theory of beams. the application of the laplace transformation method to engineering problems

THEORY OF BEAMS The Application of the

Laplace Transformation Method to Engineering Problems

2nd enlarged edition

by Dr. T. IWINSKI

Institute of Mathematics Polish Academy of Sciences Warsaw

Translated from the Polish by E. P. B E R N A T

Professor of Mechanical Engineering at the

Ahmadu Bello University, Zaria, N. Nigeria

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Copyright (c) 1958 and 1967 Pergamon Press Ltd. First edition 1958

Second edition 1967 Library of Congress Catalog Card No. 66-28689

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(3009/67

PREFACE TO THE SECOND EDITION

THE amount of material contained in the second edition is roughly doubled compared with the first edition. It was a great encouragement for the author to continue his work when the first edition of the book met with the approval of readers and reviewers and was then translated into Japanese [32].

The introductory sections of Chapter I and sections 1-9 of Chapter II of the first edition are retained unchanged, but section 10 of Chapter II, which deals with beams of variable cross-section, has been substantially expounded and appears now as Chapter III. It contains the fundamental equations of the theory and also some examples on boundary conditions, but not as numerous as those in the Chapter II. This is because the theory of Chapter II may, without much difficulty, be applied also to the corresponding cases of beams with variable rigidity.

Throughout the whole book the stress is put rather on the method used than on the comprehensive coverage of all the important cases which may be met in engineering practice. This approach has influenced the contents of Chapter IV. Although the problems selected for discussion should be of interest to all who specialize in the field of structural engineering, the choice was decided upon primarily with a view to illustrate the applications of the method (or more precisely to illustrate some important subtle variations of the same method).

vii

viii Preface to the second edition

The application of the Laplace transformation and the step functions necessitates a regular use of some transformation formulae. These formulae are tabulated, for convenience, in Chapter V.

PREFACE TO THE FIRST EDITION

THIS monograph is addressed mainly to technically minded readers. Its principal aim is to discuss the solutions of the ordinary differential equations with assigned boundary conditions which occur in the theory of beams by means of integral transforms, now called the "Laplace integrals" or "Laplace transformation".

This method of solving linear differential equations was originally published by Laplace nearly one hundred and fifty years ago (1812), and has been used by mathematicians alongside the other methods as a regular training in higher mathematics for a good many years. Engineers, on the other hand, whose mathematical training, of necessity, covers a more restricted range are still not very familiar with the method. But following the natural evolution of events, the engineer finds nowadays that he is continuously requiring more and more mathematical knowledge, i.e. more powerful mathematical tools to be able to solve the more complicated problems in ever expanding fields of engineering science. And so the engineer came in recent years to realize the usefulness of the Laplace transformation method.

The method provides a form of shorthand by means of which very quick and elegant solutions of a great many engineering problems may be obtained, often saving much of the laborious calculations involved in the classical approach. It may be objected that these solutions are in a rather symbolic form and in fact require conversion into the more practical form in order to

X Preface to the first edition

be directly amenable to physical interpretation and this, in turn, may be either a difficult or laborious process. However, the solutions can be given in standardized forms, e.g. arranged as tables similar, say, to those of logarithms or other specific functions, thus greatly simplifying the actual procedure of analysis. And once the technique has been mastered, the method of solution becomes more direct and straightforward than the "classical" methods.

E.P.B. Sheffield, September 1957

Chapter I

INTRODUCTORY INFORMATION

1. General Introduction The topics discussed in this book are taken from the field of

the theory of structures but from the point of view of the method employed for solving these problems the book may be considered as the study of some linear differential equation. The solution of several simple equations proves to be not so simple a matter. Much attention has been given to this problem both by engineers and mathematicians, but it does not appear that the subject is fully exhausted. The reason for this is the nature of the practical engineering problem: complex boundary conditions and complex loading.

The mathematical representation of these complications is the appearance of discontinuities in the functions describing the physical quantities or the geometry of the problems. Usually, this is an ordinary discontinuity (see the note in section 2 of Chapter II), which leads to the appearance of the so-called step functions.

The step functions are so characteristic for the structural problems that without them it would almost be impossible to formulate any general problem. Hence the concept of the step functions (in one form or the other) had been introduced at the very beginning of formulation of the theory of structures.

Interesting comments on this are given by J. E. Goldberg [19]. Even such classical writers as A. Clebsch [10] and A. Föppl [17] are connected with this subject. The interest in the

1

2 Theory of beams

matter has been recently revived afresh. In addition to the names listed in ref. 19 the following may be added: W. H. Macauley 125], J. Case [24], R. V. Southwell [23], A. J. S. Pippard, J. F. Baker [22], C. L. Brown [21], E. Kosko [30], S. Timoshenko [20], J. E. Goldberg [19], T. Iwinski [18], J. Nowinski [9], T. van Langendonck [7] and S. Drobot [12].

There are several good textbooks on the subject of the theory of beams. A special mention deserves the book by S. E. Mike-ladze [8] in which some very interesting methods of the mathematical analysis and of the differential equations are employed.

The lively interest in the subject in the present state of knowledge is aroused not so much by the need of solving the problems as for the sake of introducing new methods with the possibility of simplification of the solutions. The author feels that a real simplification results from the use of the Laplace transformation method or more accurately from the use of Laplace transformation together with some rationalized properties of the step functions. Such a treatment not only leads to simplification of the theory of beams, but also makes this theory more general.

Some interesting remarks on the application of Laplace transformation to the technical problems are given by the reviewer of the first edition of this book in the journal Physics To-day, July 1959, pp. 43-44. The method of Laplace transformation as well as other methods of integral transformation or of differential operators such as Carson-Laplace method [31], Ditkin-Kuznetzov method [3] or Mikusinski method [27], have been widely used for sometime in several branches of engineering (e.g. in the theory of electric circuits, in the theory of servomechanisms or in the theory of vibrations). But in the theory of beams, or more generally in the theory of deformation of solids, these methods have been introduced relatively late. Some examples on the theory of beams are included

Introductory information 3

in the excellent textbooks on the application of Laplace transformation by R. V. Churchill [2] and W. T. Thomson [26].f

In the last decade this lag has been well compensated for. A substantial contribution is due to the Polish Scientists. A reference to the Polish Scientific Publications (Archives of Applied Mechanics, Engineering Dissertations, Proceedings of Vibration Problems, etc.) will disclose that the operational methods are a useful tool in solving problems in the fields of strength of materials, theory of structures, elasticity, thermo-elasticity, plasticity and theory of heat. A special mention here deserve publications by W. Nowacki [28], [29].

The acceptance of the operational methods is understandable. They provide easy mathematical algorithms similar to those of the classical algebraic transformations; even in the more difficult, non-typical cases, the solution often reduces to the mere use of the tables of transforms.

As far as the engineers are concerned a difficulty may arise in mastering of the theory, which, for example, in the case of Laplace transformation will require familiarity with the analytical functions. However, it may be pointed out that the content of the mathematical knowledge of engineers nowadays is constantly increasing and also that it is not imperative to acquire a deeper knowledge of the theory in order to be able to use the numerous available tables of the transforms (just as the use of the mathematical or physical tables does not require the knowledge of how these tables have been compiled).

In Chapter II beams of constant rigidity or composed of sections of constant rigidity are discussed. Starting with the differential equation of the deflection curve and with a uniformly distributed loading, the solution of the equation is obtained by means of the Laplace transformation. Then, by means of a suitable definition of concentrated forces and cou-

t Compare the observations on the subject by J. E. Goldberg in Applied Mechanics Reviews, May 1959, rev. no. 2280.

4 Theory of beams

pies, the author generalizes the solution to include the case of simultaneous loading by concentrated forces and couples as well as distributed loads (uniform and non-uniform).

It is shown that the elastic curve of a beam loaded in such a general manner and resting on two or more supports is given by the relation y = W(x) + S(x), where W{x) and S(x) are step functions defined and briefly discussed in the introduction. The function W{x) is a polynomial whose coefficients depend entirely upon the nature of the supports and can be evaluated from the boundary conditions, while the second function S(x), called by the author the load function, depends solely upon the loading of the beam. If the load distribution is known, the load function can be determined beforehand independently of the step polynomial W{x) either by direct application of the Laplace transformation or by means of tables of the Laplace transforms.

In this way the problem of finding the deflection curve is transformed from the problem of solving a differential equation to that of solving algebraic equations. In short, it reduces to finding the coefficients of the polynomial W(x) by solving a system of simultaneous algebraic equations. Moreover, the use of this method does away with the necessity of different treatment of the iso- and hyper-static beams since the same general relation holds in both cases.

The chapter ends with solutions for beams on two supports under various conditions of end fixing (freely supported, built-in, etc.) as well as for continuous beams both with rigid and elastic supports.

Chapter III deals with beams of variable flexural rigidity under the action of transverse loading. The loading is most general. The elastic curve is deduced by the use of transformation. It is shown that also in this case the deflection of the beam y(x) contains two terms, i.e. y = 0(x) + S(x). The second term S(x) is, as before, a function of the load (however, much more complex because of the variable rigidity B(x)). The first

term Φ(χ) = y0 +y'0x+M^ß^x) -f Ro02(x) depends on the initial constants y0 and y'Q on the reaction of the support M0 and R0 at the cross-sections x=0 and on the two other functions which, in turn, depend on the way the cross-section varies, i.e. these two functions are in effect the shape functions. It follows, therefore, that also in the case of beams with variable cross-section the theory of Chapter II may be used. Thus the problem of the boundary conditions may be solved independently of the loading and irrespectively of the shape of the beam, which obviously renders the theory very general. The basic formulae are given either in a form which includes the transforms (this allows the use of tables) or in the usual analytical form.

The chapter ends with some examples. From these examples it is seen that the solution of any practical problem encountered by an engineer reduces simply to substitution of the numerical values into the provided formulae.

The examples solved in Chapter IV were chosen so as to point out the main features of the method used. They represent the following problems: beams with axial loading (compressive and tensile); beams on an elastic foundation (Winkler's type); beams on a non-homogeneous elastic foundation (non-homo-genity being along the beam). As for the special features of the method the following aspects are analysed: the problem of transforms for the physical quantities (loading), ways of solving for boundary conditions and derivation of the discontinuous solutions of differential equation by means of Laplace transformation.

2. Step functions and multi-step functions

Suppose we are given two functions: fx{x) which is defined and continuous in the open region (#, c), and f2(x) which is defined and continuous in the open region (c, b), where a < c < b.

6 Theory of beams

Assuming that these functions are finite at both ends of their respective regions, a new function over the whole region {a, b) can now be defined by means of them as follows:

f fAx), a< x < c, fW = r) A ( Ι · 2 · 1 }

I J2\XJ9 C < X < 0.

Alternatively, eq. (1.2.1) may be written in a more convenient form for our purposes as

fix) = <Λ(χ)>εα+</2(*)>£. (1.2.2)

The function f{x) so defined will be called the step function, and the terms such as (Α(χ))°α and (/2(x))c will be referred to as its elements. Such a function is still to be defined at the three points (a, b, c), corresponding to the ends of the intervals (a, c) and (c, b). This can be done in several ways. However, a convenient method is to make use of the left and right end limits of the elements, i.e. Λ(#+0), fi(c~~^) a n d Λ(^+0) , f2(b— 0), and to ascribe to the function f(x) at the ends of the region (a, b) the outer limitsΛ(α+0) and/2(o—0), while at the intermediate point c to treat/(x) as double valued, i.e. either /i(c—0) or f2(c+0) depending whether the interval (a, c) or (c, b) is considered.

The need for such a double valued step function at certain points arises, for example, when considering the distribution of the shear force in a transversely loaded beam. At the points of application of the concentrated loads or at the supports, the shearing force is double valued.

The graph of the step function

Αχ) = <yi>ea+<y»>bc, (1-2.3)

where yx and y2 are constant, is shown in Fig. 1. From the above discussion it is not to be assumed that a

step function must have points where it is double valued; it

is possible that/i(c-0) =/2(c + 0). In such a case, in accordance with the definition, this common limit will be the value of the step function/(x) at c, and the function is single valued over the whole region.

a T

FIG. 1.

As an example of such a single valued step function we have /(*) = <kxyQ + <kx+P(x-cfyb

c (0 < c < 6). (1.2.4)

Its graph is given in Fig. 2.

*»»x

If there are more than two elements, such a function will be called multi-step function and then we can write

/(*) = Y </i(*)>Si41 (* = *o < *i < . . . < ** = &). 2 = 0

(1.2.5)

8 Theory of beams

It will be found advantageous to retain the above notation even for such cases where one of the intervals reduces to a point. We may write then

</i(*»S=/i(<0· (1-2.6)

The symbol {fi(x))ha\, where ax < bi9 indicates that the func

tion f{{x) is to be considered only within the interval (a^ b^. Keeping this in mind, we may modify slightly the definition of a step function given in eq. (1.2.2) as follows

Ax) = £ {a < c < b). (1.2.7)

*y

k a --c H

I *&

F I G . 3.

Its graph is given in Fig. 3. It is easily seen that the elements of the step function (1.2.7) overlap. Equation (1.2.7), of course, reduces to eq. (1.2.2), viz.:

</l>5a+</2>? = </l>S+</l>S+</2>S = </l>S+</l+/2>S. (1.2.8)

When a step function, given in the form (1.2.7), has an element extending over the whole region under consideration (a, b), we shall leave out the sign < )b

a writing

Ax) = Λ(*) + <Λ(*)>2 {a^c< b). (1.2.9)

Further, if in a step function there is a zero valued element over an interval, e.g. <0>J?, we shall omit it all together; and

conversely, if in some interval in the region (a, b) the step function is not given, it is to be understood that its value is zero in that interval. Hence every sum of the form

Σ </*>£ 0·2 ·1 0) denotes a step function in the region (a, b) if only a^ a{< b or a < bi ^ b or if a{ ^ b^ and the functions f{ are determined respectively in the intervals (at 6{) and have finite l i m i t s / ^ + 0 ) a n d / A - 0 ) .

Here are some examples on the application of step functions: 1. If a beam of length /, with its neutral axis along the

Ox-axis, has a flexural rigidity Βλ in the first half of its length and B2 in the second half, then its rigidity over the whole length can be expressed as

*(*) = <*i>ii2+<*a>U. (1.2.11)

2. If the beam given above is acted on by a uniformly distributed load qx in the interval (0,|/) and in addition there is another distributed load q2 over the interval (\l, f/) while the remainder of the beam (|/, /) is unloaded, the load over the whole length may be expressed as follows:

?(*) = <<7ι>ί>,2+<<?2>?/14 (1-2.12) or

Φ) = < ι̂>?)/4 + <^ι + ^2>ί;ΐ+<^2>?^ (1.2.12α)

or alternatively

q(x) =<?ι>{(4 + <?1 + Λ>|{ϊ + <?2>β4 + <0>«Ι/4. (1.2.126

The graphs of these functions are given in Fig. 4. 3. The reciprocal of the rigidity of the above beam is

given by 1 / 1 \ l/2 / l \ l

10 Theory of beams

or

B(x) < / l \ / 4

+ < / 1 \ i / 2 / 1 \ 3 » 4 / 1 \ ' \ ß i / ; / 4 \B*/m \B*/zui

(1.2.13a)

Using relations (1.2.12ft) and (1.2.13α) the differential equation of the deflected beam

B ,,ιν (B = const.) (1.2.14)

can be written as follows:

*--<£ 1/4

+ /3±±3± V2

+ /M*\ 3Ϊ/4

1/4 \ B 2 / l l 2 + <0>Jl,4.

(1.2.14a) This form of the differential equation having a step function

on the right-hand side of the equation (in this case with intervals of constant values) is particularly suitable for solution by the Laplace transformation method. It is to be noted that the right-hand side of eq. (1.2.14) is obtained as the result of "multiplication" of two step functions, this multiplication being defined as follows:

if /=S+</8>S

and g=<gl>Ca+<g2>bo

then fg = <figl>Ca+<f2g2>bc- (1-2.15)

Whence the following relation follows:

f<g>bc = <fg>bc (1-2.16)

where / denotes a given function extending over the whole region (a, b). It will also be observed that if a step function

fix) = Σ </i(*)>fc te ^ *0

is differentiable in the region (a, b), then its derivative is

/'(*)= £<//(*)>«;- 0-2-17) i = l

The sufiicient condition that a step function be differentiable is that each of its elements must be differentiable in the respective interval, and that there must be a finite derivative at each end of each interval.

3. Some remarks on the Laplace transformation methodt

If f(x) is a function of a real variable, defined for all x ^ 0 and satisfying certain conditions (which will not be given now), then we define as the Laplace transform of the function/(x) a new function F(s) such that

F(s) = J e~8Xf(x)dx. (1.3.1)

t Full treatment of the method may be found in refs. 1 and 2.

12 Theory of beams

The integral has a definite value (i.e. converges), provided s > a, where a is a real positive number or zero. Thus, for example, if fix) = eax we must have s > a. Of course, if there is no such .y, e.g. if fix) = e*2, the Laplace transform does not exist.

We shall denote the Laplace transform by the symbol L so that eq. (1.3.1) can also be written as

F(s) = L{f(x)}. (1.3.1a)

The inversion of the Laplace transform we shall denote by L _ 1 . Thus given a transform F(s), its inversion in general leads to the function f(x) so that we can write

fix) = L^{F(s)}. (1-3.2)

On the basis of the definition (1.3.1) and certain properties of the Laplace transformation, the transforms of various common functions can be evaluated. These have, however, been tabulated and can be found in special publications, e.g. ref. 3. We quote here some formulae and theorems of which use will be made in this paper (proofs being omitted):

(1) L{afix) + bg(x)} = aL{fix)} + bL{g(x)} (1.3.3) (a = const., b = const.)

(2) L-1{aF(s)-j-bG(s)} = aL^Fis^ + bL^Gis)} (1.3.4) (a = const., b = const.). (3) L_I m=(ST (ί - ο )· (ι·3·5)

(4) L- {^} = <0>S+<^^X (s > 0). (1.3.6) (5) If a continuous function fix), defined within the region

(0, + oo)5 has (n — 1) continuous derivatives while its nth derivative has only a finite number of ordinary discontinuities,1"

t At the points of ordinary discontinuity there are definite left- and right-side limits of the function.


and if it increases at most as the exponental function eax (a > 0) when x -̂ oo? then the transform of f{n\x) is given by the following formula: L{ßn\x)} = ^^)-^-1 / (+0)- i s n - 2 / , , (+0)

j/(n-2)( + 0)-/(n-l)( + 0) (s > a). (1.3.7)

Chapter II

THEORY OF BEAMS

1. Assumptions. Differential equation of the elastic curve. Load function

Consider a beam of length / as shown in Fig. 5. We shall use the right-handed system of rectangular coordinates x, y9 z with the origin at the centroid of the left end cross-section of the beam, the x-axis along the axis of the beam and y- and z-axes taken along the principal axes of the second moment of the cross-section, the positive direction of j-axis being vertically downwards.

FIG. 5.

The vector quantities (forces, force moments, bending moments, shearing forces) with which we shall be dealing in the problems forming the subject of this paper, in the majority of cases, will be entirely in the direction of one of the axes of coordinates. Then we shall be able to treat them as scalars, a

14

Theory of beams 15

single component defining completely their magnitude (the other two components being zero). If this single component is positive, we shall speak of the corresponding vector (or vector quantity) as being positive, and vice versa. This, therefore, will be the meaning of a force or a force moment being positive or negative. The positive direction for measuring the angles will be in conformity with the universal usage, i.e. that which, for example, in the xj>-plane, which, for example, requires a right-handed screw to rotate Ox to Oy the tip of the screw moving along Oz (arrowed in Fig. 5).

The internal (elastic) force system set up at a cross section of the beam (parallel to >>z-plane) reduces to a transverse vertical force Q(x) (shearing force) and a couple M(x) (bending moment). Of the two possibilities we choose that one in which Q(x) and M(x) are defined by the effect of the left portion of the beam (i.e. to the left of the section in question) on to the right portion. In other words, if the portion of the beam to the left of the section be removed, Q(x) and M{x) would be the forces equivalent to those that the left portion exerts on the right portion through the section, so that with Q(x) and M(x) the equilibrium of the right portion would be preserved.

It can be shown that in the regions where M(x) is positive, the beam deflects so that its concavity is upwards, i.e. the elastic curve in convex with respect to the axes of coordinates (since the positive branch of j-axis is directed downwards). Similarly, in the regions where M(x) is negative, the elastic curve is concave.

With these assumptions the differential equation of the deflected beam as given by the theory of elasticity

Ely" = ±M(x), (2.1.1)

(where / is the second moment of the cross-sectional area with respect to the centroidal axis parallel to z-axis, M(x) is the bending moment and E the Young's modulus of elasticity)

16 Theory of beams

becomes Ely" = -M(x). (2.1.2)

Assuming further that no concentrated load or couple is applied within the portion under consideration, then with Q(x) being the shearing force and q{x) the load per unit length of the beam we have

*ψ-=-ΟΜ (2-1-3)

or

§=«*)· (2.1-4)

We shall now consider beams having a constant second moment throughout the whole length. In this case double differentiation of eq. (2.1.2) (with the above given relations taken into account) leads to the following differential equation

y1Y = -^q(*\ (2.1.5)

where letter B (B = const.) is used to represent the flexural rigidity El of the beam.

Let us now perform the Laplace transformation of the above differential equation. By eq. (1.3.7) we obtain for the left side

^ ) - 5 · Λ - ^ ί - ν ; ' - Λ " , (2.1.6)

where zero suffix denotes the value of the respective functions at the point x = +0, and Y(s) represents the transform of the elastic curve y(x). For the right side of eq. (2.1.5) the result of the transformation depends on the type of loading acting on the beam. Let us, therefore, consider the three cases most frequent in practice.

Theory of beams 17

Case 1

Uniformly distributed load (q(x) = q = const.) on a portion of the beam between x — a and x = b (Fig. 6).

Urn—

fy

- a — ► h

_ cr

—L 1 I

W

FIG. 6.

Transformation of the right side of eq. (2.1.5) in this case yields

oo 5

■j (e-sxq(x)dx = - | [e-**dx = - X [ e - « ] « ;

~~ sB sB6 (2.1.7)

If the beam carries several (say m) uniformly distributed loads, each of different magnitude (^) and acting on a different portion of the beam from x = at to x = b{ (Fig. 7) then instead of a single term in eq. (2.1.7) there will be m such terms, and

Γ Tl El

JLL

L· b.

FIG. 7.

18 Theory of beams

the transformation (after some simplifications) will result in

I m

Performing now the inverse transformation as indicated in eqs. (1.3.5) and (1.3.6) the equation of the elastic curve for the generalized case 1 is obtained as follows:

Λχ) = yo+y'ox+yö jr+y'o"^v + \ m 1 m

+ 4!ßi?1^< ( x~f l i ) 4 >'"~4bi?1'7' : < ( x~Z' i ) 4 > i ' · (2-L9)

The four unknown quantities y09 y'0, y'0' andy'^' appearing in this equation can be evaluated, as will be shown later, from the boundary conditions depending on the nature of the supports of the beam.

