Automatic Structural Design

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THE AUTOMATIC DESIGN OF STRUCTURAL FRAMESBy R. K. LIVESLEY

(The Engineering Laboratories, University of Cambridge)[Received 30 June 1955]

S U M M A R YThia paper considers the problem of finding the lightest structural frame of givengeom etrical form which will sup port a given set of loads. Following the developm entof a geometrical analogue an d an iterative m ethod of solution, an analytic techniqueis presented which gives an exact solution in any particular nu merical case. The

method is very suitable for use on an electronic computer, and a brief descriptionof program mes developed for the Manchester Un iversity machine is included. I t isshown th at with slight modifications these program mes can also be used to determ inethe collapse loading of a given frame.1. IntroductionTHE problem considered in this paper may be sta ted as follows: Given aset of static concentrated loads, acting at certain fixed points of a rigid-jointed plane frame of prescribed geometrical form, how should the cross-sectional dimensions of the members be chosen to produce the lightestpossible frame capable of carrying the loads ? The members are requiredto be straight and of constant cross-section throughout their length, andthe term 'frame' is used to denote a structure which resists deformationentirely by bending action within its members.To determine whether a frame will support the loads applied to it thethe ory of plastic collapse will be used (1). Th is theo ry applies to stru ctu resm ade of a ductile ma terial, and assumes tha t if the cu rvatu re of a m emberbecomes infinitely large the bending moment tends to a maximum value,called the fully plastic moment, which depends only on the section dimen-sions. If this moment is attained at a particular cross-section, a plastichinge is said to have formed, and at such a hinge a finite change of slopecan occur in the mem ber. The m agn itude of this change is inde penden tof the moment, while the direction of the moment is always such as tooppose further ro tation. Provided t h at instability and shear effects canbe neglected, a s tructu re will be on the poin t of collapse if sufficient plas tichinges exist to transform it into a mechan ism, since und er these conditionsdeformations can increase indefinitely without further increase of load.This criterion of failure implies that in any given problem a 'design' canbe specified by giving particular values to the fully plastic moments ofall the members of a frame. To simplify the analysis it will be assumed

[Quart . Joa r a . Mecb . and Applied Math. , V o l . IX , Pt . 3 (1956)]S

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THE AUTOMATIC DESIGN OF STRUCTURAL FRAMES 259in a form which expresses the moments M a in terms of r arbitrarily-chosenmoments m ( = 1, 2,..., r),

M a = a^mt+MW. (1)Multiplying (1) by the appropriate lengths gives a similar expression forthe weighted moments Xa,Xa = A^m^X^. ;2)The matrices a^, A^ are nxr rectangular matrices depending only on thegeometry of the frame, while the columns M^\ X^ depend also on theapplied loading. For the beam shown in Fig. 1, for instance, plastic hingescan form only at the points 1, 2,..., 5. If the moments at the points ofsupport B and C are chosen as redundant moments, elementary staticsgives Group 11= 2

Group 21 = 4

\ X AX sx,*i

= 124

20

000

_ 24

+(3)

Any set of values M a or Xa which satisfies equations (1) or (2) is termeda statically admissible set.

(*) (3)

Fio. 1.For any given moment system it is possible to associate with each valueof a a parameter a, denned as the suffix of the moment of maximummodulus in the group to which M a belongs. Since the weighted momentsXa within a particular group are constant multiples of the correspondingmoments M a, this definition implies that \X a\ for all values of a.The set of numbers ,a form a sub-set of the n suffixes a, each value of abeing associated with a different group. In the analysis which follows, thesymbols a and i will be used as general suffixes, repetition implying sum-mation, while other letter suffixes will denote particular variables orequations. The symbol J will be used to denote summ ation over theasub-set a, the components of this set being defined as described above.The assumption that the fully plastic moment Jl of a member is pro-portional to its cross-sectional area implies that the weight of a structure

