Equistrong heavy beam: solving the problem of Galileo Galilei Текст научной статьи по специальности «Физика»

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Равнопрочный тяжелый брус: решение проблемы Галилея

Г.П. Черепанов

Нью-Йоркская академия наук, Нью-Йорк, 10007-2157, США

В своей главной книге «Discorsi e dimostrazioni matematiche, intomo a due nuove scienze», опубликованной в 1638 г. в г. Лейден издательством Elsevier, ««отец» современной науки Галилео Галилей поставил на первое место науку о сопротивлении материалов разрушению. В ней он впервые ввел понятия о напряжениях и прочности материалов, используемые с тех пор повсеместно. Более того, он нашел идеальную форму бруса, изгибаемого поперечной силой, — одну из совершенных форм Платона; такой брус мы сегодня называем равнопрочным. Это открытие заложило основу для поиска других равнопрочных тел, что и составляет проблему Галилея. К сожалению, теоремы существования таких тел нет, поэтому каждое решение этой проблемы подобно находке золотого самородка в горной породе. В настоящей работе найдена равнопрочная форма некоторых тяжелых упругих балок, прогибающихся под действием собственного веса. Рассмотрены следующие балки: а) балка прямоугольного поперечного сечения постоянной толщины, но переменной ширины, свободно опертая на обоих концах; б) балка прямоугольного поперечного сечения постоянной ширины и переменной толщины или постоянной толщины и переменной ширины — один конец балки жестко заделан, а другой нагружен поперечной силой; в) равнопрочное крыло самолета с учетом подъемной аэродинамической силы и силы тяжести крыла с заполнителем. Найдена также форма равнопрочного стержня, сжимаемого продольной силой в задаче Эйлера. Равнопрочные конструкции обладают минимальным весом для заданного запаса прочности или максимальным запасом прочности для заданного веса.

Ключевые слова: равнопрочный брус, конструкции минимального веса, идеальные формы Платона, сопротивление материалов, оптимальные формы упругих тел, устойчивость равнопрочного стержня при сжатии

Equistrong heavy beam: solving the problem of Galileo Galilei

G.P. Cherepanov

The New York Academy of Sciences, New York, 10007-2157, USA

In his main book "Discorsi e Dimostrazioni Matematiche, Intomo a Due Nuove Scienze" published in 1638 by Elsevier in Leyden Galileo Galilei, "the Father" of modern science, put the material science and strength of materials on the first place. He introduced the notions of stress and strength that have been fundamental since then. Moreover, in unison with Plato's Theory of Forms he found out the perfect shape of a force-bent beam we call today equistrong. This discovery laid the foundation for search of other perfect elastic bodies as a continuation of Galilei's work. There are no theorems of existence for equistrong bodies so that the quest for them is like a gold-digging. In what follows, the shapes of the following heavy, equistrong beams were found out: a) beam of constant thickness and of variable width, simply supported at both ends, b) beam clamped at one end and loaded at the other end while having either constant thickness and variable width, or constant width and variable thickness, and c) equistrong shape of the profile of aircraft wings accounting for gravity and lift loads. The shape of equistrong rod at buckling under a compressive force is found in the Euler's problem. Equistrong structures possess minimum weight for given safety factor or maximum safety factor for given weight.

Keywords: equistrong beam, minimum weight structures, Plato's forms, strength of materials, optimum shapes of elastic bodies, buckling of equistrong rod

1. Introduction

Shortly before his death Galileo Galilei (1564-1642) published in Holland his main work "Discorsi e Dimostrazioni Matematiche, Intomo a Due Nuove Scienze", Leiden, Elsevier, 1638. It was done secretly from the authorities that persecuted him for his views discordant with generally-

accepted ones. In this book, Galilei introduced the notions of stress and strength that have been used everywhere since then. He advanced his principle that any body moves at a constant speed along a straight line if no force acts upon the body. In other words: "no force, no acceleration". For his follower Isaac Newton (1642-1726) born at the year

© Cherepanov G.P., 2016

Galilei died, it was a simple next step to declare his law that acceleration is directly proportional to force, with a coefficient being speed-independent. As distinct from Galilei, happy Sir Isaac hid his heretic views and was honored like a king in his lifetime; he was very loyal to William, Holland's Prince of Orange, who occupied England by his troops and became its famous king although he couldn't speak English. (For a while, Newton even served as a spy for this king). Newton used his power to ostracize unhappy Gottfried Leibnitz, the creator of modern calculus, ignored Robert Hooke who was the first to suggest the gravitation law, and undermined the role of Galileo Galilei in the discovery of the "first Newton's law" in mechanics. However, the principle of Galilei is valid up to now while Newton's law declared as an eternal, unshakeable law of nature was replaced by a more accurate law a hundred years ago.

