Chapter 3

Shells with membrane behavior

In the present Chapter the stress static response of membrane shells will be addressed. In

Section 3.1 an introductory example emphasizing the difference between bending and membrane

stresses for the simplified geometry of a spherical shell undergone to an internal pressure load

is presented. Next, in Section 3.2 the general theory of the membrane shells having an axially-

symmetric geometry together with an axially symmetric load is presented also including several

specific geometry cases. Finally, in Section 3.3 the case of a generic non-symmetric load is

carried out also considering some special application cases.

3.1 General issues on membrane v.s. bending behavior of shellstructures

In the following shell structures will be assumed having constant thickness and small compared to

the other geometric dimensions, represented by local radius of curvature of the average surface.

In order to describe the geometry of a shell structure, the geometry of the average surface and

thickness are to be described. Generally, a shell is considered to be a thin shell when the ratio

between its thickness, indicated with t, and the smaller radius of curvature, indicated with r∗,

is lower than 1/20, although in many examples of interest in the aeronautical field this ratio can

be also of order of 10−3.

In order to study the behavior of a shell structure, two different approach can be considered.

• The easiest type considers as dominant dominant behavior the membrane behavior and this

can be assumed when there are no load or structural discontinuities on the shell structure.

• The most general shell behavior include also the bending behavior and this allows also

41

42

to treat the discontinuity in the stress field caused by the effect of load or structural

discontinuity as well.

It is important to note that the most general way does not have the purpose of improving

the solution offered by the membrane approach, but it is able to deal with different problems

and, more precisely, the effects of discontinuity in the load or in the structure, that the

membrane theory can not describe.

It is also observed that the equilibrium conditions in the membrane model are sufficient to

describe and close the stress problem so avoiding the introduction of further relations like con-

stitutive or kinematic relationships.

The fundamental hypotheses at the basis of the study of a thin-walled shell in the case of small

deflections are (in analogy with the plate models):

i) the thickness is small compared to the smaller radius of curvaturet

r∗<< 1

(t

r∗<

1

20

)

ii) the deflection is small compared to the thickness of the shell,w

t<< 1

iii) the normal to the average surface plane sections remain plane and those are still normal to

the deformed average surface, so this hypothesis requires that the shear stress εxz and εyz

(z is the direction normal to the shell surface) are negligible, as well as sher the deformation

εzz

iv) normal stress σzz is negligible compared to the membrane stresses.

Figure 3.1: Part of a spherical shell subjected to a pressure

43

In order to discuss the different role of membrane in-plane stress σm with respect to the bending

in-plane stress σf , let us consider as an example the case of a portion of a spherical shell of

thickness t and radius r∗ subjected to a uniform pressure of intensity p, as indicated in Fig.3.1.

The equilibrium condition requires that the sum of vertical forces is zero and allows to derive

the force to the membrane per unit of length, Nφ, by the relation:

−2πr0Nφsinφ−∫ φ

0p cos φ̄ 2πr∗ sin φ̄ r∗dφ̄ = 0

or

Nφ = − pr02 sinφ

= −pr∗ sinφ

2 sinφ= −pr∗

2

For reasons of symmetry, which concern the structure (a spherical shell) and the load (pressure),

one has a unique value for the stress given by:

σθ = σφ = σm = −pr∗

2t

Next, one can evaluate the deformation with the use of constitutive laws, which in case of

isotropic material provide the constitutive relationship:

εm =1

E(σm − νσm) = −(1− ν)

pr∗

2Et

This deformation value corresponds to a variation of the radius of the sphere given by relation-

ship:

r′ = r∗(1 + εm)

and this variation of the radius leads to a variation of the curvature given considering a Taylor

series expansion for εm, by:

χ =1

r′− 1

r∗=

1

r∗

(1

1 + εm− 1

)≃ −εm

r∗

Then, by using the equation for εm, one has:

χ =(1− ν)p

2Et

In the case of spherical shell under consideration one can derive the local bending moment, in

analogy with the case of the plate, with the constitutive relation:

M = −D(χ+ νχ) = −D(1 + ν)χ = −D(1− ν2)p

2Et

with the bending stiffness given by:

D =Et3

12(1− ν2)

44

and, therefore, the bending moment is given by:

M = −pt2

24

Thus, one can calculate the bending stress from:

σf =M

Izmax =

12M

t3t

2= −12pt2

24t3t

2= −p

4

in which I = t3

12 and y = t2 . The relationship between the membrane stress and the bending

stress is

σmσf

=pr∗

2t

4

p=

2r∗

t

It is thus evident that the membrane stress is much more important because, in the basic

assumptions of the theory considered, the ratio between radius of curvature and thickness of the

shell is very large and consequently also the relationship between the membrane stress and the

bending stress is large as well.

