# On Galileo’s Tallest Column

Authors: Mariano Vázquez Espí; Jaime Cervera Bravo; Carlos Olmedo Rojas
Journal: Mathematical Problems in Engineering
 (2015)

---

## Abstract

The height at which an unloaded column will fail under its own weight was calculated for first time by Galileo for cylindrical columns. Galileo questioned himself if there exists a shape function for the cross section of the column with which it can attain a greater height than the cylindrical column. The problem is not solved since then, although the definition of the so named “constant maximum strength” solids seems to give an affirmative answer to Galileo’s question, in the form of shapes which seem to attain infinite height, even when loaded with a useful load at the top. The main contribution of this work is to show that Galileo’s problem is (i) an important problem for structural design theory of buildings and other structures, (ii) not solved by the time being in any sense, and (iii) an interesting problem for mathematicians involved in related but very different problems (as Euler’s tallest column). A contemporary formulation of the problem is included as a result of a research on the subject.


---

## Body

## 1. Introduction

In 1638, in hisDiscorsi e Dimostrazioni Matematiche [1], Galileo postulated the existence of the tallest column, that is, a cylindrical column, such that it attains the maximum height once the area of its cross section and the strength of its material are prescribed. Therefore, Galileo’s tallest column is in the limit of resistance only bearing its own weight. The rationale of his proof gave rise to the square-cube law, a mathematical principle that considers the relationship between the flow through the surface of a volume and the stock into the latter, in the mechanical case, for example, the stress with the weight. This principle has been very useful and applied in a variety of scientific fields, mainly in biology [2–4].In the following century, Euler [5] pointed out a very different problem, that is, to find the shape of a stable column, axially symmetric with respect to the vertical axis, such that it attains the maximum height once its volume, specific weight, and Young’s module are given, buckling due to the action of a load bear at its top.In our view, both problems are not yet completely solved nowadays, although Galileo’s problem has received much less attention than Euler’s. Furthermore, we think that Galileo’s problem is more meaningful for a theory of design of structures subjected to small limit on strains and displacements, as buildings and other structures in civil engineering [6, 7].Hereafter, we consider the solid continuum with the following standard assumptions:(1)
The scope of the analysis is the classical theory of Elasticity.
(2)
The process of deformation is isothermal and quasi-static; heat or kinetic energy is not taken in consideration into energy balancing.
(3)
We are only interested in solutions whose strains and displacements will be very small; hence equilibrium and compatibility equations approximately hold in the geometry of the undeformed body.Section2 outlines important aspects of Galileo’s problem, comparing it to Euler’s and enlightening its importance and profound meaning for a theory of structural design. Our main working hypothesis is formulated there. Section 3 deals with some clues that support our working hypothesis; that is, an insurmountable size exists for a fairly large set of structural problems, as it is the case of cylindrical columns of Galileo’s first insight into the tallest column. This section covers the main aim of this paper which is to attract mathematicians to work out Galileo’s problem, because we are architects and our mathematical knowledge is, to say the least, limited, but to solve the problem is a key point to continue the development of structural design theory. Finally, Section 4 is devoted to formulate the problem formally in contemporary terms.


## 2. Galileo’s Problem on the Tallest Column

Proposition  VII. Among heavy prisms and cylinders of similar figure, there is one and only one which under the stress of its own weight lies just on the limit between breaking and not breaking so that every larger one is unable to carry the load of its own weight and breaks while every smaller one is able to withstand some additional force tending to break it. (Galileo, 1638)Consider a cylindrical column of heightL and diameter d, of a lineal elastic material defined by Young’s Modulus E, allowable compressive stress f
, and specific weight ρ and subjected to the sole action of its own weight.Such a column will be unsafe in simple compression if the applied loadρ
L
A exceeds the column strength A
f, where A is the area of the cross section—if the base is a circle, A
=
1
/
4
π
d
2—or, which is the same, if the applied stress ρ
L exceeds f. This fact means that the height of such a column may not be greater than a characteristic length of the material, f
÷
ρ. We name this length “structural scope” of the material, A
=
f
÷
ρ. And we name “structural scope” L of cylindrical columns to the maximum height of safe columns. In this simple case, L is numerically equal to the material scope A, but generally L is related to A but is not equal to [6]. Therefore the first conclusion of Galileo can be expressed as(1)
L
≤
L
=
f
ρ
=
A
.Later, Galileo considers in which way this insurmountable limit can be increased. He envisaged two main ways: (a) to increase the material scope A, or (b) to change the shape of the column. In the latter case, he reasons—in a funny paragraph—that if the giants exist, they would have a very different aspect and proportions compared to human beings; specifically the bones of their legs would have a greater diameter/length ratio, because otherwise their weight that increases in proportion to L
3 would be greater in proportion to their strength that increases as L
2, and as a result the giants would suffer stresses—which increase as L—greater than human beings, and considering the bone material very similar in all the mammals, the giants would be unable to perform in their life as well as human beings. A few centuries later, this result could be confirmed comparing dinosaurs of different sizes but of same suborder or family (i.e.,Theropoda orTyrannosauridae) [8].As a material with infinite strength or null specific weight does not exist, it is clear that following first Galileo’s way we can only increase the insurmountable height but remaining finite. If we adopt the latter way, the main question arises: does there exist an optimal shape which has infinite height? We are looking for an answer to this question because it is a key into the theory of structural design. If the answer is “Yes,” then Galileo’s problem has a solution for any size considered. But if the answer is “No,” there exist instances of structural problems which have no solution; that is, there are unsolvable problems in structural design. Furthermore, as we will show below, near the unsurmountable size, any solution for the problem will have an unaffordable physical cost, so it would be infeasible from a practical view.Our working hypothesis is that a finite insurmountable size exists for a fairly large set of structural problems (not only for Galileo’s problem). Moreover, the optimal shape for each problem—which maximises the finite insurmountable size—is a sound reference to measure the efficiency of all other shapes with size lesser than the one of the optimal shape [7, 9, 10].In a first approximation, we can represent the physical cost of a structure by its self-weight, as many costs during the manufacturing, but not all, are approximately proportional to the self-weight of the structure: CO2 emissions, mineral resources consumption, and so forth. For a given structural problem, we define the structural efficiency r as the ratio between the useful load and the total load (i.e., the useful load plus the self-weight) required to solve that problem in a particular structure. Galileo postulated also the relationship between the size of a structure and its ability to resist a useful load. Let us consider a cylindrical column of size L
<
L. It can resist an additional useful load Q, the value of which is at most the weight difference between this column and the column of insurmountable height L. Hence, according to the previous definition, the efficiency r of such a column will be(2)
r
=
L
-
L
L
=
1
-
L
L
.Note that (2), which we name Galileo’s rule, is exact in the case of cylindrical columns, but it is not proved that it would be a general rule. The best result we get up to date is that Galileo’s rule is a very good estimate in canonical problems like bending of beams and bridges [10]. We define the load cost C as the inverse of efficiency, hence always higher than unity, C
=
1
/
r. Then, the self-weight of the column is(3)
P
=
C
-
1
Q
.As a reward, Galileo’s rule, apart from the cost, gives us a sound estimate for the self-weight that it is a required datum for the final project but unknown in the preliminary phases of design of large structures.


### 2.1. Comparison between Euler’s and Galileo’s Problems

Remember the cylindrical column of heightL and diameter d. As we saw, such a column will be unsafe if its height L  is equal to or greater than material scope A. But the column may also fail by elastic buckling. According to Landau and Lifshitz’ Course [11], the critical height for buckling is related to the diameter by(4)
L
cr
=
0.792
E
ρ
3
·
d
2
/
3
.The ratio E
÷
ρ is another characteristic length of the structural material. Whereas the scope A is its specific strength, E
=
E
÷
ρ is its specific stiffness. Let us define the geometrical slenderness of the column as the ratio λ
=
L
÷
d. Then(5)
λ
cr
=
0.792
E
d
3
,
λ
cr
d
3
=
0.792
E
3
.Therefore, the safety of a given column bearing only its own weight requires the fact that two conditions hold: (i) L
≤
A and (ii) λ
≤
λ
c
r. It is worth noting that it is always possible to satisfy the second condition, as for each height we can choose d such that λ
≤
λ
c
r. However the first condition is an absolute one, as it only depends on the properties of material. Hence, the height of a safe, cylindrical column would be lesser than or equal to L.This limit, as noted above, only could be modified in two ways: changing material’s properties or changing the shape of the column (or both). The interesting point here is that to answer Galileo’s question we must elucidate if a finite structural scopeL (related to the material scope A) exists for any shape of the column. The advances in the analysis of the solution for Euler’s problem [12–14] are useless to this aim, mainly because of ignoring the limit that strength condition imposes on the shape (condition (i) for cylindrical columns).Furthermore, as it is well-known, the classical solutions of Euler’s problems on buckling are contradicting the experimental data. “This contradiction between theory and experiments is not surprising. The ideal appearance of a phenomenon is always more or less influenced in practice by multiple causes that can deform it to the point of leaving none but a caricature. In the problems of instability, the theory considers only perfect elements, both form and structure indefinitely elastic and resistant. The test pieces, or the members actually built, are very far from perfection; the materials are inhomogeneous, and they are approximately elastic and within certain limits.” [15] As a consequence, in the engineering practice, there is no bifurcation between two different equilibria. On the contrary, as the slenderness λ approaches to its critical value λ
c
r, the failure changes continuously from simple flattening to bending with net compression. In fact, the so named critical load (or Euler’s load) is not a “load” at all but a stiffness of the column against lateral displacement, and the failure occurs for real loads numerically lesser than this stiffness. (Unfortunately, this stiffness can be expressed in load units (N), but it is better understood with stiffness unit (m
N
÷
m), showing that it is a ratio between the bending moment and the lateral displacement.) Moreover, although the buckling of real structural members (with negligible self-weight) is a nonlineal problem in a first, mathematical view, therefore candidate to a numerical solution, it is possible to overcome the difficulties and to solve the problem by a direct albeit nonexact formulation when structural strength is taken into account [16].


