Energy Theorems and Structural Analysis. J.H. Argyris

J.H. Argyris

Energy Theorems and Structural Analysis

EN ER GY THEOREMS AN D STR UC TU RA L ANALYSIS A Generalised Discourse with Applications on Energy Principles of Structural Analysis Including the Effects of Temperature and Non-Linear Stress-Strain Relations.

by J. H. ARGYR IS, D.Sc. (Eng) Professor of Aeronautical Structures, University of London, Imperial College of Science and TechnoiOKY

Co-auth or of Part II

S. KELSEY, B.Sc. (Eng) Lecturer in Aeronautical Structures, Imperial College of Science and Technology

Springer Science+ Business Media, LLC

First published by Butterworth & Co. (Publishers) Ltd.

Originally published in a series of articles in AIRCRAFT ENGINEERING Oct., Nov., 1954; Feb., March, April, May, 1955.

ISBN 978-1-4899-5852-5 ISBN 978-1-4899-5850-1 (eBook) DOI 10.1007/978-1-4899-5850-1

J. H. Argyris 1960. Originally published by Plenum Press in 1960. Softcover reprint of the hardcover 1st edition 1960

PREFACE

T

HE present work was originally published as a series of articles in Aircraft Engineering between October 1954 and May 1955. The purpose of these papers was two-fold. Firstly to generalize and extend but at the same time abo to unify the fundamental energy principles of analysis of elastic structures. Although much of the corresponding theory has been available for a number of years, to the best of the author's knowledge it has not been given before in such generality. As an example, whilst keeping within the small deflection theory the argumeats have been developed ab initio to include non-linear elasticity and arbitrary initial strains e.g. thermal strains. The first assumption introduces naturally the twin concepts of work and complementary work first put forward by Engesser. The author has attempted in this connexion to refer to all relevant and historically important papers. Since the appearance of the present articles, a few papers have been published which touch upon the same subject but suffer, unfortunately, from a rather incomplete list of references. Secondly, the writer developed in considerable detail practical methods of analysis of complex structures-in particular for aeronautical engineering applications. The most important contributions are the matrix methods of analysis. Since they are only cursorily referred to in the Introduction, it may be appropriate here to describe their use and origin in greater detail. The matrix formulation besides providing an elegant and concise expression of the theory of such structures, is ideally suited for modern automatic computation because of the systematic ordering of numerical operation which the matrix calculus affords. The necessary programming for the digital computer is simplified since it can be preprograp1med to carry out matrix operations with only simple orders as to location and size of the matrix concerned and the operation to be performed. The specific programming for a particular problem may therefore be written comparatively quickly and easily and, moreover, follows closely the algebraic analysis. As developed here, the matrix methods of analysis follow from particular forms of the two fundamental energy principles applicable to structures made up as an assembly of discrete elements. The one principle leads to an analysis in terms if displacements as unknowns (displacement method), while the second leads to an analysis in terms of forces (force method). Besides revealing more clearly the duality of the two methods, this derivation shows also the close connexion between the aproximate methods (like the Rayleigh-Ritz method) for continuous systems and the matrix methods for finite assemblies. This is particularly valuable in providing suitable techniques for establishing the basic propertiesstiffness and flexibility-of the individual elements of a complex structure where these elements have to be assigned simplified stress or strain patterns. But in stressing the advantages of a unified approach to these diverse problems, a word of caution is necessary against carrying over into the modern methods too many ideas associated with practical calculations

by the established or classical methods. The ability to tackle successfully problems in which the number of unknowns is measured in hundreds carries with it the necessity of rethinking one's practical approach if maximum advantage is to be gained from modern computational techniques. In the force method of analysis the choice of basic system and of the redundant forces must be governed primarily by the requirements of simplicity and standardization, in orJer to reduce the m:mual preparation of data to a minimum, and reduce the proba!Jility of errors. At the time of publication of the original articles it was intended to reprint them as a single volume and to follow up the Parts I and II, contained here, with further parts dealing specifically with the practical application of the matrix methods. Unfortunately it was not possible, for a number of reasons, to complete this plan and the articles have for some time teen unavailable. Sir.ce there appears to be a persistent interest in them the present reprint has been produced to meet the deficiency. Grateful thanks and acknowledgment are due to the Editor of Aircraft Engineering for permission to reprint the articles in this form. The method of reproduction has not permitted complete rearrangemc:1t of the text into book form, so that the divisions into monthly instalments are still marked by blank spaces. However, errors in the text have been corrected as far as possible, and the pages have been renumbered consecutively to make for easier reference. Grateful tha:1.ks are due to Miss J. A. Bergg for her care and skill in effecting these changes. The author would also like to thank here those correspondents who have written to point out textual errors and misprints. A list of references to further work is also appended. These are all concerned with the matrix methods of analysis whose basic theory is developed here. In particular, Ref. 6 is an expanded and developed form of part of the work which was initially planned for the original series.

1.

2. 3. 4. 5. 6.

FURTHER REFERENCES TO RECE1'11T WORK J. H. Argyris and S. Kelsey. "Structural Analysis by the Matrix Force Method, with applications to Aircraft Wings". Wissenschaft/iche Gese/lschaft fur Luftfahrt, Jahrbuch 1956, p. 78. J. H. Argyris and S. Kelsey. "The Matrix Force Method of Structural Analysis and some new applications". Brit. Aeron. Research Council, R. & M. 3034, February, 1956. J. H. Argyris. "Die Matrizen-Theorie der Statik". lngenieur Archiv, Vol. 25, No.3, p. 174, 1957. J. H. Argyris. "On the Analysis of Complex Elastic Structures". Applied Mechanics Reviews, Vol. I I, No.7, 1958. J. H. Argyris and S. Kelsey. "Note on the Theory of Aircraft Structures". Zeitschrift fur Flugwissenschaften, Vol. 7, No. 3, 1959. J. H. Argyris and S. Kelsey. "The Analysis of Fuselages of Arbitrary Cross-Section and Taper". Aircraft Engineering, Vol. XXXI, No. 361, p. 62; No. 362, p. 101; No. 363, p. 133; No. 364, p. 169; No. 365, p. 192; No. 366, p. 244; No. 367, p. 272; 1959. (To be published in book form by Butterworths Scientific Publications.)

Part I. General Theory By J. H. Argyris

T

1. INTRODUCTION

HE increasing complexity of aircraft structures and the many exact or approximate methods available for their analysis demand an integrated view of the whole subject, not only in order to simplify their applications but also to discover some more general truths and methods. There are also other reasons demanding a more comprehensive discussion of the basic theory. We mention only the increasing attention paid to temperature stresses and the realization of the importance of nonlinear effects. When viewed from all these aspects the idea of presenting a unified analysis appears more than necessary. With this present paper we set out to develop a comprehensive system for the determination of stresses and deformations in elastic structures based on two fundamental energy principles. Although much of the theory given has naturally been known for many years we believe that some of the theorems and the generality of the results are new. The loading systems considered are of an arbitrary nature and include ab initio the effect of temperature or other initial strains. Neither do we restrict ourselves to elastic bodies obeying Hooke's law but take account of purely elastic nonlinear stress-strain laws. This is possibly not of very great importance at present but may have wider applications in the future. No problems of stability will be touched upon in the present series of articles and any other considerations of large-deflexion theory are, in general, omitted. Thus the purpose is to investigate, within the small-deflexion theory, the stresses and deformations in elastic bodies not necessarily obeying a linear stressstrain law and under any load and temperature distribution. Dynamic effects are initially not considered and hence it is assumed for the present that the loads and temperature are of the quasi-static type. When investigating thermal strain effects we ought strictly to base the analysis on thermodynamic considerations. These are, however, only slightly touched upon here. As in all theoretical work, we start by discussing the exact implications and equations derived from the initial assumptions, but we do not restrict ourselves here to this aspect. On the contrary, we pay close attention to approximate methods of analysis based on the physical concepts of work and strain energy. In particular we attempt to give upper and lower bounds to overall properties of the structure such as its stiffness. No attempt is mad!! to estimate the error of stress and deformations at any particular point. This series of papers originally arose12 • 13 from lectures given by the author since 1949-50 at the Imperial College, University of London. Naturally, the scope of the present work has grown beyond the narrower concept of undergraduate teaching, but the basis of the analysis dates back to that time. It is appropriate here to point out that certain of the basic ideas originate with Engesser2 who unfortunately does not seem to have followed them up. We refer, of course, to the two complementary concepts of work and complementary work. If we consider an ordinary load displacement diagram, then, even if we restrict ourselves to small displacements, this may be curvilinear, if the material follows a non-linear stress-strain law. Work is the area between the displacement axis and the curve, while complementary work is that included between the force axis and the curve. Thus, the two areas complement each other in the rectangular area (force) x (displacement) which would be the work if the ultimate force were acting with its full intensity from the beginning of the displacement. Naturally, in the case of a body following Hooke's law, the two complementary areas are equal, but it is still useful for the purpose of analysis to keep them apart. Since writing a previous paper12 on the subject the author has had the opportunity of consulting the most interesting latest book 9 of Stephen Timoshenko. There a reference is made to the work of Westergaard, 11 who indeed has developed further the basic ideas of Engesser, but not on quite such a general basis as here. Since approximate methods figure prominently in this paper reference ought to be made to the work of Prager and Synge. They too set out to develop systematically the determination of upper and lower limits to strain energy, restricting themselves, however, to Hooke's law and excluding temperature effects. Moreover, it appears that although many of their

2

GENERAL REFERENCES (I) Biezeno, C. B., and Gramme!, R. Technische Dynamik, 1st ed., Springer, Berlin, 1939. (2) Engesser, F. Z. Architek u. lng. Verein Hannover, Vol. 35, pp. 733-774, 1899. (3) Lord Rayleigh. Theory of Sound, 2nd ed., Vols. I and II, Macmillan, London, 1892 and 1896. (4) Maxwell, J. C. Phil. Mag., Vol. 27, p. 294, 1864. (5) Mohr, 0. Z. Arch. u. lng. Verein Hannover, 1874, p. 509, and 1875, p. 17. (6) Mueller-Breslau, H. Die neueren Methoden der Festigkeitslehre und der Statik der Baukonstruktilmen, 1st ed., Korner, Leipzig, 1886. 19 ~7. Southwell, R. V. Introduction to the Theory of Elasticity, 2nd ed., Clarendon Press, Oxford,

19~8/

Timoshenko, S., and Goodier, J. N. Theory of Elasticity, 2nd ed., MacGraw-Hill, New York,

(9) Timoshenko, S. History<;{ Strength of Materials, MacGraw-Hill, New York, 1953. (10) Tretftz, E. Handbuch der Physik, Vol. VI, Springer, Berlin, 1928. _(II) Westergaard, H. M. 'On the Methods of Complementary Energy,' Proceedings Amer. Soc. C.v. Engrs., 1941, p. 190. (12) Argyris, J. H. Thermal Stress Analysis and Energy Theorems, A.R.C. 16,489, Dec. 1953. (13) Argyris, J. H., and Kelsey, S. Applications to A.R.C. 16,489, A.R.C. 16,513, Jan. 1954. Additional references are given as footnotes.

results ar~ identical_ with exist~ng _ideas _they clothed them in a language not too fam1har to engmeers. Th1s d1scuss1on of past authors' work brings us to a few points which are preferably stated now. In much of present day structur~l analy~is there seems to be an unfortunate tendency to overemphasize certam methods of analysing redundant structures and to negl~-;t more useful ideas readily available for many years. This refers particularly to Castigliano's principles which are so often set out as the basis of all ·considerations, not only in theory, but also in the actual methods of calculation. This is, in our opinion, unfortunate, even though all methods naturally lead to the same results if based on the same assumptions. For example, if we select forces as redundancies then much the best means of obtaining the basic equations for their determination is the long established O;k method of Mueller-Breslau based on the Unit L
Firstly the principle of virtual displacements may always be used to derive, for any particular structural problem, the governing differential equations and the appropriate static boundary conditions in terms of the displacements. This method, however, is not recommended in general as a substitute for the derivation from consideration of equilibrium and elastic compatibility. Next the principle of virtual work is used to derive Castigliano's theorem Part I, generalized for thermal effects. As is well known, this principle applies not only for non-linear stress-strain laws but also for large displacements. Our line of argument leads us then naturally to the principle of minimum strain energy for a fixed set of displacements and a given temperature distribution. This theorem applies also for non-linear stressstrain laws and is of great interest for approximate calculations in terms of assumed forms of displacements. It shows us that, while the strain energy is for a given set of displacements a minimum when the compatible state is also one of ·equilibrium, it is on the other hand a maximum for a given set of forces under the same conditions. These theorems were first developed for linearly elastic bodies by Lord Rayleigh more than seventyfive years ago. They are shown to apply also in the presence of thermal strain and for non-linear elasticity. In the remainder of the chapter we investigate in more detail approximate methods of analysis using the Rayleigh-Ritz procedure and it is in such applications that the principle of virtual displacements shows its greatest power. The particular form of the Rayleigh-Ritz procedure known as the Galerkin method is also discussed. It is of importance when the assumed deformations satisfy all boundary conditions. The methods indicated apply again in the presence of thermal strains and non-linear stress-strain laws. The next, Section 5, gives simple illustrations to the method of virtual displacements. The second fundamental principle is developed in Section 6. We call it the principle of virtual forces or complementary virtual work. Here we consider a state of equilibrium, apply a statically consistent and infinitely small virtual force and stress system and find, by using the idea of complementary work, the second principle. This is a necessary and sufficient condition that the position of equilibrium is also one of elastic compatibility. Again this theorem may be used to derive the differential ¢quations of any particular problem, this time in terms of stresses or stress resultants. However, our comments on the parallel method in the case of the virtual displacements are equally applicable here. It should never be used as a substitute for more physical and geometric reasoning. Next, we derive what is essentially a generalization of Castigliano's Part II theorem. Contrary to what is generally believed this theorem does apply for non-linear stress-strain laws as long as we replace strain energy by complementary strain energy, which is defined in the same way as complementary work. It is extended to include temperature effects. We proceed then with the generalization of Castigliano's principle of minimum strain energy (or least work) for non-linear stress-strain laws and thermal strains. Some interesting developments derive from this and are given in the form of maximum and minimum theorems complementary to those developed under the virtual displacement method. They do not seem to have been given previously in this form and provide a useful background to approximate methods. They show us that any assumed statically equivalent stress distribution must always under-estimate the stiffness. This is most valuable for practical purposes and is exactly opposite to the effect of assumed displacement distributions which always overestimate the stiffness. The two in conjunction give us hence lower and upper bounds to overall characteristics of the structure such as its stiffness. In this section we discuss also the Unit Load Method which, as mentioned previously, provides the basis for one of the more convenient methods for the calculation of displacements and of redundant forces. It is shown to be applicable to structures with non-linear stress-strain laws. Section 7 presents some simple illustrations of the principle of virtual forces. In the last section we develop a slightly more generalized version of the D;k method of Mueller-Breslau. These equations lend themselves readily to presentation in matrix form. Next we obtain the corresponding equations when displacements and not forces are introduced as the unknowns. A Note on the Mathematics The mathematics used in this paper is, in general, elementary and should be familiar to any university graduate. We have avoided the more formal application of the .calculus of variations which can be singularly unattractive to those more physically inspired. Chapter 3 and parts of Chapters 4 and 6 may prove, at first, rather difficult for a student. However, it is always possible to gain an understanding of the basic ide:1s by substituting simple examples (e.g. frameworks) for the necessarily more general proofs given here. The later parts of this series of papers will present a number of applic:ltions of the basic methods developed here.

2. BASIC EQUATIONS AND NOTATION body forces (e.g. gravity forces) per unif1 p II 1 volume ara ~ to a Cartesian c/>z, c/>z s.urface forces per unit surface co-ordinat-:: system direction cosines of external normal to ] Ox, Oy, Oz I, m, n surface (see FIGS. I and 2) displacements u, v, w

cp.,

~/_;_1____,..../

)::_wx

Wz ""'.,.. / L - - - - - - -

Body Forces

Stresses

Fig. !.·-Stresses and body forces

z Surface Forces

Stresses

Fig. 2.-Stresses and surface forces

a""' a •• , a.. direct stresses axu=a.z, a •• =a •• , a.,=a,, shear stresses vu Yxx=;)X'

vv Y••=S)i'

y.,.=~ +~;,

y••

g=y.,.,+y•• +y.. Y)xx• YJ••• Y)zz Y),., Y)., Y)zx Exx=y,.,~Y)xx Exu=y,.~Y),.

etc. etc.

e =Exx+Euu+Ezz dV=dxdydz dS a

0 E G

vw Yzz=vz

}As shown in FIGS. I and 2

total direct strains

=~ +~;. y.,=~: +~

l

total shear strains (

initial direct strains (e.g. thermal strains) initial shear strains elastic direct strains elastic shear strains

(I)

J

}

(2)

element of volume element of surface linear coefficient of thermal expansion (may vary with 0) rise of temperature Young's modulus } shear modulus May vary with 0 Poisson's ratio

v ay=axxYxx+a •• y •• +azzYzz+axu/'xu+a•• y •• +azx/'zx ................ (3) The corresponding explicit expressions for UYJ and aE are obtained by substituting the strains Y)xx etc. and Exx etc. respectively for Yxx etc. W work of external forces U,= ~ W+const. potential (energy) of external forces strain energy (or potential energy of elastic deformation) W*, U, *, Ui * complementary work, complement:uy potential of external .forces and complemenhry potential energy of elastic deformation Ud* complementuy pot-:::-~ti:ll energy of totli deformation From a consideration of equilibrium on an ele:nent dV=dxdydz, illustrated for the x-direction in FIG. 3 OCI,,+oa., . Oazx w =O OX 0Y -f' 0Z + X

I

(4)

3

where

o"'I/J X'I/J ()''f L1·1· ~~+~-,,+~and/, m, n are the direction cosines on the surface. 'I' oX" o)'" oZ"

This theorem can be proved by integration by parts. Examples of application: take cp=am Then

or/; or/; orf; sx=ou, ~y~Sl =0

·

J

J

Jo;~, ou·dV -- -- aa· 8yu·dV + lau· ou·dS

v

where to ou.

From a consideration of equilibrium on the surface (see

. .

FIG.

nazz+ laxz+ma.z=cfz Over part of the surface the boundary conditions may be expressed in terms of stresses or forces (static boundary _condi~ions) ~nd over the r~­ mainder in terms of displacements or strams (kmemattc or geometnc boundary conditions). Naturally, the boundary conditions may be of both types over the sat:I?-e part of the surf~ce: Consider, for example, the tube shown in FIG. 4. It ts assumed fully bmlt mat the roo_t (z=O) and _free at the tip (z=/). Ribs rigid in their own plane but freely flextble to defle_x~ons out of their plane are assumed at z =0 and z =I. The boundary condttlons are: at z=O, u=v=w=O, i.e. pure kinematic conditions; at z=l, a•• =O,

(~u) =0 uX zd

for the horizontal

walls, i.e. both static and kinematic conditions. To denote infinitesimal elements of geometric properties of the structure (e.g. co-ordinates, area, volume) we use the standard symbol. d. To denote infinitesimal increments of forces, stresses, dtsplacements, strains and work we use the symbol o. . . . Thus, dV=infinitesimal element of volume=dxdydz, oP mfimtestmal increment of force P. The symbols

J( ..... .)dV, J(...... )dS

v s denote integrations over a volume and surface respectively. The formal mathematical proof of some of the basic theorems in this paper is shortened by using Green's theor~m. * ~t and if! be two continuous functions and let also the first partial denvattves of(/> and the first and second partial derivatives of rf; be also continuous. Green's theorem states:

f

J[oo/ ox

v

is the increment of the strain

y,, corresponding

or/; or/; 81/J cp=a,., ox=ov, oy=ou, oz =0

y

v

(5)

for the vertical walls and

s

J[();~"ov+();,•ou]dV=- Ja,.oy£.dV+ Jax.[ mou+lov]ds

2)

fa,x+mavx+ nazx=cfx

( ov) oy z~! =0

8y,,~c8~=~8u

Stmtlarly take Then

Fig. ).-Internal equilibrium conditions

v

or/;+ oo/ or/;+ oo/ oi/J]dv ox oy oy oz oz

v

s

where

" --o[()u o v] _(lou ()ov oy,.- ()y +ox -- oy +ox

Note that although Green's theorem is helpful for the mathematical understanding of the present theory it is not really necessary for the physical understanding; a reader unfamiliar with these aspects of the integral calculus may omit the relevant parts. 3. WORK AND COMPLEMENTARY WORK-STRAIN ENERGY AND COMPLEMENTARY STRAIN ENERGY The analysis of the present paper is restricted to small strains which can be expressed by the linear formulae given in the notation. Such displacements and corresponding strains are obviously additive (algebraically). Thus, if u 1 , y 1 , and u 2, y 2 are displacements and strains in a deformed state 1 and 2 respectively, then u 1 + u 2, y 1 +y 2 represent also a compatible state of deformation of the body. Our assumption does not impose, however, a linear stress-strain relationship; hence if P 1 , a 1 and P 2 , a 2 are the forces and stresses corresponding to the above two states of deformation of the body, the forces and stresses corresponding to the deformed state u 1 +u 2 are not P1 +P2 and a 1 +a 2 except in the case of a linearly elastic body. In all cases, however, the stress-strain law is assumed to increase monotonically as shown in FIG. 5. In conclusion we can state that the law of superposition is assumed to hold for strains and displacements but not necessarily for the stresses. In general, we assume also that the displacements are so small that the equilibrium conditions can be written down for the undeformed body. It follows then that the question of stability or instability of equilibrium does not enter in the analysis of this paper and there is a unique solution to every problem. Consider a three-dimensional deformable body (not necessarily elastic) in equilibrium subjected to a self-equilibrating system of body forces Wx etc., surface forces cfx etc. and a temperature 0. These forces and temperature may vary with time but the variations are assumed so slow that the dynamic effects are negligible. Let, in a time interval ot, the forces increase by Owx, Ocfx etc. and the temperature by 80. The displacements increase at the same time by ou etc. There arises hence an increment of work (see FIG. 6)

oW= J[wxou +w.ov+w.ow)dV v

• See Courant, Differential and Integral Calculus, translated by J. E. McShane, Blackie and Son Ltd., London and Glasgow, 1949, Vol. II.

+sHo/xou+cp.ov+cp.ow]dS + terms of higher order.

Y,V

0

Rigid rib Fig. 4.-Kinematic and static boundary conditions

4

E Fig. 5.-Stress-strain diagram

Fig. 6.-Work and complementary work; strain energy and complementary strain energy

t

and that it is subjected both to external loads and thermal effects. In view of our initial assumption about the smallness of the strains we can write 0Yxx=O€xx+01]

Oyy

lhdy

_EY________________ ,

(12)

-

I

I

Oxx

0Yzz=O€zz+01] where OExx .... , 811 are the increments of the true elastic strains and the thermal strains e (13) 'Y/xx=1]yy=1]zz=1] = faoe

I~ I

0

e.

is the linear coefficient of thermal expansion which may vary with In view of the local isotropy of the body, thermal expansion does not give rise to any angular displacements. Hence Oyxy=O€xy, 0Yuz=O€yz. 0Yzx=O€zx • · · · · · · · · · .. · · • •. • • • • ... • • (14) Substituting Eqs. (13) and (14) into (ll) we obtain oW=oU;+s811 . . . .. . . . . . .. . . . .. . . .. . .. . .. .. . .. .. . .. .. . . . (15) and

a

Fig. 7.-Work of direct and shear stresses

The terms of higher order involve expressions -!-Swx · Su, -!-Sc/J, · Su etc, and may be neglected to the first order of magnitude considered here. Thus.

I

J ...................... (

8W=f£wx8u+w.8v+w.8w]dV

+~[c/Jx8u+c/J.8v+c/Jz8w]dS

7)

s

Note that Eq. (7) does not presume any specific force-displacement law, be it elastic or non-elastic. It is simple to derive an alternative expression by considering the additive effect of the work done by the stress resultants on each volume element dV. A perusal of FIG. (7) shows that the deformation Su, Sv, ow gives rise to an increment of work for an element dV (axxDYxx+a •• Sy •• +a •• Sy •• +axuDYxu+a•• Sy •• +a.xSy.,)dV=aoy·dV again neglecting terms of higher order. The incremental (infinitesimal) strains Sy,x etc. are those due to the displacements Su etc. Thus,

8Yxx =o~!! etc ., 8Yx• =o{~ ~} ~X ~Y +~X

={~Su ~Sv} ~Y +~X

(8)

· · · · · ·'' '·

It follows that the increment of work 8 W may also be expressed as ·sw=J(aoy)dV .. . .. . .. .. .. .. . . .. .. .. .. .. .. .. .. .. .. .. .. .. (9) v The formal equivalence of Eqs. (7) and (9) may be proved without difficulty by using Green's theorem. To this effect multiply each of the internal equilibrium conditions (4) by Su·dV, Sv·dV, Sw·dV respectively, and integrate over the body. We obtain,

J{[~ax"+ ~a.x+ ~a."'+w"']ou+[~a." +~a•• + ~ax• +w• ]ov ~x

v

~Y

~z

~Y

+ [~;;· + ~;;• +~:;· +w.

~z

JSw} dV

~x

f[aoy]dV= f£wx8u+w.8v+w.8w]dV + f£c/Jx8u+c/J.8v+c/J.8w]dS=8 W v

s

where the Sy's satisfy Eqs. (8) and are hence compatible strains. Note that where c/J, is unknown but the displacement (say u) fixed, the corresponding Su is zero and hence the relevant terms in the last relation vanish. Integrating Eq. (9) from the initial unstressed state 0 to a final state I we find the total work I

W= f { f[w,8u+w.8v+w.8w]dV+ f£c/Jx8u+c/J.8v+c/Jz8w]dS}= fWdV 0

v

s v ..................................

I

W= f {a,,Sy,,+a •• Sy•• +a•• Sy •• +axuDYxu+a•• Sy•• +azxOYzx} . . 0

where

D;=fao€

(17)

0

and aOE=axxDExx+ ........ +axuDExu+ ... +azxDEzx=OU; . . . . . . (l7a) Note that the second term in (16) and thus also the work depend on the sequence of application of loads and temperature. We assume now that the elastic properties of the material depend only on the instantaneous temperature but not on the previous history of deformation, e.g. if adiabatic or isothermal; the error involved in this assumption is indeed negligible. fhus, the expression U; is only a single valued function of the instantaneous state of elastic straining. In a closed cycle of deformation 0; is zero and SD;=dU; is a total differential. The function U; is commonly called the strain energy per unit volume; other names are strain energy function or density of strain energy. It follows from Eq. (17a) that

e

"" oUi=axxDEu+ · · · •· Hence ~Di

"

~Di"

+azxD€zx=~O€xx+ UE'xx

••••·

~Di"

+~O€zx UE'zx

~Di

(18)

Uxx=()Exx • ••• , ., Uxy=(>E'xy

We conclude that the stress-strain law is uniquely determined by the strainenergy function and vice versa. Note that the law of elasticity is arbitrary in Eqs. (12) to (18). Integrating Eq. (16) over the body we find, e fWdV= fUidV+ f [ fso'Y/]dV v

v

0

or (~

W=U;+J [ fs01]]dV

(19)

v 0

where

U;= fU;dV= f [ faOE]dV (19a) v v 0 For e =0 Eq. (19) reduces to the well-known equality W=U; . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (20) U; is the strain energy or the internal elastic potential (or potential energy of elastic deformation) of the body. If we now assume that the law of elasticity is a linear one then (21)

(10)

where -

(16)

0

v

=0

Applying now Green's theorem to the terms involving the stresses, as shown in the previous section, we find, using the surface equilibrium conditions, Eqs. (5), v

W=D;+js811

(I I)

Note that, in general, the work W done to reach a state I starting from a state 0 depends on the path chosen due to say plasticity, viscous effects, etc. In such cases 8 W is not the total differential* of the right-hand side of Eq. (10). In what follows we assume that the body is fully elastic and isotropict • We say that 6W~dWis a total differential of Wif fcdW~ 0, where the integration is taken around a closed curve; if this applies W is obviously independent of the path of deformation taken between states 0 and I. t The isotropy need only be assumed at each point; the properties of the body may vary from point to point.

and similar expressions for Euu and Ezz· Also a,.=Gy,. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Expressing (21) in terms of stresses,

(22)

v2vg]- 1:1v ..........

(23)

axx=2G[Exx+ 1 ~ 2 ve]=2G[rxx+ 1

and two similar expressions for a .•.• and a zz· The moduli E and G and the ratio v may vary with e. Substitution of the stresses given by Eqs. (22) and (23) into Eq. (15) yields

8w=8D;+f~8iv........................................

(24)

and if temperature ami loads are increased together from zero

5

-

-

EeT}

W=U;+2(l-2v)........................................

(25)

where { I _-vVe 2 -2[E,,Eyy+EwEzz+E"zzE",,- -;r(E"xy 1 · 2 2 2 } U;=G +~: •• +E"zx )] 1 2

(26)

is the strain energy as a function of the elastic strains. The following relations hold only for linear elasticity,

oO;

OUxx =E",,=y,,-TJ

oO; .. .... , oa,. =£,•......

(27)

For two-dimensional stress distributions substitute the factor 1/(1-v) for l/(l-2v) in Eqs. (24), (25) and (26). The corresponding strain energy function 0; is,

0;=G[(I~v)(~:,,+~:,.) 2 -2~:,,~:•• +t~:,. 2 ]

.•••••...••.••••••

(26a)

Parallel to the conceptions of work and strain energy two further ideas essentially due to Engesser 2 , are of particular importance to our investigations. Consider to that effect a one-dimensional force-displacement and a stress-elastic strain diagram (FIG. 6). The vertically shaded areas are obviously those of work and strain energy respectively. lt is natural to inquire if the horizontally shaded areas complementing the previous areas in the rectangular areas Pu and a~: respectively are of any importance (see FIG. 6). In fact, as is shown farther on, the introduction of these new conceptions is proved a particularly happy one when generalizing some theorems, currently assumed to be valid only for linear elasticity, to bodies with non-linear stress-strain relations. Although the complementary areas are equal for linear elasticity it is still useful in such cases to differentiate between them. It is interesting to note that in thermodynamics two similar complementary functions are used : the free energy function H of von Helmholtz and the function G of Gibbs.* In what follows we call the horizontally shaded areas complementary work and complementary strain energy and denote them by w• and U;* respectively. We generalize next our new conceptions by considering the threedimensional case. Let the actual displacements in a body subjected to body forces w, surface forces cp and temperature 0 be u, v, w. The increment of i.he complementary work as these displacements increase to u + 8u, v+8v, w+ow due to load increments 81/J, .... , 8w, . .... etc. is given by S W* = f[uSw, + v8w. + wSw.]dV + f[uoc/J, + voc/J. + wSI/J.JdS v s +terms of higher order or

8 w• = f[u8w, + v8w. + wSw.]dV + f[u81/J, + voc/J. + wScp.]dS v

s

(28)

since terms of higher order like tor/J,· Su can be neglected. It is simple, as in the case of work, to derive an alternative expression to Eq. (28) in terms of stresses and total strains. To find it, note that the increments Sa,, etc. of the stresses must be in equilibrium with the corresponding increments of the body forces 8w, . . . and surface forces 81/J, ..... There are thus six relations of the type of Eqs. (4) and (5). We write here only the two for equilibrium in the x-direction o(8u,,) o(Sa.,) o(Sa.,) "'w _ 0 ox + oy + oz +o .,-

........................

(29)

and loa,,+moa.,+nou.,=oc/J, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (30) Multiplying now each of the first set of equations by the displacements u, v, w respectively, summing and integrating over the body we obtain by applying Green's Theorem similarly to when we derived Eq. (9), and using Eq. (30)

f[u8w, +vow.+ wSw.]dV + f[uoc/J, + voc/J. + w8cp.]dS

v

s

= f[y,,Sau +y•• Sa•• +y•• Sa •• +y,.Sa,. +y•• Sa•• +y.,oa.,]dV v

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (31) where y,, etc. are the total strains. Note that where the forces, for example, 1/J,, have fixed values or/J,=O. Thus ultimately

Sw• = fyoadV . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . where

v

(32)

yoa=y,,Sa,x+ ......... +y, 11 8a,,,+ . . . . . . . . . . . . . . . . . . . (32a) Integrating Eq. (32) between the initial unstressed state 0 and a final state I we find the total complementary work w• = vf W*dV . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (33) where • The sign of the latter function is taken as that fixed by the Committee of the International Union ol Physics.

6

-

I

W*=fyoa

(34)

0

Note that as in the case of work W the complementary work depends, in general, on the chosen path of stressing between 0 and /. We assume now the body is fully elastic and isotropic as described previously and find from Eq. (34) 8W*=o0;*+TJos........................................ (35) and -

-

I

W*=U;*+ST}os 0

· · · · · · · · · · · · · · · · · · · · . · · . . . . . . . . . . . . . . . . . (35a)

where

O;*=f~:ou

(36)

0

and ~:ou=~:,,Sa,,+ ....... +~:,.8u,.+ ..... =o0;* Note that the second term in (35a) depends on the sequence of application of loads and temperature. Under the same assumptions as for the internal potential energy, U;* depends now only on the instantaneous state of stress. In a closed cycle of deformation U;* is zero and 8U;* is a total differential of the right-hand side of Eq. (36). Hence oO;• oO;* (37) £,,= oa,, ........... , ~:,.= oa,• ...........

(see also Eqs. (18)). Note that the law of elasticity is arbitrary in Eqs. (37). If 0 1* is given in terms of the stresses Eqs. (37) show that this determines uniquely the strain-stress laws and vice versa. It is natural hence to call 0; * the complementary strain energy function. Integrating over the body we find I

(38)

W*=U;*+f [ fTJos]dV v

0

where

U;*=fO;*dV=f [f~oa]dV ............................ (38a) v v 0 where U;* is called the complementary strain energy or complementary elastic potential energy. When 0 =0 then W*=U;* . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (39) In the case of linear elasticity W* = Wand U; • = U;, but it is useful to differentiate still between them. For the linear stress-strain laws ofEqs. (21) and (22), Eq. (35a) becomes, if temperature and loads are increased together from zero, w*=U;*+t

.......................................... (386)

and

-. _ _!_[L

U; - 4G l+v- 2
- * =4G 1 [(u,,+a U; I+v•• )2 - 2(a,,u•• -a,.

2)]

(40)

....... ······· ..... . (40a)

The above considerations on strain energy and complementary strain energy may be used to derive all Castigliano theorems as generalized for non-linear stress-strain relations. We postpone, however, these investigations to subsequent chapters. When the body obeys linear stress-strain relations the principle of superposition applies also to forces and corresponding stresses. Some important theorems derive from this property. Thus, if the forces on a body are increased from zero to their final value then W= W*=tf[w,u+w.v+w.w]dV+tf(c/J,u+c/J.v+c/J.w]dS I v s (41) U=U*=tf(a~:)dV

v

J

where all symbols in the brackets refer to final values. Eqs. (41) are known as Clapeyron's theorem. Another very useful theorem is due to Betti. • Assume that the body is subjected to two force systems P 1 and P 2 • Let the deflexions due to system P 1 alone be u 1 and due to P 2 alone u 2 • By applying first system P 1 and subsequently P 2 a~d _then reversing the process we prove easily-noting that the final state IS m each case the same-that W 12 = W21 • • • • • • • • • • . • • • • • • • • . • . • • • • • • • • • • • • • • • • • • • • • • • • (42) where W12 =~P 1 u 2 and W2 =~P 2 u 1 • • • • • • • • • • • • • • • • • • • • • • • • • • • • (42a) • See Num'o Cimento (2), Vols. 7, 8, 1872.

are the work done by the system of forces P 1(P2) over the displacements u2(u1) respectively. Relation (42) is known as the generalized reciprocal theorem of Betti. A special form of Eq. (42) is Maxwell's reciprocal theorem. Thus, if systems l and 2 consist each of one force (or moment) only then l. u12 =1. u21

•••. . •••. •••. . ••. •. . •. . . . . . . . . . . . . . . . •. •. •

(43)

where u12(u 21 ) is the displacement or rotation in the direction of force or moment P 1(P2) due to force or moment P 2(P1 ). 4. THE PRINCIPLE OF VIRTUAL DISPLACEMENTS OR VIRTUAL WORK We assumed in the previous section when discussing work and strain energy that the displacements Su, Sv, Sw arise from an actual variation of the applied forces and/or temperature distribution. However, this is an unnecessary restriction. We need only remember that to the first order of magnitude considered Sw and SU; are independent of the SP's and corresponding Sa's since we ignore terms of the order SP· Su. Also for the purpose in hand no variation in the temperature is called for. Hence, when finding Sw and SU; we can assume that forces, stresses and temperature remain constant while the displacements are varied to u+Su, v+Sv, w+Sw. It is only necessary that the Su, Sv, Sw's are compatible infinitesimal displacements (see Section 3, p. 4); thus they must be piecewise continuous in the interior and satisfy the kinematic boundary conditions. For example, if the u, v, ware prescribed on part of the boundary then the selected variations Su, Sv, Sw must be zero there too. Similar arguments apply if the derivatives of any of the displacements are fixed. However, where the forces and stresses are prescribed the variation of the 8u's is necessarily free. Note that there are cases when it is useful to relax even the kinematic boundary conditions when selecting the Su's. Such geometrically possible infinitesimal displacements are used extensively in rigid body mechanics and are called virtual displacements. Noting that the temperature and hence thermal strains remain constant we can restate now Eqs. (15), (19) and (19a) more generally as follows: An elastic body is in equilibrium under a given system of loads and temperature distribution if for any virtual displacements Su, Sv, Sw from a compatible state of deformation considered

8W=SU; . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

(44)

which is the standard principle of virtual displacements or virtual work. As stated here it is valid for an elastic body subjected both to loads and temperature effects. Note that Eq. (44) is also valid for large displacements but the formulation of U; is in such cases more complicated since the strain expressions are not any longer linear in the displacements and the equilibrium conditions have to be considered on the deformed element. The point made above that the virtual displacements are arbitrary as long as they are infinitesimal and satisfy the internal compatibility conditions and kinematic boundary conditions is worth emphasizing. To fix ideas consider the statically determinate framework shown in FIG. ts) subjected to a transverse force P. We apply to the system a virtual displacement Su in the form shown in the figure which allows only an elastic deformation of the upper flange 1, 2. This virtual displacement satisfies the kinematic boundary conditions

8u7"0 at A and B but bears obviously no relation to the actual displacements of the framework due to the force P. We find easily:

.

force m member l, 2: N 12 =

Pab

-7112

virtual displacement of force P: Su 1 =Su·afb 1

Suh Suh

lh

virtual elongation of member I, 2: !i/12 = - b - b =-SuiT 1

2

1 2

-,--~~~~~~~--~--~--h

t

Thus,

SW=PSu~ and SU;=( _f'7: 2)x( -t~~)=PSu~

Hence

Sw=SU; as indicated by the principle of virtual displacements or virtual work. In order to realize best some of the implications of the principle of virtual displacements let us consider again the derivation of Eq. (44) which applies when S7J =0. In accordance with the analysis of p. 5, if we multiply the internal equilibrium Eqs. (4) with the virtual displacements Su, Sv, Sw, sum the three expressions, integrate over the body, apply Green's theorem Eq. (6), and note the boundary conditions (5) we obtain Eq. (44). Again if we start from Eq. (44) we can apply Green's theorem in the opposite direction and are led to

J{[ v

()a,, ()a• ., ()a •., Jc;, i)x +~+~+w., ou+[ .....

=

c;,

c;, }

.]ov+[..... . ]ow dV

J{ [la,.,+ma•.,+na•.,-4>.,JSu+[..... .]8v+[..... .]Sw}ds s

For arbitrary virtual displacements Su, Sv, Sw this relation can only be true if each of the brackets vanishes separately; thus, starting from the principle of virtual work we have re-established the conditions of equilibrium. An exception occurs, of course, where a displacement (say u) is fixed on the surface and Su is automatically zero there. We conclude that the principle of virtual displacements Eq. (44) is a necessary and sufficient condition for the existence of equilibrium of an elastic body. Or otherwise we can state that by using virtual, i.e. kinematically possible compatible displacements, we substitute Eq. (44) for the internal and external equilibrium conditions. Note that the idea of strain energy is not necessary to the establishment of the principle of virtual work. Since the forces are assumed to be applied on the undistorted system and to remain constant during the virtual displacements we can regard S Was the variation of a potential - U,. Thus

Sw= -Su, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

(45)

and

U,= -.f[w.,u+w.v+w.w]dV-H4>,u+4>.v+4>.wJdS v

s

(46)

Thus ()U,

w.,=-7iil which is the usual definition of a potential of forces. Note that Su. is a total differential of the elastic displacement increments Su. U, is denoted as potential of external forces. We can write now Eq. (44) in the concise form S,U=O . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (47) where the suffix E indicates that only elastic strains and displacements are varied. U=U;+U.+const. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (47a) is the total potential energy of the system. Eq. (47) states that a position of elastic compatibility of an elastic body is also one of equilibrium (i.e. the body is at the true position of equilibrium) if any virtual variation of the displacements and strains whilst forces, stresses and temperature remain constant does not give rise to any (first order) variation of the total potential energy. The particular form (47) is known as the principle of a stationary value of total potential energy, if the latter is expressed in terms of displacements. Note that U; itself may be calculated from formula (l9a). Actually the stationary value of U is in our case always a minimum and this confirms our previous assertion that with the assumptions of our analysis all systems are stable. The mathematical proof that U is a minimum at the true position of equilibrium is straightforward •; the point is discussed in greater detail under (C) below. We have assumed until now that the initial strains 7J are due to a temperature variation. However, this is an unnecessary restriction and there may be strains arising from any source of self-straining. For example, in a framework they may be due to manufacturing errors in the lengths of the bars. In the more general cases of self-straining not only may the "YJ~x• "YJ•• and 7]zz be different, but there may also arise initial shear strains 7].,., 7Jvz and 7]zxo In such problems substitute

a87] =a.,.,S7]xx+a•• 87]vu+a ..S7]zz +a.,.S7].,. +a.,S"Y)vz +a,.,S7]zx . . . .

(48)

for sS7J in Eqs. (15) and (19) and in the other related expressions. Equations (44) or (47) may be used to derive the results which follow. (A) The differential equations ofthe theory of elasticity for arbitrary loading and temperature distribution or any particular structural problem in terms of the displacements; the appropriate static boundary conditions Fig. a.-Example of an arbitrary virtual displacement B

• See Biezeno and Gramme! (I), p. 74.

7

in terms of the displacements follow also from this analysis. It is important to note that in all applications it is best to form directly SU1 = f[ailE]a v v and not to evaluate first U1 and then to take its increment S. (B) Castigliano's* theorem Part I generalized for thermal effects

[~~~ 1~=

(49) const. =P, · · · · · · · · · · · · · · · · • · • • · · · · • • • • · • • • • • • • · • where P, is the force (moment) applied in the direotion of the deftexion (~otation) u,_. TJ:lls relation may be obtained immediately if we apply a virtual elastic displacement Su, solely to one external load P,. Note that Eq. (49) applies also for non-linear stress-strain laws and may also be generalized for large displacements. (C) The Principle of Minimum Strain Energy when U1 is expressed in terms of the displacements and the temperature is not varied. We arrive immediately at this theorem if we select only such virtual displacements Su which are zero at the applied forces. Then SW=O and we conclude from Eq. (44) that S.U1 =0 and U;=min. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (44a) at true position of equilibrium if only such virtual deformations are allowed that no external work is done. Hence, if we compare all possible compatible states of deformation of a body associated with a given set of displacements (not sufficient by themselves to fix completely the deformed shape of the body) then the true position of equilibrium has the minimum strain energy. This is still true if the body is subjected to temperature loading. The point is of sufficient interest to warrant some elaboration. First it may be helpful to point out that when we state that at the position of equilibrium the total potential energy has a stationary value and that this is a minimum we do not compare physically possible adjoining states. For, in stating that the potential energy has a stationary value, i.e. SU=O, we compare the true position u, with a position u+Su assuming in both cases that forces and stresses are the same. This can obviously not be true for the second position since for given forces there is a unique position of equilibrium. In fact, we mentioned that this arises due to our legitimate neglect of the higher order terms in Sa and oP. Also, when we go a step farther and state that the stationary value is a minimum we prove this by considering the influence of terms like !Sa,.,· S~:xx in U, arising from the variation Sa.,., associated with S~:"'"'' but we still keep the forces constantalthough this cannot, in general, be true. Having pointed out these aspects of the virtual displacements approach we shall, in what follows, discuss the question of the extremum of U1 from a more physical point of view. Again we prescribe certain displacements on the body and do not allow any forces P other than those arising due to and in the direction of the given displacements. The structure takes up its natural position of equilibrium from which we can deduce the value of the forces P. If we want to force the body to assume a position u+ou, v+ov, w+ow while keeping the set of prescribed displacements constant we must apply certain additional body and surface forces to push the system away from its natural configuration. The work done by these constraint forces (!'~.;oP· ou, obviously positive), produces by reason of equilibrium in the new position an equal increase in the strain energy stored. Thus, the strain energy in any neighbouring compatible configuration is greater than that for the unconstrained original position and hence the strain energy there is a minimum. An alternative way of producing a state of equilibrium different from the natural one in an elastic body under a prescribed set of displacements is the introduction (prior to the imposition of the displacements) of internal · or ext~mal constraints that do no work. For example, in a shell or plate analysis, we may assume that the middle surface is inextensible and the transverse shear strains are zero; thus, in this case we impose infinite values for E and G in the middle surface and an infinite value for G in the *See A. Castigliano, Theorie de /'equilibre des systemes

elastiqu~s,

Turin 1879.

_1I I 0

8

Fig. 9.-Stiffening and relaxation of stress-strain curve

direction: -:'nother type of constraint may be achieved by the mtroduct10n of a ngt_d support. Also the stress-strain diagram may be taken local~y to be II mstead of I (FIG. 9). In all such cases the arguments of the prevtous paragraph show that the strain energy of the constrained body for ~he given prescribed set. of displacements is greater than for the unc~:mstran:'ed body. Conversely, tf we relax any existing constraints whilst agam keepmg constant the prescribed displacements the strain energy is decreased. The relaxation of constraints may take the form of a stressstrain line III instead of I. Alternatively we may introduce a hinge in the structure. Another example is the case of a shell where we ignore the bending stiffness and admit only a membrane state of stress· a current ' procedure in wing stressing. . Thus we co~cl';lde: the strain energy of an elastic body for a given set of d1sp~acements 1s mcreased (reduced) by the imposition (relaxation) of constramts that do no wor~. ~ n el_(cephon occurs if the effect of imposition or r~moval of a constr~mt ts mi. For ex_a~pl~, in an infinitely long, thin circular shell under mternal pressure It ts Immaterial whether we take account of or ignore the bending stiffness of the walls. Also since the constraints do no work it follows that the increase (reduction) of the strain energy can only be produced by the forces P. Thus:. the fo~·ce system P which is set up at the an_d in the direction of the f?rescnbed displacements and hence also the st(ffness of the structure is mcreased (reduced) by the imposition (relaxation) of constraints that do no work. . If we consi~er now the case of an elastic body under a given set of forces mstead of displacements then we conclude immediately from the last the? rem: the displacements and hence also the strain energy in a body under a g1ven set of forces are reduced (increas!!d) by the imposition (relaxation) of constraints that do no work. Both the last two theorems illustrate two complementary effects of the action of constraints on the stiffness of a structure. The last theorem may also be expressed as follows: the strain energy of an elastic body under a gil•en set of forces is a maxim!lm when it is subjected to the least number of constraints that do no work. The immediate application of the above considerations is, of course to the effect of actual constraints on elastic structures as illustrated in 'the e~amples m~ntioned. A more important application appears in connexion wtth approxtmate analyses of deformations. Thus, if we reduce the freedom of deformation as we do in the Rayleigh-Ritz and related methods we always over-estimate the stiffness of tho structure. Hence for a given set of forces (displacements) we under-estimate (over-estimate) the corresponding displacements and strain energy (forces and strain energy). _The above theorems on the effects of constraints on strain energy and sttff~ess appear to have been given first by Rayleigh* in 1875 for linearly elasttc bodtes and no temperature effects. Our arguments indicate however, that they apply also to elastic bodies with non-linear stress~strain relationship and under thermal loading. The original principles are occasionally referred to as the static analogues of Bertrand's and Kelvin's theorems in dynamics. !ransvers~

(0) The Unit Displacement method. This method will be developed in Section 8. .
u= Uo(X, y, z)+La,u,(x, y, z) r~l

n

V=' V0 (x,y,z)+Lb,vr(x,y,z) r ''I

n

W='W (X,y, z)+Lc,w,(x,y, z) 0

r --I

I

I

I

......................

(50)

where U0 , V00 w, satisfy the kinematic conditions where these are prescribed and u,, v, w, are linearly independent functions which vanish there. a,, b,, c, are unknown constants to be determined by the Rayleigh-Ritz procedure. The elastic strains corresponding to (50) are (see Eqs. (I) and (2)) aw av au E,,=sx -YJ, Eyy=a-y -YJ, E,,=sz -YJ (51) aw au av aw au av Eu=ax+az E,"=i\Y+ox' E",=oz +s/ The chosen series satisfy the displacement boundary conditions and the infinitesimal deformations OU=Oa,· u, OV=Ob,· v,, OW=,Oc,· w, . . . . . . . . . . . . . . . . . . . . . . . . (52) *See Rei.. 3., VoL II. Jl. 94, and also 'General Theorems Relating to Equilibrium and Initial and Steady Motion , Phd. Mag., March 1875. They have been discussed more recently by D. Williams in Phrl. Mag. Ser. 7. Vol. 26, 1938. p. 617. t W. Ritz, "Theorie der Transvers~lschwingungen einer quadratischen Platte.' Ann. d. Physik, VoL 28. p. 737, 1909; see also J. reme u. angew. Math .• Vol. 135, p. 1. and Gesammelte Werke, Parts, 1911, p. 192.

obtained by variation of a" b., c, while the other coefficients and the temperature are kept constant are hence virtual elastic displacements. Note again that the chosen u, v, w functions need n'ot satisfy the given static boundary conditions. Naturally, the accuracy of the analysis is enhanced if the latter are also satisfied. To determine the coefficients a., b., c, we use the condition of the stationary value of the total potential energy in the forms DU=O or SU;=DW There are also cases when the theorem of minimum strain-energy, U; =min., is useful, see for example Part H of this series example 4. The application of DU=O ensures the average satisfaction of the equilibrium conditions. If we evaluate U in terms of (50) then for a linearly elastic body we obtain a quadratic function in a., b., c,. Condition DU=O which in this case becomes

ou ou ou oal = ...... =oa, = ...... =oa, =O ou

ou

ou

Ob 1 =

· · · · · ·

=ob, =

· · · · · ·

=obu =O

ou ocl =

......

ou =oc, =

......

ou =oc, =O

l f ............... .

J J[ oa""+Oaz"+oaM+w JvdV=O

+w u dV=O J[ oa~"'+oa.""+oau ox oy oz r .t:

v

(53)

_j leads to a set of 3n linear equations in the 3n unknowns a,,b" c,. It is, however, superfluous to evaluate first. U and then to differentiate with respect to the unknown coefficients. We can obtain directly the final equations by forming the 3n expressions S.,( U; + U,) =0, 8b,( Ui + U,) =0, De,( U; + U,) =0 . . . . . . . . . . . . . . (54) where the suffices ar, br, cr indicate that the virtual displacements are chosen respectively as in Eqs. (52). Using the first of Eqs. (52) in the first of Eqs. (54) we obtain the more explicit form

{faE,dV ~~Pu}Da,=O v

or since Sa, is arbitrary faE,dV~~Pu=O . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (55) v where E, are the elastic strains due to u, a the stress due to the elastic strains given by Eqs. (51), P the applied forces and u the displacements in their directions due to the set (u,). There are in all 3n equations in a., b,, c, which are non-linear if the stress-strain relations are non-linear. By a judicious choice of the u, v, w functions it is possible to obtain very good approximations to the deformations of the body. The number of necessary functions for a good estimate depends on the problem and on the choice of these component functions. The proof of the convergence to the exact solution with increasing n is a difficult question which cannot be considered here, see Trefftz 10, p. 130. We note only, referring to the previous paragraph, that the Rayleigh-Ritz method always over-estimates the stiffness. Whilst the Rayleigh-Ritz method can provide a good approximation to the deformations, the accuracy of the associated stresses is, in general, not as good. This is obvious if we remember that the accuracy of any approximate function is decreased with every differentiation. In two- or three-dimensional problems it is possible to improve upon the original Rayleigh-Ritz procedure by adopting a mixed technique of (A) and this paragraph. Thus, in a two-dimensional problem it is often possible to guess accurately the variation of the displacements parallel to one axis, say the x-axis, while it is much more difficult to make an intelligent assumption about the variation parallel to the other direction. We may then write the displacements u and v in the form u=u 1(x)·cp(y) } V=V (x)·l{l(y) . . . . . . . . . . . . . . (56) 1 where u1 , v1 are assumed crosswise distributions of the u, v displacements and f, if! are unknown non-dimensional functions of y. Substituting Eqs. (56) into (55) one obtains after some transformations the differential equations in f and if! together with the necessary boundary conditions. Such an analysis can yield a very accurate result. It will be illustrated on a number of examples of some complexity in Part I I to this report. Note that the Rayleigh-Ritz procedure as presented here is also valid for non-linear stress-strain relations. (F) Galerkin's method of approximate stress analysis. Consider the internal equilibrium Eqs, (4) which we write here in the by now familiar form

oayx+ oau+w ]8u ·dV=O J[Oaxx+ ox oy oz v ]1)v ·dV=O J[Oayy+Oazx+oa_,"+w oy oz ox y

where the Su's are virtual, i.e. kinematically possible infinitesimal displacements. Eqs. (57) lead to the principle of virtual work Eq. (44) if the a's satisfy not only the internal but also the boundary equilibrium conditions Eqs. (5). Assume now that the displacements u, v, w are written in the approximate form of Eqs. (50) where the a, b" c, are unknown coe:ficients. However, contrary to what we assumed in paragraph (4), series (50) are taken here to satisfy not only the kinematic conditions wb.ere prescribed on the surface but also by substitution into the stress-strain relations the equilibrium conditions where prescribed. Expressing now the stresses in the brackets of (57) in terms of the displacements and temperature 0 and selecting as virtual elastic displacements one of the 3n possibilities (52) we obtain the n equations in the 3n unknowns a" b, c,

v

0\' .