The case considered here is a special load case for a special geometry. However,the obtained

result and comment are very general.

It is observed that the thin shells are also subjected to problems of critical buckling load when

subjected to compressive loads. In order to estimate the extent of the critical stress, one can

use a relation of type:

σcr = kEt

r∗

where E indicates the modulus of elasticity of the shell material and k indicates a constant whose

value depends on several factors, but a value that may be indicative can be k = 0.25− 0.30.

It is to be noted that, due to the often very small value of the ratio between the thickness of the

shell and radius of curvature, the value of the critical buckling stress indicated by the previous

relation is very low compared to the tensile strength of the material. If one considers the case

of a spherical shell made by aluminum alloy (2024 or 7075 type) and with the ratiot

r∗= 10−3

one has:

σcr = 0.25 ∗ 70 ∗ 109 ∗ 10−3 = 17.5MPa

which is in fact a very low value compared to the tensile strength of the lightly alloy.

3.2 Axial symmetric shells undergone to axial symmetric loads

Let us consider a very special case of shells from the geometrical point of view, which are

described by an average surface obtained by the rotation of a curve, denoted as meridian curve,

45

around an axis which lies in the plane to which the curve belongs. In general, the material

lines that describe the surface of the undeformed shell are not Cartesian. It will show in the

following the two natural coordinates associated to these lines with φ (meridian coordinated

in the revolution shells) and ϑ (longitude coordinate). As indicated in Fig.3.2 ,a point on

Figure 3.2: Geometry of the revolution shell

the surface of the shell is identified by coordinates (ϑ, φ) and an infinitesimal surface element,

indicated with ABCD, is identified by meridian and parallel increment dφ and dϑ.

The curvature radius bending radii shown in Figure with r0 and r1 are related to the parallel

(ϑ direction) and meridian (φ direction) curves respectively, whereas r2 represents the distance

between the generic surface material point and the intersection between the curvature radius

46

and the rotation axis. The radii r1 and r2 are related by the relationships:

r0 = r2 sinφ

lAC = r0dϑ

lCD = r1dφ

with

dS = r0r1dϑdφ = r1r2sinφdϑdφ

Note that the radii of curvature of the shell r0, r1, and r2 which describe the geometry, are

known for specific geometry as function of the material co-ordinates (φ, ϑ).

This type of revolution shell, is considered the basis of the following considerations on the shells

as they can keep a simple geometric description while referring to structures of interest practical

in the field of aeronautics.

In the case of a shell of revolution to which is applied a load in axial symmetry, axial symmetric

load, you have as unknowns only two membrane forces, per unit of length, Nϑ, Nφ, while there

are no shear stress.

In order to determine the two unknowns membrane forces, one can consider two equations of

equilibrium in the normal direction, z, and in the direction φ. In the Fig. 3.3, is considered

in detail the shell element ABCD already indicated in Fig. 3.2. Due to the axial symmetry,

the forces to the membrane and the load does not have variations along the direction ϑ. The

external forces applied, which are forces per unit area, are indicated with pφ and pz = −pr (pz

is in the normal outward direction).

