Fig. 3.1, in which the ends of the vessel are closed. For the present, consider the ends to consist of rigid plates. The internal pressure will act on the end plates with a net force of F = π r 2p (3.2.8) and the solution (3.2.1) gives the results:

51

V = N s = p r / 2N θ = p r

h = p r2

E t 1 – ν

2

v = p rE t12 – ν s – s 0 (3.2.9-12)

The rotation and bending moments (3.2.6) are all zero in this case as well. For both cases of axial force and pressure, the quantities of interest are constant, except for the axial

displacement which increases linearly with the axial coordinate s. The value s 0 is an

arbitrary constant which determines the rigid body axial displacement.

pH

Axis Rigid end plate

Deformed cylinder wall - withoutconsideration of end constraint

Figure 3.1– Cylindrical pressure vessel with rigid end plates. The simple particular solution gives a uniform expansion of the vessel.

Note that the axial force is carried by purely axial stress, just as for a beam. Unlike a flat beam or plate with discrete supports, the pressure load is also carried by tangential stress, which is possible due to the curvature. For the beam on the elastic foundation (3.1.9), the uniform transverse load is carried by the foundation; the bending is zero. The curvature thus may be thought of as providing the effect of a continuous foundation. The solutions (3.2.2-5) and ( 3.2.9-12) are exact solutions of the equation and therefore satisfy the definition of particular solutions. However, the only arbitrary constant is the rigid body displacement, so it is not possible to satisfy end conditions withe these solutions alone. For example, the rigid plate shown in Fig. 3.1 requires that the rotation and radial displacement be zero at both ends. It is clear that the solutions

52

(3.2.2-5) and (3.2.9-12) do not satisfy these conditions. The solution must be "completed" with the complementary solution. §3.2.3 Comparison with exact solutions A comparison of the approximate shell solution with an exact three-dimensional solution provides a verification of the procedure. Rigorous error bounds tend to be either difficult to obtain and/or too conservative to provide quantitative guidance. The uniform axial stress in §3.2.1 is an exact solution, but rather trivial. More interesting is the solution for pressure loading of a thick cylinder, which is a well-known problem of plane strain. The thin shell solution from §3.2.2 used in the stress calculation (2.9.2) gives the stresses in the axial and longitudinal directions of

σ s = r p H / 2 t σ θ = r p H / t (3.2.13) We have not developed a systematic "thick shell" theory, such as that due to Reissner and Naghdi and in the recent work of Doxee (1987). Improvement can, however be gained by using (3.2.10) in the expressions (2.8.15-16) for the strain measures, then using (2.6.12-13) for the strain, and then (2.8.8-9) for the stress. The load is calculated from (2.4.15). The result for the circumferential stress due to internal pressure is

σ θ ΙD =r p ζ IDt σ θ ΟD =

r p ζ IDt

1 – t2 r

1 + t2 r (3.2.14)

while the result for external pressure is

σ θ ΙD =r p ζ ODt

1 + t2 r

1 – t2 r

σ θ ΟD =r p ζ OD

t (3.2.15)

The plane strain elasticity solution for pressure acting on a thick cylinder is

σ ζ = p

1 – b 2

r + ζ2

1 – b2

a2

(3.2.16)

53

σ θ = p

1 + b 2

r + ζ2

1 – b2

a2

(3.2.17) in which p is the outward pressure acting on the edge r + ζ = a, and the edge r + ζ = b is free. A comparison of the shell and exact elasticity solutions is shown in Figs. 3.2 and 3.3. The circumferential stress is normalized to the thin shell result (3.2.13). The behavior is what would be expected. The error in the thin shell solution is about 10 percent when r / t is 10. It is somewhat surprising that the "thick shell" correction does provide such a substantial increase in accuracy, and is seriously in error only for r / t < 2.

0 2 4 6 8 10 120.2

0.4

0.6

0.8

1.0

1.2

1.4

r/t

Stre

ss /

(pr/

t)

On ID exact

On OD exact

Thick shell

Figure 3.2 - Comparison of exact elasticity solution and the thick shell correction for the long cylindrical shell with internal pressure. The circumferential stress on the inside diameter (ID) and on the outside diameter (OD) is shown divided by the standard thin shell stress factor.