OZ

OX

r

'' '''''' ''' ''' ''

(5S)

J[o;;z +o;;z+ o;;-- +wz]w,dV=O v

This is the Galerkin *t procedure usually only given for linear elasticity. To fix ideas take the case of linear elasticity and write axx etc. in terms of u, v, w. We find easily three equations, the first of which is

J[~u+l~ 2 v ~!+~_,~ 2i~i~) ~~Ju,dV=O

..............

(59)

ou ov ow g=ox +s:Y+oz . . . . . . . . . . . . . . . . . . . .

(60)

v

where

o 2u o2u o2u

~u=ox2+oy2+oz2 and

The bracket in Eq. (59) is, of course, the equilibrium equation in the x-direction expressed in terms of displacements. The procedure in any particular structural problem is to form the equilibrium conditions in the stresses or stress resultants and express each in terms of the displacements. Next we multiply each by the corresponding u, (which may be a deflexion, slope, twist) and then integrate over the body. Thus, for a beam subjected to a distributed loading p in the y-direc.tion, the equilibrium equation in terms of the deflexion v is, assuming engineers' theory of bending to hold

::2 (EI;;~)~p=O

.................. ....................

(61)

and the Galerkin form of the virtual work equation is I

J{;;

2(

El;;~) ~p }8v·dz=O

..........................

(61a)

0

It is easy to see that for displacement functions (50) satisfying all boundary conditions the Galerkin and Rayleigh-Ritz methods must yield the same equations for a, b" c, and hence also the same deformations. We need only realize that in this case Eqs. (57) are indeed equivalent to the principle of virtual work. Hence substitution of u, v, w in Eqs. (57) must give the same result as substitution into

SU=O The advantage of Galerkin's method lies in a more direct derivation of the equations in a., b., c,. However, contrary to what is usually assumed, this advantage is small if we calculate SU directly. We note also that Galerkin's method allows only such approximate functions as satisfy all boundary conditions, while the Rayleigh-Ritz procedure requires only the satisfaction of the kinematic boundary conditions. * Timoshenko 8 mentions on p. 159 that equations of this type appear already in W. Ritz's work. See references on p. 353 and also Gesammelte Werke, p. 228. t B. G. <,;alerkin, 'Series Solutions of Some Problems of Elaftic Equilibrium of Rods and Plates,' Wjestnik lngenerow Petrograd (1915), No. 19, p. 879; see also W. J. Duncan, 'Application of the Galerkin Method to the Torsion and Flexure of Cylinders and Prisms,' Phil. Mag., Ser. 7, Vol. 25, 1938. p. 636.

z

I~

x

v

II

(57)

v fjg, 10.-Cantilever beam

9

Consider the cantilever under transverse load shown in FIG. to. To obtain an approximate expression for the deflexions v by the RayleighRitz procedure we need only select a function or sequence of functions giving dv v=az=O at z=O However, when applying the Galerkin method Eq. (6la) must satisfy also the static conditions at z=l, i.e. Shear force=Bending moment=O (d2v) dav) or ( dz 3 z~l = dz 2 z~l =O

5. ILLUSTRATIONS OF THE METHOD OF VIRTUAL DISPLACEMENTS In this section we present a number of applications of the principle of virtual work. These are not meant to give the shortest possible solutions to the problems considered but merely to illustrate the way in which the method can be applied in some simple cases. More complicated problems are investigated in Part II (a) Continuous Beam with Non-Linear Spring Support The uniform beam of bending stiffness EI shown in FIG. II carries a uniformly distributed load p and is simply supported at the ends. At the centre an additional support consists of a spring with the load displacement law P=kv[ I+ l-(~/

v.)J ............................. .....

(a I)

Thus k is the initial spring stiffness and v. is the displacement at which the spring becomes solid. With the assumption of the Engineers' Theory of Bending, establish by means of the Principle of Virtual Displacements the differential equation for the deflexion v of the beam and the boundary conditions. Find also the deflexion Vat the spring.

p

--4---El/J ·I·

z

~

Fig. 11.-Virtual displacements: Example (a) Simply supported beam with non-linear spring

The first part of the problem is, of course, trivial and the result known to any undergraduate, but we want to show here how the Virtual Displacements method can be applied in such a case. We consider virtual displacements consisting of small arbitrary additional. deflexions (Sv) of the beam from its equilibrium position. The Principle of Virtual Displacements then expresses the equilibrium condition in the form SU1 -SW=O . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (a2) The virtual displacements must satisfy certain kinematic boundary conditions, namely : Sv=O for z=O and z=2/ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (a3) and since both structure and loading are symmetrical we need only consider symmetrical virtual displacements. Hence d(Sv)/dz=(Sv)' =0 for z=l . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (a4) Strains and stresses in beam due to bending: E=Eu= -yv", a=a .. = -Eyv" . . . . . . . . . . . . . . . . . . . . . . . . . . . . (a5) and therefore the virtual strain due to Sv is SE=-yS(v")=-ji(Sv)" ... . .. . . .. . . . ... . . . . . . . . . . . . .. . .. . (a6) The increment of strain energy in the beam due to bending is thus I

I

oA

o

I

A

o

I

I

2J {Efvlv_p}Sv·dz+2Elv"(Sv)' I 0

I

2Eiv'"(8v) I +PSV~O .... (all)

0

0

But (Sv)z' =0, ( Sv) 0 =0 and (8v), = 8 V Since otherwise 8v is arbitrary we conclude that to satisfy Eq. (all) we must have (Elv"L~o=O, 2(Elv"')zd-P=O ............................ (al2) and v must satisfy the differential equation EJviv_p=O .........................•.................. (al3) Eqs. (al2) and (al3) together with the kinematic conditions (v)z~o=O and (v')zd=O . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (al4) give all the necessary information for the determination of v. Integrating Eq. (al3) and using the boundary conditions (al2) and (al4) we find finally for V the quadratic equation (I +!{i)(V/ V.) 2 -[l +!{!(!+a)+ V1 /Vo]( V/V.)+ V1 /Vo=O . .. . .. (al5) where !{i=kl 3 /6EI, V1 =Sp/4 /24 (b) Plane Redundant Framework A plane framework consisting of a single joint connected by a number of hinged bars to a rigid foundation is loaded by forces X and Y along the axes Ox, Oy respectively (FIG. 12). In addition, the bars are heated to arbitrary temperatures and have also initial strains due to errors in manufacture. Find by application of the Principle of Virtual Displacements the forces in the bars. Let u, v be the displacements of the loaded hinge, measured from the position for which all bars have the correct length and are at zero temperature. Then the total direct strain in the rth bar for these displacements is u cos Br+v sin Br Yr fr • • • • • • • • • • · · · · · · · · · · . . . . . . . . . . . . . . . . (bl) The total strain is made up of the elastic strain Er together with the thermal strain 7Jr=a0 and the initial strain 'Y]oro where the initial strain is 'Y]or=t1fr/lr · ... ·......................................... (b2) t11r being the additional length of bar (in excess of the correct length) due to manufacturing errors or other causes. Hence the elastic strain due to u, vis Er=yr-'Y)r-'Y]or

p

2J fa·SE·dA·dz=2Ef { fy 2·dA}v"(8v)"dz=2Elfv"(8v)"dz

Hence the complete virtual work Eq. (a2) becomes

u cos BrtV sin Br -('Y]r+'Y)orl

......................... .

and the direct stress in the bar is Nr_ -EEr-E[ucosBr+vsinBr Ar -Urfr - ('Y)r+'Y)or

>]

........... .

I

I

0

0

I

I -v"'(Sv) I +Jviv.Sv·dz}

r

.. x

(a7)

I

SW=2fp·Sv·dz 0

10

(a8)

0

The increment of strain energy in the spring is PSV . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . and the increment of work done by the distributed load is

(a9)

Area Ar, length lr (alO)

(b4)

If we now impose on the joint the virtual displacements Su, Sv there arises an increment of strain energy S U;, and an increment of work S W of the applied forces, where

which becomes, on integrating twice by parts, 2£/{v"(Sv)'

(b3)

Fig. 12.-Virtual displacements: Example (b) Redundant system of bars

and the work done by the distributed torque and the end torques

n

8U;="2:.Ararlr0yr

I

8 W= Jm.88dz+ T 1(88) 1 - To(88)o ................... , ..... .

r~l

=fEArlr[u cos

8r~v sin 8, ~(7],+7]or)] [Su cos 8,t8v sin 8,]

r=l

r 0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

(b5) (b6)

0

8W=X8u+ Y8v . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . By the Principle of Virtual Displacements

Applying the Principle of Virtual Work I

8U;~8W=O= J[£r8"(88)" +GJ8'(88)' -m.<88)]dz- T88 n

and therefore

(c8)

1 0

•

n

I

Integrating by parts the first term twice and the second once, we finally obtain

8U1 ~8W=O n

n

au[ u I~:r cos 8,+v I~:T sin 8,cos8,~ LEA,(7],+7]or)cos8,-x]+

I

I

j [Er8iv ~GJ8" ~m.]88dz+[GJ8' ~Er8"' ~T]88 I +£r8"(88)' I =0 0

0

0

(c9)

2

r~l

r~l

r~l

n

n

n

sv[u 2:~:r sin 8,cos 8,+v L:Etr sin 2 8r~ IEAi7]r+7]or)sin8,- y J=0 r~l

r~l

(c7)

0

r~l

For the integral to vanish, since 88 is arbitrary, 8 must satisfy the differential equation Er8iv -GJ8" ~m.=O . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (ciO)

(b7) If Eq. (b7) is to be satisfied, since Su, Sv are arbitrary, the two expressions in brackets must separately be equal to zero and hence we have two equations in the displacements u, v. Solving for u, v the stresses and forces in the bars can be calculated from Eq. (b4). The two equations are, of course, the equilibrium conditions in the x andy directions and could be derived directly by statics without recourse to the Principle of Virtual Displacements. Note that there are always only two unknowns in this approach, regardless of the number n of bars. Hence it is preferable to operate with the displacements u, vas unknowns than with forces in the bars when n>4.

which is recognized as the usual Wagner equation differentiated with respect to z. If the twist at both ends of the tube is specified then 88 must there be taken zero and the first bracketed term in (c9) vanishes also. If in addition the warping (and hence 8') is specified (e.g. built-in end) then (88)' is also zero at z=O and z=l and the remaining term also vanishes. If, however, the end z=l is free to twist and warp, (88) 1 and (88)/ are arbitrary and we have as further conditions from (c9)

(c) The Open Tube Under Torque A uniform, open tube of length I is subjected to a distributed torque m. per unit length and end torques To and Tt. With the assumption that shear strains due to restrained warping are zero (Wagner assumption) establish the differential equation for the angle of twist 8 and the static boundary conditions. Give also a series solution for the case when the ends z=O, z=l are prevented from twisting but are free to warp.

which are the necessary static boundary conditions at the free end. The first is of course the condition for equality of external and internal torque and the second the condition for zero direct stress. For the series solution, we represent the twisted shape by the Fourier series

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

}

Tt=O

GJ(8')t~Er(8)'"t~

and

£r<8")t=o

(ell)

X

. (nTTZ) 8= '\:"" L...Ja, Sin -1~

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (cl2)

n~l

and take for virtual displacements the increments of twist produced by a small variation 8a.., of the coefficient a,... Using Eq. (c8) we find

o

n=l

J.

J .

I

X

I

J

I

I

X

mTTZ nTTZ '\:"" nmTT 2 J 8U;~8W=8a.., [ GJ L...Ja,~ cos - 1- cos ~1~dz

1

J

(cl3)

(11mTT 2) 2 mTTZ . nTTZ mTTZ +Er f:;.a" f2 o s1n - 1-sm ~1~dz~ om. sm - 1-dz \.-,

which gives on integrating I

Jm. sin mTTZ -1 dz . . . . . . . . . . . . . . . . . . . . . . . . . . (c·l4) Fig. 13.-Virtual displacements: Example (c) The open tube

The warping displacement w of the cross-section is given by* w=w(d8/dz)=w8' where wis the warping per unit rate of twist. The direct strain stress a_. due to non-constant rate of twist are 0

0

0

0

0

0

0

0

0

0

E.,=~ =w8", a_.=EE_.=Ew8"

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

If m. is constant 0

Ezz

0

(cl) and (c2)

The torque due to the St Venant torsional shear stresses is TJ=GJ8' . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (c3) We obtain now virtual displacements by giving to the twist 8 an increment 88. The corresponding virtual strain 8E is given by 8E.,=w8(8")=w(88)" . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (c4) and the increment of strain energy due to the direct (torsion-bending) stresses is I

S

I

J [ Ja ..SE ..tds]dz=J£8"(88)" 0

0

0

S

I

[ fw 2 tds]dz=JEr8"(88)"dz 0

....

(c5)

0

where r is the torsion-bending constant.* The increment of strain energy du~ to the St Venant shear stresses is I

I

n

,.

fGJ8'8(8')dz=JGJ8'(88)'dz

• See Argyris, 'The Open Tube', 1954, p. 101 et seq.

.............................. AIRCRAFT ENGINEERING,

(c6)

4[2

form odd and form even a,..=O which gives for the twist distribution

I I

J

. . . . . . . . . . . . (cl5)

00

(nTTz/1) 8 -~ 41 2 m.~ [ sin(n7T)2£r] GJTT3 n~l

113

I+

T

GJ

........................ (cl6)

odd

In this case, since the assumed form of solution satisfies also the static boundary conditions 8" =0 for z =0 and I we can alternatively use the Galerkin form of the Virtual Work equation, which is in this case I

J[Er8iv -GJ8" -m.]88dz=0

. . . . . . . . . . . . . . . . . . . . . . . . . . . . (cl7)

0

and is given immediately from Eq. (c9). If we approximate our solution for 8 by retaining only the first term in (cl6) we underestimate the average angle of twist. This ties up with our statements on p. 8; for, by putting 8=ao sin (TTz/1) we apply constraints on the tube and hence overestimate the stiffness.

Vol. XXVI, No. 302, April

11

6. THE PRINCIPLE OF VIRTUAL FORCES OR COMPLEMENTARY VIRTUAL WORK T is natural in reviewing the developments of Sections 3 and 4 to inquire if it is possible to enlarge upon the conception of complementary work and strain energy in a similar way as accomplished for work and strain energy by the introduction of virtual displacements. In fact, if we consider Eqs. (28) and (36) we realize immediately that the functions S W* and S U; * are independent of the variations Su and DE associated with the force and stress increments, just as S Wand OU; are independent of the variations oP and Sa associated with the chosen Su and DE'S. Hence, when finding oW* and oU;* we may assume that displacements and strains remain constant. Also, the infinitesimal increments ,Sa, ow, ocp are arbitrary as long as they satisfy the equilibrium conditions in the interior and, where such are prescribed, on the surface. Thus, if we fix that the surface forces are not to be varied over part of the boundary we must have there ocp=O; however, where kinematic conditions are prescribed on the boundary the Scp variation cannot be assigned. It is apparent that our incremental stress system need not even be an elastically compatible one. It is only restricted by the condition that it must be statically equivalent to the load increments ow and Scp. While these increments are applied it is assumed as in Section 4 that the temperature remains constant. Such infinitesimal variations of forces and stresses which are arbitrary as long as they satisfy the prescribed equilibrium conditions we call virtual forces and stresses. Before restating theorem (35) for the more general conceptions introduced here let us consider again its derivation in the light of our new ideas. Thus, if we multiply the true displacements u, v, w by the internal equilibrium conditions (29) which the virtual stresses Sa and ow must satisfy sum, integrate over the body, apply Green's Theorem and note th~ boundary conditions (30) we obtain Eq. (35) where y,,. etc. are the total strains associated with the displacements u, v, w. Next let us apply Green's Theorem in the opposite direction by starting from the right-hand side of Eq. (35). We find that this function can only be equal to the left-hand side if the terms Yxx etc. are indeed the expressions for the strains (3) and ( 12) and satisfy the kinematic boundary conditions. Thus, we conclude that an elastic body is in an elastically compatible state under a given system of forces and temperature distribution if for any virtual increments of forces and stresses from a position of equilibrium oW*=oU;*+ f'YJOsdV . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (62) where v

I

oU;*=fEoa·dVand 7JOS=7J(0.7n+oa""+oa,z) . . . . . . . . . . . . . . (63) v See also Eqs. (35) and (38a). Eq. (62) is, in fact, a necessary and sufficient condition for elastic compatibility of the equilibrium. Theorem (62) we call the principle of virtual forces or virtual complementary work for elastic bodies subjected to loads and temperature distribution. Note that (62) applies for non-linear stress-strain laws. The above discussion indicates that there is a close parallel between the

p~inciples of virtual displacements and virtual forces. Thus, by substituting vtrtual.forces (stresses) for virtual elastic displacements (strains), actual total displacements (strains) for forces (stresses), and invariant state of straining for. invari~nt _state of forces we obtain Eq. (62) from Eq. (44). However, th1s duahty IS only complete for continuous structures which are infinitely redundant. If, on the other hand, we consider a statically determinate structure we find that while it is still possible to describe an infinite set of virtual displacements O!t associated with a prescribed set of certain of the displacements, only one stress system can exist for given external forces; hence no oa can be assigned in the latter case and the principle of virtual forces has no application. A more fundamental limitation of the principle of virtual forces appears if we want to extend ?ur th~orems to finite displacements. Here we find that it is, in general, tmposs1ble to achieve this for the principle of virtual forces while as mentioned in Section 4, no basic difficulty arises in the case of the principle of virtual displacements. However, for the usual analyses of redundant systems involving small displacement theory the principle of virtual forces with its many particular forms is t:1e most useful one since the standard proce~ure !ntrodu~es forces as unknowns. Naturally, there are many cases, especially m multi-redundant structures, where it is advantageous to introduce displacements as unknowns; here the principle of virtual displacements is the indicated method as shown in Example (b) of Section 5. . We return now to _Eq. (62) and will illustrate its validity on a very s1mple example. Constder to that effect the redundant beam of uniform flexural_ stiffness £/built-in at z =0, simply supported at z ~,I, and subjected to a umform load p (see FIG. 14). Under the assumption that the ordinary engineers' theory of bending holds and that the shear deflexions are negligible the deflexion v is given by

v=fs~/1) T3 -s(l) +2(7) 2]

As a virtual force we select oP at the centre of the beam as shown in 14(a). However, since we require only a statically equivalent stress system to equilibrate the applied virtual load we may eliminate the one redundancy and select a statically determinate beam. The two alternative choices leading to a simply supported beam and a cantilever are seen in FIGS. 14(b) and 14(c). Denoting the true deflexion at the centre v1 we have

FIG.

pf4

oW*=oP· v1 =3P· 192 £} Also since the Engineers' theory of bending applies (y 11 ,=0) I

oU;*=

I

J[j E:,Da,,dA }z=- J~:~oMdz o

A

n

where the integral in the square bracket refers to the integration over the cross-section. For the case shown in FIG. 14(c).

oM= -oPG-z) for oM=O Also since,

o~~z~t/2

for f/2
~;~=:~~ [I -sf +4GYJ l/2

DU;*=oP:~I J(£-z)[ I -s] +4(TY}z,=oP 1 :~~/ =oW* q.e.d. 0

If we apply now a force -- fooP at the free end our virtual system (c) is transformed into (a). No additional oW* arises since v=O at z=l. The additional bending moment 3M produced by - ~1 3P is

f6 ( I-l)oP.I and this is easily found not to create an additional 3 U; *. By relaxing the moment restraint at A we may finally prove without difficulty that OU;*=oW* applies also for the virtual system (b). We return now to Eq. (62) and note that since the displacements are assumed constant when the virtual forces are applied we may regard 3 W* as the variation of a potential - U, * where U, * = - f[uw,+vw 11 +ww,]dV- f[ucp,+vcp,,+wcp,]dS . . . . . . . . (64) v s Thu~, SW* = -- oU, * and U, * may be termed the complementary potent1al of the external forces. Note, however that W* is not - U * since !n obtaining W* from S W* we must, naturally, perform the i~­ tegratton for displacements varying with load. In fact, for a linear system and no temperature effects W* = - U, * /2; compare also Eqs. (64) and (46). Also, since the thermal strains are kept constant we may write the righthand side of (62) as OU;*+friosdV=o(U;*+frr~dV)

v

14.-Example of an arbitrary virtual force

12

v

~ oU,*

(65)

where 8U,*~Jy8adV

v

........................................ (65a)

but (66) U1* -t- frysdV f [ JEOa]dV f frJ.w/V v v 0 v since <1 =const. U,* we term the complementary potential energy of total deformation. Note that it is always simpler to calculate directly oU,* from Eq. (65). Particular care is necessary in evaluating U,* for as Eq. (66) U,*

shows in the first integral E is taken to vary with a from the initial to the final state while 7), s in the second integral refer only to the final values. Physically speaking we may consider U>~* as the· complementary work necessary to reach the final true state of deformation from an initial state in which we allowed free thermal expansion and destroyed compatibility. Formulae (65) and (66) may be extended immediately to the case of arbitrary initial straining by substitu1ing

7Ja -T),.,.a,.,.-t- ...... +7) .•.•,a_,., I .... I 7),...a,... Eq. (62) can now be written more concisely o,U*~o.,(U,*tU,*J

for

7JS

0 ..................................

(67)

where the suffix a indicates that only forces and stresses are varied and U*=U.,*-t-U,*

(67a) is defined as the total complementary potential energy of the system. Eq. (67) states that a state of equilibrium of an elastic body is also one of elastic compatibility (i.e. the body is at the true position of equilibrium) if any virtual variation of the stresses and forces, while displacements remain constant, does not give rise to any (first order) variation of the total complementary potential energy. This theorem we call the principle of a stationary value of total complementary potential energy if the latter is expressed in terms of forces and stresses. Actually the stationary value of U* is a minimum as we may prove without difficulty.* This point is discussed in more detail under (C) below. Eqs. (62) or (67) may be used to derive the results which follow. (A). The differential equations of the theory of elasticity (for arbitrary loading and temperature distribution), or any particular structural problem, in terms of stresses or stress-resultants; the appropriate kinematic conditions in terms of forces and stresses follow also from this analysis. It is important to note that in all applications it is best to form directly oU,*=f[yoa]dV

v and not to evaluate first U.,* and then to take its increment oU,*. (B). Castigliano'st Theorem Part II generalized for Thermal Effects and

non-Unear elasticity

[ (!Ud*] (JP r

B=const.

=u, ..................................... .

(68)

where u, is the deflexion (rotation) in the direction of the force (moment) P,. This relation may be obtained immediately if we apply one virtual external force oP, in the direction of the displacement u,. (C). The Principle of Stationary (Minimum) value of Complementary

potential energy of total deformation for internally redundant structures This may be derived from (62) if we do not apply any external virtual forces; i.e. 8c/>~=oc/>,=oc/>z=owx=ow 11 =~owz=O

while varying the stresses a. Then (69) 8.,(U,*)=0 which is our generalization of the standard principle of Castigliano of Minimum Strain Energy to include temperature effects and non-linear stress-strain laws. Note again that Principle (69) necessarily applies only to internally redundant structures, since for given external loads only one stress distribution can exist in statically determinate structures. Eq. (69) itself only indicates that Ud* has a stationary value in that particular state of equilibrium in which all the elastic and kinematic compatibility conditions are satisfied. Note, as mentioned before, that with the limitations of the present assumptions, i.e. small displacements and monotonically increasing stress-strain diagram, there is only one position possible where both the equilibrium and compatibility conditions are. satisfied. We now investigate the nature of the extremum of Ud*, which naturally requires the consideration of second order terms as in Section 4. Consider an elastic body under given loads and temperature distribution in its compatible equilibrium position. We make a series of cuts in the body but at the same time apply stresses a c acting across and along the cuts of the same magnitude as in the uncut body; these are obviously the stresses required to maintain the compatibility condition of perfect fit at • See Biezeno and Grammel', p. 75.

t A. Castigliano, Theorie de /'equilibre des systjmes elastiques, Turin 1879.

the cuts. If we impose the vii-tual stresses oa, it is apparent that since these produce corresponding deformations ou,. on the cut faces the latter are not any longer compatible. It is important to realize that the oar systems are self-equilibrating since the external loads P remain constant. Thus, in a framework we may obtain a system oa,. by cutting a redundant bar and applying a variation oN to the true force N in the bar. We now investigate the differences in complementary work W* and U>~* between the original equilibrium position of the uncut body and the new enforced equilibrium position of the cut body. Comparing Eqs. (38) and (66) we find 1

(70) v 0 In moving from the uncut (compatible) equilibrium state to the cut one we note first that the integral does not vary since 7) is constant in this step. Also the first order increment, oUd*, of Ut~* is zero since this is the condition for compatible equilibrium of the original body and no first order increment 8 W* can arise since the loads P remain constant. We are then left only with second order increments. For the complementary work this is o 2 WL1-fou,·oac·dS ... ................................. (70a) W*

=

U.t*- f
c

(where the integral is taken over the cut faces) which is the work of the virtual stresses oa c over the displacements ou c they themselves produce. This is clearly positive. The second order increment of U,* is merely o 2 U,* =Hoa· oE·dV . . . . . . . . . . . . . . . . . . . . . .. . . . . . .. . . . . . . (70b) v

since 7) remains constant; oa and OE are the stresses and strains due to oa ,. Terms (70a) and (70b) are equal and both positive. We conclude that the complementary potential energy of total deformation U,* and the complementary work W* have for given forces and temperature distribution a minimum at that position of equilibrium of the uncut body at which compatibility is satisfied. It follows that if U.t* is overestimated by assuming a statically equivalent stress system which does not satisfy all compatibility conditions and we, ignoring the latter fact, equate U<~* to W* of the applied loads Palone we cannot but overestimate the magnitude of the displacement system under the loads P. Conversely to achieve a given displacement system our calculations based on a non-compatible stress system must underestimate the corresponding load system P. Thus, the latter has its maximum for the unique equilibrium position which is also truly compatible. This may be expressed also as follows : For given displacements and temperature distribution the complementary potential energy of total deformation has a maximum when the state of equilibrium satisfies also the compatibility conditions. The above theorems may be combined to give: The stiffness of an elastic body in which the equilibrium conditions are satisfied is a maximum when the elastic compatibility conditions are all met. Thus we see that the effect of introducing assumed forms of stress distribution for the purpose of approximate solutions is the opposite to that of the method of Virtual Displacements and therefore application of both methods to a given problem yields upper and lower bounds to such aggregate quantities as stiffness. No general conclusion as to bounds can, of course, be drawn for the details of the stress distribution. The above theorems which apply also in the presence of initial strains other than those due to temperature do not appear to have been given before with this degree of generality. The Unit Load Method Assume that we require the deformation (deflexion or slope) u, at a given point and direction of an elastic redundant body subjected to given forces and thermal effects. Let the actual total strains in the structure be known and given by* (D).

Yx~=Eu+7),

Yx·y=Exy

Applying a load (force or moment), oP, in the direction of u, and using Eq. (62) we find oP,·u,=JyoadV v =HYxxOa..,..,+ ..... +Yx110a,.+ .. +yuoa.,]dV

(71)

v

where Oaxx . ... oa~•. ... are the virtual stresses due to oP T• In a linearly elastic system Sax~ etc., are proportional to oPT and Eq. (69) can be written l·u,= J[y,.p,x-1-YY·P•v+YzPzz+Yx•a,,+y,.a•• +yz,azr.]dV .... · . (71a)

v

where

au . ... a,•.... are the stresses due to a unit load. Since ii,.,.. ...

ax• . . . . need only satisfy the internal equilibrium conditions and the external one for oP,= 1 it is obviously advantageous to determine ax., . ... iix·u . ... in the most simple statically determinate basic system.

a,,.

For a non-linear system Eq. (7la) is still applicable as long as etc. are calculated in a statically determinate basic system. For only in the • Naturally. this method may .also b.> applied in the case of an initial strain system with shear strains.

13

A

:

.

'

