MODELLING OF MECHANICAL SYSTEM VOLUME 2 Episode 11 potx

Arches and shells: string and membrane forces 381then, δWp = paδq2 2π 0 2π 0R+ a sin ϕ dϕ dθ + paδq1 2π 0 2π 0R+ a sin ϕ sin ϕ dϕ dθ The factor 1/2 present in the generalized force a

subsection 8.3.2.4, bending is highly confined in the vicinity of the interface andoutside this zone of accommodation the radial displacements are very close to theasymptotic values given here.

7.3.7.3 Pressurized toroidal shell

As sketched in Figure 7.16, we consider a toroidal shell of circular cross-section,loaded by an internal pressure p, assumed to be uniform An approximate solutionmay be reasonably proposed, according to which the displacement field is thesuperposition of two distinct radial dilatations, one concerning the cross sectional(meridian) circles (radius a) and the other the parallel circles of radius lying between

R− a and R + a The resulting displacement field is thus written as
ξ = q1
i+q2
n.Furthermore, if the aspect ratio R/a of the torus is sufficiently large it can beassumed that q1and q2are essentially constant The problem can be then solved

by using the Rayleigh–Ritz or Galerkin procedure, based again on the principle

of minimum potential energy However, as the trial function is constant in thelocal frame
i,
n, it may be found more expedient to write directly the equilibriumequations in terms of q1and q2 Both methods are successively worked out below.First, the metrics of the shell is given by:

ds2= g2θdθ2+ gϕ2dϕ2= r2dθ2+ a2dϕ2

gθ = r = R + a sin ϕ, gϕ= a

Figure 7.16 Toroidal shell

Arches and shells: string and membrane forces 379

The principal curvature radii are:

1

Rϕ = dsdϕ

m = 1a ⇒ Rϕ = a1

Rθ = sin ϕr ⇒ Rθ =R+ a sin ϕsin ϕ

The displacement field is:



The equilibrium equations follow:

If the terms dependent on a/R are neglected, the stresses become independent

of ϕ The corresponding approximate values of Nϕϕ, Nθ θ are:

Nϕϕ = Nθ θ ∼ Eh

1− ν2

q2a

(1+ ν) = 1Eh

− ν

q2a



where q2

a = qR1 ⇒ Rq2= aq1whence the approximate results:

q2∼a2p(1− ν)

EhTurning now to the more refined Galerkin procedure, the Lagrangian of the systemtakes the form,

L(q1, q2)= −Es+ WpIts variation gives the equilibrium equations:

δL(q1, q2)= −δEs + δWp = 0First, the variation of the pressure work density is determined:

δwp= −p
n · δ
ξ dS ⇒ δwp= p(δq2− δq1
i ·
n) dS

dS= ra dθ dϕ, r = R + a sin ϕ,
i ·
n = − sin ϕ

Arches and shells: string and membrane forces 381

then,

δWp = paδq2

 2π 0

 2π 0(R+ a sin ϕ) dϕ dθ

+ paδq1

 2π 0

 2π 0(R+ a sin ϕ) sin ϕ dϕ dθ

The factor 1/2 present in the generalized force associated with the displacement

¯q1is associated with the term asinϕ in the curvature radius formula This term hasbeen neglected in the calculus made just above Then, the strain energy is deducedfrom the stress and the strain expressions,

ηϕϕ = ¯q2, ηθ θ ∼= ¯q1

Nϕϕ = 1Eh

− ν2(¯q1+ ν ¯q2), Nθ θ = 1Eh

− ν2(¯q2+ ν ¯q1)which give:

δes = aR1Eh

− ν2(¯q1δ¯q1+ ν( ¯q2δ¯q1+ ¯q1δ¯q2+ ¯q2δ¯q2))

 2π 0

 2π 0

dϕ dθ

= 4π2aR Eh

1− v2(¯q1δ¯q1+ v( ¯q2δ¯q1+ ¯q1δ¯q2+ ¯q2δ¯q2))Finally the equilibrium equations are:

Eh

1− ν2(¯q1+ ν ¯q2)= ap2Eh

1− ν2(¯q2+ ν ¯q1)= apleading immediately to the following approximate solution:

q1= 12 − ν paREh ; q2=1−ν2pa

2Eh

These results are more accurate than those obtained by the direct approximatesolution.