54

0 2 4 6 8 10 120

1

2

3

4

r/t

Stre

ss /

(pr

/t)

On OD exact

On ID exact

Thick shell

Thick shell

Figure 3.3 - Comparison of exact elasticity solution and the thick shell correction for the long cylindrical shell with external pressure. The circumferential stress on the inside diameter (ID) and on the outside diameter (OD) is shown divided by the standard thin shell stress factor.

55

§3.3 Complementary Solutions With zero right-hand sides and constant coefficients, the solutions of either the complex equation of second order (3.1.3), or the real equation of fourth order (3.1.9) or (3.1.10) is obtained by the standard technique of assuming a solution in exponential form. Equivalently, the matrix equation can be treated. The homogeneous equation is

– dYd s + A•Y = 0 (3.3.1)

For the cylinder with constant properties the coefficient matrix (3.1.4) is constant. A solution may be assumed in the form of a scalar exponential multiplied by a constant vector:

Y = ψ eµ s (3.3.2)

Substitution into (3.3.1) gives the relation which must be satisfied:

A – µ I • ψ = 0 (3.3.3) where I is the identity matrix. Thus µ must be an eigenvalue of the coefficient matrix and ψ must be the corresponding eigenvector. In order for a nontrivial (i.e., nonzero) solution to exist, the determinant of the complete matrix in (3.3.3) must be zero: A – µ I = 0 (3.3.4) For an n th order system, this yields a polynomial of n th order for µ. If all n roots are distinct, then each root provides a linearly independent solution of the form (3.3.2). The general solution is

Y = C 1 ψ µ 1 eµ 1 s + ... + C n ψ µ n eµ n s (3.3.5)

For the cylindrical shell, the system (3.3.1) is of the fourth order, giving a polynomial of fourth order:

E t c2 µ4+ E t / r2 = 0 (3.3.6)

It is easy to see that this is the polynomial which is also obtained from the scalar equation (3.1.10). The roots are complex-valued, as indicated in Fig. 3.4. The root in the first quadrant will be considered as the first root and written without a subscript:

56

µ1 = µ = 1

c r12ei π /4 = 1 + i

2c r12 (3.3.7)

and the corresponding eigenvector is

ψ( µ ) =

− µ2r Ε t c 2

− µ3r Ε t c 2

– µ

1 (3.3.8) The other three roots, as shown in Fig. 3.4, are the negatives and complex conjugates of the first root:

µ2 = µ µ3 = – µ µ4 = – µ (3.3.9)

µ

µ

µ

µ

Real

Imaginary

Figure 3.4 - Locus of roots µ in the complex plane. Thus the solution can be written as

Y = C 1 ψ µ eµ s + C 2 ψ µ eµ s

+ C 3 ψ – µ e– µ s + C 4 ψ – µ e– µ s (3.3.10)

57

However, Y must be real, so the complex constants must be related: C 2 = C 1 C 4 = C 3 (3.3.11) and the solution can be written as

Y = 2 Re C 1 ψ µ eµ s + C 3 ψ – µ e– µ s

(3.3.12) in which Re denotes the real part. Since the two constants have arbitrary real and imaginary parts, four real, independent constants are present in the solution. Thus (3.3.12) provides the complete set of complementary solutions required to satisfy the end conditions on the cylinder. The solution (3.3.12) consists of exponentially increasing and exponentially decreasing terms. For a semi-infinite shell one or the other coefficient must be set equal to zero for the solution to be bounded. For a shell of finite length L, with 0 ≤ s ≤ L , it is convenient to adjust the constants by introducing the new constants A and B: 2 C1 = A e µ L 2 C3 = B (3.3.13) The equation is then Y = Re A ψ µ eµ (s – L ) + B ψ – µ e– µ s (3.3.14) In the solution multiplied by B , the exponential term is unity at the edge s = 0, and is exponentially decreasing with the distance from that edge. Likewise, in the solution multiplied by A , the exponential term is unity at the edge s = L, and is exponentially decreasing with the distance from that edge. The envelope of the exponential term is

e– µ s = exp – s

2 c r12 (3.3.15)

Hence in the distance from either edge given by

s = δ = π 2 c r12 (3.3.16)