In his book, Galilei also found out the optimum shape of the rectangular beam clamped at one end and loaded by a force at the other end. The maximum tensile stress in this beam was one and same along the whole beam. Today, we call this beam equistrong. It has a minimum weight as compared to all other shapes at the same safety factor, or a maximum safety factor for a given weight of the beam. These are exceptional properties of maximum strength and maximum safety desirable for any structure design. Unfortunately, there exist only a very limited set of equistrong beams; the search and discovery of such beams constitutes the problem called the problem of Galileo Galilei.

The term "equistrong" was introduced by this author in 1962 [1]. Since then, a few dozens of equistrong shapes have been found out by many authors, see the recent article [2], for more detail. The present paper is a continuation of paper [2] which is recommended to be read at first.

2. Equistrong heavy beam with simply supported ends

In what follows, it is assumed that the beam cross-section under consideration has two axes of symmetry that coincide with axes Oy and Oz of the Cartesian rectangular frame of coordinates Oxyz, with axis Ox being along the beam, while plane Oxy is the plane of pure bending. The beam material of specific weight pg is elastic, homogenous and isotropic.

In this section, it is also assumed that the beam has a rectangular cross-section of a constant thickness 2h (in the direction y), a variable width 2a(x) in the direction z, and length 2l (in the direction x). It is supposed that the beam is subject to only the gravity force directed oppositely to axis Oy. The ends of the beam at x = ±l are simply supported so that the bending moment at the ends is equal to zero; the problem, thus, becomes statically determinate. The middle of the beam is chosen as the coordinate origin.

It is required to find the width of the beam 2a(x) if the tensile stress ox is one and the same constant along the whole surface of the beam at y = h. Let us designate this

constant stress as nCTb where <3b is the ultimate strength and n is the safety factor. This is an equistrong beam so that we have [2]

hM (x) =n<V (x). (1)

Here M(x) is the bending moment, and I(x) is the moment of inertia of the beam cross-section which is equal to I = 4/3 h3a(x). (2)

Since the problem is symmetrical with respect to x, we can confine ourselves by the interval 0 < x < l. Taking into account Eqs. (1) and (2), the equilibrium equation can be written as follows

i i 3pg(l - x)j a(t)dt - 3pgJ a(t)(t - x)dt = nhoba(x). (3)

This is an integral equation with respect to a(x). Let us differentiate it twice; as a result we get

nh^b

X2a"(x) = -a(x), where X2 = -

3pg

(4)

From Eq. (3) it follows that a(l) = 0. Hence, the solution of Eq. (4) is as follows

/ N * ■ X - l

a(x) = -A sin-, A > 0.

X

(5)

Here A is an arbitrary positive constant equal to half of the maximum width of the beam.

Substituting a(x) in Eq. (3) by Eq. (5), we can derive

l n , n \nhah

— = — or l =—1-.

X 2 2\ 3pg

(6)

The weight of the whole equistrong beam is equal to

(7)

W = — pghlA. n

The length of the equistrong beam is directly proportional to the square root of its thickness, and the width of the equistrong beam is given by Eq. (5). With a sufficient accuracy, the value of the maximum width of this beam can be up to the value of l so that the beam becomes a plate.

The total weight of the equistrong beam makes up 52% of the weight of the beam with dimensions 2lx2hx2A which thickness is chosen to match the strength and safety factor of the equistrong beam.

As to Eq. (6), let us provide a numerical example of an equistrong beam. For a steel plate of thickness 1 cm we get l = 1.5 m if nab = 20 MPa, while the amplitude of the sinusoidal shape of the plate in the plan-view can be somewhat arbitrary.

3. Equistrong heavy beam: one end clamped, and the other loaded

In this section, it is assumed that a heavy beam of constant thickness 2h, length l, and variable width 2a(x) has the clamped end at x = 0 while the other end at x = l is subject to the transverse force P. It is required to find the shape of the equistrong beam, with the maximum tensile

stress being one and the same along the whole stretched surface of the beam.

Using Eqs. (1) and (2) in this case, the equilibrium equation is written in the form of the following integral equation

r 4 2

P(l -x) + 4pghJ a(t)(t -x)dt =-n<5bh a(x). (8)

x 3

Here, the direction of force P coincides with that of the gravity force.

Let us twice differentiate Eq. (8). As a result, we get

X2a'(x) = a(x), X2 =

3pg

(9)

From Eq. (8) it follows that a(l) = 0. The solution of Eq. (9) satisfying this condition can be written as follows

x -1

a(x) = ^sh-. (10)

X

Here A is a constant.