v

B

/\1\1\--

C

A -,~r,I I

-,..,\--

\1

'1

\1

',

,..,..~

~~~--a--~2~--~~--~--~A

Unit load

8

Normal (l,m)

y,vL

- -;',

\1\ ''

~-

\

~

B

x.u

~

16.-Two-dimensional stress case, static and kinematic boundary conditions

15.-Unit load method for displacement of redundant framework

latter case w1ll Sa.,.,j'8P,=oa.,.,/oP, be the stresses corresponding to a unit load. The Unit Load Method is the most suitable tool in the calculation of structures with a finite number of redundancies expressed as stresses or stress resultants. This will be shown in some detail in Section 8. Example of the application of Eq. (71a). Consider the plane framework with a redundant support as shown in FIG. 15(a). We seek the deflexion vat joint 2 for the loading case shown. Let the actual elongation of the members due to loads P 1 •••• P 5, temperature and manufacturing errors be denoted by D./. Next we apply a unit load at 2 in the direction of v and find the forces iv in the bars. Since we need only consider a statically determinate case we ignore the support at C and are left with the very simple problem of finding N in the left-hand span only. Application of Eq. (71a) yields the simple formula I·v='J:,ND.I ............................................ (71b) where the summation extends only over the ccmtinuously drawn bars. The formula given is due essentially to Maxwell* and Mohrt who applied it to statically determinate frameworks. Actually Mohr derived this type of equation by using the principle of virtual displacements with the actual elongations taken as virtual ones and the unit load system as the actual one. Although such a procedure is in the present case of small displacements permissible, it should nevertheless be avoided since Eq. (71b) follows more naturally from the principle of virtual forces.

(74) F must, of course, also satisfy the given boundary conditions. In fact, where surface forces are prescribed we have (see Eqs. (5) and FIG. 16)

I I

o (oF oF) oy 1oy -mox =f~

J ............................

An approximate method may proceed as follows. Assume that the stress function F is expressed in the form of a finite series n

(76)

F=Fo+'J:,b,F, r= I

where F0 , F1 , • • • • • F, are known functions of which F0 satisfies the static boundary conditions (75) where these are prescribed and the functions F, vanish there. b1 to b. are constants to be determined by the virtual forces principle. The system (76) satisfies by definition all given equilibrium conditions and the increment OF=Ob,·F, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (77) may be regarded as a virtual stress system corresponding to zero increments of external loads where the latter are fixed. Since SW*=O (either forces are given or displacements are zero) the principle of minimum complementary potential enengy of total deformation is applicable here and takes the form

SUa*= J[y.,.,Sa.,.,+y.vSa•• +y.,.Saxv]dA =0 . . . . . . . . . . . . . . . . . . A

(E). Approximate method of stress analysis

We use here

Consider an elastic bodyt subjected to external loads (body and surface forces) and a temperature distribution 0. The boundary conditions are assumed to be both of the static and kinematic kind; however, where the latter are prescribed they are taken to be of the rigid kind, e.g. rigidly built-in or sliding in a rigid groove (see FIG. 16). To limit the present analysis we restrict our investigation to a state of plane stress. The solution of such problems is often expedited by the use of the Airy stress function F. Then, the stresses are given by§ X

y

a.,.,=~~- Jwxdx, a.,v=- 0~~· avv=~~- JwvdY............

(72)

which satisfy automatically the internal equilibrium conditions (4). Eliminating the displacements from the strain expressions (I) we obtain the compatibility condition for the strains, --

o~+o~o~-0 oy 2 ox 2 - oxoy................................. .

(73)

where Yxx etc. are the total strains E.,.,+'l') etc. For a given stress-strain law we cart express y.,., in terms of the stresses (72) and the temperature 0. Hence by substituting into (73) we obtain the differential equation in the unknown F, which will, in general, be non-linear. However, in the case of bodies obeying Hooke's law we obtain for a=const. the simple linear result

• .f. C. Maxwell, Phil. Mag., vol. 27, p. 294, 1864 t 0. Mohr, Zeit. Ar<-hitek. u. /ng. Ver. Hannover, 1874, p. 509; 1875, p. 17. t The presentation is restricted to singly connected domains. § See Timosbenko 8, p. 26.

14

(75)

J( ... .)dA

A

(78)

to denote the integration ff( ....•. )dxdyover"

the area of the two dimensional continuum. Substituting Sa.,., etc. in terms of (77) we find 02F, 02F, 02F,] Yu 0y2 +Yw 0x2 -yxvoxoy Sb4A=O ..................

J[

(79)

A

and since Sb. is arbitrary. o2F, o2F, o2F,J Y"'"oy2 +Y••ox2 -y""oxoy dA =0

J[

. . . . . . . . . . . . . . . . . . . . . . (79a)

A

If the total strains are expressed in terms of the stresses (72) and temperature distribution 0 we obtain from (79a) n equations for the unknowns b1 to bn. These are only linear if the body follows a linear stress strain law. In the latter case and for constant body forces Eq. (74) shows that the solution must be independent of the Poisson's ratio v if all boundary conditions are of the static type. Hence we may take v=O and (79a) becomes

o2F, o2F o2F, o2F o2F,] [o2F, o2F•]} J{[o2F ox 2 oy 2 + oy ox 2 + 2oxoy oxoy + E7J ox 2 + oy 2 dA =O

A

2

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (79b) from which we may obtain without difficulty the n linear equations for b1 to b,.. The case when all prescribed boundary conditions are static is interesting for all surface conditions are then exactly satisfied by (76). This indicates that it should be possible to express (79a) in a form similar to that given as Galerkin's method under the virtual displacement principle. In fact, if we integrate (79a) twice by parts, or better, if we apply Green's Theorem

.(Area 8

p

T..,_ b

. . (SOc)

lO~q

Areaw;,

_t__

c1z

.__,

..1!.:_

il+~·l:1

~-ckness t IOtq

'

1-.P.

~I ~z-

p

where Yt and Yu are the extreme (boundary) values of y corresponding to the same x (see FIG. 16). Thus, the coefficients of the homogeneous part of (SOc) are only constants in the case of a rectangular field. This method can yield very accurate results and is actually the one adopted in Part II.

7. ILLUSTRATIONS OF THE PRINCIPLE OF VIRTUAL FORCES In this section we present a number of applications of the principle of virtual forces to quite simple problems. Again, it is not necessarily suggested that the method is the most suitable one for the problems considered. It is only intended to show how it can be applied in these simple cases. In subsequent parts of the paper some rather more complex problems will be dealt with. In all the examples of this section linear elasticity is assumed.

17.-Virtual Forces: Example (a) Diffusion problem

(a) Diffusion Problem

The panel shown in FIG. 17 is subjected to loads P applied at the free ends of the edge members. Assuming that the sheet carries only shear stress which is constant across the width b of each half (usual diffusion assumption) obtain by application of the principle of virtual forces the differential equation for the load P. in the central stringer. Find also the displacement w of the free end of the stringer. From the equilibrium of an element of the stringer, we find for the shear flow (q ~ a .,f) in the sheet

-!(lOYx• +moy.,•)JF:ds +J[1oy•• +moy= r ox oy oy ox c

m )oFrJ I )oFr ox ds oy +( ly,,-pxu +.f[( my.,,-2y.,.

Ic<- ... .)ds

c

denotes the line integral along the boundary of the twodimensional continuum. But, on the boundary Fr =0 and also

~~ =0 and ; ; =0

=0 for

r= I

to n

............

(SO)

A

which ·shows clearly how the method of virtual forces satisfies in the average the compatibility condition (73). When the body is linearly elastic Eq. (SO) may be written as (a=const.)

(o20 o20) F o4F o4F ) J[(oox4+oy4+2ox2oy2 +Ea ox2 +oy2 ow., ow•)] FrdA =0 for r= I ton +(1-v) ( ox+ oy

............................................

(a2)

For the virtual forces we consider a variation 8P. in the stringer load. The applied forces P are maintained constant and hence to satisfy the equilibrium conditions on the free end (P,=O) we must take 8P. to be zero there

(oP,),d=O . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Otherwise the variation 8P, is arbitrary. i.e.

(a3)

The virtual shear flow in the sheet is thus

(a4)

4

and the virtual load in the edge members

A

. . . . . . (SOa)

the expression in the bracket being of course Eq. (74). Note again the independence of the solution from v when body forces are constant. The above application of the principle of virtual forces is a generalization of a method developed by Timoshenko 8 , p. 167. A thermal stress example of the above analysis is given in Part H. Naturally, the method can be extended to three-dimensional cases. W. Ritz* proposed as early as I 90S a similar procedure for the solution of St. Venant's torsion problem; this method is illustrated on an example of considerable complexity in Part II. A slightly more refined approach than that shown above may be adopted when it is possible to estimate accurately the variation f of F parallel to one co-ordinate say y while the distribution parallel to the other co-ordinate is more difficult to guess. Then, we may set

F=f(y)·cp(x) . . . .. . . .. . .. . .. . . . . .. . . . . . . .. . . . . . .. . .. . .. . . (Sl) where cp(x) is an unknown function of x. It is, of course, possible to formu-

late the analysis in any other suitable co-ordinate system. Substituting (S1), with oF=f8cp in place of F., into (79a) or (SO) (or related expressions), we obtain after some simple integrations the differential equation in cp; when there are also kinematic boundary conditions the corresponding boundary expressions for cp follow also from (79a). Consider, for example, the case of linear elasticity, zero body forces and pure static boundary conditions. Eq. (SOa) takes here the simple form,

J[Ll 2F+EaLl0]JocpdA=O ................................

A

(al)

and from the equilibrium of the free end of the panel we find for the load Pn in the edge members Pn=•P- ~"

(see Eqs. (75)). Hence Eq. (79a) reduces to the slightly simpler form,

o2y•• o2y.,.J J[o2y,., oy2 + ox2 -oxoy FrdA

q=-t~·'~-!P/ . .. .. .. .. ..... .. . .. .. .. ..... ...... .. ..

(SOb)

Using (SI) in (SOb), integrating with respect to y and noting that (SOb) must be true for any virtual variation ocp, we obtain the differential equation in cp • W. Ritz, J. reine angew. Math., vol. 135, 1908, and Ann. d. Physik, series 4, vol. 28, p. 737, 1909.

8;• ........................................... .

8Pn=-

(a5)

Since the applied forces are not varied, the virtual forces principle (Eq. (62) for 0 =0) reduces to 8U;*=0 The virtual complementary energy due to

8P. is

I

DU;*= J[%,4oP,+2;;oPB+2t1b8q]dz 0

Substituting for PB, 8PB, term by parts we find

(a6)

q, 8q in terms of P., 8P. and integrating the last

I

I

0

0

oU;*= J[i(1+ 2~)- 2 ~ 1 d;~·-~:B]oP.·dz+ch[~·oP.]=o (a7)

and therefore, since 8P. is arbitrary, we must have

d;~·-~:(1+2~)P.+~~P=O

....................... .

(aS)

· · · · · · · · ·. . . . . . . . . . . . . . . . . . . . . . . . . . .

(a9)

or 2 d2P. 2 dz2 -fL P.- -fL Pso

where 2

fL =

2A 1) Eb A+2B • P.a= A+2J!'

2Gt(l

. . . . . . . . . . . . . . . . . . . . . . . . (aiO)

which is the required differential equation in P•. Since 8P. is zero fop z=l, the remaining term in Eq. (a7) vanishes for the upper limit z=l. At the lower limit z=O, however, oP, is arbitrary and hence

15

-

:1

""q0

- -1 ...../t

.............

...-..,- --

Am

4

a••

2 ~

1 -----::1

-

..L

I \

I

I

-1.

(a)

~

I l-

--

~ y

(b)

faa.,'~ 7 ' '; A

b ~~~,,,-'

bb

- -I - ,

1s

$~

qo1

3

' - -- l~:1 ----s --

/

E00 dbcos~

~

;r~

(\..,+

l ~JOt o.x 'I Qry --- q..y

0,..

O.r

+ayy

1 I

,-t/

/·'

aa. ,'

,~

~ It""'"'~

~I ~~~'~"" ~

v

J

I

18.-VIrtual Forces: Example (b) Unit load method for twist of multicell tube

b.Ytt: X

19.-Virtual Forces: Example (c) Stresses and strains for oblique coordinates (a II)

which with the equilibrium conditions on the free end, P,=O for z=l . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (al2) gives the necessary boundary conditions. The differential equation (a9) is, of course, alternatively derived by a direct consideration of the deformations. In such case the kinematic boundary condition (all) appears as the condition of zero shear strain at the built-in end. ·Solution of (a9) with the boundary conditions (all) and (al2) gives

P,=P••

[.l

co~~~(~/z)J

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (al3)

To determine the displacement w of the free end of the stringer we apply there a unit force and since we need only consider a statically determinate system we assume the stringer alone loaded by the unit force. We find then (see Eq. (71)) I_

I·w= Ju ..E..Adz 0

and since a..A=unit load= I, Ezz=P./EA I

J

w= J,4dz . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (al4) 0

which is, of cour3e, merely the extension of the stringer under the varying end load P,. (b) Rate of Twist of Multi-cell Tube

In the uniform thin-walled tube whose cross-section is shown in 18(a), q. is the known shear flow distribution in the walls of the tube due to a given loading. Using the unit load method, find the rate of twist dO /dz of the tube. We consider unit length of the tube and apply a unit torque T= I. Since we only need consider a statically determinate system for the unit load stresses we select the single cell (I, 2, 3, 4) shown by full lines in FIG. 18(b). The unit torque gives then merely a constant shear flow FIG.

I ql =2Am . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

(bl)

around the single cell and the rate of twist is given immediately as

~~ = J~tq 1ds= 2 ~J 'fitds c

c

................................

(b2)

where the integral is obviously taken only around the single cell (1, 2, 3, 4). Eq. (b2) is, of course, a well-known result in the theory of closed thinwalled tubes. (c) Plan. Stress-Strain Relations for Oblique Co-ordinates

In a uniform isotropic plate, the stresses u •• , ubb, uab=CTba are referred to the oblique co-ordinates Oa, Ob (FIG. 19). Using the principle of virtual forces and assuming the stress-strain relations for rectilinear stresses, find expressions for the strains Eaa, Ebb and Eab in terms of the stresses. The oblique strains Eaa, Ebb, Eab are defined as the elongations in the directions Oa and Ob and the decrease in the angle () respectively of the unit parallelogram (see FIG. 19). For the stresses CTxx• u ••• CTxv equivalent to the oblique stresses, we find

16

easily from statics GXJ' :_:_-:Gaa sin 8 (cl) u •• =(ubb+CTaa cos 2 0+2uab cos 0)/sin () The rectilinear strains are I I 2(1 +v) E.1·.r= E(a.~.,. --va 1,_11 ), €,,,,~-= E(a 1111 ·-van.), E,~..!J =-E-- CTxv

(c2)

and hence the virtual complementary energy per unit thickness of the element dxdy is DU/• =dxdy[c rxDCTu+EvvDCTuu +ExyDCTry] dxdy = E [unDCTu +u•• DCTyy -v(uu8u.,+uuuDCTu)+2(l +v)CTxyDCT~•] . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (c3) Substituting for CTxx• u ••• u,. from (cl) we can now express 3U;* in terms of the oblique stresses and virtual stresses. Thus for the virtual stress 8ubb we find for the virtual complementary energy * dadb ,\ DU; = E sin y[ubb- CTaa + 2uab cos 0]8ubb .. .•.. . . .. .. . .. .. . (c4) where A=v sin 2 0 -cos 2 0 (c5) From FIG. (19) the virtual complementary work of 8ubb is seen to be 8W*=da8ub!,.Ebbdb .. .. .. .. .. . • .. .. .. .. .. .. .. .. .. .. .. .. .. (c6) and therefore from the Virtual forces principle 8W*=SU;* . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (c7) we find, using Eqs. (c4) and (c6) in (c7) l Ebb= £sin o[ubb-Au,..+2uab cos{)] (c8) 0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

Applying a virtual stress 8uaa in the same way we obtain for the strain Eaa the corresponding expression I

Eaac= Esin o[CTaa-ACTbb+2uab cos{)]

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

..

0

(c9)

0

Consider now the virtual shear stress 8uab. From Eq. (c3) we find for the virtual complementary strain energy due to 8uab da·db . 8U;*= £sin o{2[(l +v) sm 2 0+2 cos 2 0]uab+2(u•• +ubb) cos 0}8uab . . . . . . . . . . . . . . . . . . . . . . . . (clO) Calculating the complementary work of the virtual shear stress uab we find (see FIG. 19) 8W*=DCTabkaa cos 0+Ebb cos 0+Eab sin {)]da·db (ell) Thus the virtual shear stress 8uab does work not only due to the shear strain Eab, but also due to the strains Eaa and Ebb. Substituting from Eqs. (c8) and (c9) for the strains E00 , Ebb and equating 3U;* and 8W* of Eqs. (ciO) and (ell) respectively we finally obtain for the shear strain: 2(1 +v)[ cos () J Eab=-E-- uab+ 2 sin2{)(uaa+ubb) ...................... (cl2) 0

0

0

0

0

0

0

0

0

0

0

0

Note that with the strains defined as above the increment of complementary energy is 8U;*={E00 DCTaa+Eb1JDubb+[(Eaa+Eb1J) cos 0+Eab sin 0]8uab}dadb . . (cl3) as compared with the simple result for rectilinear axes in Eq. (c3).

8. METHODS OF ANALYSIS OF STRUCTURES WITH A FINITE NUMBER OF REDUNDANCIES

T

HE general theorems given in Sections 4 and 6 include, fro~ the fundamental point of view, all that is required for the analysis of redundant structures. However, to facilitate practical calculations it is helpful to develop more explicit methods and formulae. To find these is the purpose of this Section. A structure is by common definition redundant if there are not sufficient conditions of equilibrium to obtain all internal forces (stresses or stressresultants) and reactions; the number of redundancies is the difference between the number of unknown forces (or stresses) and the number of independent equilibrium conditions. Strictly all actual structures are infinitely redundant but for practical purposes it is, in general,. ne~ss~ry and justified to simplify and idealize the structure and/or stress d1stnbut10n in order to obtain a system with a finite (or even zero) number of redundancies. Such typical processes of simplification are, for example, the assumption of pin-joints in frameworks and the assumption of the engineers' theory of bending in the analysis of beams. Note, moreover, that the Rayleigh-Ritz procedure discussed in Section 6F amounts also, in fac_t, to the substitution of a finitely redundant structure for the actual elastic body. All our considerations in this Section are restricted to linearly elastic bodies but Example 2 in Part II shows how the present methods may be extended to the analysis of non-linear redundant structures. It is curious to note that, while the solution of problems in the theory of elasticity is derived very often from the differential equations in the disAdditional Notation

R, Q, p single, generalized, orthogonal force (moment) RQP corresponding column matrices single, generalized, orthogonal displacement (rotation) r, q, p corresponding column matrices r q p (true) stress and virtual (statically equivalent) stress due to unit load at i corresponding column matrices si, si strain due to unit load at i E; corresponding column matrix e; /;;. /;h direct and cross-flexibility F matrix of flexibilities /;A transformation matrix for forces B F., fv generalized and orthogonal flexibility matrices b rectangular transformation matrix for internal forces (stresses) column matrix of internal forces (stresses) 5 column matrix of strains v flexibility matrix of g element flexibility matrix of all elements f flexibility of element of unit length 4> true and virtual strain due to unit displacement at i stress due to unit displacement i ku, k 1h direct and cross-stiffnesses K matrix of stiffnesses k;h A transformation matrix for displacements K., K., generalized and orthogonal stiffness matrices a rectangular transformation matrix for strains

'·

placements, the stress-deformation analysis of engineering structures was, until a few years ago, generally based on the concept of force-redundancies. Interestingly enough, Navier, *who was the first to evolve a general method for the analysis of redundant systems, when investigating problem (b) in Section 5 used also the displacement method. The analysis of indeterminate structures on the basis of redundant forces goes back to Clerk Maxwell' and Otto Mohr 5 and was ultimately developed by Mueller-Breslau. 6 t This technique is, as mentioned in the introduction, more concise and physically more illuminating than the Castigliano approach; it derives most naturally from the unit load method.(see Section 60, Eq. (7la)). Mueller-Breslau's technique is generalized here and presented also in matrix form. The effect of temperature or other initial strains is included ab initio. Parallel to the rapid development of the force-redundant theory occasional practical problems were solved by selecting deformations as unknowns. Fundamentally this method is equivalent to the virtual displacement analysis given in Section 4. Mohrt was probably the first to use such an approach in engineering structures when finding the secondary bending stresses in frameworks of the type usually assumed to be pin-jointed. • C. L. Naviec. Resume des lecons sur /'application de Ia mecanique a l'etablissement des constructions et des machines. Paris 1826, 3 ed. par B. de St.-Venant 2 vols. Paris 1864. See also: A. Clebsch, Theorie der Elasticitaet [ester Koerper, Leipzig, 1862; French edition by B. de St. Venant: A. Clebsch, Theorie de /'elasticite des corps so/ides. avec de notes etendues de B. de St. Venant etA. Flamant, Paris, 1883. W. Thomson and D. G. Tait, Treatise on natural philosophy; I ed. Oxford, 1867.

t See also H. Mueller-Breslau, Die graphische Statik der Baukonstruktionen, 1 ed., Koerner, Leipzig, 1886. ~ 0. Mohr, Zivilingenieur, Vol. 38, p. 577, 1892; see also, Abhandlungen aus dem Gebiete der technischen Mechanik, 2 ed., Berlin, 1914, p. 407.

stiffness matrix of g element stiffness matrix of all elements K stiffness of element of unit length ao stress system of basic structure a. self-equilibrating stress systems X;, Y;, Z; redundant force (moment) S;0 , { ;0 relative displacement at cut i-redundancy in basic system due to external loads and initial strains siko influence (flexibility) coefficients of basic system for the directions of redundant forces D matrix of S;k D. column matrix of S; 0 T triangular matrix M elimination matrix ~ prescribed relative displacement (linear or angular) either inside the structure (e.g. lack of fit) or at the supports ('give' of foundations) C; force or moment in the basic system due to X;= 1 acting on an element which experiences a ~ in the direction of this ~ N (n), S (s), M (m) normal force, shear force, bending moment 0, o (rectangular) zero matrix unit matrix 1 .... j ..... h ..... m direction of external forces 1 . . . . . i . . . . . k . . . . . n redundancies a, b, ..... g, ..... .s elements of structure A', A- 1 transpose and inverted (reciprocal) matrix of A { . . . . . } column matrix k.

k

'ik

17

Following Mohr's <:~nalysis his ideas were applied to stiff-jointed frameworks, the first systematic work being that of the Danish engineer Axel Bendixen.* However, the great potentialities of the method were only discovered by Ostenfeld,t a compatriot of Bendixen. He was the first to point out the duality of the force and displacement ap;:-roach. In fact, his equations for the unknown displacements in a structure complement MuellerBreslau's equations for the redundant forces. It is regrettable that Timoshenko in his fascinating History 9 does not mention Ostenfeld's classical book. We give here a considerable generalization of Ostenfeld's ideas to include any structures under any load and temperature distribution. The 'slope-deflexion' equations of Bendixen form the basis of the method of successive approximation due to Calisevt and developed by HardyCross!! as the well-known moment distribution method. The technique used is essentially a particular example of the relaxation method of Southwell§ which has been successfully applied to a wide range of problems. In its application to elasticity and structural problems this latter method is particularly representative of the modern tendency in making practical the numerical solution of highly redundant systems and has been used in conjunction with both forces or stresses and displacements as unknowns. Further discussion of this method is beyond the scope of the present work which is not concerned with iteration methods but the reader is referred to the original literature on the subject. In this Section we make use, where appropriate, of the matrix notation. Although the complete analysis could be developed ab initio in this form it is thought preferable to give first most of the basic principles in the more familiar 'long-hand' notation. Only the most elementary properties of the matrix algebra like matrix partition, multiplication, transposition and inversion are necessary for the understanding of our theory. The reader may consult the classical work of Frazer, Duncan and Collar~] for these and more advanced matrix operations. Another modern and readable account is given in the recent book of Zurmuehl. * * The most comprehensive work to date on the formulation of aircraft structural analysis in matrix notation, anyhow on this side oftheAtlantic, isthatofB. Langefors. tt D. Williamst t presented r~cently an interesting account of some aspects of matrix operations in static and dynamic elastic problems. Before proceeding to a discussion of the general methods for the analysis of redundant structures we introduce some concepts helpful to the understanding of the following theories and their subsequent matrix formulation. A. Flexibilities Consider a cantilever beam with a plane of symmetry yz consisting of three connected segments a, b and c with bending stiffnesses for deftexions in the yz-plane (£1) 0 , (Elh and (£!),respectively (see FIG. 20). Let the corresponding shear stiffnessesfll! be (GA')., (GA'h and (GA'),. Transverse forces R 1 , R 2 and R3 are applied in the yz plane at the joints B, C and D. Since the system is assumed to be linear the principle of superposition holds and we can express the deflexions r 1 , r 2 and r 3 in terms of the loads as follows:

~"1 =fit R1 +fi2 R2+/13 R3 r2=f21 R1 +/22 R2+/2a R3

}

Fig. 20.-Fiexibilities and stiffnesses of a cantilever

. ·....................

(82)

r3=/31 R1 +/32 R2+/33 R3 where/;;, jjh are, of course, the well-known influence coefficients.§§ In fact, jjh is the displacement in the j-direction due to a unit force Rh =I in the h-direction. We call also /;; and /;h the direct- and cross-flexibilities respectively and deduce immediately from Maxwell's rrciprocality theorem (Eq. (43), Section 3) that I ·jjh ~I ·];.; . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

(83)

To find the flexibilitiesfin any linearly elastic body we may use the unit load method developed in Section 6D. Thus, from Eq. (71a),

1r...........

(84)

• Axel Bendixen. Die l\4cllwdt• tier Alplu~Gieichungen :.ur Berechnung t•on Ruhmenkonstruktionen, Springer, Berlin, 1914. t A. Ostenleld, Die Defurmationsmethod,, Springer, Berlin, 1926. ~ Calisev, K. A .• Te
18

where (see also Eq. (3)) O;E;=axxiE.rxi+ . . . . . . . . . . . . +az.x;Ezxi

I

O';Eh=a,,;Exxh+ · · ·: · · · · • · · · +a,,;Ezxh

j

~..........

(84a)

E;, a 1 (E1., ah) are the strains and stresses corresponding to a unit load at and in the direction of j (h). Under load we understand either force or moment. Similarly the f!exibilities may represent either displacements or rotations. Naturally, formulae (84) yield also linear (angular) flexibilities due to moments (forces) respectively. Note that while E;, EA must be the true strains due to unit loads at j and h respectively, a;, ah need only be virtual, i.e. statically equivalent, stresses due to the same loads. This is of great importance in redundant structures. Thus, denoting by a 1, ii,. any statically equivalent stress system due to unit loads at j and h respectively in a redundant structure we cah write Eqs. (84) also in the form,

v

l·jjh= fii;EhdV= fiiAE;dV=I fhi

v

I .......... I

If;;= fa;E;dV

v

(84b)

It is, of course, possible to substitute in the above formulae true stresses and virtual strains for true strains and virtual stresses but for reasons of logical consistency this is best avoided.* Assuming in the case of the beam shown in FIG. 20 that the Engineers' theory of bending stresses is true we find, noting that the system is statically determinate and hence a=a, .. .............. (85)

• See Section 6D, p. 14.

where M;, S; (M,., S,.) are the moments and shear forces corresponding to R; = 1 (R~> = 1). Eqs. (85) yield easily the following set of influence coefficients, (a+b+c) 3 -(b+c) 3 (b+c) 3-c3 c3 faa= 3(E/)a + 3(£/)b +3(E/)c a b c +(GA')a +(GA'h +(GA')c

J

1 [ (a+b)2(2a+2b+3c) b2 Ja2 =j"23 = 6(El)a -b 2 (2b+ 3c) + 6(Elh(2b+3c) a b +(GA')a +(GA'h f 31 =.1;.

3 = 6 (~~)J 2a+3(b+c)] +(G~')a

.................... (86)

To obtain f 22 and .1;.1 from the expression for faa. omit in th~ latte~ the terms c and b c respectively. Also to find.l;. 2=f21 omit the terms mvolvmg c in the last of'Eqs. (86). Naturally, we can also derive the influence coefficients (86) by direct integration of the differential equation for the deflected beam when shear deformations are included. A systematic method for deriving the flexibility coefficients for compound engineering structures is given later. . Influence or flexibility coefficients are of great importance in the static and dynamic analysis of linearly elastic engineering structures. In this connexion it is most appropriate to make use of the matrix notation not only for conciseness of presentation but also for the ~ystemati~ progr~m­ ming of the considerable computational work usually mvolved m practical problems. The matrix algebra is, in fact, ideally suited for the automatic digital computers now available. The matrix form of Eqs. (82) is, r=FR .................................. (87) where r and R are the column matrices of the displacements and forces

I

II r=

I~:

={r1

'2

'a},

I~:

R=

L__i

and

L

Uh2 .h~ ll

f= /21 /22 /2a fa1 fa2 faa

L

I

II

={R1 R 2 R 3 }

....

(88)

_I

................................ (89)

_I

is the so-called flexibility matrix; note that F is a symmetrical square matrix. The relation (87) is, of course, valid for any number m of displacements and rotations in any linearly elastic body. To each displacement or rotation r; there corresponds a force or moment R;. Thus, in such a case the matrices are and

~r-1~ !i.2_ ~ ·_ ·_·_·_I;~ ~ ~ ~ :-!i.-:l f=

~;~ !:_2_ ._. _·:: !:_;_. _· _·::: ~-m-1

.................. (91)

/m1/m2 • • • • /m; • • • · • fmm

_j

where the/;~> can be calculated always from Eqs. (84). The flexibilities in (91) need not necessarily refer tom different points. For example, we may choose three directions x, y, z at a particular point of a three dimensional body and define six flexibilities J.,.,,/yy,Jzz J.,.=J• .,,f•• =f•• ,J•.,=J.,. corresponding to the three forces R.,=1, R.=1, R,=1 at the same point. Similarly for a beam in which we assume that the engineers' theory of bending is true we may require the slope and the deflexion at a crosssection under transverse force and moment applied there. Three flexibilities are required for this information; note, however, that if shear deformations are included we must specify that the bending moment is applied as engineers' theory direct stresses at the particular cross-section. A characteristic property of the influence coefficients is that any/;~> in a given elastic body depends only on the points and directions j and h but not on any other directions selected for the calc].Jlation of a flexibility matrix (91). A perusal of the flexibilities of Eqs. (86) shows that it is possible to split the complete flexibility matrix F into two additive and distinctly different matrices. Thus, f=FB+Fs ...................................... (92) where FB and F8 are the flexibilities corresponding to pure bending and shear deformations respectively. The first contains only terms involving Eland the second only terms involving GA'. For example,

a3 a b hiB=3(El).J23s=(GA')a +(GA'h

.................. (92a)

Such a splitting of the flexibility matrix is extremely useful in numerical calculations, particularly when obtaining first approximations in which we neglect certain flexibilities. Thus, in a first approximate wing analysis we may neglect the rib deformability; at a later stage we can ascertain its influence by adding the corresponding flexibility matrix to the original flexibility (see Example (b) of Section 9). The method is, of course, quite general as Eqs. (84) and (84b) show, for it is always possible to write I ·fih = f [(a xx;Euh +a._.;Eyyh +a zz;E zzh) +(axy;Exyh +a·yz;Eyzh + V a,x;Ezxh)]dV ............ (93) where the first expression in round brackets gives the contribution of the direct strains to /;h whilst the second expression gives the contribution of the shear strains. Eq. (93) shows also that the splitting may be carried a step farther by considering separately, for instance, the effect of the strains Eyy or of the three shear strains Exy, Ey., E zx· Similarly, in fuselage ring analysis where we usually neglect the deformations due to shear and normal forces we can check their influence by adding the corresponding flexibility F8 and FN to the matrix FB for pure bending deformations. These matrices are, FB=[J ~;whds} FN=[J 'i~hds} fs=[J ~~~ds]

........... · (9~)

where M;, N;, S; (Mh, Nh, Sh) are the bending moment, normal and shear forces due to a unit load atj (h). It is often convenient not to operate in single loads (or moments) but in groups of loads (or moments), which are known as generalized forces. To fix ideas, consider that in the example of FIG. 20 we select as applied generalized forces the three loads QI, Q2 and Q3 given by,

t ................

I

(95)

or in matrix form Q=GR (96) where G is the square matrix, in general not symmetrical, of the coefficients G;h· We call G a load transformation matrix and assume that it is non-singular, i.e. that the determinant I G I of the coefficients is different from zero. We may solve Eq. (96) for R by premultiplying with G- 1 and obtain R=8Q where } ................ (97)* is the so-called reciprocal or inverse matrix of G. Its determination is equivalent to solving Eqs. (95) and therefore involves considerable numerical labour if the number of equations is large. In such cases approximate methods may have to be used. However, with the advent of the automatic digital computers this difficulty is no longer insuperable. We give later in this Section a systematic procedure suitable for punch-card machines for computing the reciprocal matrix but hope to return in greater detail to this and similar questions in Part III. Next we have to determine the generalized displacements q corresponding to the generalized forces Q. By definition q are obtained from the equality of the two expressions for work in the two sets of variables R, r and Q, q. Thus, in matrix notation W=tr'R=tq'Q .................................. (98) where r' and q' are the transposed matrices of the column matrices r and q and are hence the row matrices r' = [r1 '2 ra], q' = [q1 q2 qa] · · · · · · · · · · · · · · · · · · · · · · · · · · (99) Using the first of Eqs. (97) in (98) we obtain q'=r'8 or q=8'r (100) where 8' is the transpose of the matrix 8, i.e.

I

I Bn B21 Bal I 8'= I~~: ~: ~:I

I

L

........................

(101)

_I

Substituting (87) into (100) we find, q=8'FR=8'F8Q=F.Q ........................ (102) Eq. (102) shows that F.=8'F8 ........................................ (103) is the flexibility matrix corresponding to the generalized forces and displacements Q, qt. We illustrate now the application of generalized forces on a simple • In practice the generalized forces are more naturally defined directly by the matrix B ofEq. (97). t Formula (103) is also given by W. J. Duncan, Mechankal Admittances and their Applications to Oscillation Problems, A.R.C., R. & M. 2000, 1947.

19

and therefore

r---a--l---a--1----a-----j B

A

I

I

r=

I - 0·6a+ 1·5a -0·4a 2·4a+ I·Oa+0·4a

--3 · 6a ·I ·Oa--0·4a

_j

.

~~-st

l

-Nil

Q1 mode

(Q,=-6a)

I

----~-3-&Q

_.............--,-·511--L

I

1

where

1

L

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

(104)

0

8 28 54

I

_j

a3

Q2=1

fv=

for tbe assumed load transformation matrix, 1 I 11 0· 5 -1 1 -0·5 0·5

L

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

(105)

0

_j

Using Eq. (97), R=8Q=8 {0 l 0} we deduce immediately, R={R1 R 2 R3}={1 0·5 -0·5} Hence, r=FR={a/2 -a -5a} The values ~f Rand rare shown in FIG. (21). Next we find the generalized flexibility F • by straightforward matrix multiplication from Eq. (103),

10~

1 -6 -1 F 0 =8'F8=a 1126 -6 2·5 10 -1 1·5 L _j

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

(106)

We calculate now the generalized displacements from Eq. (102) as, q=F.Q=F.{O 1 0} or q={-6a 2·5a -a} Finally, we analyse the three generalized displacements q in their r-components. From Eq. (100), r=G'q =(8- 1)'q =(8- 1)' { -6a 2 · 5a -a} For the inverted and transposed matrix of 8 we obtain from Eq. (105), I

(8- 1)'=G'=

I 0·1 0·6 0·41 0·4 0·4 -0·4 0·6 -0·4 0·4

I

L 20

_j

0

0

0

0

0

0

0

0

0

0

0

0

0

5: I _j

0

0

0

(107)

I

0

0

0

0

0

0

0

0

(108)

Bm1 ...... Bm; . . . . . . Bmm

_j

I0

fvu 0

...... 0 ...... 0 fv22· . . . . . 0 ...... 0

I I0

0 ..... .fvii· ..... 0

I I0

0

I

We seek now the components of forces and the generalized displacements corresponding to the generalized force,

~,01

L

I

and is not, in general, symmetrical. However, the transformation (103), called a congruent transformation, ensures that the flexibility matrix F • is still symmetrical. Attention is drawn to the dual relationship (97) and (100). Thus, if we transform a load system R by the transformation R=8Q the corresponding displacements r are transformed as q=8'r .............................. (109) It is often required to find the set of forces and displacements P, p for which all cross-flexibilities fvih (when jc#h) are zero. These are, of course, the elastic eigenmodes corresponding to the set of displacements r 1 to rm. The load-displacements law is then given by p=FpP where F v is the diagonal matrix,

a= 6 El ........................................ (l04a)

8=

I

I

Bil ... ... B;; .. .... B;.,. L

example. Fro. (21) shows a uniform cantilever of bending Stiffness £/and negligible shear deformability. The flexibility F for forces R1 , R2 , R3 at the equidistant points B, C, Dis easily found from formulae (86) to be,

125 165 288 1

a/2

8 11 . . . . . . B1; . . . . . . B1 m

8=G- 1 =

Fig. 21.-Generalized force and displacements

F=a

I

in agreement with the previously given values of r. Each of the three columns of the intermediate expressions represents obviously the rcomponents of the corresponding q-coordinate. Fro. (21) illustrates in detail the three q-modes . Naturally, Eqs. (97), (100) and (103) are valid for any linearly elastic body and any number m of forces (moments) Rand displacements (rotations) r. The load transformation matrix takes then the form

r, =-sa

-0-61

I

I

L

I

............ (110)

...... 0 ..... . fvmm

_j

Thus, P1=fvuP1, ...... ,p;=/v;;P;, ...... ,pm=fvmmPm

.... (Ill)

It is always possible to find the unique load transformation matrix 8v

which transforms our system R, r into the orthogonal system P, p. We do not enter here into its detailed derivation since the reader can consult a number of textbooks on this subject.* Our above considerations and in particular Eqs. (84) and (84b) show that the ftexibilities are particularly simple to derive for a statically determinate structure, e.g. the beam of FIG. 20. For a redundant structure we must first find the forces or stresses in the redundant members before we can obtain the true strains €; for the unit loads. The necessary analysis is developed later but it is helpful to give here a formal matrix derivation of the flexibility F of an engineering structure, the stress distribution of which is known. To this effect we use again the unit load method given by Eq. (7la). We denote by eh, sh two column matrices for the true strains and stresses respectively, corresponding to a unit load Rh = 1 at the point and direction h. Thus, }

0

0

0

0

0

0

0

0

0

0

0

0

(112)

where the elements of these matrices may, of course, vary with x, y, z. It is always possible to write eh=fvsh ........................................ (113)

where fv is the flexibility matrix of a unit cube at the point x, y, z. Thus for an isotropic body • See Frazer;Duncan, Collar, Toe. cit. Zurmuehl, foe. dt.

I fv=

I 1/E I -viE I -v/E

I

-v/E -v/E 0 1/E -v/E 0 --v/E 1/E 0 0 0 1/G 0 0 0 0 0 0

0 0 0

0 0 0 0 1/G 0

-l

0 0 0 0 0 1/G

I I I I

........ ( 114)

I

_j L Let also 5; be a column matrix for any statically equivalent stress system corresponding to a unit load R; = 1 at the point and direction j. Thus, i;={CT~..~i1 11 u}Gzzjij;,·yiUvzi1z.-:i1 • • ••• · · . . . . • . • . . . · ... · (115) For a statically determinate system only one 5; system can be given-the true one, 5;. We derive now the flexibility coefficients};h from the unit load method as, 1·/;h= Js;'ehdV= fs/fv5hdV= J sh'f~·5;dV= I ·J,; v v v (116) I f;; = f 5/fv5;dV v Hence the total flexibility matrix F for m points and directions is, F=[/;h]=Js'fv5dV ............................. . (117) v where 5, 5 are the partitioned row matrices 5 [51 ...... 5; ...... 5m] 1 (118) 5=[5 1 . . . . • • 5; ...... sm]

j ··········

We shall apply now formula ( 117~ t~ an engineering st~ucture consisting of any numbers of simple elements JOtned together at the1r ends
f

b.,

0

0 ..... 0 f,, ..... 0

0

0 .....

0

0 ..... 0

The matrix b has submatrices with m columns. f is a partitioned diagonal matrix whose elements are the flexibility matrices f •· We denote now by b a matrix whose m columns are loading systems on the s members statically equivalent to the external loads R1 =I, R2 = I, ..... , R,. =I respectively. If we choose these systems to be also elastically compatible then, of course, b=b .............................. .. (124) Applying now the unit load method and using Eq. (122) and the transpose matrix -b' we find by an argument similar to that leading to Eq. (117) that the deflexions rat the points of application and in the directions of the loads R are given by r=b'fbR ............................ (125) Therefore the flexibility matrix F for the prescribed m directions in the complete structure is f=b'fb ............................ ( 126) The matrix operations in (126) are again congruent and thus F is indeed symmetrical. Eqs. (123) show that Eq. (126) can also be written in the form f=~.b'.f.b. . ..................... (126a) Eq. (126) is the general expression for obtaining the flexibility of a complete structure from the flexibilities of the constituent elements. The configuration of the elements is said to be in series since the assembly condition is expressed by the matrix b which derives from conditions of equilibrium. Thus, Eq. (126) may be regarded as the most genera/formulation of the flexibility matrix of a structure consisting of elastic elements in series. It is also clear why Eq. (103) for the flexibility matrix of generalized forces has the same form as Eq. ( 126). In the first case we derive generalized forces from single forces and in the second internal forces from external forces but in both cases this entails a linear transformation matrix B or b. Note also that F is in the first instance the flexibility matrix of the complete structure for the single forces and f in the second instance the flexibility matrix of the individual elements. It is seen, however, that whereas B is always a square matrix b is, in general, rectangular. Before illustrating the application of Eq. (126) we draw attention to an interesting dual relationship (see alsop. 20). Thus, Eqs. (121) and (125) prove that if the internal loads S are derived from the external load system R with the relationship S =bR .............................. <121) the deflexions r at the points of application of the R-loads are found from the internal relative displacements (strains) v from the relationship r b'v b'v ...................... (125a) Naturally, Eq. ( 125a) merely restates the unit load theorem. We stress again the fact that 6 need only be the matrix of statically equivalent stress systems. Illustration of Eq. ( 126). We observe first that Eq. (126) includes as a particular case Eq. (92) for the splitting of a flexibility matrix. This may be seen as follows. Splitting the flexibility matrix is equivalent to considering the combined effect of two or more geometrically identical structures (elements) to each of which is assigned only part of the complete flexibility of the structure (e.g. flexibility in bending or shear, or normal force). Thus, the constituent elements are in this case geometrically identical and hence the load transference matrices b, etc. are merely unit matrices I. If then, flexibilities of each of the elements are written as f., f 0 etc., we find

F cc [I I .... I ..... I]

f"

0 ..... 0 ..... 0

0

fb ••••• 0 . . . . . 0

0

0 .....

0

0 ..... 0 .....

f, ..... 0

..... 0 ..... 0

f, .....

. . . . . . . . (123)

0

. . . . . f,

-I *The usual presentation of the strain and strC''\S matrix as a 3 / 3 squar~.. 111 ~ ... !.""~ Oi n:•nsor is r.''~.. _,. suitable for our considerations her('. Se~· also ~e'-·twil uu. There too attention js drawn to the facl that i i need only be determined in the most suitable statically determinate svstem.

f,

=f,+f,+ .... +f"+ .... +f, ........ (127) q.e.d. The order of the unit matrices and the f matrices ism, the number of

assigned directions in the structure. Consider next a beam built-up by two uniform component beams a and b as shown in FIG. 22. We seek the flexibility matrix for the transverse forces R1 , Ra and moments R2 , R4 under the assumption that the E.T.B. holds and shear deflexions are negligible. We analyse first each beam separately as a cantilever built-in at the L.H.S. and subjected to transverse force and hPnding moment at the tip, the signs of which are taken to be those of the

21

where

lOtS(+) \Ela

0

A

'D~M<+>

tR1 R2Jf"""""\

\Elb

8

Element a

tR

3

b

Element

c/>a,h=(L).,

R4 SJ

cl

(132)

I

I

L

lb

. . . • . . . •• • . . . . . . . . . . . . . .

gives the flexibility per unit length of the cantilevers. The negative sign in the cross-flexibilities arises from the sign convention. The total flexibility follows as I I I f. o I .......................... (131a) f= o f"

r, la

b

_j

Applying now Eq. ( 126) the flexibility F of the complete structure is f=b'fb=b' .f.b.+b' bfbbb ..••............•..•..•. (133) In the present case ii = b since the system is statically determinate. The deformations of the structure may finally be obtained from r=FR ............................. ... (87) The expressions for the deflexions r1 and r3 derived from Eq. (133) agree with those of Eq. (86) when the shear deformations are neglected. An alternative approach to the problem is to express the loading on the component beam by the end moments (see FIG. 22b). The internal loading matrix is now,

~

.................... (134)

where I 1

_,I I

-1. -I 0 -1

-- 1

I o o

~J G -1£-/..................;...;....;,_A Ele men!

b

I

I

fla,b= i/6El

corresponding + S (shear force in the beam) and + M (bending moment in the beam). The tip deflexion and slope are fixed as positive if they are in the direction of the positive applied shear and end moment respectively. With this sign convention, MBa= -R2 -R4 -R3lb

Mcb= -R4

} ............ (128)

The loading matrices s. and Sh of the two elements are accordingly I s.={SBaMBal=b.{Rl R2 Ra R4}=b.R (129)

J............

where b.=

I

I~ L

0 -1 -lb

j

j

_j

and hence b=

II I b. I

.. .. .. .. .. .. (130) _j

· · · · · · · · . · ................ (130a)

I bb I

The flexibility matrices f., f b follow immediately from I I f,,

b

=

f3

22

(22),

12

. . . . . . . . . • . . • . . . . . . . . .

1

tcp I •. _j

b

.. ·.............................

<13t >

(136)

I

3'~' _j

(137)

where, lfla

fl =

l

o

0

flb

lI

I

.......................... (136a)

_j L It is easily seen that the flexibility matrix F of Eq. (137) is identical with

that of (133). We observe that the cantilever position of the constituent elements is derived merely by a rigid body rotation from the simply supported beams shown in FIG. 22b. No additional total virtual work is associated with such a linear transformation and the flexibility F is, hence the same. The form (136) of the flexibility of the element is important for it applies to a linear variation of any deforming stress, force or moment as long as we substitute forcp the appropriate unit flexibility. Thus, for a normal force varying linearly in a beam with constant direct stiffness EA, we can use Eq. (136) with cp= 1/ EA . B. Stiffnesses We return now to Eqs. (87) and solve them for R 1 to Rm to find equations of the type R1 =k 11 r 1 +k12r 2 + ...... +kvr;+ ...... +k 1 ,.rm I R2 =k21 r 1 +k 22 r2+ ...... +k 2;r;+ ...... +k2mrm I R;=k;lrl +k;2r2+ 1 1

I

3cfo -2cfo I

I -2cfo L

J2

FIG.

L

Hence,

bI

11-~. 1-~. I

~-~ ~+k;;r;+~. ~-~k;mr,..- ~

-----1 -=---------Rm -km r +k,. 2r + . ..... +k,.;r;+ . ..... +k,..,..r,..

L .J

I

l 1cp

I 6'~'

Fig. 22.-Break-down of cantilever for calculation of flexibility matrix

Scb=Ra

I .................... (135)

I o o I I _j L The internal flexibilities fa and fb derive in this case solely from the end bending moments and to find them we have to consider only the end slopes of the simply supported beams shown in FIG. 22b; thus, taking end slopes positive in the direction of positive moments we find

M~

MBb

-- 1

2

· · · · · · · · 0 38 )

.J

The coefficients k;i and k;h are known as direct- and cross-stiffnesses in the directions of the selected m displacements. In fact, it follows directly from Eqs. (138) that the general stiffness k ih is the force (or moment) applied in the direction j if we displace the body by rh =I whilst keeping the remaining (m -I) r's at zero. Using matrix notation the solution (138) of Eqs. (87) may be written as R=F- 1 r=Kr ............................. ....... (139)

where R and r are the column m:1trices defined by Eqs. (90) and K is the m x m stiffness matrix I I k 11 ...... k 1; ...... k 1.,. I

K=

.................. (140)

kn ...... k;; ...... k 1, 1 kml• · · ·

km;• • · · . . kmm

L _j The stiffness matrix may be determined either directly or from the identity K=F- 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . (140a) by inversion of the flexibility matrix F. Eq. (140a) shows that K is symmetrical, i.e. k ih cc k hi • . . . • . . • . . . • . . . . . . . . . . . . . . . . . • ( 141 ) This may be seen also as follows: Let a1, E1 and a\ E" be the stresses and strains corresponding to unit displacements r1 =I and rh =I respectively while all other r displacements are kept at zero. Applying now the principle of virtual work or displacements to the true state j (h) and virtual state h(j) we obtain l·k;h=fci'E 1 dV=fa 1 EhdV~cl·kh; ........................ (142)

v

v

Where ci'E 1 etc. stands for an expression analogous to Eq. (84a). This application of the pr nciple of virtual displacements, by an obvious analogy with Eq. 84, is called the 'unit displacement method'. We remarked on page 19 that the direct- or cross-flexibilities depend for a given structure only on the points and directions to which they refer. This is not so for the stiffnesses which by definition depend on the complete set of points and directions selected to describe the stiffness of the body. Thus, if we choose an additional direction m + l to augment our description of the elastic behaviour of the structure all the original k;h will in general change whilst the jjh remain unaffected. Consider again now the example of FIG. 20. A study of Eqs. (86) and (140a) shows that the stiffnesses k 1" corresponding to unit deflexions at B, C and D are considerably more complicated than the expressions for the flexibilities. However, this is not always the case. Naturally, we can calculate the stiffnesses directly. For example, the k 13 may be obtained by analysing a continuous beam built-in at A and simply supported at B, C and D at which last support there is a fixed 'give' of unity. We may solve this thrice redundant problem either with the three-moment equation or by the slope-deflexion method. Assume now that not only transverse forces but also moments are applied at the junctions of the component beams and at the tip (FIG. 23). To simplify the argument we ignore, moreover, the effect of shear deformability. The modes and stiffnesses corresponding to unit displacements in the directions, I, 2, 3, 4 can now be determined very easily. For example, for the modes r 1 =I and r 4 = l shown in FIG. 23 we find respectively,

12( ~{) + 12(~{) k31 12( ~:) k21 -6( ~:)a +6( ~:) k41- 6( ~:)I> k44=4(1!{). k24=2(~1 )b

ku =

n

0

=

b

•

=

Fig. 23.-Stiffnesses of a cantilever

is the stiffness per unit length of a bar. The stiffnesses at a point associated with the unit displacements at the same point are, in fact, already derived by the method of vittual displacements in Example 5b where they appear as the coefficients to the displacements u, v. Naturally, the problem of deriving the single stiffness k22 at the point 2 from the set ( 145) still remains. A general method for solving

R,.

r,,

R,

r,,l

1,

...... (143)

h

'

................ (144)

,d,---~..::...-~-~-~_1

The important point about this example is that it shows how easy the determination of the stiffnesses can be once we consider all possible modes of deformation a_(joints connecting simple component elements of a structure. Another example will help to clarify the argument further. Consider the symmetrical framework of FIG. 24 and assume that we seek the flexibility or stiffness at the central point 2 for vertical displacements. In the first case we must solve a thrice redundant problem and in the second a four times redundant problem with a central unit "give'. If, on the other hand, we select the complete set of stiffnesses corresponding to vertical and horizontal displacements at all movable joints then the calculations are most simple. In fact, for the typical cases shown in FIG. 24 we find by inspection, Kh Kd(h) 2 Kh k2.2=·7i+2d d ' k4.2= -71

Kd ha k7,2= -ku.2={j {j2

,

ks.2=k12.2='

Kd(h) -d (J

2

k,~-·~1<,1

k1.2 =k3,2 =ks.2 =ks.2 =k9,2 =kw.2 =0 .. (145)

,

k s.1-k 9,1--~ a

I k2,1 =k3,1 =k4,1 =ks,I =k10.1 =0 where K=EA

c

_j (145a)

Unit

stiffness of

all

honzontal

Un!l

sttffness of

all

vertrcal

Untl

&ttlfness of

all

dtagonal

bars bars bars

'Ka ')(."

'Kd

Fig. 24.-Stiffnesses of pin-jointed framework

23

this and related problem" ts given further below. It is characteristi..: that the direct calculation of the llcxibilities corresponding to ( 145) is as complicated as that for the single flexibility at 2. In both examples we see that the stiffnesses are determined most straightforwardly, in fact, practically by inspection, once we find the set of unit deformations for which it is simple to calculate the strains and hence stresses and forces. The advantage of first deriving the stiffnesses may be particularly marked in highly redundant structures but it requires, as the example of FIG. 24 shows, the consideration of many degrees of freedom which may also have its disadvantages. On the other hand flexibilities are always easier to calculate if the stresses corresponding to unit forces can be found without difficulty as in statically determinate structures. We gave in Eq. (142) a general formula for the determination of the stiffnesses. Let us consider it again in more detail. Observing first that, while ui and if must be the true stresses corresponding tor;= I and rh = I respectively, the strains Eh and €; need only be virtual strains (h and (i (i.e., compatible but not necessarily statically consistent strains*), corresponding to rh =I and r; =I respectively; this may contribute to a considerable simplification of the calculations. However, the actual practical use of Eqs. ( 142) rewritten here in the form k;1 =Jai£.1dV v .......... (146) k 1h =.fif£ 1dV = fai€hdV -=k 1.;

v

I

and

L

I kvu =f..,-' 1)11

• See Section 4 •

24

D

I

';=04

t;=

-0 4

fii~-----------R-•.._fo_•_••_>l_a_ ____,R11-o-006l/a O, mod"

(Q, =- o-oo62/Q)

,

0

Ii

..... k,,;; ...... 0

RJ

o

...... 0

q2

kp; 3 --r

I

} pn

~,I

i--

1I

0·0169

--0·00615 K =-~-0·00615 0·548 '' a -0·117 0·406 L where a is given by Eq. (104a). Hence, Q=K.q=K.(O I 0} or

-l

-o · 117

, .. • .. • ,

I

kvmm =~,--

pmm

.................... ( 152)

1

0·406 0·218

1

I

.......... (l06a)

_j

Q=l{ -0·00615 0·548 0·406} We analyse finally the generalized forces Q in their R-components. From Eq. (97) or ( 147)

r-

Io R=BQ~-. I II L_

I

Ill

I I -0·00615 I 1 0·548 I I 0·406 I .J .J L

0·5 --1 -0·5 0·5

and carrying out the multiplication, I 0 +O· 548 +0·4061 R=,-1 -0·006 +0·274 -0·406 a -0·006 -0·274 +0·203 I L j

.J

o-2Dl/a

The displacement transformation matrix A of Eq. (147a) is given by Eq. ( 107) I I I 0·1 0·6 0·41 A=G'=(B- 1 )'= I 0·4 0·4 -0·41 ............ (107a) I 0·6 -0·4 0·41 [ __ .J Hence r~Aq,=A{O 1 0}={0·6 0·4 -0·4] The generalized stiffness matrix K,, is best obtained by inversion of F,, in Eq. ( 106) and is

I

· · · • · · kvmm

Q3 mod•

Ill,= o•os/al

We illustrate now in FIG. (25) the application of formulae (147)-(149) on the simple example of FIG. (21) and for the same load transformation matrix B of Eq. ( 105), but seek here the components of forces correspond· ing to a generalized displacement

a

............ (151)

t

Fig. 25.-Generalized displacement and forces

I I

II o

_ _j

I

1

I

0

I

I

a-

c

B

v

is somewhat limited. This follows from the previous discussion which shows that stiffnesses are best found either by considering all possible degrees of freedom at the joints, in which case the determination of the k's is usually performed by inspection, or by inverting the flexibility mJtrix F. Moreover, even if we calculate the stiffnesses k for a restricted total number of degrees of freedom at the joints (e.g. example of FIG. 20) Eqs. (146) are really superfluous. Thus, in the example of FIG. 20 we have to find the true stresses a fvr a four times redundant structure, the analysis of which includes the derivation of the forces k;h and the use of Eq. ( 146) is hence unnecessary. Nevertheless, Eq. 146 is of considerable value when the elements into which the structure is broken down are characterized, not by simple loading systems (e.g. beam elements or bars) for which the k's are determinable by inspection, but by simple (assumed) displacement patterns. An example of this application is given in D of this section. Also given later is the matrix formulation of Eq. 142, which is most useful in practical cases. We now find a generalization of the concept of stiffness corresponding to the generalized flexibility given on page 19. Thus, following the argument there we introduce the generalized displacements q and forces Q defined by r·· Aq Q=A'R .............................. (147) where A=G' =(B- 1 )' . . . . . . . . . . . . . . . . . . . . . . (147a) see also Eqs. (96), (97) and ( 100). Substituting the expressions for rand R in Eq. (139) we find immediately Q=K,,q .............................. ( 148) where K0 =A'KA ............................ (149) is the generalized stiffness corresponding to the m generalized displacements q 1• Eq. (149) may naturally also be derived by inversion from Eq. (103), i.e. K.=(F.)- 1 . . . . . . . . . . . . . . . . . . . . . . . . . . (149a) The particular linear transformation B (or the corresponding matrix A) which reduces the cross-flexibilities[;h to zero nullifies also the corresponding cross-stiffness k;h· In fact, we obtain from Eq. (1o::l) P=KvP .............................. (150) where Kv is the diagonal matrix I I lkv 11 0 ...... 0 ...... 0 \0 kv22······0 ...... 0

K"=

f--·--·- a - - - + - - a A

I

I

I

-0·077/a

·954/a I I +O_ -0·138/a i

L

1

.J

Each of the three columns of the intermediate expression represents obviously the R components of Q1 , Q2 , Q3 respectively. We can check now the previously given result for r from r=FR where F is given in Eq. (104). Next we derive a general formula for the stiffness K of a structure consisting of a finite number of simple elements. The expression given is the matrix formulation of Eqs. (146) and corresponds to the flexibility matrix

F of Eq. (126). Since the analysis follows closely the arguments on page 21 we need present here only the outlines of the proof. Consider again an assembly of s structural elements joim•d together at their ends or boundaries. m displacements r are selected to describe the stiffness K of the complete structure. Let k. be the stiffness matrix of the g element due to the characteristic strains of the element arising from the displacements v. at the boundaries. Naturally, there are usually several different but equivalent possible ways of expressing the straining of the element. Let a. be the matrix, in general rectangular, which transforms the displacements r into the true strains v. of the element. i.e. (153) Then s.=k.v.=k.a.r ........................ (154) is the matrix for the forces (moments, etc.) applied on the element due to the displacements r. The internal force (or stress) matrix S of the aggregate structure is now given by S=kar .............................. (155) where S={SaSb ...... 5 0 • • • • • • S,} (!56) and a={a.ab ...... a •...... a,} .................. (157) k is the symmetrical diagonal partitioned matrix, ...... 0 0

......

o

TABLE I

Duality of Force and Displacement Methods (it is always possible to substitute a, b for i, b respectively) of

Method

Force

I

Q

Displacement

I

s

S = bR k= 0

k •......

0

................ (158)

of

r= 6'v

0

f

S)

Flexibility of complete

0

...... 0

......

k.

_j

Applying now the principle of virtual work, taking the internal forces S and external forces R as the true state and selecting as virtual state the internal strains corresponding to unit displacements ' I = I, r 2 = 1, ...... , rm = 1 respectively we find R=a'kar ............................ (!59) where a' is the transpose of a. Thus, the stiffness matrix K of the compound structure is K=a'ka .............................. (160) Eq. (160) may also be written as K=~a'.k•a• ........................ (160a) g

Since the virtual strains need only satisfy the compatibility but not necessarily the equilibrium conditions we may select for the virtual states a simpler matrix a which satisfies only the former. Eq. (160) becomes then K=a'ka ............................ (l60b) However, the application of a possibly simplified matrix a is really not required in practice. As mentioned on page 23 the stiffness matrix K is best calculated for all degrees of freedom at the joints, yielding very simple matrices. The configuration of the elements of the compound structure is said to be in parallel in Eq. (160) since the assembly condition is expressed by the matrix a which derives from conditions of compatibility. Thus Eq. (160) may be regarded as the most general formulation of the stiffness matrix for a structure with constituent elements in parallel. It is immediately apparent why Eq. (l49) which expresses the stiffness matrix for generalized displacements must have the same form as Eq. (160). In the first case we derive generalized displacements from single displacements and in the second, internal strains from external displacements. In both applications this entails a linear transformation matrix which, however, is a square matrix in the former case. Also K is the stiffness of the complete structure for the single displacements while k is the stiffness matrix of the individual members. Eqs. (153) and (159) show that there is a most illuminating parallel development to Eqs. (121) and (125a). Thus, if the internal relative displacements (strains) v derive from the external displacements r with the relationship v=ar .............................. (l53a) Then the external forces R derive from the internal forces (stresses) S with the relationship R=i'S=a'S .......................... (l59a) Eq. (l59a) restates, of course, the principle of Virtual Work.

Addition

.

(Spe-i:ial

r= r .

F.

+

Stra~n

of

Stress

on

Assembly

elements

elements

Stillness of

s k

complete

structure

K = a'ka Addition ( Sp•ci.al

a•umbly)

k•

b

~

F. =

v

V>

(for strains

Flexibilities

of uriu

Parallel

Stiffness of E'lements

structure

6'f b

F=

Q :.A'R = Kqq

R=aS

Flexibility of elements (for stresses

Force

v = ar

v

elements

KA

t

Generalized

ASSE'mbly

Stress on elements

Strain

Kq=~

Generalized

q: B'r = ~Q Series

Generalized 1 Stiffness

~ Kq=l= Kq~

t

q

r=Aq

1 l

F~exibility ~ A'B=l=B'A

BFB

Generalized

R

'Generalized Displacement

I

Generalized

K

Force +

r

Force

~=

r

I Stiffness

FK=I=KF

R:BQ Generalized

Displacem~tnl s

Displacement

F

+ Displacement Generalized

of

Method

R

Flexibility

I

...... 0

Forces

1

-r

~

--~

F

K.

Of

Still nesses

parallel

aaembly)

.riJ?-

Before illustrating applications of Eq. (160) we draw attention to the by now all too apparent complete parallel between the flexibility and stiffness approach in the analysis of structures. We may express this concisely by the tabular arrangement under the two headings: 'Methods of Forces' and 'Method of Displacements'. The analogy between the two methods is developed considerably in what follows and is shown in greater detail in TABLE 11. Illustrations to Eq. (160). Consiqer the beams I and II of FIG. (26) joined by inextensional bars which connect the set of points B, C, D and B', C', D' respectively. Let KI and K 1 I be the stiffness matrices of the upper and lower beam respectively defined for vertical displacements 'I• 'a and r 5. From the definition of the stiffness it follows immediately that the stiffness K for the displacements 'I• 'a and r 5 in the compound structure is given by K=K 1 +Ku .........................• (161) This simple result may also be derived from the general Eq. (160). For in this special case the joint displacements r etc. of the complete structure and the straining displacements vI and vI I of the component beams are the same. Thus, VI=a1 r, Vu=aur .............................. (162) where and a 1 =au=l=

We conclude,

II 00 01 01~ 1 0 0

L

........................ (163)

1

_j

25

~~

~

r:.~ c

'i.~ B

I

TI

rs~ 0

j'3

Irs

B'

c·

D'

~

f'~ 'i"

?di

B

f'3

"""' c

I

I

c•

f'S

0

Fig. 27.-Addition of stiffnesses for arbitrary structures

o·

~

Eq. (161) applies to any compound structure in which the stiffnesses are defined for at least all common degrees of freedom associated with the joining of structures I and II. Thus, in the example of FIG. (26) the common degrees of freedom are t~e vertical displacements r 1 , r 3 and r 5 • Formula (161) is, however, still tru~ if we define the stiffnesses of the upper beam and the complete structure for both the vertical displacements r 1 , r 3 and r 5 and the slopes r 2 , r 4 , r6 • Then K1 can be calculated by the methods given previously. Ku is still only definable for vertical displacement, the corresponding entries associated with r 2 , r4 and r6 being zero. Naturally, we can define the stiffness matrix K and say KI for points not connected to IJ. Again the corresponding terms of Ku are zero. FIG. (27) shows the joining of two arbitrary structures to give K= KI + KII· Note that at a joint point like (2) we must define the stiffnesses for two displacements, say the x and y-directions. Formula (161) may, of course, also be applied in obtaining the stiffness matrix of the compound cantilever consisting of elements a and b, FIG. (23). Again, we must define the stiffness for all common deflexions and slopes at the joints, assuming the E.T .B. to be true and the shear deflexions zero. The total stiffness K is then K=K.+Kb .......................... (t61b) where the elements of the split matrices may be found from Eqs. (143) and (144) for the displacements r 1 and r4 and similar equations for r 2 and r 3 • Thus, I I El 0 -6[2 0

further by comparison of Eqs. (92) and (161). The first shows the case of additive partial flexibilities for series assembly and the second, the case of additive partial stiffnesses for parallel assembly. A very simple application of Eq. (160) is given by the pin-jointed framework shown in FIG. 12 of example 5b. *Thus, the stiffness k, of the bar corresponding to unit elongation ~I,= I is, -k _ (EA), -~ k, -- r -- I, -I,

0

0

El

-6[2 0

0

L

0

[

~.

a,= cos 8, sine,

0

. ......... { 168a)

[3

[3

L

6£1 [2

4~! I

2£/ I

6EI

2£1 I

41!! I I I

-

[2

..... 0

L:?

sin

I I I I

I

.

a

I

.... • Kn/1, I _j

2

I

I kn

e, I I k _j

L

Y>

I

kxy

I

kyy

I

........ (169)

_j

in agreement with Eq. (67). ............ (164a)

_jb

where the columns and rows refer to displacements r 1 , r 3, r 2, r4 respectively. Formula (16lb) may be generalized for any numbers of component beams, and for any structure in which the joint displacements r expre3s also the straining displacements v of the elements (i.e. a= I). In such cases the stiffness matrix K can be written K=Ka+Kb+ ...... +K.+ ...... +K•............ (165) Note that the only non-zero coefficients in K. are the stiffnesses k corresponding to the displacements at the ends or boundaries of the g-element. The flexibility matrix F corresponding to (165) is F= K- 1 =(K. +Kb + ...... +K.)- 1 =(F .-1 +Fb- 1 + ...... + F,- 1)- 1 (166) The parallel between the displacement and force method is underlined

26

0

L

-6£/ [2

.... . K,/1, ... .. 0

I I:~ cos e, sine,

El El El 12 [3 -6[2 -6[2

_ 12£ / 13

El 6[2

0 L

0

I

6£[2/

-1

..... 0 ..... 0

"\'Kr e . eI L.JT; cos , sm , I

I

.................... (168)

and

_ja

12gj -12£/

J

Hence

................ (164)

4£/ 0 I I I ol 0

· · · · · · · · · · · • · . . . . . . . . . . . . . . . . . (167)

where K,=(EA), is the stiffness per unit length of the rth component bar. The transformation matrix a, for displacements in the x- and y-directions

127: 0

I

'6~

Fig. 26.-Parallel combination of cantilever. Addition of stiffnesses

K.=

+/

J---X-~~~ Jt--~

~ ~

B'

1/T/~Ti = 1/l~r

I

I

t r,

~ ~

!'5

tt;

We mentioned previously that the simplest method of calculating stiffnesses is to define them for as many degrees of freedom as are necessary to obtain simple deformation patterns of the elements of the structure. Having calculated such a stiffness matrix it will become necessary to 'condense' it-i.e. to refer it to the smaller number of displacements in which we may be interested. This changes, of course, all the stiffness coefficients, but the necessary analysis is easily arranged in matrix form for automatic computation. Let the original stiffness matrix be of the order m x m and denoted by K We want to find the matrix K referred to p-directions only, where p
0

R={R 1

.....•

Rv Rv+t ...... Rm}={RI Ru}

} .... (171)

in which we write first the p-directions required for the condensed matrix K. Ko may be expressed as a partitioned matrix as follows, • See p. 10.

X,

........................ ( 172)

where K 1 and K 11 are square matrices of order p and 111-p respectively. Eq. ( 170) can now be transformed to R1 =K 1 r 1 +K' 111 r 11 1 (173) R 11 = K 111 r 1 + K 11 r 11 Also, by definition, in the structure with stiffness defined in p directions only

J...................... I

R/·' Kr/

r

R11 =0

(173a)

Putting R 11 =0 in Eq. (173) and eliminating r 11 we find ( 174) Rlc~(K/ ·· K' ///K//-IK 1 u )r 1 and hence comparing with Eq. ( 173a) K- K 1 -K' 111 K 11 -IK 111 . . • . . • • . • • • . . . . . . . • . • . . . (175) Eq. (175) gives naturally the solution to the particular problem of FIG. (24) discussed on page 23. Another example illustrating the application of Eq. (175) is discussed under D of this Section. The above method, is, of course, the basis of the solution of partly homogeneous equations. A parallel relationship exists also in the 'forcemethod' investigation of structures. Thus, in this case, we have to find the flexibility F of a redundant structure in which we know the flexibility F, of the basic structure. The analysis is given under C below.

C.

Calculation of Redundant Structures by the Force-Method We develop now a generalization of the Mueller-Breslau 6 * technique for the calculation of linearly elastic redundant structures. Following our investigations under (A) we could easily formulate immediately the complete analysis in matrix notation. However, since the basic ideas do not appear to be generally known we think it preferable to develop them first in the more standard form. Consider a structure subject to arbitrary external loads R, temperature strains a0 and any other initial strains 7J· We assume that the system has 11 internal or external redundancies xi, X 2 , ••..•. , xi, ...... x, which may be stresses, forces, moments or linear combinations of such (generalized forces). By including the supporting body-assumed rigidin our structure we can denote all redundancies as internal. The stress distribution in the body remains statically indeterminate until an elastic analysis yields then unknown X's. If, irrespective of compatibility we assume the X; to be zero, we obtain the 'basic' (principal or null) system which is statically determinate. This procedure of obtaining the basic structure may sometimes be identified with the process of an actual physical cut of redundant members (e.g. of bars in a redundant pin-jointed framework). However, the simple idea of a cut is not always applicable to continuous structures typical of aircraft. We discuss this point later but for the sake of linguistic simplicity continue to use the expression 'cut redundancy'. Let the stress-system in the basic system be denoted by

"give" at support

-J

Th~

ao

It must obviously be in equilibrium with the applied loads. We describe it as a 'statically equivalent stress system', thus drawing attention to the fact that in its determination only statical conditions enter. We find also in the basic structure the stress systems

due to

a 1 ,a2 , •••••• ,a;, ...... all

XI= I, X2= I, ...... , X;·~ I, ...... , X,'= I respectively. The systems a I, a 2, ...... , a;, ...... , a" are obviously selfequilibrating. Since our structure is by definition linearly elastic the true stresses a in the uncut original structure can be expressed as a

a 0 + ~a;X; i=-1

........................ (176)

Similar equations may be written down for stress resultants (forces or moments). Thus, the problem reduces to the determination of the X's, which as already mentioned, need not be simple forces or moments but can be linear combinations of such (generalized forces). The Equations in the unknown X. We define the following set of deformations in the basic system. O;. Relative movement of ends of cut ith redundancy due to all external causes, i.e. loads, temperature changes, lack of fit, 'give' at the supports, etc.; i =I to n. O;k=Oki relative movement of ends of cut ith (kth) redundancy due to the self-equilibrating load system Xk = 1 (X;= I); i and k take values 1 to n. The 8-coefficients are taken positive if the relative movements are in the positive direction of the X's.

C, =

C

Force appli•d to support by structure due to X1• I.

!!• AI

•

b ( 1 + ll)

a,

b

/'\

~~~X,:l

/

Excus length of typical bar over correct length I

N, •

Tension 1n sam• bar du• to

Contribution to 610 due to give A at suppolt C and excess lengths 1!.1 of bars

X,• I

....

C,A • l: N,AI

Fig. 28.-Singly redundant, pin-jointed framework. Contribution tu from sinking or 'give' of support and excess lengths of bars

010

The 8;, are, of course, the influence or flexibility coefficients of the basic structure for the directions of the redundant forces. We use here the symbol 8 for these flexibilities since it is standard in the literature. To calculate the 8's we apply the unit load method of Section 6D. Thus, using again the abbreviations

+ ..... .

aE=auExx +UzxEzx a7] ~--au7]xx+ . . · . · · +azx7)zx

we find from Eq. (71a), 0;,= fa;(E,+7])dV v

............................. . (177)

8;;'~Ja;E;dV, o,k=fa;EkdV= fakE;dV=8ki

v

v

} .................. (3)

v

................ (178)

where ai, E; (ako Ek) are the stresses and strains corresponding to X;= 1 (Xk=1) and a., Eo are the stresses and strains due to the applied loads. Eqs. ( 178) reproduce, of course, merely Eqs. (84) for the flexibility coefficients. The total initial strains 7J imposed upon the basic system may be separated into thermal and other strains 7J.•·x=a0 +T}xxo• • • • • • · , T}xu =7]xuo• • • • · • • · • • • • • • • · (179) where T}xx•• etc. are initial strains due to say lack of fit, 'give' at the supports. The effect of the latter upon 8; 0 is best considered separately and expressed in terms of the imposed changes of length (rotations) and 'gives'. Consider, for example, the singly redundant framework of FIG. (28) and assume that the manufactured length of the bars exceeds the correct length I by 1:11. Let also each bar be subjected to a different thermal straining a0. We assume furthermore that the intermediate support gives or sinks by the amount /:1. As redundancy we select the force XI in the bar (1, 2) and denote by N_9 and NI the (tension) forces in the bars of the basic system due to the appliea loads and X1 = I respectively. The loading case X1 = l, with the corresponding force applied to support C by the structure is shown in FIG. (28). We find immediately

NI2f

8u = ~ AE ............................ (180)

• See also footnote, p, 17.

27

Applying now the unit load method to the state XI= I and the true total strains (e.+'l) and displacements in the basic system we derive (see FIG. (28) and Eq. (7la)), 0 NIN• . h ( l·810 =~NI(e.+'7)/=~ AE l+~NI(a0!+~l)+a;: I +11 ~ .... (181)

a)

Naturally, we can alternatively deduce by kinematical reasoning the contribution to 810 of the initial elongations a0 and ~/ and the 'give' ~­ However, the unit load method yields the results much more conveniently and systematically. More general formulae for the 8 -coefficients are given further below. The condition of consistent deformations at the cut ends of the n redundancies in the actual structure or application of the unit load method yields the following n equations in the n unknown X. • 811X1 +812X2 + ...... +81;X;+ ...... +81 n X.+810 =0 821 X1 +8 22 X2 + ..... +8 21 X;+ ...... +8 2 .. X,.+8 20 =0

8,.1X1 +8,.2X2 + ...... +8,.1X 1 + ...... +8,.,.X,.+8,.0 =0 .. (182) The solution of these equations determines the X and hence also the total stress distribution after substitution into Eqs. (176) To solve Eqs. (182) by elimination • is particularly simple when they are of the three-moment or five moment type; see for example the tube analysis in Section 9(b). In general, however, all unknowns may appear in each equation and we need a systematic and mnemonic method for the determination of the X. The most convenient method for this purpose is the shortened elimination process of Gausst (known also as Gaussian algorithm). This method is so well known that we need not discuss its formulation in the present 'long-hand' notation but may use immediately the matrix notation. Accordingly, we write the system of Eqs. (182) in the form DX+D0 =0 ........................ (182a) where D is the symmetrical square matrix of the 81k-cocfficients, D. the column matrix of the 81.-coefficients, X the column matrix of the unknown X 1 and 0 a zero column matrix. The Gaussian elimination process reduces the system (l82a) to TX=T0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (183) where T is a triangular matrix, i.e. a matrix whose elements above (or alternatively below) the principal diagonal are zero and T 0 is again a column matrix. For example, I

t 11 ft2

0 ...................... 0 122 0 .................... 0

Consider a system of n independent equations aX=Z .............................. (186) where a is a square matrix and Z is a column matrix. Thus, I I au · · · · · ·at; · · · · · · al" I ------------

a=

ai1

1;1 t; 2 ••••••••• • t;;

L

an 1 ••••.. ani . . . . . . ann

rl-

111 ; l 112 ••••••••••••••••••••

L

1/au

I -a,.!au

I 0 ................... 0 I

I o........ o I ....

-a.1 (a 11 0 .................. 0

L

(188)

l _j

which square matrix has, but for the first column, unit diagonal and zero cross-elements. We find, l 0

ht2 • • •• • .bli

• • • • • .bln b22 ...... b2i •..•• . bz ..

0

- - · - - - - - - - - - - b12 ••••• • b;; ..... . b;,.

