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.
Page 1 of 17
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