It is to be noted that the first four terms on the right side of eq. (2.1.9) form a polynomial of third degree. Its existence is conditioned by the form of the left side of the differential equation (2.1.5). The remaining terms (each a multi-step function), on the other hand, clearly depend solely upon the form of the function q(x), i.e. upon the type of loading of the beam. This distinction in the nature of the terms on the right side of eq. (2.1.9) will enable us later directly to write down relevant formulae for the deflection as well as for other characteristic data of a beam loaded in any arbitrary manner.

Denoting the above mentioned polynomial by the symbol W(x), eq. (2.1.9) can be written

1 m

1 m - 245 .Σ <?«<(*-*0%· (2-1-10)

Theory of beams 19

Case 2

A concentrated load P applied at x = c (Fig. 8). Instead of this concentrated load consider a uniformly distributed load q over a length a between x = c and x = c+oc9 its magnitude

P

ill—p. L« _ — i ►]

fy

X

FIG. 8.

being such that q*<x = P, i.e. q being a function of a. Now letting a tend to zero while the product q*oc is kept constant so that q -* oo? the concentrated load P may be defined as follows:

lim qoc = P. (2.1.11)

By eq. (2.1.10) the deflection of the beam with this type of loading is

y(x) = W(x) + -^q<(x-cy>*-—q((x-c-oc)*>c+a,

or using relation (1.2.8) it becomes further

y(x)= lV(x) + ̂ Bq(x-cyy+* + +ά^~' ) 4"( χ"'"α ) 4 ) έ + α · {2ΛΛ2)

The elastic curve equation is now obtained by putting a -> 0. The first term of eq. (2.1.12) does not change in the limit, since

20 Theory of beams

it does not depend upon x. The second term 1

245 <?<(*-c)*>r

is to be considered in the interval (c, c+oc) and the length of the interval tends to zero as a tends to zero. We may, therefore, put

x = c+Θχ, where Θ is a small number such that 0 ^ Θ ^ 1 and using this substitution we get

(x-cY l i m ^ ( x - c ) 4 = -024B 24* J?. q<x-

1 2 4 £ a _ 0

lim #a lim 04

= 0. ► o a

Hence the second term of the left side of eq. (2.1.12) vanishes altogether. We seek now the limit of the third term:

*o |_ l im| -^[ ( * -c )M*-c-«)«]

24B1™ qoc (x-cy-(x-c-ocy

£§«*-#-IB*-* (2.1.13)

By definition, the product q-oc in the expression for the limit (2.1.11) can be replaced by the concentrated load P, while the next factor in the limit is seen to be the derivative off(x) = x* at the point (x — c).

Hence the equation of the elastic curve of a beam with a concentrated load P is

y(x)= W(x) + j^((x-cfyc. (2.1.14)

It can be easily checked by differentiation that the meaning of the constants y0, y'0 and y'0' is always true to that implied by

Theory of beams 21

the notation, i.e. the constants are the relevant derivatives of the function (2.1.14) at x — 0; but this is not always true as far as the constant y^' is concerned. In fact, since by triple differentiation of eq. (2.1.14),

an incongruity would arise if c = 0, i.e. if the load be applied at x = 0. It might perhaps seem better to replace the symbol y^' by another, say A, and consider it to be a constant to be determined from the boundary conditions. But, for the sake of uniformity of notation, it is preferable to retain the symbol y^\ and in order to avoid the above incongruity we shall exclude the case c = 0. This restriction does not impair the generality of the theory (see section 2), while allowing a much simpler method of evaluation of the constants. We shall henceforth assume that c ^ 0 and that eq. (2.1.14) is always valid.

Λ1 jr i L C| PH 1 ^ . i

|y F I G . 9.

When the beam is loaded by n concentrated loads P{ applied at points ci (Fig. 9) the equation of its elastic curve will then have n such expressions as that in eq. (2.1.13) so that we have

y(x) = W(x) + -^liPi((x-ci)% ( e ^ O ) . (2.1.15)

22 Theory of beams

Case 3

A couple of magnitude M at a point x = d (Fig. 10). Replace the couple M by two forces — P and + P applied at d and d+oc respectively so that P.oc = M, and keeping the product constant let a -* 0, then the point moment (couple) can be defined as

lim Poc = M. (2.1.16) α - ν θ

FIG. 10.

Now by eq. (2.1.15) we can write for this case

y(x) = W(x)-^((x-dfyd + ~((x-d-ocf}ld+a9

and hence by relation (1.2.8)

y(x)= W(x)-*B<(x-dr>i+·-

When oc -*■ 0, by a similar reasoning to that employed in case 2, this equation reduces finally to

M Ax) - W(x) - TB {{X - d)'% (d*0). (2.1.17)

As in the preceding case the notation in the polynomial W(x) may be retained if we assume d Φ 0, i.e. if the case where

Theory of beams 23

Macts at x = 0 is excluded. Again, it is stressed that the generality of the method is not impaired by this restriction.

On the basis of the above simple cases (and the principle of superposition) if a beam is loaded by any arbitrary combination of uniformly distributed and concentrated loads as well as couples (Fig. 11) the equation of its deflection curve y{x) will be obtained from eqs. (2.1.9), (2.1.15) and (2.1.17) as

] m ] m

24B £ V 2 4 5 ^ 1 "

+^|/Κ^-^)3>^-^.Σ^<(^-^)2>^ (ci ^ 0, di* 0). (2.1.18)

This is the fundamental formula for the solution of problems with beams. It is to be observed that when developing this formula no assumptions were made as to the nature of the supports of the beam. We conclude, therefore, that the formula, after some suitable further transformation which will be shown later, may be used for any arbitrary type of beam, a single span as well as a continuous beam, provided the loads are as specified above, i.e. transverse loads or/and couples.

0 ♦ t t JL M; VM

FIG. 11.

24 Theory of beams

There are four unknown constant quantities in eq. (2.1.18), viz., y0, y'0, y'0' and y^' which for any particular case have to be evaluated from the boundary conditions, i.e. fixing conditions, and the procedure is the same both for the statically determinate and statically indeterminate beams. Further, with this method there is no need to differentiate between iso- and hyper-static beams which gives a more general character to the treatment of the problem of bending. This will be shown clearly in examples given further in the paper (see section 2 and following).

Let us denote the second term of eq. (2.1.18) by I r a 1 m

(2.1.19) This is seen to be a multi-step function, and it will be called the load function. With this notation eq. (2.1.18) simplifies to

y(x) = W(x) + S(x). (2.1.20)

To find the deflection at any point of the beam x = x0 (e.g. at the mid-span x0 = 1/2) we must evaluate S(x0), which will not be a difficult matter if we observe that an expression like

<(*o-*i)n>L, is to be taken as equal zero whenever the value of x0— a{

becomes negative as well as when it is equal zero. In practical application of the formula one cannot fail to notice this, and the terms with negative values under the power sign will be left out automatically.

In this way, having started with the differential equation (2.1.5) of the elastic curve, we have succeeded in reducing the problem to one of solving for the coefficients in the polynomial W(x) and evaluation of the load function S(x), the latter, for the most frequent cases, being given by eq. (2.1.19).

Theory of beams 25

However, this is not the most general form of the load function since beside concentrated loads and couples it comprises only uniformly distributed loads. In practice we may encounter the loads which are distributed non-uniformly. In these cases q(x) itself is a step function of x and in consequence the load function S(x) will contain an additional term, namely

Sx(x) = - L " 1 T*L{l(x)} (2.1.21)

Also the flexural rigidity of the beam need not necessarily be constant over its whole length. The beam may be stepped and so have only portions of constant rigidity. In such a case we can subdivide the regions over which the integration of the Laplace transforms is to be carried so as to take account not only of the discontinuity in the load on the beam but also the jumps in the rigidity. The result will be a similar load function S(x) only having more elements in it.

If we recall that there are available very extensively worked out tables of the Laplace transforms, it will be seen that the evaluation of functions S(x), and hence the elastic curves, including the cases of beams with a rather involved loading, is not a difficult proposition, and can be successfully dealt with even by technicians not very skilled in mathematical analysis.

In our further discussions, if no detailed specification of the loading is given, it is to be understood that the general case of distributed loading q(x) is implied. The question of evaluation of the load function S(x) for harder examples, for the time being, will not be considered.

For the reasons given above (when developing the load function) it will be assumed that the load function S(x) satisfies the following conditions:

S"(0) = S'"(0) = 0.

26 Theory of beams

2. The elastic curve of single-span beams

By eq. (2.1.20) of the preceding chapter the problem of finding the equation of the elastic curve has been reduced to:

(1) evaluation of the relevant load function S(x) for any given loading of the beam,

(2) writing down the relevant polynomial W{x). The first part of the problem has been dealt with in section 1

and the expressions for the load function were developed by means of the Laplace transformation for cases sufficiently general from a practical point of view, i.e. including the case of a stepped beam loaded simultaneously by uniformly distributed loads, concentrated loads and couples. In this section we shall show how to determine the polynomial W{x) for single span beams. The coefficients of this polynomial depend both upon the loading and the fixing of the beam. We shall consider some standard types of supports as found in practice and hence general forms of the deflection curves will be given, applicable to any type of loading, as defined by the load function S(x), but valid only for the specified type of the supports. These formulae will enable us to solve any particular problem simply by substituting the appropriate numerical data in them.

(a) CASE 1. BEAM RIGIDLY BUILT-IN AT BOTH ENDS (FIG. 12)

The boundary conditions for this type of supports are (1) y0 = 0, (3) y(l) = 0, (2) y'0 = 0, (4) / ( / ) = 0.

Using the first two conditions the equation of the elastic curve is

y(x) = y'a~+y'o'^+s(x). (2.2.2) The two remaining unknown coefficients in this equation can be found from the following two equations obtained by the

Theory of beams 27

UK· f

Wr-*< 1_J!

> Ü »-X

FIG. 12.

conditions (3) and (4)

3/J>O+/2X;'= -7^ (0 ,

2iy;+/Vo" = -2S'(/). These equations yield

Jo = 72 [« '( /)-35(0],

>£'=£ [25(0-«'(/)].

(2.2.3)

(2.2.4)

If we confine ourselves only to the types of loading for which the load function S(x) was developed in eq. (2.1.19), then after S(l) and S'(l) had been evaluated and their values substituted in eqs. (2.2.2) and (2.2.4) the equation ofthe elastic curve is as follows:

I r a I r a

• +24B^1qi<ix~aiY'>la~24B^1

qi<(x~biy>lb'+

+ jg .Σ Pi<(*~Cif>lc< - jg .Σ M^x-difYä,, (2.2.2c)

28 Theory of beams

and the equations for y'0' and y'^' are then

yö Bl Σ« [(l-af-il-bd3]-

4/ [(/- f l i )4-(/-^i)4] \ + t p i (l-cd2-

-jQ-Cif £ΜΛ 2 ( / - ^ ) - γ ( / - ^ ) 2

y'o" «\Σ* [('-«^-('-W (2.2.4a)

B/2

-27[(/"-«04-(/-^)4]| + .ΣΛ 3 ( / - C i ) 2 -

/ (/-cO3 P

i = l eii-dd-jii-dd2

Sometimes, for the computational reasons, the above equations will be more convenient when slightly modified as follows:

1 ( 1 m

yö = BP 1 2 .Σ ? i [C- «()3( /+3«i)-( /-^)3( /+ i t —1

n

Σ 1 = 1

+ 3^)] + X PiCi(l- Cif - X Afi(/- i/0 ( 3 4 - / ) [ , P

.Σ i = l

i f i m J ° " = ~ i? I y £i i [ ( /_ ai)3 (/+ad ~ {l~bif (/+

(2.2.40)

+ *i)] + X Pi(/-c i)2(/+2ci)-6 X M ^ Z - J o f · P

.Σ 1 = 1

(b) CASE 2. SIMPLE BEAM FREELY SUPPORTED AT BOTH ENDS (FIG. 13)

For this type of supports the boundary conditions are (1) y0 = 0, (3) y (i) = 0, (2) K' = 0, (4) / ' ( / ) = 0. (2.2.5)

Theory of beams

0 T

dj

I \)JX. » '

FIG. 13.

Hence the equation of its elastic curve is

y(x) = y'0x+y'0"~+S(x), (2.2.6)

and the simultaneous equations defining the coefiicients y^' and y'0" become

6y'0i+yO"i3= -65(/), ^"/3 = -PS"(l). ( ' ' )

Whence

^--^[^"(θ-θ^θ] , (2.2.8)

y'0"= -jS"(l).

Again for S(x) as defined by eq. (2.1.19) the equation for the elastic curve in this case is

x3 1 m

y(x) = y'0x+y'0" -g- + ^ {Σ <?*<(*-fli)%-

I m I n

- 245 .Σ i i<(^*, ) 4 )U^.E/ i ( (^^>l ·

^f^ax-ddVä,, (2.2.6a)

30 Theory of beams

and for the constant y'0 and y'0"

- ß i ) 2 - ( / -^) 2 ]}+ | Σ PiW-Cif-P(i-cd]-

- y .Σ Mi [ c - ^ - j ] } ' (2·2·8α>

Jo 1 f m

,X?i[(/-e<)2-(/-*<)i]+ 2BI )fr

+ 2 Σ Λ ( / - ^ - 2 έ A/A i=l i=l J

fcj CASE 3. CANTILEVER BEAM PROPPED AT THE FREE END

In order to facilitate the application of the results developed here to practical problems, as well as to make easier the comparison with the results obtained by other authors, two possible variants as given in Fig. 14 and Fig. 15 (alternatives a and b) will be considered; these variants are mutually interchangeable by a simple linear substitution.

cr V/j

>

Mi

-b|-

di

FIG. 14.

Theory of beams 31

'//, w σ I Mi

di

ty FIG. 15.

Alternative (a). In this case the boundary conditions are

(1) y0 = 0, (3) y(l) = 0, (2) y'0' = 0, (4) / ( 0 = 0.

Hence the equation of the elastic curve is

(2.2.9)

y(x) = yOx+y'o"'^+s(x). (2.2.10)

The unknown coefficients can be found from the simultaneous equations

6 (2.2.11)

2y'o+PyO" = -2S'(0, as

6yi+Py'o" = -jS(i),

1 y'o= 2j [» '( /)-3S(0],

^" = -^-[s(/)-/s'(/)]. (2.2.12)

Assuming the load function S(x) as given by eq. (2.1.19) we have respectively

32 Theory of beams

t t I % y(x) = ^ + ^ " ^ - + 2iB .Σ <?i<(*-«i)4>a(

1 m 1

l A 25 ^

2 \ ( Σ Midx-diT)

^ = ^{ | i ? 1 [y[( ' -« i )M/-W

-|/[(/-βί)4-(/-6ί)*]]+ίΣ/1

-^ ( / -q^ j - . lMJl i / -^ ) -^ / -^

Jo' = - i i j ^ i [[(^-«i)3-(^^)3]-

)2

-^[( / -^-( / -^)4 ] +.ΣΛ 4/

Τ ( / - 0 3

1 f 1

i = l 3( / -c f ) 2 -

- χ ^ i = l

6(l-di)-1Q-dif

Jo - 4 ^ ]Ί2 .Σ <?* [(/-«i)3(/+3ai)-(/-6i)3(/+

+ 3^)] + £ PiCi(l-cty- Σ Mtf-ddQdi-t) i = l i=l

1 f l m

Λ>" - - l W j 4 X^iW- ̂ i)3(3/+ «0 - (/- ̂ i)3(3/+

+ ̂ )]+ J A(/-^)2(2/+q)-3 £ M^P-df)

(2.2.10a)

(2.2.12«)

(2.2.12ft)

Alternative (b). Interchanging the supports at the ends of the beam we get respectively for:

Theory of beams 33

the boundary conditions (1) j 0 = 0, (3) y (t) = 0, (2) y'0 = 0, (4) y"(f) = 0;

the deflection curve

the evaluation of constants ^ ' and ^ "

WP+y'J'P = -6S(/),

whence the unknown coefficients are as follows:

y'o = jp[(Ps"(i)-6S(i)i

y'o'=jp\2S(Q-PS"iQ].

(2.2.13)

(2.2.14)

(2.2.15)

(2.2.16)

With S(x) as defined by eq. (2.1.19) the respective equations are then

y0 = 1

Σ Qi 2BP \& [ ( /_α . )2_ ( /_ έ . )2 ]_

+ ^Pi[P(l-ci)-i = l

P

Jo 3

25/3 Σ Vi

t = l

J_ 12

(2.2.16a) [(/-*,.)«-(/-ft.)4]

1 2

-P(l-cd Σ Μ ^ / - ^ ) 2 - / 2 ] !

( / - C i ) 3

34 Theory of beams

(d) CASE 4. CANTILEVER BEAM RIGIDLY FIXED AT ONE END

Alternative (a). The right end rigidly fixed (Fig. 16). We have in this case: boundary conditions

(1) /0 ' = 0, (3) y(l) = 0, (2) / 0 " = 0, (4) y'{l) = 0;

P:

(2.2.17)

H-Qi-

f

-bi

k-£

I

- ► x

V/.

(2.2.18)

(2.2.19)

FIG. 16. deflection curve

unknown y0 and y'0

y0 = ISXQ-SW, y'o = S'{1).

Hence with S(x) as defined by eq. (2.1.19) respective equations become

1 f m r 7 yo=-g ll^jW-aif-d-bd^-

-ί(/-^)3]-.Σ Mi Γ/(/-4)-ί(/-4)2]}, (2.2.19«) 1 (Ί m

+ z i=l i=l J

Theory of beams 35

The first of these formulae may also be written as follows

1 f l Jo = j j2-iiE^[(/-«i)3(3/+«i)-(/-*i)3(3/+fti)] +

+ T Σ Ptf-cWU+cd-l Σ MlP-dfX. (2.2.1%) 0 1=1 ^ 1=1 J

Alternative (b). If the cantilever beam is rigidly fixed at the left end (Fig. 17) then we have:

FIG. 17.

boundary conditions

(1) j 0 = 0, (3) / ' ( / ) = 0, (2) y0 = 0, (4) / " ( / ) = 0;

deflection curve

jW^öy+JÖ'^+SC*);

(2.2.20)

(2.2.21)

unknown coefficients y'0' and y'0"

y'0' = IS"'(l)-S"(l), y'0" = -S'"(I),

which with the load function S(x) as defined by eq. (2.1.19)

(2.2.22)

36

become

yö

Theory of beams

1 B i = l i = l

Jo =

= 1 i = l ' '

m n (2.2.22a)

(e) CASE 5. SIMPLY SUPPORTED BEAM WITH OVERHANGS (FIG. 18)

Expressing the effect of the supports on the beam by the reactions Rc and Rd, the equation for the elastic curve is as

to· ^r ty

FIG. 18.

given by eq. (2.1.20) the only difference being that the load function contains two more terms due to concentrated loads Rc and Rd. Hence, taking also into account relation (2.1.18), we have

y(x) = W(x) + ^((x-c?yc+^«x-dfyd+S(x)

(2.2.23) or denoting by

SR(x) = S(x) + ̂ ((x-cfylc+f^(x-dfyd, (2.2.24)

j(x) = W(x) + SR(x). (2.2.25)

There are six constants to be evaluated: four unknown coefficients in the polynomial W(x) and two unknown reactions

Theory of beams 37

Rc and Rd. To find them we have the following six boundary conditions:

(1) y(c) = 0, (4) / '(/) = 0, (2) j ( J ) = 0, (5) j ; " = 0, (2.2.26) (3) y'n' = 0, (6) / " ( / ) = 0.

The problem can, therefore, be solved. Applying conditions (3) and (5) we obtain

Λχ) = yo+yOx+^-<(x-cTyc+^§<(x-dryd+s(x), (2.2.27)

and on differentiating this equation three times the last two relations so obtained are

/'(*) = R£<x-cyc+-Bd(X-dyd+s"(X),

/ R \ l / Α , \ · ( 2 · 2 · 2 8 )

These are freed from constants y0 and ^ , and using the boundary conditions (4) and (6) the following equations are obtained:

^-(l-c)+~U-d)+S"(l) = 0,

T+Ri+S'"^ = °> from which the unknown reactions are found to be

(2.2.29)

Rc^^-[(l-d)SffXl)-SfV)l

Rd = ^-d[V-c)S''V)-s'V)l (2.2.30)

Now y0 and y'0 can be evaluated. Applying the boundary conditions (1) and (2) and using eq. (2.2.25) the follow-

38 Theory of beams

ing simultaneous equations are obtained:

y0+y'0c+SR(c) = 0, y0+y'0d+SR(d)=0.

(2.2.31)

whose solution yields

Jo =

JO =

1 c—d

1

[dSR(c)-cSR(d)],

cd[SR{d)-SR(c)l (2.2.32)

These constants can also be expressed by means of S{c) and S(d) since if c < d

then SR(c) = S(c),

(2.2.33)

whence

Jo

y0 =

1 c—d

1 ~c^d

dS(c)-cS(d)- cRc(d-cf 6B ]·

RJid-cf 6B + S(d)-S(c) \.

(2.2.32a)

In actual computation it may prove advantageous to substitute for Rc and Rd the respective expressions as found in eq. (2.2.30) in which case we get eventually

Jo = dS(c)-cS(d) c(c-d)

J o = ~

c—d 6 S(d)-S(c) d-c

d-c 6

[(l-d)S'"(l)-S"(l)l

[(l-d)S'"(l)-S"(l)l

(2.2.32b)

Theory of beams 39

Thus the problem of the simply supported beam is completely solved. As an example let us consider an overhanging beam with loading and dimensions as given in Table 1, taken from a very useful book by S. Neumark (see ref. 6, p. 147).

TABLE 1.

I

1 (*) 2 3 W) 4

c{ (cm)

0 60 100 200 260

(/=) 310

Λ (kg)

110

150 280

90

630

CiPi

0

15,000 56,000

27,900

98,900

The load function in this case is

*(*) = £ Ä «*-**)%. i = l

whence s"<*>= Σ-ί-<*-*>'.'

i = l 4 / P . \ I

*"'<*> = Σ < ι τ > 1 = 1 \ - (-i

and

i= l i = l

Now using relations (2.2.30) we find the reactions at the supports

40 Theory of beams

Rc = -jz-c Σ pi(ci~d) = 20Ö(98,900-~260X630) i = i

= -324-5kg,

Rd = -^t^ii-Ci-c) = - ^ ( 9 8 , 9 0 0 - 6 0 x 6 3 0 )

= -305-5 kg. From the equation of the elastic curve (2.2.27) taking into

account relation (2.1.2) the following equation for the bending moment is obtained: Mix) = - [Λο<*-^+Λ,Κ*-4>&+ΣΛ<*-^ ]

i = l = — [ 1 IOJC—324- 5<JC—60>gJ° + 1 50<JC — 100>f Jg +

+ 280<*-200>iä8-305-5<jr-260>ü8].