Substituting the function a(x) in Eq. (8) by that in Eq. (10), we find

P + 4pghXA = 0. (11)

From Eqs. (10) and (11) it follows that

ï P J - x

a( x) =-— sh-

(12)

4pgXh X

This is the sought shape of the equistrong heavy beam in this case. It becomes a plate in the plan-view if Psh( //X) >> >> 4p g X h2.

4. Equistrong heavy beam of constant width and variable thickness

In this section, we assume that a heavy beam has a constant width a but a variable thickness 2h(x) while its end at x = 0 is clamped and the other end at x = l is loaded by a transverse force P. The tensile stress at the unknown surface of the beam is assumed to be equal to along the whole surface which is to be found. By taking into account the own weight of the beam, this problem generalizes the original problem of Galilei.

In this case, the equilibrium equation is written in the form of the following integral equation

P(l - x) + 4pga J h(t )(t - x)dt = - nab ah2( x). (13)

x 3

Here, the direction of force is assumed to coincide with the direction of gravity.

By differentiating this equation twice, we come to the following differential equation

h( x) = 2SA f h ^

dx 1 dx

f

8 =

3pg

(14)

The first integral of this equation can be written as follows

dh

B3

381 — I = h .

1 dx J h2

Here B is an arbitrary constant.

(15)

The solution of Eq. (15) satisfying the boundary condition h(l) = 0 following from Eq. (13), has the following form

i - x = V3SBI1 B

m , ,-

where I(m) = J t/VI + t3 dt.

(16)

When x ^ l, the second term in Eq. (13) becomes small as compared to the first term so that we get:

h = I

3P

(l - x), x ^ l.

(17)

2 V

Substituting this value of h into Eq. (15) when x ^ l provides

f D f/3

B = 3 81/3 4

(18)

Equations (16) and (18) represent the final solution to the problem.

When the transverse force P is zero, the beam is subject to only the gravity. In this case, the unique solution of Eq. (13) is as follows

h=l28(l - x)2

This is the thickness of the heavy equistrong beam.

(19)

5. Equistrong wing of aircraft in transport aviation

As it was found in [2], the swept delta wing is equistrong for some special shapes. This result is valid only if the own weight of the wing can be neglected which is close to the reality for jets and fighters. In transport aviation, the own weight of wings with a filler can be essential. Let us take this weight into account under some assumptions.

The neutral plane of bending of a wing is assumed to coincide with the Oxz plane where the direction of flight is along the z axis, and the wing is attached to the fuselage at x = 0 while the wing end being at x = l. Generally, for swept wings the neutral axis of bending is different from the x axis and slanted under an angle with it.

Let us assume that:

(i) Every cross-section x = const of the wing profile is a plane of only the bending so that torsion and axial tension can be neglected.

(ii) All the cross-sections of the wing are self-similar so

that

/2 = f (x)/20, h = f (x)h0. (20)

Here lzo and h0 are some scaling units of length dimension, f(x) is the dimensionless coefficient of variable similarity to be found, lz is the chord of the wing profile at the cross-section x = const in the direction z, and h is the distance of the most stretched spot of this cross-section from the neutral plane of bending.

(iii) The area and moment of inertia of the cross-section x = const of the wing profile can be written as

A = Q1hlz, I = 92lzh3. (21)

Here 6j and 62 are some shape-dependent dimensionless coefficients of this cross-section.

Let us consider horizontal flight under positive angles of attack so that the gravity force acts along the y axis while the lift force acting in the opposite direction. The tensile stress at the most stretched spot of the wing profile is supposed to be one and the same in all cross-sections x = const throughout the equistrong wing.

Under these assumptions, the equilibrium equation for the equistrong heavy wing can be written as follows

J№„, V/c)p«V2Zz0f(t) - pgW0f2 (t)]x

x

x (t - x)dt = nab 02lz 0h2f 3( x). (22)

Here pa is the air density, V is the flight velocity, V/c is the Mach number, a0 is the angle of attack, x is the dimensionless coefficient depending on the flight regime, see [2]; and pg is the specific weight of the fill-in stuff in the wing.

In Eq. (22), the first and second terms describe the moment of lift and gravity forces while the right-hand term providing the elastic reaction.

Let us rewrite Eq. (22) as follows

f 2(x) = J[af (t) -pf 2(t)](t - x)dt,

(23)

a = -

XPaV ¿

■ P = -

pg0i

ntfhe2 hû " n^h e2 h

Here a and P are some known coefficients of dimension m 2.

Differentiate twice the integral equation, Eq. (23), which results in the following equation

( f2df I = af (x)-Pf2(x).

dx I dx

This equation can be transformed to the following form

(24)

3W^(Wf2) = af -pf2, W = df. df dx

(25)

Solving this first order differential equation provides the following result

df dx

f4 + Ci

=a f4-p f5

4 5 '

(26)

According to Eq. (23), f(l) = 0; hence, Ci = 0, and we have

f I a--2 Pf (x). dx V 6 15

(27)

By solving this equation and using the boundary condition f(l) = 0, we get

f (x) = -J0(/-x)2 + J|(/ -x).