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260 R . K. LXVESLEYis proportional to the quantity ^Jtl, summed for all the individualmembers or groups. For any arbitrary set of mo me nts m t, equation (1)defines a statically admissible set M a, from which the values of a formingthe a-set can be picked out by inspection. These values specify th e p ointin each group at which the moment is greatest, and hence the points inthe structure where the particular moment system M a is most likely toinduce plastic hinges. The fully plastic m om ents of any str uc tura l designcapable of supporting the moments M a must clearly satisfy inequalitiesof the type uf ^ \M a |, one such inequality being associated with eachgrou p. The lightest of these structures will be th e one in which eq ualityholds in every case, i.e.Jt = | J4 | , o r Jn= \Xa\, (4)and this s truc tur e will be term ed a critical design. It will hav e a t least oneplastic hinge in each group of points, and the weight will be proportionalto the function 0, where, from (4),


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THE AUTOMATIC DESIGN OF STRUCTURAL FRAMES 261space, equation (1) associates with each point in that space a momentsystem Ma, and hence a particular critical design and value of 0. Thecomplete space may be divided into 'regions', a region being defined as

\

F I G , 2.

the set of all points for which the moments of maximum modulus 3f5 occura t certain fixed places in the structure and have constant sign. Throughouta region the sub-set a remains unchanged, so tha t the function 0 definedby (6) is linear in the variables m ;. In general, the regions are convexpolyhedra, their boundaries being hyper-planes on which either two


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262 R . K. LIVESLEYmoments within a group have equal modulus (greater in value than anyother m om ent in the group) or some mom ent of maxim um modulus changessign. Movement across a hyper-plane from one region to another impliesa change in either the sign or the position of a moment of maximummodulus, and hence in the coefficients of the linear function 0{Tn t). Inthe example shown in Fig. 1, for instance, the m^-space is two-dimensional,and is divided into regions as shown in Fig . 2. Lines of con stant 0 areshown on the figure, together w ith the positions and signs of the do m inan tmoments in the various regions.Before discussing methods of finding th e min imu m of the weight function0, it is necessary to prove tha t this function ha s, in fact, a unique minimumvalue. Since, from physical considerations 0 is always positive, and iscontinuous throughout the m (-space, it is sufficient to prove that, for anyconstant A for which the surface 0 = A exists, the surface is bounded andconvex, arid that 0 > A at any point external to it .F rom equa tion (5) it follows th a t 2 l-^sl = A on the surface, and hence

dtha t \X$\ ^ A. Since the weighted m om ents X^ are linear functions of thevariables m it the latter must also be finite, and the surface is thereforebounded.To prove that the surface is convex, consider the hyperplane (7)

where /Z indicates the set a associated with some region R which the surfaceintersects, and X^ indicates the set of quantities Xp, at some point in R.Let v indicate the components of the a-set at some point S on the hyper-plane, where S lies in a region R' different from R. Then in R the hyper-plane defined by (7) is coincident with the surface 0 = A, while at S,|X P | ^ \Xji\, with inequality in at least one case. It follows from (5) that0B = |Z P | > A. Thus 0 ^ A at all points on each hyper-plane forming

Vpa rt of the surface 0 = A, so th at all the co nditions required for uniquenessare fulfilled.Fro m the linear natu re of the function 0, it is app aren t th at the m inimumwill always be atta ine d at th e ve rtex of some region, where r hyper-planesintersect. If two or more vertice s give rise to c ritical designs of equalminimum weight, then all points belonging to the convex set generatedby these vertices will correspond to designs of th e same weight, bu t othe r-wise the minimum will occur at a unique poin t. In th e example shown inFig. 2, the minimum vertex is the point E, corresponding to the solution

M , = 2f, 34 = M 3 = M t = M& = $; 0 = 7*.