7.3.7.4 Spherical cap loaded by its own weight

The problem is sketched in Figure 7.17 In local coordinates the loading isgiven by,

fς = −ρhg cos ϕ; fθ = 0; fϕ = ρhg sin ϕBecause of the symmetries, the solution is independent of θ and V = 0 The localequilibrium is then described by,

(Nθ θ − Nϕϕ)cos ϕ− sin ϕdNdϕϕϕ = ρhgR(sin ϕ)2

Nθ θ+ Nϕϕ = −ρhgR cos ϕand Nϕϕverifies:

2 cos ϕNϕϕ+ sin ϕdNdϕϕϕ = −ρhgRwhich can be integrated (though integration is not obvious) to give:

Arches and shells: string and membrane forces 383

by two horizontal planes at z and z+ dz is readily found to be 2πR sin ϕR dϕ; thecorresponding weight is−ρh
g2πR sin ϕR dϕ So, the total weight is:

P = −ρh
g

 ϕ 02πR sin ϕR dϕ= −2πR2ρhg(1− cos ϕ)

as it must be balanced by the vertical component of support reactions, we get:

2π R2ρhg(1− cos ϕ) = −2πR(sin ϕ)2Nϕϕ(ϕ) ⇒

Nϕϕ(ϕ)= −ρhgR1(sin ϕ)− cos ϕ2 =1−ρhgR

+ cos ϕThe normal stresses are plotted in Figure 7.18 versus ϕ Nϕϕ is negative, hencecompressive, whatever the ϕ value may be However, Nθ θ becomes tensile in therange ϕ > 52◦ This less intuitive result caused many flaws and even failures inmasonry domes because masonry resists compressive stresses much better thantensile stresses For instance, the dome of St Peter basilica in Rome was foundlargely cracked during the seventeenth century The empirical solution implemented

by the architects – which was perfectly satisfactory – has been to reinforce the domeperimeter support by a masonry ring [COT 90]

Once the stresses are known, it is possible to calculate the strains and the placements The analytical solution is not very simple and requires one to assumespecific boundary conditions The interest of such analytical solutions, which holdfor particular boundary conditions, is to obtain approximate solutions for various

dis-Figure 7.18 Normal stresses plotted versus the ϕ angle

actual boundary conditions which are not too far from the assumed ones, providedSaint-Venant’s principle can be invoked This point is illustrated below.

If the material is linear elastic, the strains in the spherical dome are:

ηθ θ = E1(Nθ θ− νNϕϕ)=WR cot ϕ+UR

ηϕϕ = E1(Nϕϕ− νNθ θ)=R1 dWdϕ +UREliminating U gives,

This equation has the general form W cot ϕ− dW /dϕ = f (ϕ) and the solution is

W = sin ϕ' f (ϕ)/sin ϕ dϕ + Cste

2(1+ ν)

Log 1+ cos ϕ

2(1+ cos ϕ)− cos ϕ

+ W cotgϕ

In Figures 7.19, the reduced value of U is plotted as a function of ϕ for anhemisphere and a dome defined by a 3π/4 base angle The tangential displacementalong the meridians is set to zero but the other components are left free, which is