= 2.5 r t12 (For ν = 0.3) (3.3.17)

the solution decreases to about four percent of its value at the edge. We call δ the decay distance. If the length of the cylinder is greater than δ, the solution is substantially

58

simplified, since the solution from one edge is negligible at the other edge. This means that the problem of satisfying the boundary conditions uncouples into two equivalent semi-infinite problems. The conditions at the edge s = 0 are satisfied by the constant B without regard for the constant A, and, likewise, the conditions at the edge s = L are satisfied by the constant A without regard for B. If δ < L, the shell is referred to as long, otherwise short. It is important to realize that the case of the shell being long is the more common occurrence. From (3.3.17) the ratio of decay distance to radius can be computed:

δr =

1.12 for r / t = 50.35 500.11 5000.035 5000 (3.3.18)

Thus a shell with length greater than the radius will be long; for a very thin shell, a length rather small in comparison with the radius is sufficient for the shell to be long. An awareness of this decay distance is very important in numerical calculations. To capture correctly the behavior of the solution in the edge region a finite element or finite difference mesh must be small in comparison with the decay distance δ. Typically, the mesh should be:

FEM mesh spacing < 14 ( r t )1/2 (3.3.19)

A prevalent mistake in the use of finite element codes is that the mesh for a thin shell does not satisfy this condition, with resulting errors in the results which can be large. So far as we are aware, none of the commercially available, and widely used, finite element codes provide any warning to the user about this condition. For the static problems under present consideration, a fine mesh satisfying (3.3.19) is necessary only in the edge region of width δ. Away from the edge region in the shell interior, a coarse mesh can be used when the surface loads are reasonably smooth. For problems of dynamics and shell stability, however, significant activity may occur everywhere in the shell which requires the mesh (3.3.19). This requires for the thin shell a large number of elements, in comparison with seemingly similar beam or flat plate problems. §3.3.1 Edge Stiffness Matrix The complementary solution (3.3.14) provides the self-equilibrating solutions necessary to satisfy the edge boundary conditions. It is often convenient first to compute the relation between the force and displacement quantities on the edges. For linear problems the relation can be placed in a matrix form. In Fig. 3.5 a cylindrical shell is shown.

59

hH

M

s

s

M s

H, h

t

L

χ

, χ

Figure 3.5 - Self-equilibrating edge loads on a cylindrical shell.

At the edge in the positive s - direction, the radial force is in the same direction as the radial displacement and the moment is in the same direction as the rotation. At the "back" edge, the force quantities change direction.

§3.3.1.2 Long Shell When the shell is long L ≥ δ , the effects at each edge are uncoupled. In the vicinity of the lower edge s = L, the solution from the upper edge is negligible, and the solution (3.3.14) reduces to

Y =

r M s

r H

χ

h

= Re A

– µ2r Ε t c 2

– µ3r Ε t c 2

– µ

1

eµ (s – L )

(3.3.20) in which the arrays have been written out. At the edge, the exponential term is unity. The relation between the force and displacement quantities is provided by the edge stiffness matrix:

60

r M sr H L

= KL•χh L (3.3.21)

in which the subscript L denotes values at the edge s = L. From the last two rows of (3.3.20) and the value of µ from (3.3.7), the edge displacement and rotation can be obtained in terms of the constant:

h L = Re A

χ L =– Re A + Im A

2 c r12 (3.3.22-23)

Solving for the constant gives

A = h + i h + χ 2 c r

12

L (3.3.24) The first two rows of (3.3.20) give the moment and radial force:

r M sL = E t c Im A

r Η L = E t c Re A + Im A

2 c r12 (3.3.25-26)

Substituting in the values of A from (3.3.22) gives the coefficients of the stiffness matrix:

r M sr H L

= E t c

2 c r12 1

1 2

2 c r12

•χh L

(3.3.27) In the vicinity of the top edge of the shell s = 0 in Fig 3.5, the term in (3.3.14) multiplied by the constant B is significant. The algebra is the same with the result for the upper edge stiffness given by

61

r M sr H 0

= E t c

– 2 c r12 1

1 – 2

2 c r12

•χh 0

(3.3.28) The diagonal terms are negative because the force and displacement quantities at the upper edge are in the opposite direction. Frequently it is convenient to use the inverse flexibility matrices, at the lower edge s = L :