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THE AUTOMATI C DESI GN OF S T R U C T U R A L F R A M E S 2834. An iterative method of steepest descents

In the geometrical analogue described above, each vertex is defined byr m oment-equalities, each equa lity relating two mo ments belonging to th esame group . The problem is therefore t o find th e partic ula r set of equalitieswhich defines the vertex at which G(m1) is least, and the set of weightedmoments Xa appropriate to that ver tex.Both the methods described in this paper attain the minimum vertexby tracing a path through n^-space from some initial trial point wij1'. Ineach case th e p at h consists of a series of discrete steps, and it is convenient,before each new step is take n, to make the trial point which has just beenreached into the origin of coordinates. For any trial point m[k) equation(2) defines a ' trial solution' X\ so th a t if new independent variables xi aredefined with the point mj*' as origin, equation (2) can be writtenXa = A^Zi+Xp. (8)In the same way equation (6) becomes

f ^ (9)In the analysis which follows, the variable xi will be used in the senseindicated above, representing a displacement from the current trjal point.Since the solution of the problem can be obtaine d direct from the last termof the sequence X\ X\..., X..., it is unnecessary to keep a record ofthe absolute coordinates mt of each poin t. W hen a new solution X+1) isfound, the shift of origin is achieved merely by substituting it for the oldsolution X in (8).The simplest method of tracing a path towards the minimum vertex isby moving a suitable distance in th e 'direction of steepest descent* a t eachtria l po int. Differentiating (9) at the point mj*' gives

6 = ^AaiBgaXf=pit say, (10)where the set a is determined by inspection of the vector XK It is clearthat the components of the gradient vector^ are constant throughout theregion containing m[k) . If 0 is some positive number, then movement to anew point mj*+ 1) defined by

x, = m[k+ -m?> = -6p itrepresents a ' s tep ' 8p t in the direction of maximum decrease of 0. Thetrial solution associated with the new point is given byand th e process can clearly be repeated indefinitely. Th e path w ill eventua llyreach a state of random variation about the minimum vertex, the speed of


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264 R . K . LIVESLEYconvergence and the accuracy of th e solution depending on the p aram eter 6which controls the step length.The method outlined above is extremely simple to programme for anelectronic computer, and has been extensively tested on the Manchestermachine by Jen nings (4). In th e simplest of these experim ental program mesthe sequence of operations is as follows:

(1) The computer is supplied with the matrix A^, the vector X\ andthe coordinates of the first trial point mj 1' . In ad dition, a num ber isfed into the machine whose binary digits specify the division of themoments into groups.(2) The trial solution X is formed according to equation (2), and thisis substituted for the original vector X\ thus shifting the origin tothe point m\x).(3) From the trial solution the components of the sub-set a are deter-mined, and the vector p t calculated according to (10).(4) Th e new tria l solution is calculated from (11), the p ara m eter 6 beingchosen by the operato r and fed to the com puter b y means of switcheson th e m achine's control panel.(5) The new solution and th e value of 0 are printed if required, and theprogram me retu rns to step (3), with X*,1' replaced by X (*\ After kcycles the solution X> is thus replaced by X+r> .This iterative scheme proceeds automatically, apart from the manualsetting of 6, which may be altered a t any time during the calculation. Thetime required for a single step v aries with r and n in the exam ple shownin Fig. 2 each step took about 2 seconds.The chief difficulty experienced with the programme was in adjustingthe parameter 6, and in deciding when a solution reasonably close to the

minimum had been reached. I t is clearly desirable to commence with afairly large step and to decrease 0 as the minimum is approached, but itwas found in practice that a premature reduction often resulted in a falselimit po int being attain ed. An autom atic metho d of controlling the stepsize would have been preferable, but no criterion was found which couldbe applied in all cases.An other disadvan tage was the time factor. Experience showed tha t inmany problems the steps were so misdirected that many hundreds were

required to reach a satisfactory solution. I t is ap paren t th a t when th eweight function G has gradient vectors in two adjacent regions which arealmost in opposite directions, the trial point may alternate continuouslybetween the two regions without making much progress in the rightdirection. This situation ma y then deceive the machine operator into

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THE AUTOMATIC DESIGN OF STRUCTURAL FRAMES 265thinking that the step length should be reduced, in which case a falsesolution will be reached.