## 2.1. Comparison between Euler’s and Galileo’s Problems

Remember the cylindrical column of heightL and diameter d. As we saw, such a column will be unsafe if its height L  is equal to or greater than material scope A. But the column may also fail by elastic buckling. According to Landau and Lifshitz’ Course [11], the critical height for buckling is related to the diameter by(4)
L
cr
=
0.792
E
ρ
3
·
d
2
/
3
.The ratio E
÷
ρ is another characteristic length of the structural material. Whereas the scope A is its specific strength, E
=
E
÷
ρ is its specific stiffness. Let us define the geometrical slenderness of the column as the ratio λ
=
L
÷
d. Then(5)
λ
cr
=
0.792
E
d
3
,
λ
cr
d
3
=
0.792
E
3
.Therefore, the safety of a given column bearing only its own weight requires the fact that two conditions hold: (i) L
≤
A and (ii) λ
≤
λ
c
r. It is worth noting that it is always possible to satisfy the second condition, as for each height we can choose d such that λ
≤
λ
c
r. However the first condition is an absolute one, as it only depends on the properties of material. Hence, the height of a safe, cylindrical column would be lesser than or equal to L.This limit, as noted above, only could be modified in two ways: changing material’s properties or changing the shape of the column (or both). The interesting point here is that to answer Galileo’s question we must elucidate if a finite structural scopeL (related to the material scope A) exists for any shape of the column. The advances in the analysis of the solution for Euler’s problem [12–14] are useless to this aim, mainly because of ignoring the limit that strength condition imposes on the shape (condition (i) for cylindrical columns).Furthermore, as it is well-known, the classical solutions of Euler’s problems on buckling are contradicting the experimental data. “This contradiction between theory and experiments is not surprising. The ideal appearance of a phenomenon is always more or less influenced in practice by multiple causes that can deform it to the point of leaving none but a caricature. In the problems of instability, the theory considers only perfect elements, both form and structure indefinitely elastic and resistant. The test pieces, or the members actually built, are very far from perfection; the materials are inhomogeneous, and they are approximately elastic and within certain limits.” [15] As a consequence, in the engineering practice, there is no bifurcation between two different equilibria. On the contrary, as the slenderness λ approaches to its critical value λ
c
r, the failure changes continuously from simple flattening to bending with net compression. In fact, the so named critical load (or Euler’s load) is not a “load” at all but a stiffness of the column against lateral displacement, and the failure occurs for real loads numerically lesser than this stiffness. (Unfortunately, this stiffness can be expressed in load units (N), but it is better understood with stiffness unit (m
N
÷
m), showing that it is a ratio between the bending moment and the lateral displacement.) Moreover, although the buckling of real structural members (with negligible self-weight) is a nonlineal problem in a first, mathematical view, therefore candidate to a numerical solution, it is possible to overcome the difficulties and to solve the problem by a direct albeit nonexact formulation when structural strength is taken into account [16].


## 3. On the Existence of a Finite Height for Galileo’s Tallest Column

Our epistemological situation confronting Galileo’s problem is analogous to the situation that algorithm designers are confronting when using the well-known Theory of NP-Completeness [17]. We, the structural designers, do not know if a finite insurmountable size exists for the problem at hand; hence we cannot know in advance if our problem is solvable or not. But if we believe that this limit exists, we can manage at least a rough estimate of its value and, armed with this knowledge, take a decision about the solvability of the problem. Indeed, if we know the size limit that different types of structures can reach for our problem, we can evaluate approximately the relative merit of each type and select the most promising one for the actual size of our problem. So the existence (or not) of a finite height for Galileo’s tallest column is a key point for our everyday work.Let us consider the two main approaches to the problem: first that such a limit does not exist because it is easy to find the corresponding shape and second that such a limit probably exists because it is very hard to find out any shape that can overcome a given finite limit on its height.


### 3.1. The Known Solutions with Infinite Height Are Unfeasible

Although Euler’s tallest column and related problems have received very much attention up to date, some researchers while studying that problem spent a few minutes to study, too, problems related with that of Galileo. This is the case of Karihaloo and Hemp [18] that study the “maximum strength design” of structural members. In their approach, all cross sections of a structural member attain the maximum allowable stress for the given material; therefore the solution is also referred to as “constant maximum strength design.”Let us examine with some detail the constant maximum strength design of Karihaloo and Hemp for tension members; see [18] (Section 2.1) and Figure 1. Consider a cable of length L and cross-sectional area A
(
y
), hung on its top edge (y
=
L) and when its bottom edge (y
=
0) is a free boundary. If the gravity and the external load are acting in the negative direction of y-axis, the condition of the constant maximum strength is(6)
f
A
=
∫
0
L
ρ
A
d
y
;
d
A
A
=
d
y
A
.The solution is(7)
A
=
A
0
·
exp
⁡
y
Aas it can be checked obtaining its derivative and comparing the result with (6). If the cross section is circular, the radius r
(
y
) gives us the contour of the member as follows:(8)
r
y
=
r
0
·
exp
⁡
y
A
=
r
0
·
exp
⁡
y
2
A
.This solution can have an infinite height with constant stress and bear a useful load at the bottom edge (y
=
0, Q
=
A
0
f). Anyhow, its volume grows exponentially with its size (A
0
A
·
(
e
x
p
(
L
/
A
)
-
1
)), so in practice it is an “intractable” solution—in the same meaning that the term is used in algorithm complexity theory [17]—with a load cost as follows:(9)
C
L
=
Q
+
ρ
V
L
Q
=
1
+
1
ρ
exp
⁡
L
A
-
1
.Figure 1
Constant maximum strength cable: classical solution.
(a)
3D-solution
(b)
2D-solutionFurthermore, the solution is not feasible from the point of view of equilibrium because only the equilibrium in the directiony is considered for obtaining a constant stress σ
y. Let us consider the 2D-case for the sake of simplicity; see Figure 1(b). As the border A
B (or C
D) is stress-free, the tangent in any point is a principal direction (σ
a), as the orthogonal direction is (σ
b
=
0). As σ
y
=
f, we have(10)
σ
a
=
f
+
σ
x
2
+
f
+
σ
x
2
2
+
τ
2
,
0
=
f
+
σ
x
2
+
f
+
σ
x
2
2
+
τ
2and hence σ
a
=
f
/
c
o
s
2
(
α
) if α is the angle between the principal stress direction and coordinate axes. This value is greater than allowable stress f for all α. In fact, as L increases, cos
⁡
α
→
0 exponentially, and σ
a grows in the same way. As a consequence, the classical solution is not a feasible one for common failure criteria, that is, as Von Mises criterion. This error is common to all solutions obtained making use of Bernoulli-Euler theory or Navier hypothesis, as these models are not applicable in the case of variable cross section beams—a well-known drawback; see, for example, [19]. These solutions do not prove in any way that an infinite size for columns or beams would be feasible. Of course, these solutions are almost exact when the size L is much lesser than the material structural scope A, as the exponential function grows very slowly when its argument is very small, and for small size the solutions obtained pass fairly well through experimental checks [20]. But they are useless to answer the question pointed out by Galileo, because to this issue we must explore sizes of the same order of magnitude that material scope, that is, large structures for which the equilibrium equations must be completely fulfilled.