Figure 7.19 Analytical displacements of two spherical caps

Arches and shells: string and membrane forces 385

unrealistic However, these plots show clearly that the radius of the parallel circles

is reduced in the upper zone ϕ > 52◦thus inducing a compressive hoop stress andenlarged in the lower zone, thus inducing a tensile hoop stress

note –Finite element solution

A finite element solution which accounts for both the membrane and bendingeffects shows that a clamped boundary condition does not deeply modify the ana-lytical solution except in the vicinity of the clamping In Figures 7.20a, 7.20b and7.21a, 7.21b, the deformed shapes of a masonry dome with the base either on slid-ing or clamped support can be compared The radius is 20 m, the thickness 10 cmand the weight 400 tons To make the deformed shapes clearly visible, the actualdisplacements have been multiplied by the factor 20000 The left-hand side plotshows the deflection of a meridian and the right-hand side plot shows that of an

Figure 7.20a Hemispherical cap provided with sliding supports at the base

Figure 7.20b Membrane stresses Nϕϕ, Nθ θ

Figure 7.21a Hemispherical cap clamped at the base

Figure 7.21b Membrane stresses Nϕϕ, Nθ θ

angular sector Computed deflection in the case of a sliding support is found to bevery close to the analytical solution which discards bending effects, except at theimmediate vicinity of the base, see Figure 7.20a This close agreement also holds sofar as the membrane stresses are concerned, except near the top and the base of thedome, where bending stresses are present, see Figure 7.20b As could be expected,the effect of clamping the dome at the base is to increase further the importance ofbending near the base which becomes significant over a much larger zone than inthe case of a sliding base, see Figure 7.21a Outside this perturbed zone, the domeresponse remains essentially the same in both support configurations

7.3.7.5 Conical shell of revolution loaded by its own weight

The structure is shown in Figure 7.22 The shell is shaped as a conical frustum of

revolution around the axis Oz; the cone half angle is denoted ψ, the shell thickness

Arches and shells: string and membrane forces 387

Figure 7.22 Truncated cone loaded by its own weight

is h and p= ρgh is the weight per unit area of the shell The bottom of the frustum

is supported in the Oz direction We are interested in determining the membrane

stresses

The position of a current point is defined by the two parameters θ and ξ = OP The top and the bottom of the cone are defined by ξ= ξ0and ξ = ξ1respectively.The coordinates of a point are:

r = ξ sin ψ; x = ξ sin ψ cos θ; y = ξ sin ψ cos θ; z = ξ cos ψ

So, the coefficients of the metrics are given by:

(ds)2=(dx)2+ (dy)2+ (dz)2

(ds)2=(sin ψ cos θ )2+ (sin ψ sin θ)2+ (cos ψ)2 (dξ )2

+(sin θ )2+ (cos θ)2 (ξsin ψdθ )2(ds)2=(dξ)2+ (ξ sin ψdθ)2

then,

gξ = 1; gθ = ξ sin ψThe principal curvatures are given by:

The meridian is a straight line, so its curvature is zero It has to be stressed that theparallel circles are not the principal curvature lines; the latter are defined by theintersection of the frustum surface by planes which are orthogonal to the meridianlines As determined just above, their curvature radius is Rθ = ξtgψ The equilib-rium equations are obtained from [7.39] Because the problem does not depend on

θthey can be written as:

−Nξ ξ

ξ −∂ Nξ ξ

∂ ξ = p cos ψ + p sin ψ tgψ = cos ψp ⇒ ∂(ξ Nξ ξ)

∂ ξ = −cos ψp ξIntegration is also immediate and the top base being assumed to be free,

we get:

Nξ ξ = −p(ξ

2− ξ2

0)2ξ cos ψAccordingly, the shell is found to be everywhere in a compressive state, both inthe meridian and the circumferential directions As ψ → 0 the structure tends to acylindrical shell, so Nθ θ → 0, as suitable In contrast, if ψ → π/2 the structuretends to a plate loaded by its own weight (bending only) and the membrane solution

is obviously meaningless Finally, it can be noted that, as for the dome, Nξ ξcan becalculated by balancing directly the meridian stresses and the weight