Hence the moments at various sections are as follows: (1) M (0) = 0, (2) M (60) = -110X60 = -6600 kg cm, (3) M(100) = - [110x100-324-5(100-60)] = 1980 kg cm, (4) M(200) = -[110X200-324-5(200-60)+150(200-100)]

= 8430 kg cm, (5) M(260) = -[110Χ260-324·5(260-60)+150(260-100)+

+ 280(260-200)] = -4500 kg cm. It will be noted that putting c = 0, d = / all formulae of this

case reduce to those of case 2, as they should. It can easily be shown that when c — 0 and d — / then

JV-0 , y'0^ j7[PS"-6S(l)l

and the two constants become identical with those of eqs. (2.2.5) and (2.2.8). As regards the terms

^ < ( x - c ) 3 > £ and - § < ( x - J ) 3 > i ,

Theory of beams 41

appearing in eq. (2.2.27), it is seen that the second term vanishes as d -> /, and the limit of the first term is

R v'"xs

lim —-<Ot-cF>1 = - ° c T o 6 f l U * C)?c 6 '

since by eq. (2.2.30)

lim Rc = —?S"(0-

Thus we have proved that eq. (2.2.27) transforms into eq. (2.2.6) and the constants given by eq. (2.2.8) can be deduced from the formulae for a simply supported beam with overhangs by taking the relevant limits as shown above.

Alternatively, if we put that only c — 0, then the equation of the elastic curve is that for the beam hinged at one end and simply supported at x = d (Fig. 19).

0 ΓΠ t „ ^ f h

FIG. 19.

Performing the necessary computation the following results are obtained for: elastic curve

tix) = y'o*+yö'i+^i <(.x-<V>l*+SW, (2.2.34) reaction Ri

6 ' 6B

B Ä„ = T[S"(/)-/S"'(/)]. (2.2.34a)

42 Theory of beams

coefficients of the polynomial W(x)

y'o = -S-f-~[«-d)s"V)-s"(i)l

yo,=^i(l-d)S"'0)-S"(l)]. (2.2.35)

(f) CASE 6. PROPPED CANTILEVER WITH AN OVERHANG.

Beam rigidly built in at one end and propped at a point x = c (Fig. 20).

jB_L_r ty

FIG. 20.

The equation of the elastic curve in this case is

Χχ) = yo+y'ox+yö^+y'o'^-+^((x-cn+S(x), (2.2.36)

with the boundary conditions

(i) y0 = o, (2) y'0 = 0, (3) / ' ( / ) = 0, (4) / " ( / ) = 0, (5) y(c) = 0.

(2.2.37)

Theory of beams 43

Applying first two boundary conditions in eq. (2.2.36), differentiating three times, and then substituting into the final relation the remaining three boundary conditions leads to the following equations:

y'0"=-S"'(f)~,

3c2y'0· + c3y'0'' + 65(c) = 0, (2.2.38)

y- + ly'0" + S"([) + ̂ (l-c) = 0,

whose solution is

y* -^i^c-i)srf\i)-es{c)^c^s'\i)i

y'o" = ^ 3 [c\l-c)S"\l) + 2S{c)-c*S'\l)l (2.2.39)

R = ^ [(c-3l)S'"(l)-6S(c)+3c2S"(l)l

When c-+l this case reduces in the limit to the case 3(b), i.e. the cantilever beam with the prop at the free end, and it is easy to prove that the above formulae reduce to those obtained before.

(g) CASE 7. SIMPLE BEAM WITH TERMINAL FORCES AND COUPLES

In all the cases discussed above we have been assuming that the load function S(x) satisfies the conditions S"(G) = S'"(0) = 0 which is equivalent to the assumption that no concentrated forces or couples are applied at x = 0. We shall show now how the above formulae can be. used for other types of loading which do not satisfy these conditions. Consider a simple beam freely supported at both ends; its load between the supports being given by the load function S(x) and, in addition, having

44 Theory of beams

ΡΛ and Mi at the left end and P2 and M2 at the right end (Fig. 21).

To avoid unnecessary complications let us reduce the problem to the case 2 discussed above by assuming that the forces

M.

-ε * ty

p;

M,· M2

; - & * -

FIG. 21.

Pl9 P2 and the moments Ml9 M2 are applied not at the ends of the beam but in between the supports at some points dx and d2

(these later on will be made to tend to zero and / respectively). With this new loading the function S(x) is replaced by another S(x) such that

Mx Af, " M < ^ - r f i ) 2 > d . - ^ < ( * - < 4 ) % + £ ( * ) . (2.2.40)

Now the function S(x) satisfies the conditions 5"(0) — = 5'"(0) = 0 , therefore the elastic curve is given by eqs. (2.2.6) and (2.2.8) if S(x) be replaced by 5(x). We then get

x3

y(x) = y'ox+y'a'-£-+s(x);

y'0 = ~[PS"([)-6S(l)l (2.2.41)

y'·' = -jS"(l).

Theory of beams 45

Differentiating eq. (2.2.40) twice and solving for S(x) and S"(x) at x = I we find

-¥l(l-dtf- ^{l-d2f + S(D, (2.2.42)

5"(0 =^«-d1)+^(l-d2)-^-^+S"(l).

Hence the relations (2.2.41) may be also written as

y = /o*+K'-£+||<(*-4)%+§<(*-

y0 = ̂ W,(/)-6,s(o+|[p1(/-rf1)+p2(/-rf2)-^M1-M2\-~[P1{l--d1f^P2{l-d2f--

-3M^/-^2-3M2(/-rf2)2]j ,

*>" = - | iS"( / )-^[P1( / - rf 1)+P2( / - i / 2 ) -M1-M2] .

(2.2.43)

With these formulae the equation of the elastic curve for the original problem will be obtained by replacing dx and d2 by their limiting values, i.e. dx = 0 and d2 = I.

Butnow5"(0) 5* 0andS'"(0) ^ 0 (since dx = 0) hence instead of the constant j ^ ' we shall use the symbol A. Also, it will be noted that when d2 = / the force P2 and couple M2 are acting on the left side of the cross-section at the support x = /, and in

46 Theory of beams

accordance with our notation they are negative, i.e. we shall write — P2 and — M2. Taking this into account we get

, . , , -r x3 Ρχ x3 M-, x2 , _. . y(x) = y x+ÄT+-£ -g—g1 Y+S(X);

y'o = ii{ PS"(l)- 5S(l)+4 [PJ- Mx+Mt] -B

-jlPjp-m^ }· Ä = jsrw-^m-M^M,].

Simplifying and using the substitution

A^-Ls-m-^^Mi, these become eventually

Mi x? x3 Λ \

1

A = -i

PS"(l) - 6S(l) + ̂ (2MX+M2)

ls»it)-¥^.

J |

(2.2.44)

(2.2.45)

For this type of supports the forces Px and P2 acting at the ends of the beam, as could have been foreseen, do not affect the equations of the elastic curve. Formulae (2.2.45) are of great importance and will be found useful in section 6.

3. Determination of static quantities for a single-span beam

Once the elastic curve of a beam is known the problem, as far as statical requirements are concerned, is solved, since all the relevant data are then determined, e.g. the reactions at the

Theory of beams 47

supports and the fixing moments (if any), distribution of the bending moment along the span of the beam and hence its greatest values, the maximum deflection, the slopes at the supports, etc. The main advantage of knowing the elastic line is that it enables us to determine the bending moment and the shearing force distribution along the beam.

It should be noticed that the problems of finding some of the above quantities is really equivalent to evaluating the coefficients in the polynomial W(x). If the origin of the reference system be placed at one end of the beam, e.g. left end, then y0 is the deflection of that end, and y'0 is the slope at the left support provided this slope is so small that it is legitimate to assume

Similarly, y'0' represents, to a scale, the bending moment at the support which is at the origin, since

M0= -By'^ (2.3.2)

and y'n' is proportional to the reaction at that support, because

Ro = By'o'. (2.3.3)

To determine the maximum deflection / we need to find where dy/dx = 0, i.e. to solve the following equation

/o+/o'*+Ji" Y+S'(X) = 0 (2.3.4)

and then to substitute the result into the equation for the elastic curve (2.1.20). The above equation cannot be solved in general terms because the function S'(x) is a step function. In order to solve eq. (2.3.4) we must either be able to deduct (or guess) beforehand in which interval the maximum deflection is likely to occur, or to arrive at that knowledge by successive trials. In the former case, we retain only those elements of the

48 Theory of beams

step function S\x) that refer to the interval in which we expect the maximum deflection to be, and eq. (2.3.4) reduces then to an ordinary equation (in general algebraic and not of a higher than third degree). In the latter case, we can subdivide the region (0, /) into as many intervals as there are elements in the step function S'(x), obtaining in this way, instead of one equation (2.3.4), a set of equations but each free from step functions. We then solve these equations (or only those which are most likely to contain the maximum deflection). For the intervals in which no extremal deflection exists, the above equations have no real solutions. In practice the computation is really a simple one.

If the deflection at any arbitrary point x = c is wanted (e.g. at the mid-point) then by a simple substitution in the equation of the elastic curve we obtain

y(c) = W(c) + S(c) (2.3.5)

which is an elementary problem. The formula for the bending moment is given by eq. (2.1.2)

and can be obtained by differentiating twice the elastic curve equation (2.1.20). Thus we get

M(x) = -B[y'0'+y'0"x+S"(x)]. (2.3.6)

Putting in this equation x = 0 and x = / (the sign of the result in the latter case is to be reversed) we find the bending moments, on the left support

M0 = - B[y'0' + S"(0)] = - By'0', (2.3.7)

if S"(0) = 0, and on the right support

M(l) = B[y'0'+y'0"l+ S"(l)] = By"(I). (2.3.8)

These equations represent also the fixing moments at the supports, if no couples are applied at x = 0 and x = I.

Theory of beams 49

Differentiating the expression for the bending moment and using the relation (2.1.3) we get the shearing force

Q(x) = *OJ"+ £"'(*)]. (2.3.9)

Again putting x = 0 and x = I (with a change of sign in the latter case) the shearing force on the supports is found as follows, for the left support

*o = * |>;" +S"'(0)] = By'0", (2.3.10)

if S""(0) = 0 and for the right support

R(l) = -B[/0" + S"'(l)] = -By"\l). (2.3.11)

If no concentrated loads are applied at x = 0 and x = I, the above equations also represent the reactions at the supports. Frequently we need the slopes at the supports (approximately equal to the angles of rotation of the corresponding end cross-sections). For example, for the support on the right we can find its slope from the formula

β(/)« y'(D = y'o+y'oi+yö'j+sv). (2.3.12)

Thus all the characteristic statical data for a single span beam of flexural rigidity B = const, with an arbitrary loading have been defined by means of the load function S(x) and its first three derivatives S'(x)9 S"(x), S"'(x) at the points x = 0, x = I.

4. Beams on three supports Consider a beam rigidly built in at both ends and propped

at x = c (Fig. 22). This problem can easily be reduced to that of a single-span beam discussed in section 2. The effect of the prop can be taken into account simply by the introduction of the concentrated load R at Λ: = c representing the reaction of

50 Theory of beams

:JiV-f Λ

FIG. 22.

the prop. In other words, the load function S(x) receives an additional element,

R 6B Φ-cfy k3\l

since, due to the additional force R at x = c, this term will appear on the right side of the differential equation of the elastic curve as given by eq. (2.1.15). Therefore in this case

SR = -^<(x-c)*>l+S(x). (2.4.1)

Hence by eqs. (2.2.2) and (2.2.4) the relevant relations are as follows:

y =y'o^r+yO"^-+sR(x);

y'o = - 7 2 - ^ ( 0 - 3 ^ ( 0 ] , (2.4.2)

y'0" = -p[2SR(l)-lS'R(l)l

With the load function as given by eq. (2.4.1) we have

Theory of beams 51

S'n(x) = ^<(x-cfylc+S'(x),

SR(l) R 65

(l-cf+S(l), (2.4.3) SR(c) = S(c),

and substituting these in eq. (2.4.2) we obtain

y = y'o ^-+yO"~+-^<(.*-c?yc+s{x),

y<>

y'o (l-cf+S'W

Ό'=ψ |/[^(/-c)2+5'(/)]-3[-^(/-c)3+>S(/)

or alternatively

y = yö ~+y'o" ^-+-^<(x-cfyc+s(x),

y'o = ψ [-J c(l-cf+2lS'(l)-6S(l)] ,

y'o" = ψ \-^(l-cf (2c+l)+ l2S(D-6lS'(i) 1

}· (2.4.4)

(2.4.5)

In the above equation there is only one unknown quantity, i.e. reaction R. It may be evaluated from the condition y(c) = 0, i.e. that there is no deflection at the support. Whence we get the following equation for finding the reaction R:

3cX' + c*y'<;' + 6S(c) = 0, (2.4.6)

in which R appears at y'0' and y'0", these in turn being defined by the last two relations of eq. (2.4.5). Solving this equation

52 Theory of beams

we get finally

Ύ = C3(/!.C)3 [cVl-2c)S(l) + cV(c-l)S\i)-l3S(c)]. (2.4.7)

Also, in this case, the equation of the elastic curve is fully determined by the load function S(x) and the parameter c defining the position of the prop. Eliminating R from the formulae for y'0' and y^' in eq. (2.4.5) by means of eq. (2.4.7) the following alternative forms for these constants are obtained which may prove useful in practice:

" - 3S(/) S'jf) 3lS(c) y° ~\l-c) I c\l~cY

(2.4.8) v"· = 9 5 ( / ) . 3 y ( / ) , 3(2c + QS(c) y° cl(l-cy I + c\l-c) '

Applying an analogous reasoning and performing computation similar to that used above, the formulae for a two-span beam with all possible types of end fixing could be derived from the expressions already established for the corresponding single-span beams. However, it will be more convenient to develop the relevant formulae for a continuous beam with s intermediate supports, and then, from these formulae, by putting s — 1, the required relation for a two-span beam will follow automatically.

It is to be noted that when c — 0 or x -*■ I the reaction R -+ — oa. We record therefore a seemingly paradoxical conclusion that a support placed at an unsuitable point seems to endanger the structure instead of increasing its safety.

5. The elastic curve of continuous beams

The equation of the elastic curve of a continuous beam with its ends rigidly built-in (Fig. 23) can be derived by the same method which was employed when a similarly supported beam

Theory of beams 53

with two spans was discussed. Let us consider such a continuous beam with built-in ends having s intermediate supports at x — c-i, x — c< 2> X = C r X cs, where cx < c2

< cr < cs. Replace the effect of the support at an arbitrary station x = cr by its reaction Rr exerted on the beam, the magnitude of this reaction, at this stage, being still unknown. Con-

m ^

cr

ΊΓ IT—rr cr

■ * rr V//.

?y

FIG. 23.

sequently, the load function S(x) must also be replaced by SR(x) which by eq. (2.1.15) will be

SR(x) = S{x)+t^«x-c,ryCr. (2.5.1)

By so doing the problem is reduced from that of a continuous beam with s props and the loading as defined by the load function S(x) to that of a single-span beam with the loading as given by the load function SR(x). Whence the equation of the elastic curve follows at once, for we may use the formulae developed for a single-span beam provided that function S(x) be replaced by function SR(x). Thus the problem is formally solved. Therefore, by eq. (2.2.2) the elastic curve in this case is

y = y0 r +K Ί Γ + Σ Ä<(*-'r)»>l+s<*), 6B (2.5.2)

54 Theory of beams

and for evaluation of the constants y'0' and y'0" we have by eq. (2.2.4) the following two relations:

y'o

y'o"

-ΗΊ s'(i)+ Σ r = l

- 3

=H2

£(/-*/ —

i^O+E^CZ-O3"

[w+i*B~«-^] -/l s'(0+ Σ -

r = l -

—

| ( / - c ^

(2.5.3)

With these results ready all that remains to be done is to find the s reactions Rrs substitute their values into eq. (2.5.3) and then into eq. (2.5.2) and finally to revert to the original load function S(x), that is to consider this original load function as the true loading on the continuous beam in question. In this way the first term W(x) in the equation of the elastic curve for a single-span beam transforms, when the beam is continuous, into

^χ) = γο+/οχ+/ο^+ϊο -+£§«*-*>% (2.5.4)

(the nature of the supports not being specified). It is seen that W(x), a polynominal in the case of a single-span beam becomes a multi-step function in the case of a continuous beam. When this is taken into account the expression for the elastic curve does not differ any more from that for a single-span beam so that we may write

y(x) = W(x) + S(x).

Reactions Rrs may be now obtained from the following s conditions

Λ*ρ) = 0 0 > = l , 2 . . . . , J ) , (2.5.5)

Theory of beams 55

which in this case become r2 r3 s R

0=1,2 , ...,s),

yö^-+yO"^-+s(cp)+ Ziw<(cP-crf> = o (2.5.6)

where symbol < ), as specified before, denotes that only the positive elements (cp — cr) are to be retained.

From eq. (2.5.6) after some simple transformations, we obtain

Z/rH(/-02[3^r(/-^)-S(/-0] + /3<(S-03>} = BHp

(p= 1,2, . . . , * ) , (2.5.7) where

# p = 6[4(3/-2cp)5(/)-/c |( /-^)y(/)-/3S(cp)] (2.5.8) (/> = 1,2, . . . , * ) .

This represents a system of s simultaneous equations with s unknown reactions. The coefficients of the unknown reactions are some parameters determined by the total length (/) of the beam and by the location of the props (cr) as measured from the end of the beam which is placed at the origin of the coordinate system. The right-hand sides of eqs. (2.5.8), on the other hand, are some functions of the above parameters, of the flexural rigidity and of the load function S(x) as well as its derivatives at the points x = / and x — cr.

Thus the problem of finding the elastic curve of a continuous beam reduces finally to that of solving a system of simultaneous linear equation with numerical coefficients, which in the case of only a few intermediate supports is a fairly easy task.

For practical purposes, it is possible to solve the relevant systems of equations for some typical arrangements of supports even without having to specify in detail the loading on the beams, obtaining thus ready formulae for various types of continuous beams used in practice. We shall not deal with

56 Theory of beams

these here, restraining ourselves entirely to the above discussed case. If wanted, the elastic curve for a continuous beam with other types of supports can be derived in a similar way from the corresponding formulae in section 2.

It can be easily checked that putting s — 1 the formulae of section 5 reduce to those of section 4.

6. Theorem of three moments

Using the equation of the elastic curve for a single-span beam, developed in section 2, it is quite easy to derive an equation relating to each other the values of the bending moment

* r - l -e Or-I 3 Γ _i_ T )V xr +l

y r- i yr

FIG. 24.

at any three adjacent supports of a continuous beam. This equation, known as the equation of three moments, involves in it, for any given continuous beam, only three unknown statical quantities at a time, and for this reason it may render the computational procedure by far simpler when the number of supports is large. Suppose we are given a continuous beam with the number of spans s > 1. In accordance with the practice employed in the previous examples, we shall assume that every intermediate support can exert only a vertical (transverse) reaction. Let Fig. 24 represent any arbitrary span (but excluding the two end spans), designated as lr, cut transversely out of the above beam at supports Or_1and Or. The single span beam so obtained is a simple beam. Let the coordinate axes for it be xr, Or_v y Λ as shown in the figure. The load on this new

Theory of beams 57

simple beam is the same that was there before the span was singled out from the continuous beam. The relevant load function for this load will be denoted by Sr(x). In compliance with statical requirements the beam selected in this manner will remain in equilibrium and its deflection curve will remain unchanged, if the end sections of the beam are loaded by: (a) the resultant of all the external forces acting on the continuous beam to the left of the respective end section and (b) the resultant moment of these forces with respect to the given end section. Let the resultant force and moment at the left end section be denoted by Pr_1 and Mr_1 respectively, and those at the right end section by — Pr and —Mr?

The deflection curve of such a simple beam was determined in section 2 (case 7) and is given by eq. (2.2.45) and, if the change in notation is taken into account, we get the following relations for the rth span

yr-i(Xr) = Λ - ΐ ( ° Κ J~ Y + Ar-^+Sr(Xr),

JV-i(0) = ~[PrS';(Jr)-6Sr{lr)+l^{2Mr^+Mr)

Ar= -^. %'&) — &,-Mr_J.

(2.6.1)

' r ^"'T

It is evident, therefore, that if for a given span of a continuous beam the loading on the span as well as the bending moments at the supports are known, the deflection curve of the span is fully determined, and hence all other relevant data can be found.

The slope of the beam at the left support is given by the second relation of eq. (2.6.1). Differentiating the equation of the elastic curve and putting xr = lr the slope at the right sup-

t The negative sign must be assigned, for both the resultant force and the moment are acting on the left-hand side of the right end section.

58 Theory of beams

port is obtained as

y'r-iUr) = ^[3/r^(/r)-/?s;u)-3^r(/r)]-1 [Mr_xlr Mrlr\ (2 6 2)

- * (■

Applying now eq. (2.6.1) to the (r + l)th span the slope at the left support of the (r + l)th span is

y'M = T i ! - [ ? + i 5 ; /+ i ( / r + i ) - 6 5 r + 1 ( / r + 1 ) ] + o/r+1

+ 1 / ^ ^ + ^ Γ + Ι / Γ + Ι \ ( 2 e 6 J ) JL (Mrlr + 1 , ^ r + l ' r - n \ ^ 3 f 6 ;

and since the deflection curve is continuous, the slope of the curve must also be continuous at every point of the beam, hence at the rth support we have

y'r-LOr) = y'AP), (2-6.4) or by eqs. (2.6.2) and (2.6.3) we get

M r _ 1 / r +2M r ( / r + / r + 1 ) + M r + 1 / r + i = - f - [65 r ( / r ) + 2/2s; ,(/ r)-ir

-6lrS'r(lr)]--^-[Pr+1S'r'+1(lr+1)-6Sr+1(lr+l)] (2.6.5)

( r = l , 2 , . . . , * ) . This is the equation of three moments. The right-hand side of this equation is defined by the load function Sr(x), the rigidity B and the lengths of the spans lr and /(Γ+1).+ In the classic form of the equation of three moments the right-hand side is usually somewhat different. Instead of the load function Sr(x) quoted here, the products Qrar and ^(r+1)»b(r+1), are used, where Qr

denotes the area of the bending moment diagram of the corresponding span (considered as a simply supported beam and separated from the rest, i.e. omitting the end moments), while

t See, for example, ref. 5, Pt. II, p. 147.

Theory of beams 59

ar and br denote the horizontal distances of the centroids of the bending moment diagrams from the left and right supports respectively.

With this notation the classic form of the equation of three moments is Mr_1/r+2Mr(/r+/r+1) + Mr+ Jr+1

_ 6Qrar 6Qr+1br+1 (2 6 6) Ιγ / r + 1

In this form the equation of three moments is closely allied to the area-moment method which is widely used when solving hyperstatic problems, and also to the method of superposition in which the hyperstatic problems, by means of some simplifying assumption, are reduced to the statically determinate problems, which, when solved, will give the statically indeterminate quantities.

In order to be able to make use of eq. (2.6.6) its right-hand side must first be determined, and this, in turn, necessitates the knowledge of the bending moment diagrams in the relevant rth and (r + l)th spans, both spans being treated as independent simple beams carrying "their own" loading only. This last stage can be solved using the equations of statics only; nevertheless, it is seen that the whole procedure is fairly complicated.