In order to evaluate the force balance in the normal direction, it is necessary to determine the

components the forces applied in this direction and one has the following equation (see Fig. 3.3):

pzr0r1dϑdφ+Nφr0dϑdφ+Nϑr1 sinφdϑdφ = 0

which becomes:

Nφr0 +Nϑr1 sinφ+ pzr0r1 = 0

If one divides by r0r1 and considers that r0 = r2 sinφ, one obtains:

Nφ

r1+

Nϑ

r2= −pz (3.1)

By imposing the equilibrium in the direction φ of the applied forces, one has

∂

∂φ(Nφr0) dφdϑ−N0r1dφdϑcosφ+ pφr1dφrodϑ = 0

47

Figure 3.3: Element of revolution shell: forces balance

or, considering loads axialsymmetry and thus including the ordinary derivative with respect to

48

φ,1

d

dφ(Nφr0)−Nϑr1cosφ+ pφr1ro = 0 (3.2)

Now, considering Nϑ as given by Eq. 3.1, substituting this to Eq. 3.2 multiplied by sinφ, one

Figure 3.4: Revolution shell under the angle φ

obtains:

d

dφ(Nφr0) sinφ+Nφr0 cosφ = −pφr1r2 sinφ sinφ− pzr1r2 sinφ cosφ

The left-hand term of this equation can be written as:

d

dφ(Nφr0 sinφ) =

d

dφ

(Nφr2 sin

2 φ)

and then is obtained

Nφ(φ) = − 1

r0(φ) sinφ

∫ φ

φ1

r1(φ̄)r2(φ̄) (pz cos φ̄+ pφ sin φ̄) sin φ̄dφ̄− F̄

2πr0(φ) sinφ(3.3)

1Note that, instead of this second equation of equilibrium, one could use the one obtained by considering aportion of the shell seen under an angle φ as indicated in Fig.3.4. If one indicates with F the vertical resultantof all the external loads that are applied to the shell and it is considered that for reasons of symmetry the forcesNφare constant on the edge, one has the balance equation:

2πr0Nφ sinφ+ F = 0

and therefore:

Nφ = − F

2πr0 sinφ

This equilibrium equation is only a different global way of expressing the equilibrium condition locally given byEq. 3.2.

49

where φ1 is the extreme value of the coordinate φ in correspondence of which the external load

boundary is applied and so giving a positive vertical force downwards and equal to F̄ . Indeed,

in this case with pz = pφ = 0, one has

Nφ = − F̄

2πr0 sinφ

The equilibrium equations in the direction of z and φ allow to obtain the unknowns Nϑ and Nφ

and from these related stresses. The membrane stresses, in the case of a shell of revolution with

axisymmetric load, are derived from the equations of equilibrium according z and seconds φ.

Cases of special shell geometries are now considered in the following.

3.2.1 Spherical pressured shell

In this case r1 = r2 = r∗ and the equilibrium in normal direction z becomes (pφ = 0):

Nφ +Nϑ = −pz ∗ r∗

Consider an internal pressure of intensity p such that pz = −p. Moreover, the balance equation

in φ becomes:

Nφ =p

r∗ sin2 φ

∫ φ

0r∗

2cos φ̄ sin φ̄dφ̄ =

pr∗

2

Therefore

Nφ = pr∗

2

Nϑ = −pzr∗ −Nφ = pr∗ − p

r∗

2= p

r∗

2

The membrane stress is then:

σφ = σϑ =N

t= p

r∗

2t

From the constitutive equations in the case of isotropic material, the strain is obtained by:

εφ = εϑ =σϑE

(1− ν) =p

E

r∗

2t(1− ν)

Note that we can also calculate the variation of the radius of the spherical shell with:

r′ = r∗(1 + εϑ) = r∗[1 +

p

E

r∗

2t(1− ν)

]

50

Figure 3.5: Conical shell

3.2.2 Conical shell

In this case indicated in Fig.3.5, the angle φ is constant and the radius r1 goes to infinite. From

the condition of equilibrium in z direction:

Nφ

r1+

Nϑ

r2= −pz

one has:

Nϑ = −r2pz = −pzr0

sinφ

while the condition of equilibrium occurs only at the end of the contribution due to possible

boundary terms:

Nφ = − F̄

2πro sinφ

3.2.3 Cylindrical shell of circular section

This case can be traced back to the previous conical shell if you put the angle φ = π2 . From the

equation equilibrium in the direction z one has (r1 → ∞ e r2 = r0 = r∗):

Nϑ = −pzr0 = −pzr∗

and by the condition of equilibrium in the direction φ one has (note that no integral contributions

of Eq. 3.3 are present for this geometry):