### 3.2. Trying to Refute the Existence of a Finite Height

In 2010 (unpublished work), we tried to refute the existence of a finite structural scope whatever the structural material of Galileo’s column is. Our try was naive and unsuccessful. However we think it can help others to understand the difficulties of the problem and perhaps gives clues to better searches for a complete solution. To be short, let us consider a 2D-version.We have an elastic linear material as before. We choose Von Mises criterion as the failure one; hence in any point of a feasible body the following expression hold:(11)
σ
c
x
,
y
=
σ
x
2
+
σ
y
2
+
3
τ
2
≤
f
.Let us suppose that we are able to determine a suitable stress field for an instance of Galileo’s problem, defined over all the x
y plane. This field implicitly defined the shape of a body for which (11) holds. The support line must intersect in some point of some solution of(12)
σ
c
x
,
y
=
f
.Let a and b be the principal planes of the stress tensor. Then the curves that can form the stress-free contour of the column will be solutions of one of the following equations:(13)
σ
a
x
,
y
=
0
,
σ
b
x
,
y
=
0depending on if we are looking for a compressive or tensile solution. Furthermore, these curves intersect in some point of solution (12) and also the solution of(14)
σ
a
x
,
y
=
±
f
,
σ
b
x
,
y
=
±
f
.Indeed, if in (13) we select the direction a, we must select now b in (14) and vice versa. This intersection point will be named “base-end vertex” hereafter. Selecting an appropriate set of arcs from solutions of (13), with the additional condition that they define a closed region on the plane together with the support line, we get the shape of a solution for this instance of Galileo’s problem and we can determine its height relative to the material scope; see Figure 2.Figure 2
A feasible Galileo column from a given stress field.If we would be able to explore completely the set of all possible stress fields—and, of course, we are not—with the maximal height obtained we could answer—“Yes” or “No”—Galileo’s question.


#### 3.2.1. Generating Subsets of Stress Fields

To check the possibility of our idea we recall on well-known Airy’s function. Whatever Airy’s stress functionΦ is, for which the biharmonic equation holds, it can be defined as(15)
Φ
=
Re
⁡
Z
¯
Ψ
Z
+
X
Z
,where Ψ and X are analytical functions in C [21]. The function Φ satisfies both compatibility and equilibrium equations, and the displacements can be calculated without integration of the stress functions, from the complex potentials Ψ  and  X.The stress field is(16)
σ
x
=
∂
2
Φ
∂
y
2
-
ρ
y
,
σ
y
=
∂
2
Φ
∂
x
2
-
ρ
y
,
τ
x
y
=
-
∂
2
Φ
∂
x
∂
y
.Using the Hooke law with this field, the displacement functions are(17)
u
x
,
y
=
1
E
-
1
+
ν
∂
Φ
∂
x
+
4
·
Re
⁡
Ψ
x
,
y
-
ρ
1
-
ν
x
y
+
θ
0
y
+
u
0
,
v
x
,
y
=
1
E
-
1
+
ν
∂
Φ
∂
y
+
4
·
Im
⁡
Ψ
x
,
y
+
ρ
2
1
-
ν
x
2
-
y
2
-
θ
0
x
+
v
0
,where ν is Poisson’s modulus.The complex potentials considered were simple polynomials as follows:(18)
Ψ
0
=
A
p
+
i
·
B
p
,
Ψ
1
=
A
p
+
i
·
B
p
+
C
p
+
i
·
D
p
·
Z
,
Ψ
2
=
A
p
+
i
·
B
p
+
C
p
+
i
·
D
p
·
Z
+
E
p
+
i
·
F
p
·
Z
2
,
⋮
=
⋮
Ψ
n
=
P
n
0
+
i
P
n
1
·
Z
n
,
X
0
=
A
c
+
i
·
B
c
,
X
1
=
A
c
+
i
·
B
c
+
C
c
+
i
·
D
c
·
Z
,
X
2
=
A
c
+
i
·
B
c
+
C
c
+
i
·
D
c
·
Z
+
E
c
+
i
·
F
c
·
Z
2
,
⋮
=
⋮
X
n
=
C
n
0
+
i
C
n
1
·
Z
n
.The coordinate origin will be at the top of a column (or at the bottom of a cable). Some boundary conditions must hold always as follows:(19)
symmetry:
σ
x
x
,
y
=
σ
x
-
x
,
y
,
σ
y
x
,
y
=
σ
y
-
x
,
y
,
τ
x
y
x
,
y
=
-
τ
x
y
-
x
,
y
,
origin:
σ
x
0,0
=
σ
y
0,0
=
τ
x
y
0,0
=
0
.We considered also different supports as follows:(20)
base line support:
v
x
,
L
=
0
,
base point support:
v
0
,
L
=
0
.With a concrete selection of boundary conditions and support geometry, an instance of the problem is defined. We considered several Standard Problem Definitions as follows.Mountain
or
Peak. A shortening of the shape can be measured with v
(
0,0
)
>
0. Consider(21)
σ
x
x
,
y
=
σ
x
-
x
,
y
,
σ
y
x
,
y
=
σ
y
-
x
,
y
,
τ
x
y
x
,
y
=
-
τ
x
y
-
x
,
y
,
σ
x
0,0
=
σ
y
0,0
=
0
,
support line:
y
=
H
,
v
x
,
H
=
0
,
u
0
,
y
=
0
,
u
0
,
H
=
0
.MountainBis. A shortening of the shape can be measured with v
(
0
,
H
)
>
0. Consider(22)
σ
x
x
,
y
=
σ
x
-
x
,
y
,
σ
y
x
,
y
=
σ
y
-
x
,
y
,
τ
x
y
x
,
y
=
-
τ
x
y
-
x
,
y
,
σ
x
0,0
=
σ
y
0,0
=
0
,
support line:
y
=
H
,
∂
v
x
,
H
∂
x
=
0
,
u
0,0
=
0
,
u
0
,
H
=
0
,
v
0,0
=
0
.MountainIII
or
Summit. A shortening of the shape can be measured with v
(
0
,
H
)
>
0. Consider(23)
σ
x
x
,
y
=
σ
x
-
x
,
y
,
σ
y
x
,
y
=
σ
y
-
x
,
y
,
τ
x
y
x
,
y
=
-
τ
x
y
-
x
,
y
,
σ
y
0,0
=
0
,
support:
y
=
H
,
∂
v
x
,
H
∂
x
=
0
,
u
0,0
=
0
,
u
0
,
H
=
0
,
v
0,0
=
0
.The trial stress field is derived fromΨ and X, which are defined through some constant parameters to be determined. In this way, Φ
(
x
,
y
) is completely defined. Let us name p to the set of parameters to be determined for the expressions of Ψ, X, u
(
x
,
y
), and v
(
x
,
y
). These parameters, once determined, will define the body (or bodies) generated by the trial stress field.


#### 3.2.2. Getting a Feasible Body

Once a trial stress field is selected in algebraic form, (19) and (20) must hold simultaneously for all (
x
,
y
). Let us represent this condition with the set of problem equations. One has(24)
P
=
0
.It is worth noting that, into (24), all the equations needed for satisfying (19) or (20) for all (
x
,
y
) must be included. For example, if one of (20) is(25)
A
2
-
C
B
x
2
+
L
-
D
A
x
y
+
E
3
-
A
B
C
y
2
=
0then the following three equations will be included into (24). Consider(26)
A
2
-
C
B
=
0
,
L
-
D
A
=
0
,
E
3
-
A
B
C
=
0
.Solving (24) for all (
x
,
y
), we obtain a set of relationships between the parameters of p that we can write as(27)
p
=
Q
q
,where q is the set of independent parameters for a given Φ such that Φ
=
Φ
(
x
,
y
,
q
). It is worth noting that some (or all) parameters in q can disappear from the stress or displacement fields, because only derivatives of Φ are present in the expressions of these fields.If the stress field depends on some components ofq, these components can be chosen freely as equilibrium or compatibility equations will hold for arbitrary values of q. This fact means that the given Φ represent a family or set of solutions, not a unique one.Remeber thatσ
c is the comparison stress, and it must be calculated to assert if the solution is feasible or not; that is, σ
c
≤
f. The given or unknown line of the support (straight or curve) must intersect some solutions of(28)
σ
c
x
,
y
,
q
=
f
.This requirement is necessary if the body attains its maximal resistance.Leta and b be the principal planes of the stress tensor. Then the curves that can form the contour free of the stress will be solutions of one of the following equations:(29)
σ
a
x
,
y
,
q
=
0
,
σ
b
x
,
y
,
q
=
0
.We must choose one of the two, depending on if we look for a column (compression) or a cable (tension). The solutions would be curves of the form F
(
x
,
y
,
q
)
=
0.Selecting support curves with free contour curves in such a manner that they form a closed domain, a set of shapes is determined. For each shape, the safety criterion must be imposed in any interior point. In this phase, some parameters ofq could be dependent on others, and in this case a new reduction of the number of independent parameters results in the following:(30)
q
=
R
r
.If all the parameters become determined now, the stress field corresponds to a unique shape. Otherwise, we have a family of shapes depending on arbitrary values of r.Perhaps, the main problem is to determine the base-end vertex mentioned before. If the curves that define the shape can be obtained in an explicit form (y
=
f
(
x
)), they can be managed directly. Otherwise, the shape will be defined by inequations, whose sign can be determined for each function considering the sign of its value in (
0
,
L
/
2
), where L is the size used in (20), if the origin was specified as a point of the free contour (null surface stress) in (19).Anyhow, it is always possible simply to draw the curves of the contour of the bodyσ
a
=
0 and the support line for an arbitrarily chosen value of the size L. Generally, the last one will intersect the former and σ
b
=
±
f in two different points. But these two points may be the same base-end vertex of the contour; the first point mentioned is the vertex of the base where the stress-free contour ends; the second one is the vertex where the own base ends. Hence, adjusting the value of L in such a manner that the two points be the same, we will find out the correct base-end vertex and the height of the shape by a simple albeit tedious trial-and-error method.