0

bn2••••••bni••••••bnn

l

· · · · · ·.... (l87a)

- - --- -- - - --- --

_j

where b1 k=a1 k Ia 11 an dbtk=a 1k -a;tatk -.............. (189) au Next we eliminate the elements of the second column of b, except for b" which we reduce to l, by premultiplying with <2 I 0 .................. 0 0 .................. 0

I

_j

The solution of Eqs. (183) and (184) is straightforward by substitution, starting from the first of Eqs. (183) or in matrix language it is simple to find the inverse T- 1 ofT and write X=T- 1T0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . (185) For the automatic computation techniques now available and, in particular, for the punched card machines it is usually preferable to modify slightly the Gaussian elimination process and obtain directly the inverse of D and hence also the column of X. This method is known as the Jordan technique and we restrict here our discussion to this process. • Natur.ally we may .also use iteration techniques but such methods are n3t discussed in this series of papen. t See: Simultaneous linear Equation• and the determination of the eigenvalues, National Bureau of Standards, Applied !t'fath. Series 29 (1953), and 0. He~k: 'Ueber ~en Zeitaufw'!J'd fiUr das Bereclmea von Detemunanten und fuer das Anftoesen von lmearen Gletchungen.' DISsert. Techri. Hnchsch. Damutadt, 1946. Zurmuehl, loc. cit.

28

1

_j

o...................... ~I

1 0

Inn

.............. (187)

I

b=M 1a= [

.. (184)

li

To eliminate in the first column of a all its elements but au and to reduce the latter to l we premultiply a with the matrix

I

0 ...... 0

..... . ain

-- --- -------

L

T=

••••• • aii

0 ...... 0

0

L

~b,.2(b22

0 ................ 0 1

_j

.... (186a)

We obtain

I

1 0

0 L

where

and

I

Cta •.•..•••.•. . Cln

c23

•..•...••.• . C2n

----0 0 C;a - --- c,.

C=

c2 ~~:=b2 ~~:/b22

0 1

............ ci,

3 ........... . c.,.

0

I I I I

I~ = 1/El

•••

0

••••••

(187b)

I

_j

9 for i*2 .............. (18 a)

b;2b2k

c;k=bik-~

f

X,

The procedure will by now be clear. Thus, at the (g-1) 1h = f1h elimination step we obtain a matrix g of the form I I

1\

.............. (187c)

;\

G

g=

L

1 0

0. 0000.0 l. 0000.0

............ (187d)

h=M.g=

where

0

0 .......... . hAll. ..... . hAn

0

0 ........... h,.A ..... . h,.n

L

__l

Kk•g•"' h•m=K.m Ig•• and h~~:m=g~~:m--g--

••

for k*g .......... (189b)

and so on until the last premultiplication with I :-n1 n/n,.,.

••

L

0

••••

0

fx.

-!'/-

_j

where I is the unit matrix with f columns and 0 is a zero matrix with f columns and (n-/) rows. G is a rectangular nx(n-/) matrix. For the next step, i.e. to obtain the h matrix, we premultiply with M. which has a gth column (188b) { -gl.fg•• -g2ofg••. 000000.1/g••. 0000000-g,..fg•• } resultThe cross-elements. zero and elements and otherwise unit diagonal ing h-matrix is of the form

I

fx,

Y,

0

1/n,.,.

__l

(where I has (n-1) rows) gives M,.M,._1 . • . . . . M; ...... M1a=l .................. (190) Eq. (190) then yields the inverse matrix a-1 =M,.M,._1 . . . . . . M; ...... M1 . . . . . . . . . . . . . . . . (191)* From which we find our unknown X=a- 1 Z ............................ (192) In practice it is usually preferable not to determine a- 1 explicitly but to perform first the multiplication M1 Z and continue then premultiplying to find X directly. It is apparent that if in the above procedure any hM becomes zero the elimination process cannot b: continu~d. An interchange of rows is indicated but this is obviously inconvenient for automatic computation. Clearly, if a;;>>a;~~:

no hM can vanish. Moreover, the condition (193) • As a matter of interest we point out that tba actual operations on the dieital computer to obtain a-• do not follow exactly the typical matrix multiplication rules as implied by formula (191).

(a)

(b)

t

X,

I~"

(c)

Y,

-.,_x%:

6./
(d)

Fig. 29.-Continuous beam. Good and bad choices of basic system and redundancies

is necessary to avoid serious accumulation of round-off errors in the most important digits. For if aii <,a;k the limited number of digits of the machine is not sufficient to ensure reliable computations. Thus, requirement(l93) is seen to be essential for well conditioned equations. Our Eqs. ( 186) are also ill-conditioned if two or more columns are nearly linearly dependent, e.g. {a1 ;a2 ; . • . . . . a;; ..... . a,;}~C{a1 A:02 k· •... . akk· ..... a,.d .. (194) The one diagonal element g •• will inevitably become very small and grave errors will again arise. Naturally, in actual structural problems two columns could never be exactly linearly dependent for otherwise this would indicate that we overestimated by one our number of redundancies. Nevertheless, a bad choice of redundant forces or moments may give an approximate linear dependence which would yield a result grossly in error. If it is found that our initial choice of redundancies leads to an illconditioned set of equations then we can always obtain more suitable equations by introducing as unknowns appropriate linear combinations Y of the initial unknown X. Such a transformation may be represented always by X=BY .............................. (195) where B is a non-singular square matrix n x n. If X are initially single forces or moments then Y represents groups of forces or generalized forces. Such groups of forces were first introduced by Mueller-Breslau• guided by pure physical reasoning and this is still the best method of finding them. The transformation (195) may be introduced directly into Eqs. (186a) yielding DBY+D.=O or (196) where D is determined by matrix multiplication from 00000000.. 000000000000000000 (196a) D=DB compatibility conditions at the cuts the Physically, Eqs. (196) express of the original unknowns X in terms of the new unknowns Y. If the transformation matrix B is unsymmetrical then the resulting matrix will also be unsymmetrical. Although, in general, the simple substitution of (195) into (186) can lead to a slight improvement of ill-conditioned equations the effect is usually small. The next obvious step is to express the compatibijity condition (186) in terms of the generalized displacements at the cuts corresponding to the generalized forces Y of ( 195). Following our discussion on generalized displacements and ftexibilities on p. 19 the generalized compatibility equations are derived by premultiplying Eqs. (196) by B' as B'DBY + B'D. ~. 0 or (197) D.Y+D •• =O where D.=B'DB and o •• =B'D•........................ (l97a) It is evident that the column matrix o •• and the symmetrical matrix D. would be obtained directly by selecting ab initio the generalized forces Y as redundancies and deriving the corresponding S;. and S;k from the standard formulae given. In many cases it is best, when the equations are ill conditioned, to select a different basic system and corresponding redundancies Y. Naturally, the latter are again statically related by a transformation matrix B with any previous choice X of redundancies. • Mueller·Breslau, lo<'. dt. p. 17.

29

We illustrate the above considerations on ill-conditioned equations on a very simple structural example. FIG. (29) shows a uniform, continuous beam of three equal spans simply supported at A, B, C, D-a twice redundant system. We discuss four alternative choices for the two redun· dancies. (a) X1 and X 2 are taken as the reactions at the supports A and B. The D matrix for this system is

D=

I

1: I 3u 312

I 9

24

9 4 L _j A remarkably bad choice since 312 > >322 (b) X1, X 2 taken as the reactions at the intermediate supports Band C. Then,

_aaq, ~~ 8

D- 18

7l

7 8

L

_j

Still a bad choice since all 3's are of the same order of magnitude. (c) Y1 and Y2 are generalized redundancies formed from system (b) by the transformation matrix

B=

Ill

-0·~

-0·5

1

L

I

(b)

(a)

System

Y1 = 1:

"'1

System

r,

=1 :

M, • Y,

System

r, = 1:

M, • Y,

t,. t, = [ t13 = t" =

.................. (198)

t,=

_j

Orthogonal redundancies

Non-orthogonal redundancies

,.r , =

:Z'

r:r·

ds}

s1

=

N, = coso&,

s2

= - sino&

N, =- sin-3,

S,

= - cos-3

N1 =

=1•

0,

0

= 0 for elastic centre C.

ds

f EI

Y,Y, ds

+

f

1 1 ) sin-3 cos-3 (&A' - EA ds

The D. matrix may be obtained either directly or from Eq .. (197a) and is I

4

1

= 0 for "principal axes" Cy2

I

D=~~

I L

tl

•

_j

'ik

Y;=-~ for i=l ton ............................

we restrict ourselves to bending deformations and assume the E.T.B. to hold, an orthogonal set of redundancies is obtained by referring them to the principal axes (Cy2 Cy3 )of the ring neutral axis weighted with the ring flexibility c/J =I/ El (see FIG. 30). The origin C of this system which is, of course, the centroid of the elastically weighted ring is known as the elastic centre. The transformation matrix B relating the orthogonal set of redundancies Y to a set X consisting of bending moment X1 , normal force X 2 and shear force X 3 at some point is in the notation of FIG. 30,

a~~-~

4

and clearly a scalar multiple of the D matrix of (c). The final system (d) is recognized as a particular case of the well-known Three-Moment Equation of Clapeyron. Since the basic system approximates more closely to the actual system than that of (c) it is clearly the mosl suitable choice of all. The differences between the above four systems become even more pronounced when the number of spans is increased. Moreover, the 3w coefficients of choices (a) and (b) tend, for a large number of spans, to become linearly dependent. This discussion shows how important the choice of the redundant forces is for the convenient numerical solution of a problem. An extreme case of simplicity is achieved if in all equations only one unknown appears. The particular set of redundancies Y1 , Y 2 , Y3, - - - - , Y;,---- Yn for which this condition is satisfied is called orthogonal. Then all but the corresponding direct influence coefficients are zero, i.e. =0 if ;,*' k . . . . . . . . . . . . . . . . . . . . . . . . . . (199) where we introduce the symbol ' for 3 to emphasize the special nature of this system. Eqs. (182) take now the simple form, ';;Y;+,;.=O for i=l ton ........................ (200) and hence (200a)

~ii

This system Y; may always be obtained by a particular linear transformation X=B,Y .............................. (201) However, the computations involved in determining B, are usually more laborious than the direct solution of Eqs. ( 182) if these are well conditioned. Nevertheless there are structures in which it is simple and hence advantageous to find the orthogonal set of redundancies. This is particularly so if physical and not mathematical considerations indicate how to find them. For example, this is so in arches and singly connected rings where, if

30

Cy3

Fig. 30.-0rthogonal and non-orthogonal redundancies-elastic centre of singly connected ring

1 4 L _j The improvement over (b) and (a) is immediately apparent. (d) X1 and X 2 are the bending moments at supports B and C. This choice of unknowns is statically identical to Y1 and Y2 of (c), but the basic system is different. The D matrix is

I 4

,

I 0

Ya

Y2-

cosO -sin~ -sin 0 -cos()

I

II

.............. (202)

_j L In practice it is nearly always worthwhile to find the elastic centre and eliminate two of the cross-flexibilities but determination of the principal axes, unless possible by inspection, is not usually worth the trouble. A further point is that the elastic centre concept is still valid if deformations due to normal and shear forces are included whereas the principal axes requirement becomes more complicated. Interestingly enough this solution was first given by Mohr* more than seventy years ago, but it appears not to be universally known, for otherwise it would not have been necessary to rediscover it so many times. A more recent derivation and application of an orthogonal set of redundancies (in general infinite) is the system of self-equilibrating eigenloads developed by Argyris and Dunne t for their general theory of tubes in bending and torsion. Particular forms of the 3;k and 3; 0 coefficients We return now to Eqs. (177) and (178) for the 3-coefficients and give, following our expressions (180) and (181) for a pin-jointed framework, some further explicit formulae for more complex structures. Stiff-jointed plane framework. We assume the Engineers' theory of bending stresses to be true and introduce the special notation: N., s., M. normal force, shear force, bending moment in basic system due to applied loads. N;, S;, M; normal force, shear force, bending moment in basic system due to X;= 1 where i= 1 ton. s coordinate along axis of beam. temperature at neutral line of cross-section. 0 D.0/h temperature gradient across depth h of beam ; positive if giving rise to thermal bending strain of the same sign as that due to positive bending moment M. • 0. Mobr, Z. Architek u. lng. Ver. Hannover, Vol. 27, 1881, p. 143. See also the 1enenlization and tabular presentation of this method in: J. H. Aqyris and P. C. Dunne, Structural Analysis (Vol. I of Handbook of Aeronautics), Pitman, 1952, Table 17.1. Both the deformations due to normal and shear forces are included in the latter analysis. t See J. H. Aqyris, P. C. Dunne, 'The ~ral'lbeory', ete., J.R.Ae.S., Vol. LI (1947), Feb., Sept., Nov. and Vol. Lm (1949), May, June.

tx, r·~·l

-

r ·'

X1 ~~X 1

o..-

I

-~ ::-=t::A. I I

II

A.

-1 N

tx, •

•..!..

a N, • -h

:~ . 1

U

'

a~

-o ..

JJ::h=_

·· -1C::::====T=:r . 1 N

a =-

I

n~c · h

A.~

1-

c,IRJ

:fc. • J '1 I~ 1

X2 •1

1

X1

X,

Normal force in horizontal member due to X, =I

A0

A.

c,. c,. c,. c,.

Force (moment), due to X, sl, acting on the element which experiences a A, In the direction of this A

Manufactured shape of panel

X3

Redundancies

Shear flow, due to X1 =I, acting on the element which experiences the A , in the direction of this

A

Contributions to 0 10 due to A

0,0 = fe,Adf,

=-f-A,

o,. = 0

0., • IC1A • ~ A.+ o.A1 + fA• + o.A. • *A• + fAn

Fig. 32.--3; 0 due to initial shear strain arisin1 from incorrect manufacture

020 • Ic,A • ~ A.- (a+ blA1 + fA.+ t.A.

direct and tangential forces per unit length g in the basic system due to X;= I acting on the element which experiences a a in the direction of this ~. t thickness. We deduce from Eq. (I 78),

c ni• cti

Fl1. 31.--3~ 0 due to initial strains In ri1id jointed frame and manufacturin1 errors and 'live' at supports

prescribed relative displacements (linear or angular) either inside the structure (e.g. lack of fit) or at the supports ('give' of foundations). force or moment in the basic system due to X;= I acting on an element which experiences a ~ in the direction of this ~. EA, GA', EI direct, shear, bending stiffness of beam. We deduce immediately from Eq. (I 78), see also Eq. (94), a;k=

JN;Nk EA ds+ JS;Sk GA'ds+ JM;Mk -ads

.................... (203)

Also from Eq. (177) or directly from the unit load method,

J

J

a~0 +l::C;~ S;.= JN;N. EA + N;a0ds+ M;-h-

................. . (204)

For pin-jointed frameworks we omit the terms involving M and S. On the other hand in stiff-jointed frameworks we can, in general, omit the terms involving Sand often also the terms inN. FIG. (31) illustrates on a twice redundant system how the contribution of the prescribed displacements to S;. is calculated. X1 and X2 are the chosen redundancies and ~"' ~b. a. and ~ are linear or angular imposed displacements arising from errors in the manufacture of the frame.

a

Two-dimensional stress distribution We restrict ourselves here to a presentation in cartesian co-ordinates x, y but the formulae are not restricted to stress distributions in flat plates. They are applicable to stress-states in any curved membrane which, by definition, does not allow for any variation of the stress over the thickness. Hence, x, y are, in general, orthogonal curvilinear co-ordinates; for example, in a cylindrical membrane y may be measured along the generator and x along the periphery of the cross-section. The following special notation applies. ~ prescribed direct (n) and tangential (t) relative displacements either inside the membrane (lack of fit of parts, slip of rivets) or at the boundaries ('give' at supports). g distance along part which experiences~ .. and/or~~

... at

tJ

Correct shape of panel

c, = T1 . c, • - TI . c, = o

Prescribed deformation ("give")

Contributions to 0., and 0 20 due to prescribed displacements A

~

I

Prescribed shearing deformation due to incorrect manufacture

C11

A. A1

I

L - _I

life,.= -(a+ b)

LOADING X2 • I

LOADING X, • I

r - -,

AL :_ ----- ~

A1

I

a c.. • =h

\~!= :

o.,

;-~::.::.:~~-·-~~A.

h

1I

N,

LI

S;k

=i:J J[

Jtdxdy

a,,;O"xxk +a•• ;O"vvk -v(a,,;O"vvk +a••;O"xzk)+ 2(1 +v)a,.;a,.k

Also from Eq. (177),

.................. (205)

............ (206)

The immediate application of the above formulae is to major aircraft components like wings and fuselages. Their matrix formulation is discussed further below. FIG. (32) illustrates how on a thrice redundant beam with shear carrying web the contribution of a prescribed displacement ~ to S;. is calculated. ~ is in this case an initial shearing displacement of a panel due to error of manufacture. It was assumed in all our above considerations that the basic or cut system is statically determinate. However, nothing in the theory so far given restricts us to such a choice. We can select in a structure with a total number of redundancies n a statically indeterminate basic system with (n-r)(r
31

be found in the simplest statically determinate system. Thus, if we introduce the notation a;=Statically equivalent stress system in redundant basic system due to X;= I we may write, 3; 0 =J(1;(t: 0 +t:,1)dV .................. (177a) v 3,,=fa,t:;dv. 3/A= fa,t:kdV=Jah-t:,dV=3k, ................ 078a) v v v The introduction of a,· instead of a, in Eqs. (177) and (178) may shorten the analysis greatly. Naturally, Eqs. ( 178a) are again identical with formulae (93). The above method presumes that the strains e and e,1 in the redundant basic system are known. Such information may be available either from previous calculations or the literature. Alternatively, we may have to analyse first the basic system for the external loads (and/or imposed strains) and the r X,= I by the method given previously. From a mathematical point of view the selection of a redundant basic system means that we solve the problem of 11 equations with n unknowns in two steps involving respectively the solution of r equations with r unknowns and (11-r) equations with (11 -r) unknowns. This method is particularly useful if we have available information on the stress distribution of the redundant basic system or if the number 11 is very high. Consider, for example, FIG. (33) showing a fuselage ring with transverse beam under uniform load p. The loading is equilibrated by tangential shear flows q applied by the fuselage to the ring. The structure is six times redundant and as redundancies we select the two groups X 1 , X2 , Xa and X 4 , X.,. X6 at the intersection of the axis of symmetry with the upper part of the ring and the transverse beam. Due to symmetry of loading and structure X:1 =X6 =0 and hence the problem reduces to finding the remaining four redundancies. We may solve the system by direct application of Eqs. (182), which in the present case take the form, 311 X1 +3 12 X2 _L3u X 4 ..:..3~;; Xr, +3 10 ~~o S21X, +322X2+324 X4 _L32r, Xr, +32o=O S41 X1 +3 42 X2+3 44 X 4 ~'-3 4 ,,X,, +3 40 =0 r ................ (207) 3,,1 X1 +3r,2 X2 -+-3,, 4 Xt...:. 3,;r,X,-, +3r, 0=-0 where the 3,k and 3,, are calculated with formulae (203) and (204) in which the integrals extend over ring and transverse beam. Note that if the unknowns X 1, X2, Xa are referred to the elastic centre of the ring the co~fficient 312 vanishes. Having solved Eqs. (207) we find N, S, M in the actual structure from, N ·- N0 +N1 X1 _t_N2 X2 +N4 X 4 +N;;Xr, } ................ (208) S=· S0 +S1 X1 +S2X2+S4 X4 +S;X; M=·M 0 +M1 X 1 +M2 X2 +M4 X 4 +M5 X 5 where M 0 , M 1 are defined on page 30. Alternatively to this standard method we may solve the problem by cutting only the beam at the axis of symmetry. Then the structure is only twice redundant (X4 and X5 ) with a basic system that is itself twice redundant. We denote by

1

m,H

nfa So

m 1, 11 1, s1 (i =·4 or 5) the normal force, shear force and bending moment in the new basic system due top and X 1 =I respectively and assume that they are known. The stress distribution in the actual structure is then found from M=m 0 +m 4 X 4 +m;X5 I N=n 0 +n 4 X 4 +n 5 X 5 ~ ••.•••.••••.••.. (209) S=s0 +s 4X4+s 5 X,; j The equation of compatibility in the unknowns X 4 and X 5 are now of the form 'uX4+,4aX5+,4o=--O } ................ (210) ';4X4 +';;;X5 +'•o=O where we write ' instead of 3 to stress that these coefficients are different from the correspondrng 3's in Eqs. (207). To find the ''s we apply Eqs. ( 177a) and ( 178a) in the new basic system and remember that the virtual stresses due to X1 =· l may be selected in a statically determinate system. Thus, omitting the contributions of the normal and shear forces for convenience of printing, we find m 1mk JM;mk Jm;Mk r ,,k= -----erds= El ds= ~ds=~:,kt .......... C2ll)

I

' "= Jm~7"ds= JM~7°ds

.......................... (212)

If, in addition, any initial strains YJ are imposed on our structure giving rise to moments m'l in the basic system the corresponding contribution

to ''" becomes r Jm;m'l <..to= El ds= JM;m'l ~ds ....................... . (212a) The effects of the normal and shear forces may be included without difficulty, the terms following immediately from Eqs. (203) and (204) and the arguments leading to the contribution of the moments.

~

x,-

'

~

X,

f x;

J

~x~~x:~ '~

(b)

(a) Statically

basic

determinate

R•dundant

basic

syste-m

sys1Pm

Fig. 33.-Statically indeterminate basic system. Doubly connected ring It is useful to write down the 3,k and 3," for a two-dimensional stress distribution when the basic system.is redundant. Thus, with the definition of a, given previously we find,

3,. ~-i:

JJ[a~,.,a_,.,k +a.",;a•••k -- v(a_,.,.,a••k +a••;a.•. •. k) +2( I +vla.ruta.r•k] tdxdy

............ (205a) and a similar formula with a and a interchanged. Also,

=iJ I[

a_,._,.,a_,_,.,_L(],,,,;a.,,,, --~·(a,. ,;CJyyo·H1•• ;a_,.,..)+2( I +v)a.ryiCJ.ryo]tdxdy .................. (206a) where a_,.,., etc., may be here not only the stresses due to external loads but also due to the imposed initial strains YJ· (Remember that the latter cannot develop freely in the statically indeterminate basic system and give rise to some strain-stress state e,,, a,,) The use of a redundant basic system arises continuously in wing theory. Thus, following Ebner and Koller* and Argyris and Dun net it is customary in wing analysis to express the actual stresses in the form, a=a.+a. . ........................... (213) where the stress system a -the choice of which is at our discretionsatisfies both the external and internal equilibrium condition and is therefore the statically equivalent stress system, a, are the self-equilibrating stress systems (in general, infinite in number), necessary to ensure external and internal compatibility. In our present terminology a. is the basic stress system and a. the redundant stress systems which for practical purposes are approximated to a finite number. In fact,

3,,

0

n

a,=a1 X 1 +a2 X2 + . . . . +a,Xn=:Ea;X; i=l

.......... (214)

It is advantageous in the selection of the basic system a. to try and satisfy the two, at times conflicting, requirements of simplicity and not too great difference from the exact system a. For it is obvious that small a,-systems are highly desirable from both the theoretical and practical point of view. Now for wings with not too small an aspect ratio an excellent choice for the direct stresses of a. is given by the Engineers' theory of bending for beams since it combines simplicity with reasonable accuracy. If the wing forms a single cell tube we deduce the shear stresses from the boom load gradients, the undetermined constant of integration being found from the overall torque equilibrium; thus, in this case the basic system a. is statically determinate. If on the other hand the wing is an N-cell tube we see that whereas there is still only one torque equilibrium condition there are N undetermined constants of integration. To calculate them we must introduce conditions of deformation and those are the equality of rate of twist of all cells. Hence, our basic system is redundant, the degree of redundancy being N -l. The solution to this problem is reproduced in Example (a) of Section 9. General considerations on the calculation of the redundant self-equilibrating stress systems a. are given later in this Section in matrix form (see also Example (b) of Section 9). The example of the tube is useful also to illustrate another point. We stated on p. 27 that the basic structure is obtained by cutting redundant members but mentioned that in continuous structures the idea of a physical • Loc. cit. p. I. p. 30.

t Lo..-. cit.

cut is not always applicable. Thus, in the case of the tube in the last paragraph, when obtaining the basic stresses we do not actually cut any redundant member but rather select the engineers' theory of bending direct stresses and the associated shear flows as statically equivalent to the applied bending moment. In general, all members are found to be load carrying. The stresses a. only exceptionally satisfy the elastic compatibility conditions-for example, due to warping varying parallel to the axis of the tube and to rib deformability. We may give some physical reality to the basic structure in which a. is true by releasing in the actual structure the warping restraints at every cross-section and by assuming the ribs as rigid; the former idealization does no doubt require a complicated mechanism for its realization. The idea of selecting a. as any suitable statically equivalent stress system without reference to actual cuts may, of course, also be applied to frameworks. The use of a redundant basic structure is important also from a further point of view. Consider the wing of FIG. (34) the main portion I of which is swept and attached to some root structure II. It is assumed that the ribs of I are taken perpendicular to a longitudinal axis approximately parallel to the spars. The necessity may arise of investigating alternative angles of sweep obtained merely by changing the root structure II. Thus, in FIG. (34) we show two alternative arrangements. In such instances it is obviously advantageous to have as much as possible of the stress analysis in common in the two alternative calculations. To this purpose we release at the junction ofl and II all redundant forces or groups of forces X 1 to X,. appearing there. The tube I is then connected to tube ll by some statically determinate arrangement and this new structure is taken as the basic system. The scheme of the analysis is as follows. Analyse first Tube I for all external forces and also for X 1 = I to X n = 1 respectively. This investigation involves, of course, the solution of a highly redundant system. Irrespective, however, of the form tube H takes the analysis of tube I remains unaffected. Next we analyse tube II for the external forces (which include the statically determinate reactions P from tube I), and also for X 1 =I to X n = 1 respectively. Again this may involve the solution of a redundant problem. Finally we can write down equations of the type ( 182) for the unknowns X1 = 1 to X n = 1 and note that the S-coefficients are in each case 8=81 +8u . ........................... (215) Hence if we change structure II only S11 but not S1 is altered-an obvious advantage. The solution of Eqs. (182) yields ultimately the stress distribution in the actual wing.

Fig. 34.-Aiternative root structures for swept-back wing. Redundant basic systems

33

H

AVING discussed in the standard longhand notation the main ideas and methods for the calculation of redundant structures on the basis of forces as unknowns we now turn our attention to the matrix formulation of the analysis. Consider a system consisting of s structural elements with a total number n of redundancies which may be forces (stresses), moments or any generalized forces. We select a basic system by 'cutting' a number r of redundancies where r
Example for the b0 and~ matrices. ~10. (35a) shows a five times redundant structure, assumed symmetrical, subJected to the loads R1 and Rt· Due to symmetry of loading and structure the system is effectively three times redundant. For the basic system we sc:Iect the statically determinate structure of FIG. (35b). The b0 and b1 matnces for half the structure including the central vertical member (11) are found easily as

>;-I L

0 -a/h

a/h

afh 0 0

0 -a/2h a/2h afh 0 0 I

2

3

4

5 6

0

0

(218)

-d/h

0

0

-d/2h-d/2h 0 7

8

9

0 1/2

1

IO

11

and 0 -afd ~-afd .,;= 0 -a/d 0 L (219)

0

-a/d 0

0

l

l .J

-hfd-hfd 0 0 -h/d-2h/d

0

0

-1

-1

0 0

0

0

0

0

0

I

2

3

4

5 6

7

8

9

IO

11

.J

~Jb.re the numbers under the columns refer to the numbered bars of FIG.

Additional Notation X,Y, Z column matrices of X,, Y1, Z 1 respectively. column matrix of initial strains (displacements). H rectangular matrix of forces (moments) C1 (seep. 31). c B,C B., C. 1 d h

0 ~. ~..

L1,, 11,.

areas of actual longitudinal and transverse flanges. areas of effective longitudinal and transverse flanges. length of longitudinal flanges between nodal points. length of transverse flanges between nodal points. height of web. area enclosed by cell. areas of cover panel and web panel respectively. 2 x 2 matrices.

L=[L1,], 1=[11,] P, Q effective longitudinal and transverse flange load respectively.

k., k.,, k1 U, W 8o ~.

ii1

Partial stiffnesses due to shear strains in sheet, direct strains in sheet and direct strains in flanges respectively. column matrix of kinematically indeterminate joint displacements U, W respectively. column matrix of strain of elements due to unit r's and U=O. column matrix of actual and kinematically equivalent (virtual) strain of elements due to unit U's when r=O respectively.

X,

C=i,.'k~ =~'k~

Ce=~'kaor=~'kaor

J

34

column matrix of initial stresses.

Flp. 35 (cr and b).-5tatlcally Indeterminate pin-jointed framework. Illustration of b 0 and b 1 matrices

Following our previous analysis of structures when the basic system is redundant we introduce now the matrix ~ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (220) to denote any suitable statically equivalent stress (force) matrix corresponding to X={l 1. ....... 1. ....... I} Thus, b1 corresponds to the stress system cr; introduced before. When the basic system is statically determinate only one 61 can be found, 61 = b1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (220a) We define also by 60 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (221) any suitable matrix statically equivalent to R={l I. ....... 1. ....... 1} b0 may, in fact, be determined in a different statically determinate system from b1 . Next we derive the matrix equation for compatibility of deformation in the actual structure. Denoting the relative displacements at the r cuts of the basic system due to loads Rand the r redundancies X; by Vr the compatibility condition is (222) . . . . . . . . . . . . . . . • . . . . . . . . . . . . . . . . . . Vr=O where v, is a column matrix with r elements. To express Eq (222) in terms of Rand X we note from Eq. (122) that the relative deformations v (these may be elongations of bars or flanges, shearing angles of plates), at the ends or boundaries of the s elements are, v=f5=fb0 R+fb1 X . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (223) f, the flexibility matrix of the s elements, is the partitioned diagonal matrix of Eq. (123). We find now vr directly from the argument leading to Eq. (125) as

--

Basic system twice redundant )

structure

Complete

R/2~ (12

~J[~ x,

Cantilever

and hence (224) These are the required equations in the r unknown X;, and are, in fact, equivalent to formulae (182). The symmetrical square matrix D=b1'fb1 (to use the notation of Eq. (182)) is the flexibility matrix for the directions of the r unknown X; in the basic system. Also in the notation ofEq.(l82a) D0 =b1'fb0 R . . . . . . . . . . . . . . . . . . . . . . . . . . . . (225a) Eqs. (224) are the most general formulation in matrix algebra of the equations for the r unknown X; in a structure with a redundant basic system. Solving for X we find X=- (b1 'fb1 )-1b1'fb0 R . . . . . . . . . . . . . . . . . . . . . . . . . . . . (226) Substituting (226) in (217) we determine 5 solely as a function of the R's. Thus, (227)

X.

elements

beam

Fig. 36.-Doubly connected ring. Analysis with redundant basic system

Simple example of Eq. (224)

Consider the symmetrical fuselage ring with transverse beam and central load R shown in FIG. 36. As in page 32 we select as a basic system the structure with the beam cut at the centre. For the components s of the basic system we take the two statically determinate cantilever beams and the closed ring. It is assumed that we know the stress distribution and hence the flexibilities due to the pairs ofloads applied to the ring (FIG. 36). The basic system is thrice redundant but due to symmetry X 3 =0. The load transformation matrices b0 and ~ are I

I

I

_,boB bubOR L _j

••

0

•••••••••••••••••••••••••

(231)

where

Comparing (227) with Eq. (121) we can write S=bR where

I

(227a) Naturally, it is always possible to substitute b1 ' for 61 ' in Eqs. (224) to (227a). However, the introduction of the statically determinate matrix ~ when the basic system is redundant simplifies the calculations, often considerably. We can apply now Eq. (227a) to derive the flexibility

0

I

1/2

boa=

L

0

bOR=

where . . . . . . . . . . . .

(230)

r,~~l

••

0

•••••••••••

(231a)

1/2 L _j

_j

and

-l~l

F

of the actual structure for the m points and directions of the applied loads. Eq. (126) gives f=b'fb . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (126) Forb we may use b = b0 or even simpler b = b0 We obtain F = b0'f[b0 - b1 (61 'fb1 )-1 b1 'fb0 ] or

fo=60'fb0 =b0'fb0 . . . . . . . . . . . . is the flexibility of the basic system for the loads R.

element

Ring

bl-

L

0

••

0

0

••

0

••

0

•••

0.

0

•••••

(232)

blR

_j

where I

I1 btB=

lo

Ol

IO

I

I

L

I

I 01

I

_j

b1R=

1 0 0

L

~~

•

0

•••

0.

0

0

•••••

(232a)

QlI _j

Note that in the present case b1 = b1 since the loadings on the two elements of the structure are statically determinate. The flexibilities of the elements for the forces and moments may be written as

35

~u '~lI

[2 I EI -2EI O [2

fB=

-2EI 0

and hence

L

fu

fB= fm /22 /23

[3

3EI 0 I 0 EA

f=

fal /82

faa

L

~B 0

L

ol

..J

........

(233)

......................

(234)

fR

..J

It is now possible to write down the equation for X, ~'fb1 X+ ~'fb 0 R = 0 Returning to the general argument we investigate now in a system with a total number n of redundancies the effects ofinitial strains e.g. those due to temperature rise, excess length of bars due to manufacturing errors, 'give' of foundation at supports. Assume that the column matrix of the relative displacements at the ends or boundaries of the s elements due to 71 is (235) H ························ We assume first that the basic system is statically determinate: then the elements of Hare merely the integrated effect of the imposed 71· For example, in a pin-jointed framework subjected to temperature rise the elements of Hare (235a) H ={(a0/) 0 • • • • • • • • • • • • • • • • (a0i),} If the bars are of an excess length fl./ due to inaccurate manufacture, then \.hese form directly the elements of H. Independently of the ,nature of H, however, the corresponding relative displacements at the cut redundancies are simply b1'H and the equation for the n unknowns X is b1'fb1 X+b1'H=0 . . . . . . . . . . . . . . . . . . . . . . . . . . (236) Note that in the present case b1 =b1 since the basic system is taken to be statically determinate. When the deformations arise from p 'gives' fl. at the foundations it is advantageous to express H as . . . . . . . . . . . . . . . . (235b) H ={!l.1 •.•••..••.•..••• fl.,} Then for ~' we must substitute the matrix c' where

C=

_·_·: ~: ~ ·_·_ :~~-~·-·: ~: ~ ·_ ~~"l :~2 C,l ........ C;, ........ C;p

r----------------------

.......

0

••••••

(237)

Cross

s~ction

A-A

Fig. 37.-Aircraft wing structure. Geometry and definition

and since x is usually small in comparison with X approximate methods may often be used in the evaluation of the right hand side. This technique is immediately applicable to the correction of a completed stress analysis when subsequent small changes in the flexibilities of some elements are introduced by modifications to their cross-sectional dimensions. When the number of equations is too large for performing the matrix operation on a digital computer then we may apply the following method which is basically identical with the idea of a redundant basic system. Assume that in a structure with a number n of redundancies we select first a basic system with t=n-r redundancies and that we write down the r equations in X; in the standard form (224), DX+Do=O . . . . . . . . . . . . . . . . . . . . . . . . . . . . (224) If the number r is still too large for handling by the digital computer we solve the problem in two steps. Eqs. (224) are first put in the partitioned form

c,.l ........ c,., ........ c,.,

..J L the n rows of which are the forces due to X;= 1, applied by the structure to the foundations in the directions of the gives fl.. Eq. (236) becomes now, . . . . . . . . . . . . . . . . . . . . . . (236a) ~'f~X+c'H=O If we select a redundant basic system we cannot derive the elements of H immediately from the prescribed initial strain since the latter are not free to develop in a redundant structure. In this case unless we have the necessary fnformation from previous calculations we must first analyse the basic system by the method of the previous paragraph. Having found the column matrix H the r equations in the r unknowns take the form b1 'fb1 X+b1'H=0 . . . . . . . . . . . . . . . . . . . . . . (236b) where we may write ~ for ~ since the basic system is now redundant. The systematic solution of (224) and related equations was discussed on page 28 but there are a few further points arising in practical calculations which are best investigated here. Thus, we mentioned on page 20 that it is often possible and justified to neglect certain part flexibilities of the elements; for example, in a ring analysis we can usually ignore the direct and shear flexibility. This applies not only to the evaluation of the external flexibility F but also to the determination of the internal redundancies X. We write now the D and D0 matrices in the split form D=Da+Dband D0 =Doa+Doo . . . . . . . . . . . . . . . . . . . . . . (238) where the suffices a and b refer to the two flexibilities into which we separate the total flexibility of each element. An approximate solution X a to the unknown column X is then obtained by ignoring the flexibility b. Then, DaXa+Doa;;=O or X 0 =-Da-1 Doa . . . . . . . . . . . . . . . . . . (239) Occasionally we may require subsequently the correction x to Xa to find the true column X, X=Xa+x . . . . . . . . . . . . . . . . . . . . . . . . (239a) Substituting (238) and (239a) into Eq. (224) we derive easily . X=-D-1 (D 00 +DbXa)

36

(240) where the number of rows in the matrices with suffices I and II is p and

r-p respectively and Db Du are square matrices. Eq. (240) gives D/XJ+Du/Xu+Do/=0

I

~ . . . . . . . . . . . . . . . . (240a) DuiXI+DuXu+Dou=O j We split next the column matrix Xu into matrices Xu=x+y ......................... . (241) where x satisfies the equation Dux+Dou=O or X= -Du-1 D0 u.............. (242) Hence (242a) . . . . . . . . Du 1 X 1 +Duy=0 or y=-Du-1 Du1 X 1 Substituting for x and y into the first of Eqs. (240a) we find D1 X 1 +D u/[ -Du-1 D0u-D11- 1 Du1 X 1 ]+D01 = 0 or X1=- (DJ-Du/Du-1 DuJ)-1(D0 I-Dm'Du-1 Dou) ..... (243) from which we deduce y and hence X. Eqs. (243) are identical with the elastic compatibility equation for a basic system with n- p redundancies. The matrix form (224) of the equation of compatibility is particularly suitable to illustrate the transformation (seep. 29), X=BY .............................. (195) when the equations are ill-conditioned. Thus, by substitution of(l95) into (224) we find ~'fb1 BY +b1 'fb0 R=0 and premultiplying by 8' B'b1 'f~BY +Bif»1 'fb0 R=0 or b 2'fb2Y +~'fboR=O . . . . . . . . . . . • . . . . . . . . . . . . . . . . (244)

where (see also Eqs. (196) and (197)), b2 =b1 B and b2 =~B . . . . . . . . . . . . . . . . . . . . . . . . . . (245) are merely the matrices for the true and statically equivalent stress systems in the basic system due to Y={l 1. ............... 1} The form (244) of the equations of compatibility may, of course, be written down directly when starting ab initio with the group unknowns Y. Application to a typical aircraft structure We present now a detailed investigation of a type of system characteristic of aircraft wings. Consider to this purpose the structure shown in FIG. 37 which consists essentially of an orthogonal or nearly orthogonal grid of spars and ribs covered with sheet material. Longitudinal and transverse flanges may be placed at the intersections of spar and rib webs with the covers. In addition the covers may be stiffened with further longitudinal and/or transverse flanges. The cross-section is assumed arbitrary and the spars may taper differently in plan view and elevation but the angle of taper 20 is taken to be so small that cos 20R:> I and sin 20R:>20. The analysis is not restricted to structures with continuous rib and spar-webs, covers and flanges and includes hence any kind of minor or major cut-out. The geometry considered excludes swept-back wings with ribs parallel to the line of flight. On the other hand swept back wings with ribs perpendicular to the main wing axis can be analysed by the present method as long as we are given the necessary information for the triangular rootsection. Delta wings may also be investigated by our theory as long as the grid of ribs and spars conforms to the geometry stipulated here. Naturally, many of the restrictions imposed limit the applicability of the- method. Indeed we intend our analysis only as an exploratory and tentative first attack on the more general problem. We hope to return to this and similar points in later publications. The problem of finding the stress distribution in the shell type of structure considered is strictly infinitely redundant. Hence 'it is necessary to introduce for practical calculations considerable simplifications. First we adopt the standard assumption in wing stress analysis of a membrane state of stress, i.e. we exclude any bending of covers and flanges normal to the surface of the wing. For the very thin wings now coming into prominence this idealization is open.to grave doubts and will no doubt have to be reconsidered in future. An essential characteristic of our theory is the assumption that the longitudinal and direct stresses vary linearly between the nodal points of a three-dimensional grid of lines traced on the wing cover. This system of lines should, in general, be at least as fine as the grid of spars and ribs whose intersection with the covers forms the best minimum set of grid lines. The latter grid will often be sufficiently close if we are dealing with a multi-web structure and ribs at not too great a distance. However, many instances occur where it is necessary to select additional nodal points between which the direct stress is taken to vary linearly. For example, we may choose points intermediate between spar webs on the rib stations if the spacing of the spars and the sheet thickness of the cover are large. Similarly, if the structure has few ribs we may have to introduce new transverse stations in order to reduce the spacing of the grid in the longitudinal direction. In either case there need not be an actual longitudinal or transverse reinforcement along the new grid lines. We call the surface enclosed between two adjoining grid lines in the z- and sdirection a field, and denote by 'bay' a part of the wing structure which lies between two cross-sections taken through adjoining grid lines running in the s-direction (see FIG. 37). The assumption of a linear direct stress distribution along the edges of an orthogonal and flat field yields from overall equilibrium conditions a parabolic shear flow distribution along the edges. Naturally, neither the linear direct stress nor the parabolic shear flow variation are, in general, exact and violate the internal and boundary compatibility conditions of the field. This is not serious as long as we keep the spacing of the grid lines reasonably close. Moreover, we simplify further the problem by neglecting the quadratic and linear terms in the shear flow and considering it to be constant within each field. We note that for non-orthogonal grid lines (tapered structure) the uniform shear flow offends against the equilibrium conditions even if the direct stress is constant between adjoining nodal points. The errors introduced by the assumption of uniform shear flow are, however, practically insignificant for the reometry of structure considered here when the nodal point distances are small. If the direct stresses along the grid lines were known we could calculate the fraction of sheet area to be added to the reinforcements to form the equivalent or effective flanges. This applies to the cover, spar-webs and rib-webs and yields an idealized structure in which the fields are only shear carrying and the direct stress carrying ability is concentrated in flanges; an assumption widely used in aircraft practice. Neglecting the Poisson's ratio effect and assuming the same material for flanges and sheet material cover, the fraction of sheet cross-sectional area to be added to the flanges varies between 1/6 and 1/2 if the fields are flat; the former value applies when the field is in pure bending in its own plane and the latter when it is under uniform stress. Since the stress distribution is unknown we can at best only estimate the effective areas of the flanges but may use an iteration process if the first guess proves inadequate. However, the latter procedure is really clumsy and lengthy and a direct method, obviating the guessing of flange areas would evidently be useful, in particular at the root or other

Fig. 38.-5lgn convention for flange loads and shear flows

Unstiffened cut- out

I

bay

12~CD~~~~--L Unstiffened cut-out

0

Longitudinal flang~s continuous at junction

•

Longitudinal flanges interrupted at junction

}

~ = 12

N=3 Fig. 39.-Geometry of typical bay for determination of number of redundancies

structural or loading discontinuities where the stress distribution is more difficult to estimate and the Poisson's ratio effect more pronounced. Such a method is given here at the end of this sub-section but at first w~ d~velop the theory under the assum;Jtion that the effective flang~s areas are known and that they are constant between two adjoining nodal points.* For the webs, when considering torque and lift loads, it is alwtys su n:ient to add I /6 of the web cross sectional area to the longitudin:ll and transv~rse flanges at the intersection of the spars and ribs with the cover. We summarize now the main assumptions underlying the idealized structure selected for analysis. Thus, our system consists of an orthogonal or nearly orthogonal grid of spars and ribs with top and bottom covers. Effective flanges of constant area betw~en adjoining nodal points and carrying only direct stresses are assumed placed along the grid lines in longitudinal and transverse directions. For the tim;: being w~ assum;: that the flange areas are known. The direct stresses and hence also the flange loads are taken to vary linearly between nodal points. All sheet material for covers, spars and webs is assigned a purely shear carrying role and a constant thickness within each field. The angles of taper of spars in plan view and elevation are assumed to be small. The shear deformability of covers, ribs and spars is included ab initio in the analysis. For the stresses and loads in the various elem;:nts we adopt the si~n convention illu;trated in FIG. 38. Naturally, the idetliZltions and sim;Jiifi:ations introduced are strictly only necessary for the calculation of the redundtncies. The btsic or statically equivalent stress system may and should preferably be d~termined in the (cut) actual structure. Degree of redundancy of idealized structure We proceed next to the enum;:ration of the redundancies in our idealized structure. In addition to the sim;>lifications introduc~d previously we ignore here the bending stiffness of the flanges for displac~m~nts tang~ntial to the wing surface. This is, no doubt, su.nciently accurate for th~ present exploratory analysis. The wing structure supported at the root and free at • Strictly, the latter assumgtion is only necessary when finding the Oexibilities of the Dances for the calculation of the redundancies.

37

the tip is assumed stiffened by ribs at least at the root and the tip. These ribs need not necessarily consist of a web with flanges but may take the form of a &tiff-jointed frame or ring. However, independently of the design of the ribs we may always substitute an equivalent shear web with flanges. The wing structure is subdivided into a number of bays of which we show a typical intermediate one in FIG. (39). The cross-section at the junction nearer to the tip may be stiffened by a rib carried across some or all cells; FIG. (39) indicates also those longitudinal flanges which are continuous across the same junction. It should be noted that if there is a change of transverse slope of the cover at a longitudinal flange the latter must be connected to a spar web. We use the following notation : fl=number of longitudinal effective flanges which are continuous across the junction, i.e. are not interrupted there. N =number of closed cells stiffened by ribs at the tip end of the bay. Then the number of redundancies arising from the geometry of the bay is {1-3+N-l Hence, in a tubular structure of the type shown in FIG. 37, free at the tip and either fully built-in at the root or with prescribed displacements there at all longitudinal flanges, the total number of redundancies is . . . . . . . . . . . . . . . . . . . . . . . . (247) n=~ [@-3)+(N-l)]

Root. with

rib

Tip

Fig. 40.-Multi-web wing without intermediate ribs

bays

> s

If certain of the flanges are not held at the root section the number of redundancies reduces accordingly. For example, if the root-section is at the aircraft centre line and the wing is subjected to anti-symmetrical loading the number of unknowns reduces by p,- 3, {3, being the number of longitudinal flanges at the root. The number in the square brackets in (247) can, of course, vary from bay to bay since effective flanges may be interrupted at such stations. Also the number N of stiffened cells may be made different in each bay by the addition or removal of spar webs. However, when f3 and N are the same in all bays and all the flanges are held at the root formula (247) becomes simply . . . . . . . . . . . . . . . . . . . . . . . . (247a) n=a[f3+N-4] where a=number of bays. If the sheet cover is missing between two adjoining longitudinal flanges in a bay and the cut-out is not provided with a stiff-jointed closed frame to restore partially the lost shear stiffness of the sheet then the corresponding cell is open in this bay and by definition is not included in N. Similarly, if there is no rib or equivalent frame in a cell at the section considered this cell is excluded from the numbering for N. Note that spar webs need not be continuous throughout the length of the wing and may be discontinued at any junction. Formula (247) still remains valid. If the cross-section is singly symmetrical the n redundancies of Eq. (24 7) split into two groups :

n1 = I;[(~-1 )+(N-t)J bays

na= I;(~-2) bays

}

stiffened

stress

panel

system

X: 1

(Flat panel)

(248)

where f3 is defined as in the case of the wing. When M fields are removed without being replaced by stiff-jointed frames the number of redundancies reduces by M. The next step in our investigations is the discussion of suitable selfequilibrating systems which may be chosen as redundancies. Consider first the simple case of a rectangular flat panel shown in FIG 41. For the redundancies we may select n systems of the type X= 1 illustrated in the

38

Rectangular

Self.equillbratmg

of which n 1 applies for the lift and torque loads and n2 for the drag loads. If all cells are closed, with the same number N in all bays and effective flanges are only placed at the corners of the cells, then f3=2(N+1) and from (247a) the total number of redundancies is n=a(3N-2) . . . . . . . . . . . . . . . . . . . . . . . . . . (249) which formula again assumes that all the flanges are held at the root. Of considerable importance in modern aircraft structural practice are the multispar syst~ms with .few, often only two, end ribs. A typical wing of the latter type IS shown m FIG. (40). To analyse this structure we subdivide it into a number of bays whose length should not exceed say five times the spar pitch. Effective flanges will by virtue of our idealization process be acting at the junction of these bays although no ribs are provided th~re. For such a system the number n of redundancies when there are no cut-outs in the sheet, when all spars are continuous for the full length of the wing and all flanges are held at the root, is given by n=(N-l)+l+a({3-4)=N+a({3-4) . . . . . . . . . . . . . . (250) The last system to be considered is a flat panel which is of special importance for diffusion investigations (see FIG. 41). It is assumed built-in at z=O or held with prescribed displacements and free at the other three edges. Here the number n of redundancies when there are no unstiffened cut-outs is simply II=~ (/3-2) . . . . . . . . . . . . . . . . . . . . . . . . . . (251) bays

z

Longitudinal

flange

loads

Transverse

flange

toads

Fig. 41.-Rectangular stiffened panel. Unit self-equilibrating stress system of type X=l

figure. All information as to flange loads and shear flows is given there. The corresponding equations (182) or (224) for the unknown X; are easily seen to be reasonably well conditioned. Naturally, we can further improve the conditioning by introducing group loads X=BX . . . . . . . . . . . . . . . . . . . . . . . . . . . . (252) wh~re B is a suitable square matrix. We do not enter at this stage into the chmce of B but hope to discuss these points in Part III. When the panel is symmetrical about its middle line it is preferable to combine the %-systems into symmetrical and antisymmetrical groups. In a wing structure of the type investigated previously we can describe three simple types of self-equilibrating internal systems. They are shown in FIGS. 42, 43 and 44 and denoted by X=l, Y=1, Z=l respectively. The first is the generalization of the %-system used in the flat panel and the second and third may be considered as slightly modified four boom load systems taken in the longitudinal and transverse directions respectively. The longitudinal four-boom load systems are applied extensively in standard wing analysis.* The three figures are self-explanatory and give all flange loads and shear flows associated with the unit systems. Note, however, that the effect of taper is neglected except that we introduce the true local dimensions in the evaluation of the self-equilibrating • See J. H. Argyris and P. C. Dunne, 'The General Theory, etc.,' J.R.Ae.S., Vol. LI February, September, November 1947 •

Multi-web

wing

Multi- web

wing

Area •·•

1 I + I q =•-·...:L-..:.1

11 / 2 h, + h 2

q =---11, (h,•h,)

Central cross section Self.equilibrating

str~ss

system

X= 1

p.~!;_, dld1 h

Self-equilibral1ng

stress

system

Y= 1

-I

___.1iJ{"+,-,.,~"""'

Longitudinal

flange

loads

Fig. 42.-Multi·web wing. Unit self-equilibrating stress system of type X= I

systems. This ignoring of the influence of taper gives rise only to small errors as long as the lengths I of the bays are reasonably small and the angles of taper restricted to the order of magnitude mentioned initially. The expressions for the shear flows in the ribs for the Y- and Z-systems are also approximate, being derived for an equivalent rectangular rib. Again the error introduced by this assumption is for practical purposes insignicant. The conditions of equilibrium for the X- and Z-systems yield a load in an 'effective' vertical flange at the intersection of the ribs and webs. This flange load may always be neglected. We enumerate now the number of X, Y, Z systems, independent within their own group, which can possibly be applied. We find easily, with the notation of Eq. (247) that we can use for each bay ({J -4) X-systems, N Y-systems and (N -I) Z-systems Thus there are more systems than we require, the difference from the number n of redundancies being obviously linearly dependent systems. Evidently the Z-systems (N-l) are independent of the X- and Y-systems and hence must all be chosen as redundancies. To complete the number n of unknowns we may use ({J -4) X- and one Y-system However, it is preferable to apply more longitudinal four-boom ( Y) systems since they are better conditioned, and to reduce accordingly the number of X-systems. Thus, if we introduce all N Y-systems we have to adopt ({J- 3)- N X-systems The last number reduces to N -I when the effective longitudinal flanges are only placed at the corners of the N-cells. It is often advisable to improve the conditioning of the D-matrix by the introduction of group loads X and Y where X=B1 X Y=B2Y . . . . . . . . . . . . . . . . . . . . . . . . . . (252a)

Longitudinal

flange

loads

Fig. 43.-Multi-web winJ. Unit self-equilibrating stress system of type Y- I. (Longitudinal four-boom tube)

When the cross-section is singly symmetrical the number of redundancies for lift and torque loads reduces to n1 given by the first of Eqs. (248). The N - l Z-systems must still be included in the analysis for such loading cases. If, in addition, we use all N Y-systems the necessary number of X-systems becomes

f32-l-N

and is zero when the effective flanges are arranged merely at the corners of the N-cells. For the multispar wing of FIG. (40) with ribs only at the root and tip the n redundancies of Eq. (250) may be selected as ( N- l) Z-systems at the root one Y-system at the root and ({3 -4) X-systems at each junction of bays and at the root. The Y-system may involve a considerable length of the tube and if the latter is tapered a more accurate estimate of the longitudinal variation of the flange loads may become necessary. Having selected a suitable system X, Y, Z of redundancies we can write down the b1 matrix with the information given in FICiS. 42, 43, 44. To obtain the b0 matrix we may use any suitable statically equivalent stress system in the actual or idealized structure, but preferably the former. It was mentioned on page 32 that it is advantageous to select a basic stress system which, while being simple, approximates as closely as possible to the true stress system and reference was made to the method of example (a) of Section 9. Nevertheless, if the work in finding such a b0 matrix proves excessive it may be preferable-since the choice of b0 does not affect the conditioning of the D matrix-to sacrifice the closeness to the true stress system and to select a b0 as simple as possible. Thus, we can calculate a b0 matrix for a basic system in which the spars act independently; a choice differing, in general, widely from the final b matrix.