Nφ = − F̄

2πr0= − F̄

2πr∗

51

In the case of cylindrical shell, closed at the ends, and pressurized with an internal pressure p

as in the case of a fuselage one has:

p = −pz

F̄ = −πr20p

Thus, the axial stress in the circumferential direction is given by

σϑ =Nϑ

t=

pr0t

=pr∗

t

whereas in the axial direction

σxx = σφ =Nφ

t=

πr20p

2πr0t=

pr02t

=pr∗

2t

The circumferential axial strain may be obtained:

εϑϑ = εϑ =1

E(σϑ − νσφ) =

1

E

(pr0t

− νpr02t

)=

pr0Et

(1− ν

2

)and the axial strain:

εxx = εφ =1

E(σφ − νσϑ) =

(1

2− ν

)pr0Et

In the case of a cylindrical fuselage with a radius r0=4m, with a thickness t = 2 · 10−3m and

with a pressure p=105Pa, the axial stress components are

σϑ =pr0t

= 200MPa

σφ =σϑ2

= 100MPa

with corresponding strains

εϑ =pr0Et

(1− ν

2

)= 2.43 · 10−3

εφ = 0.57 · 10−3

3.2.4 Spherical dome loaded by its own weight

Consider a homogeneous semi-spherical dome of radius r∗ and with ai t thickness, simply sup-

ported to the boundary and subjected to its own weight, as shown in Fig.3.6. M is the total

mass of the dome, S its total area and g the acceleration of gravity. Thus, the weight force per

unit area in the gravitational field direction has an intensity q =Mg

S. Thus, the components of

the applied load are:

52

Figure 3.6: Geometry of a spherical dome

pϑ = 0

pφ = q sinφ

pz = q cosφ

applying the Eq. 3.3, one has:

Nφ(φ) =−1

r∗ sin2 φ

∫ φ

0r∗2q(cos2 φ̄+ sin2 φ̄) sin φ̄dφ̄ = −r∗q(1− cosφ)

sin2 φ= − r∗q

1 + cosφ

From the condition of equilibrium according with z:

Nϑ

r∗+

Nφ

r∗= −pz

one obtains:

Nϑ = −pzr∗ −Nφ = −q cosφr∗ +

r∗q

1 + cosφ= r∗q

(1

1 + cosφ− cosφ

)By the membrane stress components Nϑ, Nφ are obtained:

σφ =Nφ

t= − r∗q

t(1 + cosφ)

σϑ =Nϑ

t=

r∗q

t

(1

(1 + cosφ)− cosφ

)

In the case of the hemispherical dome one has for σφ a behavior as a function of the angle φ

that starts from the value for φ = 0:

σφ = −r∗q

2t

53

and reaches φ = 90o

σφ = −r∗q

t

Regarding σϑ it has a course always in function of the angle φ that starts from:

σϑ(0o) = −r∗q

2t

and arrives at:

σϑ(90o) =

r∗q

t

as indicated in Fig.3.7, σϑ changes sign and then vanishes for a value of the angle φ obtained

Figure 3.7: Performance of stress σϑ e σφ

by:

1

1 + cosφ= cosφ

φ = 51o 50′

Since the stress components have been determined, the deformations can be obtained by:

εϑϑ =1

E(σϑϑ − νσφφ)

εφφ =1

E(σφφ − νσϑϑ)

54

Consider as an example the case of a hemispherical cement dome with the following data:

q = 1.725KN/m2

r∗ = 35m

t = 75 · 10−3m

E = 20GPa

σϑcomp = 21MPa

The σφ is always compression, while the σϑ may either be compression, when φ < 51o 50′ or

traction, when φ > 51o 50′. The maximum value of compression is given by:

σφcomp =qr∗

t= 23 · 103 · 35 = 0.805MPa

As can be seen the maximum value of compression is much smaller than that shown for the

material. However, as regards the tensile behavior, this is determined from σϑ that has the

same value of the compressive stress.

The maximum tensile strength is known by the specific characteristics of the material and, in

the case of cement, it can be very limited. Regarding the compression, the critical load could

be estimated as limit case by the formula:

σcr = KEt

r∗

where the coefficient K can be put K = 0.25 in order to obtain for this example:

σcr = 0.25 · 20 · 109 · 75 · 10−3

35= 10.71MPa

also in this case is σφcomp ≪ σcr.