#### 3.2.3. An Example in Detail

Definitions. Consider the following:(i)
Complex potentials:Ψ
=
Ψ
5
+
Ψ
6
X
=
X
3
+
X
4 see (18).
(ii)
Problem and support:MountainBis (see (22)).
(iii)
p
T
=
{
θ
0
,
u
0
,
v
0
,
C
30
,
C
31
,
C
40
,
C
41
,
P
50
,
P
51
,
P
60
,
P
61
}.
(iv)
q
T
=
{
C
31
}.
(v)
Stress function:(31)
Φ
x
,
y
=
-
y
4
-
6
x
2
y
2
+
x
4
6
ν
+
6
C
31
+
ν
-
1
ρ
24
ν
+
24
H
-
3
x
2
y
-
y
3
C
31
.
(vi)
Stress tensor:(32)
σ
x
x
,
y
=
12
ν
+
1
y
C
31
-
2
ν
-
1
ρ
y
H
+
6
ν
+
1
x
2
-
y
2
C
31
+
1
-
ν
ρ
y
2
-
x
2
2
ν
+
1
H
,
σ
y
x
,
y
=
-
2
ν
+
1
y
6
C
31
+
ρ
H
+
6
ν
+
1
x
2
-
y
2
C
31
+
1
-
ν
ρ
y
2
-
x
2
2
ν
+
1
H
,
τ
x
y
x
,
y
=
6
ν
+
x
C
31
H
-
y
+
1
-
ν
ρ
x
y
ν
+
1
H
.
(vii)
Displacements:(33)
u
x
,
y
=
6
6
ν
+
1
x
y
C
31
+
ν
-
ρ
x
y
H
+
6
-
3
ν
-
1
x
y
2
+
ν
+
1
x
3
C
31
+
1
-
ν
ρ
3
x
y
2
-
x
3
6
E
H
,
v
x
,
y
=
-
6
3
ν
+
1
C
31
-
1
-
ν
ρ
y
2
-
x
2
H
+
6
ν
+
1
3
x
2
y
-
y
3
C
31
+
1
-
ν
ρ
y
3
-
3
x
2
y
6
E
H
.Bodies. There is a family dependent on parameter C
31  of  X. This parameter defines the abscissa of the base-end vertex.(i)
Vertices:V
1
=
(
f
(
C
31
)
,
H
); V
2
=
(
0,0
); V
3
=
(
-
f
(
C
31
)
,
H
)
. For a normal steel (see Table 1), the real domain of f is approximately (
-
0.43
;
0.23
) with roots in the extremes of the interval and in 0.1. Consider(34)
f
steel
C
31
=
10
136
C
31
2
-
20
C
31
+
1
-
300
C
31
2
-
3
10
3
C
31
-
3
.
(ii)
Stress-free arcs: they do not have analytical expressions. Drawing them, it is clear that there are solutions only forC
31
∈
[
-
0.165
;
0.1
].
The base width varies from 0,9 up to 1,4A. The height of all shapes is constant, only depending on material properties; see Figure 3.
(iii)
Insurmountable sizeL: 0.5
A.Table 1
Shapes of  “mountain” found.
Normal  steel,  dimensionless  properties:ν
=
0.25
;
ρ
=
f
=
1
;
E
=
808
Problem
PotentialsΨ
,
X
L
÷
A
b
a
s
e
÷
A
Shape
M
ountain
Ψ
2
;
X
1
0,389
6,96
Triangle
Mountain
Ψ
4
;
X
4
0,50
1,53
Ellipse segment
MountainBis
Ψ
5
+
Ψ
6
;
X
3
+
X
4
0,5
f
(
C
31
)
Pseudoexponential
Peak
Ψ
3
;
X
=
0
8
/
11
f
(
P
21
)
?
Parabolic  segment  base32
2
/
11
3  for P
21
=
0Figure 3
Stress fields and the corresponding body. (a)σ
a
=
σ
m
a
x. (b) σ
b
=
σ
m
i
n. (c) Von Mises stress σ
c. The origin of coordinates is up in the centre of the figures, and there the stress tensor is null. In (a), the three curves that define the body—from the stress field and satisfying the Von Mises criterion—are drawn. The red curve is σ
m
a
x
=
0, that is, the stress-free surface of the body. The pale blue curve is σ
c
=
f; the body cannot extend beyond. The pale rose curve is σ
m
i
n
=
-
f; the support line cannot extend beyond. The three curves intersect in a unique point. The support can be the horizontal line for this point; in this way, it can be assured that the stress at the support does not exceed the allowable stress. The insurmountable size of the body is the absolute value of the ordinate of the intersection point, A
/
2 in this example.
(a)
(b)
(c)


#### 3.2.4. Provisional Conclusion

After our research, we can claim nothing about Galileo’s question, in spite of the fact that all shapes we found have insurmountable size lesser than Galileo’s column; see Table1. But at least we have tried to do the best, trying to refute our own conjectures. Maybe it is possible to search on all the set of complex potentials with methods of high mathematics, which of course are beyond our knowledge.


## 3.1. The Known Solutions with Infinite Height Are Unfeasible

Although Euler’s tallest column and related problems have received very much attention up to date, some researchers while studying that problem spent a few minutes to study, too, problems related with that of Galileo. This is the case of Karihaloo and Hemp [18] that study the “maximum strength design” of structural members. In their approach, all cross sections of a structural member attain the maximum allowable stress for the given material; therefore the solution is also referred to as “constant maximum strength design.”Let us examine with some detail the constant maximum strength design of Karihaloo and Hemp for tension members; see [18] (Section 2.1) and Figure 1. Consider a cable of length L and cross-sectional area A
(
y
), hung on its top edge (y
=
L) and when its bottom edge (y
=
0) is a free boundary. If the gravity and the external load are acting in the negative direction of y-axis, the condition of the constant maximum strength is(6)
f
A
=
∫
0
L
ρ
A
d
y
;
d
A
A
=
d
y
A
.The solution is(7)
A
=
A
0
·
exp
⁡
y
Aas it can be checked obtaining its derivative and comparing the result with (6). If the cross section is circular, the radius r
(
y
) gives us the contour of the member as follows:(8)
r
y
=
r
0
·
exp
⁡
y
A
=
r
0
·
exp
⁡
y
2
A
.This solution can have an infinite height with constant stress and bear a useful load at the bottom edge (y
=
0, Q
=
A
0
f). Anyhow, its volume grows exponentially with its size (A
0
A
·
(
e
x
p
(
L
/
A
)
-
1
)), so in practice it is an “intractable” solution—in the same meaning that the term is used in algorithm complexity theory [17]—with a load cost as follows:(9)
C
L
=
Q
+
ρ
V
L
Q
=
1
+
1
ρ
exp
⁡
L
A
-
1
.Figure 1
Constant maximum strength cable: classical solution.
(a)
3D-solution
(b)
2D-solutionFurthermore, the solution is not feasible from the point of view of equilibrium because only the equilibrium in the directiony is considered for obtaining a constant stress σ
y. Let us consider the 2D-case for the sake of simplicity; see Figure 1(b). As the border A
B (or C
D) is stress-free, the tangent in any point is a principal direction (σ
a), as the orthogonal direction is (σ
b
=
0). As σ
y
=
f, we have(10)
σ
a
=
f
+
σ
x
2
+
f
+
σ
x
2
2
+
τ
2
,
0
=
f
+
σ
x
2
+
f
+
σ
x
2
2
+
τ
2and hence σ
a
=
f
/
c
o
s
2
(
α
) if α is the angle between the principal stress direction and coordinate axes. This value is greater than allowable stress f for all α. In fact, as L increases, cos
⁡
α
→
0 exponentially, and σ
a grows in the same way. As a consequence, the classical solution is not a feasible one for common failure criteria, that is, as Von Mises criterion. This error is common to all solutions obtained making use of Bernoulli-Euler theory or Navier hypothesis, as these models are not applicable in the case of variable cross section beams—a well-known drawback; see, for example, [19]. These solutions do not prove in any way that an infinite size for columns or beams would be feasible. Of course, these solutions are almost exact when the size L is much lesser than the material structural scope A, as the exponential function grows very slowly when its argument is very small, and for small size the solutions obtained pass fairly well through experimental checks [20]. But they are useless to answer the question pointed out by Galileo, because to this issue we must explore sizes of the same order of magnitude that material scope, that is, large structures for which the equilibrium equations must be completely fulfilled.