It can be shown that

Qr = ±[2SVr)-lrS"Vr)l

2 [3lrSXlr)-PrS"(lr)-3S(lr)] ar — , (2.6.7)

3 [2S'(/r)-/rS"'(/r)] b = 6S(lr)-PrS"(lr)

3[2S"(/r)-/rS"(/r)] and hence checked that both forms of the equation of three moments (2.6.5) and (2.6.6) are mutually equivalent.

60 Theory of beams

From eq. (2.6.5) it is evident that the number of these relations is equal to the number of intermediate supports (s), and the bending moments at the supports can be calculated by solving the simultaneous s equations. As regards the end moments M0 and M(l), these must be given by the boundary conditions. If the ends of a continuous beam are freely supported then M0 = M(l) = 0 and two of eq. (2.6.5) greatly simplify. If one or both ends be rigidly built-in then M0 or/and M(l) will be found from the condition that no rotation occurs at the built-in ends. Hence there will be one or two additional equations:

2M0/1+M1/1 - -f-lllS^l^-eS^l

Mjs+1+2M{f)ls+1 = - * [6S8+1(/,+1) + ( 2 · 6 8 )

-f 2l8+1S8+1(l8+1) — 6l8+ iSs+1(/s+i)].

Thus, in each case, we have a sufficient number of equations to determine the unknown moments at each support, if we wish to solve the problem of the continuous beam by means of the theorem of three moments. It will be observed that by finding the bending moments over all supports, the problem of the continuous beam is virtually solved as well, for by eq. (2.6.2) the elastic curve in each span is completely defined, thus giving the full picture of static behaviour of the beam.

The bending moments at various sections are then given by

Mr_x(xr) = Mr^-^lMr^-Mr-BSWrft-BS'r'iXr),

(2.6.9) while the shearing force equation is

ßr-i(*r) = T [Mr-i-Mr-BSMr) + BlrS'r"(xr)l (2.6.10) ir

Theory of beams 61

The reactions at the end supports will be obtained directly from the equation of the shearing force as

*o = ßo(0) = ) - [M„ - Mx - BS'MJ], (2.6.11)

R(f)= - ß . + i(/.+ i)

= -^-[M{l)-Ms+BS's'+1(ls+1)-Bls+1S'a'Uh+i)l 's+1

(2.6.12) As for the reactions at the intermediate supports, these are

given by the magnitude of the jumps in the diagram of the shearing force at each support; we have, therefore,

*r = ß r ( 0 ) - ß r - l ( / r ) and on substitution of the corresponding values found from eq. (2.6.10) this becomes

Rr = -—S"(lr) — BS"'(lr) — - £Γ+Ι(Α·+Ι) +

Mr-Mr_1 Mr+X-Mr n*i*\ H . (2.6.13)

h 'r + 1

7. Single-span beams on elastic supports

So far we have been considering beams with "perfectly rigid" supports, i.e. neither a deflection nor a rotation of the beam at the support being possible. Analytically, the perfect rigidity of the supports is expressed by the following conditions: y0 =· 0, y'0 = 0, y(l) = 0, etc. In this section we shall discuss the case of a single-span beam having elastic supports which can both sink and rotate.

To start with, we take a simpler case where only a rotation of the transverse section of the beam is possible, excluding any sinkage of the supports.

With the small rotations which usually are likely to occur, it is permissible to assume that the angle of rotation is propor-

62 Theory of beams

tional to the fixing moment, hence the following two boundary conditions are obtained:

(1) y'o=-^, (2) / ν ) = - ψ , (2.7.1)

where C denotes modulus of rigidity of the support, and M0 and M(l) are the fixing moments on the left and the right supports respectively.1" In addition to the above boundary conditions we have also

(3) y0 = 0, (4) X/) = 0. (2.7.2) If the loading on the beam with elastic supports so defined is given by the load function S(x) then, taking into account the boundary conditions (1) and (3), its elastic curve is

y = -^x+yÖY+yö'^+s(x). (2.7.3)

Since by eq. (2.3.2) M0=-By'0',

eq. (2.7.3) can be transformed into

y = Ky'0'x+yi' γ + y'ö' ^-+S{x), (2.7.3a)

where K = B/C. The remaining two boundary conditions (2) and (4) furnish the following two simultaneous equations from which the constants y'0' and y'0" can be determined:

3l(l+2K)y'0' + Py'0" = -6S(/),

2(l+K)y'0' + iy0'' + 2 ^ - = -2S'(Q. ( 2 ' 7 ' 4 )

Replacing M(l) by its value as found from eq. (2.3.7) and simplifying we get further

31(1+ 2K)y'0' + Py'0'' = - 6S(l), W+2K)yö + l(l+2K)yO" = -2S'(l)-2KS"(I), ( ' ' °}

t See ref. 4, p. 147.

Theory of beams 63

where S(x), in accordance with the previous practice, is assumed to satisfy the conditions S"(0) = *S""(0) = 0. Hence we obtain finally

„ _ 2[KPS"(l) + PS'(l)-3(l+2K)S(l)] Jo —

l(l+2K)(6K+l) ,„ = 6[2S(l)-lS'(l)-lKS"(l)] ( 2 , 7 , 5 )

y° ~ P(l+6K) As an example let us find the fixing moments MQ for a beam

built-in elastically at both ends and loaded symmetrically at x = 1/2 by a concentrated force P.1" We have then

*»-£<K)X and, as it may easily be checked,

PP PP PI W = W SV)=JB> S"W = -2B-

Since M0=-By'0'9

hence by the first relation of eq. (2.7.5) we find PI I

M0= — 8 I+2K'

It is seen, therefore, that the fixing moment M0 depends not only on the load P but also on the parameter K, i.e. the ratio of the flexural rigidity of the beam B to the modulus of rigidity of the support C.

We note that the above result may be considered to be a general solution applicable both to freely supported and to rigidly built-in ends. Indeed, if the modulus of rigidity of the supports C — oo? i.e. if K - 0, then eqs. (2.7.3a) and (2.7.5) reduce to eqs. (2.2.2) and (2.2.4). If, on the other hand, C — 0,

t The example is taken from ref. 5, Pt. II, p. 147.

64 Theory of beams

i.e. JK* -► oo (which corresponds to the loss of rigidity of the material at the support), then

isv) Jo =■

Γ 2S(l) K — IS" (I)

P (H that is y'0" is identical with that of eq. (2.2.8), as it should be. It is seen directly from the first relation of eq. (2.7.5) that y'0' in this case is equal zero. We need yet to determine the factor Ky'Q' which appears in the equation of the elastic curve.

Ky'o

ftT(0 + ^ - 3 ( - ^ + 2 ) S(l)

"67

(HK) [PS"(l)-6S(l)l (2.7.6)

It will be observed from eq. (2.7.3a) that Ky'0' appears only as the coefficient with the variable x in the first power, therefore in the case of C = 0 it represents y'0. Comparing the result (2.7.6) with the first relation of eq. (2.2.8) the perfect agreement between them is self-evident.

In a likewise manner, by introducing two coefficients C0

for the rigidity of the left support and C for the rigidity of the right support, we can develop the elastic curve equation for the beam built-in at both ends with different rigidities of the supports, and by computing the appropriate limits we can obtain another set of formulae.

Next we consider a single-span beam whose end clamps are capable not only of rotation but also of sinking (Fig. 25). In addition, we shall assume that the rigidity of each clamp is different. Let C0 denote the modulus of rigidity (in rotation) and D{) modulus of elasticity (in sinking) of the left support,

Theory of beams 65

and let C and D be the corresponding moduli for the right support.

With such elastic supports the end conditions are as follows:

(2.7.7) (i) y0 =

(2) y0 =

' i 1

■ - £ ■ <3>

'-ΤΓ- <4>

v. m Γ

xo / ( / )

^ ' • • o

R(l) D

M(l) C

T.SL * .

- 1 - '

'Vr-I 1

yr

FIG. 25.

The flexural rigidity of the beam being denoted as before by B, we introduce the following additional symbols:

*° = ^Γ< « = £> Ν° = ΊΓ0> Ν = Ί>· ( 2 · 7 8 )

Omitting the computational procedure we arrive finally at the following relations:

y = -Nf>yi''+x*y'ox+y'o^-+y'o,Jt-+s(x);

y0

y0

,, = 61(1+ 2K) [NS"Xl) - S(l)] + 2(/3 - 6N0 - 6N) [£"(/)+KS"(l)] ~ l*+4P(K0+K)+ 121(N0+N)+ 12PK0K+ 12(K0+KXN0+N)' = -6l(l+2K0) [S'(l)+KS"(l)]-12(l+K0+K)[NS'"([)-S(r)~]

1*+4Ρ(Κ0 + Κ)+ 121(Ν0+Ν)+ 12PK0K+ 12(K0 + K)(N0+N)' (2.7.9)

Now letting the relevant parameters K0, K, N0, N approach their limiting values (zero or infinity respectively) we can obtain the elastic line equations for various particular cases of beams with elastic or perfectly rigid supports as well as for the cantilever beam.

66 Theory of beams

8. Continuous beams on elastic supports

The deflection curve of a continuous beam resting on elastic supports may be obtained in a similar manner to that employed when discussing continuous beams with perfectly rigid supports (section 5). The difference between the two cases is manifested only by the different boundary conditions. Taking into account the elasticity of the intermediate supports we must not overlook the fact that the end supports are elastic as well. For this reason, when reducing the present problem to the fundamental case of a single-span beam, the basic relations from which we start will be as given in eq. (2.7.9) and not those of section 2.

In other words, for a continuous beam on elastic supports, we assume the boundary conditions as follows:

(a) For the end supports—conditions as in eq. (2.7.7). (b) For the intermediate supports:

y(cp)= - ^ (/>= 1,2, . . . , * ) , (2.8.1)

where Dp is the modulus of elasticity of the/?th support. When formulating eq. (2.8.1) it has been assumed that sinking of the supports is not large and may, therefore, be taken as directly proportional to the reactions at the corresponding supports.

On the basis of the above discussion, the elastic curve and the constants y'0' and y'0" for a continuous beam on elastic supports can be derived directly from eq. (2.7.9) as follows:

y = -No/o'+Ko/Jx+y'^ + y-^+Snix);

„ 6l(l+2K)[NS'm-SR(i)] + 2(P-6N0-6N)[S'R(l)+KS'A(f)] 0 l*+4P(K0+K)+ 121(N0+N)+ \2PK0K+ 12(K0+K)(N0+N)' „, _ -6l(l+2K0)[SR(i)+KS'^l)]-l2(l+K0+K)[NSR\r)-SR(l)] 1 /*+4P(K0+K)+ W(N0+N)+ 12PK0K+ 12(K0+K)(N0+N) '

(2.8.2)

Theory of beams 67

where SR(x) = S(x)+}f ^ <(*- crfyCt. (2.8.3)

In the above equation there are s unknown reactions at s intermediate supports and to determine them we have the following set of simultaneous linear equations (with constant coefficients):

3cp(cp+2*oK' + ( 4 - 6 W + 6 % +

+ Σ ΐ < ( C P - <V)3> + 6S(cp) = 0 (p = 1, 2, · · ·, s). (2.8.4) r=l -o

y'0' and j ^ " appearing here are to be replaced by their respective values as found from eq. (2.8.2).

9. The theorem of five moments The equation of five moments can be derived without diffi

culty from the equation of the elastic curve of the single-span beam hinged to the elastic supports. And the elastic curve equation for this case follows from eq. (2.7.9) on substitution K0 = Kx= K-+ °°. Thus we have

y'o = -6^{-6W(/)-[6(Aro +^)-/3]yX0 + 6W5'''(/)},

,„ _ _S"(l) Jo - 7 .

(2.9.1)

If the above single-span beam is additionally loaded at the left support by a concentrated force Px and a couple M1 and at

68 Theory of beams

the right support by P2 and M2 (Fig. 25), then from eq. (2.9.1) the following relations are obtained:

y = y*+y^

N.

Mi x2 . x3 „ , x

Jo

Jo

^"(/H/4-^+M2 5 5 5

- ^ ; / 3 5 " ( / ) - 6 / 5 ( / ) + 2 / ^ + Ρ 4 - 2 -1

- 6 Λ Γ 0

— 6iV

B B

s"w+lT-f+f s"(i)-is'"w+i%-^+^

(2.9.2)

A = — 5"(/) + 5 5

One of these relations gives the deflection y0, and the other the slope y'0, at the left support; the corresponding formulae for the right support are

• Λ N s-lO-is-XD+iZ-^+f

yV) = jp i6PSV)-2PS"(l)-6lS(l)-P^-

- 2 ^ - 6 * . s-VHl^-f+f (2.9.3)

-6N s"(/)-/S"'(0+5/-Jx+:r' The next step is to follow the procedure of section 6.

Retaining the notation of that section, and taking into account the following self-explanatory conditions

Jr- l( ' r ) = Jr(0), y'r-ldr) = y'M,

(2.9.4)

Theory of beams 69

the eqs. (2.9.2) and (2.9.3) will enable us to derive the equation of five moments for a continuous beam on elastic supports with any arbitrary loading which is as follows:

« r - l ' r 1 l\

'--3!¥(Τ-+Γ) ir \ 'r—1 ' r /

h \lr ' r + l / J L r

+ 3^ r ( i + 7 i - \ i+ 3 ^ ± i l 3 f P + [ / r + 1 - 3 7 ^ ( i + r L ) -

\ ' r ' r + 1 / ' r + 1 J L ' r + 1 \lr ' r + 1 /

' r+1 \ ' r + 2 ' r + 1 / J ' r + l ' r + 2

B . 6 M r l + 6s;(/r) - 2/rs;f (/r)+6 - S r + l ( / r + 1 }

- / r + 1 o r + 1 ( / r + 1 ) —3 Nr_

/,

lr+l

S'Alr) S;i!(/,_!) +

+s;i'i(/r-i)

+s;"(/r)

+ 3iVr|

, Ny+1 ' r +1

\ 'r ' r + 1 / •S* n ( * r + l ) ^ r V'r)

' r + 1 ' r

Sr + 2{lr + 2) *J r+ l ( ' r+ l )

+

'r+1 'r + 1 + Sr+1(lr+1)

(2.9.5)

The right-hand side of this equation is defined by the load function of the three consecutive spans. It is not difficult to prove that the theorem of five moments enables us to determine the fixing moments at all the intermediate supports, if the moments at the end supports are known. The knowledge of the moments at the supports is the only requirement for finding the elastic curve for a given beam, and hence for solving completely the problem in hand.

In fact, if in eq. (2.9.2) the suffix r (denoting the number of the span in the beam) be introduced, the equation becomes a

70 Theory of beams

set of simultaneous equations giving, in turn, the elastic curve for each span of the continuous beam. The moments Mr_1

and Mr appearing in it are furnished by eq. (2.9.5). The concentrated loads Pr_1 and Pr9 as yet not determined, can be found from the following relations:

Pr

B 1

1

\+1(/r+1)+ Mr

B Mr B

s;'(/r)-/rs;'U)- Mr-x B

Mr (2.9.6)

Chapter III

THEORY OF BEAMS WITH VARIABLE FLEXURAL RIGIDITY

1. Derivation of the differential equation of the deflection curve Shape function. Examples

In the more general case when the flexural rigidity of a beam B(x) = El is variable (Fig. 26), the theory of such beams (with the assumptions defined in Chapter II) reduces to the solution of the differential equation

Consider a more general type of loading than that discussed so far. Assume that the beam is also subjected to a distributed bending moment whose magnitude for unit length is m{x). If

72 Theory of beams

this moment is uniformly distributed and acts on a segment (oii9 ßj then it may be defined by the symbol <jn^. With this additional assumption it will be found directly from the definition of the bending moment

M{x) = Mo-Rox-—^ ii<(*-*i)2>£ +

1 rii n2 n2

+-2Γ Σ ii<(*-*i)a>S+ Σ »«ί<*-«ί>ϊ-Σ'«ι<Α-Α>Λ-i = l i = l i = l

n3 n4 f f

i= l i= l J J 0 0

(3.1.2) where M0 and i?0 denote respectively reactive moment and the reactive force of the left support.

On performing the transformation of both sides of eq. (3.1.1) the following equation is obtained:

s*Y(s)-sy0-y'0 = -L i^X . (3.1.3)

From the known formulae defining Laplace transforms (1.3.1) it will be found that

B(x) f " ) B(x) [ » ) B(x) 1

2! >\kqiL\\ B(x) /a< + 1 ». \/{x-btf χ , ^

** r ( ν * - - α ί \ ~

i-i . x f i W

Beams with variable flexural rigidity 73

-L{mL~1{7L{^}}+

+L\*. In developing the transform of the right-hand side of eq.

(3.1.2) use has been made of the Borel's theorems on convolution (5.5.10) so that the integrals have been replaced by the corresponding products of the transforms; e.g.

X

W^-{«(*)}}= [{x-t)q{t)dt. 0

Dividing now both sides of eq. (3.1.3) by s2 and applying then the inverse transformation L"1 the deflection curve is obtained as

y = yo+yix+Mo01(x)+Roo2(x)+s(x)9 (3.1.4) where the following substitutions have been made:

s2 ) B(x) I , r

l \ l · (3.1-5) 02(x) = L - M - i - L l X

B(x) and the symbol S(x), as before, denotes the load function, which in this case is equal to

-ΜΜ<5?>:}}+

74 Theory of beams

♦MW<s&>;}}* ^-i-W<w>:}|-

To obtain the boundary values it will be convenient to link the reactions M0 and R0 with the value of the derivatives of y(x) at the point x =0. As defined earlier in this book the shear force Q(x) in this case is given by

nx nx n3

Q(x) = R0+ £ qi(x-ai>- - £ ? i<*-*1>S:+ Σ <pi>e~+ i = l i= l i = l

x

+ {q(t)dt, (3.1.7) 0

where q(x) may be a step function. Comparing eqs. (3.1.2) and (3.1.7) the following relations

are obtained: 1. If the loading does not include a distributed bending

moment, then M'(x)= -Q(x); (3.1.8)

2. If there is a distributed bending moment of magnitude m(x)> then

M\x) = -Q(x) + m(x); (3.1.9)

with the proviso that m(x) denotes a step function defining the total distributed bending moment [uniformly distributed m„


or variable m(x)]. From eqs. (3.1.1), (3.1.8) and (3.1.9) it follows that

M0 = -Β&Ϊ (3.1.10) and

Κο = ΒΜ + Βύ%'. (3.1.11) While developing these relations it has been assumed that M0 = M(0) and R0 = g(0), i.e. that there is no load at the cross-section x = 0 which corresponds to the assumption S"'(0) = S'"(0) = 0 in Chapter II.

The load function (3.1.6) is a generalization of the function (2.1.19) and will reduce to it when i?(.x;)= const., and when the terms resulting from the additional loading are added to it. Likewise, eq. (3.1.4) is a generalization of eq. (2.1.20); however in the case when B = B(x) the function

W{x) = yo+y'oX+MtfP^ + R^x) (3.1.12) is no longer a polynomial.

It is seen from eqs. (3.1.5) that the functions Φ^χ) (i = 1, 2) do not depend on the loading of the beam; they depend on the flexural rigidity B(x) = EI(x) and hence, if E = const., on the way the cross-section varies along the beam. For this reason the functionsΦ{{χ) {i = 1, 2) will be called the shape functions.

These functions [as well as the other functions of eq. (3.1.6)] can be obtained for any specific problem from the tables of transforms; in certain cases, particularly after some skill in the use of the tables has been acquired, this becomes a simple mechanical operation not much more difficult than the use of other mathematical tables. The procedure will be illustrated on examples further on in the book. Often the form of the functions of eq. (3.1.6) (although fairly complicated) will suggest clearly the programme for the solution.

EXAMPLE 1. As the first example consider the load function S(x) for a distributed load given by q(x) = (q0IP)(l ~x)2 when the rigidity of the beam diminishes parabolically with lengths

76 Theory of beams

as given by

B(x) =^2(6l2-4lx + x2).

The formal solution in this case is

There are four operational steps, each requiring the use of the tables of Laplace transforms as well as some elementary computation as follows:

1° Ψ-21Χ+**} =£-%+!, (5.2.2)

Px2 llx3 2x* x2

3° HU=£*> (5-2-2)

2! 3! T 4! ~ 12 2!

Ϊ2j" ~ 12?'

4 ÄW - Bo L | 6 j 5 | - Bo - 6 4 , - 4 ! 5 o . p.2.1)

The numbers in brackets denote here the transformation formulae in the tables of Laplace transforms (Chapter Y).

Needless to say, not all computations are as simple as those above. Difficulties will arise in cases when the required transformations are not included in the available tables of transforms. These difficulties can be overcome, of course, if one has the full knowledge of the theory of the Laplace transformation; otherwise one can replace these functions (which are inconvenient from the point of view of transformation) by some other functions which are either their equivalents or approximations. However, if the transformation fails altogether the problem of determining the required relations may be tackled by the mathematical analysis, i.e. expressing the formulae containing

L and L""1 symbols by their analytical counterforms without these symbols. This is possible as a direct result of Borel's theorem defining the product of the transforms of two functions. The corresponding formulae are listed in the tables of Chapter V and numbered (5.5.10)-(5.5.14).

Some of the functions so modified are quoted below and their derivation is left to the reader as an exercise in the Laplace transformation method.

·*>--** {Η*5)Ητ (3U3) o

^ ^ Η ^ Ι Η ^ · (3U4)

-'{M^xiW^r^. <31I6)

L"{^L{w)L-l{iL{^}}\ = X t

= [{^dtUt-r)q{T)dx, (3.1.18) 0 0

X t

= i^dt(m(t)dt. (3.1.19)

78 Theory of beams

The above relations have been obtained by S.E. Mikeladze [8], using the method of mathematical analysis.

As represented by eq. (3.1.4) the deflection curve is most general and is applicable directly to beams with variable rigidity. In fact, following the procedure of Chapter II, we can derive solutions for any particular type of end supports for single-span beams as well as continuous beams employing only relatively simple mathematics. Such a general solution for a given type of supports (but without defining the type of loading) represents a great saving of effort. A further gain may be derived from tabulating the shape functions Φ{ ( /=1 , 2). Since the latter are independent of both the load functions S(x) and the boundary conditions (i.e. they do not depend on y0, y'0, M0 and R0 or any additional reactions appearing as in the case of continuous beams) it is possible to determine them beforehand for some typical shapes of beams most frequently occurring in engineering practice and then to tabulate these results.

In order to derive the load functions S(x) not only the load itself must be considered but also the change in the rigidity of the beam must be taken into account. But in spite of this it is still very useful to work out the tables of S(x) for some typical load distributions and some typical changes of the cross-sections.

2. Deflection curve for single-span beams

To illustrate the theory some examples will be considered. It may be helpful to the reader if we take the same cases which have already been discussed in Chapter II, since it will simplify the treatment. Thus, for example, neither the sketches nor the procedural details need be repeated here afresh. However, one must remember that the thick line which used to represent on these sketches the beam of a constant rigidity, will


now represent the axis of the beam before deflection. The beam (of a variable rigidity) itself is not represented.

