Table of contents

Table of contents1Introductory notions2Symmetrically loaded shells of revolution in membrane theory. the governing equation3Axially symmetric deformation6Reinforcing rings8Application :9Numerical application :12Bibliography17

SYMMETRICALLY LOADED SHELLS OF REVOLUTION IN MEMBRANE THEORY. THE GOVERNING EQUATION.AXIALLY SYMMETRIC DEFORMATION.REINFORCING RINGS.

Introductory notions Thin shells with small deflections By thin shells it is understood that the thickness is much smaller than the radii of curvature . By small deflections it is understood that the shell deflections are small compared to the shell thickness . Thin shells versus platesThin shells have a great capacity to withstand forces distributed on their surface.The thickness of a shell may be much smaller than the thickness of a plate for the same covered surface.They are recommended for large spans (18 40 m ) h in-plane forces and bending forces ( bending moments,shear forces normal to the midsurface and twisting moments).Notes : 1) Membrane forces are independent of bending forces and they are completely defined by the conditions of static equilibrium.2) Membrane theory is exact for thin shells having no abrupt changes in thickness,slope or curvature.3) Bending theory comprises the membrane solution,corrected in areas with pronounced discontinuity effects due to edge forces or concentrated loadings.

Symmetrically loaded shells of revolution in membrane theory. the governing equation

Shells that have the form of surfaces of revolution find extensive application in various kinds of containers,tanks and domes.A surface of revolution is obtained by rotation of a plane curve about an axis lying in the plane of the curve.This curve is called the meridian, and its plane is a meridian plane.An element of a shell is cut out by two adjacent meridians and two parallel circles,as shown in figure a) below. The position of a meridian is defined by an angle , measured from some datum meridian plane; and the position of a parallel circle is defined by the angle , made by the normal to the surface and the axis of rotation. The meridian plane and the plane perpendicular to the meridian are the planes of principal curvature at a point of a surface of revolution, and the corresponding radii of curvature are denoted by r1 and r2, respectively. The radius of the parallel circle is denoted by r0 so that the length of the sides of the element meeting at O, as shown in the figure, are r1d and r0 d = r2 sin d . The surface area of the element is then r1r2 sin d d .

The load is distributed on the shell surface and it may be divided in three components in a point on the shell : X on the tangent at the parallel circle ; Y on the tangent to the meridian ; Z on the normal to the meridian ; X,Y,Z [F/L^2]

From the assumed symmetry of loading and deformation it can be concluded that there will be no shearing forces acting on the sides of the element. The magnitudes of the normal forces per unit length are denoted by N and N as shown in the figure. The intensity of the external load,which acts in the meridian plane, in the case of symmetry is resolved in two components Y and Z parallel to the coordinate axes.Multiplying these components with the area r1r2 sin d d, we obtain the components of the external load acting on the element. In writing the equations of equilibrium of the element, let us being with the forces in the direction of the tangent to the meridian. On the upper side of the element the force which is acting is The corresponding force on the lower side of the element is

From these two expressions, by neglecting a small quantity of second order, we find the resultant in the y direction to be equal to The component of the force in the same direction is The forces acting on the lateral sides of the element are equal to N r 1 d and have a resultant in the direction of the radius of the parallel circle equal to N r 1 d d . The component of this force in the y direction is - N r 1 cos d d . Summing up the forces, the equation of equilibrium in the direction of the tangent to the meridian becomes (a) . The second equation of equilibrium is obtained by summing up the projections of the forces in the z direction. The forces acting on the upper and lower sides of the element have a resultant in the z direction equal to N r 0 d d. The forces acting on the lateral sides of the element and having the resultant N r 1 d d in the radial direction of the parallel circle give a component in the z direction of the magnitude N r 1 sin d d . The external load acting on the element has in the same direction a component Z r 1 r 0 d d . Summing up the forces we obtain the second equation of equilibrium : N r 0 + N r 1 sin + Z r 1 r 0 = 0 (b)From the two equilibrium equation the forces N and N can be calculated in each particular case if the radii r 0 and r 1 and the components Y and Z of the intensity of the external load are given. Instead of the equilibrium of an element, the equilibrium of the portion of the shell above the parallel circle defined by the angle may be considered :

If the resultant of the total load on that portion of the shell is denoted by R, the equation of equilibrium is 2 r 0 N sin + R = 0 ( c )This equation can be used instead of the differential equation (a), from which it can be obtained by integration. If equation (b) is divided by r 1 r 0 it can be written in the form : ( d )It is seen that when N is obtained from (c) , the force N can be calculated from eq. (d). Hence the problem of membrane stresses can be readily solved in each particular case.