χh L

= 1E t c

2

2 c r12

– 1

– 1 2 c r12

• r M sr H L

(3.3.29) and at the upper edge s = 0:

χh 0

= 1E t c

– 2

2 c r12

– 1

– 1 – 2 c r12

• r M sr H 0

(3.3.30) The details for any prescribed loading of the edges can be worked out from the solution (3.3.14). In the case of the long shell, the solution significant at the edge s = L is (3.3.20). The boundary conditions for a radial load and zero moment loading at the edge are

At s = L , Η = Η L Μ s = 0 (3.3.31) The edge flexibility coefficients (3.3.29) give the radial displacement and rotation:

χ L = –

r H LE t c h L =

r H LE t c 2 c r

12

(3.3.32) The constant A can be evaluated from (3.3.24):

62

Re A = h L Im A = 0 (3.3.33) so the distribution in the shell interior obtained from (3.3.20) is

h = h L e– ξ cosξ

σ θ D=N θt = σ 0 e

– ξ cosξ

σ s B =6 M s

t 2= – 3

1 – ν2

12σ 0 e

– ξ sinξ

(3.3.34-36) in which the reference stress is the circumferential stress on the edge:

σ 0 =

E h Lr =

H Lt

2 rc

12

(3.3.37) and a dimensionless edge zone coordinate, which increases with the distance from the edge, is introduced:

ξ = L – s

2 c r12 (3.3.38)

Note that the reference stress (3.3.37) is equal to the average transverse shear stress applied to the edge multiplied by a factor of the square root of radius to thickness. In the example of the cantilevered beam with a shear load (2.8.1), the tangential stress is of the order of magnitude of the length to thickness larger. So as the beam becomes short or the shell becomes thick, the assumption that the transverse shear stress is negligible in comparison with the tangential stress becomes less valid. The meridional bending stress (3.3.36) is zero at the edge and has a maximum amplitude at the point ξ = π / 4:

σ s B

σ 0 max

= 3

1 – ν2

12e– π / 4 sin π4 = 1.82 ∗ 0.322 = 0.568

(3.3.39) Thus the maximum bending stress is smaller than the circumferential direct stress.

63

The results (3.3.34-39) are for the semi-infinite shell. Solutions for the shell of finite length can be obtained from (3.3.14) which are almost as simple as (3.3.20) if the load is divided into a parts which are symmetric and nonsymmetric about the center point of the meridian. Then the exponential functions in (3.3.14) are replaced with suitable hyperbolic functions. Fig. 3.6 shows the results for symmetric loading of the ends (both radial loads outward or both radial loads inward). It is clear that the solution for the semi-infinite shell is quite good when the length is greater than the decay distance, defined by (3.3.16).

0 1 20

1

2

L/decay distance

Semi-infinite

Ring

Fact

ors

Radial displacement

Shell

shellShell bending stress

Figure 3.6 - Effect of shell length for cylinder with symmetric radial loads on the ends. When the length is greater than the decay distance, the semi-infinite solution is valid; when the length is smaller than the decay distance, the elementary ring solution is valid. The factors consist of the displacement and stress normalized to the values for the semi-infinite shell.

More details of the transition from semi-infinite shell to ring are shown in Figs. 3.7-9, in which the cylinder radius, thickness, and magnitude of the symmetric end loads are held the same while the length is varied. For the length equal to 10 decay distances, Figs. 7a,b show the very localized end deformation and stress. The length equal to the decay distance in Figs. 8a,b is a transition. For a slight decrease in length in Fig. 9a,b, the displacement calculated from the ring approximation is not too much in error, but the significant bending persists. The displacement amplitude in these and the following figures is greatly exaggerated to show the behavior. In reality, substantial nonlinear effects are significant when the displacement is larger than the shell thickness.

64

Figure 3.7a - Meridian for cylindrical shell with symmetric radial

end loads (L = 250, r = 100, t = 1, H L = 1000). The self-

equilibrating end loads have a very local effect for this case of a long shell L / δ = 10.

Figure 3.7b - Stress resultants for cylindrical shell with symmetric

radial end loads (L = 250, r = 100, t = 1, H L = 1000). Significant

variation of the stress occurs within the decay distance.