3.2.5 Open dome at the top

Consider again a dome with a radius r∗ and thickness t, but open at the top for a corner φ0

with a reinforcement ring and a load per unit length, P (ie from Eq. 3.3, F̄ = 2πr∗P sinφ0)

which acts on the ring as shown in Fig.3.8. While still account that, as in the previous case, a

weight force per unit of spherical area acts equal to q =Mg

S, the components of the weight are:

pϑ = 0

pφ = q sinφ

pz = q cosφ

55

Figure 3.8: Open dome at the top

One has, by applying l’Eq. 3.3:

Nφ(φ) =−1

r∗ sin2 φ

∫ φ

φ0

r∗2q(cos2 φ̄+ sin2 φ̄

)sin φ̄dφ̄− F̄

2πr∗ sin2 φ

=−r∗q

sin2 φ(cosφ0 − cosφ)− P sinφ0

sin2 φ

Then one get:

Nφ

t= σφ = −qr∗

t

(cosφ0 − cosφ

sin2 φ

)− P

t

sinφ0

sin2 φ

From:

Nφ +Nϑ = −pzr∗

one has (con pz = q cosφ):

Nϑ

t= σϑ =

pr∗

t

(cosφ0 − cosφ

sin2 φ− cosφ

)+

P

t

sinφ0

sin2 φ

One can then calculate the deformation εϑ from:

εϑ =1

E(σϑ − νσφ)

and you can evaluate the deformation of the radius r′ = r∗(1 + εϑ).

56

3.2.6 Pressure Tanks

Many pressure vessels are constructed with a cylindrical body and end caps that are shells of

revolution (can be considered as a special case of semi-spheres or ellipsoids).

In all cases these end caps must withstand an internal pressure p is constant and perpendicular

to the wall. One has

pϑ = 0

pφ = 0

pz = −p

It is interesting to point out that, in all these cases, one can obtain Nφ by Eq. 3.3 as

Nφ(φ) =−1

r2 sin2 φ

[∫ φ

0r1(φ̄)r2(φ̄)(pz cos φ̄+ pφ sin φ̄) sin φ̄dφ̄

]which becomes:

Nφ(φ) =1

r2 sin2 φ

∫ φ

0r1r2p cos φ̄ sin φ̄dφ̄ =

p

r2 sin2 φ

∫ r0

0r̄0dr̄0

In fact:

r2 sinφ = r0

dr0 = ds cosφ ; ds = r1dφ ; dr0 = r1 cosφdφ

Then:

Nφ =p

r2 sin2 φ

r202

=p

2

r22 sin2 φ

sin2 φ=

p

2r2

While from the equilibrium equation:

Nφ

r1+

Nϑ

r2= p

one has:

Nϑ = pr2 −Nφr2r1

therefore:

Nϑ = pr2 −p

2

r22r1

= pr2

(1− r2

2r1

)= pr2

(2r1 − r2)

2r1

57

As a special case for the hemispherical shell, one has

r1 = r2 = r∗

and then:

Nφ =p

2r∗ σφ =

p

2

r∗

t

Nϑ = pr∗(2r∗ − r∗)

2r∗=

p

2r∗ σϑ =

p

2

r∗

t

In the case of an ellipsoidal cap, Fig.3.9, one has:

Figure 3.9: Ellipsoidal cap

r1 =a2b2

(a2 sin2 φ+ b2 cos2 φ)32

r2 =a2

(a2 sin2 φ+ b2 cos2 φ)12

There are two typical situations for φ = 0 (corresponding to the point 0) and φ = π2 (corre-

sponding to the point A,A’): Per φ = 0

r1 =a2

b; r2 =

a2

b

58

then:

Nφ =p

2r2 =

p

2

a2

b

Nϑ = pa2

b

a2

b

2a2

b

=p

2

a2

b

σϑ = σφ =p

2

a2

b

1

t

Per φ = π2

r1 =b2

a; r2 = a

then:

Nφ =p

2a σφ =

p

2

a

t

Nϑ = p · a(1− a · a

2b2

)= p · a

(1− a2

2b2

)σϑ = p · a

(1− a2

2b2

)1

t

Note that Nφ is always positive, while Nϑ can be negative ifa2

2b2> 1. The case of the sphere

goes back to a = b = r∗.