(a) CASE 1. BEAM RIGIDLY BUILT-IN AT BOTH ENDS (FIG. 12)

The boundary conditions for this type of support are given by eq. (2.2.1). Hence the deflection curve as given by (3.1.4) becomes

y = Mo01(x) + Ro02(x) + S(xy (3.2.1)

The conditions (3) and (4) lead to the following linear equations for M0 and R0.

Μ0Φ1(/) + ^0Φ2(0 = -S(0 , Mo0i(O + i?o^2(O = - m (3.2.2)

from which the two reactive quantities of the left support can be determined, viz.

S(l)0'2(t)-S'(t)02(l) Φ1(1)Φ,

2(1)-Φ[(1)Φ2(1)9

0 Φ1(1)Φ2(1)-Φ'ι(1)Φ2(1) ·

With the general relations (3.2.1)-(3.2.3) established for the beam built-in at both ends, the problem for any particular type of loading, shape and dimensions of the beam reduces to that of defining:

1° the load function S(x) by using eq. (3.1.6), 2° the shape function Φ{{χ) (i = 1, 2) by using eqs. (3.1.5), 3° the reactions M0 and R0 by using eqs. (3.2.3).

It may easily be shown that relations (3.2.1)-(3.2.3) reduce to the corresponding relations (2.2.2)-(2.2.4) when B{x) = const.

80 Theory of beams

(b) CASE 2. SIMPLE BEAM FREELY SUPPORTED AT BOTH ENDS (FIG. 13)

(3.2.4)

The boundary conditions in this case are: (1) y0 = 0, (3) y(l) = 0, (2) M0 = 0, (4) M(l) = 0.

By (3.1.1) the above conditions are equivalent to those of (2.2.5). Hence the deflection curve is given by

y = y'ox+Ro02(x) + S(x), S"(l)

Ro Φ'·{1)' 1

(3.2.5α)

(3.2.56)

y'0= -j[S(l)+Ro02(l)]. (3.2.5c)

In this case there are only two functions to be determined: S(x) and Φ2(χ), from which the constant R0 and y'0 may be evaluated.

EXAMPLE 2. As the next example consider the beam shown in Fig. 27.1, The rigidity of the beam B(x) is a step function

|

WA

Ty

a

ς .

.;, ; i J

- — _ _

P

f

^w.

FIG. 27. t This example is taken from ref. 7, p. 147.

with intervals of constant values

Ä(x) = <B1>g+<Äa>S+<Äi>S

and the only loading is a point force P acting at a point x = c, where a < c. By eq. (3.1.6)

S(*) = A L - i | ^ L { < x _ c > i } j . (3.2.6)

If the beam were infinitely long (i.e. if / = <*>), by eq. (5.3.3) we would obtain

P f e~cs) P

We are interested in S(x) within the region (c, I) only, hence

S(x) = -^-{(x-cfylc. (3.2.7)

The above procedure, which is typical for the Laplace transformation, follows from the definition of the transform F(s) as the definite integral defined by (1.3.1) in an indefinite region. We shall refer to this method frequently further in the book and shall extend the functions from the region (0, /) to the region (0, ©o). The rigorous treatment requires a different notation for the functions (0, /) and (0, <»)· While noting this requirement for the sake of clarity, we shall, however, for convenience use the same symbol in both cases whenever there is no likelihood of confusion.

As is seen from eqs. (3.2.5) in the case considered it suffices to determine only one function of shape. By (3.1.5) we find

•*»- i - i {± i {<£>><*>!}}· (3·2·8)

82 Theory of beams

From eqs. (5.4.5) and (5.4.4) we have then

^ r \ T i f l Γ l ( l e~as ae~as\

1 /e~as ae~as e~ls fe~is\]|

Limiting Φ2 to the region (0, /) all terms with e~ls drop out, which leads to

where

^ / Λ r i ί α 0 /e~as aße-as\]

B1 ' ^1^2

Consequently, as a result of relations (5.2.1) and (5.3.3), we obtain

Φ2(χ) = ?£—Ιφ-αγ + 3α(χ-άΤ>*α. (3.2.9)

The reader who took the trouble of following the line of argument leading to eq. (3.2.9) will find a rewarding satisfacy tion from the knowledge that the function (3.2.9) will be vere useful in many other examples, whenever the shape of the beams is of the type shown in Fig. 26.

Thus the problem of example 2 is formally solved. Indeed the deflection line is given by eqs. (3.2.5) and the function S(x) and Φ2(Χ) of the equation are given by relations (3.2.7) and (3.2.9). Hence utilizing eqs. (3.2.5) we find

P{l-c) R0 = I

y'o = ^p^[^l3-ß(l-af-3aß{l-af-l(cc-ß)(l-cy].

It is to be noted that in order to determine reaction R0 it is not really necessary to use eq. (3.2.5Z?). The problem of example 2

is statically determinate and i?0canbe found directly by considering the conditions of static equilibrium.

EXAMPLE 3. When c < a, i.e. when the force P acts at a point within the first portion of the beam where the rigidity is Bl9

then the problem is already partly solved, since the shape function Φ2(λ) is given by eq. (3.2.9). We need to find only the load function S(x). By eq. (3.1.6) this is given as

x — c B(x)/C

^ ί ? 1 « ' - » } ^ 1 ?M<*-'M)}=

* 1

P__ ~B~2

(L{(X)?}-CL{(\)?})

(L{(xya}-cL{(\ya})

+

(3.2.10)

Employing now relations (5.4.4) and (5.4.3) we can determine the four transforms of eq. (3.2.10) and multiplying by l/s2 we get

S(x) P_[ ΒΛ

- + ■ ce~ ae~ ce

+ -ce~ P sa

L-1 - + -ae~ s°

-Is

" +

le -Is ce - + -ce~

Performing the apparent reductions and dropping the terms containing e~ls (which define the function S(x) in the region (/, oo)} hence of no interest in the present case), we obtain

84 Theory of beams

By the virtue of relation (5.3.3) on performing the necessary simplifications and on making use of the substitutions of eq. (3.2.8) it follows that

Six) = j{«<(x-c)yc-ß^-af + 3(a-c)(x-a)^

(3.2.11) which is the last step in solving our problem. Indeed the deflection curve is then given by eq. (3.2.5<z) while the shape and the load functions are given by (3.2.9) and (3.2.11) respectively. On the basis of these relations we find

S(/) = j[<l-cy-ß(l-ay-3ß{a-c){l-ari

S"(Q = P(«-ß)(I-c)9

Φ2(0 = j{*P-ß[(l-aY + 3a(l-a)%

*ί'(0 = ("-/% (3.2.12) whence

Ro = -^η^, y'o = -j[SV) + WM-

EXAMPLE 4. We shall now consider a numerical example.f Let c = //3, a = 1/2, Βλ = B, B2 = 3B. We can determine the following quantities and functions:

R - 1P Ro- - y , 1

P 17 PP

^) = h [<H)X-K(-ι i / \ v 1

t This exan iple is taken from

5

ref. 8, p. 147.

R 2 ß=3B

- 2 ) +

(3.2.13)


, . P fl7/2 x3 1 / / / \ 3 \ l

+\"Ϊ7('-7)* + ϊΐ(Λ:-τ) 2 \ I

1/2

(c) CASE 3. CANTILEVER BEAM PROPPED AT THE FREE END

Alternative (a) (Fig. 14). In this case the boundary conditions are as follows:

(1) Jo = 0, (3) y(I) = 0, (2) M„ = 0 or j ; ' = 0, (4) / ( / ) = 0.

These conditions lead to the following linear equations: yil+Ro02(l)= -S(l), yo+R&W) = -S'(l).

Whence the deflection curve is y = y'ox+Ro02(x)+S(x),

_ S'WM-SWM yo - ΙΦ·2{ΐ)-Φ2{1) ' ( i , 2 - 1 4 )

_ S(l)-lS'(f) *° /Φ;(/)-Φ2(0"

Alternative (b) (Fig. 15). The boundary conditions are

(1) y0 = 0, (3) j(/) = 0, (2) ^ = 0, (4) AfCO = 0 or / ' ( / ) = 0.

These two constants are determined by the equations

Mo01(l) + Ro02(l) = -S{1),

Μ„Φ;'(/) + W ( 0 = -S"{1),

86 Theory of beams

and the equation of the deflection curve becomes

y = Mo01(x)+Ro02(x) + S(x),

M° Φ1(ΟΦ^'(/)-Φ;'(/)Φ2(/)' l J

Λ ° 0,(1)0','([)-Φ[\1)Φ2(1) ■

(d) CASE 4. CANTILEVER BEAM

Alternative (a) (Fig. 16). The right end rigidly built-in. The boundary conditions in this case are

(1) M0 = 0, (3) y{l) = 0, (2) Ä0 = 0, (4) / ( 0 = 0.

It is left to the reader to check that in this case the solution is exactly as given by eqs. (2.2.18) and (2.2.19). This is a particularly simple case, and there is no need to determine any shape functions Φχ{χ) at all. But it is to be noted that the load function S(x) for this case is different from the corresponding function in Chapter II, because it depends upon the flexural rigidity B(x).

Alternative (b) (Fig. 17). The boundary conditions are

(1) y0 = 0, (3) M{1) = 0, (2) j,i = 0, (4) Ä(/) = 0.

For beams with a variable rigidity the condition (4) cannot be replaced by the relation y"'(l) = 0. However, it can easily be shown that the conditions (3) and (4) are equivalent to

(3') / ' ( / ) = 0, (4') /"( / ) = 0.

The latter two conditions can now be used directly in our method.


As a result of the above we obtain the following set of simultaneous equations, for this example

MjDi'{l) + Ro0'%'(l) = -S"(l), MW(J) + RW(J)= -S'"(I),

and the equation of the line of deflection is

y = Μ0Φ1(χ)^Ά0Φ2(χ) + Ξ(χ), (3.2.16) where

M° Φ-{1)Φ-\Ϊ)-Φ-\1)Φ-{ΙΥ

The above equations for the alternative (b) as compared with the corresponding equations of the alternative (a) are much more complicated. They have been developed here simply in order to show that solution of any practical problem based on the approach (b) would be more difficult and would take a longer time to solve. It is therefore not recommended. It is preferable to reduce it to the alternative (a) by placing the origin of the coordinate at the other end of the beam. This is a good example of how an appropriate choice of the coordinates simplifies the solution.

(e) CASE 5. SIMPLY SUPPORTED BEAM WITH OVERHANGS (FIG. 18)

Referring to the explanations given in Chapter 11 for an analogous case only the final expressions will be quoted here.

The boundary conditions are

(1) Xc) = 0, (4) M(/) = 0, (2) Xrf) = 0, (5) Ä0 = 0, (3) M0 = 0, (6) R(l) - 0.

88 Theory of beams

The condition (4) may be replaced by (4') / ' ( / ) = 0,

and consequently, utilizing this relation, the condition (6) may be replaced by

(6') / " ( / ) = 0. Supports inside the region (0, /) will produce the reactions

Rc and Rd and these, in turn, will lead to appearance of the following terms in the equation of the deflection curve

R^\W/X-C

RdL-ipL

Using now the substitutions

s*"\\B{x)/t

i f /χ—ά B(x)/d

0,c{x)=L-^\^L{/X' s» \\B(x)/c (3.2.17)

the deflection curve becomes

y = J o + ^ + ^c^2cW + ^d^2dW + ^ W . (3.2.18)

To evaluate the four unknown constants there are two sets of simultaneous equation, each set containing only two unknown. Performing the necessary differentiation and utilizing the boundary conditions (4") and (6') we find

Äe<J£(Q + * Ä ( / ) = S"(l),

Whence the reactions are determined by the following relations:

s"WZ(D-s'"W&i) ( J ·2 · }

d Φ·ή(.1)Φ'ΐά{1)-Φ£(!)Φ'ΐα(1) '

Next from the conditions (1) and (2) we obtain

yo+yOc = -SR(c), yo+y'od= -sR(d),

where the symbol SR(x) denotes the following functions: SR(x) = Rc02c(x) + Rd02d(x) + S(x),

whence finally _ dSR(c)-cSR(d)

yo — — c-d (3.2.20)

_ SR(d)-SR(c) y° J^d '

(f) CASE 6. PROPPED CANTILEVER WITH AN OVERHANG (FIG. 20)

The prop being at a distance c from the built-in end. The boundary conditions are

(1) y0 = 0, (4) Jl(/) = 0, (2) y'0 = 0, (5) j<c) = 0. (3) M(/) = 0,

The conditions (3) and (4) are equivalent to

(30 / ' ( 0 = 0, (4') / " ( / ) = 0.

Using similar notation as in the previous case the deflection curve is given by

y = ΜοΦ^χ) + Ro02(x) + R02c(x) + S(x). (3.2.21)

The three unknowns M0, R0 and jR can be determined from the following equations:

Mo0['Xl) + Ro0'2'\O + R&2c\l) = -S'"(l), Μ,Φλ (c) + R0O2 (c) + iW>2c(c) = -S (c), (3.2.22) Μ0Φ[' (O + W {l) + R0Z(!) = ~S"(l).

90 Theory of beams

(g) CASE 7. SIMPLE BEAM WITH TERMINAL COUPLES (FIG. 21)

(Mx being at the left end and M2 at the right end of the beam.) The problems will be solved in the same way as that used for the corresponding case in Chapter II, i.e. assuming first that Mx and M2 are applied at some points xx = dx (d± ^ 0) and x2 = d2 (d2 ^ Ϊ) and then making xx tend to zero and x2

tend to /. In accordance with eqs. (3.2.5) and (3.1.6) the equation of

the deflection curve is

y = y'ox+Ro02(x) + S(x), (3.2.23) where

S{x) = ,S(x) + M10 l d l (x) + M20l d 2(x), S"(0

* o = - Φ2'(0 '

y'*= -jW+WM- (3·2·24)

The meaning of the symbols Φ1ά{(χ) being

*-w - - L i > K « r * > x } } < ί = , · 2 ) ' (3.2.25)

The function S(x) and its derivatives can now be replaced by its component terms as defined by (3.2.24). Hence by eqs. (5.1.3) and (3.1.5) we find

Φ^ = - < ^ ) Χ </=ι·2>·

φ'Λΐ) =

B(l) B(l) '

I m


Employing these substitutions and also making dx -+09d2-+ /, eqs. (3.2.23) and (3.2.24) transform to the following:

y = γ'0χ+Μ1Φ1(χ) + Κ0Φ2(χ) + 5(χ),

R0= -l-lBWiO-M^Mzl

y'o = - γ [ 5 ( / ) + ΑΤιΦι(/) + Α Λ ( 0 ] .

Thus our problem has been solved. On the basis of the relations developed for the cases 1-7 and

on employing also the methods explained in Chapter II, it is possible to solve other problems of beams with variable cross-section, in particular the additional cases discussed in Chapter II.

Thus, for example, the deflection curve of the continuous beams with various boundary conditions can be obtained, or the equation of three moments can be derived. The solutions will contain the functions S(x) and Φ(χ) as defined above. One of the advantages of the theory so developed is the uniformity of the method and its generality. But the problem is so large that it exceeds the scope of the present work.

Chapter IV

PROBLEMS WITH MORE COMPLEX LOADING

1. The transform of the load function

The methods described in this book have been used by the author and J. Nowinski between 1953 and 1955 in solving some problems of structural mechanics (see refs. 18, 9, 10 and 11). This chapter is based on the above work and its subject matter is the theory of beams carrying a transverse loading as well as axial loading or beams on an elastic foundation.

It has been shown by J. Nowinski [9] and S. Drobot [12] that some generalization of the theory may be achieved if the transform of the load function containing the point loads and moments is employed, the concentrated load being treated as the limit of the distributed loading (see section 1 of Chapter II) and the evaluation of the limit being performed on the elements of the transform of the load function not on the load function itself.

The method will be explained while considering a large family of differential equation for beams with a constant flexural rigidity (B = const.) in the following form:

By^+Ci/' + Ciy = φ), (4.1.1) where

ni n2

Φ) = Σ <«i(*)>«'- Σ <XW>£· (4.1.2) i = l i = l

In the above equations q{{x) and m^x) define respectively a distributed load and a distributed moment (per unit length).

92

Problems with more complex loading 93

These functions, in general, are multi-step functions. The symbols C1 and C2 denote some constants.

When Q = C2 = 0, the problem is that of simple bending of beams. When only C2 = 0, the equation represents bending of beams of constant flexural rigidity under transverse and axial loading, and when only Cx = 0, the problem is that of beams on an elastic foundation (Winkler's type).

Let us find the transform of eq. (4.1.1), introducing at the same time the terms defining concentrated loads, which do not figure in eq. (4.1.1). Assuming, in the manner explained in Chapter II, that a force P acting at a point x = c may be replaced by a uniformly distributed load q over a length a (Fig. 8) we shall obtain by eq. (5.7.3)

1 p — 0(.S p—CS

Making now α-^Owe find that a force P is represented in the transform <P(s) of the function <p(·?) by a term Pe~cs. A similar procedure (Fig. 10) applied to a concentrated moment M leads to a term sMe~~ds in the transform.

By virtue of eq. (5.4.2) to each element {q^x)}^ in the function φ(χ) there corresponds a transform

Qll(s)e~aiS - Qi2(s)e-b<s, (4.1.4) where

Qn(x) Ξ 4i(x+a{) and qi2(x) = qx{x+fy). (4.1.5)

If in a region (a0, b0) there exists a constant loading q0, eq. (4.1.4) becomes

p— aos p—bos

?o— 4o——- (4.1.6) s s

which follows from the relation (5.4.3). It remains yet to find the transform of the expression

<m-(*)>fj. By reference to relation (5.4.1) it will be seen that

94 Theory of beams

the transform of such a function is of the form

sMn(s)e-«is - sMi2(s)e-ß*s - mfai + 0)e~^s + m^ - 0)e~^s. (4.1.7)

Similarly, as in the previous case the symbols M{fs) (j= 1, 2) represent respectively the transforms of the functions

m^{x) = mi(x + <x.i) and mi2 = m^x+ßi). (It is to be noted that here a symbol M with a double index such as M^ denotes the transform, not a concentrated moment.)

The last two terms, however, should be omitted. If they are retained, it will be observed that the resulting equation will not agree with (4.1.1) which represents a particular group of problems dealt with in Chapter III (see also eq. (3.1.6)). It can easily be checked that these two terms appear in eq. (4.1.7) because in the formulation of the problems there is the derivative m'(x), not the function m(x) itself. But it is possible to formulate the problem in such a way that the differential equation will contain the function m(x), not its derivative. For example, in the case of a beam with transverse loading these two terms will not appear at all if the equation is written in the form

{By")' = Q-m{x).

On the other hand, on differentiating both sides of eq. (4.1.1) we get

Bf+cj"+c2y = g <?ί(*)>2{- Σ Wi'{x)>% and the transform of this expression would contain, among the others, the terms q{a{+0) and q{bi—G), which would have to be dropped.

A different method of solution which does not give rise to the above terms at all, will be given in section 7.

If, on a certain segment of the beam (α0, β0), there acts a uniformly distributed moment of intensity m0, then in the

transforming of the function φ(χ) there appears the following sum:

-m0e-a<>s + m0e-ft>s. (4.1.8)

Thus we have derived all the terms of the transform of function φ(χ) and the result is

*(*) = Σ ?<« I ^ - — I - Σ moi[e-«»-e-'O,'] + " l

Σ i = l

S

n3 v4 n5

i = l i = l i = l

- β ί 2 ( φ - Μ ] - Σ [sMi^e—i'-sMizWe-to]. i = l

(4.1.9)

2. Beams with transverse and axial loading

As a particular case of eq. (4.1.1) consider the beam with a transverse loading and subjected to an axial force P. The corresponding differential equation of deflection is

B(y1Y ± p*y") = φ), (4.2.1)

where p2 = P/B and the plus-minus sign denotes a compressive or tensile force respectively.

If the transform of the right-hand side of eq. (4.1.1) is available, we can easily obtain the transform of eq. (4.2.1)

n w . A B C D 0{s) BY(s) = _ + - + _ _ _ + _ — _ + s ^ s2 ^ s(s2±p2) ^ s\s2 ±p2) τ s2(s2 ±p2)'

(4.2.2)

The symbols A, B, C and D denote here some constants which can be determined from the known boundary conditions.

96 Theory of beams

(a) COMPRESSIVE AXIAL FORCES

From the tables of Laplace transforms [see eqs. (5.2.1), (5.2.3), (5.2.4), (5.3.4), (5.3.5), (5.3.6)] we find

By(x) = A + Bx+C(\ — cos px) + D(px—sinpx) + S(x). (4.2.3)

In this equation

«νΛ - v „ /cosp(x-aoi)-l (x-a0i)2\l

i W _ A,qoi \ J* + 2p2 / ~ 1=1 \ r tr / aQ.

- Σ v°i i = l

cosp(x-boi)-l (x-boif\l

V Λ

where

co(s) =

~Σ -^r<^-"aoi)--sin/?(x--aoi)>io j4-i = l ^

+ Σ -Jr<P(x-ßod-s™ P(x-ßoi)>lßoi + i = l ^ n8 p.

i = l P

- Σ ^< ! ~ c o s Pi . x -dd^+L-^ i s ) } , i = l ^

Ä Γ Qn(s)e-a,s Qi2^)e-biS

- Σ i = l

i2(52+^2) J2(52+p2)

φ2+/>2) φ2+/>2)

(4.2.4)

(4.2.5)

When the loads are constant, the above relations give the solutions directly. But when the loads are non-uniformly distributed it is necessary in each particular case to refer to the tables of transforms. Tf these tables do not contain functions of the type M^{s) and Q^{s) (j = 1, 2) the last terms of the function S(x) can, on the basis of Borel's theorems (5.5.2) and

(5.5.3), be expressed in the form X

I T » = I £ f [>(*_,)-sin/>(%-0] <?4(0>Si A -

1 η β Γ - ^ Σ [ l - c o s ^ - / ) ] < m ( 0 > 2 | * (4.2.6)

and the corresponding integrals can be evaluated either exactly or by some approximate method.