39

Tr,.nsverse

II ange ·-·· . ,

1

~e '

c

1 1

a

h /

Shear

/

field --

/

D f

'

1

Multi- web

wing d

2

~-Longitudinal

1

k

'

'

1

g

flange

//

b

'

Fig. 45.-5imple panel to Illustrate b 0 and b 1 matrices

Consider the simple grid shown in

FIG.

45 and denote by P0 (P1), Q0 (Q1),

q0 (q,) the longitudinal flange loads, transverse loads and shear flows

corresponding to some R = 1 (X;= 1). The corresponding columns in the b0 and b1 matrices are ( P oa1 P.oa2PObi P Ob2pOcl P oc2POdlpOd2QO
and

{Pial Pia2Pibl Pib2pi c!Pi c2Pidlpid2QieiQie2Qif1Qif2Qi o1Qiu2qihqik}

q=•t/m.,• n.,>

. . . . . . . . . . . . . . . . . . . . . . . . (254) respectively. To find the D and 0 0 matrices it only remains to give the flexibility matrix f of the elements. We write it in the partitioned form associated with the b0 and b1 matrices of Eqs. (253), I I o 0 0 I f1 0 ft

0

q:+t/(Q,•Q,)

Self-equilibrating

Transverse

Fig.

stress

flange

system

z=

0

0

0

0

(255)

0

lo

1

fr 0 0 0 _j L where the suffices have the same meaning as in Eqs. (253). The matrices f 1 and f 1 are themselves partitioned diagonal matrices, the sub-matrices being the flexibility matrices of the longitudinal and transverse flange elements respeetively. Since the flange loads vary linearly and the effective flange area of each element is assumed constant within each element the flexibility of the flange elements is that given on p. 22. Thus, for the grid of FIG. 45 the f 1 and f 1 are, I I I I fa o o o 0 f. 0 0 0 fb 0 0 (256) fz= 0 0 0 fc f, 0 0 _j L 0 0 fd _j L typical sub-matrices being ,I I I I

lo

loads

wing. Unit self-equilibrating stress system of type Z= I. (Transverse four-boom tube)

44~-Multi-web

We write now the b0 and b1 matrices in the partitioned form:

lb 3EBeb

lb 6EB.b

lb 6EB.b

lb 3EB.b

where the suffices /, t, s, w and r denote matrices for the longitudinal flange loads, transverse flange loads, shear flows in the fields of the cover, shear flows in the webs and shear flows in the ribs respectively. Since the flange loads are assumed to vary linearly between nodal points we need at least two entries in the b matrices to describe the loads in each flange. As such we use the loads at the ends. (nodal-points) of each flange element and denote them by the suffices 1 and 2 where 1 is the end first met when we proceed along the +z or +s direction. These two associated loads are entered in the assigned column of the appropriate sub-matrix of b0 or b1 in two consecutive rows, the first of which always corresponds to the end 1. Since the shear flow is assumed constant in each field only one entry appears for a field. In the b1 matrix we arrange the columns in three groups, the first referring to the X, the second to the Y- and the third to the Z-systems. It is of the utmost importance to organize ab initio a rigid and consistent system for the setting up of the b0 and b1 matrices.

d.

d.

3EC••

6EC••

d.

d.

f.=

fb=

(253)

40

0

f=

q:-1/2Q,

0

(257)

3EC••

6EC••

_j _j L L The flexibility matrices f., fw, fr are diagonal matrices with elements 4>/Gt, 4> .. /Gfw and O.jGtr respectively. 1 1 ·0 .......... ·0· ·O o~· 0·

r I

r

··········

, fw=

·0 0· .......... ·0· L

.:J

4> I 0· .. ·0 ___2£_0·. ·0 Gtw

·0

o............ o· L

1 _j

and hence using the second equation of (261) X= -D-1 b1 'fb0 R-D-1 ~,.'H ................... . where

·0· .......... ·0

0·

(262)

D=~'f~

The stress matrix 5 follows as, 5=[b 0 -b1 D- 1 b1 'fb0 ]R-~D-1 ~,.'H . . . . . . . . . . . . . . . . (263) The expression in the square bracket is the matrix b which we write in the partitioned form

·~ 1:: l·..................... ~64) L

·0 0· .......... ·0· L

_j

(258) where<}), <})w, Q are the areas of the shear fields in the wing surface, webs and ribs respectively and t, tw, t, are the corresponding thicknesses. We have now all the information to form the matrices D=b/fb1 and D 0 =b1 'fb0 R and can hence solve the system of equations b1'fb1{X Y Z}+b1'fb0 R=O . . . . . . . . . . . . . . . . . . . . (259) for the unknowns X, Y, Z. Finally, we find the true flange loads and shear flows of the idealized structure from 5=b0 R+b1{X Y Z} . . . . . . . . . . . . . . . . . . . . . . (260) It will then be necessary to translate the results (260) into stresses of the actual structure. Finally the flexibility F of the structure at the points and directions of the loads R may be determined from Eq. (229). When the number of equations (259) is too large to be dealt with by our digital computer we may proceed in two (or more) steps by the method given on page 36. Essentially this introduces into our analysis a redundant basic system. If initial strains H are imposed on the structure in addition to the loads R we have merely to add the column matrix b1 'H

on the left hand side of (259). Thus the analysis includes inter alia the complete calculation of wings under thermal loading. A new approach to the problem of cut-outs We emphasize that our above analysis is valid in the presence of any kind of cut-out stiffened or unstiffened by closed frames as long as the overall geometry and idealization conforms with the initial assumptions. Nevertheless, when we have a structure which is essentially continuous with only minor unstiffened cut-outs it may be worthwhile to apply an artifice which avoids the lack of uniformity in the pattern of the equations inevitably associated with cut-outs. Moreover, it is the idt>.al method of finding the alteration in the stresses due to the subsequent introduction of cut-outs in our system without having to repeat all the computations ab initio. The method is as follows. To preserve the pattern of equations disturbed by missing shear panels or flanges we eliminate the cut-outs by introducing fictitious shear panels or flanges with an arbitrary thickness or area. Naturally, it is usually preferable to select for the latter dimensions those of the surrounding structure. To obtain nevertheless the same flange loads and shear flows in our altered structure as in the original system initial strains are imposed on the additional elements of such a magnitude that their stresses become zero. The effect of the fictitious elements is thus nullified whilst the uniform pattern of our equations is retained. Let the column matrix of the unknown initial strains, in the additional elements only, be H In the new structure (i.e. without the cut-outs) we determine the flexibility matrix f and the matrices b0 and b1 which we write in the partitioned form bo=

~~boabo,. ~~

bl =

~~bl•bl,. ~~

..............

(261)

L _j L _j where the sub-matrices with the suffices g and h refer to the forces in the elements of the original structure and the fictitious new elements respectively. Denoting the column matrix {X Y Z} simply by X and taking the initial strains in the original structure as zero the Eqs. (259) in the unknown X become, II ~'fb1 X+ b1'fb0 R+ b1' ~ = 0

I I L_j

_j

To find now the column matrix H we put the stresses in the additional elements to zero. Thus, the matrix 5 must be

s~ I~ L

l·.............. .......

(26SJ

_j

where 5. are the true stresses (forces) in the original structure. Applying Eqs. (261), (264) and (265) in (263) we find I I I I I I 5. = b. R- I1 bl• o-1 b1,.'H . . . . . . . . . . . . (256)

I o I Ibh II

Hence

L.J

L.J

I

bv,

L

I

_j

or

H = (blhD- 1b 1,.')- 1 b,.R . . . . . . . . . . . . . . . . . . . . (267) The true stresses in our actual structure are thus 5.= [b.- b 1 .D-1 b1 ,.'(bthD- 1 bt~o')- 1 b,.]R . . . . . . . . . . . . (268) which solves our problem completely. As mentioned already the method is ideally suited for finding the alteration of the stresses in a structu!e through a subsequent introduction of cut-outs, such as access doors which usually seem to materialize at a late stage of design. Another particularly meful application of the new approach may be found ih the analysis of fu~­ lages with window-openings. Naturally, the degree of redundancy IS increased by the 'filling-in' of the cut-outs but this is of no importance for the automatic computations envisaged here.

A more refined wing stress analysis The above general method of wing stress analysis suffers from the serious defect mentioned initially that the effective flange areas have first to be guessed since the stress distributi~n on. which t~ey depend_is. unknown. It is certainly feasible to apply an IteratiOn techmque but this IS not only . . . necessarily lengthy but als~ rather uninspiring. To obviate these difficulties we develop a method which ehmmates the determination of the effective flange areas and works directly with effective flange loads. The method has the further virtue that it takes full account of the Poisson's ratio effect which may be important at the root and at other structural and loading discontinuities. The addition of 1/6 of the web area to the flanges is always sufficiently accurate for lift and torque loads and is retained here. Hence, our problem is restricted to the wing surface alone. . . The previously introduced assumptions that the loads are earned m the idealized structure by a grid system of effective flange loads and fields purely shear-carrying form also the basis of the new method. We assume also that both the direct stress distribution and the effective flange loads vary linearly between consecutive nodal points. However, our analysis does not presume that the so-called effective flange areas-whiC:h do not enter into our developments-are constant between nodal pomts. ~he shear flow is again taken to be constant within each field. When replacmg the linearly varying direct stresses across a grid line by efftJ7tive flange loads at the nodal points we introduce the additional ass~p.hOJ?. that. the actual flange areas and thicknesses do not vary across this grid hne. Smce in wing structures plate thicknesses a~d pos~ibly flange ar~s may vary just there it is suggested to take fo~ this particular c~lc~lauon the mean values of areas and thicknesses on either Side of the gnd hne; on the other hand when there is a cut-out on one side of the grid line or the flange is interrupted the corresponding values should be taken as zl?ro. These simplifications are not necessary for the purpose of the analys~s but ease the problem of notation; moreover they do no! affect s~nously the accuracy of the computations. Contrary to our previous practice of n~;J:m­ bering the flange elements with letters we numb:Jr here only the nodal pomts with numerals. We detive now the equation connecting the effective flange loads at nodal points in the z and s directions of the idealized structure a!ld the direct stress distribution in the plate material. It is more convement to fix a particular point and for this purpose we select the point 9 in the gridsystem shown in FIG. 46.

41

The conditions of equilibrium for the actual and idealized systems yield in conjunction with Eq. (269)

I

P 9 ==a,9[B 9 +2(A 8 ,9 +A 9,10)] +a , 8 A8 ,9+a,,10A9.lo

J I J ....... .

+a.,92v(A 8 ,9 + A9 , 10) +a, 8 vA 8 , 9 +a.,10 vA 9,10

Qg=O'.g[Cg+2(A3,& +As.ts)] +a,aA3,9 +a•tsAs.ts +a,g2v(Aa.s +A9,I5) +a,avAa.9 +a,IsVAs.ts

(270)

(271)

These equations are expressed more concisely in the form, I

i

I a,s I

Points

Nodal

a,s

I

I

+ Lg,s

L _j

+ L9.3

Structure

Actual

and

=L9.91

Stresses

~~a,a'l aR3 L _J

i

I~

I

L

_j

I

._I

I

+ L9.1f>

I

I O'zto I I + Ls.to a,to I a.,s I

I a,g

I (272)

i O'ztr.ll

I

a.~ 15

L

_j

where the matrices L are as follows: I 8g+2(As.s+As.w> Ls.s= 2v( Aa.9 + A9,15)

I

Dire-ct stress ·

in

...~ f-lange'- , _,.-..;vEEu ~

a..,-

a:.=

L

Direct stress in

sheet

E:(E 11+VE:z.)=

I

u:,

I

f
Ls.3=

0 I vAa,9 L

A

0

l ...

3.9

_j

The matrix L9 ,10 (L9 ,15 ) is obtained from L9 , 8 (L9 , 3) by substituting 10(15) for 8(3). Equations corresponding to (272) may be written down for any other nodal point. We see immediately that

o,

(274)

and that,

Idealized

Structure

and

Equivalent

Flange

(273a)

(274a)

when r and s are not adjoining nodal points of the grid. We deduce also that Eqs. (273a) are the general formulae of the L-matrices for adjoining nodal points in the z- and s- directions respectively. Consider now the column matrices for the flange loads and stresses at all p nodal points 5={P1 Q1 P 2Q 2 . . • . . . . . P 9 Q 9 •.•••••• P,Q,}

Loads

}

..

(275)

From the set of equations of the type (272) we find, 5=Ls . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . where L is the symmetrical partitioned matrix

(276)

Purely shear carrying

Lt2· · · · · · · · Lt9· ·······Ltv Fig. 46.-Equlvalent flange loads in idealized structure

L92 . . . . . . . . Lgg ....... . L9 ,

The following notation is used: all values apply to nodal point 9. effective flange loads in z and s directions (actual) area of longitudinal flange+ ! (spar-web area) (actual) area of transverse flange+ ! (rib-web area) C9 Young's modulus of sheet and flanges respectively E., E longitudinal and transverse flange stresses a, 9, a.
LP2 ........ Lvs ........ L,,,

. _1_ E.' _1_ E.' A3,9= 0 E 13.913.9• A8 ,9= 0 ed8 ,9t 8 ,9 effective ~ange areas due to plate in pure bendmg Note that the material of flanges and sheet is assumed to be different. We have,

~(a.,+va.,)

a,9'=E,'(E.,+vE,9)= ~'(u.,+va,9) 42

}

.. . .. .. .. .. .

_j

From (274a) the submatrices with suffices referring to non-adjoining nodal points are zero. Solving Eq. (276) for s we find s=L-15............. ... . . . . . . . . . . . . . . . . . . . . . . . . . (278) and hence also s I e=e=.EL- 15 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (279)

E.:= 1~·v2 effective elastic modulus of sheet

and a,9'=E,'(E,9+vE.,)=

(277)

(269)

where e is the column matrix of the flange strains at the nodal points, i.e. (279a) • • • • • • • • • • • • e={EztEst· ......... E,gE.g .......... E,,E8 ,} Thus, once we have determined the effective flange loads the flange stresses and strains follow from Eqs. (279) and the direct stresses in the sheet from Eqs. (269). No guessing of effective flange areas is involved in this procedure but we have on the other hand to invert the matrix L with 2p rows and columns. It is apparent that if we knew the effective flange areas B., c. at the nodal points we could immediately write down the inverted matrix as a diagonal matrix whose elements are the unit flange flexibilities at the nodal points. In fact, then

l

I

I

EB.1 0

···························0

0

0

EC.1

0

0

0

0

0

0

0

0· ....... ·0 _I_

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0· ............ ·0

EBet~

E~et~

0· ............ ·0

(277a)

O· ....... ·0

0· ...................... ·0 _I_ 0 EB.'P 0······ .................... ···0 _I_ EC.'P

I

L

~. ~ 1:

·0

I

I

I

I

I b11 I

b1=

II bl. I

(280)

I blW

I bl, L

We seek next the contribution of the flange strains e 1 and e 0 to the matrices D and D0 of the relative displacements S;k and 310• For this purpose we apply a self-equilibrating unit load system X;=l (which may be also a Y- or Z-system) and denote by P 13 and P 19 its longitudinal effective flange loads at the points 3 and 9 and by Q18 and Q19 its transverse effective flange loads at the points 8 and 9. LetS~ be the relative displacement at the points and directions of X 1 = I due to some given flange strains E. and E •• The contribution to S; of the straining of the flanges (3, 9) and (8, 9) is then 9

9

3

8

31 = .......... + f P;E,dz+ f Q1E.ds+ and since we assume that both flange loads and strains vary linearly between nodal points we find,

-

1

11/3

S;= .... +[P;aPt9].,

l/6llI I

1

1

d/3

Eia I

Ez9

1/61/3 L :.J L _ja.9

d/6

1

1 1

I sl EEs 89

j+[Q;sQ;9]·

I

d/6 d/3 L _j L _j s.9

+ ....

(282)

The complete expression of S1 is arranged by pairing the terms involving the longitudinal and transverse strains at the same nodal point. Thus, showing only the typical terms involving {EzsE89},

3;= .. · · +{[P;aQ;a]J9,a+[P;sQtsl~.s+[P;9Q;9]~,9+ ~[P,loQ;lO]~.lo+[P;lsQtlsl~.1s{::J + where the I matnces are, ~.9=

r

3•15/3

L

0

(283a)

111~1~-~

I=

I ~.1· L

IPl

0

I

·_· ~-~ ·_· ~~·~·-· ~-~ ·_· ~-~ •:.'P

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

.I.J.9·

0

l'P9

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

(285)

l;,,p

···p

_j

we can express Eq. (282) concisely as S;=[P;1 Q;1 . . . . . • . . P;9 Q;R· ....... P;pQ;p]le . . . . . . . . (286) Observing that the matrix [P;IQ;1....... . P;9 Q; 9 ••••••.• P;pQ;,] is the ith row of the b11 matrix and noting Eqs. (281) we find that the contributions of the flange strains to the D and D0 matrices are L _, L-' D1 =b11'1 E b11 D0,=b11'1E b 01 R.................. (287) We conclude that the flexibility of the flanges at the nodal points is given by (288)

_j

where the submatrices bot, b 11 give the flange loads at the nodal points. The rows in bot, ht1 are arranged in p pairs corresponding to the p nodal points; the first and second row in each pair refers to longitudinal and transverse flange loads respectively. Returning now to Eq. (279) we apply it in the basic system for the external loads R and the redundancies {X Y Z} respectively. We "find with the notation of Eq. (280) the flange strains

-

I

~.a=

_j L L The matrix 19 ,10(19 ,15) is obtained from ~.s~.a> by substituting 10(15) for 8(3). It is simple now to write down in Eq. (282) the terms for any other pairs of strains (E.E,). We deduce immediately that, 1.,=1.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (284) and that all lr., matrices are zero when they do not refer to adjoining points. Moreover, matrices (283a) are typical for adjoining nodal points in the z and s directions respectively. Introducing the matrix

_j

Consider now the b 0 and b 1 matrices of the basic system. We emphasize that the b 0 system can in the present method only be derived from pure static considerations since the effective flange areas are unknown. Thus, b 0 cannot be derived for a multicell wing from an Engineers' theory of bending cum Bredt-Batho analysis unless we assign arbitrary areas to the flanges. To find, in fact, the b 0 system in this example it is probably best to select a set of longitudinal flange loads in equilibrium with the applied loads and derive a statically consistent set of shear flows and transverse flange loads in equilibrium with the applied shear forces and torque. Contrary to the system adopted in the previous method we write here the statically determinate b 0 and b 1 matrices in the partitioned form

:l

I ll9.a/6 0

(282a)

Note the structural similarity between the I and L matrices. It is particularly pronounced when v=O. The total flexibility of the elements of the structure is now I I 0 0 f, 0 f=

0

f.

0

0

0

0

f.,

0

•

lo

0

0

•••••••

0

•••••

0

••

0

••

(289)

0 0 fr _j L where the flexibilities of the cover, webs and ribs are as before. We find for the matrices D and D 0 • D=b1'fb1 and D 0 =b1'fb0 R . . . . . . . . . . . . . . . . . . . . (225b) where b 1 and b 0 are arranged as in Eqs. (280). Using Eqs. (225b) in (259) we solve the problem completely. The refinement introduced by the L matrix need not of course extend over the complete wing but may be restricted to the root and other marked changes of structure and loading. For the rest of the structure it may still be sufficient to estimate the effective flange areas and to use the simple form (277a).

D. The Analysis of Structures by the Displacenrent Method The analogy between the developments for the flexibilities and stiffnesses given under A and Band summarized in TABLE 1 shows clearly that parallel to the analysis of structures with forces as unknowns there must be a corresponding theory :with d~formations as unkno~s. ~s ~entioned in the introduction to this section Ostenfeld • when mvestigatmg frameworks was the first to draw attention to such an analogy. In fact, his equations are the exact counterpart of the classic S;k equations given by MuellerBreslau for forces as unknowns. In more recent times Southwellt and his pupils have used his relaxation technique to solve the elasticity equations in the finite difference form with displacements as unknowns for a great number of problems. Hofft has applied the latter method to diffusion and related problems in aircraft structures and has solved also the corresponding equations directly. Lately Williams II has outlined an analysis of wingstructures of the standard or solid type by introducing the deflexions at a finite grid of points as unknowns ; his technique, which is intended for use in combination with the automatic digital computer, neglects however the shear deflexions, which may have an important influence. • toe. cit. p. 43.

(283)

t toe. cit. p. 43.

~ N. J. Hoft and Paul A. Libby, 'Rec:ommendations for numeric&! solution of reinforced-panel and fuselage-ring problem. N.A.C.A. Rep. 934 (1949). llloe. cit. p. 43.

43

Naturallly, a theory using displacements as unknowns would only be of value if it could show some concrete advantages. It is clear that such an advantage may possibly arise when the stiffnesses are simpler to calculate than the flexibilities, which is, as we have seen previously, very often the case. In particular in the egg-box structure, characteristic of aircraft wings, the stiffnesses k;~ are much easier to find than the influence or flexibility coefficients S;h· Another obvious advantage arises when the number of unknowns is smaller for the displacement analysis. This may occur in frameworks. especially the stiff-jointed type with few degrees of freedom at the joints. The equations in the displacements for stiff-jointed frameworks are almost invariably well conditioned; a further point in their favour, not only for iteration techniques but also for the direct solution. On the other hand in continuous structures, like wings and fuselages, this is not the case. Here, in fact, the equations in the displacements are nearly always ill-conditioned and it then becomes necessary to introduce generalized or group displacements as unknowns in order to improve the conditioning. This is a pronounced drawback of the displacement method when applied to aircraft structures. Furthermore, in such continuous systems the displacement method will usually involve a considerably greater number of unknowns than the force analysis in order to achieve a comparable degree of accuracy. It is apparent then that the choice between the two parallel techniques must be made on an ad hoc basis after careful consideration of the possible advantages and disadvantages of each method for a particular problem. It would, however, appear that at least with the present types of construction the force method is to be preferred for aircraft structures. Before proceeding to the general development of the displacement analysis we introduce first a simple example to familiarize ourselves with the ideas. Consider the framework shown in FIG. 47, symmetrical both in structure and loading. The number of unknown forces or moments when the engineers' theory of bending is assumed to hold is evidently six. On the other hand, if we neglect the deformations due to shear and end load, two deformations alone, the rotations r 1 and r 2 at the stiff joints. suffice to specify completely the deformation of the system. The analysis may proceed as follows (moments and rotations are taken as positive if in the anticlockwise sense). We freeze first the joints, i.e. put r 1=r2 =0 Then, due to the loading on the upper member, moments M 0 are applied at the joints and are. with the notation of FIG. 47 P1l1 2 P1l1 2 M21o=+12 M12o=-u

' s---T h

"

% "

J

system

Frozen

Displacement

Analysos -

Unknowns

2

r; and r,

Fig. 47.-Displacement analysis of stiff-jointed frame

We obtain from Eqs. (292) (292a) where

(290)

M

230-

12

(291) Consider next the system with free joints and no transverse loading subjected to the loading by the joint-moments P2l2 2 Pt/1 2 P1l1 2 R1=-M1=+u • R2=-M2=u - 12 .......... (291a) The superposition of this and the previous case yields the true solution of the given system under the transverse loading. To analyse the second problem we apply Eqs. (138) which take the form k21r1 +k2'J!2=R2

~ J

......................

The stiffnesses k;A are easily found as (see also Eqs. (144)) k _ZEit -k k _ 4£/A .. 4Eit 21 12- /1 ' h + /1 11 k

(293a)

- _p2/2:

}

4EIA 4Eit . 4Eit

22=JI+t;+T

(292)

(293)

which assumes that the horizontal beams and supportilll struts have the constant bending stiffness £/1 and £/A respectively.

Having the rotations r 1 and r 2 and using the stiffnesses of the individual elements contained in (293) we easily derive the actual moments in the structure. Thus, introducing again the usual sign convention giving positive bending moment when upper fibres are in compression, we find for the bending moment M 12 at the junction (I) of element (I, 2) M12=

_P:r- 4~11r1-Z~l1,.2

. • •• • • •• . • . • •• • •• • ••

This method forms also the basis of the Hardy-Cross or the more general Southwell relaxation technique in stiff-jointed frameworks. We develop next the general theory of the displacement method. We introduce immediately the matrix notation and assume that the structure consists of a finite numbers of elements whose stiffnesses k, due to relative displacements at the ends or boundaries of each element, are known. In order to show most clearly and concisely the striking analogy between the force and displacement methods we present them side by side in the following TABLE 11; most of the information with respect to forces has been given previously under C. The complete duality between the two theories is, of course, a direct consequence of the twin principles of virtual displacements and virtual forces from which they derive most naturally. We believe that the analysis has not been given previously in this generality. It includes ab initio any effects of initial strains like temperature, lack of fit and •give' at the foundations. The great advantage obtained in deeper insight and new theorems and applications by developing the theory on the most general lines is too apparent to need stressing.

TABLE II A COMPARATIVE PRESENTATION OF STRUCTURAL ANALYSIS BY THE FORCE AND DISPLACEMENT METHODS METHOD OF DISPLACEMENTS

METHOD OF FORCES External forces Flexibility Displacements See also

44

TABLE I

R F

r=FR FK=I

(294)

Joint displacements Stiffness Forces See alSO TABLE

I

r

K

R=Kr

KF=I

TABLE II (continued) Unit Load Method

Unit Displacement Method

Given the true strains £ in a structure the kinematically related displacement r at a given point and direction can be calculated from

Given the true stresses a in a structure the equilibrating force Rat a given point and direction can be calculated from

(295a)

......................

a

I. R= JiadV

l·r=fa£dV v

v

where is a virtual or otherwise statically equivalent stress system due to unit load in given direction. Statically equivalent stresses ignore compatibility conditions. (See also Section 6 and FIG. 15).

Cf Eq. (84b)

................................

(295b)

where € is a virtual strain system due to unit displacement in given direction. In what follows we denote virtual strains as kinematically equivalent strains. Kinem"ltically equivalent strains ignore equilibrium conditions. (See also Section 4 and FIG. 8.)

Cf Eq. (146)

StaticaUy Determinate System

Kinematically Determinate System

Internal forces (stresses) on elements determined from (296a) . . . . . . . . . . . . . . . . . . . . . . S=bR where matrix b is obtained by static reasoning alone.

Internal relative displacements (strains) of elements determined from v =ar . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (296b) where matrix a is obtained by kinematic reasoning alone by displacing one joint at a time whilst keeping all others fixed. k Stiffness of individual (unassembled) elements 1nternal stresses, S=kv=kar . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (297b) External forces, R=a'S=a'kar (298b) Stiffness, Ko=a'ka (299b)

Flexibility of individual (unassembled) elements Internal strains, (297a). . . . . . . . . . . . . . . . . . . . . . . . v =fS =fbR External displacements, (298a) ..................... . r=b'v=b'fbR Flexibility, F0 =b'fb (299a) ..................... .

C.f Eqs. (153), (154), (159), (160)

Cf Eqs. (121), (122), (125), (126)

Statically Indeterminate System

Kinematically Indeterminate System

In the relation S=bR (296a) b cannot be determined by statics alone. Flexibility of structure needs to be considered, entering as compatibility conditions. On the oth.!r hand if the internal strains v are known the kinematically related external displacements may be derived from

In the relation v = ar . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (296b) a cannot be determined by kinematics alone. Stiffness of structure needs to be considered, entering as equilibrium conditions. On the other hand if internal stresses 5 are known the equilibrating external forces may be derived from

r=b'v (300a) where b is merely a statical~•· equivalent (virtual) matrix due to unit R's.

R =ca'S . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (300b) where a is merely a kinematically equivalent (virtual) matrix due to unit r's. Hence stiffness

Hence flexibility (301a) Eq. (300a) is a special form of the Unit Load method (Principle of Virtual Forces). FIG. 48a illustrates the matrices b and ii on a particularly simple example of a singly redundant system. R =1

t

'

K=a'ka . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . (30lb) Eq. (300b) is a special form of the Unit Displacement method (Principle of Virtual Displacements). FIG. (48b) illustrates the matrices a and i on the same example as in 48a.

FIG.

R=1

f

r:~--,r~~

1

I

~

·-~~.1--..-~tl----,--~-..--b-.. -~

True

system

lbl = 113 3 1 -2 1 3 -3 3 3 -3 3 }

Statically

Trul!

•quivatent systl!m

[b)= 113 3 3 0 3 3 -3 3 3 -3 3}

Fig. 48(a).-True and statically equivalent stress systems in singly redundant, pin-jointed framework

)5l5li,'>

(a]=~ { 3

displacl!ment

systl!m

3 1 -2 1 3 -3 3 3 -3 3 }

.{- ----- ,'\- -.---- ., ;. -T '

'

\

'

"-~~--:;-

Kinematically

l!quival•nt

system-1

lal={ooooooo'j-'t oaf

Kinl!matically equivalent system-2

oo = {13 o o o o o o o o o o 1

Fig. 48(b).-True and kinematically equivalent displacement systems In pin-jointed framework

C/. Eqs. (125a), (126)

Cf Eqs. (l59a), (1"60b)

45

TABLE ll (continued) Problem a Given a set of forces R, determine a set of statically indeterminate forces X necessary to satisfy the compatibility conditions. Find also the displacements r in the directions of R. Complete force matrix (302a) . . . . . . . . . . . . . . . . . . . . . . {R X} By putting X= 0 we obtain the so-called basic system which is statically determinate within limits of idealization. Stresses in basic system (303a) . . . . . . . . . . . . . . . . . . . . . . So=b0 R Stresses due to X (with R= 0) (304a) . . . . . . . . . . . . . . . . . . . . . . 5 1 =b1 X where b 0 and ~ are obtained from statics alone. True stresses in actual structure (305a) . . . . . . . . . . . . . . . . . . . . . . 5=5o+51 =b0 R+b1 X Strains of elements (306a)

......................

v=f5=fb0 R+f~X

Compatibility condition in actual system at points of application of forces X (307a) . . . . . . . . . . . . . . . . . . . . . . b1 'v=b1 'fb0 R+b1 'fb1 X=O or (308a)

where (309a)

Hence (310a) True stresses (296a) where (3lla) . True strains (297a) . . . . . . . . . . . . . . . . . . . . . . Displacements r due to R (300c)

where (312a)

and

......................

..................

S=bR

v=fS=fbR r=b'v=b'fbR=FR

F= F0 -b0'f~(b1 'fb1 )-1 b1 'fb0

F0 =bo'fb0 is the flexibility of the basic system since we may choose (313a) . . . . . . . . . . . . . . . . . . . . . . b=b0 Cf. Eqs. (217), (223), (222a), (224), (225), (226), (227a), (228), (229), (230)

Problem a Given a set of joint displacement r, determine the set of kinematically indeterminate joint displacements U necessary to satisfy the equilibrium conditions. Find also the forces R in the directions of r. Complete displacement matrix {r U} . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (302b) By putting U = 0 we obtain the so-called basic system which is kinematically determinate within limits of idealization. Strains in basic system v0 =a.or... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (303b) Strains due to U (with r=O) v 1 =a1 U . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (304b) where ao and a 1 are obtained by kinematics alone. True strains in actual structure v=v0 +v1 =a.or+a1 U . . . . . . . . . . . . . . . . . . . . . . . . . . (305b) Forces on elements S=kv=ka.or+ka1 U . . . . . . . . . . . . . . . . . . . . . . . . . . (306b) Equilibrium condition in actual system at non-prescribed displacements U ~'5=a 1 'ka.or+a 1 'ka1 U =0 . . . . . . . . . . . . . . . . . . . . (307b) or (308b) CU+C.=O where (309b) Hence U = -C-1C0 = -(a1 'ka1)-1at'ka.or (310b) True strains v=ar . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (296b) where (311 b) True stresses 5=kv=kar . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (297b) Forces R due to r R=a'5=a'kar=Kr . . . . . . . . . . . . . . . . . . . . . . . . . . . . (300d) where K=K 0 -a.o'k~(a 1 'ka 1 )- 1 a 1 'ka.o . . . . . . . . . . . . . . . . . . (312b) and Ko=a.o'kao is the stiffness of the basic system since we may choose a=ao . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (3f3b)

Problem b Given a set of displacements r find forces R, stresses 5 and strains v From Eqs. (300c) (314a) . . . . . . . . . . . . . . . . . . . . . . r=FR Hence (315a) .. .. .. .. .. .. .. .. . .. .. . R=F-1 r (316a) . . . . . . . . . . . . . . . . . . . . . . 5=bR=bF-tr (317a) . . . . . . . . . . . . . . . . . . . . . . v=f5=fbF-1 r Once F is known the question of statical determinacy or indeterminacy is irrelevent in this problem.

Problem b Given a set of forces R find joint displacements r, strains v and stresses 5 From Eqs. (300d) R=Kr . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (314b) Hence r=K-1 R . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (315b) v=ar=aK-1 R . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (316b) 5=kv=kaK-1 R................................ (317b) Once K is known the question of kinem:ttical d~terminacy or indeterminacy is irrelevent in this problem.

Problem c Given a set of initial strains H imposed on· free unassembled elements due to temperature, lack of fit, 'give' at foundations, find stresses 5 and total strains v when forces R=O. Total strains of elements (318a) . . . . . . . . . . . . . . . . . . . . . . v=fb1 X+H

Problem c Given a set of initial stresses J imposed on elements with frozen joints (i.e. all joint displacements zero) due to temperature, lack of fit, 'give' at foundations, find strains v and stresses 5 when displacements r=O. Total stresses on elements 5=k~ U +J . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (318b) Note that the column matrix U must here include all unknown joint displacements. Equilibrium condition in actual system in the directions of U ~'5=at'k~U+a1 'J=0 ...................... (319b) Hence,

Compatibility condition in actual system at points of application of forces X (319a) . . . . . . . . . . . . . . . . . . . . . . b1 'v=bt'f~X+b1 'H=O Hence, (320a)

and (32la) (322a) Note,

......................

...................... ......................

Cj. Eq. (236)

46

X=-(~'fb1 )- 1 b1 'H

5= -~(b1 'fb1 )-1 b1 'H V=-f~(bt'f~)-lbt'H+H H=-fJ

(320b)

and Note,

v=-a 1 (~'k~)- 1 ~'J ............................ 5=-k~(~'k~)- 1 ~'J+J . . . . . . . . . . . . . . . . . . . . . .

J=-kH

(32lb) (322b)

TABLE ll (continued) Problem d Assume that we write the total set of forces (including the statically indeterminate forces) in the partitioned form

l~l

(323a)

L

_j

Problem d Assume that we write the total set of joint displacements in the partitioned form

~~l L

(323b)

_j

in which Z is known in terms of R and X. We set now the modified problem (a): Given the set of forces R determine the set of forces X necessary to satisfy the compatibility conditions. Here in the basic system obtained by putting X= 0 the stresses (303a) . . . . . . . . . . . . . . . . . . . . . . S0 =b0 R are completely known although the system is statically indeterminate.

in which W is known in terms of rand U. We set now the modified problem (a): Given the set of displacements r determine the set of displacements U necessary to satisfy the equilibrium conditions. Here in the basic system obtained by putting U = 0 the strains v0 =aor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (303b) are completely known although the system is kinematically indeterminate.

Similarly we know the stresses (304a) . ... . . . .. . . ... . . . . . . . . 5 1 =b1 X when R=O True stresses in actual structure, (305a) . . . . . . . . . . . . . . . . . . . . . . 5=50 +51 =b0 R+b1 X strains in elements (306a) . . . . . . . . . . . . . . . . . . . . . . V=f5=fb 0 R+fb1 X Compatibility condition in actual system at points of application of X (324a) . . . . . . . . . . . . . . . . . . . . . . b1 'v=b1 'fb0 R+b1 'fb1 X=O where b1 is a set of stresses statically equivalent to unit X's (and R= 0) preferably found for Z = 0. In the latter case the rows of b1 are the same as the corresponding rows of b1 of Problem (a). Thus, (325a) . . . . . . . . . . . . . . . . . . . . . . X=- (b1 'fb1 )-1 b1 'fb.,R True stresses and strains S=bR, v=fbR (296a) where (326a) . . . . . . . . . . . . . . . . . . . . . . b = b 0 - b1 (b1 'fb1 )- 1 b1 'fb0 Displacements r due to R (see Eq. (300a)) (327a) . . . . . . . . . . . . . . . . . . . . . . r =b0 'v = FR where (328a) and (329a) is the flexibility of the basic system. The matrix b0 is a set of stresses statically equivalent to unit R's preferably found for Z = 0 and X= 0. In the latter case 60 is identical with b0 of problem (a).

Similarly we know the strains vl =ax u . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (304b) when r=O True strains in actual structure, v=v0 +v1 =aor+a1 U . . . . . . . . . . . . . . . . . . . . . . . . . . (305a) forces on elements S=kv=kaor+kaxU . . . . . . . . . . . . . . . . . . . . . . . . . . (306a) Equilibrium condition in actual syst~:n at diJ;:>hc::m::nts U a1 'S=a1 'kaor+i/kaxU=O . . . . . . . . . . . . . . . . . . (324b) where a1 is a set of strains kinematically equivalent to unit U's (and r= 0) preferably found for W = 0. In the latter case the rows of i.1 are the same as the corresponding rows of a 1 of Problem (a). Thus, u ~ -(a1 'ka1)-1a 1 'kaor . . . . . . . . . . . . . . . . . . . . . . . . (325b) True strains and stresses v=ar , S=kar . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (296b) where ...................... (326b) a=ao-al(al'kax)-lil'kao Forces R due to r (see Eq. (300b)) (327b) R=a0 'S=Kr where (328b) and Ko=ao'kao .. . . .. .. .. . .. . . . . . . . .. . . .. . .. .. . .. . . (329b) is the stiffness of the basic syst;:m. The matrix is a set of strains kinematically equivalent to unit r's preferably found for W = 0 and U = 0. In the latter case lio is identical with ao of problem (a).

a.l

Cf Eqs. (226), (227a), (229), (230)

Condensation offlexibility matrix The calculation of the flexibility matrix F given under problem (a) can be developed concisely as a condensation of the complete flexibility matrix for the forces Rand X. This matrix may be written as

I Fn/

Fn

Condensation of stf(f'ness matrix The calculation of the stiffness matrix K given under problem (a) can be developed concisely as a condensation of the complete stiffness matrix for the ...~isplacements r and U. This stiffness may be written as I

I I

I

I

(33la) ........................ f=F 1 -fu/Fu-1 fu 1 Naturally this condensation may be performed in two or more stages and is then equivalent to the method of problem (d).

Ku/

Ku,

Ku

L

_j

where 1(11) is for forces R (X) only and was denoted by F0 (0) in problem (a). Evidently FlllR=D0 • The flexibility matrix F of the actual structure under the forces R is then

I

K1

I I

_j

where I(ll) is for displacements r (U) only and was denoted by K0 (C) in problem (a). Evidently Kmr=C 0 • The stiffness matrix K of the actual structure for the displacements r is then K=K 1 -Ku/Ku-1 Km . . . . . . . . . . . . . . . . . . . . . . . . (33lb) Naturally this condensation may be performed in two or more stages and is then equivalent to the method of problem (d).

Cf Eq. (175)

47

TABLE II (continued) Elimination and rigidification of structural elements Assume a set of initial strains, written as column matrix H, in the structural elements to be removed, of such magnitude as to give zero stress in resultant system. Write the b and b 1 matrices of the complete structure in the partitioned form I I I I b. I b= (332a) bl- bu I b,. bl,. I L _j L _j where the suffix h refers to those elements that are to be removed. •••••

0

••

I

I

-I

.

I

We find (333a)

......................

H = (b11,D- 1 b 1 ,.')- 1 b,.R

and hence forces in the new structure (334a)

..........

S.={b. -b11 D- 1 b 111 '(blhD- 1b 111 ')- 1 b 11 }R

In this process the number of statically indeterminate forces X has been reduced to a degree depending on the number of elements removed. In the inverse process of making infinitely rigid certain of the structural elements we have merely to put f,. =o for the affected elements. The number of statically indeterminate forces remains the same.

Rigidification and elimination of structural elements Assume a set of initial stresses, written as column matrix J, in the structural elements to be made infinitely rigid, of such magnitude as to give zero strain in resultant system . Write the a and a 1 matrices of the complete structure in the partitioned form I I I I I a, I I at• i (332b) a= I at= I I a,. I al,. L _j L _j where the suffix h refers to those elements that are to be made infinitely rigid. We find J =(a1 ,.C- 1a 111 ')- 1 a,.r . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (333b) and hence strains m the new structure v. ={a. -a1 .C-1&t,.'(a1,.C- 1a 1,.')-• a,.}r . . . . . . . . . . (334b) In the process of making elements infiniteiy rigid (stiff) we introduce kinematic relations between displacements and hence reduce the number of unknown displacements U accordingly. In the inverse process of eliminating certain of the structural elements we have merely to put k,. = o for the affected elements. The number of kinematically indeterminate displacements remains the same.

.

I

I

•

•••••••••••

0

•••••

Cf. Eqs. (264), (267), (268)

Generalized Di~placements Generalized displacements given by

Generalized Forces Generalized forces given by II IR (335a)

II I r I I I I I I u I L_j

lx

.. .. .. .. .. .. .. .. .. .. ..

II

Ir I IL_j o\

•••••

0.

0

•••

0

••••••••

DX+D0 =0

CU+Co~O

................................ (336b)

where

where

(337b)

(337a)

Then

Then (338a)

S=bR where

where

v =ar

........................................

(339a)

and the flexibility of the actual structure for the forces (340a)

R is

and the stiffness of the actual structure for the displacements

f=f0 -B0 'b0 'fb1 B1 (B1 'DB1 )-1 B1 'b1 'fboB0

K=Ko-Ao'ao'ka 1A 1(A1 'CA1)-1 A 1 'a1 'kaoAo

(339b) (340b) (340d)

(340c)

is the flexibility of the basic system under the forces

r is

(338b)

where,

where

48

(335b)

_j L The equation for the unknown U is

L_j

The equation for the unknown X is (336a)

I I I Ao 0 I I I I I I 0 AlI

R

is the stiffness of the basic structure for the displacements r

T

HE application of the general theory with displacements as unknowns to frameworks-both of the pin-jointed and stiff-jointed type-is straightforward. For the stiff-jointed system the method is particularly simple when direct and shear deformations are ignored. In fact, for all frameworks the determination of the matrices C and C 0 is trivial once we consider all possible degrees of freedom of the joints. See for example, the systems of FIGS. (23), (24) and (48) investigated on pp. 23, and 45, which show clearly how elementary the matrices a and stiffness k are when we break up the structure into its simplest constituent components. We need not therefore concern ourselves any more here with frameworks, and we turn our attention to the membrane type of system characteristic of aircraft applications. Essentially, a major aircraft structure like a wing consists of an assembly of plates (fields) stiffened by flanges along their edges. The field may be a curved and/or tapered surface but we ignore here both these effects and consider only rectangular flat elements of constant thickness. For convenience the element formed by the plate (sheet) and its fouredge members is denoted by the term unit panel. It is assumed that the flange areas are constant along each edge.

panels reasonably small. Nevertheless, it is inevitable that the stress distribution derived from an approximate deformation analysis should, in general, be less accurate than the one obtained from the approximate force method in the same grid system. We denote for the purpose of the analysis of the unit panel the displacements parallel to the s and z axes by u and w respectively and introduce also the local coordinate system ~. '· Consider now the state of strain and stress arising from a unit displacement va=l ............................... . (341) Following our assumption the internal displacements are given by

w,,= j( 1-~)

(34Ia)

where the suffix 3 indicates that these displacements are due to v3 =l. The strains and stresses in the sheet are*: Uz.a=

'

-GTtJ

l

Uzz3~~( 1-~) where £'=£/(l-v 2 ) Strain .: 1 and load P 1 in flange 8 1 I E1.a~T

,

I P 1 ,3 =B17

................

(342)

(342a)

all other flange strains and loads are zero. Similar formulae are obtained for the strains and stresses due to any other v;=• I. To derive the stiffnesses we apply the unit displacement method Eq. (295b), which takes here the form, •

l.k;n'~ J:J~ U;End~d' k.J '

-

where the integral extends over sheet and flange. For example, for the stiffnesses associated with va =I we obtain, _ E'dt £81 Gtl k.la ___ 3/- I +6d

k":l7

E'dt

k2.1~c --6{

.bL'-=---

lk.,

: v,=

..bz.~_

1

____,

Fig. 49.-5tiffnesses of unit panel

k4.1-=

Gtl + 0 - 6d

E'dt £81

Gtl

E'dt

Gtl

+7J +

0 -3d

and

k

vE't Gt

vE't Gt 4 +4

k63=+4+4

_ _ vE't _ Gt

vE't Gt

k _

;;a--

We determine first the stiffnesses k;n of the unit panel shown in FIG. 49 for unit displacements in the z- and s-directions at the four corners or nodal points of the idealized system. The stiffness of our element is hence an 8 x 8 matrix. As in the case of the force method it is necessary for the practical evaluation of the k;A to introduce simplifying assumptions which are, naturally, concerned here with the state of deformations. Thus, we assume that the displacements vary linearly between the nodal points. Although thi!> idealization offends against the equilibrium conditions its effect upon the stiffness is not pronounced as long as we keep the unit

I (343)

kaa=+ 31 +1 +3d

L,,

v, = 1

I

73 -

4

4 ' k63=+4-4

J

}

(343a)

It is simple now to write down the stiffnesses corresponding to any other

unit displacement. For convenience we express the total stiffness matrix in the form k=k.+kd+k, . . .. .. .. .. . . .. .. .. .. .. . . (344)

* Contrary to our usual notation subscripts are used here to denote stresses and strains due to •it displacements.

49

where the suffices s, d, .f indicate the partial stiffnesses for shear strains and direct strains in sheet, and direct strains in flanges, We find Gtl

Gtl

"""3d-3d Gtl

-"""3d

Gtl Gtl 6d - 6d

Gtl

Gtl """3d- 6d

Gtl Gtl 6d . -- 6d

Gt ~

Gtl Gt 6d -. -;r-

Gtl Gtl 3d . . 3d

Gt -4-

Gt

4

Gt

Gt

-~--4

Gtl 3d

Gtl

Gt

3d

-6d -6d

Gtl

Gtl

Gt

-TiT-

Gtl

Gt

Gt.

Gt

6d Gtf - 6d

Gt

Gt

Gt

.~ 4

4-

4 - 4 ·444

Gt

Gt

4-4

Gtd ~

sion in the direction of the height of th:' w~b. The corresponding formulae may be obtained by putting 1':, ~ 1·,, and 1· 7 ~ I'H which apply when C1 and c2 are infinite. The stiffnesses k., and k<~ contract to 6 >< 6 matrices and are

Gtd

Gtd

Gtd

Gtf

(345)

61_1l_M_

Gt

Gt

Gt

Gtd

Gtd

Gtd

Gtd

Gt

Gt

Gt

Gt

Gtd

Gtd

Gtd

Gtd

Gt

Gt

v

E'dt

&

E'dt

-4--4

Gtd v-

E'dt

I vE't

E'dt

E'dt

E'dt

E'dt

E'dt

v-v-31 E'dt

E'dt

vE't

vE't

vE't

E'dt

vE't

vE't

v£ 't

Gtl

Gtl

vE't

vE't

vE't

Gt

Gt

vE't

vE't

vE't

vE't

vE't

vE't

E'lt

E'lt

E'lt 6d

E'lt

-6d

vE't

vE't £'/; 4-3d

E'lt E'lt 3d - 6d-

E'lt

(345a)

6d

vE't

vE't

0 0

---,EB1

vE't

0

--,-

0

0

EB2

E'lt

E'lt

-3d 3d

_I

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

EC1

EC1

0

0

EC1

EC1

-,EB1

0

y-y

0

0

0

0

---d

0

0

0

0

0

0

E'lt 6d

0

0

0

0

E'lt

4 - 6d

0

0

0

d

0

0

0

EC2

EC2

0

ECn

EC2

Gt

Gtd

Gtd

(346)

Gt

E'dt

E'tlt

v-

61

E'dt

E'dt

E'dt

E'dt

6/

7

2

E'dt

31

E 'tit

E'clt

7J E'dt

E'dt

61 3/

Gtd

-,-

0

0

0

0

I)

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

(346a)

No contribution of the flanges is called for when evaluating the stiffnesses of the webs since k is best included in the top and bottom panels. Further simplification of the stiffness matrices for the webs is possible when the top and bottom panels of our wing structure are identical. Then for vertical loads alone the horizontal displacements in the two covers are antisymmetrical and the stiffness matrices (346) and (346a) may be contracted to 4 > 4 matrices. We illustrate now the application of the unit panel stiffnesses to the diffusion problem shown in Fl
a'ka

where k is the stiffness matrix of the unassembled unit panels and may be written in the diagonal partitioned form (345b)

y-y

d

Gt

2

Gt

Gtd

E'dt

vE't

vE't

Gt

-_:r-

-,-. -,-

-i--

E'dt

vE't

-4-4 4

Gt

2

-..,-

E'dt

y·-·4 4 . 4 4

4 - 4 - 4 v-Td

Gt

Gt

E'dt

E'dt

Gt

2-2

-T

Gtl

6r 31 y-

E'lt E'lt 3d -3d

k,

0

0

k,,

.............. 0 0

k=

.......... 0

k,,

..............

0

...... 0

0

0

.............. 0

. ,0

d

In assembling the panels of FIG. 50 to form a wing structure the stiffnesses (345) and (345a) may be simplified considerably when applied to rib and spar webs. Thus, for these cases we can always neglect the expan-

50

Gtl

61 11

E'dt

0

-3d

Gt

6d .Jd 3d---r -..,-

v

4-4 4-4

E'dt

EB1

Gtl 3d

v£ 't

E'dt

--,-

Gtf

61

61--4 4 - 4 4

4

L

6d

v

v

-61-31 61

vE't

6rl

-T

&-v-61 4·-4 4 - 4

-y-(;1

4

6rl

4 - 4 - 4-31--61 31 61

L

E'dt

Gtf

-~"

Gt

4--4- 4 - 4 61 31-61-31

Gt ---4-

Gtl

2

Gt

.4

Gtl

k,

(347)

4E'dt 2EB 4Git kts.ts=-y+-/ + 3d

k, is the (8 x 8) stiffness matrix for the unit panel g as derived previously,

see Eq. (344). • To preserve the symmetry of formulation, the stiffnesses

of the reinforcement elements are included with the panels. Thus each of the areas B or C is split in two and B /2 or C /2 associated with the panel on each side. For the boundary member, of course, the whole area must be included with the panel it bounds.

(350)

E'dt 2Git k2s.1s =ks.ts ="3/- 3d

All the remaining k's associated with 15 are here zero due to symmetry. If R is the column matrix (50 rows) of forces applied at the nodes then the displacements r are given by r=K-1 R

Naturally, loads may not be applied at all nodes (joints) in which case it may be desirable partially to solve the problem by eliminating the displacements where forces are not applied and to use the condensed matrix. (See TABLE II.)

Fla. 50.-Rectanaular stiffened panel. Assembly of stiffness matrix from unit panels

Since the terms in the a matrix are either unity or zero their formation is particularly simple. Writing a in the partitioned form .................. (348)

a={a,.a, ........ a, ........ a.}

a, is the sub-matrix of 8 rows and 50 columns relating the displace-

Finally we apply the unit panel to the assembly and analysis of the egg box type of structure illustrated in FIG. 51 where upper and lower plates are connected together by longitudinal and transverse webs. Any stiffeners on the plates are assumed for the present example to be along the lines of web-plate intersections. The structure is taken to be symmetrical about the horizontal middle surface and we consider the application of vertical loads only. With these assumptions it is only necessary to specify three displacements at each web intersection : the vertical displacement and the two rotations of the web intersection line (FIG. 51). In many cases the webs may be too widely spaced for the assumed linear variation of displacements between them to give satisfactory accuracy. It then becomes necessary to introduce further grid lines intermediate between the actual webs, the displacements being defined at all nodal points formed by grid line intersections. Where such nodal points do not lie on a web then obviously we define there only the two rotations, since vertical displacement does not affect the cover plates. Naturally, further lateral and longitudinal reinforcement of the plates can lie along the extra grid lines. The analysis of such a structure under vertical loads follows that given under Problems (a) and (b) in TABLE n. Thus, we designate the vertical displacements as r and take the rotations as the redundant displacements U.