## 3.2. Trying to Refute the Existence of a Finite Height

In 2010 (unpublished work), we tried to refute the existence of a finite structural scope whatever the structural material of Galileo’s column is. Our try was naive and unsuccessful. However we think it can help others to understand the difficulties of the problem and perhaps gives clues to better searches for a complete solution. To be short, let us consider a 2D-version.We have an elastic linear material as before. We choose Von Mises criterion as the failure one; hence in any point of a feasible body the following expression hold:(11)
σ
c
x
,
y
=
σ
x
2
+
σ
y
2
+
3
τ
2
≤
f
.Let us suppose that we are able to determine a suitable stress field for an instance of Galileo’s problem, defined over all the x
y plane. This field implicitly defined the shape of a body for which (11) holds. The support line must intersect in some point of some solution of(12)
σ
c
x
,
y
=
f
.Let a and b be the principal planes of the stress tensor. Then the curves that can form the stress-free contour of the column will be solutions of one of the following equations:(13)
σ
a
x
,
y
=
0
,
σ
b
x
,
y
=
0depending on if we are looking for a compressive or tensile solution. Furthermore, these curves intersect in some point of solution (12) and also the solution of(14)
σ
a
x
,
y
=
±
f
,
σ
b
x
,
y
=
±
f
.Indeed, if in (13) we select the direction a, we must select now b in (14) and vice versa. This intersection point will be named “base-end vertex” hereafter. Selecting an appropriate set of arcs from solutions of (13), with the additional condition that they define a closed region on the plane together with the support line, we get the shape of a solution for this instance of Galileo’s problem and we can determine its height relative to the material scope; see Figure 2.Figure 2
A feasible Galileo column from a given stress field.If we would be able to explore completely the set of all possible stress fields—and, of course, we are not—with the maximal height obtained we could answer—“Yes” or “No”—Galileo’s question.


### 3.2.1. Generating Subsets of Stress Fields

To check the possibility of our idea we recall on well-known Airy’s function. Whatever Airy’s stress functionΦ is, for which the biharmonic equation holds, it can be defined as(15)
Φ
=
Re
⁡
Z
¯
Ψ
Z
+
X
Z
,where Ψ and X are analytical functions in C [21]. The function Φ satisfies both compatibility and equilibrium equations, and the displacements can be calculated without integration of the stress functions, from the complex potentials Ψ  and  X.The stress field is(16)
σ
x
=
∂
2
Φ
∂
y
2
-
ρ
y
,
σ
y
=
∂
2
Φ
∂
x
2
-
ρ
y
,
τ
x
y
=
-
∂
2
Φ
∂
x
∂
y
.Using the Hooke law with this field, the displacement functions are(17)
u
x
,
y
=
1
E
-
1
+
ν
∂
Φ
∂
x
+
4
·
Re
⁡
Ψ
x
,
y
-
ρ
1
-
ν
x
y
+
θ
0
y
+
u
0
,
v
x
,
y
=
1
E
-
1
+
ν
∂
Φ
∂
y
+
4
·
Im
⁡
Ψ
x
,
y
+
ρ
2
1
-
ν
x
2
-
y
2
-
θ
0
x
+
v
0
,where ν is Poisson’s modulus.The complex potentials considered were simple polynomials as follows:(18)
Ψ
0
=
A
p
+
i
·
B
p
,
Ψ
1
=
A
p
+
i
·
B
p
+
C
p
+
i
·
D
p
·
Z
,
Ψ
2
=
A
p
+
i
·
B
p
+
C
p
+
i
·
D
p
·
Z
+
E
p
+
i
·
F
p
·
Z
2
,
⋮
=
⋮
Ψ
n
=
P
n
0
+
i
P
n
1
·
Z
n
,
X
0
=
A
c
+
i
·
B
c
,
X
1
=
A
c
+
i
·
B
c
+
C
c
+
i
·
D
c
·
Z
,
X
2
=
A
c
+
i
·
B
c
+
C
c
+
i
·
D
c
·
Z
+
E
c
+
i
·
F
c
·
Z
2
,
⋮
=
⋮
X
n
=
C
n
0
+
i
C
n
1
·
Z
n
.The coordinate origin will be at the top of a column (or at the bottom of a cable). Some boundary conditions must hold always as follows:(19)
symmetry:
σ
x
x
,
y
=
σ
x
-
x
,
y
,
σ
y
x
,
y
=
σ
y
-
x
,
y
,
τ
x
y
x
,
y
=
-
τ
x
y
-
x
,
y
,
origin:
σ
x
0,0
=
σ
y
0,0
=
τ
x
y
0,0
=
0
.We considered also different supports as follows:(20)
base line support:
v
x
,
L
=
0
,
base point support:
v
0
,
L
=
0
.With a concrete selection of boundary conditions and support geometry, an instance of the problem is defined. We considered several Standard Problem Definitions as follows.Mountain
or
Peak. A shortening of the shape can be measured with v
(
0,0
)
>
0. Consider(21)
σ
x
x
,
y
=
σ
x
-
x
,
y
,
σ
y
x
,
y
=
σ
y
-
x
,
y
,
τ
x
y
x
,
y
=
-
τ
x
y
-
x
,
y
,
σ
x
0,0
=
σ
y
0,0
=
0
,
support line:
y
=
H
,
v
x
,
H
=
0
,
u
0
,
y
=
0
,
u
0
,
H
=
0
.MountainBis. A shortening of the shape can be measured with v
(
0
,
H
)
>
0. Consider(22)
σ
x
x
,
y
=
σ
x
-
x
,
y
,
σ
y
x
,
y
=
σ
y
-
x
,
y
,
τ
x
y
x
,
y
=
-
τ
x
y
-
x
,
y
,
σ
x
0,0
=
σ
y
0,0
=
0
,
support line:
y
=
H
,
∂
v
x
,
H
∂
x
=
0
,
u
0,0
=
0
,
u
0
,
H
=
0
,
v
0,0
=
0
.MountainIII
or
Summit. A shortening of the shape can be measured with v
(
0
,
H
)
>
0. Consider(23)
σ
x
x
,
y
=
σ
x
-
x
,
y
,
σ
y
x
,
y
=
σ
y
-
x
,
y
,
τ
x
y
x
,
y
=
-
τ
x
y
-
x
,
y
,
σ
y
0,0
=
0
,
support:
y
=
H
,
∂
v
x
,
H
∂
x
=
0
,
u
0,0
=
0
,
u
0
,
H
=
0
,
v
0,0
=
0
.The trial stress field is derived fromΨ and X, which are defined through some constant parameters to be determined. In this way, Φ
(
x
,
y
) is completely defined. Let us name p to the set of parameters to be determined for the expressions of Ψ, X, u
(
x
,
y
), and v
(
x
,
y
). These parameters, once determined, will define the body (or bodies) generated by the trial stress field.


### 3.2.2. Getting a Feasible Body

Once a trial stress field is selected in algebraic form, (19) and (20) must hold simultaneously for all (
x
,
y
). Let us represent this condition with the set of problem equations. One has(24)
P
=
0
.It is worth noting that, into (24), all the equations needed for satisfying (19) or (20) for all (
x
,
y
) must be included. For example, if one of (20) is(25)
A
2
-
C
B
x
2
+
L
-
D
A
x
y
+
E
3
-
A
B
C
y
2
=
0then the following three equations will be included into (24). Consider(26)
A
2
-
C
B
=
0
,
L
-
D
A
=
0
,
E
3
-
A
B
C
=
0
.Solving (24) for all (
x
,
y
), we obtain a set of relationships between the parameters of p that we can write as(27)
p
=
Q
q
,where q is the set of independent parameters for a given Φ such that Φ
=
Φ
(
x
,
y
,
q
). It is worth noting that some (or all) parameters in q can disappear from the stress or displacement fields, because only derivatives of Φ are present in the expressions of these fields.If the stress field depends on some components ofq, these components can be chosen freely as equilibrium or compatibility equations will hold for arbitrary values of q. This fact means that the given Φ represent a family or set of solutions, not a unique one.Remeber thatσ
c is the comparison stress, and it must be calculated to assert if the solution is feasible or not; that is, σ
c
≤
f. The given or unknown line of the support (straight or curve) must intersect some solutions of(28)
σ
c
x
,
y
,
q
=
f
.This requirement is necessary if the body attains its maximal resistance.Leta and b be the principal planes of the stress tensor. Then the curves that can form the contour free of the stress will be solutions of one of the following equations:(29)
σ
a
x
,
y
,
q
=
0
,
σ
b
x
,
y
,
q
=
0
.We must choose one of the two, depending on if we look for a column (compression) or a cable (tension). The solutions would be curves of the form F
(
x
,
y
,
q
)
=
0.Selecting support curves with free contour curves in such a manner that they form a closed domain, a set of shapes is determined. For each shape, the safety criterion must be imposed in any interior point. In this phase, some parameters ofq could be dependent on others, and in this case a new reduction of the number of independent parameters results in the following:(30)
q
=
R
r
.If all the parameters become determined now, the stress field corresponds to a unique shape. Otherwise, we have a family of shapes depending on arbitrary values of r.Perhaps, the main problem is to determine the base-end vertex mentioned before. If the curves that define the shape can be obtained in an explicit form (y
=
f
(
x
)), they can be managed directly. Otherwise, the shape will be defined by inequations, whose sign can be determined for each function considering the sign of its value in (
0
,
L
/
2
), where L is the size used in (20), if the origin was specified as a point of the free contour (null surface stress) in (19).Anyhow, it is always possible simply to draw the curves of the contour of the bodyσ
a
=
0 and the support line for an arbitrarily chosen value of the size L. Generally, the last one will intersect the former and σ
b
=
±
f in two different points. But these two points may be the same base-end vertex of the contour; the first point mentioned is the vertex of the base where the stress-free contour ends; the second one is the vertex where the own base ends. Hence, adjusting the value of L in such a manner that the two points be the same, we will find out the correct base-end vertex and the height of the shape by a simple albeit tedious trial-and-error method.