Figure 3.8a - Meridian for cylindrical shell with symmetric radial

end loads (L = 25, r = 100, t = 1, H L = 1000). This is the case of

length equal to decay distance L / δ = 1. The center of the meridian experiences only a slight effect of the end loading, as far as displacement is concerned.

Figure 3.8b - Stress resultants for cylindrical shell with symmetric

radial end loads (L = 25, r = 100, t = 1, H L = 1000). Even though

the radial displacement and circumferential stress are small at the center, the meridional bending resultant is at a maximum for this case of length equal to decay distance.

65

Figure 3.9a - Meridian for cylindrical shell with symmetric radial

end loads (L = 15, r = 100, t = 1, H L = 1000). This is the case of

length less than decay distance L / δ = 0.6. A substantial ring rigid body displacement of the meridian occurs.

Figure 3.9b - Stress resultants for cylindrical shell with symmetric

radial end loads (L = 15, r = 100, t = 1, H L = 1000). Although

significant bending persists, the circumferential stress is closer to the ring value of uniform stress for this case of length slightly less than decay distance.

66

The case of loading on the shell ends which is nonsymmetric with respect to the center point of the meridian produces similar results. Fig. 3.10 shows the variation in the rotation at the edge and in the maximum meridional bending stress. As in the case of symmetric loading, the length equal to the decay distance marks the transition between semi-infinite shell and short ring behavior.

0 1 20

1

2

3

4

RotationShellRingFa

ctor

s

L/decay distance

Shell bendingstress

Figure 3.10 - Effect of shell length for nonsymmetric radial loads at each end of the cylinder ( one load outward, one inward ). Similar to the symmetric load case, the response is that of a ring or semi-infinite shell depending on whether the length is less than or greater than the decay distance.

§3.3.3 Center Radial Force If a line load in the radial direction acts at any point of the meridian of a long cylinder which is more than one decay distance from either end, then there is little effect from the ends on the solution in the neighborhood of the load. The shell can be regarded as two semi-infinite shells joined together. Because of the symmetry, only one of the semi-infinite shells need be treated. For the upper shell s ≤ 0, the conditions at the edge are At s = L, χL = 0 HL = q / 2 (3.3.40) where q is the intensity of the radial line load. The symmetry requires that the rotation under the load be zero, and each shell carries half of the total applied load. This conditions (3.3.40) are mixed, so either the edge stiffness or the inverse flexibility matrix

67

must be partially inverted. For the present system this is an easy matter. Since the rotation is zero, the stiffness matrix (3.3.27) gives the relation between edge load and displacement:

r M sL = k 12 h L = E t c h L

r Η L = k 22 h L = E t c2 h L

2 c r12 (3.3.41-42)

so the radial displacement and moment under the load are

h L =r H Lk 22

= r 2 c r12

4E t cq

M sL =k 12k 22

H L =2 c r

12

4 q (3.3.43-44)

The distribution for a long shell is shown in Figs. 3.11a,b.

Figure 3.11a - Meridian for cylindrical shell with a radial line load at the center (L = 250, r = 100, t = 1, q = 1000, L / δ = 10). The deformation is localized to a decay distance on either side of the load.

Figure 3.11b - Meridional stress (SsOD) and circumferential stress (SthOD) on the outer surface (OD) for a long cylindrical shell with a radial line load at the center (L = 250, r = 100, t = 1, q = 1000, L / δ = 10).

68

§3.4 Total Solution for Pressure Vessel with Rigid Ends The particular solution for surface loading and the complementary solutions for edge loading of the cylindrical shell are discussed in the preceding sections. In this section, these are put together for a total solution of a problem of practical interest. In Fig. 3.1 is shown a cylindrical shell with thick end plates. Because the flat plate carries the pressure by bending, end plates of the same material as the shell must be relatively thick in order to be regarded as "rigid". Later the realistic flexibility of the end plate and other more efficient shapes for the end closure will be analyzed. For the present assumption of rigid ends, the boundary conditions for the cylinder are