The strains of the elements are here identified as the displacements of the unit panel defined in FIG. 49 and can therefore be written (Eq. 305b).

ments defined for the unit panel g (FIG. 49) to the displacements as defined for the complete system of FIG. (50). Superimposing the unit panel on panel g of the complete assembly we find that the directions I, 2, 3, 4, 5, 6, 7, 8 of the unit panel coincide with 15, 25, 17, 27, 16, 26, 18, 28 respectively and the sub-matrix a, is thus 1 .... 15

16

17

18 ...... 25

26

27

28 ...... 50

0 ..... 1

0

0

0 ....... 0

0

0

0 ....... 0

0 ..... 0

0

0

0 ....... 1

0

0

0 ..... 0

0

0 ....... 0

0

0

o....... o I I

0 ..... 0

0

0

0 ....... 0

0

0

0 ....... 0

0

I

··=

0 ..... 0 1

o..... 0 o o o....... 0

I

0 ..... 0

1 o..... o L

I

0 ....... 0

I

I

o....... o I 0

o....... o 1

0

0 ....... 0

I ....... 0

0

0

0 ....... 0

o o o....... o

0

0

1 ....... o I

0

0

v=aor+a1U

(349)

_j

All remaining columns in a, are zero. As typical terms in the complete stiffness matrix the stiffnesses associated with direction I 5 are, for a uniform panel with t, d, I, B and C, the same in each bay:

Fig. 51.-EII·box type of structure. Analysis by displacement method

The equation for the unknowns U, which is here the condition of equilibrium of the moments corresponding to the rotations U at the joints is (TABLE n, Eqs. (310b)), or

a 1'ka1U

+~'kagr =

0

U =- (a1 'k~)- 1 a1 'kagr

The stiffness for displacements r only is then • Sutllx s refen here to the number of unit panels and should not be neu due to lhear stresses in (344).

~onlused

with suffix s for stitr-

K = &o'k&o---&o'ka1(a1'ka1)- 1a 1 k&o

(312b)

51

from which the displacements r under loads R are found as

I

l2k.

r=K-1R

0

The total strains of the elements due to R are then

I

0 ........................ 0 0

1

................ 0

(316b)

v=ar={a 0 -~(~'k~)- 1 ~'kao}K- 1 R

~covers

from which the stresses in the unit panels are calculate<.i. Due to the simple geometry of the structure the matrices a 1 are again quite straightforward. Thus writing ao in the partitioned form

ao.

ao={aoo~~o~,

........ aog· ....... ao,}

it is apparent that for the cover plates the ao's are all zero since vertical displacements r can cause no strain in the plates• (with U = 0).

k=

I

aw=

0 ..... 0 0 ..... 0 0 ..... 0 0 ..... 0

0 ..... 1 0 ..... 1

0 ..... 0 0 ..... 0 L

I

~.=

L

L

0

0

0

0

0

0

0

0

h/2

0

0

0

0

0

0

0

0

0 .... . h/2

0

0

0

0

0

0

0

0

0

h/2

0

0

h/2

0

0

0

0

0

0

0

0

0

h/2 ..... 0

0

0

0

0

0

0

0

0 h/2

0

0

0

0

0

0

0

0 h/2 .....

16

0

17 18

I

12

h/2

0

0

..... -h/2

0

0

0

0

h/2 .....

0

0

-h/2 .....

0

0

0

0

0

0

0

0

0

0

0

I

,J

}

ao'kao=Ko

........... . (356) ~'kao=C Thus the complete stiffness matrix for the displacement column { r U} is a 1 'k~=C

(353)

I

I .............................. .

_I

I~ ~I

L

(357)

,J

Ko is clearly the set of vertical forces R which arise due to unit r displace-

l

ments when U is zero. Evidently only the webs are involved and we find easily as a typical example the vertical forces at the joints due to r 5 = 1

_j

t......

_ 2Ghtw 2Ght, o.s.s- 1 + d

I

(358)

Ghlw ko.s.s =ko.2.s =- -~•••••••••••••••••••

0

(354)

Similarly C is the set of moments arising at the joints due to unit displacements (rotations) U. By using the stiffnesses of the unit panel (or by carrying out the matrix multiplication a1 'k~) we find for the moments due to U9 =1 2E'h 2dt EBh 2 E'h 3 tw 2Gh 2lt Ghltw ce,e=-3-1- +-~-+~+3d+ -3Ca,e=Cts.9=

_j

All other columns are zero. The matrices for the web plates may of course be reduced to six or even four rows by using the assumption of zero vertical direct strain and the antisymmetric character of the U displacements (see also Eqs. (346) and (346a)). However, we retain here the full eight displacements of the unit panels to show the simple formation of the a matrices with a completely standard unit panel. The stiffness matrix k of the unassembled unit panels is written again in the diagonal partitioned form: • This, of course, is no longer true if the webs are tapered in depth, when an appreciable part of the shear force carried here by the web is equilibrated by the vertical component of the direct stresses in tbe covers.

52

k.

The factor 2 is introduced for the cover plates to take advantage of the symmetry of the structure by including the unit panels of the lower cover with their opposite numbers in the top cover. The k., kb etc., are of course the stiffnesses of the unit panels discussed earlier. The method of formulation of the matrices ~'k~, etc., given above is probably the most convenient for use with the automatic digital computer since the various terms in the constituent matrices are reduced to their simplest and most standard forms. However, it is instructive to consider directly the components of a1 'k~, etc., and gain some physical insight into their formation. We call

k

11

.•..................•... o

0

0

0

I

L

_j

12 .... . 15

(355)

l

o

(352)

and for the web plate f

at,=

0 ...

webs

11

10

k,

.......... 0

I\

I 0 ..... 0 0 ..... 0 0 ..... 0 0 ..... 0 0 ..... 0 0 ..... 0 1 ..... 0 1 ..... 0

10

.... . h/2

j

6 ..... 9

Likewise the a 1 matrix for unit panel c of the top cover is ..... 9

I I0 I

For the web/, the 8of matrix is easily seen to be 1 ..... 5

II

(351)

. . .•. .•. .. . .••. .

E'h 2dt

EBh 2 E'h 3 tw

---:rr- ---zr-'U/ +

Gh 2lt Ghltw 6d +-6-

l

E'h2dt Gh 2lt

cu,9=c7,9=~---g

Ct7,g=Cta,9=cs.9=c1.9=

2

E'h dt Gh lt ---.zr-I2d 2

vE'h 2t Gh 2t -cts.g=cu,g·= -c2,9=cs,9= -g-+-g-

f

(359)

j

Finally C is the set of moments arising at the joints due to the vertical displacements. Again only the webs are involved, and we obtain easily _ _ Ght,

l

c12.s= -cs.s=r _ _ Ghtw Cts.s= -ca.s=z

J ...................... (

360)

In some problems it may be possible to neglect the shear deformations of the webs. Quite obviously this introduces kinematic relations between the rotations and the vertical displacements. However, the above breakdown of the structure into the simple unit panels is then not the most suitable. A suggested method for the setting up of the K matrix for this case, based on the theory of bending of plates, has been given by Williams.* 9. ILLUSTRATIONS TO THE ANALYSIS OF REDUNDANT STRUCTURES BY THE FORCE METHOD In this section we present two very simple applications of the force method developed in Section SC. The first example shows how to determine the statically equivalent stress system in an N cell tube typical of a wing structure.t With the direct stresses distributed according to E.T.B. we find using the sik method the corresponding shear flow distribution for the multi-cell cross-section with the assumption that the ribs are rigid in their plane. Naturally, the axial constraint stresses and the effect of rib deformation remain to be investigated but the statically equivalent stress system derived here is particularly useful, being, in general, a reasonable first approximation when the structural design has to be based merely on a statically equivalent stress system. This, of necessity, has been the approach in most cases up to the present owing to the difficulty of computing highly redundant systems. Although the problem may, with some justice, be described as trivial in relation to the powerful analytical techniques of Section 8 it is astonishing to see what an unfortunate treatment it often receives~ven today. The second example analyses-first by the S;k method-the axial constraint stresses in a four flange tube with shear carrying walls and deformable ribs under arbitrary loading at the rib stations. The solution of the same problem is also obtained by the matrix method and the effect of a cut-out is investigated by the H matrix device of p. 41. The 'exact' flexibility of the structure is derived and compared with that given by E.T.B. and Bredt-Batho. A thermal loading is also investigated by the matrix method. These two problems are only meant as preliminary illustrations of the force method. Thermal applications of the S;k method are investigated in Part II. More complicated structures, particularly suitable for showing the power of the matrix formulation of the theory, will be analysed in a later publication. (a) Shear Flow Distribution in a Multi-cell Tube Due to E.T.B. Direct Stresses Consider the uniform cylindrical and multi-cell tube of the type shown in FIG. 52 subjected to shear forces s., Sx through and a torque Tn about the point 0. Find the corresponding distribution of shear flow if the direct stresses are given by the engineers' theory of bending; thus axial constraint stresses due to restrained warping are ignored. Instead of referring the loads to the arbitrary point 0 we may alternatively give the point D through which the resultant of all transverse forces is acting, i.e. T D ~ 0. The direct stress a due to bending moments M"" M. about the axes Gx, Gy through the section centroid G and parallel to Ox, Oy, is

- y

-x

X

y

a=Mxr+M.1 .................... .................... .. where

M __ Mx-MuU:r

11 /fy) x- l-[(/;ry} 2 jf,.J.]'

M

_M11 -M,,(I.,yjl;r) . - l-·[(/;ry)2/fxf.J

(al)

(a2)

are the effective bending moments for the chosen axes which are, in general, not the principal axes of the cross-section. Physically M,., M. are the combinations of M:r, M. which give rise to pure bending strains about Gx, Gy respectively. We could alternatively restrict ourselves to principal axes of the cross-section but in practice, unless these are obvious, it is preferable to use Eqs. (al), (a2). They are not only more convenient from the computational point of view but permit also the retention of parallel axes Ox, Oy at all cross-sections of a wing regardless of the change of directions of the principal axes. The condition of equilibrium in the z-direction of an element dsdz of a wall gives,

oa oq os+t•oz=O

. . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

(a3)

where q=ta. 8 =shear flow in the wall. (,=effective direct stress carrying thickness of the wall (i.e. including an allowance for the stringers). Similarly, we find from the equilibrium of an element dz of a typical flange g placed at a web-cover intersection (see FIG. 52),

(a)

Loading

~· I s,Dxb

(b)

Basic system for calculation of Dxb

~

~

s

( Dxl 1 DxM-11 P.M ~

(c)

SM..-

1~

P..M•1

1~ £?. H~ l .V-1

DxM Distribution

Fig. 52.-Shear flow distribution in multi-cell tube. Sign conventions and equilibrium conditions (lay

By(!;:, ~-q.~

--q.,,+qY+=O

.. .. .. .. .. .. .. .. .. .. .. .. .. .. .. ..

Eqs. (a3) and (a4) yield, except for a constant of integration in each of the

N cells, a shear flow distribution q whose resultants in they and x directions

are s. and S" respectively. Since there remains one further equilibrium condition, To= fqpods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (a5) for torque about 0, the degree of redundancy is (N- I); (the integration in (a5) extends over all walls and the normal Po is taken positive (negative) if movement along s leads to an anticlockwise (clockwise) rotation about 0). It follows that the shear flow distribution in a single-cell tube under prescribed transverse loading is statically determinate once we stipulate that the direct stresses are distributed as per E.T.B. For the analysis of the general case of an N cell tube we find it more convenient to use a slightly different approach. Thus, for the moment, we prescribe instead of the torque equilibrium condition, the rate of twist

c/>=8'=c~ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

*lac. cit. p. 17. J.R. Aer. Soc., Vol. Ll, September 1947, p. 770, and J. H. Argyris and P. C. Dunne, Structural Analysis, Part II of Handbook of Aeronautics, Pitman 1952.

(a6)

in all cells. The prescribed c/> may be considered as an initial 'give' experienced by the ribs maintaining the shape of the cross-section and is subsequently determined from the torque equilibrium equation. Our modified problem has now N redundancies. The basic system is obtained by cutting the wall in each of the N cells and the unknowns X[, Xu, ......... , X.~· are then the shear flows at the cuts. They are determined from the compatibility equations, N

~DMrXr+DMo=-cO

r=J

t See J. H. Argyris and P. C. Dunne, 'The General Theory, etc.', Part V.

(a4)

which express the conditions of zer-o relative warping

(a7)

oat the cuts.*

.*We denot.e here the warping by the unconventional symbol 6 to apply directly Eqs.(282)intheir ongmal notation.

53

The shear flow distribution qb in the open tube forming the basic system is obtained from Eqs. (a3) and (a4). Integrating (a3) with respect to s and using Eq. (a2) we find - Dxb -Dub qb=S.T+SxT . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

where

-

S. and

V

X

s.~~Sx

_

Sx-Sifxu/1,) 1-[(1,..) 2/fx/y)

1-[(1.,.)2/f.,/y]' Sx s

..

0

0

(a8) (a9)

Dxb=- JYt.,ds, D.b= -- fxt,ds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (aiO)

To determine completely the Db distributions from 'Eqs. (a 10) we require also the equilibrium conditions of the type (a4) at each joint of spar web and cover. Using (at) and (a8) in (a4) we have I (Dx+- Dxd- Dx- +By)bo =0

?-

.. . . . .. .. .. .. . .. .. .. ..

j The positive directions of q and s are indicated on (Dv+- D.d- D._+ Bx)b• =0

FIG.

Warping due to twist - 6.,., =- 2 Q""'

52.

Fig. 54.-Warping in Mth cell of basic system due to rate of twist cp

Choice of Basic System

The reduction of the multi-cell tube to an open section can, of course, be achieved in a variety of ways. For example, we may cut the upper or lower cover in each cell or we may cut N of the vertical walls (see FIG. 52). However, consideration of the form of the 8's shows that the compatibility equations (a7) are very much simpler for the former choice. We confirm this immediately by applying the unit load method for the calculation of the 8's which measure the relative warping at the cuts. Thus, if we apply unit shear flows at each of the cuts of the open section in FIG. 52 we produce merely constant shear flow around each of the individual cells. The Redundant Shear Flows

s.r'

~
Transverse loading

s, through

0

~=0

I9 • Gb..-.--"

Transverse loading S, through S,.=O

••••••••••••••••••••••••••••• •••



....

(al3)

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (al4)

and Eqs. (al2), (al3), (al4) yield for the cross-sectional function D:x: in theM cell I Dx=Dxb+D.rM external walls web between M-1 and M cells

D.,=Dxb+D.,,M_ 1 -D.rM

All cross-terms but 8.11. M- 1 and 8.u. :u n vanish since only unit shear flows in adjoining cells act over a common wall. We find .\1 8.u• .\1-1 = O.11-b .11 = - f3M-1• -G--

. . . . . . . . . . . . . . . . . . . . . . (a21)

0.H.M+1 =0 .11+1 ..I t-- _{3.\1,GM+l where !3.\1-1• M =

J~'

f3 :lldl H

~ Jo/

.. .... .. ..... ... .. ... .. .. ...

~

(a22)

M,M+I

M-I,M

are the integrals extending over the common walls (spar webs) of cells M -I, M and M, M +I respectively. The minus sign arises since the shear flows due to unit redundancies have opposite signs in the common walls. Determination of Redundant Shear Flows

web between M and M+l cells q=qb+qM-qM+I We may always put the total shear flow in the form -.D., -D. q=S. lx +S., I.

0

J

where the DxM and D. 111 are unknown. For the basic system of FIG. 52 the shear flows in the M cell of the actual system can then be written as: I q=qb+qM external walls

J~

term the stgn of qb must be reversed over this wall. Similarly, we obtain for the relative warping 8Motp due to cp (see FIG. 54). 8.\1otp =-2D.Mcp ............................. ............. (al8) The standard derivation of &i_s. (al7) and (al8) is by kinematics. Thus Eq. (a 17) is obtained by integration of the shear strain expression corre: sponding to zero rate of twist. Also, we find (al8) from the condition of zero shear strain along the middle line of the wall.* The unit load method also yields directly the coefficients of the unknown X's. Thus, the relative warping 8.11 M due to unit shear flow in the M cell is

M

s.

q=qb+q 111 _ 1 -q 111

denotes _integration over the M cell. It must be emphasized that the sign

conve~tiOn used for qb and s in the basic (open) system is in opposition to sM m t~e left-hand wall of the M cell and hence in evaluating the above

f3 .·11 •c• dstM . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (a20)

0

If we consider now the total transverse loading s. and s. as acting through D and split it into the two component loads and S.r (see FIG. 53) we can express the statically indeterminate shear flows in the form

webbetweenM-landMcells

f( ... . )dsM

M

11 ={3M _ Jds. 0.11.11 ~~ G . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (al9) Gt where

Fig. 53.-Effective shear forces for bending about non-principal axes

xM=qM=s.D;"'111 +S.,D/.111

-

unit shear flow

(all)

s.

and S'x by We denote the unknown rates of twist associated with the M'th compatibility Eq. (a7), whtch expresses the condition of zero relative warping at the M'th cut is obviously

c/>. and c/>x re~pectively. For the loading due to

.~. 1" f3 s.{ tdsM-2n. G'I'•'S_7., JD.rh 11

M- 1 ,

s.

MD,., 111_ 1

M

+{3.11DrM -~{3.11, .II +I Dr, .11+1} =0

(al5)

We obtain hence the set of N equations in the N unknowns D, 111

web between M and M+l cells D.,=D.,b+DxM-Dx.M+I j Similar equations may be written down for D •.

J

f3£Dxl-{3h nDxn = - Dtdsl

The 8 Coefficients for the Basic System o/FIG. 52 The 8Mo coefficients consist of two parts 8Mob and SMotp corresponding to

l

+2D.l Gcp.i y

.....................................................................................................................................................

the shear flow qb in the basic system and the initial 'give' cp (see also Eq. (177)). We have: 0 Mo =0 Mob +8 Motp

Application of the unit load method yields immediately for 8Mob OMob=

J~;dsM

............................. .............

(al7)

-{3,\'-1•

NDr,

54

=-

JDtdsN+2D.NGcp,~

N

•

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (a23)

M

where s 111 is the circumferential anticlockwise co-ordinate in the M cell and the integral

.V-1 +f3NDxN

• _See e.g. J. H. Argyris, 'The Open Tube',

Apr1l 1954, p, 101.

AIRCRAFT ENGINEERING,

Vol. XXVI, No. 302,

These formulae are usually derived more simply from the condition of equal rate of twist* .P=

~~= 2iLaHdsM

....................................

(a24)

Similarly

M

in all N cells in the actual system. In this approach we specify initially zero relative warping at all cuts and express the compatibility condition by the equality of .p in all cells. On the other hand in our present method we specify initially the same rate of twist .p in all cells and express the compatibility by the condition of zero relative warping at the cuts. The solution of Eqs. (a23) and the corresponding ones for Du.lt is straightforward and may be put in the form

I I

J

· · · · · · · · · · . · ·. . . . . . . . . . . . . . . . . (a25)

2'La 111n 111 I

twist. They yield hence the shear flow distribution qE---commonly known as the engineers' theory of bending shear flows-due to transverse shear forces acting through the shear centre E, · q E may be written

-d.

"'

v

(a26)

qE=Svy+Sq

where the cross-sectional functions d, (d.) are obtained from Eqs. (a 15) with dxM (d.M) in place of DxM (D.M). The co-ordinates Xg, YE of E., can and Sr be determined from a consideration of the two loading cases through £, shown in FIG. 55. We find

s.

I

I

I j

N

-IxvYE+IvYE=- fdvPods=- fDvbPods-2LduMnM I

. . . . (a27)

where the integrals extend over all walls and the normal Po is taken positive (negative) if movement along the positive s direction produces anticlockwise (clockwise) rotation about 0.

•

I G.P.of = .r

-

JDubP nd5 -

Transverse loading

S, through S,=O

shear centre Es

·shear centre Es

S, through S, = 0

Fig. 55.-Effective shear forces applied through the shear centre Es

The shear centre allows us to define the loading alternatively by the shear forces through and the torque TE=-S.(xE-Xn)+Sx(yE-Yn)=To-SuxE+S,.yg .......... (a28) about£,. To obtain the total redundancies Dx 111 and D. 111 we have still to determine the rates of twist .P. and .p,. Observing that the distribution coefficients a 111 are the same in the two equations (a25)-a direct result of the rigid diaphragm assumption which allows us to displace the transverse forces anywhere along their lines of action-we can combine the indeterminate and S"' into a single set shear flows due to

s.

Su

A..

Ix

'S,.

A..

I.

A..

A..

A..

qB.11= jaMG'~'"s +/aMG'I'"'s =aMG('f'x+'I',J=a.,1G'I' ;t

y

y

...... (a29)

X

where

dO

cf>={{z=.f»,+.P• ..........................................

(a30)

is the total rate of twist due to the given loading. To derive the rate(s) of twist we may apply the equilibrium condition about D (see FIG. 53) fqpnds=O .............................................. (a31) where the sign of Pn is defined as for Po· Applying Eq. (a31) to the loading we obtain

s.

fDxbPnds+2'LdxMnM+2'LaMnM · G.f».t=O

or

•

*See foe. cit. p. 129, 'The General Theory'. See also derivation of Eq. (al4) by the Unit Load method in Section 7.

2~d"_,,n .11 I

N

Tf;=G.f»2'Lawfl.w ........................................ (a34) I

or G.f» =

Tg

--.-,N---'-'--

.......................................... (a34a)

2'LaMnM I

Hence the redundant q BM due to torque Tf; are given by: Tf:

(a35)

q ll Jf =c a .II ____,N___,:__

2'La.un.,r I

The shear flow is, of course, constant in each wall between two consecutive joints. Thus, we have in the M'th cell external walls qll=qBM I web between M-1 and M cells

qn=qno.~r- 1 -qllM

t .. ..

(a36)

web between M and M +I cells qu =q RM -q B• 111+ 1 J The distribution q 8 is known as the Bredt-Batho shear flows in a multicell tube. The special case of Eqs. (a23) corresponding to pure torque is now written more conveniently in the form ~~qB

Transverse loading

N

where the integrals extend over all walls. Having derived the complete set of redundancies we can calculate using formulae (a14) and (al5) the shear flow distribution in the actual system under any given transverse loads. If we have found the shear centre £, and the torque TE it is more convenient to proceed as follows. Thus, using the equilibrium condition (FIG. 55) Tg=fqpgds ............................................ (a33) and remembering that the engineers' theory of bending shear flows have no torque contribution about £." we obtain with Eq. (a29)

dx 111 , d. 111 are the values of the redundancies corresponding to zero rate of

-d,

. . . . . . . . . . . . (a32)

N

,-_8,, uqB u =znl a.p

-~ .v-1• sqn, s-1 +~ 1\·qn .v =2n.vGr/> . . . . . . . . . . (a37) which are most simply derived from the rate of twist formula Eq. (a24). Expressing the solution as in Eq. (a29) we find the unknown G.f» from Eq. (a34a). Note that the total shear flow q of Eq. (al4) may be written as q=qE+qn .............................................. (al4a) where the engineers' theory shear flows qg are given by Eq. (a26). With the chosen basic system Eqs. (a23) and (a37) are particularly well conditioned since the diagonal coefficients are predominant and at the most only three unknowns are involved in a given quation. Direct solution by elimination is quite easy and may be performed by slide rule even for a high degree of redundancy; application of the relaxation technique is superfluous. An additional virtue of the basic system of FIG. 52 is that the Dxb distribution is quite close to the final dx distribution and the values of the redundancies are small. In finding on the other hand the d. 111 redundancies there is a conflict in the choice of the basic system. For when we cut the external wall in each cell the Dub distribution is vastly different from the final one although the equations are as well conditioned as for the D, 111 since the matrix D is the same in both cases. Naturally, the criterion of good conditioning of the equations is always the most important one. The alternative basic system in which the webs are cut gives a D.b distribution close to the actual d. but the equations are not so easy to solve since each equation involves all the unknowns. To satisfy both the above requirements it is necessary to make composite cuts, i.e. to cut both upper and lower external walls and instead of the condition D.b =0 at a cut to take say equal and opposite values of D 11 b at the two cuts of each cell. In any case the d. distribution is, in general, of small importance and a more approximate solution is acceptable.

Generalizations of the Above Analysis The method given above for the determination of the statically equivalent stress system in uniform cylindrical tubes may be generalized and applied to tubes with conical or non-conical taper and with sheet thicknesses and boom areas varying lengthwise. The angle of taper 28 is taken, however, to be so small that cos· 28R:>! and sin 28R:>28.

55

from Eqs. (a27). The engineers' theory of bending shear flows for transverse forces through the flexural axis are now

Q.d., Q., d.

qE=p T,.+p

T.

.. . .. . .. . .. . . .. .. . .. . . .. . . .. . .. . .. . . ..

(a45)

where d.,, d. are obtained for the root dimensions from Eqs. (a23) for

c/>.=c/>.,=0.

If the shear centre E, has been found we may calculate at any crosssection the torque T E of all applied transverse forces about the flexural axis. It is then preferable to calculate the Bredt-Batho shear flows qB of the total statically equivalent shear flow q=qE+qB .............................................. (a4la) by a slightly modified version of the method on p. 55. Thus, Eqs. (a37) for the redundant q 8 M become

(3IQBI-f3J.IIQB/l=2QI pi/JGcp T, is the torque about axis V-A

-(3 N-1• NQB, N-1 +f3 NQBN

=20.Npi/JGcp

(a46)

The solution of which may be written

QBM=a.wGpi/Jcf> . .. . . .. . . .. . .. . . .. . . . . . . . . . . . . . . .. .. . . . .. . (a47) Next we deduce from the equilibrium condition about the flexural axis

Fig. 56.-Geometry and loading for conical tube

1. Conical and Cylindrical Tubes with Similar Distribution of Material at All Cross-Sections Consider the conical tube shown in FIG. 56. At the root the sheet thicknesses for direct and shear stresses are t. and t respectively and the (effective) boom areas B. At any intermediate section the corresponding thicknesses and areas are given by

I

J

y

M~

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (a38)

x

Mv

+ p21/J, lv

fx

N

TE

(a48)

2p2~aMQM I

Hence the shear flows QBM are

QBM=aM

TE N

(a49)

2p 2 ~aMQM I

where p=r/r. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (a39) rand r. being the distances from the apex V of the current and root crosssections respectively; for cylindrical tubes p = 1. 1/J and 1/J. are nondimensional functions of z or p. In what follows under the present heading (l) all cross-sectional dimensions, areas and functions refer to the geometry of the root-section. As in the previous analysis we assume that the direct stresses are given by the engineers' theory of bending and write them in the form

(J= p2!fJ,

Gpi/Jcf>

(a40)

Finally, we derive the shear flow q 8 in all walls from Eqs. (a36). If the torque is given initially about an arbitrary axis VA we can calculate TE with the formula TE=TA-pQ.(xE-XA)+pQx(YE-YA) ...................... (a50) where xA, YA are the co-ordinates of the point A at the root. 2. Conical and Cylindrical Tubes with Arbitrary Variation of Boom Areas and Wall Thicknesses

Here we investigate conical or cylindrical tubes with an arbitrary lengthwise variation of skin thicknesses and boom areas. The engineers' theory shear flows are not any longer proportional to Q. and Q., and Eq. (a42) does not apply. The concept of a flexural axis, either straight or curved, is also not any longer strictly true. Under this heading all cross-sectional dimensions, areas and functions are based on the current cross-section.

where Mx, Mv are given by Eqs. (a2). Note that y, lx are based on the root geometry. Since the stresses (a40) act along the generators they give rise to shear resultants at the apex V in the y and x directions which are easily found to be

r

M~

and

r

M.

.

respectively.

Hence the shear forces resisted solely by shear flows q are M~

My

Q.=S.+-;:-. Q.,=S.,+,- . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (a41) If now the transverse forces are applied through a line VD where D is a point at the root the shear flow q at any cross-section may be expressed as (see also Eqs. (at4))

q=~· ~"'+~"' ~: ·· · ·· · · · · ·· · ·· · · ·. · ................... where Q., Thus,

(a42)

Q., are determined from Eqs. (a9) with Q., Qx in place of s., S.,.

Q~= ?v_[(~~j~iL~:~· Q.,= ?:_[(~~j!j/~~:~

············. ....... (a43)

The cross-sectional distribution functions Dx. D. are obtained at the root cross-section by the method of the previous analysis but with

p 2 1/Jcf>.~.

(p 2 fcf>.,~)

in place of c/>.t.

(

cp.,;:) ....

(a44)

in Eqs. (a23), (a24), (a25), (a29), (a32). Note that p21/Jcf>J,JQ., etc., are constants for the type of loading considered as the transformed Eqs. (a32) show. The flexural axis is in the present case a straight line VE, through the apex and the shear centres at all cross-sections. We define this axis by the shear centre E. at the root. The co-ordinates XE, YE are found as before

56

Fig. 57.-Equilibrium condition in element of conical tube

For the subsequent analysis we require the modified forms of the internal equilibrium conditions (a3) and (a4). Thus, we deduce immediately from the geometry of FIG. 57 the equilibrium condition on a conical element on the surface - ~(rf) +~q =0

or

~r

~w

(a51)

where Hence where

f =at, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (a52) ~q I ~(r() . ~s=r ~=Df

.................................. (aSia)

~f f ~f f Df="iir+-,.=-sz+-,.

.............................. (a53)

In practical calculations we find '?!f/(Jz numerically by finite differences

The equilibrium condition Eq. (a4) for a flange B. at the intersection of cover and web becomes here

d~. -q._ -q.d+qo+=O

..................................

(a54)

where P. =B.a. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (a55) To find the statically equivalent stresses for an arbitrary system of transverse loads defined by the shear forces (and associated bending moments) S.(M.,) and Sz(M.) and the torque

Top and Bottom Covers

TA

about an axis VA we proceed as follows. First we equilibrate the applied bending moments by the E.T.B. direct stresses -Y - xa=M.,T.,+Mvf.

........................................

(al)

The co-ordinates x, ji from the centre of gravity and the corresponding moments of inertia may vary arbitrarily from section to section. The statically equivalent shear flows are calculated as before in two parts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (a56) q=qE+qB where qE are in equilibrium with Q., Q., with the imposed condition of zero twist throughout the tube and additional shear flow necessary to balance the applied torque.* The solution for an N cell tube follows closely the method given initially and the selection of a basic system derives from similar considerations. The calculations must, of course, be repeated for every section to be analysed. Calculation of qE: The shear flow qb in the basic system is obtained at all sections we want to analyse by integration from (a5la) s

............................. ...............

qb=JDfEds

(a57)

0

where !E is calculated from Eqs. (al) and (a52). The necessary constants of integration to compute the shear flow in the webs, etc., are derived from equations of the type (a54). TheN unknown shear flows qEM at the cuts of each section are determined by the same method leading to Eq. (a23). We find

I~bdS[

f3IqEI-/3[,JiqEJI=-

I

-f3M-l•MqE,M-1 +f3MqEM-/3M, M+lqE, M+l = - J~bdsM M

- f3N-l> NqE,

N-1

+/3 Nq EN

=-

J~bdsN

. . . . . . (a58) Note that all dimensions are based on the current cross-section. Having solved these equations we calculate the shear flow qE with Eqs. (a13). N

Calculation of qB: The torque T Q of the shear flows q E about the VA axis is N

TQ=fqEPAds=JqbpAds+1'2:.Q.MqE M ........................ I

(a59)

where p A is the normal from the point A to the tangent at the wall; note the usual sign convention. Hence the torque T 11 to be resisted by the shear flows qB is ............................. .............

TB=TA-TQ

The shear flows and (a29) with -1.= G'I'

N

QB

Tn

Length

\
Th1l~iss

•• 5 6 7

1. 10

2.'

3. 8

30·0

30-0

30·1

30·2

17

20

20

16

25

B

B

2·5

1-6

1-6

1'1

1·1

,..

•

5

7

&

Cell

1

n

lll

Il

y

Yl.

12

10

Area

350

551

100

540

310

210

3

Area

I

12

12

I

•

0

7

5,6

0

(em~)

10

65·7

14·6

1·0

1·0

Fig. 58.-Geometrical data for six-cell tube

The above analysis of a statically equivalent stress system in tapered or cylindrical multicell tub;:s could b;: used to write down the b0 matrix. Using the information develop!d in S:!ction SC for the b 1 and f matrices we can calculate next the axial constraint stresses in the structures considered here. However, when an 'exact' analy3is is to b;: performed by the matrix method in what is essentially a single continuous operation the choice of the b 0 matrix should b;: guided entirely by the requirement of simplicity. Hence, in such cases th:! sy3tem of indep;:ndent spars is greatly to be preferred to that of the multicell tube in the present examp.le: see also Example 9b. Numerical Example To illustrate the application of the foregoing analysis we determine now the shear flow distribution in the single symmetrical six-cell section of a uniform cylinder illustrated in FIG. 58. It was found convenient to use the metric system in this example. The direct stress carrying ability of the walls has been replaced by effective flange areas at the web-cover intersections (augmenting the actual flange areas there) so that the walls carry only shear flow which is constant between flanges (t.=O). This procedure enormously simplifies the work and is quite accurate enough for practical computations although naturally additional effective flanges may be introduced at intermediate stations (see Section SC, p. 37). It should be noted that since the distribution of direct stress is known (E.T.B.) the appropriate effective flange areas can be calculated explicitly without any need for guesswork. Strictly, different areas have to be calculated for bending about Gx and Gji but since in practice the horizontal shear force S., is of much less importance it is usually sufficient to apply throughout those calculated for bending about Gx. For a singly symmetrical section under a shear force normal to the axis of symmetry we are obviously only concerned with the D., (or d.,) distribution. Due to the assumption of t,=O the (constant) values of D.,b in the upper and lower walls are zero and the values in the webs can be written down by inspection using the equilibrium equation (all) for the flanges. The Dxb distribution is shown diagrammatically in FIG. 59a. In FIG. 59b we illustrate also the Dxb distribution for the alternative basic system mentioned previously in which the multi-cell tube is reduced to an open section by cutting N vertical webs. No further comment should be necessary to point the moral of this illustration.

Table I of Example 9a Values of JD~bdsMft and coefficients {3 M

Cell M

(a61)

1'2:.aMQIII I

• qE may be regarded as quasi-engineers shear flow; see also J. H. Argyris and P. C. Dunne, Joe. cit., p • .53 (Handbook).

3 •• 7. I

2 9

(a60)

3. Tubes with Non-Conical Taper For tubes with non-conical taper we can find the statically equivalent stress system !E and qE+qB by the method under (2). However, when finding the torque T B to be carried by the qB shear flows we must make allowance for the torque Tp carried by the direct flow IE and the boom loads. Thus ...... , ............................. .. (a62) TB=TA-TQ-TP

2. 3 1.1

1 10

at any section may now be determined from Eqs. (a37) ..........................................

1. 2 9.10

Boom l
Nose Rear cell cell

Webs

I

II

III

IV

v

350

555

600

540

390

471

763

•

490

466

JD

J'b

d.~.'i t

106

125

125

100

+68

+120

+120

t96

\ 7200 1 780J 1

o

1

210 ----

908

40-l

D_,._b_f3M_._~_r+_l_____ 72~~500~\_1500~-- 9l0J M

VI

--"!.'I -+50

_ 3120---

-5~0:) 1 -6.1-80 \ -3120 57

With D,b having (constant) values in the inter-cell webs only the six compatibility equations are formed easily and systematically by means of the arrangement in TABLE 1. All values in the table are obtained directly from the dimensional data of FIG. 58 and the D,b distribution of FIG. 59a. The equations for the unknown D.,M are therefore (see Eqs. (a23)) 763D., 1 -106Dxo

(a)

GOOD CHOICE

(b)

BAD

= -7200+ 700Gc/>fy = -7800+ lllOGc/>~y

-106D., 1 +471Dxo-125Dxa

-125Dxa+466D,. -100 D,s

=

0+1200Gc/>~ .JJx + 5400 + IOSOG'I'S y

-lOODx<+ 404Dxs-62·5D.,.=+6480+ 780Gfl

s.

-62·5Dxs+ 908Dxo=+3120+ 420Gc/>&_

Sv (a63)

· · · · · · .. · ·........................

Also the equilibrium Eq. (a3l) for torque about Dis 700Dx1 + 1110Dxo+ 1200Dxa+ l080DX4 + 780Dxs+420Dxa+94600=0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (a64) Writing the solution of Eqs. (a63) in the form

D.,M=d.,M+aMGc/>~v

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (a25)

Grpl.,/S. is determined from (a64) and the results are as in

TABLE

n.

CHOICE

Fig. 59.-Dxb distribution in six-cell tube for alternative choices of basic system

Table II of Example 9a Values of dxM and DxM (em. 3) M

dxM

aM

DxM

I

-12·17

1·46

-20· 13

II

-19·62

3·90

-40·92

III

-

1·15

4·52

-25·85

IV

+15·79

4·20

-

v

+20·71

3·08

+ 3·89

VI

+ 4·88

0·68

+ 1·16

~

10.000 kg.

7·16

I

f-z2·3cm-;

q in kg/em. Fig. 60.-Final shear flo;v distribution in six-cell tube

l"'

From Eq. (a64) G s.=-5·46 Adding these DxM to the Dxb we obtain the total Dx distribution due to Sv through D. Since Dxb is zero in the outer walls we have simply (see also Eqs. (al5)) for outer walls D.,=DxM for web between M and M+1 cells D,=Dxb+D.,M-Dx•M+l The total actual shear flows under the 10000 kg. load at Dare Dx 10000 q=Su I.,= 7992 Dx=1·25D., . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (a64) in kg./cm., the actual values being shown diagrammatically in

58

FIG.

60.

Shear Centre Since the present cross-section is singly symmetrical, the shear centre lies on the axis of symmetry and the first of Eqs. (a27) (with 1.,.=0) gives immediately the XE co-ordinate. Alternatively we note that the shear flows dx have a resultant s.=I., through the shear centre and we can modify slightly Eq. (a64) for torque about D to give 700d.,1 + 111 Od.,2 + 1200d.,3 + 1080dX4 + 780dx5 +420dx6 +94600=/.,(xE-XD) Therefore

(b) The Four-Boom Tube with Deformable Ribs and some more General

Structures In this example we determine, using the force method of analysis, the 'exact' stress distribution in the idealized four-boom tube shown in FIG. 61. The investigation is carried out first by the Swmethod and illustrated in a numerical example. Subsequently, we analyse the same example by the general matrix method of Section 8C. We show also how the matrix formulation may be used with advantage in the more interesting case of a six-boom tube with or without intermediate spar web. It is hoped that these simple applications of the matrix method will condition the reader to the new ideas and show him their power and basic simplicity.

Consider the cylindrical tube of FIG. 61 with a singly symmetrical trapezoidal cross-section the flange cross-sectional areas Band the wall thicknesses t of which may vary arbitrarily length-wise. Loads are applied only at the rib positions 1, 2, ........ , i, ........ n in the form of shear forces

l:· z

Rat• Rbl, ...... .. , R.;, Rbt• ........ , Ran• Rbn

and moments Mato Mbto ........ , Ma;. Mb; • .. . ... . . , Man. Mbn

at the front (a) and rear (b) spars. Following the general discussion on the idealization of aircraft structures given in Section 8 C (pp. 'J7and 40) we assume that the walls carry only shear stresses (the direct stress carrying ability is allowed for by suitably increasing the flange areas). The shear flow is hence constant in any field of each bay since changes in the z-direction can only be brought about at the ribs. It follows then that the end loads in the flanges vary linearly between ribs and that a knowledge of the flange loads at the rib positions suffices to determine them everywhere. Having found the flange loads the corresponding shear flows are determined easily from the flange load gradients and the condition of equilibrium with the applied shear force and torque. In the tube shown in FIG. 61 there are n ribs including those at the free and built-in ends. At each rib position there are four flange loads and only three equilibrium equations are available for their determination. Hence, nothg that the flange loads at the tip are zero, the degree of redundancy is (n -I). This trivial result is confirmed by the general Eq. (247a) on p. 38 by substituting N= I, f1=4, a=n. In selecting the basic system many choices are open to us. We may, for example, make a single cut in one of the flanges at each rib statiori to reduce the structure to a statically determinate three-flange basic system. Here, however we calculate the statically equivalent stress system by the E.T.B. and the Bredt-Batho theory of torsion, a more general example of which was investigated in Section 9a. In this choice, instead of making a single cut we have, in fact, cut all the flanges to allow the relative warpi?g consistent with the statically equivalent stresses while at the same tune the direct stresses are transmitted across the cuts ; • see also the discussion in Section 8C, pp. 32 and 33. The redundancies then consist of self-equilibrating flange load systems at the (n -1) rib stations. A suitable and symmetrical measure of such a system is the boom load function P introduced by Argyris and Dunne.t We prefer to use instead a slight variant of P~ the Y-system introduced on p. 39. The (n -I) redundant Y are determmed from the compatibility conditions of warping at the (n -I) rib stations. As a further alternative procedure we could, of course, choose as basic system the very simple structure consisting only of the two spars acting independently, i.e. we cut the top and bottom covers of the tube. However, • Actually, we may consider this basic system as .also derived by a single cut from the given system . Thus if at the cut Bange we - l y the correspondmg E.T.B . Bange load and allow there any out of place' movements, the stress distribution in the other structural elements is obviously that of the E.T.B. and Bredt-Batho theories. t Joe. cit. p. 38.

Fig. 61.-Four-flange tube. Geometry and loading.

except for very flexible covers and ribs our previous choice is much closer to the final correct solutions. The latter method is hence to be preferred when the design work has to be checked solely by a good statically equivalent stress system as is, in general, the case when no automatic digital computer is available. Subsequently the exact distribution, if found at all, is obtained only after completion of the design work-usually by a more clumsy version of the O;k method. On the other hand when we use the matrix formulation in conjunction with a digital computer and derive the complete stress distribution as a single process it is preferable to select the simplest possible b 0 matrix (see also the discussion on p. 39). Then the basic system formed by the independent spars is obviously indicated. An important advantage of this system is the absence of rib stresses if external loads are only applied in the plane of the spars. In multispar construction the simplicity of the basic system consisting of independent spars in comparison with the multi-cell system of Example (9a) is even more striking. If it is necessary to taKe into account external loads applied at intermediate points in the ribs and/or if spars are interrupted we may select as basic system for the matrix type of analysis the grid formed by spars and ribs (without covers}-still a very simple structure in which to find the b 0 matrix. (I) Analysis by the O;k Method* The statically equivalent stress system

Following our previous discussion we select here the E.T.B. (or quasi E.T.B.) and Bredt-Batho stresses as statically equivalent stress system. The flange loads are calculated at the rib stations with the effective areas B; there. Thus, the m'th flange load at the i'th rib is, Pom;=M.,l'mi Bmi lcr.i

........ • • • ... • •;. • • • • • • • • • • • • (bl)

where M.,; is the bending moment at the i rib due to the applied loads Ra • The particular case of the four-Bange single-cell tube under a given loading has also been treated by W. J. Goodey in 'Two-spar Wing Stress Analysis', AIRCRAFT ENGINEERING, Vol. XXI, No. :U7, p. 287, September 1949; No. 288, p, 313, October 1949; No. 289, p. 358, November 1949. 'J'here. the author uses the Castigliano technique to formulate equations analogous to the three- and f!Ye-joiat equations given here with essentilUiy the same basic idealizations but with tbe effect of taper induded. A comparison of the theoretical results with experimental strain measurement• is aloo JliYeo.

59

l

n;~

l

"' ,., ' o•2

Fig. 63.-Fiange loads due toY; = I and Y;+ 1 =I (See Fig. 4l)

,.,

i,,

Fig. 62.-Rib shear flow In four-flange tube

and Rb. The corresponding shear flows in the walls, observing that they are statically determinate in a single cell tube, may be expressed in ·the i bay as TAi q .; = q..4;+2n .. . ... . . . .. ............. . . .. . .. ... (b2) where q...4, is calculated from the flange load gradients

(d~;"') i ' ba..

p omi-tom,

i-1

. . . . . . . . . . . . . . . . . . . . . .

(bJ)

with the condition of zero torque about the chosen reference point A and T.t is the torque about this axis (see p. 56). There appears little virtue in single-cell uniform tubes or tubes with similar longitudinal variation of stress carrying material to select A at the shear centre £ ., i.e. to write q. as q . = qg+qB •. • . . . ............... . .. ... ... . . ,(b4) (see alsop. 56). Similarly it is of no advantage in tubes with arbitrary longitudinal variation oft and B to split q. as in (b4), in a quasi-engineers' theory shear flow qE for zero rate of twist and a Bredt-Batho shear flow qB for a torque TB to be found by Eq. (a60). Naturally, in multi-cell tubes it is on the other hand preferable to use the presentation (b4) for the calculation of q •. Note that since we assume that flange loads vary linearly between rib stations, systems (bl) and (b2) only represent E.T.B. at any cross-section between rib station if the flange areas are constant or vary similarly within each bay. To complete the statically equivalent stress system we have to calculate the rib shear flows. Consider to this effect FIG. 62 where rib i is shown isolated under the action of the equilibrating forces; in this it is assumed that the moments M a;, M bi are applied as flange loads ± Mad h., ± Mb;/hb For a trapezoidal rib the shear flow is not, of course, constant but resolving forces horizontally for the upper or lower half of the rib we find the average shear flow qor; = q o,i-l -q 0 ;-

Ma;

M bt

7l; - 71; ......... ... . . ...... . . (b5)

where q 0 ,;_ 1 and q 0 ; are the shear flows in the covers of the i - I and i bays respectively. The self-equilibrating stress systems Y; We show in FIGS. 63 and 64 the flange loads and shear flows due to Y, = I and note that they extend only over the bays (i - I) and i; these figures reproduce in a more convenient form for the present purpose the information given in FIG. 43. The system Y is connected to the boom load function P of Argyris and Dunne by

U.+ ~J

Y= P

. . . . . .. . . . . ... .... ...... . . .. (b6)

The Y-systems are determined from (n - 1) equations of the type (282a) which take here the form n- 1

~ o,kYk + o;. = O .. ... . . ... . . . . . . ... . . . .. . ... (b7)

k=l

where 8,. is the relative warping at the rib i due to the chosen statically equivalent stress system. Having the Y redundancies we find the 'exact' stress distribution by superposition of the statically equivalent stress system and that due to the Y;'s.

60

,.,I

w,., .~h!h n. • • I,,, \

w~,::::- h.•/\

4•2

Fig. 64.-Shear flows due to Y; = l and Y; +t = I (See Fig. 4l) Arrows indicate actual direction of shear flows for positive Y systems

The coefficients O;o and Otk We write all 8 coefficients in the split form

0= 8, + 8. + 8, .... . .. .. .... . .... .. ..... . ..... . (b8) where the suffices f, s, r refer to the flanges (shear), walls and ribs respectively. The coefficients 0; 0 For tubes with similar longitudinal variation 1/J, of flange areas Bin each bay O;o1 is zero since the direct stresses associated with flange loads (bl) are at any cross-section those of E.T.B. and hence generate only planar displacements of the flanges. On the other hand for arbitrary variation of flange areas system (bl) does not represent, in general, E.T.B. and there arises then a term 0; 01 which may be calculated from

.,

O;ot =

~ "'--

m= l

[ JP;,.P J.. .... .. ...... .. . ..... i+ l

i-1

om ~dz m

( b9)

where P;,., P ... are the (linearly) varying flange loads due to Y, = I and external loads respectively. The integration in (b9) is restricted to bays (i - 1) and i since the influence of Y, is restricted to this region. In general, the contribution of (b9) to 8;. is negligible and could only become significant in a bad design. The general formula for 8,., follows immediately from the unit load method as,

s••.=iJ Jq1'ds dz

.... . . . . .. . ... . ... .. ... . (b!O)

where q; are the shear flows due to Y, = I and the subscript s denotes inte-

gration over the walls. Again we need only consider bays (i- I) and (i) and we obtain immediately with the notation of FIG. 64 Slo• =

I

G

where

4

(btl)

~ W,.,,.+l{ [q om,m+1,8m,m+1]i- [q om,m +1,8m,m+di-1} m-1

,\'m,m+l ,8 m,m+1=t-···························· (b12) m,m+l

The idealization of constant flange areas in each bay has been discussed on p. 37. Proceeding now to the contribution of the Sib we note that its general expression is ik.•= IIqiqkd Gt s d z .......................... (b20)

s

and derive easily from FIG. 64 that

s

and the subscripts i and (i -I) indicate that the expressions in the square brackets have to be evaluated in the i and (i -I) bays respectively.

t.1 -·1··"

TA

+ 20

4 m~ .£w,8]m,m+1·

..... (b 13)

where

This formula may be simplified considerably if we refer the torque to a point S whose co-ordinates x 8 , y 8 are

I

J

4

~ £qAw,8]m,m+1 j Xs-XA=-20 m~l 4 S .... [ ,8] Y m~l W m,m+1

'Y= For the last term

............ (b14)

s. is the shear force. Eq. (b13) reduces then to

.... (b20a)

4

~ (Wm.mn) 2,8m.mt-1 .................... (b21)

m

I

of sik we find from FIG. 64 that it is zero but for k-i 2, i --1, i, i+ 1, i+2

n (w 41 ) 2 d I 1 St.i-2,r=S;-2.ir=(-Gt) ,-,-=2(h a +h) (-Gt) ,-,r r i-1 i-2 i--1 b r r i-1 i-2 i-1

.

~0

4

~ 'Yi-1

G /;_ 1

sikr

We obtain,

Ys-YA=O

where

-

t-t.tR--

I 'Y, S;.;,,.,=8q1.i .• =c-c; 7;

Note Using Eq. (b2) in each of the expressions in the square brackets we find 4 4 m ~ 1Wm,m+1q om,m+1,8m,m+1 =;,~,}qAw,8]m,m+1

-s

m~.£wq.,8]m,m+1= 2 0G' ....................... (bl3a)

where

r.~

is the torque about S and

0=

4

(b15)

~ [w,8]m,m+1

m=l

The coefficient S1., of Eq. (btl) can now be written concisely as 1 S; •• = 2oG [Ts181- r.~. t-10t-1l ............... . (blla) S 1 is, of course, the zero warping centre* for the bay i of the tube. The expressions given above for its co-ordinates allow for the effect of different taper of the flanges (within the present assumptions). For the uniform bay Eqs. (b14) reduce to the well known result* 4

~ [Dr.,Bw]..,,m+l

Xs - XA

cc

-

r.m~"l

2.1.~~ 4 -'-------

.............. (b14a)

(b22) Finally we find the total S1k coefficient from S;k =Sikf+Stks+otkr This concludes the derivation of all formulae for the S-coefficients in the Eqs. (b7) for the unknown Yi; Solution of the latter system of equations yields the total stress distribution in the tube. We illustrate our analysis on a numerical example further below. I::jfect of Cut-outs

~ [,8w]m,m+1

m=l

where D,. of Eq. (al4) is obtained by statics, seep. 53. The last contribution to S1• is that of the ribs. FIG. 64 shows that

stor =o[(q.,ql•) G,t,

1-1

+(q.,q;,) +(q.,q;,) G,t, ; G,t, 1+1

J

Rib k removed

I f"\ ha+hb w 41 =-ha+hb and .1.~=-2-d

1

G,1 and t, 1 are the shear modulus and thickness of the i'th rib.

The total value of 010 is now given by s~.=S;.,+St •• +S;., ........................ (b17)

h. '1

h.

The coefficients Slk FIGS. 63 and 64 indicate that sikf and sib can only arise for k=i-1, i, i+l The contribution S1kt of the direct strains in the flanges may be expressed as

slkt=IY;Yk

~ ~8m2 dz= JY;Ykdz ....................

m=l

(bl8)

where wm Y1(z) and w,. Yk(z) are the linearly varying flange loads due to Y1 = I and Yk =I respectively and 4 w,.2 = ~ EB · ............................. (bl9) m=l

k-2

Yk. 1

system when rib k is removed

Fig. 65.-System

Yk-t =I

when rib k is removed

m

If we assume that the flange areas and hence are constant in each bay we find immediately,

ll ....

(bl8a)

J • See AJ'IIYris and Dunne loc. cit. p. 84 and also Argyris and Dunne, Stress Distribution in Conical Tubes. puhli•hPd by R.Ae.S. as Vol. Ill of Stressed Skin Data Sh-.ts.

If a rib k is removed bays k --I and k merge into a single bay and the calculation proceeds by the previous analysis observing that the number of unknowns is, of course, also reduced by one. FIG. 65 shows schematically the flange loads due to Yk-" the influence of which extends now over the three bays k- 2, k --I and k; there is no system Yk. If in the (k ~I) bay a wall is removed it is again possible to use our analysis with few alterations. Thus, (a) the statically equivalent stress system Pmo• q. in the open tube bay k is calculated with the engineers' theory and torsion-bending theory. • • See J. H.

Ar~yris.

'The Open Tube',

AIRCRAFT ENGINHRING,

Vol, XXVI. No. 302. p. ll2.

61

"

--~.

10

__::..:.-,... ••

All loads rn lb.

Pa·-lf

(IIHO')

'•xact '-defarmablt ribs

'•11ac:t' · rigid ribs

5

E T.B

..,

F"lan;. I oa
J50 ' -

~ q,,

1------

140

OMI

-,

--r-

-- -·-

.,

-- - -- -

---~~-=-.:.-=~-~:.....

~-

...; (

'._ _ _ _ _ _ _J ~ r-----i-----~-----4------1------+-

...