### 3.2.3. An Example in Detail

Definitions. Consider the following:(i)
Complex potentials:Ψ
=
Ψ
5
+
Ψ
6
X
=
X
3
+
X
4 see (18).
(ii)
Problem and support:MountainBis (see (22)).
(iii)
p
T
=
{
θ
0
,
u
0
,
v
0
,
C
30
,
C
31
,
C
40
,
C
41
,
P
50
,
P
51
,
P
60
,
P
61
}.
(iv)
q
T
=
{
C
31
}.
(v)
Stress function:(31)
Φ
x
,
y
=
-
y
4
-
6
x
2
y
2
+
x
4
6
ν
+
6
C
31
+
ν
-
1
ρ
24
ν
+
24
H
-
3
x
2
y
-
y
3
C
31
.
(vi)
Stress tensor:(32)
σ
x
x
,
y
=
12
ν
+
1
y
C
31
-
2
ν
-
1
ρ
y
H
+
6
ν
+
1
x
2
-
y
2
C
31
+
1
-
ν
ρ
y
2
-
x
2
2
ν
+
1
H
,
σ
y
x
,
y
=
-
2
ν
+
1
y
6
C
31
+
ρ
H
+
6
ν
+
1
x
2
-
y
2
C
31
+
1
-
ν
ρ
y
2
-
x
2
2
ν
+
1
H
,
τ
x
y
x
,
y
=
6
ν
+
x
C
31
H
-
y
+
1
-
ν
ρ
x
y
ν
+
1
H
.
(vii)
Displacements:(33)
u
x
,
y
=
6
6
ν
+
1
x
y
C
31
+
ν
-
ρ
x
y
H
+
6
-
3
ν
-
1
x
y
2
+
ν
+
1
x
3
C
31
+
1
-
ν
ρ
3
x
y
2
-
x
3
6
E
H
,
v
x
,
y
=
-
6
3
ν
+
1
C
31
-
1
-
ν
ρ
y
2
-
x
2
H
+
6
ν
+
1
3
x
2
y
-
y
3
C
31
+
1
-
ν
ρ
y
3
-
3
x
2
y
6
E
H
.Bodies. There is a family dependent on parameter C
31  of  X. This parameter defines the abscissa of the base-end vertex.(i)
Vertices:V
1
=
(
f
(
C
31
)
,
H
); V
2
=
(
0,0
); V
3
=
(
-
f
(
C
31
)
,
H
)
. For a normal steel (see Table 1), the real domain of f is approximately (
-
0.43
;
0.23
) with roots in the extremes of the interval and in 0.1. Consider(34)
f
steel
C
31
=
10
136
C
31
2
-
20
C
31
+
1
-
300
C
31
2
-
3
10
3
C
31
-
3
.
(ii)
Stress-free arcs: they do not have analytical expressions. Drawing them, it is clear that there are solutions only forC
31
∈
[
-
0.165
;
0.1
].
The base width varies from 0,9 up to 1,4A. The height of all shapes is constant, only depending on material properties; see Figure 3.
(iii)
Insurmountable sizeL: 0.5
A.Table 1
Shapes of  “mountain” found.
Normal  steel,  dimensionless  properties:ν
=
0.25
;
ρ
=
f
=
1
;
E
=
808
Problem
PotentialsΨ
,
X
L
÷
A
b
a
s
e
÷
A
Shape
M
ountain
Ψ
2
;
X
1
0,389
6,96
Triangle
Mountain
Ψ
4
;
X
4
0,50
1,53
Ellipse segment
MountainBis
Ψ
5
+
Ψ
6
;
X
3
+
X
4
0,5
f
(
C
31
)
Pseudoexponential
Peak
Ψ
3
;
X
=
0
8
/
11
f
(
P
21
)
?
Parabolic  segment  base32
2
/
11
3  for P
21
=
0Figure 3
Stress fields and the corresponding body. (a)σ
a
=
σ
m
a
x. (b) σ
b
=
σ
m
i
n. (c) Von Mises stress σ
c. The origin of coordinates is up in the centre of the figures, and there the stress tensor is null. In (a), the three curves that define the body—from the stress field and satisfying the Von Mises criterion—are drawn. The red curve is σ
m
a
x
=
0, that is, the stress-free surface of the body. The pale blue curve is σ
c
=
f; the body cannot extend beyond. The pale rose curve is σ
m
i
n
=
-
f; the support line cannot extend beyond. The three curves intersect in a unique point. The support can be the horizontal line for this point; in this way, it can be assured that the stress at the support does not exceed the allowable stress. The insurmountable size of the body is the absolute value of the ordinate of the intersection point, A
/
2 in this example.
(a)
(b)
(c)


### 3.2.4. Provisional Conclusion

After our research, we can claim nothing about Galileo’s question, in spite of the fact that all shapes we found have insurmountable size lesser than Galileo’s column; see Table1. But at least we have tried to do the best, trying to refute our own conjectures. Maybe it is possible to search on all the set of complex potentials with methods of high mathematics, which of course are beyond our knowledge.


## 3.2.1. Generating Subsets of Stress Fields

To check the possibility of our idea we recall on well-known Airy’s function. Whatever Airy’s stress functionΦ is, for which the biharmonic equation holds, it can be defined as(15)
Φ
=
Re
⁡
Z
¯
Ψ
Z
+
X
Z
,where Ψ and X are analytical functions in C [21]. The function Φ satisfies both compatibility and equilibrium equations, and the displacements can be calculated without integration of the stress functions, from the complex potentials Ψ  and  X.The stress field is(16)
σ
x
=
∂
2
Φ
∂
y
2
-
ρ
y
,
σ
y
=
∂
2
Φ
∂
x
2
-
ρ
y
,
τ
x
y
=
-
∂
2
Φ
∂
x
∂
y
.Using the Hooke law with this field, the displacement functions are(17)
u
x
,
y
=
1
E
-
1
+
ν
∂
Φ
∂
x
+
4
·
Re
⁡
Ψ
x
,
y
-
ρ
1
-
ν
x
y
+
θ
0
y
+
u
0
,
v
x
,
y
=
1
E
-
1
+
ν
∂
Φ
∂
y
+
4
·
Im
⁡
Ψ
x
,
y
+
ρ
2
1
-
ν
x
2
-
y
2
-
θ
0
x
+
v
0
,where ν is Poisson’s modulus.The complex potentials considered were simple polynomials as follows:(18)
Ψ
0
=
A
p
+
i
·
B
p
,
Ψ
1
=
A
p
+
i
·
B
p
+
C
p
+
i
·
D
p
·
Z
,
Ψ
2
=
A
p
+
i
·
B
p
+
C
p
+
i
·
D
p
·
Z
+
E
p
+
i
·
F
p
·
Z
2
,
⋮
=
⋮
Ψ
n
=
P
n
0
+
i
P
n
1
·
Z
n
,
X
0
=
A
c
+
i
·
B
c
,
X
1
=
A
c
+
i
·
B
c
+
C
c
+
i
·
D
c
·
Z
,
X
2
=
A
c
+
i
·
B
c
+
C
c
+
i
·
D
c
·
Z
+
E
c
+
i
·
F
c
·
Z
2
,
⋮
=
⋮
X
n
=
C
n
0
+
i
C
n
1
·
Z
n
.The coordinate origin will be at the top of a column (or at the bottom of a cable). Some boundary conditions must hold always as follows:(19)
symmetry:
σ
x
x
,
y
=
σ
x
-
x
,
y
,
σ
y
x
,
y
=
σ
y
-
x
,
y
,
τ
x
y
x
,
y
=
-
τ
x
y
-
x
,
y
,
origin:
σ
x
0,0
=
σ
y
0,0
=
τ
x
y
0,0
=
0
.We considered also different supports as follows:(20)
base line support:
v
x
,
L
=
0
,
base point support:
v
0
,
L
=
0
.With a concrete selection of boundary conditions and support geometry, an instance of the problem is defined. We considered several Standard Problem Definitions as follows.Mountain
or
Peak. A shortening of the shape can be measured with v
(
0,0
)
>
0. Consider(21)
σ
x
x
,
y
=
σ
x
-
x
,
y
,
σ
y
x
,
y
=
σ
y
-
x
,
y
,
τ
x
y
x
,
y
=
-
τ
x
y
-
x
,
y
,
σ
x
0,0
=
σ
y
0,0
=
0
,
support line:
y
=
H
,
v
x
,
H
=
0
,
u
0
,
y
=
0
,
u
0
,
H
=
0
.MountainBis. A shortening of the shape can be measured with v
(
0
,
H
)
>
0. Consider(22)
σ
x
x
,
y
=
σ
x
-
x
,
y
,
σ
y
x
,
y
=
σ
y
-
x
,
y
,
τ
x
y
x
,
y
=
-
τ
x
y
-
x
,
y
,
σ
x
0,0
=
σ
y
0,0
=
0
,
support line:
y
=
H
,
∂
v
x
,
H
∂
x
=
0
,
u
0,0
=
0
,
u
0
,
H
=
0
,
v
0,0
=
0
.MountainIII
or
Summit. A shortening of the shape can be measured with v
(
0
,
H
)
>
0. Consider(23)
σ
x
x
,
y
=
σ
x
-
x
,
y
,
σ
y
x
,
y
=
σ
y
-
x
,
y
,
τ
x
y
x
,
y
=
-
τ
x
y
-
x
,
y
,
σ
y
0,0
=
0
,
support:
y
=
H
,
∂
v
x
,
H
∂
x
=
0
,
u
0,0
=
0
,
u
0
,
H
=
0
,
v
0,0
=
0
.The trial stress field is derived fromΨ and X, which are defined through some constant parameters to be determined. In this way, Φ
(
x
,
y
) is completely defined. Let us name p to the set of parameters to be determined for the expressions of Ψ, X, u
(
x
,
y
), and v
(
x
,
y
). These parameters, once determined, will define the body (or bodies) generated by the trial stress field.


