P

2p r0 cos a

From Equation 2.93, one obtains Ny = 0.

In the case of a conical surface in which the lateral forces are symmetrically distributed, the membrane stresses can be obtained by using Equations 2.92 and 2.93. The curvature of the meridian in the case of a cone is zero and hence r1 = 1; Equations 2.92 and 2.93 can therefore be written as

2p r0 sin f and

sin f

If the load distribution is given, Nf and Ny can be calculated independently.

For example, a conical tank filled with a liquid of specific weight g is considered as shown in Figure 2.46 The pressure at any parallel circle mn is

For the tank, f = a + (p/2) 19 and r0 = y tan a. Therefore,

Ny is maximum when y = d/2 and hence, g d2 tan a (Ny)max = 4 cos a

The term R in the expression for Nf is equal to the weight of the liquid in the conical part mn0 and the cylindrical part must be as shown in Figure 2.45. Therefore,

R = —J py3 tan2a + py2 tan2a(d — y)] g = —pgy2(d — 3y) tan2a

Hence,

2 cos a

Nf is maximum when y = 4 d and

3 d2g tan a

16 cos a

The horizontal component of Nf is taken by the reinforcing ring provided along the upper edge of the tank. The vertical components constitute the reactions supporting the tank.

Consider an element cut from a shell by two adjacent meridians and two parallel circles as shown in Figure 2.47. In the general cases shear forces Njy = Nyj and normal forces Nj and Ny will act on the sides of the element. Projecting the forces on the element in the y direction, we obtain the governing equation as

Similarly, the forces in the x direction can be summed up to give q . . 0n0 .

qj (roNj0)+ -qy- ri + Ne^ri cos j + Xrori = 0 (2.95)

Since the projection of shearing forces on the z axis vanishes, the third equation is the same as Equation 2.93. The problem of determining membrane stresses under unsymmetrical loading reduces to the solution of Equations 2.93 to 2.95 for given values of the components X, Y, and Z of the intensity of the external load.

2.6.6 Cylindrical Shells

It is assumed that the generator of the shell is horizontal and parallel to the x axis. An element is cut from the shell by two adjacent generators and two cross-sections perpendicular to the x axis, and its position is defined by the coordinate x and the angle j. The forces acting on the sides of the element are shown in Figure 2.48b.

The components of the distributed load over the surface of the element are denoted as X, Y, and Z. Considering the equilibrium of the element and summing up the forces in the x direction, we obtain

0Nx , 0Njx , , —— r d j dx + d j dx + Xr d j dx = 0 0x 0 j

The corresponding equations of equilibrium in the y and z directions are given, respectively, as n xj r dj dx + -—- d j dx + Yr dj dx — 0

The three equations of equilibrium can be simplified and represented in the following form

In each case we readily find the value of Nv. Substituting this value in the second of the equations, we then obtain Nxj by integration. Using the value of Nxj thus obtained we find Nx by integrating the first equation.

2.6.7 Symmetrically Loaded Circular Cylindrical Shells

To establish the equations required for the solution of a symmetrically loaded circular cylinder shell, we consider an element, as shown in Figure 2.48a and Figure 2.49. From symmetry, the membrane shearing forces Nxj = N jx vanish in this case; forces Nv are constant along the circumference. From symmetry, only the forces Qz do not vanish. Considering the moments acting on the element in Figure 2.49, from symmetry it can be concluded that the twisting moments Mxj = Mjx vanish and that the bending moments Mj are constant along the circumference. Under such conditions of symmetry three of the six equations of equilibrium of the element are identically satisfied. We have to consider only the equations obtained by projecting the forces on the x and z axes and by taking the moment of the forces about the y axis. For example, consider a case in which external forces consist only of a pressure normal to the surface. The three equations of equilibrium are dN

—— a dx d j = 0 dx a dx dj + Nj dx dj + Za dx dj = 0 (2.97)

The first equation indicates that the forces Nx are constant, and they are taken as equal to zero in the further discussion. If they are different from zero, the deformation and stress corresponding to such constant forces can be easily calculated and superposed on stresses and deformations produced by lateral load. The remaining two equations are written in the simplified form dQx 1

These two equations contain three unknown quantities: Nv, Qx, and Mx. We need, therefore, to consider the displacements of points in the middle surface of the shell.

The component v of the displacement in the circumferential direction vanishes because of symmetry. Only the components u and w in the x and z directions, respectively, are to be considered. The expressions for the strain components then become du w . .

By Hooke's law, we obtain

Eh Eh w du

From the first of these equation it follows that du w dx a and the second equation gives

Considering the bending moments, we conclude from symmetry that there is no change in curvature in the circumferential direction. The equations as for plates, we obtain the circumferential direction. The curvature in the x direction is equal to — d2w/dx2. Using the same

where

is the flexural rigidity per unit length of the shell. Eliminating Qx from Equations 2.98 we obtain d2 Mx 1

from which, by using Equations 2.101 and 2.102, we obtain d2 ( d2w\ Eh .

All problems of symmetrical deformation of circular cylindrical shells thus reduce to the integration of Equation 2.103.

The simplest application of this equation is obtained when the thickness of the shell is constant. Under such conditions Equation 2.103 becomes d4w Eh D + ^ w = Z

Using the notation

Equation 2.104 can be represented in the simplified form d4w „4 Z . .

The general solution of this equation is w — ebx(C1 cosbx + C2 sinbx) + e-bx(C3 cos bx + C4 sinbx) + f (x) (2 .106)

Detailed treatment of shell theory can be obtained from Timoshenko and Woinowsky-Krieger (1959).

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