(b) TENSILE AXIAL FORCES

Utilizing relations (5.2.1), (5.2.6), (5.2.7), (5.3.7), (5.3.8) and (5.3.9) the solution of eq. (4.2.3) is

By{x) = A + Bx+C(l-coshpx) + D(smhpx-px) + S(x), (4.2.7)

where

Σ -ττκ coshp(x-aoi) 1 \ -i=l P \ L / aoi

-ϊ$(«**-«-^->>' -i=i ^ \ L / bQi

~ Σ ^ < s i n h / ? ( X - a O i ) - / ? ( ^ - a O i ) > i o i + i = l ^

+ Σ -§L<sinhJp(x-/30i)-jP(x-^0i)>^{ + i = l -̂

713 -Pi + Σ ^ <sinh/?(x-cO-/>(*-q)>c,-- Σ ^icoshpix-dd-iy + L-ifaas)} (4.2.8)

i = l P

98

and

Theory of beams

ωι(ί) Ä Γ Qn(s)e-a<° Qi2(s)e-b<s i d s\s*-p*) s\s*-p*)

"· Γ Μα(Α)ε-α·δ Mi2(s)e-ß<s

kl *( s(s2—p2) s(s2—p2)

The corresponding expression to eq. (4.2.6) is

(4.2.9)

X

1 we f ί.-Ηωιί*)} = T3 Σ tsinh ** - ' ) - /** - ' ) ] <?(0>ίί * -

P i= l J 0-i

1 n« f - - , £ [ c o s h p ( x - 0 - l ] < m i ( 0 > S * . ^ i= l J

(4.2.10)

3. Beams on elastic foundations (Winkler's type): general solution

If the beam rests on an elastic foundation of Winkler's type (which is capable of transmitting not only pressures but also tensions), then eq. (4.1.1) has the form

yW + AVy __ φ(χ) B

(4.3.1)

where λ is a known coefficient. Having found the transform of &>(s) as given by eq. (4.1.9)

the transform of eq. (4.3.1) is determined at once as

Y(s) 4λ + ί 4 (■ yS+ytf+y'0''+yi"+^p Ψ)- <«· 2)

From the tables of transforms [eqs. (5.2.9)-(5.2.12)and(5.3.10)-(5.3.12)] we obtain the following equation of the deflection

curve:

+ % : [/«(*) - / s W ] + S(x). (4.3.3)

In the above equation the meaning of various symbols is as follows:

fi(x) = c o s λ* c o sh λχ, fz{x) = cos λχ sinh λχ9

f2(x) = sin λχ cosh λχ, /4(x) = sin λχ sinh λχ, (4.3.4)

4 κ

4B' The quantity defined by symbol κ is called the coefiScient of yielding of the foundation.

The load function for this case is

s<*) 4 ! ? o i }k,+ i = l

+τ Σ PftMx-cd-Mx-cd)1«-t—1

9 32 η4

£ ΜΚ/4(χ-^)>^+^-1{ω2ω}> (4.3.5) where

ι = 1

„ ίΛ - ν Γ Qii(.s)e~a's QiÄß)e-ßtS 1 ~ i=i L 5(4λ4+54) 5(4λ*+54) J

_ *· r sMj^e-««* sMi2(s)e-e>* 1 £ [ 5(4λ4+ί4) 5(4A*+J4) J· l ;

100 Theory of beams

Similarly, as in the previous cases, eq. (4.3.6) may be expressed, if required, in the form

λ nß

i = l J

x

Λ(*-0-Λ(*-0<ίι(0>ί,Λ-

- Ί Γ . Σ i/*(Ä-0<«i(0>S*. (4-3.7) ΐ = 1 J

which follows from relations (5.5.6) and (5.5.7). When developing eqs. (4.3.3) and (4.3.5) it was assumed that

the beam coincides with the axis Ox along the region (0, /) and that at the cross-section x — 0 there is neither concentrated force nor moment applied and also that any distributed moment is acting over the region whose boundaries do not coincide with the origin of the coordinates. As a result of these restrictions the four constants of eq. (4.3.3) can be expressed as derivatives of y^\x) at the point* = 0. Otherwise these constants would have to be expressed, say, by the symbols A9 B, C and D and evaluated from the boundary conditions for each individual case.

4. Free beam on an elastic foundation

The general solutions derived in the preceding section will now be applied for solving a boundary problem, viz. that of a simple beam freely supported by an elastic foundation. In the analysis which follows the only additional relations needed are those between the functions of eq. (4.3.4) and their derivatives.

It can easily be checked that the following six relations exist between them as expressed by six equations (4.4.1) and the


enclosed Table 2.

Λ ( 0 ) = 1 , Λ(0) = 0, / ,(0) = 0, /4(0) = 0,

fl(x)+nw-fl(x)-fl(x) = i, Ά(χ)Μχ) = f2(x)Mx)- (4.4.1)

TABLE 2.

ί

1 2 3 4

/,(*)

A(x)

AM

AM

f'M)

λ[Α(χ)-Α(.χ)]

λ[/ι(*)-Λ(*)] Α[/,ω+Λ(*)]

//'(*)

-2λ»Λ(*) 2λ«/,(*)

-2Ay2w 2λ2Λ(*)

//"(*)

-2λ°[/3(χ)+Α(χ)] 2A»[/iW-/iW]

-2λ«[Λ(*)+/4(*)] 2λ ' [ / 3 ω- / 2 ω]

The four constants of eq. (4.3.3) should be determined from the boundary conditions of the beams. For the case of a simple beam, resting freely, these boundary conditions are

(1) y'0' =0, (3) / ' ( / ) = 0, (2) y'0- = 0, (4) / " ( / ) = 0. (4.4.2)

Using these above relations as well as those of Table 2 and performing some simple transfigurations the equation of the deflection curve assumes the form

y(x) = y«Mx)+|f- [/*(*) +/,(*)]+s(*);

y° = -2^D{2Xfi(.0S"{D-[Mi)-M0]s'"ii)},

i y0 X*D {Ml)S"V) - A[/2(/) +/,(/)]S"(Q}, (4.4.3)

where D = sinh2 XI— sin2 //.

102 Theory of beams

5. Simple beam with terminal forces and couples resting on an elastic foundation

With the help of eq. (4.4.3) we can write down the equation of the deflection curve for a beam resting on an elastic foundation when acted on by concentrated forces and moments at the cross-sections x = 0 and x = /. As the first step it is necessary to transform eq. (4.4.3) so as to show clearly that there are two concentrated forces and two moments. To this end we shall single out the corresponding four terms of the load function S(x) by expressing it as follows:

S(x) = 4 Σ Pi<Mx-cd-fz(x-cd>1»-X i = 1

2A2 2

- — Σ MiWx-dd^ + Six). (4.5.1) X i = i

Next we seek S"(x) and S'"(x) using the relations of Table 2; thus we find the following expressions for S"(l) and S"'(l):

17? 2

S"W = ΊΓΣ PilMi-cd+fiV-cd]-i = l

-^iM*/i(/-«*)+s"(/), i = l

4λ4 2

κ i = l

4λ5 2

—— Σ MiW-di)-f&-dd] + S''Xl). (4.5.2) i = l

If we now substitute the above expressions for S(x), S"(l) and £'"(/) in eq. (4.4.3), an equation is obtained which contains two forces and two concentrated moments acting on the beam at points cl9 c29 dx and d2 respectively. In the limit, when

c1-^0, c2-+l, d1-*0 and d2^l, (4.5.3)

eq. (4.4.3), which was developed for a beam without the terminal loads, transforms into equation of the deflection curve for a beam with Px and Mx and x = 0 and P2 and M2 at x = /.

This limit analysis leads to the following results: (1) Terms containing P2 and M2 in eq. (4.5.1) vanish so that we obtain finally

IP ?λ2

■*(*) - -7T[Λ(*)-Λ(*)]-— ^ ! / i W + S(jc). (4.5.4)

(2) When computing the boundary values S"(l) and £""(/) it is helpful to use relations (4.4.1). Further noting that for c2 = d2 = I both the force P2 and the moment M2 act on the "left" side of the cross-section x = I, it is necessary to change the signs in the terms containing P2 and M2 (in accordance with the sign convention for the shearing force and bending moment). Thus

s"V) - ^ [MQ+f*(Q¥i~fiWi+^ M2+S"®9

4λ4 4λ4 4λ5

S"XQ - —/ ι (0Ρ ι - — Λ + —M1[/1(0-/»(0] + S/'/(/). (4.5.5)

(3) Substituting in the relations of (4.4.3) for £"'(/) and S'"(l) their limit values (4.5.5) and denoting fß) = /J the following expressions for y0 and y'0 are obtained:

y0 = 2&Ö lwJS"(t)-(fa-fs)S"'(0+

+ ~ (ΛΛ +ΛΆ -fik +fifz)Pi + ̂ - <J* -fJP* ~

- — [2fJl+(f2-f3f]M1+— /4M21 ,

(4.5.6)

104 Theory of beams

— {fzfi-hU-fifz-fJt^ — ift+MMX.

Making further the following substitutions:

p X*D' 7 2A2Z> '

2A3£> ' φ ~ 2A3Z> '

and using relations (4.4.1) it can easily be shown that

0 1 sin A/ sinh A/ A2 sinh2 A / - sin2 A/'

1 s in 2 A/+sinh 2 A/ y o i2 „;

6 =

2A2 sinh2 A / - s i n 2 A/'

1 sinh A/cosh A/—sin A/cos A/ 2A3 sinh2 A / - s i n 2 A/

_ 1 sin A/ cosh 11— cos A/ sinh A/ ψ ~ 2~A3 shüVÄT^shi 2 ! / '

_ 1 sin A/cosh A/+cos A/sinh A/ Ψ~Ί sinh2 A / - s i n 2 A/ '

„ _ 1 sin A/ cos A/+ sinh A/ cosh A/ A sinh2 A/—sin2 A/ (4.5.8)


From the above relations the deflection curve results directly as

v' λΡ y(x) = yJi(x)+^[ft{x)+Mx)]+-£[Mx)-A(xy]-

-^^Mx)+S(x);

γ9 = βε»Φ-φ!?"®+»*±+φ*ί-γψ+βψ,

y'0 = ßS'"(D-y,S"(l)-V^-ß^+i>^-f^, (4.5.9)

where the bar over S has been omitted throughout. We are able now to determine y(l) and / ( / ) . With the help of the relations (4.4.1) we find

y(l) = bS"\D-VS"(!)-9^-b^ +ß^-y^+S(l),

(4.5.10)

We can also determine expressions for M(x) and Q(x), i.e. for the bending moment and the shear force distribution along the beam. These expressions follow from eqs. (4.5.9) and differential equations (3.1.1) and (3.1.9). They are

Mix) = 2^oA(*)-4ä«yi[fa(*)-/2(*)]-

-^i[Mx)+A(x)]+fi(x)M1-BS"(x),

Q(x) = -^yo[Mx)+Mx)]-^y&(x)+fi(x)Pi-- *[/»(*)-f2(x)]M1+BS'"(x)+m{x), (4.5.11)

106 Theory of beams

where in the case of the most general type of loading 1 nt

S"M = 2p£ Σ ?oi</4(*-tfoi)>L-i = l

I m

1 ^2

~ -TyTE Σ m0i(MX-a0i) +/2C*-a0i)>i0i + i = l

1 n2

+ -2Tg Σ ^οί<Λ(χ-^)+Λ(^-ω>^·+ ΐ = 1

1 ^̂ i = l

X

1 n* 1 Γ -]B Σ ^i</i(^-*)>iii+23[B [ /e(^-0 +

0

X

+Λ(*-0]?(0 Α-— |Λ(χ-ίΜ0Λ, o

1 ™i £'"(*) = 2 ^ Σ aoi<Mx-aoi)+f2(x-aoi)>laoi-

i = l 1 ri!

- 2TÖ Σ ^οΚΛί* ~ *ot) +ΛΟ - bodYboi -i-1

I n2 J n2

~1Γ Σ moi(fi(x-*od)lot + -j Σ moi(fi(x~ßoi))ßoi + i = l i = l

+ "F Σ ̂ </i(*-Ci)>i,-4- Σ M^Mx-dd-i = l i = l

Λ; Χ

-Λ(*-4)>&,+4- |/ι(*-0?(0Λ-4 f[/s(*-0-0 0

-Mx-t)]m(t)dt-~m(x). (4.5.12)

The functions q(x) and m(x) are in general multi-step functions. In relations (4.5.10) there appears the first derivative of the load function. It can easily be shown that it is equal to

1 n i

i = l 1 nx

+ WB Σ ?Oi</s(^-*Oi)-/2(^-*Oi)>lo#-i= l

1 * «

i = l 1 n2

+ 2λ^β Σ ™οΚΛ(*-Α«)>Α,ι + i= l

1 n3 I nA

+ 2T2Ö Σ Pi<f*(x-cd%—üB Σ MiiMx-diH i = l i = l

+/3(*-4)>&, + 2 ^ |Λ(^-0?(0 A -0

~ 2 l S [ [ / ί ^ - Ο + Λ ^ - θ Μ θ * · (4-5.13)

This concludes the solution of the problem. Many other problems of beams on elastic foundation of Winkler's type can be solved by employing a similar procedure.

6. Beams on non-homogeneous elastic foundations whose elasticity varies in a stepwise manner

In general, problems dealing with non-homogeneous foundations will have to take into account the non-homogeneity extending vertically, i.e. towards the interior of the foundation as well as the horizontal non-homogeneity. The latter type occurs, for example, with the railway tracks passing from the embankment over the relatively rigid abutment on to the

108 Theory of beams

bridge where the foundations are partly on the piles and partly on the ground or on the piles of a variable density.

Consider this last case assuming that the elastic properties of the foundation are similar to those of Winkler's type and also that they change abruptly in a stepwise manner in the direction of the beam axis. That is the foundation is non-homogeneous along the beam but within each particular segment, the elastic properties remain constant. To solve this problem we shall make use of the results developed in sections 4 and 5.

(a) EXPOSITION OF THE PROBLEM

Consider a beam resting on a non-homogeneous foundation of Winkler's type which can be subdivided in the axial direction in C?-f-l) homogeneous segments (Fig. 28). Let the

FIG. 28.

coefficient of yielding for each successive homogeneous segment of the foundation be equal to «r (r = 1, 2, . . . , £ + 1).

Assume that the deflection of the beam on respective segments is given by the function y^-^Xr), and that the pressure of the foundation is given by the function p(r_1)(^r) (r = 1, 2, . . . , j + 1) (Fig. 29), then for the Winkler's type of foundation we have

y(r-l)(xr) — p(r-\)(xr) ( r - 1,2, . . . , j + l ) . (4.6.1)


Assume further that the flexural rigidity of the beam Br = ErIT is also varying in stepwise manner. Without impairing the generality of the discussion it may be taken that the changes in Br occur at the same points where the elastic properties of the foundation change. For, if at some point only

i rX r

FIG. 29.

the rigidity of the beams changes without simultaneous change in the elasticity of the foundation, such a case may be derived from the previous one by simply equating the coefficients of yielding of the two adjacent segments of the foundation.

With these assumptions the problem reduces to that of finding the deflection function y{r_1)(x1) for each segment of the beam on a homogeneous foundation. Once this function is determined all the relevant static quantities can be derived from it.

(b) METHOD OF SOLUTION

Consider a segment of the beam between the cross-sections Or_1 and Or and supported by an elastic foundation whose coefficient of yielding is κν (Fig. 29). Discard the left portion of the beam and replace its effect by the shear force Pr__1 and the bending moment Mr__x both acting on the cross-section Or_r Similarly, the effect of the right portion of the beam will be replaced by the corresponding shear force Pr and bending moment Mr at the cross-section Or. In this manner we are left with the problem of a beam supported by a homogeneous

110 Theory of beams

elastic foundation the beam having, in addition to its initial loading, terminal forces and couples at each end Or_1 and Or. It follows, therefore, that to solve the problem it is necessary:

(1) To obtain the equation of the deflection curve for a beam on a homogeneous elastic foundation with an arbitrary loading along its span and terminal forces Pr_1 and Pr and terminal couples Mr_x and Mr at the end sections Or_1 and Or respectively.

(2) To determine the values of Pr_v Pr, Mr_1 and Mr from the condition that the deflection curve must be a continuous and regular curve at all points yr(x) where there is a change in elasticity of the foundation or in rigidity of the beam.

(c) EVALUATION OF THE SHEAR FORCES AND BENDING MOMENTS AT THE POINTS OF DISCONTINUITY

Equations (4.5.9) will give the complete solutions of the fundamental problem of a beam on non-homogeneous elastic foundation with segments of constant elastic properties, if for each segment (denoted by suffix r in Fig. 29) we are able to determine the two shearing forces Pr_1 and Pr and two bending moments Mr_1 and Mr at the cross-section Or_1 and Or respectively which represent the effect of the remaining (discarded) portions of the beam.

Divide the beam into segments of constant flexural rigidity and supported by (for each segment) homogeneous elastic foundation. Equation of the deflection line for each of these segments is given by relations (4.5.9); all that is now necessary is only to allocate appropriate suffixes to various quantities appearing in these relations and to assume that S(x) denotes the load function consistent with the loading for each segment. The elements of eq. (4.5.9) which contain the two point forces and moments represent the effect of the discarded portion of the beam (left and right) on the deflection line. Consequently,


eqs. (4.5.9) applied to each segment of the beam will enable one to write down a set of functions y^-^Xy) defining jointly y(x) for the whole beam.

In order to determine the unknown Pr_v Pr, Mr_± and Mr

adopt the notation of Fig. 29 where to a segment denoted by a number r corresponds a coefficient of yielding of the foundation κ , the length of the segment lr, the system of coordinates yr-v Or_v xr and four functions (4.3.4) in the form fir(xr) (i = 1, 2, 3, 4). Denote the parameters defined by the relations (4.5.7) by ßr, γν, etc., respectively; the symbol Sr(xr) will denote the load function corresponding to the actual loading on the rth segment of the beam. For the sake of brevity denote by Sr = Sr(lr) the value of this function at the point xr = lr Similarly, take S(

rj) = S[j)(lr) to represent its derivatives.

Assuming now that on the whole length of the beam there are s points of discontinuity Or (r = 1, 2, . . . , s) for the foundation which points divide the beam on i + 1 segments, we deduce that to solve the problem we need to determine 2s unknowns: Mr and Pr (r = 1, 2, . . . , s). These unknowns can be determined from 2$ conditions of continuity of the deflection curve y(x). Since the load carried by the beam does not rupture it at any point and since the axis of the unloaded beam is a straight line, hence the axis of the deflected beam must be a continuous and regular curve possessing a continuous first derivative at all points. In particular this condition must be satisfied at the points of abrupt changes of elasticity of the foundation Or or at the points where the rigidity of the beam changes abruptly. This gives us the following 2s conditions:

Jr-l(/r) = JY(0), y'r-iVr) = y'M (r = 1, 2,. . . , *), (4.6.2)

which are sufficient to determine the unknown Pr and Mr

(/· = !, 2, ...,s).

112 Theory of beams

Indeed, on substituting in the conditions (4.6.2), the relevant values of the deflection function and its derivatives, as defined by eqs. (4.5.9) and (4.5.10), the following set of 2s linear equations is obtained in which there are 2s unknown forces Pr and bending moments Mr:

ίδν + 1 ] δΛρ Ψτ+1 p

\Br+1 BrJ r Br+1

, / r r+ i Yr\p \ Ργ+1 pr \Br+1 Br) r # r 4 1

— Pr+l^r+l~Wr+l^r+l~yr^r + # r o r — Sr

( r = 1,2, . . . , * ) . (4.6.3)

To solve these linear but non-homogeneous equations it is necessary to know how the beam is fixed at both its ends, i.e. the values of P0 , M0, P s + 1 and Ms+1 must be prescribed beforehand.

(d) THE APPROXIMATE EQUATION OF FIVE MOMENTS

Let us compare the above problem with the analogous problem of a continuous beam on rigid supports. It is known that in the latter case there exists the equation of three moments. The equation relating these three moments does in fact correspond to the set of equations (4.6.3) in the above problem. Because of the existing analogy between the two cases it would be possible to call each of the equations (4.6.3) an "equation of six static parameters" (bending moments and shearing forces). However, there are only two relations and six unknown


quantities; alternatively, these can be reduced to one equation with five unknowns but such a step does not represent any gain, only an increased computational difficulty.

By a similar line of argument it may be concluded that each of equations (4.6.3) corresponds to the equation of five moments which exists in the case of a beam on elastic supports.

In spite of the similarity indicated above between the three cases, we must not overlook the difference existing between them. In the case of the theorem of three moments we have to determine only s unknowns (i.e. the bending moments) because the forces P acting through the supports do not affect the line of deflection. In the case of a continuous beam on elastic supports there are 2s unknown moments and forces to be determined. The theorem of five moments is therefore a definite advantage because it represents a simplification of the problem resulting from the elimination of one set of s unknowns so that we are left only with a set of s equations with s unknowns.

In the case considered here of the beam on an elastic non-homogeneous foundation such a separation of equation into two sets, one of which contains moments only and the other forces only, is not possible at all. This can be seen clearly from comparison of eqs. (4.5.9) and (4.5.10) with the corresponding equations (2.9.2.) and (2.9.3).

From the equation of the deflection line for a single-span simple beam resting on two elastic supports (which leads to the equation of five moments) it is seen that a force acting through one of the elastic supports will not influence the deflection of the other support, i.e. the deflection of the other end of the beam.

It is precisely this mode of deflection so characteristic for the simple beam and the appearance resulting from it of only one shearing force in the equations of the elastic curve j>(0) and y{t) which enables one to determine this shearing force at

114 Theory of beams

the support directly from the equation ;;r-i(^r) = Λ ( 0 ) (assum

ing the moments to be known). Consequently, this leads to the formulation of a system of s

equations with s unknowns, i.e. to the equation of five moments.

On the other hand, it will be seen from (4.5.9) and (4.5.10) that equations for the deflections y(0) and y(l) contain both forces, i.e. Px as well as P2. As a consequence of this a force acting at one end of the beam which rests freely on an elastic foundation produces a deflection not only at the point of its application but also at the other end of the beam. It is possible, therefore, to say figuratively that a disturbance of the equilibrium caused by the action of a force at one end of the beam is transmitted through the foundation over the whole length of the beam which results in a deflection of the opposite end of the beam as well. This is reflected in the fact that the equation JV-i(^r) = Λ·(0) contains three unknown forces, and this is the reason why the number of equations (4.6.3) cannot be reduced. From the above discussion it follows, however, that it would be possible to develop an approximate equation of five moments for a beam on a non-homogeneous elastic foundation of the type considered. This possibility is governed by the magnitude of deflection at the unloaded end of the simplified model of the free beam with a point load at the other end. Let us therefore examine closer the magnitude of the above deflection. From eqs. (4.5.9) and (4.5.10) it appears that if S(x) = 0, P2 = 0 and Px 5* 0 then

P P y0 = δ -— = . * (sinh XI cosh XI—cos XI sin XI),

P —P y(l) = — φ-~ = ^ o J p (sin A/coshA/— cos A/sinh λΐ). B 2λ DB

(4.6.4)


It can be seen from these relations that when the products XI are not very small (i.e. when the beam is not very short) the deciding factor as far as the magnitude of y is concerned is the term sinh λΐ cosh λΐ representing the exponential function equal to e2Uj4. It may be taken therefore, that in this case | j 0 | is substantially larger than |j>(/)|. Thus there is sufficient justification to neglect in eqs. (4.5.9) and (4.5.10) the terms φΡ2/Β and —φΡχ\Β respectively for the most frequently encountered practical cases and to obtain in this way from the first relation in eqs. (4.6.2) [or from the corresponding first relation eqs. (4.6.3)] an approximate expression for the shearing force at the points of discontinuity, and from the second relation in eq. (4.6.2) or (4.6.3) and approximate equation of five moments.