7.3.7.6 Conical container

A container full of liquid is shown in Figure 7.23 It is uniformly supported atits top z= H = L cos ψ and full of liquid whose density is ρ; the cone half-angle

is ψ The equilibrium equations are:

−ξsin ψ1  ∂(ξsin ψN∂ξ ξ ξ)− Nθ θsin ψ



= fξ = 0

−ξsin ψ1

%1

Arches and shells: string and membrane forces 389

Figure 7.23 Conical container full of liquid

From the second equation, ξ2Nθ ξ = a ⇒ Nθ ξ = a/ξ2

At the top, the stresses must verify the condition Nθ ξ(H )= 0 ⇒ a = 0 ⇒

Nθ ξ ≡ 0 everywhere within the shell, as could be anticipated

The meridian stress is given by:

+ b,

Nξ ξ = ρg sin ψξ 2 −ξ

23

+bξ

The constant b must be zero because the stress must remain finite at the apex, that

is as ξ → 0 So we get:

Nξ ξ = ρg sin ψ 2 −ξ

23



Finally, the stress maxima are found to be:

Nθ θ = ρgξ(L − ξ) sin ψ ⇒ (Nθ θ)max= Nθ θ(L/2)= ρgL

2sin ψ4

Nξ ξ = ρg sin ψ 2 −ξ

23



⇒ (Nξ ξ)max= Nξ ξ(3L/4)= 3ρgL

2sin ψ16Max Nθ θ

Nξ ξ



=43Thus the shell is in a compressive state everywhere, both in the meridian and thecircumferential directions On the other hand, here again, the hoop stresses arefound to be larger than the meridian stresses, though by about 30% only

Chapter 8

Bent and twisted arches and shells

To deal with general loading conditions, it is necessary to include bending andtorsion into the equilibrium equations of arches and shells; which leads to inter-esting coupling effects between various elementary modes of deformation In thecase of arches, in-plane bending is found to be coupled with tangential stretchingand out-of-plane bending with torsion In the case of shells, all the elementarymodes of deformation are found to be coupled together, except in a few partic-ular loading cases The problem of shell vibrations was first attacked by SophieGermain in the early nineteenth century However, the basic development of thethin shell theory is due to Love in 1888 As already indicated in the precedingchapter, Love’s model is based on simplifying assumptions which extend in a nat-ural manner those already used to model straight beams and plates The thin shelltheory was the object of various refinements during the twentieth century As aresult, there exists a wide variety of shell equations However, all of them arebasically of the Love type, differing only by the approximations made to deal withthe metric coefficients Gα, Gβ Furthermore, such differences turn out to be oflittle practical importance Therefore, presentation is restricted here to the Lovemodel Particularization to cylindrical shells of revolution gives us the oppor-tunity to discuss a few problems of practical interest and the validity of varioussimplifications of Love’s equations, in relation to the specificities of the loadingconsidered

8.1 Arches and circular rings

The relations [7.4] to [7.9] of Chapter 7 are extended to describe the motionsout or in the arch plane, see Figure 8.1 For this purpose, a local frame defining adirect Cartesian coordinate system is needed The unit vectors are
t, tangent to theneutral fibre,
n1normal to the neutral fibre, in the arch plane and pointing towardsthe extrados, and
n2=
t ×
n1 The local displacement vector is written as:

curvi-Gs =1+ςR1; G1= G2= 1 [8.4]

Figure 8.1 Arch: local frame and global displacements

Bent and twisted arches and shells 393

8.1.2 Tensor of small local strains

The results [8.2] and [8.4] are substituted into the general expression [5.87]

of the strain tensor Here, the non-identically zero components are firstwritten as:

8.1.3 Pure bending in the arch plane

8.1.3.1 Equilibrium equations

It is possible to formulate the equilibrium equations of an arch by taking intoaccount all the terms present in the strains [8.7] Nevertheless, it is more convenientand instructive to start by investigating each basic mode of deformation separately,namely the in-plane flexure, the out-of-plane flexure and the torsion The modelsderived by this manner have the advantage of simplicity and can be used as aguideline for formulating afterwards more refined coupled models