At s = 0, χ 0 = 0 h 0 = 0 At s = L , χ L = 0 h L = 0 (3.4.1-2) The total solution is written as Y = Y P + Y C (3.4.3) where the subscript P indicates the particular solution of section §3.2 and the subscript C indicated the complementary solution of section §3.3. The total solution must satisfy the conditions (3.4.1-2):

χ 0 = χ 0 C + χ 0 P = 0h 0 = h 0 C + h 0 P = 0

χ L = χ L C + χ L P = 0h L = h L C + h L P = 0 (3.4.4-5)

When the shell is long, the effect of the edges are considered independently. At the lower edge, the stiffness matrix (3.3.25) gives immediately the edge values of the resultants of the complementary solution:

r M sr H L C

= E t c

2 c r12 1

1 2

2 c r12

•– χ– h L P

(3.4.6) Since the rotation of the particular solution due to pressure is zero and the radial displacement is given by (3.2.11) the edge bending stress is

69

r M s = E t c – h P = – c r2 1 – ν2 p

σ s D=6 M s

t2= – 3

1 – ν2

121 – ν2 σ 0 = – 1.55 σ 0

(3.4.7-8) in which the numerical value is for Poisson's ratio of 0.3, and the reference stress is the circumferential direct stress from (3.2.10):

σ 0 =

p rt (3.4.9)

The meridional stress at the inner and outer surfaces at the edge is then

σ s OD = σ s D+ σ sB = − 1.05 p 0

σ s ID = σ s D– σ sB = 2.05 p 0 (3.4.10-11) The behavior in this problem is typical of shells. The great expanse of the shell carries the pressure load very efficiently, but a price in some bending stress must be paid in the edge zone. In this example, the maximum stress is at the edge on the inner surface (3.4.11), which is twice the magnitude of the circumferential stress of the particular solution. Failure of a shell is often initiated by this peak stress in the edge zone. For the distribution near the edge since the edge rotation is zero, the constant A from (3.3.22) is just:

A = – h L P (1 + i ) (3.4.12)

which used in (3.3.20) added to the particular solution (3.2.2-5) gives the following results for the direct and bending stresses:

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σ s D=σ 0

2

σ θ D= σ 0 1 – 1 – ν2 e– ξ cosξ + sinξ

σ s Β= – σ 03

1 – ν2

12

1 – ν

2e– ξ cosξ – sinξ

σ θ Β= ν σ s B (3.4.13-16) In Figs. 3.12a,b is shown the results for the deformation of the meridian and the stress.

Figure 3.12a - Meridian in a cylindrical pressure vessel with thick end plates ( L = 250, r = 100, t = 1, δ / L = 10, p = 10 ). All bending deformation occurs in the edge regions of width equal to the decay distance δ.

Figure 3.12b - Stress on the inner and outer surfaces in a cylindrical pressure vessel with thick end plates ( L = 250, r = 100, t = 1, δ / L = 10, p = 10 ).

§3.5 Symmetric Solution for Short Shell In the preceding calculations, it has been assumed that the shell is sufficiently long in comparison with the decay distance, so that the edge bending from one edge is negligible at the other. Numerical results for the short shell have been discussed in Figs. 3.6, 3.8, 3.9, and 3.10. In this section, the formulas for the edge stiffness of a shell of any length will be obtained. The problem is simplified by considering symmetry. For this it is convenient to measure arc length from the center point of the meridian. If a

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solution is desired for which the radial displacement is an even function of the arc length s, then the constants in the solution (3.3.12) are chosen to be equal:

C3 = C1 = A

2 Cosh µ L2

(3.5.1) and the solution is:

Y = Re A

Cosh µ L

2

– µ2 r E t c2 Cosh µ s

– µ3 r E t c2 Sinh µ s

– µ Sinh µ s

Cosh µ s (3.5.2) At the edge s = L / 2, the stiffness matrix is:

r Ms

r H = E t cIm z

1 Re z

Re z z2 • χ

h L (3.5.3)

in which

z = µ Tanh µ L2 (3.5.4)

the inverse flexibility matrix is:

χ

h L = 1

E t c Im z

z2 – Re z– Re z 1

• r Ms r H L

(3.5.5)

For the solution for which the radial displacement is an odd function of the arc length, the hyperbolic functions Cosh and Sinh are interchanged in the equations (3.5.1-5).