In the case in which is a = 2b = r∗ one has:

Perφ = 0

Nφ =p

2

4b2

b= 2pb = pr∗

Nϑ =p

2

a2

b= pr∗

Per φ =π

2

Nφ =p

2a =

p

2r∗

Nϑ = pr∗(1− 4b2

2b2

)= −pr∗

In the case of a very flattened cap, for example, a = 10b = r∗ one has:

For φ = 0

Nφ =p

2

100b2

b= 50pb = 5pr∗

Nϑ =p

2

100b2

b= 5pr∗

59

For φ = π2

Nφ =p

2r∗

Nϑ = pr∗(1− 100b2

2b2

)= −49pr∗

3.2.7 Toroidal tank with circular section with internal pressure∗

Consider now a toroidal tank of circular cross section, radius of the torus b, subjected to an

internal pressure p. If reference is made to the part of the shell defined by the angle φ as

Figure 3.10: Toroidal tank with circular section

indicated in Fig.3.10, for the balance of vertical forces one has:

2πr0Nφ sinφ = πp(r20 − b2)

whence

Nφ =p(r0 + b)(r0 − b)

2r0 sinφ

But one has the positions:

r0 − b = r∗ sinφ

r0 + b = 2r0 − r∗ sinφ

and then it has:

Nφ =pr∗(r0 + b)

2r0

From:

Nφ

r1+

Nϑ

r2= −pz

60

putting

pz = −p

r1 = r∗

r2 sinφ = r0

is obtained

Nϑ = r2p−r2r∗

Nφ = r2p−r2p(r0 + b)

2r0

namely:

Nϑ = r2p−r2p(2r0 − r∗ sinφ)

r0=

r2 sinφpr∗2r0

=pr∗

2

Finally, the stresses are:

Nϑ

t= σϑ =

pr∗2t

Nφ

t= σφ =

pr∗

2t

r0 + b

r0=

pr∗

2t

(r∗ sinφ+ 2b

r∗ sinφ+ b

)

Consider as an example a toroidal tank of circular cross section built in light alloy with the

following data:

r∗ = 200mm ; p = 5 · 105Pa

b = 1.5m ; t = 1 · 10−3m

One has

σϑ =5 · 105 · 0.22 · 10−3

= 50MPa

σφ =pr∗

2t

(r∗ sinφ+ 2b

r∗ sinφ+ b

)by:

φ = 0o σφ =pr∗

t= 100MPa

φ = 90o σφ =pr∗

2t

(r∗ + 2b

r∗ + b

)= 94MPa

61

If r∗ ≪ b it can be written:

σϑ =pr∗

2t

σφ =pr∗

t∼= 2σϑ

εϑ =1

E(σϑ − νσφ) ∼=

1

E(σϑ − 2νσϑ) =

σϑE

(1− 2ν)

εφ =1

E(σφ − νσϑ) ∼=

1

E(σϑ(2− ν)

Therefore the toroidal tank pressurized behaves in a similar manner to cylindrical tank with a

radius equal to r∗.

3.3 Axial-symmetric shells undergone to not axial-symmetricload

In the more general case the shell of revolution is subjected to a load condition which does not

have axial symmetry; in this case, the unknowns are still Nϑ and Nφ and more cutting forces per

unit length Nφϑ and Nϑφ in the case of thin shell are equal and therefore have three unknowns

which can be determined by imposing the three equilibrium equations according to the axes z,

ϑ, φ. The load per unit area has the three components pz, pϑ, pφ. Fig.3.11 indicates the element

of the shell ABCD with the forces.