Let us consider first the in-plane flexure The relevant components of the globaldisplacement field are Xs, X1, ψ2 Therefore, the local strain field is reduced to:

∂s − ς1

∂ψ2

∂s +X1R

Bent and twisted arches and shells 395

The bending strain [8.12] differs from that obtained in straight beams by anadditional term due to the finite curvature of the neutral fibre in the non-deformedstate It is also important to be aware that the arch axial displacement Xs does notvanish The equation of equilibrium and the boundary conditions can now be easilyestablished The variation of kinetic energy is:

δ[Eκ] =

 s2

s1ρS( ˙Xsδ ˙Xs+ ˙X1δ ˙X1) ds [8.13]and the variation of strain energy is:

in which M2designates the bending moment about the flexure axis
n2

The equation of transverse equilibrium and the related boundary conditions arefound to be:

The tangential equation reduces to:

ρS ¨Xs = F(e)

The left-hand side of [8.16] comprises an inertial term only since no axial globalstress results according to the ‘pure bending’ model Finally, if the arch material islinear elastic, the stresses are found to be:

The equation of transverse vibration is:

Equation [8.18] differs from the straight beam case by two additional ness terms which arise as a consequence of the finite curvature of the neutralfibre

stiff-8.1.3.2 Vibration modes of a circular ring

In agreement with the usual notations used for structures of revolution, thedisplacement field and the curvilinear coordinate are rewritten as:

Xs = V ; X1= U; X2= W ; ds = R dθ [8.19]

R being the radius of the ring neutral fibre For mathematical convenience, ringcross-sections are assumed to be circular; their radius being denoted a Theequations of free vibration are:

If the ring is free to move in its own plane, the sole condition to be fulfilled by

Uand V is a 2π periodicity Therefore, it is natural to search for modal solutionssuch as Un(θ; t)= cos(nθ)eiωntor Un(θ; t)= sin(nθ)eiωnt, or more generally anylinear superposition of these two orthogonal families Adopting the cosine family

is equivalent to setting the axis θ= 0 at an antinode of vibration According to thischoice, the first equation [8.10] implies necessarily the following mode shapes:

Un(1)= ancos(nθ ); Vn(1)= −annsin(nθ )

Un(2)= ansin(nθ ); Vn(2)= +ann cos(nθ )

[8.21]

Again, anis an arbitrary constant used to normalize the mode shapes Hereafter it

is set to one Substituting [8.21] into the radial equation [8.20] leads to:

−ω2nρS+EIR4(n2− 1)2



Bent and twisted arches and shells 397

From [8.22] the natural frequencies of the in-plane bending modes are given by:

fn =n

2

− 12π R2

[8.23]

ηis interpreted as a slenderness ratio For a ring of circular cross-section of radius

ait is readily found that η = 2R/a It is noticed that the modes n = 1 refer to afree in-plane translation of the ring, as already discussed in Chapter 7 section 7.5

In Figure 8.2 two mode shapes of the first family (antinode at θ = 0) are plottedand the corresponding frequencies are specified for a steel ring R= 1m, a = 1cm.Full and dashed dotted heavy lines refer to two vibration patterns separated by half

a period It can be verified that the radial and tangential nodes are not located atthe same angular positions

The ratio between the frequencies [8.23] and the breathing mode frequency(formula [7.23]) is:

fn

As η is usually much larger than one, the result [8.24] indicates that the bendingstiffness of the ring is much less than the axial one On the other hand, the tangentialequation produces another rigid body mode It is immediately identified as a freerotation around the ring axis

note –Correction of the natural frequencies due to the tangential vibration

Figure 8.2 Natural modes of vibration of a circular ring: in-plane bending