Outer Cylinder Linked to a Meniscus Shell

The theory of cylindrical shell allows determining the flexure of the outer cylinder of a vase shell. We assume that the effect of the pressure load q on the cylindric element N +1 can be neglected because its axial length or thickness is such as tzN+1 / rN < 1.

Let x, z be a local frame in a normal section of the cylinder and passing through the common z-axis of the shell meniscus elements (Fig. 6.6).

A simple case is when the radial thickness tx is constant and q = 0. Referring to Timoshenko and Woinowsky-Krieger [29], the radial displacement u in the x-direction is given by

Fig. 6.6 Vase shell geometry and link of the outer cylinder to the meniscus shell

Fig. 6.6 Vase shell geometry and link of the outer cylinder to the meniscus shell

where the bar superscript holds for the cylinder. Using the notation

4 Nd 'Nlx we obtain the simplified form d4ii 4

The general solution of this equation is

H = eKz(cicoskz + c2sinKz) + e—Kz(c3cosKz + c4sinKz), (6.42)

in which constants c1 c2 c3, and c4 are determined by particular boundary cases at the cylinder ends.

The transverse shearing forces, usually denoted Qr for a cylinder, correspond in fact to z-plane radial tensions Nr in the meniscus shell frame z , r. Similarly the bending moments Mr about a tangential direction of a z-cut of the cylinder correspond to Mr in the meniscus frame. Hence, we use hereafter the notation Nr and Mr for referring to those quantities. They are determined by

dz2 dz dz3

where Mr is always given the positive sign convention (cf. Sects. 1.13.8 and 3.2). Since dw/dr = du/dz at the meniscus-cylinder junction, local equations (6.37) write for the cylinder variables as du H

— = a Mr,N+1 + p Nr,N+1, — = jMr,N+1 + $ n,n+1 , (6.44a)

dz rN

a dMr,N+l\ dz J ' dNr,N+\ dz J ' 7 rN dMrN+1 ' rN dNr,N+1 '

Taking the origin z = 0 at the external end of the cylinder, each case (6.38a) and (6.38b) for the boundary conditions at this end allows us to determine the ci constants as follows.

• Articulated and movable cylinder external end: In the local frame x, z of the cylinder, the boundary conditions (6.38a) for an articulated and movable external end write

After substitutions of the first and third u derivatives, solution of the two relations leads to c3 = c1 and c4 = c2. (6.46)

At the cylinder end linked to the meniscus the continuity conditions are determined by the bending moment Mr,N+1 and the radial tension Nr,N+1. In the meniscus frame, equations (6.44) ensure the continuity of the slope du/dr and of the relative radial displacement U/rN.

Let us introduce a second subscript M to denote a Ci coefficient set relative to a unitary bending moment Mr,N+1 = 1 while Nr,N+1 = 0, so the ci write ci<M. From the axial ordinate z = tz,N+1 of the junction we also introduce the quantity

From equations (6.43), we obtain respectively

- c1 M (eT - e-T) sin t + c2 M (eT - e-T) cos t = , (6.48a)

2Dk2

c2,M [eT(sin t - cos t) - e-T(sin t + cos t)] = 0, (6.48b)

so that all coefficients c1mm = c3mm and c2,M = c4,M are known by solving the above system. This allows us to determine du dMrN+1 d ( dû

ex( c1; M cost + c2 , M sin t) + e ^ c3 , M cos t + c4, M sin^, (6.49a)

{d^j = k [c1>MeT(cost-sint) + c2,MeT(cost + sint) -

dMr,N+i\ dz c3,M e-T( cos t + sin t) + c4,M e-T( cos t — sin t) ], (6.49b)

from where the 7 and a coefficients in matrix (6.36) are straightforwardly obtained from (6.44b).

A second resolution for Mr,N+1 = 0 and unitary radial tension Nr,N+1 = 1, now by use of c,N coefficients, provide two equations similar to (6.48a and b) except

Fig. 6.7 Left: Flat closed shell made of identical vase shells. Right: Curved closed shell

Fig. 6.7 Left: Flat closed shell made of identical vase shells. Right: Curved closed shell that the righthand terms are 0 and 1 /2Dk3, respectively. Solving them provides the values of the two partial derivatives with respect to Nr,N+i in (6.44b) which therefore determine 5 and ¡, respectively.

• Built-in and movable cylinder external end: As seen above, a built-in and movable external end of the cylinder will allow analysis of the half part of a closed shell. In the local frame x, z of the cylinder, the boundary conditions (6.38b) for an external end write dw dr du dz

After substitutions of the first and third u derivatives, solution of the two relations leads to c3 = c1 and c4 = - c2. (6.51)

Hence, from (6.42), the radial displacement of the cylinder can be represented by

U = 2c1 cosh kz cos kz + 2c2 sinh kz sin kz , (6.52)

from where we now note that the flexure is even, thus in accordance to a closed form made of two strictly identical vase shells. Similar determinations to the latter boundary case, with unitary bending moments and radial tensions at the meniscus-cylinder junction for the determination of ci,M and ci,N set values, thus provide the a, 5, y, and 5 coefficients.

Note: The above conditions strictly apply to the analysis of a closed shell made of two identical vase shells linked together at the plane base of their outer ring because the linked bases remain plane during flexure (Fig. 6.7-Left). This form may be called a flat closed shell.

If the mean curvatures <R1 >, <R2> of two vase shells are similar but with opposite sign, then the radial displacements ¿¡1{0} and U2{0} may have opposite signs. Now when two such vase shells are linked together to form a closed shell, this effect introduces a perimeter bending moment during loading (Fig. 6.7-Right). We have seen in Sect. 2.1.1 that a uniform bending moment acting on the contour of a plain plate only generates a pure Cv 1 mode (parabolic flexure); this result also applies to a moderately curved plain shell. Hence, for a curved closed shell, final calculation by finite element analysis is appropriate.

• Other external end boundary cases: It must be underlined that the two above latter cases accurately apply to the simplest implementations for practical execution conditions. Other boundary conditions such as, for instance, a built-in and immovable external end of the cylinder are set by

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