The first expression becomes then

BrB^ r + l Br

r r+ l drS'r'' - TVS" + ßr —-£-^ - yr ~ + Sr -

Br r

~~Hr+lSr + l + 99r+l*Sr+l + >Y+l~7i P r+1 # r + l

Mr+il Br+l\

where BI+1 = Brdr+1 + Br+16r,

while the equation of five moments is

M ßr-lßl M \Vr , ßrVr-1 ßrVr jBr-l , Br+1\ , Mr-2~^1~Mr~1[Yr

+ ~W^~^\B^

(4.6.5)

(4.6.6)

,βτϊν+Ι + MT Br Br+l

ß2rBr-i r2

rBr+1

+ 2 VrYr+l Ύ* + 1Βτ ß$+lBr+2 Br+iBr Br+1B[+1 Br

r+1

ßr+lTr-i-l I Br Hr+lVr+l I

Br+1 \

Br ±\ +

ΒτΒ'_ν

Mr+i

Br+iyr+

►r+l BrBrr

Wr+l , ßr+l7r

+

Brr+1 +Br

r$2 ) ^ Brr\\

Br+i

+ Mr+2

Brr+1

P r + l P r + 2

$rS'r' + S'r -

116 Theory of beams

_i^i=i.[if_1s;:'1_yr_1s;i1+sr_1-i8rls;'+9,r5;"]-

- y A ^ i r r + l g r [ ^ ; , , - y ^ ; ' + ^ - i 3 r + i ^ ; i + y r + i + ^ ' i ] +

^ ~ör+2—[dr+lSr+l~Vr+lSr+l + $r+l ~~Pr+2*^+2+ ^ + 2 ^ + 2 ] ·

(4.6.7)

Having found the moments from the above relations the unknown P r ' s are determined from eqs. (4.6.5) or if a greater accuracy is required from eqs. (4.6.3).

EXAMPLE. AS an application of the above theory we shall find the deflection curve, equation of five moments and the shear force for a beam of constant rigidity, resting freely on a non-homogeneous foundation consisting of three homogeneous segments (Fig. 30). Assume the following data: q = = 20 kg/cm, P = 4000 kg, B = (4/3) x 109 kg cm2, xx = κ3 = = 80 kg/cm2 and κ2 = 8000 kg/cm2.

Since M0 = M3 = 0 and P0 = P 3 = 0 while B(x) = B = const, over the whole length of the beam, hence in this case equations of the six static quantities (4.6.3) reduces to a system of four simultaneous equations with four unknown quantities, viz. (γ2~7ι)Μ1-β2Μ2-(δ2+δ1)Ρ1-φ2Ρ2

= / ?ü3 2 Si l -y 2 5 i " - a 1 Si " + y 1 S i r - S 1 ] , β2Μ1+(γζ-γ2)Μ2-φ2Ρ1-(δ3+δ2)Ρ2

= B[ß3Sä-cpzSz"-ö2S2" + y2S2 - S 2 ] , -(^2+^1)Μ1+ψ2Μ2-{-(γ2-γ1)Ρ1+β2Ρ2

= B\ß1SX'-y>2S'2'-VlS'1" + #1S'1'-Si]9

ψ2Μλ - (08 + §2)M2 - ß2Px + (y j ~ V 2)^2 = B\ßtS's"-VzS'z'-Y2S'2" + #2Si'-S&

(4.6.8) These unknowns are determined in the following order:

(a) Obtain the load functions and their derivatives for the three segments of the beam in accordance with eqs. (4.3.5), (4.5.12) and (4.5.13). These are:

Si (*i) = - r [i - Λ Μ , si(*i) = ~ ^ [fM-ΜχύΙ

s2 (Jc2) = ^ - [ i - / 1 ( J f a ) ] _ X / i - / 1 ( x a - ^ ) \ 4 0 0 / 8 , κ2 κ2 \ V 3 7/200/8

Si (*2) = ~^·[/8(*2)-Λ(*ί)] + 9Aa / , / 200\ , / 200\\«°/»

+-ί-<Λ(Ι·-τ-)-Λ('·~Γ)>11„· s - w - M ^ - M / , , / , - » » ^ - " .

κ 2 «2 \ \ J 7 / 200/3 S2"(*2) = M l [/3(χ2)+/2(χ2)]-

-¥<Α(-")+Μ-")>Ζ· * « - * < Λ ( * - · » ) - Α ( ^ ) > - .

^ Ι \ \ J / / 100/3 (4.6.9)

(b) Compute the magnitudes of λχ — λ3 and λ2 from the last relation of eq. (4.3.4).

118 Theory of beams

F I G . 30.

(c) Compute the magnitudes of the trigonometric and hyperbolic functions1" appearing in eqs. (4.5.8) and (4.6.9) at the points x{ — lt = 400/3 cm.

(d) Compute the values of the constants in eqs. (4.6.8) using the relation (4.5.8) and the results of (4.6.9). Leaving out the details of the above computations the results are tabulated in Table 3.

TABLE 3.

Segment I

Quant.

ßl Vi <5> Ψι Ψι »1 Si S[ S'i S{"

Magnitude

11 067X10-« 50 976X10-1 65 309X10-1 52 184X10 23 371X10 68 025X10-3 13 300X10-2 19 5 3 0 X 1 0 - 5

57 927X10-7 12 635X10-« 16 8 6 0 X 1 0 - ' °

Segment II

Quant.

K

y2

Ψ2 Ψ2 »2 s2 s2 Si' s2"

Magnitude

34 996X10-« - 1 5 354X10-3

40 855X10-2 11 671

- 2 0 925X10-2 - 5 6 225X10-5

28 590X10-3 - 2 8 449X10-7 - 5 0 682X10-7 - 3 4 769X10-« - 1 1 9 5 5 X 1 0 - 9

Segment III

Quant.

ßs Vs <53

Ψ3 V>3 #3 s3 s3

s:t-

Magnitude

11 0 6 7 x 1 0 - « 50 976X10-1 65 309X10-1 52 184X10 23 371X10 68 025X10-3 13 300X10-2 49 645X10-5 14 752X10-« 28 505X10-« 22 5 2 8 X 1 0 - 1 0

Table 3 contains the values of all constants appearing in eqs. (4.6.8). Putting these values into the equations they reduce

t Five figure tables of ref. 13 were used in computing these functions.


to the following system:

- 6 1 223X 10~1Λ/1+15 354X10-3M2-53 351X10Px+ + 20 925X10-2P2 = - 3 2 952X104,

- 1 5 354Xl0~3M1+61 223XlO"1Ma+20 925χ10~ 2 Ρ 1 -- 5 3 351Xl0P 2 = 12 359X105,

- 1 6 159Χ10-ΪΛ/1-56 225Χ10-5Μ2-61 2 2 3 X 1 0 - ^ -- 1 5 354Χ10~3Ρ2 = - 1 4 123,

- 5 6 225X 10- 5 Mj-16 159X10-2M2+15 354X10-3ΡΧ + + 61223Χ10-ΧΡ2 = - 1 0 526X103.

(4.6.10)

The solution of these is

Mx = - 4 0 391 kg/cm, Px = 1078-9 kg, M2 = - 3 9 502 kg/cm, P2 = -2768-2 kg.

Having found the bending moments and shearing forces at the end sections of each of the three segments of the beam we can now use eqs. (4.5.9)-(4.5.11) to obtain the formal equations for the deflection curve, bending moments and shear forces. We shall determine here only the deflection curve. The required equation is

for the segment 00±

j0(*i) = 0-2847 1 / 1 ( Χ 1 ) - 0 · 0 9 3 2 6 5 Γ / 2 ( Χ 1 ) + / 3 ( Χ 1 ) ] +

+ 0-25[l- / 1(x1)] ,

for the segment Oy02

jj(x2) - 0-025547/1(x2)-0-017961 [/2(x2)+/3(x2)] + + 0-0047199 [/2(x2) -/„(jca)] + 0O12367/4(x2) +

/ / 200W 400'3

+ 0-0025 [1 -MxJ] - 0-0025 / 1 _yi (*a - _) \ , \ \ J / / 200/3

120 Theory of beams

for the segment 0203

= 0·036640Λ(*3)+0-077362 [/2(χ3) +/3(x3)] -0·38294[/2(χ3)-/3(χ3)+0·12095/4(χ3)

Ι\<ΚΙ*Λ/*{ 1 0 0 \ r l lOOW4»»/·' °'55334\Λ('--)-/·('ϊ-~)>1„,

J^fe)

(a)

(b)

(c)

+ (

f50000 kgcm *

+ 400/3

+ 55311 kgcm

FIG. 31. (a) Deflection curve, (b) Shear force diagram, (c) Bending moment diagram

In all the above computations the starting figures contained five decimal places and the results in (4.6.11) contain at least three significant decimal places. Figure 31 shows the deflection curve, the bending moment diagram and the shear force diagram. The computed values are marked off on the diagrams by small circles.

In order to demonstrate the effect of the non-homogeneity of the foundation in Fig. 32 the corresponding diagrams are shown for the same beam resting on a homogeneous foundation, of κ = 80 kg/cm, taken from the book by M. M. FILONENKO-BORODICH. (See ref. 14, p. 147)

If the approximate relation of eqs. (4.6.5)-(4.6.7) be used for solving the above example, then the following values of bending moments and shearing forces would be obtained instead of those found in eq. (4.6.11):

Mx = - 4 1 224 kg/cm, Px = 1089-6 kg, fi

M2= - 3 9 515 kg/cm, P 2 = -2788-4 kg.

To find these values it is necessary only to solve two simultaneous equations containing two unknowns.

For the reasons given previously (the length lr of the segment being small, only 400/3 cm) the example is not really suitable for the approximate method of solution, but even so the values of (4.6.12) are not far off compared with those of (4.6.11) obtained by the exact method.

7. On the non-continuous solutions in the theory of structures

As applied to the solution of differential equations the Laplace transformation is based on the fundamental relation which usually is expressed in the form

L{/(n>(;t)} = snF(s) - s^A+0) - sn~2f\+0) /<»-i>(+0) (4.7.1)

122 Theory of beams

80017 kgcm

FIG. 32. (a) Deflection curve, (b) Shear force diagram, (c) Bending moment diagram

When the solutions of the differential equations are the continuous functions with (n — 1) first derivatives then the application of the above relation does not present any difficulty. But, as is known, in the theory of structures the required solutions and their derivatives most frequently are non-continuous.

In the preceding sections it has been shown how eq. (4.7.1) can be employed for obtaining some non-continuous solutions


of the diiferential equations in the theory of beams. To this end we have made use of a fairly straightforward method of substitution, by means of which a concentrated force was replaced by a distributed load and then the limit analysis was applied either to the solution of the differential equation (Chapter III, section 1) or to the transform of the equation (Chapter IV, section 1). In this chapter we shall present an alternative method for solving the diiferential equations of the theory of structures while still employing Laplace transformation. This method will enable us to take into account not only the discontinuities created by the loading but also the discontinuities due to the geometry of members of the structures, e.g. small curvatures of some parts, non-uniformity of the cross-sectional areas, etc.

We shall evoke for the purpose a theorem on the Laplace transformation which, in fact, represents a generalization of eq. (4.7.1).1" This theorem deals with the derivatives within the range where the function has a discontinuity of the first order. When referring to the derivatives at the points of discontinuity (or at the end of a region) we must distinguish between the left-side and right-side derivatives. The gist of the theorem is as follows:

Suppose that a function/(JC) as well as its successive derivatives / '(*), /"(*), . . . , /{η~χΧχ) are all continuous in every region (0, Γ), where T is any arbitrary positive number, except in a limited number of points x = hl9h29 . . . , hn (some of these values or all of them may be equal to one another) at which points these functions have respectively only one discontinuity each, the discontinuities being of the first order and of the magnitudes /(Α1+0)-/(Α1-0), / , (Λ2+0)-/ , (Λ2-0), . . . , /(n"1)(An+0)-/(n"1)(An-0). Suppose further that all these functions do not increase faster than the function eax (a > 0) when x tends to infinity. If for such a function its/(n)(x) has in

t The proof of the theorem is given in ref. 2.

124 Theory of beams

the given region at the most only a limited number of discontinuities then there exists the transform L{f{n\x)) and it is given by the relation

L{f(n\x)} = snF(s)-sn-1f(+0)-sn-Y(+0)- ... ■ ■. -f(n-1)(+0)-sn-1ehiS[f(h1+0)-/(Αχ-Ο)]-

-sn-2e-h>°[f'(h2 + 0)-/'(A2-0)]- ,_e-/ l»s[-y-(r f-l)(An + 0 ) _ - / ^ ( A n - O ) ] (*=►*). (4.7.2.)

In order to demonstrate the features of the method of solution of differential equations based on eq. (4.7.2), we shall obtain the deflection curve for a beam whose flexural rigidity B(x) varies in a stepwise manner. This may be as a result of either the abrupt changes in the cross-sectional area or in the Young modulus when the beam is composed of different materials. We shall assume arbitrary type of supports for the beam, that the beam consists of n segments of different rigidity and that it carries the most general type of loading (Fig. 33). The differential equation for such a beam is

νιν^νΛ _ /q{x)-m\x)\^ t /q{x)-m\x)\^ , " W " \ *i / o + \ ί - " / · / " · '

.+ /q(x)-m\x)\l

Bn (4.7.3)

Qn-1

M iPi

W R R M " B i - B :

W~^ di

ty

W:

m;(x),

mi ^ } (—-AT— C

FIG. 33.

where y{x) denotes deflection of the axis of the beam, q(x) denotes the distributed load, m(x) the intensity of the distributed moment and B{ the rigidity of the beam over a segment (Qi-v Qi)·

Without impairing the generality of the solution we shall assume now, for simplicity, that the beam is acted upon only by one concentrated force Pt, one concentrated moment Mi

BS 3 Q I -

_bi_

M t qiCx)

•Bbi-

F I G . 34.

(applied respectively at the points x = ci and x — d{), one distributed load q{(x) acting over the region (ai9 b^) and one distributed moment acting over the region (ai5 β^. As regards the function qt(x) and m^x), we assume that they are sufficiently regular for the purpose of the required operations, and in particular that they can be extrapolated outside their boundaries of action and that they possess Laplace transforms. Further, we shall assume that the region (ai9 bt) and (at, ßt) do not occur over those portions of the beam where the rigidity changes abruptly. If, however, a distributed load over a region, say, (aiy b^ does extend over a portion where the rigidity changes, then such a region can be subdivided into subregions of constant rigidity and the case reduces to the previous one, but instead of the single element function <#i(X)>£* we shall have to consider the sum of the elements (see Fig. 34), viz.

<?i(*)>2 = <* ( * )> ! + <*(*)>$' · (4.7.4)

126 Theory of beams

With the above assumptions the right-hand side of eq. (4.7.3) may be represented by two terms only, and the equation becomes

yiv,x) = /*W\b'_ /mM\ß\ ( 4 7 5) y w \Bai/at \ BH/:t *·'*>

We are seeking now the solution of this differential equation, i.e. such a function y(x) which at the point x=0 has the following prescribed values: y0, y'0, y'0\ y0".

In addition the function y(x) itself and its first three derivatives y\x), y"(x) and y'"(x) have to be discontinuous at some points within the region (0, /). This discontinuity is of the first order, i.e. at the point of discontinuity the function y{l\x) has a finite jump of magnitude

y{i\hi+i + 0)-/Hhi+1-0) (i = 0, 1, 2, 3). (4.7.6)

The required solution of eq. (4.7.3) which will satisfy all the above conditions can be obtained by means of Laplace transformation, provided that the points of discontinuity as well as the values of the jumps (4.7.6) are known beforehand. In such a case the values of y0(i), together with the values of eq. (4.7.6), form the "boundary" conditions for the problem.

It will be noted that in this problem the magnitudes of the jumps of the function y(x) itself and of its first derivative may be assumed arbitrarily provided only that the assumed values both for the deflection of the axis and for its slope are rather small (Fig. 35) so as to comply with the geometrical conditions of the problem (see ref. 15).

We shall take, therefore, the following "boundary" conditions for the geometry of the problem

y{hi + 0)-y{hi-0) = Ai9

y'(n + 0)-y'(n-0) = Ci.

On the other hand, the discontinuity of each of the remaining two derivatives y'\x) and y"\x) cannot be defined arbitrarily;


they are determined by the bending moment M(x) and the shear force Q(x) respectively through the following two equations :

M(x) B(x) >

nt = Q(x)-m(x) B

(B = const).

l l B,

+y ΪΡΓ

FIG. 35.

r*

I JC

(4.7.8)

It should be observed that a discontinuity in y"(x) at any point is simply due to a concentrated moment Mx applied at that point; similarly, a discontinuity in y'"(x) is due to a concentrated force P{. As a result of the assumptions adopted earlier it follows that to obtain the correct solution of eq. (4.7.3), the following two conditions must be satisfied:

/ ' ( 4 + 0 ) - / ' ( 4 - 0 ) = - - 1 ,

/ " ( q + 0 ) - / " ^ (4.7.9)

where Bdi and BC{ denote the rigidity of the beam at the points of application of the concentrated loads.

128 Theory of beams

From the second relation of eqs. (4.7.8) it will be seen that a discontinuity in the third derivative will occur also in the case when a distributed moment m(x) is not acting over the whole length of the beam, i.e. when its diagram is as shown in Fig. 33. With our assumptions there will therefore be two more conditions:

//// , Λ\ ttu ΛΧ ra(oq + 0) y,f(ai + 0)-y,,Xcci~0) = - - - ^ A

(4.7.10) /'m+o)-^m-o) = - ^ 0 ) ,

where Ba, denotes a constant rigidity of the beam over the region (aif ft).

Further points of discontinuity in the derivatives y"(x) and y"\x) will occur whenever the rigidity of the beam B(x) changes abruptly.

Assume that at the points of a sudden change of rigidity, x = Qi there is only a bending moment M{qi — 0) — M ^ + O ) and a shear force Q(Qi~0) = Öfe + O). This last assumption is, in fact, equivalent to the assumption that at those points there is no concentrated force or moment acting on the beam. If these, however, do occur then they would have to be included in eqs. (4.7.9). Taking into account the last assumptions, we shall obtain from eq. (4.7.8) the following two final conditions:

/ ■ t e + m - z - f a - w - - 1 ^ " - * ^ , # i # i + l

/"(ft + 0)- /"( f t -0) = gfa-0>(* - *+ i ) .

(4.7.11) It is noted that in eqs. (4.7.7), (4.7.9) and (4.7.10) the magni

tudes of the jumps at the points of discontinuity are determined, while the magnitudes of M{oi — 0)and β(ρ{—0) appearing in eqs. (4.7.11) are unknown. These unknown quantities

can most conveniently be determined after eq. (4.7.5) has been solved.

Incorporating all the above "boundary" conditions the transform of the left of eq. (4.7.5) will be found directly from eq. (4.7.2) on putting n=4 and/(x)— y(x). The resulting equation is L{ylv(x)} = MW-styo-fy't-syX-yX'-Aife-k»-

_ c-s*e-v<* + sMie~d'S _ Pie~CiS + ™»(«i + 0)g-*'s _

Bdj Bei Ba.

AH.-QS,-0)e-i,'s ,,, „, Bi-Bi+l

- o ( G i - 0 ) % ^ - + 1 e - ^ . (4.7.12)

The transform of the right side 0(s) is found employing eqs. (4.1.5) and (4.1.7) (all terms being included)

φ φ = Qn(s)e-"i* Qi2(s)e-^ sMil(s)e~^ [

Bai Ba. B<*i sMi2(s)e-h* mi(o:i + 0)e-^s mi(ßi-0)e-^s

Bai Ba. B*i (4.7.13)

Comparing the right-hand sides of these two equations (after dividing throughout by s*, reducing and ordering) the following relation is obtained:

Yis) = ^+4+^+y^+Ai^+Ciqi_ s s* s* s* s si

Mi e~d's Pt e~c' Qu(s)e-a's

Qi2(s)e-b<s Mu(i)e-«'« Mi 2( i )e~^ Bais* Bais3 BXis*

Mfa-OXBi-Bj+Je-«» Q(Qi-0)(Bi-Bi+1)e-^ BiBi+1 s* + BiBi+1 s* ■

(4.7.14)

130 Theory of beams

Hence, using eqs. (5.2.1), (5.3.3) and (5.5.1a) the required solution of eq. (4.7.5) becomes

y M ( r o ) ( j ? r i ? m ) / ( * - a > 2 V £ 4 Α + ι \ 2! / /

n-l 8 f e - 0 ) ( A - ^ - f l ) / ( ^ - g i ) 3 V , « M

s ω _ 3 Aft / ( * - 4 ) 2 V . £ Λ /i*-ctf\l ,

a,· A·

n« 1 f - Σ ÖTTT (*-o2<Wi(o>S*- (4·7·15)

The form of the above equations has been generalized by reverting to the original conditions, i.e. that the beam is acted upon by any arbitrary number of different types of loads and that it consists of n segments of different rigidity.

The transformation of eq. (4.7.14) into eq. (4.7.15) requires several elementary operations appropriate to the Laplace transformation method, which are omitted.

Function y(x) contains four constants of integration j 0( i )

and its form depends upon the geometry of the beam and of the supports.

Let us now determine the bending moment and the shearing force for the beam. Deriving from eq. (4.7.15) the third derivative y'"(x) we get

y'"(Qi-o) = y'o,'+s,1"(ei~o)+'g * i~ / j + 1 ö(e,-o)

j=i ^ i ^ i + i

(i = 1,2, . . . , / ! - 1 ) (4.7.16) and from the second relation of eq. (4.7.8) we have

/■■fa-,) = gto-<»-'"fe-°> p.,,2,.....-,). (4.7.17)

Equations (4.7.16) and (4.7.17) will enable us to determine in turn Q(Q±— 0), ο(ρ2—0)> · · ·> o((?n-i~0) and hence, by means of the inductive proof, to deduce the following equation for the shear force βίρ^—0) at the cross-sections where the rigidity of the beam changes:

ß ( f t - 0 ) = Biy'0" + BiSl'XQi-0) + ni(Qi-0) + i-l

j = l S["(Qi-0) + ̂ ( g j - 0 )

Bi ( i = l , 2 , . . . , « - l ) . (4.7.18)

A similar procedure can be employed to determine the bending moment at the cross-sections x = ρί—0. From eq. (4.7.15) we find

y"(Qi-o) = yO'+/o"<H+s['(ei-o)-ZBi~fi+1 M(Sj-0)+

+ Σ1%#±-1δ^-0)(ρί-ρί) (/= 1, 2, ..., n-1) j t i ^ » j + i

(4.7.19) while from the second relation of eq. (4.7.8) it follows that

y"(Qi-0) = - M ( ^ ~ 0 ) (Ϊ = 1, 2, . . . , n - 1 ) . (4.7.20)

Noting further that the values Q(Q{ —0) are determined with the help of eq. (4.7.18), we shall be able to find in turn M(QX —0), Μ(ρ2 —0), . . . , and hence by induction the equation

132 Theory of beams

for the bending moment at the cross-sections of abrupt changes of the rigidity of the beam:

M(Qi-0) = -{B1yf0' + Bieiy'o" + BiS'1\Qi-0) +

S,i\Qj-0) + (Qi~Qj)(s'l"(gj-0) +

miQj-O)

+ 1* (**-**+1) j = l

+" ( i = 1,2, . . . , / ι - Ι ) .