Axially symmetric deformation

In the case of symmetrical deformation of a shell, a small displacement of a point can be resolved into two components : v in the direction of the tangent to the meridian and w in the direction of the normal to the middle surface .

Considering an element AB of the meridian, we see that the increase of the length of the element due to tangential displacements v and v + (dv/d)d of its ends is equal to (dv/d)d. Because of the radial displacements w of the points A and B the length of the element decreases by an amount w d . The change in the length of the element due to the difference in the radial displacements of the points A and B can be neglected as a small quantity of higher order.Thus the total change in length of the element AB due to deformation is :

Dividing this by the initial length r1 d of the element, we find the strain of the shell in the meridional direction to be

Considering an element of a parallel circle it may be seen that owing to displacements v and w the radius r0 of the circle increases by the amount v cos w sin . The circumference of the parallel circle increases in the same proportion as its radius; hence , r0=r2 sin =>

Eliminating w from the eqs above, we obtain for v the differential equation (1)The strain components and can be expressed in terms of the forces N and N using Hookes law :

Substituting in (1) we get (2) :In each particular case the forces N and N can be found from the loading conditions and the displacement v will then be obtained by integration of the differential equation. Denoting the right-hand side of this equation by f() we get General solution : where C is an integration constant to be determined from the condition at the support. For the application in this report,a spherical sphere loaded by its own weight, with r1=r2=R and the appropriate boundary conditions,we get:

Reinforcing ringsDue to the fact that in practice, the support is on the vertical direction, there appear bending moments. The bending theory of thin shells of revolution in axisymmetrical loading shows that the values of the bending moments decrease quickly. This means that the membrane theory is good for the most part of the shell, except in the support vicinity. In order to resist these bending moments, a reinforcing ring is done on the support edge, by increasing the thickness of the shell in this local area. If necessary, concrete rings will have reinforcements to resist tension .

Application : Hemispherical roof dome supporting its own weight

=> 2r N sin + P = 0 (eq.1) P =

dA = r d R d sin = => r = R sin => dA = R2 sin d d

= > Z = q cos => Y= q sin

P = = = (eq.1) => = 0

r1 = r2 = R Z = q cos =>

Due to the fact that in practice, the support is on the vertical direction, there appear bending moments. The bending theory of thin shells of revolution in axisymmetrical loading shows that the values of the bending moments decrease quickly. This means that the membrane theory is good for the most part of the shell, except in the support vicinity. In order to resist these bending moments, a reinforcing ring is done on the support edge, by increasing the thickness of the shell in this local area. Equal and opposite compression forces(those acting in the membrane) are transmitted to the reinforcing ring.

Vring = Rs => check global equilibrium on the vertical : OK

Numerical application : For a concrete shell

self-weight of the dome

radius of the spherical dome

The state of stress will be computed for the same data using the finite element approach in SAP :

Fig.1 3D view of the spherical roof dome

The type of element used for modeling was the thin shell type, having a thickness of 20 cm . The only load case used was the DEAD one, since we are interested in the state stress only due to the own weight of the element.

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Fig.2 N efforts

Fig.3 N efforts

Numerical results comparison between the two approaches it can be seen that the results are very close one to the other

0o30o60o90o

Analytic solutionN (kN)-75-80.3-100-150

N (kN)-75-49.5125150

SAP solutionN (kN)-87-80.79-99.99-157.7

N (kN)-77.9-49.2725.32140.47

Bibliography

Lecture notes Timoshenko S., Theory of plates and shells , McGraw Hill Book Company