Rib stahon

Fig. 66.-Fiange loads and shear flows In extended Yk system when web or cover Is removed in (k - I) bay

(b) the unknown system Y~:-1 and Y~~; merge into a single system which we denote by Yt and illustrate in FIG. 66. In this system the flange loads are constant in bay k due to the kinematic freedom of our open tube. Observe that the system of (n - 2) unknown Y;'s consists now of Y1, Y2, ... . .... , Y1:-2, Yk, Y "+b ... . .... , Y n-1

Fig. 67.-Numerlcal example of 81~~; method. Flange loads and shear flows In spar 'a'

All other 8;;. and 81, , are given by formulae (b20a). Formulae (b21) for 8;;, etc. take for i= k the form, d 1 1 8k-s,kr=2(ha+h&) (Grtr)i:-2 l~c-alk-2

(c) there is no term 81:-1,o· In general, the coefficients 81• 1 are again either zero or negligible. Should they be important they may be calculated with Eq. (b9) noting that for 8~~;.1 the integral extends from ribs k - 2 to k + 1 and that in the open bay P. includes the torsion-bending flange loads. The terms may be obtained from formula (b11) or (blla) but for 8~:-1 .•, which is zero and 8~~;., which is calculated from (see also FIG. 66) 1 4 8~:o,=-G ~ Wm,m+1{[qom,m;-1,8m,m+1Jk-[q..,,m+1,8m,m+lh - z} · • · · • • (bllb)

8 _

Similarly, the

8"+1•kr =

8,.,

m= l

8

1. .

are found from Eq. (b16) except for

8 _ tj_{_1 [(qor) ~~;.,. - 2

1~:- 2

G.t. 1:-1

-(qor) Grtr

t-2

8kor which is,

] -J. [(q.,.) - (qor)]} It G,t, "+t G,t, 1c

(d) there are no terms 8~:-1,i· Under the assumption that flange areas are constant within each bay we obtain

(bl8) All other 8;;1 and 8;,1 are unaffected except that O;,k-1,1 =0 Formulae (b20a) for 811, etc. become in the case of i= k,

8~:-s,to= -J

'r

i:-2

k- 2

8t-1,h = O

~ _G _!['Yk-2+'Yk] 1~:-a /~

Otk•-

62

ll

(

J

.... (b20b)

8

J

d _1 [-1-- c--~ +-~ )+ 1 _1 2(ha + h&) /k- 2 (Grtr)k-2 /1:-2 /k-3 (Grtr)i:-1 /1:-2

_

k- 2 , kr -

8k-1,kr = O d

kkr - 2(ha + hb)

{ [ - 1 - + -~-Jc-1 )2+ (G,trh- s (Grtr)i:-1 /k-2 +

[(Gr~r)k + (G,t~)"+J (~) '}

d I[ 1 I I ( I 1 )] 2(ha + hb) /;, (G,t,)k /;,+(G,t,h+1 "J;,+/"+1

· · · · · · · · (b22a)

The remaining 8, coefficients are derived as before from (b21). It will be seen later under the matrix analysis in (3) how much more convenient the use of the H matrix device is for the treatment of cut-outs. This concludes the long-hand analysis by the 8;~~; method. In what follows we illustrate it by a numerical application. 2. Numerical Example of the S;k·Method Consider the singly symmetrical cylindrical tube with six bays shown in FIG. 67 under the action of rib loads Ra1 and Rbi· All.bays have the same length 1= 20 in. The overall cross-sectional dimensions are given in FIG. 67 from which we derive, using FIGS. 63 and 64, thew-functions as, W1=+1/6, W2 = - ·1 /6, W3 = + 1/10, w,=-1/10 in.- 1 w 12 =+5/48,w 23 = - l/16,w 84 =- 3/80,w 41 =-1/16 in.- 1 .. (b24)

The ratio GjE, of shear modulus of walls to Young's modulus of flanges, is taken as G/£=0·385 in all bays. TABLE I gives the flange areas Bm and moment of inertia lx at the rib stations, the wall thicknesses t,.,m+1 in each bay, and the thicknesses t, and shear moduli Gr of the ribs. For the calculation of the influence coefficients etc., we assume constant flange areas in each bay and compute these as the mean values of the flange areas at the rib stations. Having calculated the ratios flm.m+l of the walls and the functions and 'Y of Eqs. (bl9) and (bll) we determine easily the coefficients 8,kt and 8,ks of Eqs. (bl8a) and (b20a). These are also given .in TABLE I. TABLE II presents the statically equivalent stress system from which we derive the term 8,., of Eq. (bll). • With the information given in TABLES I and II we can obtain the redundancies Y1 to Y6 for the case when the ribs are taken as infinitely rigid. The relevant equations are in matrix notation 14. 108 - 3 . 560

0

-3·560

29·103

-3·116

0

-3·116

31·635

0

0

0

0

0

0

0

I

0

0

0

0

0

0

0

-2·380

--2·380

0

0

39·583

-1·9381

-1·938

44·742

0

0

I 0

I I (q or/G,t,)!

-1 +2 -1

0

0

0

0

I

0 -1 +2 -1

0

0

0

0

0 -1 +2 -1

0

0

0

0

0 -1 +2 -1

0

0

0

0

0

I

1

L

_j

-8·33~

16·800

I -8 · 334

............ (b24)

the solution of which is in lb. in. Y 1 = -19573,Y2 =422, Y3 =8915, Y 4 = -485, Y5 = -737, Y 6 = -349 .. (b25)

L

0

0

(q or/G,I,)s

_j

0

2 -1

0

0

0

0

2 --1

0

0 -1

0

0

0

2 -1

0 -1

0

0 0

0

0

0

0

0

2 -1

0

2 -1

0

0

2·083

0

44· 133 -10·712

(G,t,h

0

I

0

2·083

52·081 -10·270

2·083 -10·270

57·240

I

L

_j

Ys

I

_j

(continued on next page)

O (G,t,h

0

I -I

0

0 O

-1

0

0

0

0

2 -1

0

0

0

2 -1

0

0

2 -1

0

0 -1

(G,t,)a

0

0

(G,t,)4 0

0

0 -1

0

0

0 -1

0

0

0

0 -1

I o

0

0

0

0

(G,t,);

0

0

(G,t,)s

_j

0

0

O (Grtr)1

L 2·691

- 4·774

2·083

0

0

0

4·774

11·023

8·332

2·08~

0

0

2·083

8·332

12·498

8·332

2·083

0

0

2·083

8·332

12·498

8·332

2·083

0

0

2·083

8·332

12·498

8·332

0

0

0

2·083

8·332

12·498

-

2·083

47·892 -10·974

I

0

0

0 -1

I I

0

2·083

2·083 -10·974

Next we consider the effect of rib deformability. Eqs. (b22) for 8;kr are easily calculated directly but it may be convenient also to use their matrix formulation for their evaluation (see also Eq. (258) on p. 41 ). We have

L

I

.............................. (b27)

2·083 -10·712

0

141·8

=10-" i

0 -1 +2 --1

40·126 -II ·448

2·083 -11·448

0

(q or/Grlr)J

2247

_j

272·2

0

I I (q or/G,t,)al

We may now compute the total 8;k and 8;. coefficients to find the redundances Y; for deformable ribs. The equations are

- 364·2

L

I (q or/G,/,)21

174

479

I

0 -1

I

-2821

Ys

I

-1

I

582

1

-1

I

_j

I

2818·7

L

0

(q ;,..(G,fr)1

=10-6

-2·642

541·8

0

+I -1

-

-2·642

I -2776·4

I

I 1151 I

35·394

0

Similarly we find for the contributions 8;or ofEq. (b16) using the values of (qor/G,t,); in TABLE II.

I I

L

2 -1 2

0 -1

_j

_j

.......................................... (b26)

• The small term 6iof arising due to the dissimilar taper of front and rear spar flanges has heen ignored here (See p. ~0.)

63

TABLE I of Example 9b Cross-sectional dimensions and 3;kf, 3;ks coefficients

Rib

2

(in. 2) B 3 =B, (in. 2) 1., (in.') tr (in.) G, (lb./in. 2) Mean B 1 = Bz (in. 2) Mean B3 =B• (in. 2) t 12 (in.) t 34 (in.) t 23 = t n (in.) ~12 ~34 ~2a=f3n
B 1 =B2

3i,i-1.! X 10"} Eq.(bl8a) 3iifxl0 8 3i,i+I,f X

10 8

3i,i-t,s X 10 8 } 3;;. x 10 8 8 Eq.(b20a) 3;,;+1•• X 10

5

4

3

2

Bay

7

6

5

4

3

6

0·35 0·45 0·25 0·40 0·40 0·40 26·3 28 ·1 24·5 0·036 0·036 0·036 2·50 2·50 I 2·50 0·40 0·65 0·50 0·65 0·30 0·60 I 0·40 0·40 0·40 I 0·40 0·60 I 1·00 0·036 0·048 0·036 0·048 0·036 0·048 0·036 0·080 0·064 0·080 0·036 0·064 0·048 0·048 0·048 0·048 0·048 I 0·048 167 167 167 125 125 I 125 278 278 156 156 I I 125 I 125 501 501 501 501 501 I 501 0·2350 0·1425 0·1887 0·1186 0·1610 0·1054 6·118 5·490 I 5·945 I 6·118 5·446 I 5·446 6 297 5 371 3 960 3 516 4 753 28·271 23·333 20·247 17·427 14·952 7·031 6·297 5·371 4·753 3·960 3·516 -7·076 7·133 8·235 7·076 8·013 16·248 16·469 14·208 14· 151 15·146 7·076 -8·235 -7·076 -8·013 -7·076 -7·133 0·65 1·20 71· 7 0·08 3·85

I

I

0·55 0·40 29·9 0·036 2·50

0·65 0·40 31·7 0·036 2·50

0·65 0·80 51·7 0·036 2·50

I

I

I

I

I

I

I

TABLE II of Example 9b Statically equivalent stress system and 3; 0 ., coefficients k.ib Bay Ra (lb.) Rb (lb.) s. (lb.) Ta (lb. in.) about spar 'a' M, (lb. in.) X 10 3 ·--Pol ='Po2 (lb.) Poa- --Po• (lb.) q 012 (lb./in.) q o34 (lb. /in.) qo2a=qon (lb./in.) q 0 , (lb./in.) Eq.(b5) 'L.w{3qo

3;., x 10 6 Eq. (btl) (q ;,_;G,tr) X WS

I I I -3473·0

I 1

6300 75600

77·0 429·2

4 150 1000

I

6

5

600 400

I

1450 5100 I 3600 I 2450 I 70800 I 52800 I 28800 I 19200 89 161 40 263 I

I

I

I i . . . . . . . . . . . . (b28)

I

_j

y 1 = -21898, Y2 =297, Y3 = 11066, Y. = -1243, Y 5 = -721, Ys= -834 (b29) Having the Y; we can find the total st~ss dis_tribl!tion in the given structure. Thus, for the flange loads at the nb statiOn 1, Pm;=Pomi+ Y;Wm .......... · · · · · · • • • • • • • • • • (b30)

and for the shear flows in bay i .......... (b30a) Typical results for the flange load P and shear flow q are plotted in

FIG.

67.

3. Application of the Matrix Force Method to the Same Example The long-hand method of formulating t~e equations for the re~un­ dancies in the analysis of the four-boom smgle-cell tube JUSt descnbed,

I

I

I

I

I

I

200 700

38lJ I 9870 1920 9920 4910 10570 5950 2850 32510 I 20350 I 10160 87·0 1 -2t6·9 171· 9 243·4 134·9 -133·2 -216·3 -87·6 -377·2 -424·8 -21·8 5·5 -12·8 -137·2 --184·0 27·3 184·0 I -50·8 I 138·7 I I 9·0 I I 10689 I 8603 I - 2249 I 847 I 201 272 -542 I 364 2819 2776 303 I -100 597 I -564 I -1541 I

I

7 6

450 100

I

which yield (in lb. in.)

64

I

5

4

3

750 750

1000 200

I

-1712·4

L

I

3 2

652·2! 4511·3

I

2

550 2400 II 440 840

0 0 0

I

41·8 -21·9 -19·9

I

747

7·1 142 79-

I

I

19·9

I -221

leading to the three or five-joint equations, provides a practical and powerful method for the determination of the primary stresses in such a structure. However, in more complicated cases, with more than one redundancy arising in each bay, the more systematic technique of the matrix formulation and solution is the only natural tool. Even for the four-boom tube the analysis under a general loading and the calculation of the flexibilities requires the use of matrices for its practical realization . Therefore, for comparison with the long-hand solution and to illustrate the way in which the method is applied in practice we analyse the same tube as previously under an arbitrary set of forces applied at the rib stations and calculate also the flexibility matrix for the points at spar-rib intersections. Finally, we apply the H-matrix device of p. 39 to find the stress distribution when one web field is removed. Although the matrix method is naturally associated with the use of automatic digital computers and large numbers of redundancies it also proves extremely useful in smaller problems with more elementary methods of computation. Thus, the calculations described here were all performed on a conventional desk type calculating machine and it may be of interest that the complete calculation-up to and including the flexibility matrixoccupied a total time of about 40 man hours. The applied loads R; are numbered consecutively from I to 12 as shown in FIG. 68 which gives also the numbering of the elements. As mentioned in the introduction to (9b) we consider as basic system the two spars acting independently as beams and it is obvious that the top and bottom covers and the ribs are unloaded in such a system. The rib booms are loaded neither in the basic system nor due to the redundancies and need not be considered at all. Due to our choice of basic system it is apparent that in the matrix a considerable number of zeros will arise and to obviate carrying these we advance the partitioning of bo in Eq. (253) a stage further and write,

r r

I

b.z

bola

0

0

bolb

b ••

b= 0

0

bow

~ bOT L

7 ....... . 12 ,-_A__---,

1 ........ 6 ,----A-----,

- -- - -

I

..... (b31)

0

bon· a

Edernal loads

I

I - - - - - - - - - ,_ - - 0 I 0

_j

bnwb

L

Basic system

where the numbers above the columns refer to the loads R;. Suffices a and bare used to denote front and rear spars respectively. In addition, we take advantage wherever possible of the symmetry of the structure by writing terms for the top surface only. Thus b. 1• is the matrix of longitudinal flange loads in the top flange of spar 'a' due to loads R 1 • ••••• R 6 and has only 12 rows, i.e. one for each end of each element of flange (see discussion on p. 40). Numbering of elements is from root to tip in each case (see FIG. 68). The six redundant force systems of the Y-type are applied as in the previous analysis and the matrix b 1 is partitioned in a similar way to b •.

for

b0

matrix

Numbering of elements

.......................... (b32) blu:a Y, Y,

The corresponding form for the flexibility matrix f of the unassemblcd elements is

f=

flu

0

0

0

0

0

0

f,b

0

0

0

0

0

0

(

0

0

0

0

0

0

fwa

0

0

0

0

0

0

fwb 0

0

0

0

0

0

. . . . . . . . . . . . (b33)

Y- systems for b: matrix (se-e F1gs. 63 and 64)

r

where f,., f," are obtained from Eqs. (257) and f., to f, from Eqs. (258). Carrying out the various matrix multiplications we can put the products b 1'fb. and b 1'fb 1 of Eq. (259) in the forms, bl'fb.= [[bl'fb.],.

[bl'fb.J,b]

_j_

[[bl'fb.],.."

...... (b34)

[bl'fb.],b]

bl'fbl = [bl'fbr], a+ [b1'fb1J1 b + [bl'fblJ., + [bl'fbdwa+[bl'fbl]wb+[bl'fbl]r ...................... (b35) In the same way, the product F.=b/fb, which is the flexibility matrix of th~

basic system under the external loads (see Eq. (230) is

r

+

I [bo'fb.],." I L

o

,,

Fig. 68.-Four-flange tube. Key diagram for matrix method of analysis

f,

1 0

[bo'fb,J,.."

I

bola·=

.... (b36)

_j

The above partitioning of the various matrices and the computation of the tlexibilities as the sums of the component f!exibilities is not of course strictly necessary. However, it usually happens that the number of structural elements is considerably greater than either the number of external forces or the number of redundancies and hence controls the order of matrices to be handled. For example, in the present problem the full b 1 matrix has 49 rows whereas in the scheme used above the highest order matrix is 12 x 12 determined by F •. Although the reduction is necessary here for manual computation it is obvious that such a method is required even with automatic computation for actual aircraft structures where the degree of redundancy may be of the order of 200 or so. With the extremely simple basic system chosen the b. sub-matrices can now be written down by inspection. As a typical case we find for the top flange of spar a the matrix b. 1• of tensile flange loads due to R

20

-6

I I I

I I I

L

2

3

4

5

6

2

3

4

5

6

0

2

3

4

5

0

2

3

4

5

I

I

JJ

}2

0

0

2

3

4

0

0

2

3

4

0

0

0

2

3

0

0

0

2

3

0

0

0

0

2

}4

0

0

0

0

2

I

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

}6

}3 ........ (b37)

J5

_j

The numbers in italics over the columns of the matrix refer to the external loads R1 • . • . . . R 6 and the numbers opposite the pairs of rows are the number of the elements of the flange 'a' in each pair of rows; the first one is for the end of the element nearer the root and the second row for the end away from the root. (See FIG. 68 and p. 40)

65

Fig. 69 (right).-Four-flange tube. Flange loads and web shear flows due to single load R3 = 1.000 lb. Application of H matrix to cut-out problem.

For the web of spar 'a' the matrix b •• ,., of shear flows due to the loads R in the basic system is 2

r-

3

4

5

6 1

0 1

0

0

0

0

0

I

0

0

0

0

I

0

0

0

0

bowa=-~·1

I

I L

''

2

(b38)

4

-1

0

0

0

0

0

0

0

0

}1

0

0

0

0

1

0 0

0

0

0

0

0

0

0

I

0

0

0

0

0

0

0

0

0

0

0

(2

•••

I

••••••

(b39)

L

\

with cut-out

\

I

q_,

Upper flange loads

-so . - 4Q

-so - 4Q

2o 4o so

_j

I

5

6

-1

0

0

0

0

1

-1

0

0

0

2

0

0·. l -I

0

0

3

0

0

0

-I

0

4

0

0

0

-I

5

0

no cut-ou' \

J6

4

o o o o

\

- 2o

15 j

7

8

2 ·00 I ·001

0 0

I ·00 2 ·001

0 0

9

0 0 0

•••••

12

I

I

I

6

13·331·66l

0 01

0 0 lt·66 3·331 0 0 I ---· -1---·---1----~----1 I 0 0 15·00 2·50 1 0 0 1

(b40)

I

1 0 0

2 f1b=3x ws"

I

I

12·50 5·001

0 0

1

I o o

I ----1 15·00 2·501 o o 1

I 0 0

12·50 5·001

,----~----1----

The corresponding matrices for web 'b' and the top cover differ from Eq. (b40) only in the scalar factor which, instead of 10/6 x 320, is

I

I

0 0

I

1

1----~----~----~----

1 and

11

10

----~----~----]

••••••

I

\

-so 0

r4

3

0

\

(lb./In)

'1o 0 .

I

0

\

10oo

'i4o

J3

'

0

'12o

2

0

''

\

Numbers over the columns refer here to the Y systems. For the web of spar 'a'; I

',,',,, ''

q_,

I

0

0

''

P,

(lb/.n)

I.

0

0

0

Cut-out web (w3)

6

12'3456

0

''

- 3ooo .

Corresponding matrices for spar 'b' are identical with those of Eqs. (b37) and (b38) except that they are multiplied by the scalar factors 20/10 and -1 flO instead of- 20/6 and --I /6 respectively. Using FIG. 68 which shows the part of the structure affected by each selfequilibrating stress system in combination with the information of FIGs. 63 and 64 on the Y-type system the b 1 sub-matrices can now be formed. We find for the top flange of spar 'a'

0

''

(lb.)

5

0

''

o o

I 0 0

} ............ (b41)

15·00 2·50 1 o o 12·50 5·001 0 0

1----~----~----

1 0 0 15·00 2·50 I

The sub-matrix b 1r for the shear flows in the ribs has been given already in Eq. (b25) and need not be repeated here. The flexibility matrix f of the unassembled elements of the structure is straightforward, using Eqs. (256), (257), (258) and (b33). As an illustration, we obtain for the upper flange of spar ,b', using the data of TABLE I:

66

L

0 0

12·50 5·00 J

. ............. (b42)

where the numbers in italics over the columns refer to the flange elements. Note that the only non-zero elements appear in the diagonal submatrices.

TABLE III of Example 9b Flange loads and web shear flows, due to R 3 =1000 lb., in tube with or without cut-out

I I

Upper flange load in spar 'a' (lb.) Upper flange load in spar 'b' (lb.)

I

Shear flow in web 'a' (lb./in.) Shearflow in web 'b' (lb. fin.)

I

Rib station Tube without cut-out Effect of cut-out Tube with cut-out Tube without cut-out I Effect of cut-out Tube with cut-out

I

1

L

Bay Tube without cut-out Effect of cut-out Tube with cut-out Tube without cut-out I Effect of cut-out Tube with cut-out

6·6384 1·3916 0·0347 -0·0558 1·3916 2·9840 0·7156 0·0249 2·5828 0·5494 0·0347 0·7156 -0·0558 0·0249 0·5494 2·3167 -0·0132 -0·0244 0·0088 0·4661 -0·0003 -0·0053 -0·0184 -0·0007 I

L

b,a=lo- 3 x

r L

11·320 2·389 0·063 -0·095 -0·023 -0·001

3 -971 I 1762 791 -1418 1 -1057 -2475

I 1

I

2 -84·0 -36·4 --120·4 -29·8 13 ·1 -16·7

I 1

I

I

I

I

5 -30 -396 -426 18 238 256

4 323 -1532 -120) -194 919 725

3 -102·9 102·9 0 -23·0 -37·0 -60·0

I

I I

I

-2·098 -0·841 0·430 -0·017 -0·021 -0·003 I

2

3

4

-0·0132 -0·0244 0·0088 0·4661 2·0885 0·3578 5

27·415 43·613 59·689 14·956 30·050 45·396 2·581 14·176 28·202 -0·102 1·938 13·520 -0·128 -0·179 1·477 -0·019 -0·103 -0·227

-0·0003 -0·0053 -0·0184 -0·0007 0·3578 1· 8112

6

75·572 60·395 42·159 27·196 13·604 1·593

I

4

11·0 -35·5 -24·5 -4·0 12·8 8·8

1

6 -17 -15 -32 10 9 19

5

6 -0·5 -0·5 -1·0 0·2 0·2 0·4

-1?:; -12·3 ~:~

I

4·4

1·989 4·178 4·919 4·000 2·118 0·063

2·389 5 ·244 6·586 6·000 4·118 2·000

7

0 0 0 0 0 0

I

_j

I

_j

8

7

IO

9

I2

11

I 91·509 -2·595 -6·598 -10·780 -14·873 -18·934 -22·993 75·470 -0·575 -4·087 -8·894 -13·867 -18·802 -23·72l 56·034 -0·022 -0·986 -5·700 -11·782 -17·980 -24·142 40·097 0·021 -0·036 -1·275 -6·494 -12·950 -19·434 28·050 0·005 0·033 -0·033 -1·242 -6·227 -12·012 14·328 0·000 0·007 0·041 0·013 -0·755 -4·942

R °

..•. (b43a)

_j

2

I

L

1

-95·9 -16·3 -112·1 -25·5 5·9 -19·6

I

I

-1·447 0·398 0·011 -0·016 -0·004 -0·000

1

I

2

-1658 599 -1059 -3005 -359 -3364

-1·704 -3·413 -5·123 -6·832 -8·541 -10·251 0·389 0·789 1·189 1·589 -0·006 -3·419 -6·838 -10·294 -13·676 -17·094 0·012 0·978 2·044 3 ·111 0 -0·006 -3·514 -7·075 -10·636 -14·198 0 0·100 1·586 3·253 0 0 -0·053 -4·233 -8·307 -12·381 0 0 0·181 2·000 0 0 0 0·104 -4·815 -9·815 0 0 0 0·181 o o o o -0·081 -6·173 o o o 0

L

I

I

1 -2726 78 1 -2648 -4364 -47 1 -4411

I

I

I

I

3

-2·726 -1·658 -0·971 0·323 -0·030 -0·017 2

4

-3·385 -2·434 -1·966 -1·080 0·246 -0·038 3

5

-4·071 -3·268 -2·974 -2·134 -1·066 0·266 4

6

8

7

IO

9

I2

11

I -4·749 -0·433 -1·100 -1·797 -2·479 -3·156 -3·832 -4·088 -0·096 -0·681 -1·482 -2·311 -3·134 -3·954 -3·994 -0·004 -0·164 -0·950 -1·964 -2·997 -4·024 -3·182 0·004 -0·006 -0·213 -1·082 -2·158 -3·239 -1·992 0·001 0·006 -0·006 -0·207 -1·038 -2·002 -0·945 0·000 0·001 0·007 0·002 -0·126 -0·824

_j

5

-120·15 -101·78 -95·87 -92·22 -87·62 12·11 -102·21 -83·99 -77·11 -71·69 0·83 13·97 -102·93 -90·20 -88·74 -0·38 0·14 11·03 -103·94 -95·87 -0·12 -0·57 -0·40 8·87 -104·11 -0·00 -0·10 -0·53 -1·18 8·30

6

7

8

9

IO

11

I 2 3 4 5 6 0.

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

(b44a)

I2

I -83·13 -10·52 -13·08 -9·82 -5·24 -0·69 3·81 -65·44 -2·88 -16·15 -16·63 -10·86 -4·28 2·18 -87·88 -0·22 -4·95 -23·05 -27·55 -26·20 -2~·52 -89·35 0·08 -0·36 -6·47 -27·35 -35·02 -38·66 -99·70 0·03 0·14 -0·38 -6·54 -28·49 -36·82 -95·20 0·00 0·04 0·21 0·07 -3·93 -25·74

_j

wi

w2

w3

w4

w5 w6 00

00

00

00

(b44b)

Note The numbers above columns refer to the loads R1 to R12 • In b 1a the numbers opposite rows refer to values at the rib stations. In the b., and b11 matrices two values were used at each rib station to denote the (equal) load in the adjoining flange element sections. Here in baz these two values are merged. In b,a the numbers opposite rows refer to values in the web fields. The additional factor 2 is introduced to take care of the lower flange since b. and b1 contain only the loads for the top flange. This applies also for the flanges of spar 'a' and for the covers. Carrying out the multiplications and assembly indicated in Eq. (b35) we obtain the complete matrices b1'fb. and b1'fb1 and hence, on inverting b1'fb1o the column of redundancies Y = -(b/fb1) - 1 b1'fb.R ........................ (b43) These results are printed in full above except for b1'fb1 which is identical with the matrix D of the S;k coefficients in Eq. (b28) and need not be repeated here. The calculation of the reciprocal matrix (b 1'fb1) - 1 was performed using the Jordan technique described on p. 28 and required approximately 8 hours for completion-by an unpractised and comparatively unskilled computor.

The complete set of flange loads and shear flows due to the forces R is obtained now with the b. and b1 matrices already given and the solution (b43) as 5=bR=[b.-b1(b 1'fb1) - 1 b1'fb.]R .................. (b44) Limitations of space prevent the presentation of the complete matrix, but as typical results we show above the flange loads in the upper flange of spar 'a' : bza=(b 0 za -bua(bt'fb1)- 1bt'fb.J ................ (b44a) and the shear flows in the web of spar a : • • • • . • • • • • • • b,a=(b,owa-btwibt'fb1)-lbl'fb 0 ] (b44b) Calculation of the flange loads and shear flows for the particular set of

67

forces R 1 used in the previous analysis gives results in agreement with the values calculated there. Such differences as arise are due to the neglect of the terms 8;.1 in the O;k analysis. (See footnote on p. 63). As a further illustration, the 11ange loads and shear flows given in TABLE III due to a single load R3 =I ,000 lb. are easily obtained from the third column of the final b matrix of Eq. (b44.) These values are also depicted graphically in FIG. 69.

The numerical result for the Y 8 matrix is printed in full below. It should be emphasized that this matrix gives the complete solution for a general set of forces R. It only remains to premultiply with bh to obtain the flange and shear flows. Thus, taking only a single force R3 =1,000 lb. we obtain from the third column the additional flange loads and shear flows given in TABLE III due to the removal of the web w3. Adding these to the values already given for the structure without the cut-out, we obtain the total values for the structure with cut-out. FIG. 69 illustrates the results graphically. This particular example of the application of the H matrix has been worked out rather more fully than is strictly necessary in practice in order to illustrate the way it works. In fact, the final result for the stresses in the structure with cut-out is given directly by Eq. (268) on p. 41. However, the amount of work involved is astonishingly small, consisting of a few simple matrix multiplications and a single inversion. In more general cases where several elements are removed at the same time the additional matrix to be inverted is of order m x m where m is the number of stresses (loads or stress flows) which have to be nullified.

E.ff'ect of a Cut-Out To show now the application of method of treating cut-outs developed on p. 41 we determine the effect of removing the web of spar 'a' in the third bay (i.e. element w3). This, of course, converts this bay into a four-boom open tube which, if taken into account initially would have required the introduction of the special extended Y system over the cut-out discussed on p. 62 and destroyed the simplicity and regular pattern of both b 0 and b1 matrices. By our artifice of imposing initial strains in the cut-out element until the stress there is zero we eliminate all such special considerations. Moreover, when the cut-out is introduced subsequent to the initial stress analysis, our method does not require any additional special stress systems but works solely with the existing information. Since the web carried only shear stress, the matrix of initial strains has here only one row and the result is obtained with remarkable speed and simplicity, the additional matrix to be inverted being of order 1 x 1. The shear flow in the element to be eliminated, i.e. b 11 , is given by the third row bwa (Eq. (b44b)), (printed in full on p. 67) and the matrix b 111 of the shear flow in this element due to unit redundancies is similarly the third row of b 1 wa (Eq. (b40)), i.e., 1 b 11,= 192 [0 0 1 -1 0 0] .................. (b45)

The Flexibility Matrix

To conclude the present example we calculate now the complete flexibility matrix of the structure for the points and directions of the forces R 1 to R 12 • In Eq. (312a) for the flexibility matrix, F F=bo'fb.- bo'fb 1(b 1'fb 1) - 1 b 1'fb. fb., b.'fb 1 and (b 1'fb 1) - 1 b1 'fb.

Now the equation for the set of Y H redundancies due to initial (shear) strain H in this element is (Eq. (236) or (262)) DY 8 +b 111 'H=O Y n= -D- 1 b111 'H .................... (b46) and therefore the additional stress (shear flow) in the element due to H is b11,Y 8 =-b 111 D- 1 bu'H ................ (b47) Adding this to the stress in the element already calculated, b,R,and equating the resultant to zero, we find, using the above b1 11 and the D- 1 already calculated, the required initial strain H =(b 11,D- 1 bu')- 1 b 11 R 192 2 3 · 801 X 10 8 b "R ............ · ........ · (b48) Hence from Eq. (b46) the additional Y 8 system required to nullify the stress in the cut-out element is

I

and therefore we only have to perform two further matrix multiplications and one addition. The complete matrix obtained is given in full below, Eq. (b50a). The numerical values given there (as in all other numerical matrices in this example) have been rounded off to fewer significant figures than were used in the actual computation. Although of course the F matrix is symmetrical, this characteristic was not used to shorten the computations and all values were calculated independently, the symmetry being then used to give a check on the numerical work. The same applies, naturally, to the computations of any other symmetrical matrix, like D. The whole question of numerical accuracy and avoidance of computational errors is obviously one of extreme importance in the practical application of these methods, and it is essential to have comprehensive methods of testing or checking calculated results. We hope to discuss these aspects more fully in Part III of the series with special reference to the use of automatic digital computers. For comparison with the present analysis, the flexibility matrix was also computed on the assumption that direct stresses in the flanges are given by E.T.B. i.e.-the statically equivalent system as used in the Otk long hand analysis. Thus F.=b.'fb. . ......................... (b51) where b. is the corresponding statically equivalent matrix. Particular attention is drawn to the fact that Eq. (b51) includes all shear strains associated with b,. The usual artificial separation of deflexions into those due to bending shear and twist does not even arise in these calculations. In the following table we compare some of the/; 11 calculated by Eqs. (b50) and (b51).

l

0·0238 0·1817 0·5350

b"R ................ (b49)

-0·4650 -0·1203 -0·0047

L

_j

Increments of redundancies Y 8 due to cut-out, I

1

2

3

4

I

1

21·54 27·62 35·67 41·42 48·36 55·32 4·43 11·27 18·42 25·43 32·38 39·32

L

68

6

7

8

9

10

1/

12

I -0·004 -0·064 0·471 0·412 0·406 0·402 0·001 0·023 0·105 0·125 0·120 0·112 -0·029 -0·488 3·591 3·147 3·096 3·066 0·008 0·173 0·804 1·961 0·914 0·856 -0·085 -1·435 10·573 9·266 9·115 9·027 0·023 0·509 2·367 2·830 2·691 2·519 0·074 1·247 -9·189 -8·053 -7·922 -7·846 -0·020 -0·442 -2·058 -2·459 -2·339 -2·189 .R 0·019 0·323 -2·378 -2·084 -2·050 -2·030 -0·005 -0·114 -0·532 -0·636 -0·605 -0·567 0·001 0·013 -0·092 -0·081 -0·079 -0·079 0·000 -0·004 -0·021 -0·025 -0·023 -0·022

L Exact flexibility of structure in in./lb.,

F=lo- 8 x

5

. ................... (b50)

we have already available the products

2

27·62 73·68 103·4 132·9 164·0 194·8 10·82 36·50 67·23 98·25 129·0 159·7

3

35·67 103·4 192·-l 265·9 342·3 419·1 17·30 65·22 136·1 212·8 289·4 365·9

_j

4

41·42 132·9 265·9 423·3 576·0 722·2 23·71 94·30 210·9 355·1 50~·3

653·1

.................. (b49a)

5

48·35 164·0 342·3 576·0 843·6 1090·3 30·05 122·9 285·1 501·9 747·7 997·3

6

7

55·32 4·43 10·82 419·1 17·3() 722·2 23·71 1090·3 30·05 1510·6 36·42 36·42 10·81 151·6 17·21 359·3 23·53 649·3 29·88 9)6·9 36·25 1373·0 42·61 19~·8

8

9

11·27 18·42 36·50 67·23 65·22 136·1 94·30 210·9 122·9 285·1 151·6 359·3 17·21 23·53 49·90 78·73 78·73 157·7 107·4 232·1 306·2 136·1 164·9 380·4

10

25·43 93·25 212·8 355·1 501·9 649·3 29·88 107·4 232·1 383·8 530·6 677·5

11

32·38 129·0 289·4 504·3 747·7 995·9 36·25 136·1 306·2 530·6 788·1 103.6·9

12

I 39·32 159·7 365·9 653·1 997·3 1373·0 42·61 164·9 380·4 677·5 1036·9 1424·7

1

_j

.................. (b50a)

TABLE IV of Example 9b Comparison of •exact' and approximate ftexibilities /;h (in.flb.) Exact x 10 6 Approx. x 10 6 Error

/2,2 73·68 49·61 -33%

!t.4 41·42 29·32 -27%

!t,l

21·54 10·50 -51%

e·c

As would be expected, the errors are much greater for points near to the root. Towards the tip they become negligible but it should be remembered that the structure analysed has quite a high aspect ratio. The example shows all too clearly that for modern low aspect ratio wings it will be necessary not only to take account of the shear strains associated with E.T.B. but to base the calculations of flexibility on the exact stress distribution. Moreover, as shown by our simple example the computations leading to F appear in the matrix method as a natural final step of the stress analysis. Thermal Stresses We consider now the stresses which arise in the four-flange tube due to non-uniform temperature. For this purpose we assume that the upper flange of spar 'a' has the temperature distribution shown in FIG. 70. The variation is taken to be linear between nodal points and hence is defined by the values at the nodal points. Since also the flange loads of the Y systems vary linearly between these points the argument on p. 43 leading to the second of Eqs. (287) may be used here too. Thus in place of the matrix e (Eq. 281) of the direct strains at the nodal points due to the loads R in the basic system we introduce here the thermal strains (a8) at the same points. As transverse flanges do not enter into the present considerations the matrix I simplifies considerably and observing that' I is the same in each bay, becomes for the affected flange I

I

2

0

0

4 0 I

1=6'

L

0

0

0 4

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

4 4

40 0

P• (lb X 100)

q•• (lb/in)

so

P.

b

Temperature distribution on upper flange of spar 'a'

(lbxlOO)

Q., (lb/in)

zo 0

'2o -40

·so ·eo I

I

I

I I 0

0

0

•

0

0

0

0

0

(b58)

Fig. 70.-Four-flange tube. Flange loads and shear flows due to temperature rise of upper flange 'a'

and taking a=23 x 10- 6 per degree centigrade we find I

I

I

2

a81

D,e=(b 11

I

37

L

a8a

.)'l.l

ae~

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

(b59)

a8 5

I

aG&

IL

a07

0 0

(buaY=~·

L

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

42

00

00

••

00

00

00

••

00

(b59a)

_j

Using the inverted matrix D- 1 of Eq. (b43) we obtain the column Y a of redundancies due to the given temperature distribution as (in lb. in.). Ya={ -41340 -36860 -28740 -18210 -13980 -11220} .. (b6l) from which premultiplying by the appropriate b 1 matrix, we derive the flange loads and shear flows. These are given graphically in FIG. 70. 4. Examples of More Complex Structures

_j

01> 0 2, •••••• 0 7 are the temperatures at the nodal points (see FIG. 70). In b 1az the two equal values in the elements on either side of the nodal points are merged into one so that btaz gives here single flange loads at the nodal points. Thus I

I

57

a82

I

35 66

_j

Hence, the matrix D.e is from Eq. (287). I

I

............

web shear flow

I

4

'

upper flange load

0

0

0

so

I

0

0

120

/a,a !t2,12 1424·7 192·4 154·3 1415·1 -20% -0·7°1,o

/2.4 132·9 106·1 -20%

0

I

I

Having investigated at some length the relatively simple example of a four-tlange tube both by the sik and matrix methods we turn our attention to some more complex and representative aircraft structures where the matrix method alone is indicated. In what follows we discuss mainly the selection of the appropriate self-equilibrating stress systems for the setting-up of the b1 matrix; (see also the more general analysis on pp. 38 and 39). We start with a two-spar six-flange tube, the two additional flanges being introduced to represent more accurately the direct stress carrying ability of the covers. FIG. 71 illustrates a structure of this type fully supported at the root with six bays and no cut-outs. The total number of redundancies is easily seen to be n = 6 x 3 = 18 which may be confirmed also by Eq. (247a) on p. 38 ; thus, n=a(,B+N-4)=6(6+1-4)=18 (b62) The calculation of the b. matrix is straightforward and the very simple choice of two independent spars made in (3) is here too the most advantageous. For a cross-section with no axis of symmetry we may form the b1 matrix for the following redundant systems: 00

00

_j

00

00

00

00

(b60)

00.00

00

00.00

69

Su:·flange s1ngte cell tubw

Singly symm•tric~l s iJC·flange stngle cell lubf'

Two-bay segment

r -~y5tem apph•d to corner flanges

bet F;gs 63 and 64.or 43)

i-sysltm (equal •net opp05 iU• X~systems

on covvrs (2.)) •nd (4.1 l)

Fig. n.--Singly symmetrical six-flange single-cell tube. Assemby of X and Y systems

X- §yt.lem for cover {4,1)

(sn Fig 41)

Fig. 71.--Slx-ftange, single-cell tube. Assembly of X and Y systems

6 Y systems applied to the corner flanges, 6 X systems in the top cover and 6 X systems in the bottom cover

s

I

(b63)

as indicated on FIG. 71; see also FIGS. 41, 43 (or (63) and (64)) for a full description including shear flows of these systems. If the cross-section of the tube in FIG. 71 has an axis of symmetry the number of redundancies for lift loads and torque is

S••-flang~

two-cell tube

n1 =6 X2 = 12 As redundant systems we choose in this case 6 Y systems applied to the corner flanges

} (b64) 6 X systems An X system consists of an antisymmetrical combination of two equal and opposite X systems in the top and bottom covers as illustrated on FIG. 72. These coupled X systems were not mentioned on p. 38. Next we discuss the six-flange tube with intermediate web shown in · FIG. 73. There are again six bays and for the moment we assume that there are no cut-outs. When the root is fully supported the degree of redundancy is from Eq. (247a) n=a(,B+N-4) = 6(6+2-4) = 24 ........... . .... (b65) The b. matrix may be obtained simply from the basic system consisting of three independent spars. As redundant systems we select,

X-system (see Fig. 42)

6 x 2 = 12 Y systems applied to the adjoining four flange tubes } 6 Z systems

(b66)

Two - bay segment

6 X systems in the top (or bottom) cover The structural elements affected by these systems are shown schematically in FIG. 73. Fxos. 42, 43 (or (63) and (64)) and (44) describe fully the X, Y, Z systems. Naturally, half of the Y systems may alternatively be based on the tube formed by the four corner flanges. Assume now that in one bay the intermediate web is missing and that no stiff-jointed closed frame is provided to restore partially the lost shear stiffness of the sheet. The number of redundancies is obviously reduced by one. The b. matrix may be determined for the basic system consisting of the spars and the two ribs shown in FIG. 74. Of the redundancies given in system (b66) we lose

70

Y- systems.

( ue Figs. 63 and 64. or 43)

Z-syst~ms

(see Fog. " )

Fig. 73.-Six.ftange, two-cell tube. Assembly of X, Y and Z systems

iJ.

Basic system Missing rib

X-system (see Fig. 42)

Missing web

Z-system over two bays (see Fig. 44)

Extended Y-system (see Fig. 66)

Y-systems over three bays (see Fig. 65 )

Fig. 76.-six-flange, two-cell tube. Modified Y and Z systems when rib of one cell Is removed

Y-system applied to comer flanges

Extended X-systems (see Fig. 75)

Fig. 74.-slx..ftange two-cell tube. Modified X and Y systems when Internal web removed i-2

systems. On the other hand it may be best to stick to the choice (b68) for the bay with the missing web and to work also in the other bays with Y systems applied at the four comer flanges. If the cross-section has an axis of symmetry and we consider only torque loads we lose through the missing web in one bay five of the redundant systems (b66), i.e. one Z system in the affected bay } (b67a) four Y systems at the adjoining rib stations The required four new systems may be taken as, one stretched Y system two standard Y systems applied to the four comer flanges at the adjoining ribs

one stretched X system (equal and opposite stretched X systems at the top and bottom covers) Again it is better to adopt the Y systems applied at the four corner flanges at all rib stations. Another internal cut-out may occur through a missing rib, for example in the right hand side cell (see FIG. 76). Here we lose in the affected two bays five of the redundant systems (b66), i.e. the two adjoining Z systems I

+

~

three Y systems applied at the affected and adjoining two (b69) ribs in the same cell J The required four new systems may be chosen as (see FIG. 76) two Y systems extending over three bays and applied at the} adjoining ribs in the right hand side cell (see FIG. 66) (b70) one Z system extending over the affected two bays

Fig. 75.-Extended X system applied at top cover

the one Z and one X system in the affected bay

} (b67) and the four Y systems acting at the adjoining rib stations We have a net loss of six standard redundant systems and need hence five new systems. As such we choose (see FIGS. 74 and 75), one stretched Y system (as shown in FIG. (66)) two standard Y systems applied to the four comer flanges at the adjoining ribs

(b68a)

(b68)

two stretched X systems at the top and bottom covers (as shown in FIG. 75) Alternatively we could select two stretched Y systems (two adjoining four flange tubes) but since these are linearly related to the two above Y systems applied at the comer flanges it is then necessary to drop one of the latter Y

one X system j Note that the latter X system is in addition to and in the opposite cover to the one chosen in system (b66). If the cross-section has an axis of symmetry the X system of (b70) is replaced by an X system. Finally, we consider the case of a cut-out in the cover (FIG. 77). We lose here in the affected bay three of the systems } two Y systems at adjoining ribs (b7l) one Zsystem the X system of (b66) may be assumed applied to the opposite cover and is hence not lost. Our choice of two new redundancies is one stretched Y system in the affected cell one stretched X system at the affected cover

}

(b72)

71

('

Cut-out tn cover

Cut-out in cover Extended

Y-system

(see Fig . 66)

' Extended

--

X-system

{see F1g 75)

Fig. 77.-Six-flange, two-cell tube. Modified Y and Z systems when one cover is removed

The above discussion of the effect of missing shear fields and the choice of appropriate redundancies may be applied with little alteration to the more general multi-cell structures considered in 8c. However, we require in addition for tubes wi•h three or more cells the stretched Z systems illustrated in FIG. 78 when an intermediate web or cover is missing. The above discussion shows that every type of cut-out requires special consideration and inevitably disturbs the pattern of the b" and b 1 matrices. We emphasize also that particular care is necessary in the selection of the necessary number of redundancies at cut-out regions to avoid the danger of linear dependency. It is evident from the previous example of a cut-out that all these complications are avoided by the use of the new technique on p. 41. With the help of this powerful method we may carry out the stress analysis in the continuous structure (always a simpler and smoother process) and derive hence (by the indirect use of initial strains in the elements to be removed) the stress distribution in the actual structure with cut-outs. Naturally we introduce in this approach additional redundancies but this should be, in general, of little or no importance when a digital computer is used. In concluding Part I we point out again that the matrix force method developed on p. 37 et seq. of Section 8C and illustrated here on some very simple problems is directly applicable to delta wings and in fact to any wing configuration as long as the grid of spars and ribs is orthogonal or nearly so. Emphasis in the examples has been on the force method of analysis since it is, in general, more suited to aircraft applications. Although the a., a 1 matrices of the displacement method (Section 80) may be slightly simpler to form than the b0 , b1 matrices of the force method 1 it is on the other hand easier to write down the f than the k matrix of the unassembled elements. Possibly the greatest advantages of the force method in wing and fuselage problems lie in the better conditioning of the Eqs., the much greater accuracy in the derived stresses and the smaller number of unknowns.

72

I

;;.

Extl'nded !-system

Fig. 78.-Extended Z system in multi-cell tube when one cover is removed

Part II. Applications to Thermal Stress Problems and St. Venant Torsion By J. H. Argyris and S. Kelsey

1. INTRODUCTION

W

E follow up and amplify the general theorems and principles

developed in Part I by presenting here a number of applications.

It is hoped that the problems considered may prove of more than

academic interest. Three of them are concerned with stresses due to thermal straining-a current fashion in the aircraft structural field-and were originally presented in an A.R.C. paper.* The first deals with the stresses arising in a circular ring due to a peripherally varying temperature rise on the inside of the ring, which is linearly elastic. Solution in terms of the bending moment and normal force in the ring is obtained easily by a straightforward application of the D;k-method described in Section 8 of Part I. The second example is concerned with the same problem but the material of the ring is taken to have a non-linear elastic stress-strain relation. The analysis is based on the unit load formulation of the virtual forces principle (Section 6 of Part I). It is of some interest that this not too trivial problem is amenable to a complete solution without excessive algebraic and numerical work. The use of the general principle of virtual forces for thermal stress problems is illustrated in the third example. An approximate solution is obtained for the stresses in a rectangular plate due to a temperature distribution symmetrical about the longitudinal axis of the plate and uniform along the length. In the fourth example, the twin principles of virtual displacements and virtual forces are both used to obtain approximate solutions to the St. Venant torsion problem for thin section bars. General differential equations for the approximate warping and stress functions are derived and applied to rectangular and double-wedge cross-sections. Use of the two methods gives upper and lower bounds to the torsion constant and an estimate of the maximum shear stress. The closeness of these bounds and comparison with some exact solutions indicate that the results achieve a high degree of accuracy.

The method of calculation follows closely Section 8 of Part I. From the symmetry of structure and loading, the shear force in the ring on the axis of symmetry is zero. Hence as the basic or statically determinate system we select the structure shown in FIG. Ia. As redundancies we choose the bending moment X 1 and normal force X 2 at the elastic centre,* which in this case is the centre of the ring. Forces and moments in basic system due to X 1 = l and X 2 = l respectively : System X 1 =I;

}

I

System X 2 = I; M 2 =rcos 1/J, N 2 = -cos 1/J

0o

71'

71'

-m
l

~ ............

J

J

(3)

}

(4)

Hence total moments and normal forces: M=M 1 X 1 +M2 X 2 Straining of basic system due to temperature 0 (see FIG. lc). Relative rotation c/J tds of two cross-sections at distance ds apart due to temperature gradient 0 I h across the section • See J. H. Argyris and P. C. Dunne, Structural Analysis (Part II of Handbook of Aeronautics, Vol. I) Pitman, 1!152, Table 17.1.

h

2. FIRST EXAMPLE Thermal Stresses in a Circular Ring A uniform circular ring of symmetrical cross-section and mean radius r is subjected on the inside circumference to a temperature distribution 0=y(l+cosmi/J)for

(2)

(I)

0 =0 over the remainder of the circumference the temperature varying linearly across the depth h of the section and falling to zero on the outside circumference (FIG. Ib). The bending moment and normal force distribution in the ring are to be found with the assumption that: (a) the material follows Hooke's law; (b) the bending deformations of the ring may be calculated by the engineers' theory of bending for straight beams; and (c) the deformations arising from the normal force are to be considered but the shearing deformations are to be neglected.

(a)

Fig. I.-First Example: thermal stresses in circular ring (a) basic system (b) temperaturedistributlon (c) thermal straining of ring

Additional Notation A : area of cross-section of ring

I: moment of inertia of cross-section of ring M: bending moment ( +ve if producing tension outside) N: normal force ( +ve tension) • See Ref. (13) of Part I.

(c)

73

a0

~eds= --,ds

..........................................

(5)

The negative sign arises since positive bending strain ~ corresponds to increased length on the outside of the ring. Increase of length 7Jds at centroid (at mid-height of section) a0 (6) 7JdS='J: The two redundancies X 1 and X 2 are calculated from the compatibility equations (182) given in Part I which take here the form SuX1+3 10 =0 I

j ······················

(7)

and hence

_ slO _ 32o X1- -3u' X2- -322 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

(8)

From the definition of the 3-coefficients or equations (203) and (204) of Part I s,,= ~2ds+ ~~ds ................................... . (9)

J

J c

S,.,=

M,. =

a§.EI h

c

JM,~etfs+ JN,7Jds

..................................

(10)

c

for p = 1 and 2. The sign f( • .• . )ds denotes integration around the circum-

Fig. 2.-Bending moment and normal force distribution in circular ring due to thermal strain, m= I

c

ference of the ring. Using (2), (3), (5) and (6) in (9) and (10) we find: dlfl 2TTr I Su =r El= El I

J c

J

c

and

(II)

+n/m

a0.rJ TTra0 0 3 10 =- 2h (I +cos mlfl )dlfl=- ---,;zjl -n/m

c )J

+n/m

a0.r 2

h 3 20 =- 2h 1 +z,.

a0.r 2 ( ·

= --h-

1

-n/m

h) m2 +lr m2-1 El

x1 = -sll =a0o2mk and X2=

~

(I +cosmlfl) cos lfldlfl

.

SID

m 7T

Thus 310

1 j

Mro = a§.EI h

I

....................................

-~::=a0o!~ m';~ 1 sin f[z } !~ ....................

(13)

(14)

where

h I f3=2r and K= r2A . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

(15)

Substituting (13) and (14) into (4) we obtain the final bending moment M and normal force N in the actual structure as M- a0oEI[ 1 2 ~ sin (TT/m) 1 +{3 - 2mh + m2-1 TT/m I+K- cos 'f'·'·] a0oEI N=- TTrh m2-1 m2

.

SID

+{3 m 11+K cos lfl 7T

which for m=1 reduce to a0oEI[ 1 +fJ M=-u~ 1 +1 +K cos lfl

J

............

(16) (17)

(l6a)

N- _ a0oEI 1 +fJ 2rh I +K cos lfl · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · (l7a) It is often possible to neglect the deformations due to normal force which appear only in 3 22 . To obtain the corresponding formulae we put in Eqs. (16) to (l7a) 1+{3 l+K=J .... • ............ • .. • • • • ... • •. • ........ •........ (18)

On the other hand no difficulty arises if the effect of shear deformations on 3 22 has to be considered. • This term is, however, usually negligible. Next we calculate the bending moment set up in the ring if the thermal straining is completely prevented. Thus : • See Aqyris and Dunne, loc. cit.

74

e= e.(1+cos31jl)/2

(12)

Fig. ].-Bending moment and normal force distribution in circular ring due to thermal strain, m=3

a0El a0 0 El Mt= -~~El=-h- =211 (l +cos mlfl) and at !fl=O a0 0 El Mto=-h- ············································ Hence

MM

~

7T

~

M 1 = z-0 + cos m'f') for -iii<. 'f' <.

m ... · ·· · ·· · ·· · ·· · ·· · · 7T

(19)

(20)

Comparison of (20) with (l6a) shows that in the case m = 1 the thermal straining is completely prevented if we ignore the deformations due to normal force. Using Eq. (19) in (16) and (17) we find finally for the bending moment and normal force: M _ _I_ Mto--2m

[t +2m~sin (TT/m) 1 +{3 -l TT/m I +K 2

Nr ~ _ ~ sin (TT/m) 1 +{3 .t. Mto ~ mL-} 7T 1 +K cos 'f'

cos ·'·] 'f'

··············

(2 1)

......................