## 3.2.2. Getting a Feasible Body

Once a trial stress field is selected in algebraic form, (19) and (20) must hold simultaneously for all (
x
,
y
). Let us represent this condition with the set of problem equations. One has(24)
P
=
0
.It is worth noting that, into (24), all the equations needed for satisfying (19) or (20) for all (
x
,
y
) must be included. For example, if one of (20) is(25)
A
2
-
C
B
x
2
+
L
-
D
A
x
y
+
E
3
-
A
B
C
y
2
=
0then the following three equations will be included into (24). Consider(26)
A
2
-
C
B
=
0
,
L
-
D
A
=
0
,
E
3
-
A
B
C
=
0
.Solving (24) for all (
x
,
y
), we obtain a set of relationships between the parameters of p that we can write as(27)
p
=
Q
q
,where q is the set of independent parameters for a given Φ such that Φ
=
Φ
(
x
,
y
,
q
). It is worth noting that some (or all) parameters in q can disappear from the stress or displacement fields, because only derivatives of Φ are present in the expressions of these fields.If the stress field depends on some components ofq, these components can be chosen freely as equilibrium or compatibility equations will hold for arbitrary values of q. This fact means that the given Φ represent a family or set of solutions, not a unique one.Remeber thatσ
c is the comparison stress, and it must be calculated to assert if the solution is feasible or not; that is, σ
c
≤
f. The given or unknown line of the support (straight or curve) must intersect some solutions of(28)
σ
c
x
,
y
,
q
=
f
.This requirement is necessary if the body attains its maximal resistance.Leta and b be the principal planes of the stress tensor. Then the curves that can form the contour free of the stress will be solutions of one of the following equations:(29)
σ
a
x
,
y
,
q
=
0
,
σ
b
x
,
y
,
q
=
0
.We must choose one of the two, depending on if we look for a column (compression) or a cable (tension). The solutions would be curves of the form F
(
x
,
y
,
q
)
=
0.Selecting support curves with free contour curves in such a manner that they form a closed domain, a set of shapes is determined. For each shape, the safety criterion must be imposed in any interior point. In this phase, some parameters ofq could be dependent on others, and in this case a new reduction of the number of independent parameters results in the following:(30)
q
=
R
r
.If all the parameters become determined now, the stress field corresponds to a unique shape. Otherwise, we have a family of shapes depending on arbitrary values of r.Perhaps, the main problem is to determine the base-end vertex mentioned before. If the curves that define the shape can be obtained in an explicit form (y
=
f
(
x
)), they can be managed directly. Otherwise, the shape will be defined by inequations, whose sign can be determined for each function considering the sign of its value in (
0
,
L
/
2
), where L is the size used in (20), if the origin was specified as a point of the free contour (null surface stress) in (19).Anyhow, it is always possible simply to draw the curves of the contour of the bodyσ
a
=
0 and the support line for an arbitrarily chosen value of the size L. Generally, the last one will intersect the former and σ
b
=
±
f in two different points. But these two points may be the same base-end vertex of the contour; the first point mentioned is the vertex of the base where the stress-free contour ends; the second one is the vertex where the own base ends. Hence, adjusting the value of L in such a manner that the two points be the same, we will find out the correct base-end vertex and the height of the shape by a simple albeit tedious trial-and-error method.


## 3.2.3. An Example in Detail

Definitions. Consider the following:(i)
Complex potentials:Ψ
=
Ψ
5
+
Ψ
6
X
=
X
3
+
X
4 see (18).
(ii)
Problem and support:MountainBis (see (22)).
(iii)
p
T
=
{
θ
0
,
u
0
,
v
0
,
C
30
,
C
31
,
C
40
,
C
41
,
P
50
,
P
51
,
P
60
,
P
61
}.
(iv)
q
T
=
{
C
31
}.
(v)
Stress function:(31)
Φ
x
,
y
=
-
y
4
-
6
x
2
y
2
+
x
4
6
ν
+
6
C
31
+
ν
-
1
ρ
24
ν
+
24
H
-
3
x
2
y
-
y
3
C
31
.
(vi)
Stress tensor:(32)
σ
x
x
,
y
=
12
ν
+
1
y
C
31
-
2
ν
-
1
ρ
y
H
+
6
ν
+
1
x
2
-
y
2
C
31
+
1
-
ν
ρ
y
2
-
x
2
2
ν
+
1
H
,
σ
y
x
,
y
=
-
2
ν
+
1
y
6
C
31
+
ρ
H
+
6
ν
+
1
x
2
-
y
2
C
31
+
1
-
ν
ρ
y
2
-
x
2
2
ν
+
1
H
,
τ
x
y
x
,
y
=
6
ν
+
x
C
31
H
-
y
+
1
-
ν
ρ
x
y
ν
+
1
H
.
(vii)
Displacements:(33)
u
x
,
y
=
6
6
ν
+
1
x
y
C
31
+
ν
-
ρ
x
y
H
+
6
-
3
ν
-
1
x
y
2
+
ν
+
1
x
3
C
31
+
1
-
ν
ρ
3
x
y
2
-
x
3
6
E
H
,
v
x
,
y
=
-
6
3
ν
+
1
C
31
-
1
-
ν
ρ
y
2
-
x
2
H
+
6
ν
+
1
3
x
2
y
-
y
3
C
31
+
1
-
ν
ρ
y
3
-
3
x
2
y
6
E
H
.Bodies. There is a family dependent on parameter C
31  of  X. This parameter defines the abscissa of the base-end vertex.(i)
Vertices:V
1
=
(
f
(
C
31
)
,
H
); V
2
=
(
0,0
); V
3
=
(
-
f
(
C
31
)
,
H
)
. For a normal steel (see Table 1), the real domain of f is approximately (
-
0.43
;
0.23
) with roots in the extremes of the interval and in 0.1. Consider(34)
f
steel
C
31
=
10
136
C
31
2
-
20
C
31
+
1
-
300
C
31
2
-
3
10
3
C
31
-
3
.
(ii)
Stress-free arcs: they do not have analytical expressions. Drawing them, it is clear that there are solutions only forC
31
∈
[
-
0.165
;
0.1
].
The base width varies from 0,9 up to 1,4A. The height of all shapes is constant, only depending on material properties; see Figure 3.
(iii)
Insurmountable sizeL: 0.5
A.Table 1
Shapes of  “mountain” found.
Normal  steel,  dimensionless  properties:ν
=
0.25
;
ρ
=
f
=
1
;
E
=
808
Problem
PotentialsΨ
,
X
L
÷
A
b
a
s
e
÷
A
Shape
M
ountain
Ψ
2
;
X
1
0,389
6,96
Triangle
Mountain
Ψ
4
;
X
4
0,50
1,53
Ellipse segment
MountainBis
Ψ
5
+
Ψ
6
;
X
3
+
X
4
0,5
f
(
C
31
)
Pseudoexponential
Peak
Ψ
3
;
X
=
0
8
/
11
f
(
P
21
)
?
Parabolic  segment  base32
2
/
11
3  for P
21
=
0Figure 3
Stress fields and the corresponding body. (a)σ
a
=
σ
m
a
x. (b) σ
b
=
σ
m
i
n. (c) Von Mises stress σ
c. The origin of coordinates is up in the centre of the figures, and there the stress tensor is null. In (a), the three curves that define the body—from the stress field and satisfying the Von Mises criterion—are drawn. The red curve is σ
m
a
x
=
0, that is, the stress-free surface of the body. The pale blue curve is σ
c
=
f; the body cannot extend beyond. The pale rose curve is σ
m
i
n
=
-
f; the support line cannot extend beyond. The three curves intersect in a unique point. The support can be the horizontal line for this point; in this way, it can be assured that the stress at the support does not exceed the allowable stress. The insurmountable size of the body is the absolute value of the ordinate of the intersection point, A
/
2 in this example.
(a)
(b)
(c)