If you consider the balance of forces in the direction ϑ it has that the contribution of external

forces is given by the component pϑ for the area element r0dϑr1dφ; it has then the contribution

due to variation according to ϑ di Nϑ that results∂Nϑ

∂ϑr1dϑdφ; it then has the component due

to shear forces acting on the sides AB and CD of the element of the shell which form an angle

dϑ and then have the resulting indicated that the direction tangent to a parallel; finally, one

must consider the contribution of the cutting forces acting on the sides AC and BD which is

given by:

Nϑφdr0dφ

dφdϑ+∂Nϑφ

∂φr0dφdϑ =

∂

∂φ(r0Nϑφ) dϑdφ

the balance equation in ϑ direction results:

Nϑφr1 cosφdφdϑ+∂Nϑ

∂ϑr1dφdϑ+

∂

∂φ(r0Nϑφ) dϑdφ+ pϑr0r1dϑdφ

which finally becomes:

∂

∂φ(r0Nϑφ) +

∂Nϑ

∂ϑr1 +Nϑφr1 cosφ+ pϑr0r1 = 0 (3.4)

62

Figure 3.11: Shell of revolution with not axisymmetric load

You must proceed in a similar manner with regard to the balance equation according with φ

in which compared to the case of axial symmetry must consider the term∂Nϑφ

∂ϑr1dϑdφ which

derives from the difference of the cutting forces acting on the sides AB and CD of the element

of the shell and is obtained:

∂

∂φ(r0Nφ) +

∂Nϑφ

∂ϑr1 −Nϑr1 cosφ+ pφr0r1 = 0 (3.5)

Finally, as regards the third equation of equilibrium in the z-direction is observed that the

contribution of shear forces is zero and the equation is then seen previously in the axisymmetric

case:

Nφ

r1+

Nϑ

r2= −pz (3.6)

We thus have the three equilibrium equations that allow the calculation of forces in membrane

Nϑ, Nφ, Nφϑ in the case of a shell of revolution with a not axisymmetri load.

63

3.3.1 Cylindrical shell of circular section

It is now considered a particular case from geometrical point of view: the cylinder of circular

section. An element of this type of shell is shown in Fig.3.12 and is constituted by two genera-

trices and two planes normal to the cylinder axis away dx; considering a load px,pϑ, pz you have

as unknowns the forces per length unit Nx, Nϑ, Nxϑ and their increments.

Figure 3.12: Cylindrical shell of circular section

Considering the force balance in the x direction, one has

pxdxrdϑ+∂Nx

∂xdxrdϑ+

∂Nϑx

∂ϑdxdϑ = 0

For the balance according to ϑ one obtains:

pϑdxrdϑ+∂Nϑ

∂ϑdxdϑ+

∂Nxϑ

∂xdxrdϑ = 0

And finally, according to the balance along z one has:

pzrdxdϑ+Nϑdxdϑ = 0

64

This yields the system of equations:

Nϑ = −pzr∗

∂Nxϑ

∂x+

1

r∗∂Nϑ

∂ϑ= −pϑ

∂Nx

∂x+

1

r∗∂Nxϑ

∂ϑ= −px

We note that the equations of this system can be solved in sequence starting from the first;

in fact, once assigned to the external load, one can derive immediately in finite terms the first

unknown Nϑ and then the other two unknowns for integration by the relations:

Nxϑ = −∫ x

x0

(pϑ +

1

r∗∂Nϑ

∂ϑ

)dx̄+ f1(ϑ)

Nx = −∫ x

x0

(px +

1

r∗∂Nxϑ

∂ϑ

)dx̄+ f2(ϑ)

where the functions f1(ϑ) e f2(ϑ) should be evaluated depending on the boundary conditions of

the problem.

3.3.1.1 Cylindrical tank

Consider a cylindrical tank with a constant section of radius r∗ supported at the ends, but free

to move in the x direction, with a length l as indicated in Fig.3.13. The tank is completely filled

with a liquid of specific weight γ = ρg (ρ is the mass density and g the gravity acceleration) the

pressure of which depends on the angle of position indicated with ϑ.

Figure 3.13: cylindrical duct

The pressure in the tank is determined by the law which considers the hydrostatic pressure equal

to the weight of the liquid column at the point considered; in the case of the tank, the height of

65

the column varies from zero, corresponding to ϑ = 0, up to 2r∗, corresponding to ϑ = π. The

external forces per unit area, acting on the tank are then:

pz = −γr∗(1− cosϑ)

pϑ = 0 (3.7)

px = 0

dove γ := ϱg.From the first of 3.7 one has:

Nϑ = −pzr∗ = γr∗

2(1− cosϑ)

By using the expression of Nϑ one has:

1

r∗∂Nϑ

∂ϑ=

1

r∗γr∗

2sinϑ = γr∗ sinϑ

and the second of the equations 3.7 becomes:

Nxϑ = −∫ x

0γr∗ sinϑdx̄+ f1(ϑ) = −γr∗x sinϑ+ f1(ϑ) (3.8)

By using the expression of Nxϑ one has:

1

r∗∂Nxϑ

∂ϑ= −γr∗x cosϑ

r∗+

1

r∗df1dϑ

Then the third equation 3.7 becomes:

Nx(x, ϑ) =γx2

2cosϑ− x

r∗df1dϑ

+ f2(ϑ)

The condition of the tank boundary that leaves freedom of movement in the x direction is

translated into:

x = ± l

2Nx = 0

γl2

8cosϑ− l

2r∗df1dϑ

+ f2(ϑ) = 0

γl2

8cosϑ+

l

2r∗df1dϑ

+ f2(ϑ) = 0

Summing and subtracting these two equations are obtained

f2(ϑ) = −γl2

8cosϑ

df1(ϑ)

dϑ= 0

66

From the second of these equations is derived:

f1(ϑ) = cost = c

It is observed from the equation 3.8 that the function f1(ϑ),and then the constant c, represents

the value of Nxϑ in the central point of the tank, but since it is not present on a load of torsion,

this shear force must be zero and therefore the constant c must be zero. The solution of the

problem is then given by the equations:

Nϑ = γr∗2(1− cosϑ) (3.9)

Nxϑ = −γr∗x sinϑ

Nx = γcosϑ

2

(x2 − l2

4

)By constitutive equations in the case of an isotropic material, the deformation is obtained:

εxx =1

E(σxx − νσϑϑ)

which can be written in terms of forces per unit length as:

εxx =1

Et(Nx − νNϑ)

From the integration of the kinematic equation is then obtained:

u(x) =1

Et

∫ x

0(Nx − νNϑ) dx̄ =

1

Et

∫ x

0

[γcosϑ

2

(x̄2 − l2

4

)− νγr∗

2(1− cosϑ)

]dx̄

while stresses are to be determined by the equations 3.9:

σϑϑ =γr∗

2

t(1− cosϑ)

σxx = γcosϑ

2t

(x2 − l2

4

)σxϑ = −γ

r∗x

tsinϑ

from which:

σϑϑmax =2γr∗

2

t(ϑ = π)

σxxmax = γl2

8t(x = 0 , ϑ = π)

σxϑmax = −γr∗l

2t

(x =

l

2, ϑ =

π

2

)

67

3.3.2 Wind load on a spherical shell of revolution∗

In order to consider the load conditions, the pressure on one side of the shell and of depression

on the opposite side, only the normal component pz is taken into account although, in reality,

there could be also friction contribution. Assuming that the wind acting in the direction of the

meridian plane ϑ = 0 the components of the pressure due to the wind are:

pϑ = 0 ; pφ = 0

pz = p sinφ cosϑ

where p is indicated by the static pressure of the wind. In the case of a spherical shell under

the action of the wind, one has:

r1 = r2 = r∗

r0 = r∗ sinφ

The equations 3.4, 3.5 e 3.6 become in the lodged loading conditions:

∂

∂φ(r0Nϑφ) +

∂Nϑ

∂ϑr1 +Nϑφr1 cosφ = 0

∂

∂φ(r0Nφ) +

∂Nϑφ

∂ϑr1 −Nϑr1 cosφ = 0

Nϑ

r1+

Nϑ

r2= −p sinφ cosϑ

With the use of the third of these equations that can be written as:

Nφr0 +Nϑr1 sinφ = −pr0r1sinφ cosϑ

located in the first two equations, we can eliminate Nϑ from them and the obtained system is:

∂Nφ

∂φ+

(1

r0

dr0dφ

+ cotφ

)Nφ +

r1r0

∂Nϑφ

∂ϑ= −pr1 cosϑ cosφ

∂Nϑφ

∂φ+

(1

r0

dr0dφ

+r1r2

cotφ

)Nϑφ − 1

sinφ

∂Nφ

∂ϑ= −pr∗ sinϑ

Nϑ = −pr0 cosϑ− Nφr0r1 sinφ

These equations allow to derive the membrane forces in a shell of revolution that is subject to

wind load.

68