(28)

Equations (20) and (28) provide the sought profile of the equistrong heavy wing of aircraft subject to lift and gravity.

As seen, the solution led us to the amazingly simple equation where the first term describes the effect of the gravity and the second term the effect of the aerodynamic lift force. It is a rare case of linear superposition in a non-linear mathematical problem with unknown boundaries.

When the effect of the gravity can be neglected, the first term in Eq. (28) disappears, and we come to the delta wing shape discovered in [2].

Equation (28) imposes some limitations on the maximum span of an aircraft wing that can't surely be longer than l*

l* = ^Vta. (29)

However, from the design requirements the maximum span of an aircraft wing should be even less than l*/2.

6. Euler's problem of buckling for equistrong rods

Let us study the problem of buckling for a rod of an rectangular cross-section 2ax2h and of length l where l >> a >> h, being loaded by compressive force P applied along the rod. Designate as v = v(x) the deflection of the rod/beam in the plane of bending relative to the equilibrium position v = 0, with the ends of the beam at x = 0 and x = l always being at rest. This deflection is a low-frequency, small perturbation of the initial rest position that satisfies the following equation

4Eah

3 d2 v dx2

+ 3Pv = 0, 0 < x <

(30)

Here E is Young's modulus of the rod/beam.

The function satisfying Eq. (30) and the boundary condition at x = 0 is

3P

v = v0sin(kx), where k2 =--. (31)

4Eah3

From here, satisfying the boundary condition at the other end of the rod Leonhard Euler found the minimum value of force P buckling the rod 4n2 3

PE = —rEah . (32)

E 3l2

It does not depend on the magnitude of the maximum deflection v0 and so it is valid even for infinitesimal perturbations.

However, this result takes place only in linearized problems. Generally, the stability and collapse depend on the magnitude of perturbation which is the case for equistrong beams at buckling.

We accept the following formulation of the problem: find the thickness h = h(x) of the rod/beam which is equistrong for a given buckling force P and for a given maximum deflection v0. For equistrong beams, in this case we have

3Pv = 4^abah2. (33)

From Eqs. (30) and (33), it follows that

vV2= -X, x=^/anK. (34)

dx2 E V 3P v 7

The solution of this equation satisfying the boundary conditions can be written as follows

(2 + v1/2)Vl - v1/2 = 2(1- 2x), (35)

0 < x < 1/2 /, x = x/l, v = v/v0, whereby the following condition equation should be met

8v3/4 = 3/TX. (36)

According to Eq. (35), the deflection of equistrong beam/rod increases from zero at x = 0 to the maximum value of v0 at x = //2; for / > x > //2 the deflection is symmetrical.

Using Eqs. (33), (34), and (36) we can find the shape of the equistrong rod/beam and the criterion equation connecting P and v0

h l3PvCx), 32^3/^ = 3/2VMnab )3'2. (37)

V natfb

Here v(x) is given by Eq. (35).

As a reminder, by the limitation of the beam bending theory the following requirement should be satisfied

anv0ab << 3P. (38)

Within this limitation, from the solution, Eq. (37), it follows that the greater is the admissible maximum deflection, the less is the critical load and the thinner is the equi-

strong rod. According to Eqs. (37), the value of h is directly proportional to 1/ v0.

This solution demonstrates the paramount role of restraining ties in nonlinear stability problems.

7. Conclusion

Besides the problems solved in the present paper, the foregoing analysis and methods can be used in many other problems of beams, plates, and arcs so that the list of equi-strong structures might be considerably increased in near future. This fact provides a great moral motive to this author.

The simplicity and urgency of these problems solutions make us recommend their inclusion in curricula of any courses on mechanics of materials.

8. Dedication

The present paper is written to honor Victor Evgenievich Panin, the founder of the Journal of Physical Mesome-chanics, who celebrated his eighty fifth birth day last year. I appreciate very much his quest for uncommon paths in science.

References

1. Cherepanov G.P. Inverse elastic-plastic problems of anti-plane strain // J. Appl. Math. Mech. - 1962. - V. 26. - No. 5. - P. 784-788.

2. Cherepanov G.P. Optimum shapes of elastic bodies: equistrong wing of aircraft and equistrong underground tunnel // Phys. Mesomech. -2015. - V. 18. - No. 4. - P. 391-401.

Поступила в редакцию 09.07.2015 г.

Сведения об авторе

Genady P. Cherepanov, Prof., Hon. Life Member, The New York Academy of Sciences, USA, genacherepanov@hotmail.com