## 3.2.4. Provisional Conclusion

After our research, we can claim nothing about Galileo’s question, in spite of the fact that all shapes we found have insurmountable size lesser than Galileo’s column; see Table1. But at least we have tried to do the best, trying to refute our own conjectures. Maybe it is possible to search on all the set of complex potentials with methods of high mathematics, which of course are beyond our knowledge.


## 4. Formal Definitions of Galileo’s Problem

After all, we have two main hypotheses that we outline informally as follows.Conjecture 1.
A finite insurmountable size exists for a fairly large set of structural problems (not only for Galileo’s problem) when the self-weight and stress limit are taken into consideration.The second one is suggested for the results of our search in Section3.2, and it is stronger than the first one.Conjecture 2.
In case of original Galileo’s problem, the insurmountable size for a solid column (without holes of any kind) is equal to the material scope; that is,L
=
A.Refuting any of both hypotheses consists in showing a given problem—including support, boundary conditions, and failure criterion on stresses—and a shape family for structures that can solve it and that includes a shape of infinite size.


### 4.1. The 2D Galileo Problem

For the sake of simplicity let us consider a 2D-universe.Problem 1 (see [1]).
To find ay-symmetrical body of maximal height, placed in the semiplane y
>
0 and supported in the y
=
0 line, only bearing its own weight, of a homogeneous, linear elastic material defined by Young’s Modulus E, Poisson’s modulus ν, allowable compressive stress f
, and specific weight ρ and subjected to displacement constraints v
(
x
,
0
)
=
0 and u
(
0
,
y
)
=
0 and to some criterion C on stresses to be fulfilled over all the body that can be expressed as(35)
C
:
σ
C
x
,
y
≤
f
∀
x
,
y
∈
body
.The support line can be a surface with friction following Coulomb theory. Hence the tangential stress is subjected to(36)
abs
τ
x
,
0
≤
μ
σ
y
with
μ
≥
0
.There are not additional fundamental constraints on shape, but someone can be imposed for convenience.The original Galileo column is simply a rectangular domain of heightA and whatever width w. The principal stresses are σ
b
=
σ
y
=
-
f
(
1
-
y
÷
A
) and σ
a
=
σ
x
=
0. Further, the Von Mises stress is σ
V
M
=
f
(
1
-
y
÷
A
).If anyone can envisage a general proof of our conjectures, the related problems would be directly solved. In any case else, the general formulation of the problem can be stated as to find out a shape with infinite size or height, proving in this way that our working hypothesis is false but being the shape of maximum scope determined. We think on this class of problems as good candidates for some topology or shape optimisation methods [22–26].


### 4.2. Other Related Problems

As the general formulation can be hard to attack with available methods, we can suggest some alternative problems which in our view could be equivalent (or at least approximately equivalent) to the original.LetV
0
=
A
w be a given volume in the 2D-universe. We can consider the problem of finding a shape with this given volume that maximises the height of the figure subject to the same stress tensor constraint.Perhaps the stress constraint can be replaced by minimising the (maximum or mean) Von Mises stress in the volume, being the latter unbounded, and the total height of the figure fixed to a given valueA. With this problem it should be the case that we will get solutions with maximum absolute Von Mises stress lesser than f; hence with appropriated scaling we will get a solution higher than Galileo’s column.Another approach arises from considering the calculus of the maximum scope of a shape as a limit case. Let us consider a useful load at a heighty
=
L
>
0 as a uniform load p along a width w
0. The problem is now to find a shape of minimal weight in equilibrium with p and its self-weight with the stress constraints as above. One additional constraint on the shape will be that it must lie into the region limited by y
≤
L and y
≥
0. If this problem can be solved, the structure scope L will be the limit of L when p
→
0 or w
0
→
0. Obviously, a solution is Galileo’s column of constant width equal to w
0, but is there another one? The useful load can be defined too as P
=
∫
-
w
0
/
2
w
0
/
2
‍
p
(
x
)
d
x,  where  P is a given constant. In this case the function p can be viewed as a design variable, or its integral over w
0 can be viewed as an additional constraint on stresses.More equivalents formulations can exist or can be proposed following these lines.We think that a minimum compliance approach is not equivalent to problems in Galileo’s realm due to self-weight. But it could be the case that minimum compliance objective leads to useful solutions that after appropriate scaling provide that the stress constraint be fulfilled.


## 4.1. The 2D Galileo Problem

For the sake of simplicity let us consider a 2D-universe.Problem 1 (see [1]).
To find ay-symmetrical body of maximal height, placed in the semiplane y
>
0 and supported in the y
=
0 line, only bearing its own weight, of a homogeneous, linear elastic material defined by Young’s Modulus E, Poisson’s modulus ν, allowable compressive stress f
, and specific weight ρ and subjected to displacement constraints v
(
x
,
0
)
=
0 and u
(
0
,
y
)
=
0 and to some criterion C on stresses to be fulfilled over all the body that can be expressed as(35)
C
:
σ
C
x
,
y
≤
f
∀
x
,
y
∈
body
.The support line can be a surface with friction following Coulomb theory. Hence the tangential stress is subjected to(36)
abs
τ
x
,
0
≤
μ
σ
y
with
μ
≥
0
.There are not additional fundamental constraints on shape, but someone can be imposed for convenience.The original Galileo column is simply a rectangular domain of heightA and whatever width w. The principal stresses are σ
b
=
σ
y
=
-
f
(
1
-
y
÷
A
) and σ
a
=
σ
x
=
0. Further, the Von Mises stress is σ
V
M
=
f
(
1
-
y
÷
A
).If anyone can envisage a general proof of our conjectures, the related problems would be directly solved. In any case else, the general formulation of the problem can be stated as to find out a shape with infinite size or height, proving in this way that our working hypothesis is false but being the shape of maximum scope determined. We think on this class of problems as good candidates for some topology or shape optimisation methods [22–26].


## 4.2. Other Related Problems

As the general formulation can be hard to attack with available methods, we can suggest some alternative problems which in our view could be equivalent (or at least approximately equivalent) to the original.LetV
0
=
A
w be a given volume in the 2D-universe. We can consider the problem of finding a shape with this given volume that maximises the height of the figure subject to the same stress tensor constraint.Perhaps the stress constraint can be replaced by minimising the (maximum or mean) Von Mises stress in the volume, being the latter unbounded, and the total height of the figure fixed to a given valueA. With this problem it should be the case that we will get solutions with maximum absolute Von Mises stress lesser than f; hence with appropriated scaling we will get a solution higher than Galileo’s column.Another approach arises from considering the calculus of the maximum scope of a shape as a limit case. Let us consider a useful load at a heighty
=
L
>
0 as a uniform load p along a width w
0. The problem is now to find a shape of minimal weight in equilibrium with p and its self-weight with the stress constraints as above. One additional constraint on the shape will be that it must lie into the region limited by y
≤
L and y
≥
0. If this problem can be solved, the structure scope L will be the limit of L when p
→
0 or w
0
→
0. Obviously, a solution is Galileo’s column of constant width equal to w
0, but is there another one? The useful load can be defined too as P
=
∫
-
w
0
/
2
w
0
/
2
‍
p
(
x
)
d
x,  where  P is a given constant. In this case the function p can be viewed as a design variable, or its integral over w
0 can be viewed as an additional constraint on stresses.More equivalents formulations can exist or can be proposed following these lines.We think that a minimum compliance approach is not equivalent to problems in Galileo’s realm due to self-weight. But it could be the case that minimum compliance objective leads to useful solutions that after appropriate scaling provide that the stress constraint be fulfilled.


## 5. Conclusion

Galileo’s problem has theoretical interest in mathematics and very practical interest in the structural design theory. It would be a benchmark problem for topology or shape optimisation methods. Each different stress constraint or material model (e.g., plasticity) leads to new instances of the problem. In 2038 it will be 400 hundred years since Galileo formulated the problem.We will appreciate any insight into it.



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*Source: 649341-2015-08-12.xml*