(22)

Numerical Applications The distribution of bending moment and normal force periphery for a ring with r/h=l2 (i.e. {3=1/24) and K/{12=1 FIG. 2 for the case m=1 and in FIG. 3 for m=3. Values of -Nr/Mto are plotted radially with the centre line of the ring

around the is shown in M/M 10 and as base.

3. SECOND EXAMPLE Thermal Stresses in a Circular Ring with Non-Linear Elasticity Consider again the ring of the first example subjected to the sam:! temperature distribution. The cross-section of the ring is the idealized I-section shown in FIG. 4 and the material is assumed to obey a non-linear stress-strain relationship of the form E

·=i+~(a) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

(23)

1·0 --o-- - - . - - - - - - - - -

0·8

--0--~:>---iJ---

0·6 Fig. 4.-Second Example: idealized 1-section

where cp (a) is an odd function of a (same stress-strain diagram in compression and tension). Moreover, cp =dcp/da =0 for a =0 which ensures the usual linear relation for very small stresses (see FIG. 5). The bending moment and normal force distributions in the ring are to be found with the assumptions that : (a) Only the flanges of the section carry direct stresses, which are taken at any cross-section to be constant in each flange. (b) The direct strains in the flanges derive both from bending moment and normal force. (c) Shear deformations of the web are negligible.

Additional Notation A 1, A,: area of inner and outer flange respectively A : total direct stress carrying area P1, P,: loads in inner and outer flange respectively The structure is twice redundant as in the first example and as basic system we take once more that of FIG. Ia. Also as redundancies we select again the bending moment X 1 and the normal force X 2 at the centre. To find the unknowns X 1 and X 2 it is convenient to work in terms of flange loads and strain~. Flange loads due to X,,= I : M,,.· p ,., = N,,A,+M,, P,,,= N,,A ---;r-1 -h A h ........................ ..

I

I

....................................

(25)

and for system X 2 = I : P, 2 -

(1' +!,) cos 1{1, P,2- - (1' -;;) cos 1{1

(26)

Hence the total flange loads P, and P, can be expressed as:

P,- P, 1 X 1 +P, 2X 2=a,A, Substituting these expressions for the stresses in Eq. (23) the elastic strains• in the flanges are found to be E;-

EA;

+'f'

A;

__ PnX 1 +P, 2X 2 ..1..(P, 1 X 1 +P, 2X 2)

E,-

EA,

+'f'

The total strains are y 1 - E; -!-a(-1

A,

l

(28)

r

(31)

are the relative displacements of the point of application of X 1 and X 2 due to free thermal expansion of the inner flange. Using Eqs. (25), (26) and (28) for the flange loads and strains in (30) we find 11 X 1 +o 12 X 2 + f
o

..

o2lxl +o22x 2 + f
.-

(32)

o o 211 -

(33)

o

are the usual displacement coefficients arising from the linear part, a/E, of the strain, E, and are the same as in the first example. In fact, 812 =8 21 =0 due to the selection of the elastic centre as origin of X 1 and X 2 and Eqs. (32) simplify to o 11 X 1 +J(Pqcp,+P, 1cp,)dl·+o 111 - 0 .-

(34)

S22 X 2 + J
-

r2( m2 sin -7T aH°h - I + A. -' -") --. A r m2 - I m

~

I

............

(35)

) which are essentially the same as in (12), but for a small difference in 8 20 arising from the geometry of the cross-section chosen. Furthermore 27Tr 7r1· 3 £7(1 +K)

................................

(36)

..................................

(37)

where

,.

f(E;P 12 + E,P, 2+a0P; 2)ds=O

,-

f(E 1Pil + E,Pn)ds+8 10 =0 ('

,-

c

8 11 = £/' o22 =

Noting that in the actual system no relative movements take place at the cut ends and applying the unit load method as given by Eq. (71a) in Part I we obtain the following two equations f{E;Pq +E,P, 1 +a0Pq)ds=0

J(E 1P 12

2·0

Fig. 5.-Non-linear stress-strain curve

where o 10 =Ja0P; 1ds, o 20 =Ja0P, 2ds . . . . . . . . . . . . . . . . . . . . . . . . . . . .

820 =

y,- E,

or

1·5

('

(27)

_ PnX 1 +P, 2X 2 ..J..(PqX1 +P; 2X 2)

0·5

(24)

Note that M,. is taken about the centroid G of the cross-section. Using the expression for N,, and M,, of Eqs. (2) and (3) in (24) we obtain: for system X 1 =I:

Pq=-h; P, 1 =+r,

0·2

+ E,P, 2 )ds +8 20 =0

}

(29)

Eqs. (36) are identical with (II) in the first example. Non-linear terms of Eq. (34): Using Eqs. (24) and (27) for P,, P, and Eqs. (2) and (3) for M,, N,, the arguments of the cp-function in (34) are conveniently written in the form

c/J,= c/J{AI;h[ -X 1 -

l

(30)

J

• Dinrtional suffices in the strains arE' omitted siner direet bendinte strains only auly!i!iis.

K-A~·2and/-A,~,It2

OCl'ur

in thl'

X 2 r cos 1{1 (1

cp,=c/J{)h[ X1 +X2r cos 1{1

+~1 ~)]}

(I -1' ~)]}

(38)

Substitution of (38) into (34) shows that for a given cp-function we obtain after integration with respect to 1{1 two non-linear equations in the unknown X 1 and X 2 . The solution of these equations determines the complete stress distribution in the ring.

15

e =0.,(1+ cos!JI)/2 0·1

P=O

p=1 P=2

3·0

2{)

Fig. 6.-Second Example: thermal stresses in non-linearly elastic ring. Values of redundancies x 1 and x2

a0.,EI M,o • - h Fig. 7.-Bending moment and normal force distributions in non-linearly elastic ring due to thermal strains

We select now the specific a-E relationship

..................................

E=i[, +C~r-lJ or aE (a)" cp(a)=-3 a.

(40)

where a. is an arbitrary reference stress (a. may be defined as the stress at which the total strain is twice the linear strain component) and n is an odd integer. By varying a. and nit is possible to obtain a versus E diagrams with a wide variety of shapes (see FIG. 5). We assume also that A 1 =A,=A/2 then, K=I/Ar 2=(h/2r)2=f32 The cp-functions of Eqs. (38) take now the form

lr ............

(41)

....................................

(42)

cp;=-p"[x 1 +x 2(1+f3> cos cp.= where

pn[x 1 +x2(1-{J)

Jr1

X!= Mto

and

Jr2r

X2= Mto

COS

,Pr~

j

,Plni

in which Mt •• the moment at ,P=O for fully restrained thermal expansion, is given by (19), and Ea0. h p= 2a. 'fJ=z, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (43)

l

Substituting from Eqs. (25), (26) and (41) in the integrals of (34) we find

!. P; 1 cp;ds=p"~ 1-f [x +x (1 +fJ) cos ,P ]nd,P JP, cp,ds=p"Jz ~J [x +x (1-{J) cos ,P ]"d¢' 1

1

2

1

2

c

I (44)

c

which can be integrated without difficulty since n is an integer. Taking n=3 Eqs. (44) reduce to the simultaneous cubic equations 1 2x 1 +p 2x 1[2x 12 +3(1 +f3 2)x 22 ]-;n =0 2(1

+f3 2)x 2+p 2x 2 [3(2+f3)x 1 2+3(1

+6f3 2 +{3 4)x 2 2] -

-(l +fJ) 2m 2 sin (1rjm) =O m 2 -1

1r

l

which form= l and {3=0 (2r»h) take the particularly simple form

76

2x 2+p 2x 2[6x 1 2 +3x 2 2]-1 ~ 0 The linearly elastic case is represented by p =0. Then x 1 =x 2 = 1/2 which values may also be obtained from Eqs. (13), (14) and (19).

(46)

Numerical Results Eqs. (46), which have only one pair of real roots, have been solved for values of p between 0 and 3 and FIG. 6 shows values of x 1 and x 2 plotted against p. The roots of the simultaneous cubic equations were obtained by an iteration method, successive values of x 1 and x 2 being calculated, using the. previous approximations for the terms in the brackets. For p =0 · 5, successive values of x 1 and x 2 , calculated in this way are: X1 0·5 0·433 0·452 0·449 0·449 x2 0·5 0·390 0·416 0·410 0·412 The initial approximation in this case was the linearly elastic solution; for further values of p, extrapolation of previous results gives satisfactory starting points. Since successive approximations are alternately greater and less than the final values of x 1 and x 2 it is quite easy to improve the natural convergence of the process by a little judicious anticipation. Direct solution ofEqs. (46) was also obtained. Eliminating x 2 an equation cubic in p 2x 1 2 and x 1 is obtained, which can be solved for either parameter and the corresponding values of p. x 1 and x 2 calculated. The bending moment and normal force distribution around the ring is plotted in FIG. 7 for p=O (linear elasticity), p=l·O, and p=2·0, in the non-dimensional form M/Mto and Nr/M 10 • As would be expected, the effect of the non-linear stress-strain curve (in which da/dE decreases with a) is generally to reduce the maximum bending moment produced by a given temperature rise and also to smooth out the variation around the ring. 4. THIRD EXAMPLE Thermal Stresses in a Rectangular Plate

Consider a thin rectangular plate of length 2a and width b with a temperature distribution 0=0.g(y/b) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (47) symmetrical about the longitudinal axis of the plate and constant along the length (see FIG. 8). The stresses in the plate are to be found with the assumptions that they are constant across the thickness (plane stress) and that the material is isotropic and follows Hooke's law.

c

c

}

(39)

Additional Notation

(45)

2a, b: length and width respectively of plate x, y: cartesian co-ordinates with respect to the centre of the plate x'=a-x g=x/a, 7J=y/b, 'o;=x'/b: non-dimensional co-ordinates a0 =a0.g(7J): thermal strain

~..F(y/b)

We apply the principle of virtual forces in the form of Eq. (80b) of Part I, i.e.

~,. (Vyy}y.o

JJ[.6. 2F+Ea.6.0]SFdxdy=0

0

b/2 :------'L--- - ' - - - - ' ' - - - ------"'---- ---'---'-------'

t------a--~

_l

-1/2

J

I

I ......................

(48)

a .•• =a.,.=O is set up. To obtain the stress distribution in a plate of finite length, we can apply a stress-system -a.,., at the two ends x= ±a and superimpose the resulting (diffusion) stress-system on (48). We shall, however, adopt here a slightly different approach in order to illustrate the application of our principle of virtual forces in the presence of thermal strains. Guided by the solution (48) for an infinite plate, we assume that it is possible to express the longitudinal stress in the form a.,.,=Ea0J.(TJ)tP(~) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (49) where +1/2

f.(TJ)= Jg(TJ)dTJ-K(TJ)

.. .. .. . .. .. .. . .. . .. .. .. . .. .. . . .. .. ..

(50)

-1/2

and tfo is a function only of x. We know that Eq. (49) cannot, in general, represent the true distribution since the cross-wise distribution will also vary with x. However, for a smooth function g(y) it can be expected that (49) will yield a sufficiently accurate answer. The distribution (49) can be represented by the stress function F=Ea0,)J2J(TJ)t/J(~) .. .. .. . .. . .. .. .. .. . .. .. .. .. .. .. . .. . . .. (51) which gives the stresses o2F do/ au= 0 yz=Ea0.dTJ 2t/J(~)=Ea0.J.t/J(~) . . . . . . . . . . . . . . . . . . . . . . (52) o 2F

bdl dt/J

a.,.=-oxoy=-Ea0•ad1Jd~

o2F b2 d2t/J a •• = 0xz=Ea0.-(i21 de

. . .. . .. . . . . . . . . . . . . . .. . . . . . .

(53)

................................

(54)

The a.,.,, a•• distributions are symmetrical and the a~• distribution antisymmetrical with respect to the x-axis. From Eq. (52) and the conditions (a••)•=±b; 2=0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (55) (a.,.)._ 0 =0 (antisymmetrical distribution) . . . . . . . . . . . . . . . . . . (56) the function I is : '1

I(TJ)=J

'I

J!.(TJ)dTJdTJ .................... ................

where the virtual force system SF is here due to the increment Stfo of the function tfo 8F=Ea0b 218t/J . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (59) Substituting for F and 0 from Eq. (51) and (47) respectively, and noting that d 2g d 4/ dTJ2=-dTJ4 ···················· ···················· ···· (60) we obtain from Eq. (58)

J J[ d~d +I +1'2

d 4 t/J

a2 d 2t/J d 21

a4

d'l]

+27)2 de dTJz+ b'
(61)

I -·1/2

J[ +I

g, /., 1: functions of TJ A, B, C: constants tfo : function of ~ or ' {J 1, 2 =JL±iw: roots of auxiliary equation Ji=J-Lbfa, iii =wbfa The thickness of the plate is taken as unity. The problem can be solved exactly but we shall derive here an approximate solution using the principle of virtual forces in the form discussed in Part I, Section 6E. Consider first a plate of infinite length for which the stress distribution is well known.* Thus, for the symmetrical temperature distribution considered here, the self-equilibrating stress system +1/2

(58)

which may be written, on integrating with respect to TJ

Fig. 8.-Third Example: thermal stresses in rectangular plate; transverse distribution of temperature and stresses

axx=Ea0{J g(TJ)dTJ -g(TJ)

.. .. . . .. . . . . . .. . . . .. . . .. . .. . ..

(57)

-1/2 0

J

d4tfo a2 d2t/J a' A d~c2b2B de+IJ4C(t/J-I) Stfod~=O

.. .. . .. .. .. . ..

(62)

I ([2dTJ, C= I dTJ 4dTJ .. .. .. .. .. .. .. .. -1/2 -1/2 TJ -1/2 Note that A, B and C are all positive. Eq. (62) can only be satisfied for an arbitrary Scp if d4t/J a2 d2tfo a 4 Ad~ 4 -21J2 B d~ 2 +l)4C(t/J-1)=0 .. .. . . .. . .. . .. . .. . . . . . ..

(63)

-I

where the coefficients A, B, C are given by

J+ I;2

A= f2dTJ, B= -

·rl·l/2do/

r+

I''Jy

which is the required differential equation in tfo. The particular solution of (64) is t/J= l . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (65) which is, of course, the solution for an infinite plate and the complementary function can be written in the form t/J =K 1 cosh {1 1 ~ + K 2 cosh {1 2 ~ + K 3 sinh {1 1 ~ + K 4 sinh {1 2 ~ . • • • (66) where 8±(82-AC)vz]vz (67) flt·z-b A

·-i![

are the roots of the auxiliary equation a2 a4 A{J4-21)2Bfl2+pC=0 . .. . .. . . .. . . . . .. . . .. . . . . .. . . . . . .. .

(68)

A B C are functions solely of the temperature distribution and are ind~pe~dent of a/b. Thus, the length/width ratio of t~e plate _ente~s only as a common factor in the arguments of the hyperbolic functions m (66). The tfo functions must obviously be symmetrical with respect to the centre of the plate (~ =0) and therefore K 3 =K4 =0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (69) At the ends of the plate (~ = ± l) the stresses axr. a.,. are zero and therefore

tfo= ~~=0

for~=± I

....................................

(70)

The boundary conditions (70) determine the constants K 1, K 2 and the t/J-function is found to be J. t _ _{1 1 sinh {1 1 cosh {1 2 ~ -{1 2 sinh fJz cosh fJ1~ ( 71 ) 'l'(s)-l {J 1 sinh{J 1 cosh{J 2 -{J 2 sinh{J 2 cosh{J1 ·········· In general AC>B2 and the roots {1 1, 2 are therefore complex. They can be written in the form {J 1 , 2 =JL±iw . . . . . . . . . . . . . . . . . . . . . . . • . . . . . . . . . . . . . . . . . . . . (72) where JL=bii=bnc

v/~+~)r 12

8)]

I

~

. .. . .. .. .. . .. .

112 a_ a[l(~ /C w=bw=b 2 VA-A j Substituting (72) into (71) we obtain JL[cosh JLO +t) sin w(l-~)+cosh JL(l-~) sin w(l-r~)] tP = I JL sin 2w +w sinh 2JL w[sinh JLO +0 cosw(l-~)+sinhJL(l-~) cosw(l +~)] JL sin 2w +w sinh 2JL •••••••••••••••••••

• See Timoshenko and Goodier, Th~ory of Elasticity, 2nd Edition, McGraw-Hill, New York 1951, p. 399. •

(64)

0

0

••••••

0

•••

0

••

(73)

(74)

which, at the centre of the plate (~=0) reduces to

77

1 _ 211- cosh IL sin w +w sinh IL cos w (75) IL sin 2w+w sinh 211For the discussion of Eqs. (74) and (75) it is best to consider a particular form of temperature distribution. Take, for example, the function 1 + cos 2mrT} . K(T))= where n IS odd . . . . . . . . . . . . . . . . . . . . . . . . (76) 2 .JJ ) _ 'f'\ 0 -

which is shown in FIG. 8 for n= i. Then " __ cos 2n7TTJ Jo2

I

1 +~~~:1TTJ

l()

lt----+--

(77)

....................................... .

(78)

and the constants A, B, C are, from Eq. (63) 3 1 1 A=128n«1T«' B=32n21T2' C=g ........................... .

(79)

=

1·2 ...---.-----. ----

also iJ.=1Tn[t(v3+ 1)]11 2= 1· 351Tn, =1Tn[t(v'3-1)]11 2=0· 6981Tn .. (80) A perusal of Eqs. (74) and (75) on the basis of the arguments (80) shows that even for the lowest value (n=1) the function !fo, which is zero at g= 1, increases rapidly with decreasing g towards the value 4> =I for an infinite plate. In fact, not only does it reach it but, due to the influence of the trigonometric terms, it exceeds, very slightly (,..._,1·002) the infinite plate value if a> 1 · 2h for n = 1. The above discussion indicates that, for plates with alb> 1, the end effects for g= + 1 and g= - l are practically independent of each other. Thus, for such plates the stress distribution between g= 1 and g=0 may be taken to be the same as in a semi-infinite plate. Introducing the co-ordinate

y

w

x'=a-x

- 0·4 r---+r---t----t---+---+---+-----l

from the free end and putting

x' a-x

-0·2

a

{=b="=o-e>b

we find that for a-+oo, Eq. (74) reduces to

r/>(~)= 1 ;isinw{:wcosw{e-il; ....................... .

-06(81)

which is the appropriate solution for the semi-infinite plate. Note that, for the particular temperature distribution of Eq. (76), ji and w are proportional ton and hence lj>({) is solely a function of n{. The corresponding complete stress system is :

~_tf21 -I.-

cos2n7TTJ[1-ii sin 2

Ea0.- d7J 2 'f'-

w{+w cos w{ -ii:] w

e

·J. • 2n 7TTJ w -2 +11-z . -r -~<· df d'l' sm Ea0. = dTJ d{=- ~ sm we, e •

----w-

Un

u1J11 tP!fo +(l+cos2nry) C.02+;i2[_ -r _ . _,.1-;;· Ea0. =1 d{2= 81r2n2 -w w cos w.,-IL sm we, e '

I

..................................

Jl

(83)

Maximum stresses: For any T} the transverse direct stress (u••) reaches its greatest value at the end of the plate ({=0) and is given by un .~ /c 2 Ea0. =f(fi. +w 2)=f'V A The maximum value (at TJ =0) is (

)

Uww

-2

-2

max.=~ t~ =1·155

27T n For the maximum shear stress (u.,.) YJ=l/4, {=0·218n and thus (u ..) •...,

(uo:w)max. =0· 302 (u.. ).=<>

We mentioned on page 77 that it is possible to find the stress distribution in a finite plate by superposing on (48) the diffusion stress-system arising from -u.,., applied at the two ends x= ±a. In aircraft structural theory such diffusion analyses are often based on the simplifying assumption Ew=O. It is interesting now to compare the results obtained by the

78

Method of Virtual Forces

I

Assuming

£ 11 =0

Fig. 9.-Third Example: therm!'l stresses in rectan1ular plate; lon1itudinal var1ation of stresses

I~

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (82) ji and ware defined in general by Eq. (73) and for the particular temperature distribution of (76) by Eq. (80). The stress system (82) is illustrated in FIGS. 8 and 9 which show the transverse and lengthwise variations respectively of the stresses for the case when n = 1. u .. is the longitudinal direct stress u:u in an infinite plate

u.. = -Ea0. cos f'7TTJ

I

I

la~ter method with t~ose derived from our principle of virtual forces. To this purpose we consider for the temperature function (76) the plate under an end load stress-system

-u = + Ea 0 cos 2n7TTJ 0

..

2

...............................•

(84)

app~ied at x~ ±a. Now, in .a plat~ ~ith Ev.• =O, the s~ress system (84) is an etgenload, I.e. the cross-wise vanat1on of Its stresses IS invariant with x. •

The eigen-value ,\ corresponding to (84) is simply ,\=~1rn/b

and the diffusion coefficient JLD=M(G/E)

112 =21Tnba(G)112 E

The lengthwise variation of the u.,.,-stresses is obviously in this case cosh ILDg cosh IL» Thus, the stresses in a plate subjected to temperature function (76) are when Evv-O• easily found to be ' .!!E-.=_ cos 2n7TTJ(t- cosh IL»g) 2 cosh ILD 112 ~= _ sin 2n7TTJ sinh ILDg Ea00 E 2 COShJLD

Ea0.

..5L = Ea0.

_

(Q) (Q) I+ cos 2n7TTJ

E or for a long plate

2

l

J ... . ....... (8~

cosh IL»g cosh ILD

p. ~~;' Argyris, Dunne, 'The General Theory, etc.', Part V, J,R.Ae.S .. Vol.

Ll, November 1947,

__!!iu_ _

Ea0o-

-(Q) E

112sin

2mr7J .-- • 2 e 11 Ir

a.. GE l+cos2mr7J _-. rae.=-2 e

(86)

1'D·

where

fLD ~~ 27Tn(~) 112

For G/£=0·385 we have: liD = I . 2047TII which is in reasonable agreement with the value of IL in (80). However. the corresponding stresses are very different as may be seen from FIG. (9) for n = 1. It appears that the assumption €•• =0 overestimates considerably the rate of diffusion of the end load system -au in the range 0· 7> ~>0.

5. FOURTH EXAMPLE St. Venant Torsion of a Thin Section Bar In treating the torsion of uniform bars of thin section, approximate expressions for the maximum shear stress and the stiffness may be obtained by assuming the distribution of shear stress across the thickness to be identical with that for a very wide strip of constant thickness equal to that at the point considered in the cross-section. Thus for a section as shown in FIG. 10 we obtain the linear variation of a.x acro!.s the thickness dB a •.,=-2yGdz (90) with the maximum value dB a,.,max.=tGdz and the torsion constant

J=Hy0 3dx

..........................................

(91)

To obtain a more accurate solution we assume still that azr varies linearly across the thickness, but determine the variation over the width by application of the energy theorems enunciated in Part I. A solution is obtained firstly by application of the Principle of Virtual Displacements with an assumed transverse warping distribution to correspond to the linear au distribution and secondly by application of the Principle of Virtual Forces with the assumed stress variation. The two methods provide upper and lower limits respectively for the St. Venant torsion constant J (see Sections 4 and 6 of Part 1). These two analyses are fully worked out for singly symmetrical double-wedge sections and the results illustrated in a number of figures. The closeness of the upper and lower limits prove the methods to be very accurate. This is underlined by the coincidence of the approximate and exact solutions in three cases. Considerable effort has been devoted in the past in developing many approximate methods for the solution of the St. Venant torsion problem: some of them are discussed at the end of this example. Particularly relevant to our present analysis is the approximate solution for the isosceles triangle developed by Duncan, Ellis and Scruton* which is, in fact, identical with our solution of this case based on the Virtual Forces Principle. t

Additional Notation a, al> b, c, !=dimensions of cross-section (see FIGS. 10 and 15) a, a 1 =a/b, a 1 /b respectively Ox, Oy=axes in plane of cross-section (Ox along axis of symmetry of cross-section) Oz=longitudinal axis of bar Yo= y-eo-ordinate of section boundary g, g1, ?]=non-dimensional co-ordinates (see nu. 15 and Eqs. (142) and (143)) B' =dB /dz =rate of twist of bar about z-axis T=applied torque about z-axis J= T/GB' =~St. Venant torsion constant w=wB'=warping displacement of cross-section «!>=stress function /, /1> cf>, 1> 1 =functions of x n=a,,max./GtB'=maxi mum stress parameter • W. J. Duncan, D. L. Ellis and C. Scruton, 'The Flexural Centre and the Centre of Twist of an Elastic Cylinder', Phil. Mag., Series 7, Vol. 16, 1953. p. 231. t After completion and submission for publication of the present analysis the authors' attention was drawn to: E. H. Mansfield, The torsional rigidity o.fsolid cylinders of double-wedge section, A.R.C. Report, 1954, to be published in the R. & M. series. In tbis paper the same upper and lower limits are obtained for the torsion constant of a double-wedge section by a basically identical method. The stress distribution is, however, not discussed.

Fig. 10.-Fourth example; thin solid section in torsion

For the general background of the St. Venant theory of torsion of solid sections, the reader is referred to: Timoshenko and Goodier8 , Theory of Elasticity, 2nd edition, ch. II, McGraw-Hill, 1951. (a) Principle of Virtual Displacements To obtain the linear au variation we assume that the warping displacement w of the cross-section can be written in the form II' _c, ~.()' =yj(X). B' (92) • ' • •• • •. • • . • ' • •. •. •• • • ' • • • •• • •• • • •• • where j(x) is a function of x only and is to be determined.* Then in terms of Eq. (92), the shear strains and stresses are:

a u =G€

g

,,cGB'[~W-)']=,GB'y[ df.-l] ~X ~

a,.=-GE,.=GB'[~ +x] =GB'[f+x]

(93)

....................

(94)

We consider virtual displacements defined by an increment 8/ of the function f. For such virtual displacements B' =dB/dz remains constant and therefore no virtual work is done by the applied torque T. Thus, 8W is zero and Eq. (44) of Part I reduces to the principle of minimum strain energy Eq.(44a) of Section 4C which takes here the form: 8U;=~Jf[au8E,r+a, 11 8E, 11 ]dxdy -~0 . . . . . . . . . . . . . . . . . . . . . . . . (95) where the virtual shear strains due to the increment of are from (93) and (94)

8€,,,~B'y8(:fx) O'_,.t!!.Jt>

-~ . . . . . . . . . . . . . . . . . . . . J

b€,11- 8'8.f

(96)

Substituting from (96), (93) and (94) into (95), integrating with respect to y between ±Yo, we find

1{'~\X -I Y~f/+yo(f+x)b{ }dx=O

(97)

x,

and integrating the first term by parts

(98) Since 8/ is an arbitrary virtual increment off d [y}(df --- --I )] v
Y·/(:fx --

I)= 0 at the limits x

1

and x 2

•••• . ••••• ••••••••

(99)

(100)

giving the necessary boundary contitions at the ends of the cross-section. These latter arise from the virtual work of the stresses on the edges of the strip and are, in fact, the equilibrium conditions there in terms of the assumed transverse stress distribution.

Torsion Constant J To determine J by the Principle of Virtual Displacements, we consider a virtual increment 8B' of the rate of twist of the bar. Then from Eq. (44) of Part I 8U,: -8W=O . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (101) where 8W=T8B' . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (102) and 8U;=Jf[a,x8€u+a, 118€:u]dxdy . . . . . . . . . . . . . . . . . . . . . . . . . . (103) In calculating 8U,. due to the increment 80' we note that part of the virtual strains arises from the corresponding increment of the warping displacements which are proportional to B' (cf. Eqs. (92), (93) and (94)). But the increment of strain energy due to any virtual warping displacements is zero (Eq. 95). Hence for the virtual strains giving 8U; in (101) we can omit the part due to warping and take simply ::::::::::

}

.. .. .. .. .. .. .. .. .. . .. . .. .. . .. .

(104)

• See Part I, Section 4E.

79

Using Eqs. (93), (94), (104), (103) and ( 102) in ( 101 ), and integrating with respect toy, we obtain for the torque T due to the shear stresses

J[ ~_Y]\1; x,

T=2GO'

-I )+xyo
(105)

which gives, on integrating by parts the first two terms, x,

oU;*=G(0')2;{

- J{fx(l~os

x,

Substituting from the differential Eq. (99) for Yo(f+x), integrating by parts and noting the boundary conditions (100), we finally obtain (106) x,

a(

df) _ _I_ -~Jx, Yo l - dx dx

J~G0'-3

(107)

x,

~~) +Yoa[Ji(Yo1xo) -1 ]c/> }ocf>dx}

(121) ................................ Since ocf> is zero at the ends of the section the first term vanishes and there remains only

oU;* - -G(0') 2

and therefore the torsion constant is

i2~os~~Scf>+y/Jx·c/>ocf> I

·H {fx(l~o5t)+y.{fx(Yo1x0)--l]c/> }ocf>dx

ow• =OT·O' =G(0') 2· Hyo 3ocf>dx

x,

(122)

From Eq. (119) for S T, the virtual complementary work is (123)

(b) Principle of Virtual Forces To obtain a shear stress system satisfying the internal equilibrium conditions, we express the shear stresses in terms of a stress function
Substituting now Eqs. (122) and (123) in Eq. (116) we conclude that in order to satisfy (116) for any arbitrary ocf>, the function c/> must satisfy the differential equation

The equilibrium conditions on the boundary will also be satisfied if* (Aoq)
and must be zero at the ends of the cross-section.

T=ff[a.wx-a,,y]dxdy

which gives on integration by parts with respect to x and y for the first and second terms respectively and noting the boundary condition (109) (110) T=2JJ(x) .. . ... . . .. . .. . . . .. . ... . .. . . . .. . . where c/>(x) is a function of x only.t This satisfies (109) at the upper and lower boundaries and gives our assumed linear transverse distribution of azx· Substituting from Eq. (Ill) in (110) we find for the torque

T=G0'Hyo 3 c/>dx

......................................

(112)

............................. .......

(113)

The stresses and strains corresponding to (Ill) are from Eq. (108)

a,,=~C:=-2yc/>·G0'=GEzx

..........................

dyO,J.] G0'de/> ()
. . .. . . ..

(114)

(115)

In the general equation of the Virtual Forces principle (Eq. (62) of Part I, for 0 =0). (116> sui• -ow•=o . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . in which (117) oU;* = ff(E.,oa._,+E •• oa•• ]dxdy . . . . . . . . . . . . . . . . . . . . . . . . . . and the virtual stresses oa•.,, oa•• are obtained by giving to the function c/> a virtual increment ocf>. Such virtual stresses automatically satisfy the internal equilibrium conditions. We note moreover that since zero at the ends of the section. From Eqs. (114) and (115), we find for the virtual stresses due to ocf>

oa•..,= -lySe/> . GO'

].1.

........... .

(124)

APPLICATIONS TO PARTICULAR SECTIONS We now apply the equations developed in the previous section to obtain approximate solutions for the shear stress distribution and torsional stiffness of two particular types of cross-section: first the simple rectangular section and second a double wedge section formed by two isosceles triangles having a common base. Solutions are obtained by both methods and the results compared with one another and also with relevant exact solutions. 1. Rectangular Section (a) Solution by the Method of Virtual Displacements with Assum?d Transverse Warping Variation Consider a rectangular section of width s and thickness t. Taking the origin of axes at the centre of the section, we find for the differential Eq. (99) in this simple case d2[

d~2 -p,2f== ph

whence

J=Jo,=Hyo 3 c/>dx

a[

5 dyo) -I 'I'+ Yo 3 =0 d ( Yodx d (2yo dx +Yo dx T de/>) dx

........................................

where JL 2= 12/t2 • • • • . • • • . . • • . . • . • • • • • • . • • • • • The boundary conditions (100) are here

(125)

. • . • . • • • . • • • . • . •

(126)

dx =I for x= ±s/2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

(127)

and the appropriate solution is therefore 2 sinh JLX f=ji:coshJLs/2-x ················ ···················

(128)

d{

Stress Distribution: Substituting for /in Eqs. (93) and (94) ,

[ cosh JLX

az,=GO · ly cosh JLS/2 -I

J

II

(129) ?~GO'.~ sinh JLX JL cosh JLS /2 •• The distribution of shear stress according to Eq. ( 129) is illustrated graphically in FIG. II.

a

J

I

J ......... .

(118)

sr=Go'Hyo3ocf>dx . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

(119)

and from (112), the virtual torque due to ocf> is

Substituting for the strains from (114) and (115) and for the virtual stresses from (ll8), integrating with respect toy between the limits ±Yo, we obtain from Eq. (117) the virtual complementary energy

sui• =G(0'>2H [2~os;t fx>+ y.'!fxo fxSc/>>+Y.ac/>ocf> Jc~x

Assumed C1z• distribution

(120) • See Timoshenko and Goodier•, and also Part I, Eqs. ( 5). t See Part I, Section 6E.

80

tis • 0·20 Fig. !I.-Rectangular s~tion; shear stress distribution according to Virtual Displacements solution

1·0

---

Virtual Forces Virtual

m·

......... .........

_jJ

---Virtual Forces

~s--C' 0·4

0·2

0·4

-Virtual Displacements o

tis

0·6

0·8

Exact

Solutions

1·0

Fig. 12.-Maximum shear stress for uniform torsion of rectangular section

The maximum stress is reached at the mid-point of the longer sides and is amax. 1 Gre'=n=l-cosh Oa · · .. · · · · · · .. · · .. · · · · · .. · · · .. · · · · · · <130) where 8a=p.s/2=v3s/t . .. . . . .. . . . . . .. . . . .. . . .. . . . .. . .. . . .. . . (131) Eq. (130) is plotted as the broken line in FIG. 12 with t/s as abscissa. We note that in the extreme case of a square section (s/t=l) n=0·657, which is 2 · 7 per cent less than the exact value 0 · 675. * An obvious error in the detailed stress distribution is the constant a .. with y leading to unbalanced stresses on the boundary.

0·2

0·4

t/s

0·6

0·8

t·O

Fig. 13.-Torsion constant J for uniform rectangular section

Torsional Stiflness: The general expression (I 07) for the torsion constant gives, with the /-function of Eq. (128) 3 +s/2

J=t_J[l- cosh p.x ]dx 3 cosh p.s/2 -s/2

=

s~ll-ta~~ OaJ

(132)

The broken curve in FIG. 13 shows the variation of 3J/st3 with t/s according to (132). In accordance with the general Principle of Virtual Displacements this approximate solution must always overestimate the stiffness. (b) Solution by the Principle of Virtual Forces with Assumed Form of Stress Function The differential Eq. (124) becomes in this case

~~->..2cf>= -,\2

......................................

(133)

where

A2 =10/t2

(134)

and the boundary conditions are cp=O for x=±s/2 .. .. . . . .. . . . .. . . . . . . . . . .. .. .. . .. . .. . From (133) and (135) we find easily cosh Ax tP = 1 cosh As /2 · · · · · · · · · .. · · · · · .. · · · · · · .. · · .. · .. · · .. ·

(135)

( 136)

0

Assumed C1zx distribution

1 .JL

L.........l.... Gt~

tis • 0·20

Fig. 14.-f\ectangular section; shear stress distribution according to Virtual Forces solution

where

8b=As/2=V~s/t

(139)

The full line in FIG. 12 shows the variation of the maximum stress parameter n with tfs. Torsional Stiffness: Eq. (113), in conjunction with the cp function (136), gives for the torsion constant

J=s~l 1 _tan8~eb]

..................................

040

(137)

which is shown as the full line in FIG. 13. We find remarkably close agreement with values given by the exact solution.* Thus for a square section, which can hardly be considered as thin, Eq. (141) gives the value 0·419st3 /3 which is only 0 · 7 per cent below the exact value. In accordance with the general Principle of Virtual Forces our present approximation must underestimate the stiffness.

FIG. 14 illustrates these distributions for a section with s/t=5. We note that although the a,., distribution is generally similar to that of FIG. 11, the a •• distribution now satisfies all the boundary conditions. For the maximum shear stress (at the mid-point of the longer sides) we find amax. 1 Gt8' =n=l-cosh 8b · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · (138)

2. Double Wedge Section Since the centre of twist for unrestrained torsion is arbitrary, we assume for convenience that it is at the centre of the common base of the two isosceles triangles which form the section (See FIG. 15). We introduce also the non-dimensional co-ordinates g=(x+a)/a, g1 =(a 1 -x)/a 1 . .. . .. . .. . . . . .. .. . . . . .. . .. .. (142) measured inwards from the ends of the section and 7J=y/b . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (143)

Stress Distribution: Substitution for cp from Eq. (136) in Eq. (114) gives for the stresses =-2y[l-cos\.\x]·G8' cosh 1\S /2 ) , sinh Ax (}' t2 2 ( a •• = 4 -y 1\ cosh >..s/2 · G

a

""'

• Timosbenko and Goodier', foe. cit.

I;

............ · ...... ·

• Timosbeoko and Goodier', /oc. cit.

81

r_·

1---------

c

Fig. IS.-Double wedge section

0

Then for the left-hand part of the section x=a(' -1), y.=,b . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . and for the right-hand part x=a1(l-,1), y.=,1b ·...............................

I _Q_ I Gt .:j

....I..,JIU...,!. L . J •

I - !....... . . . . . . ...........

(144) (l44a)

tic • 0·20

0/C •

0·50

Fig. 16.-5hear stress distribution in symmetrical double wedge section t/C=0·2; approximate solution by Virtual Displacement method

(a) Solution for Assumed Warping Functions: For this type of cross-section we have separate expressions for the function fin the two triangular parts of the section. Writing f and / 1 for the function in the left ( -a
+a.

+

0

0

-a

-a

Iy.3(~ -I )at I - J{~[y.3 (1x -I)

J-y.(J+x) }otdx

a1

IY. 3 (~-t)otii-J{~[y. 3 (~-I)]-y.(f1 +x)}ot1 dx=O 0

0

................................ (145) To satisfy Eq. (145), the functions/and/1 must, as before, satisfy the differential Eq. (99) and the conditions at the ends of the section

[Yo3 (~-t)l=-a =[Y.a(~1 -t)l=+a,

=0

..........

(146)

In addition, however, we see that we must have

[y.3 (~ -I )atJz..o -[y.3 (1x1 -I )ot1].,.. 0 =0

[y.s(~-1 )l_o -[y.s(~I-1 )l=o =0

(150)

Eqs. (148) and (150) are the boundary conditions at the junction of the two parts of t~e section, giving the compatibility and equilibrium requirements respecttvely. With x and Yo from (144), the differential Eq. (99) becomes, for the left-hand part of the section do/ df ' 2iif2+3'a,-3a 2/=3a'(l +a 2)-3a 2a .................. (151) where a=a/b The complete solution of ( 151) is a [ /1-1 -/1-1 f=as_ 1 A' +B' _,(a 2+l)+(aLl) ........... .

J

(152) (153)

where fP=l +3a 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (154) Similarly we find for the right-hand part of the section (substituting from (144a)) the differential equation i:. 2do/1 i:. d/1 !>1 d' 12+Js 1~~ -Ja1o/1 = -3a 1, 10 +a 1 2)-i-3a 12a . . . . . . . . (ISla) and the solution

J

a1 [ /1,-1 -/1,-1 /1=a 12_1 -A1'1 -B1'1 +' 1(a 12+1)-(a 1 2-l) . . where a 1 =a 1 /b, {J 12= I +3a 12 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

(I53a)

(154a)

Boundary Conditions: In terms of the non-dimensional co-ordinates, the boundary conditions (146), (148), (150) may be written:

82

l

--

I

.. I ...

1 ..sL I

I

Gt~

(Oyz)y.o

-·~---------~--~~tic • 0·10

(147)

Since we assume a priori that the displacements are compatible, it follows that at the junction of the two regions f,._o=/ltz-o · ·.... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (148) and therefore (149) (ofl-o=(o/1)-.o ···································· which gives in Eq. (I 47)

0

I

a/C • 0·10

Fig. 17.-5hear stress distribution in single wedge section t/c=O·I; approximate solution by Virtual Displacement method

d'

l

d'

(155)

![df] +_!_[df1] =0 a 1 o,= 1 o=l a 1 The first of these gives immediately B=B 1=0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (156) and substitution from (153) and (153a) in the last two of (155) leads to A _ 2a({J1 2 +fJ 1-2)+a 1(fJ2+fJ 1-2) I ({J 1+2)[a({J1-I)+a 1({J-I)] (157) A -2a1(fJ2+fJ-2)+a(fJ12+fJ-2) ............. . 1({J+2) [a(fJ 1-I)+a 1({J-I)] For a doubly symmetrical section (a=a 1), Eqs. (157) reduce simply to A=A 1=2 ................ ·~·.......................... (158) and when a 1 =0, i.e. the section consists of a single isosceles triangle A=2(l +fJ)/3 . . . . . . . . . . . . . . .. . .. . . . . . . . . . . . . . . . . . . . . . (159)

I

Stress Distribution: From Eqs. (94) the stresses corresponding to the general solution (I 53) are a [ A (160) O"zv=-2a2-J' l-2g bG8' ..................... .

fJ-2]

O"zz=-

2[

a 3A ~:.fJ-2] 8 , 2a2-)7J l-2(1+fJ)S bG

................ ..

(161)

The maximum stresses obviously occur on the boundary of the section

(7]=,). Resolving there into tangential and normal components we find

(162)

TABLE I Comparison of Values of Maximum Stress Parameter n with Exact Soluti,.,.ns

and O'zn =

aazy-U.rz

(l +a2)1/2

(fl-2)

a 2A fl-1 , =(a2-l)(l+a2)112 P+t ~ bGB #0

.. . .. . ... ... ..

In an unsymmetrical section (a#a 1) the maximum stress is always on the longer side and hence in general we use in Eq. (164) the greater value of a. In FIG. 18 the broken lines show values of the maximum stress parameter n calculated for the cases a=a 1 and a 1 =0 and plotted against I fa. TABLE I gives some particular results for sections where exact solution is available. Torsional St!ffness:

Eq. (107) takes here the form 1ta=a

J~a( t-1 ~)d~+a 1 J ~~a( I+i I

0

1 1 ~ )d~ 1

0

....... .

(166)

J

6A1

,

(f1t+0(f1t+2)J

(167)

where A, A 1 are given by Eq. (157). For a very thin strip, i.e. when a, a 1 become infinitely large, the limiting form of (167) is (168) where t=2b and c=a+a 1 (see FIG. 15). Eq. (168) is the result given by the rough approximation of Eq. (91); it is exact for the infinitely thin section. For a doubly symmetrical section (a=a 1 ) the values of A and A 1 of Eqs. (158) lead to the simple result 12J <11 +5)([1-1) eta =
Exact Solution

1·00

0·-4n•

Virtual Displacements Solution (a)

1·00 Exact

0·-461 (-3·5%)

0·-4-43 (+l·3%)

Virtual Forces Solution (b)

1·00 Exact

0·-485 (+2·0%)

v'3/-4=0·-433 Exact

Square

Right Angled Isosceles Triangle

Equilateral Triangle a=Vl, a,=O

a=l.a 1 =0

0·32lt

v'3/-4=0·-433•

1=0·375 (+ll·5%)

t. ==0·3125 ( -60~)

• Timoshenko and Goodier'. Ch. II, pp. l66 and ln. t Galerkin, B. G. 8ui/.I'Acad:miedesSciencesdeRussie, p. Ill, 1919; Kolossol, G. Compt. rend., Vol. 178, p. 1057, 1924.

TABLE II et'

l IT With Exact Solutions

Comparison of Values of 1 Infinitely Thin Section

a=a 1 =-l

Equilateral Trian&le a~ v'l, a,==O

Richt Ancled Isosceles Triancle ao:.~ I, a 1 =0

1·00

0·-422•

0·300·

0·1567t

Virtual Displacements Solution (a)

1·00 Exact

.,\=0·-437 (+3·7%)

0·338 (+ll·6%)

... ==0·1B7S (+19·7%)

Virtual Forces Solution (b)

1·00 Exact

~~=0·406

0·300 Exact

,;:,~0·1563

t/c-->0 Exact Solution

Square

(-3·7%)

(-0·16%)

• Timoshenko and Goodier'. Ch. II, pp. l66 and ln. t Galerkin, B. G. 8uii.I'Academiedes Sciences de Russie, p. Ill, 1919; Kolossol, G. Compt. rend., Vol. 178, p. 1057, 1914.

which with the values off and / 1 from (153) and (153a) reduces to 3J aa 2 [ 6A a 1a 12 [ 2ba=aLJ I-~ +a1LJ I

a=a 1 =1

(163)

Expressions analogous to (160)-(163) are obtained for the right-hand part of the section and need not be quoted here. The distribution of shear stress is illustrated in FIGS. 16 and 17 for some particular cases. As in the previous example, our assumed form of warping displacements, yielding a constant u ·~withy, leads to unbalanced shear stresses u zn on the boundary. The maximum u,. stress is given by u,.max. a(a 2 +1) 1 12 [2(a 2 +1) Jl/(fl-2) (164) n= GtO' =J(fi+2)(fi-1) Afi(fi-1) and occurs at fl-2 2(a 2 + I) (165) ~ = AfiCfi -1)

I

Infinitely Thin Section t/c= l/(a 1 +a,)-->0

(169)

Similarly, we find for the isosceles triangle 12J fi2-t eta =(/3+2) 2

(170)

Eqs. (169) and (170) are plotted in FIG. 19 against the ratio 1/a and in TABLE u particular results are given for comparison with the corresponding exact solutions and with the results of the Virtual Forces method. (b) Virtual Forces Solution With Assumed Form of Stress Function

As in 2(a) we have separate expressions for cp and cp 1 in the two triangular parts of the section. Following the argument used there and noting that at the junction of the two parts the two functions must be equal cf>x=o=c/>~o:r=O

· · • • · · • · · · • ·• · ·•..•.•••..•.•.•.•••••...•

(171)

0·8

0·6

c/12 0·4

Virtual Forces

0·2

0·2

Virtual Displacements 0

Virtual

Exact Solutions 0

0·4

b/a

0·6

08

1·0

Fig. 18.-Maximum shear stress for uniform torsion of double wedge section

Forces

Virtual Displacements Exact Solutions

0·8

1·0

Fig. 19.-Torsion constant J for double wedge section: comparison of results by Virtual Displacements and Virtual Forces methods

83

..JL

0 1 _g_ Gt:5

0

Gt.f/

1.. 1.._,......_'""'-'-_._.'-''.._,wl

f/C : 0·10

ale • 0·50

f/C • 0·20

Fig. 20.-Shear stress distribution in symmetrical double wedge section t/c=0·2; approximate solution by Virtual Forces method

a/C = 1·0

Fig. 21.-Shear stress distribution in single wedge section tjc--0· 1: approximate s&lution by Virtual Forces method

and hence the virtual increments are also equal [Sc/Jl.--o=[Sc/J1lx-o

.............................. ......

(172)

we deduce from (121) and (116) the additional condition [ 2yo 5

5

dc/J+

=[2yo5 dc/>1+

4-1..]

4-1..

J

dx Yo '+' x-o 5 dx Yo '+' 1 x-o which may be written more concisely as !!._(Yo5/2J..)J [ dx '+' x-o -[!!_(Yo5/2J.. dx '+' 1)] x-o =0

............... ·.·.

(173)

Eqs. (171) and (173) provide the necessary additional boundary conditions for this section. Eq. (171) is the equilibrium condition (equality of au) and (173) is a compatibility condition. From Eq. (115) we see that it is not possible to satisfy the compatibility condition (equality of *'••) for all values of y because of the discontinuity in dy 0 /dx. Eq. (173) expresses the requirement for the virtual complementary work at the junction to be zero and is thus an averaged compatibility condition. The differential Eq. (124) becomes here, on substituting for x and Yn from (144), for the left-hand part of the section*

ct!J!

c2t!:.P. de+ 5~d~+2~( 1 _

~

__ 2a ~ 2 ....................... . a2)-1..'+'-

(174)

the complete solution of which may be written a2 [ 6-2 c/>=a:2_ 1 1-cg

-ng -6-2] ....................... .

(175)

Similarly we find for the right-hand part of the section a 12

[

cP1=a 12_J 1-c1~1

6,--2

.. 1),-2]

t/c : 0·05

(176)

Boundar.y Conditions: Since must be zero at the ends of the section on boundary)

<-1

+_! [_!!__
g=g 1 =0 (equilibrium

=0 ··············

(173a)

<,~1

For the functions cP and c/> 1 of (175) and (175a) the boundary conditions (171) and (173a) yield _ a 2(S 1 +l)+aa 1 (S 1 +2)+a 1 2 C- 5a(S 1 +2)[a(2S 1 +1)+a 1(2S+1)]

C 1 = 5al(S+2)[a(2S 1 + 1)+a 1(2S+ I)] When a =a 1 Eqs. (178) give simply

I I

J ··········

5

C=Cl=2S+I ....................................... .

and when

r-

_I

Stress Distribution: From Eqs. ( 108) or ( 114) we find, for the function c/J of ( 175), the stresses

b-2] bG8'

l

a2 [ u,.= -2a2 _ 1l] 1-C~ -2a2a_ 1g{ 1-

s;l-l~ 2 -( 1-Dl] 2]

and at the boundary (lJ =g) a2-=r-~ 1-C~

}bG8'

/j-JJ bG8,

Uzn=O

~

(181)

I ~ J

(182)

I J

The maximum value of u .. is given by

D=D 1 =0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (177) Eq. (173) becomes, on substituting for x and Yo from (144) and (144a)

![.!!. (g512c/J)] a d~

ate • 025

I

Fig. n.-Shear stress distribution in unsymmetrical double wedge section t/c=0·05, afc=0·25; approximate solution by Virtual Forces method

Uz.,=2-

S2=(3+5a 2)/2, S 12=(3+5a12)/2

(178)

a(a 2 + 1)1! 2

[

I

]1/(1)-2)

which is easily shown to satisfy the compatibility equation (179)

a 1 =0

............................................... .

Uz 8 max.

(183) n= Gt8' = 52(S+2)(S-I) C(S-1) Similar expressions are ebtained for the right-hand side of the crosssection and need not be given here. Again we find that, in general, the maximum stress occurs on the longer sides of the section. In FIGS. 20, 21 and 22, some typical distributions given by Eqs. (181) and (182) are shown graphically. Note the discontinuity in Uz• at the junction of the two regions. This appears because the variation with y is fixed by the assumed form (Ill) of the stress function which cannot represent adequately the effect of the sudden change of slope of the boundary. It does not arise in the case of the single wedge (a 1 =0) for which we obtain more accurate results. In fact, for an equilateral triangle (a 1 = v'J, S =3) our assumed form for the stress function gives the exact solution. Thus, the stress function* for this case is (184)

(180)

• This equation, _in slightly different form, was used lor the isosceles triangle by Duncan Ellis and Scuton, /oc. Cit. '

84

Oyz'

a(a2+J)ll2 [

where

C=l

~~==------~-~--

r""'==..:J::;;..,..,

Uzy=

(175a)

-D1g1

•

0 2<1>

ox 2 + 0oy2<1>2 +2GB ' =0

..................................

(185)

Values of the maximum stress parameter n are plotted in FIG. 18 against I fa for comparison with the results of the previous method and TABLE 1 gives some particular results for cases where the exact solution has been given. • See Timoshenko and Goodier!, Joe. cit.

0-9

0·8

J

cto/12 0·6

Fig. 23.-Torsion constant j for general double wedge section

Torsional Stiffness:

Using Eq. (115) we find for the torsion constant 3J

I

I

8ba=alg 3 cfodg+at bgt 3cPtdgl

.............. .. ............

which with Eqs. (175) and ( 175a) reduces to

J

a 12 4C a2 [ 12J 8ba=aar::-t 1- 3+2 +atUT!

[

4C 1 1-sl +2

J

......... .

(186)

(187)

where C and C 1 are given by Eqs. ( 178). Eq. (187), which is very similar in form to (167), yields the same limit as a and a 1 become infinitely large. When a=a1 we find a 2(2S+9) 12J 5 ct3 =2 (2S+I)(S+2) 2 and when a 1 =0 a2 12J 5 ct 3 =i ~ Eqs. (188) and (189) are plotted in some particular results are given.

(188)

(189) FIG.

19 as the full lines and on TABLE 11

Accuracy of Results In an approximate solution it is always desirable to be able to assess the order of accuracy achieved. In the two methods applied in this example, the method of Virtual Displacements gives an overestimate whereas the method of Virtual Forces gives always an underestimate for the stiffness. Thus, by using both methods we have succeeded in establishing upper and lower bounds for the stiffness. Some idea of the relative accuracy of the two methods can be obtained from TABLES 1 and n by comparison with results of exact solutions. Note, however, that these are for sections well beyond the intended range of application of the solutions developed here and the basic approximations are not really suitable for such sections. Nevertheless the agreement appears to be quite good. In general the Virtual Forces method appears to give the most accurate solution, in particular for the triangular section where the general formula (189) for J gives the exact value at both ends of the range lfa~o and a= v3. In FIG. 23 we show finally the torsion constant J calculated by the Virtual Forces method for the general double wedge section. Note that FIG. 23 shows minimum values of stiffness. For t/c
85