(4.7.21)

If in eq. (4.7.15) the values of M ^ - O ) and ß f e - O ) are replaced by the corresponding expression obtained from eqs. (4.7.21) and (4.7.18) respectively, then, after a suitable grouping of the terms, the following equation is obtained for the deflection curve:

y(x) = yo+y'ox+yö 2! +*ιΣ tBi-Bi^/ix-Qd2

2! + r - n , Y \ I

+

^ BiBi+1

(4.7.22)

where S(x) = 5Ί(Λ·) + 5 2 (Χ) and

(4.7.23)

The values of Mff( and P appearing in the above equation are defined as follows:


M0 = Bi — Bi + 1

Bi

X

JJ1si'(ft-o)+X(5i--Bi+i)x

/η(ρ,-Ο) 5ί ' (ρ4-0) + ( ρ 4 - ρ ί ) 5 ί " ( ρ , - 0 ) + -

5,· )]}' Ρ ρ ( = g i * i + 1 ^ 4 5 ί " ( ρ 4 - 0 ) + « ( & - 0 ) +

* i

3 = 1

(4.7.24)

On comparing eq. (4.7.22) with the corresponding equation for a beam of constant rigidity it is seen that in the case of the beam whose rigidity changes in a stepwise manner these appear additional terms, characteristics for the concentrated forces and moments acting at the cross-sections where the rigidity changes. The magnitude of this fictitious "loading" is determined by the function S2(x).

Further comparison of equations (4.7.22) and (3.1.5) shows that the shape function for the beam whose rigidity B(x) consists of n segments of constant but different values has the form

Φι(*)

φ2ω

« Bi-Bi+x/{x-qtf

1_ X3 n-l Bj-Bi+1 Ax-Qif n , BiBi 2!

+

(4.7.25)

The method employed in this section as seen against the previous methods commends itself on account of its generality and its mathematical exactness. It enables us to define the elements of the function S2(x) resulting from the geometrical

134 Theory of beams

discontinuities of the beam, i.e. the terms MAi /(*T3l\l p*< Ax-Qif\l

Bi+1\ 2! /Q; Bi+1\ 3! /ei

as well as the magnitudes of the fictitious moments MQi and forces PQ,. Thanks to this there is no need to formulate separate solutions for each different problem depending upon the relative position of the points of discontinuity in the geometry of the beam ρ1? ρ2> Q& · · · > Qn

a n ^ the discontinuities in the loading of the beam ci9 dp etc., which, of course, makes this method so general.

We shall now demonstrate how the general solution may be employed for solving particular problems.

EXAMPLE 1. Consider the problems solved in section 2 of Chapter III. It can be easily checked that if we put n = 1 in all the relations of this section then the second relation of eq. (4.7.25) reduces to eqs. (3.2.8).

Again from the second part of eq. (4.7.15) we find

In the case when a^ci—cy we get *S"'(a)=0, S"'(a)=0 and hence S2(x)=0. But when c^a

S»(a-0) = ^ - l £ > , S"'(a-0) = ~ , Bx Bx

and therefore by (4.7.24) we obtain _P(a-c)(B1-B2) _ Ρ^Βχ-Β,)

Ma~ B1 ' Γα~ Bx

while from (4.7.23) _ Ρ(ΒΧ-Β2) /{α-ό){χ-ά? t (x-af\i

S^X) ~ BlB2 \ 2! + ~3!-> ·

It is easily seen that the sum S{x) = S1(x) + S2(x) leads to the expression (3.2.11).

In the case when c—a the corresponding expressions can be obtained from either of the two above cases, i.e. either from the case when a < c o r when c < a.

To solve for the boundary values the following relation between the reactions of the supports and the derivatives γψ of eq. (4.7.22) will be useful:

(1) reactive moment at the left end of the beam

M0 = -B^'-M, (4.7.26) (M1 is the applied moment at the cross-section x — 0);

(2) reactive force at the left support Ro = B1y'0" + m(0)-P1 (4.7.27)

(Pt is a concentrated force applied at the cross-section x = 0 and m(0) the intensity of the distributed moment at that section);

(3) reactive moment at the right end of the beams Ml = (Bny"(l)-Mn (4.7.28)

(Mn is the moment applied at the cross-section x = /); (4) reactive force at the right support

Ri = -Bny"'<J)-m(!)-Pn (4.7.29) (Pn being the concentrated force applied at the cross-section x = I and m(J) the intensity of the distributed moment at that point).

EXAMPLE 2. As another example of application of the developed equations we shall solve the boundary problem of the beam shown in Fig. 33 rigidly built-in at both ends. For this case the boundary conditions are

J>o = 0 , y'0 = 0, y(l) = 0, / ( / ) = 0. (4.7.30) Consequently, the first two constants appearing in the equation for the deflection curve (4.7.22) are determined directly

136 Theory of beams

from the conditions of (4.7.30) and the remaining two constants y'0' andjo' 'can t>e determined using the last two conditions of (4.7.30), i.e. from the following system of simultaneous equations :

L £ l ßißi+l i

+yö'\p + *i " f Β^Φ±λ (3/f Qi + If) L t=l W + l

= -65(/),

2yö l+Bl +

+K P + BiY^-JT^VhQi + lt) & -ßi-öt+i

-2S'(Q.

(4.7.31) The solution of these equations yields

n - i / ? . _ D ^ Ι ^ + ^ ι Σ »« l + 1h(2i-id •65(/)

- 6 S ' ^ + £ , " | ; ' ^ ^ i l / ? U , (4.7.32)

where Z> is the principal determinant of eqs. (4.7.31) whose magnitude is

D = ΜΙΒιΣ ^ i ± i / i [ / ( 2 / - 3 / i ) + 2/?] +

+ *? "I V / i + 1 J r / i + 1 A/J[6/(/i-/J)+/j(4/i-3/i)]. (4.7.33)

In eqs. (4.7.32) and (4.7.33) the symbol l{ denotes the difference in lengths, i.e. l{ = l~q{.

Thus the deflection line of a beam fixed at both ends, consisting of n segments of different rigidity and loaded transversely is defined by eqs. (4.7.22), if the polynomial appearing in them does not contain the first two terms with the constants

!K 2-25 2-25

Mn 3-00 2B 3·00 3-00 - * # M,

FIG. 36.

y0 and y'0, while the constants y'0' and y'0" are as given by eqs. (4.7.32).

As a numerical illustration consider the case shown in Fig. 36f with the following data: q = 1 ton/m, n = 3, / = 9 m, Qi = 3 m, ρ2 = 6 m, lx = 6 m, /2 = 3 m.

For this case the second relation of eq. (4.7.15) gives the following load function:

S i (* )=24^<(*-2 -25)*>$ .„ -1

25 2 4 B

1 + 4 8 * « * - 3 ) 4 > 3 - 4 8 *

<(x-3)<>» +

<(*-6)*>8 +

+ 2*5 « χ - 6 ) 4 > β - 2 i 5 «*-6·75)*>».75 . (4.7.34)

t This example is taken from ref. 8.

138 Theory of beams

To obtain this relation it was necessary to solve three integrals, all of the same type, by means of the following relation:

a a a

where a I.25-^-<(x-3)3>»+

+ ^<(*-3)3>!-T 21^<(*-6)3>9

6+

+ -~ {{x-€fy%-^ <(x-6-75)3>t.75,

S'i ( x ) = 2^<(*-2·25)2>2·25-^B<(*-3)3>l+

+ ^ <(^-3)2>39- 4^<(^-6)2>g+ (4.7.35)

+ ^ -<(χ -6) 2 >»-^<(χ -6 ·75) 2 >»

S["(x) = ^<x-2-25yi25-~(x-3y3 +

+ -i-<x-6>»--i-<x-6-75>».75.

From eqs. (4.7.24) we find

6·75 J

9 , . 2 2 5 „ 3 Λ _ ^ 5 " 8 ' (4.7.36)

M = M — - P — —— P = 1 32 ' 2 ~~ 6 4 ' ~~ 4 '

whence the function S2(x) of eq. (4.7.22) is

3 -4£g<(*-3)3>39+4g5 -4-Β((^)3)ΐ+^<(Χ-β?>ι,

(4.7.37) so that the load function S(x) = S1(x)+Sz(x) is determined.

We can now determine the following: (1) from eg. (4.7.33)

(2) from eqs. (4.7.34), (4.7.35) and (4.7.37)

(3) from eqs. (4.7.32) 33 ..,„ 9

Jo - iB, y0 - AB-

Thus all the functions and all the constants of eqs. (4.7.22) have been determined. The deflection line of the beam is then given by , , 1 [33 2 3 3 75 /(x-3)2\°

_ 75 /(x-6)V 3 /(x-3)V 64\~ 2! /„ 4 \ 3 ! A 3 /(x-6f\9 /(χ-2·25)4\9

+ 4 \ 3! / e + \ 4 ! Au

_ /(x-3)^\« 1_ /(x-3)\»_ j. /(x-6)*y \ 4! / , 2 \ 4! Λ 2 \ 4! A

+ x-6)4\9 /(x-6-75)4\J

4' A \ 4! Λ (4.7.38)

140 Theory of beams

The consecutive derivatives of this function will enable us to define respectively the slope of the curve, the bending moments M(x) and the shearing force Q(x) over the whole length of the beam, in accordance with eqs. (4.7.8).

Further, eqs. (4.7.26) and (4.7.27) and the already found values of y^ and y^" will enable us to find the fixing moment and the reaction at the left support. They are M0 = = -4-125 ton-m, R0 = -2.25 tons.

As regards the reactions of the right support, these are determined by eqs. (4.7.28) and (4.7.29); but in this example, because of the symmetry, they are equal to M0 and RQ respectively.

Chapter V

TABLES OF TRANSFORMS

TABLES OF TRANSFORMS: L{f(x)} = J e~'x f(x) dx = F(s)

No.

(5.1.1)

(5.1.2)

(5.1.3)

(5.2.1)

(5.2.2)

(5.2.3)

(5.2.4)

(5.2.5)

(5.2.6)

i5.2.7)

Transform F(s)

snF(s)-sn-1f(+0)-- s M - V ' ( + 0 ) - . . .

-i/<"-2>(+0)-/<»-1>(+0)

m s

1^F^ 1

sn

1 s(s2+a2)

1 s2(s2 + a2)

1 s3(s2 + a2)

1 s(s2-a2)

1 s\s2-a2)

Function/(JC)

/<*>(*) X

I m dt 0

ΐ [-{>>}] xn~l

(rT=Äy.

xn

—r-(l-cos ax) a2

—- (ax — sin ax) a3

cos ax — 1 JC2

ä* + 2Ö2~

— (cosh αχ —1)

sinh OJC x a3 a2

141

142 Theory of beams

Contnd.

No. Transform F(s) Function f(x)

(5.2.8)

(5.2.9)

(5.2.10)

(5.2.11)

(5.2.12)

(5.2.13)

(5.3.1)

(5.3.2)

(5.3.3)

(5.3.4)

(5.3.5)

(5.3.6)

(5.3.7)

(5.3.8)

(5.3.9)

1 s*(s2-a2)

1 4a*+ s*

4a* + s4

■s2

4a* + s*

4a* + s* 1

s(4a* + s*) e~a'F(s) e-as

s e-as

Sn

s(s2 + a2) e-b8

s2(s2 + a2) e-bs

s3(s2 + a2) e-bs

s{s2-a2) e-bs

s2(s2-a2) e-bs

s3(s2-a2)

cosh ax -1 2a2

-—r- (sin ax cosh ax -4a3 v

— cos ax sinh ax)

J_ 2a' J_ 2a

- sin ax sinh ax

(cos ax sinh ax +

-f sin ax cosh a*)

cos «A: cosh ax

1 4a4 (1 — cos ax cosh due)

(f(x-a)ya

r

\ ( l l - l ) ! / a

— <l-cosfl(x-6)>

(Λ(Λ: - b) - sin #(Λ: - />)>*

/ c o s Α(Λ: - fr) - 1 (x - b2)\°° \ a4 + 2a2 / *

^-<COSh « ! ( * - « - 1 ) T

< sinh a (x — b) x — b\°° e» a 2 / *

/ c o s h a(x-b)-l __ (x - b)2\ « \ tf4 2α2 / »

Tables of transforms 143

Contnd.

No.

(5.3.10)

(5.3.11)

(5.3.12)

(5.4.1)

(5.4.2)

(5.4.3)

(5.4.4)

(5.4.5)

(5.4.6)

(5.5.1)

(5.5.1a)

Transform F(s)

e-b» 4a*+s*

se~bs

4 a 4 + 5 4

e-bs s(4a*+s4)

sFê-"8 - sF2{s)e~bi -~f(a + 0)e~us +/(& - 0)e~bt

Fx{s)e-as-F2{s)e-bt

e-OS e-bs

s s e~a$ ae~a* e~bs be~bs

s2 s s2 s 1 e~b8 be~bs

s2 s2 s

s

F&lGê-o'-Gê-1»]

Gê-o'-Gê-1» sn

Function f(x)

-j-jj (sin a(x — b) cosh a(x—b)-

- cos a(x — b) sinh a(x — b))%°

—2 (sin a(x - b) sinh a(x - 6))6°°

- Τ - Ϊ ( 1 - cos <*(*-

-b) cosh a{x-b))^

</'(*)» where fx(x) =f(x + a)

{f(x))i where fx(x) = f(x+a)

/2(*) =/(* + *) <nt

<*x«

<*>Äo

where fi(s) = f(x+a)

]f(x-t){g{t))badt

a

where #i(x) = g(x + a) g2(x) = g(x + l>)

X

^ j j i x - i ) " - 1 ^ ) ) » . * a

144 Theory of beams

Contnd.

No. Transform F(s) Function f(x)

(5.5.2)

(5.5.3)

(5.5.4)

(5.5.5)

(5.5.6)

(5.5.7)

(5.5.8)

(5.5.9)

G1(s)e"a8-G2(s)e-bs

φ2+/?2)

Gi^e-*'- G2(s)e~bs

s\s2+p2)

G 1 ( j ) e " , " - G , ( j ) g - * ' φ 2 -p2)

C i ( j ) g - W - G 8 ( j ) g - » ' Α2(52-/>2)

G t ( s ) g - W - G , ( j ) g - * 4/74 + 54

j [ G 1 ( s ) e - < w - G , ( j ) g - > a ] 4 /7 4 +J 4

F(s)G(s)e-

— G(s)e~at

s11

- i j [1 - cos p(x - 0 ] {g{t))ldt a

x

— f [p(x-t)-s'mp(x-a

-t)](g{t))idt x

\ [[cosh p(x-t)-\\{g(mdt a

x

— J [sinh p(x -1) -p{x -a

-t)]{g{t))ldt x

-— [sin p(x -1) cosh p(x -a

— t)—cos p(x-t) sinh p(x--t)]{g(t))ldt

X

— J [sin p{x - /) sinh p{x -a

-t)](eV))b.dt

lf{x-t)(.g{t-a))?dt a

x

(-^ϊ)ϊ|(*-<)"-1<*(ί-«)>Γ<Λ

Tables of transforms 145

Contnd.

No.

(5.5.10)

(5.5.11)

(5.5.12)

(5.5.13)

(5.5.14)

Transform

F(s)G{s)

i^w) ±L{(gix))r}

XL"1/—L

F(e)

{*(*)}}}

Function f(x)

X

$f(t)g(x-t)dt 0

or X

lf{x-t)g(t)dt 0

a;

$(X-t)g(t)dt 0

X

J(*-o<*w>r*

X t

j(x-t)f{t)dtfg(r)(fr 0 0

X

$(x-t)f(t)dt $(t-T)g(T)dr 0 0

REFERENCES

1. M. A. LAVRENTIEV and B. W. SHABAT, Theory of Functions of Complex Variable (in Russian), Moscow, 1951.

2. R. V. CHURCHILL, Modern Operational Mathematics in Engineering, New York, 1944.

3. W. A. DITKIN and P. I. KUZNETZOV, Handbook of Operational Mathematics (in Russian), Moscow, 1951.

4. W. WIERZBICKI, Structural Mechanics (in Polish), Warsaw, 1949. 5. M. T. HUBER, Engineering Mechanics (in Polish), pts. I, II and III,

Warsaw, 1951. 6. S. NEUMARK, Engineering Mechanics (in Polish), Warsaw, 1948. 7. T. VAN LANGENDONCK, A funcäo «ressalto» e sua aplicagao a prob-

lemas da Estätica das Construcoes, Revista de Engenharia do Rio Grande do Sul, Sept.-Dec. 1954.

8. S. E. MIKELADZE, Some Problems of Structural Mechanics (in Russian), Moscow, 1948.

9. T. IWTNSKI and J. NOWINSKI, The transforms of differential equations of the theory of structures (in Polish; summ.: Engl., Russ.), Archives of Applied Mechanics, 2 (1954).

10. T. IWTNSKI and J. NOWINSKI, Equations of six statical quantities for beams resting on unhomogeneous elastic foundation composed of segments of constant modulus (in Polish; summ.: Engl., Russ.), Archives of Applied Mechanics, 3 (1954).

11. T. IWINSKI, Discontinuous solutions of the equations of structural statics obtained by means of the Laplace transformation and illustrated by example of a beam of sectionally constant rigidity (in Polish; summ.: Engl., Russ.), Archives of Applied Mechanics, 3 (1955).

12. S. DROBOT, Some remarks in application of the operational calculus to the statics of structures (in Polish; summ.: Engl., Russ.), Archives of Applied Mechanics, 1 (1954).

13. F. TÖLKE, Praktische Funktionenlehre, Berlin, 1950. 14. M. M. FILONENKO-BORODICH et al., A Textbook of Strength of

Materials (in Russian), Vol. 2, Moscow, 1949. 15. W. WIERZBICKI, Beams having a horizontal projection in the shape

of a broken line (in Polish;summ.: Engl., Russ.), Engineering Dissertation, 11 (1954).

16. A. CLEBSCH, Theorie der Elastizität Fester Körper, 1862, p. 389. 147

148 References

17. A. FÖPPL, Festigkeitlehre, 5th edition, 1914, p. 124. 18. T. IWINSKI, On the application of the Laplace transformation to

static problems in building (in Polish; summ.: Engl., Russ.), Archives of Applied Mechanics, 1 (1953), pp. 107-64.

19. J. E. GOLDBERG, The application of Heaviside's step-function to beam problems, Proceedings American Society of Civil Engineer's, July 1953.

20. S. TIMOSHENKO, On the treatment of discontinuities in beam deflection problems, Quarterly of Applied Mathematics, 3 (1945), p. 182.

21. C. L. BROWN, The treatment of discontinuities in beam deflection problems, Quarterly of Applied Mathematics, 1 (1943), pp. 349-51.

22. A. J. S. PIPPARD and J. F. BAKER, The Analysis of Engineering Structures, 2nd edition, Arnold, London, 1943, pp. 31-34.

23. R. V. SOUTHWELL, An Introduction to the Theory of Elasticity, Oxford, 1936, sections 194-196.

24. J. CASE, Strength of Materials, 2nd edition, Arnold, London, 1932, p. 169.

25. W. H. MACAULEY, Note on deflection of beams, Messenger of Mathematics, 48 (1919), p. 129.

26. W. T. THOMSON, Laplace Transformation, Prentice-Hall, 1950. 27. T. MIKUSINSKI, Operational Calculus (in Polish) 2nd edition, Warsaw,

1957. (English Translation available.) 28. W. NOWACKI, Theory of Creep (in Polish), Warsaw, 1963. 29. W. NOWACKI, Thermoelasticity, Pergamon Press, London, 1963. 30. E. KOSKO, On the treatment of discontinuities in beam deflection

problems, Quarterly of Applied Mathematics, 2 (1944), pp. 271-2. 31. J. R. CARSON, Electric Circuit Theory and Operation Calculus, New

York, 1926. >

32. T. IWINSKI, Theory of Beams, Morikita Publishing Co., Ltd, Tokyo, 1965.

INDEX

Approximative equation of five moments 112,115

Axial force 95

BAKER, J. F. 2 Beam

on elastic foundation 100,102 on non-homogeneous elastic

foundation 107 on three supports 49 rigidly built-in 26,79 with compressive axial forces

96 with overhangs 36, 53, 87 with tensile axial forces 97 with terminal couples 90 with terminal forces and coup

les 43 Bending moment 15, 48, 60, 72

on support 47,48 BorePs theorem on convolution

73,145 BROWN, C. L. 2

Cantilever beam propped at free end 30, 85 rigidly fixed 34, 86 with overhangs 42, 89

CARSON, J. R. 2 CASE, J. 2 CHURCHILL, R. V. 3,11 CLEBSCH, A. 1 Coefficient of yielding of foun

dation 99 Concentrated load 19

Concentrated moment 22,93 Continuous beam 52

on elastic supports 66

Differential equation of elastic curve 15 ,16 ,71 ,92 ,95 , 98, 124

Distributed bending moment 71, 74

DITKIN, W. A. 2,12 DROBOT, S. 2, 92

Elastic force 15 Elastic supports 61, 64 Equation

of deflection curve 23,53,62, 65 ,73,96,97,99,132

of five moments 69 of six static parameters 112 of three moments 59

FILONENKO-BORODICH M. M. 121

FÖPPL, A. 1 Foundation of Winkler's type 98

GOLDBERG, J. E. 1,2,3

Inverse transformation 12

KOSKO, E. 2 KUZNETZOV, P. I. 2 ,12

149

150 Index

LANGENDONCK, T. VAN 2 Laplace transformation 11, 12

of derivatives 13 of discontinuous derivatives

123 Laplace transform of function

11,12 concentrated load 93 concentrated moment 93 distributed load 17,93 distributed moment 94 elastic curve 18,72,95,98 inversion of 12 load function 95

LAVRENTIEV, M. A. 11 Load function 4, 24, 73, 96, 97 Load per unit length 16

MACAULEY, W. H. 2 Maximum of deflection 47 MIKELADZE, S. E. 2, 78 MIKUSINSKI, J. 2 Modulus

of elasticity of support (in sinking) 64

of rigidity of support (in rotation) 62

NEUMARK, S. 39

NOWACKI, W. 3 NOWINSKI, J. 2, 92

PIPPARD, A. J. S. 2

Reaction at support 47, 49, 75 for continuous beams 55, 61

Rigidity of beam 16,71

SHABAT, B. 11 Shape function 5, 75 Shearing force 16,49, 60, 74 Slope at support 47,49 SOUTHWELL, R. V., 2 Step functions 4, 6, 8

derivative of 11 elements of 6 multi- 7 multiplications of 10 sum of, 8

Tables of transforms 12, 140 THOMSON, W. T. 3 TIMOSHENKO, S. 2

Young's modulus 15

MADE IN GREAT BRITAIN

theory of beams. the application of the laplace transformation method to engineering problems

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