These factors render the iterative approach unsatisfactory except insimple problems with only two or three redundancies. An alternativemethod developed by the present author which avoids these convergencedifficulties will now be described.5. A modified method of steepest descents

As before, the method requires a trial solution X> derived from an initialtrial point m^K Subsequent work falls into two distinct stages. If R is theregion to which the initial point belongs, then the first part of the analysisis concerned with finding the solution associated with some vertex of R.This corresponds to the 'basic feasible solution' of linear programming.The second part of the work involves movement from vertex to vertexuntil the minimum is attaineda procedure very similar to that whichoccurs in the 'simplex' method (7).5.1 . Attaining a vertex

In the iterative scheme described in 4 the path always follows the lineof steepest descent appropriate to the region containing the current trialpoint. In the method now to be described, the path follows the line ofsteepest descent until it reaches a bounding hyper-plane of the initialregion R. It is then constrained to lie within this boundary, following amodified line of steepest descent until it intersects another hyper-plane,after which it moves through the sub-space common to both. Each inter-section imposes one more constraint on further movement, a vertex of Rbeing reached when a total of r constraints have been imposed.

At the initial trial point mj1' equation (10) defines the gradient vec*/or p{associated with the region R. The line of steepest descent is given by

* < = -Opt, (12)where 6 is now considered as a variable which may take any positive value.Writing Pa = A^pf, substituting for xi from (12), and taking the pointmjx) as origin in equation (8) gives

Xa = 0Pa+X>. . (13)This equation defines the variation in each Xa with movement down theline of steepest descent. As 6 is increased from zero, the point x{ definedby (12) traces a straight line through R and eventually this line mustintersect a hyper-plane separating R from an adjacent region R'. Asmentioned in 3, a hyper-plane is defined by an equality between two

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266 R. K. LIVESLEYmoments in a grou p. Let the one dividing R from R' be denned by thee q U a U t y Z M = Xy, (14)where /i and v are particular values of a, and where

\Xli\-=\Xf\>"\Xv\ inR,\X V\ = |Z A | > |X M | in R'.In other words, the quantity X^ is the dominant one of its group in R,while in R' it is deposed by Xv. The equality (14) will of course have adefinite positive or negative sign in any particular problem.The complete line defined by (12) intersects a large number of hyper-planes denned- by equations similar to (14). To find th e correct equ alityit is necessary to determine, for the set a appropriate to R, the value of "associated with each of the equalities

^ 5 = Xa (5 *


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TH E AUTOMATIC DESIGN OF STRUCTURAL FRAM ES 267the suffix I is not strictly necessary to the analysis, but is a convenientway of keeping th e equ ations well-conditioned. I t corresponds to th eprocess of taking the largest pivot in the elimination method of solvingsimultaneous equations. Multiplying (19) by A^ and adding to (18) gives

It will be seen that the coefficients of x l in (20) are identically zero, so th a tth e ope ration has th e effect of replacing the variable x t by the new variableyj. Eq uation (20) m ay be w rittenXa = A'^+XP, (21)

where x\ indicates the set of variables xv x% ,..., yu..., xr, and A'ai indicatesthe" matrix obtained by subtracting the appropriate multiples of A& fromth e oth er columns of the original m atrix A^, as in equa tion (20). The columnAtf itself remains unaltered, although it is now associated with the newvariable y t.It is easy to verify from (20) that

4 * = A'vt (i * I), (22)where the sign is the same as the sign of the equality (14), and it followst h a t X ^ ^ y , , (23)all term s in th e other variables dropping out. Th us any variati on fromthe new trial point in which y t is kept equal to zero does not affect theequality (14), and therefore corresponds to a movement in the hyper-plane defined by th a t equatio n. In pa rticula r a new direction of steepestdescent p\ denned according to the equations *M (24)= 0 ( = I ) Jlies in the hyper-plane, where the same set a m ay be used, as before, sincethe point just reached can still be regarded as belonging to B. From thenew gradient components a new set of parameters P"a = A'aip[ may befound and the whole process repeated. In this way a new trial solutionXW is reached and a new variable ym introduc ed. Th e choice of ym is madein the same way as before, with the restriction that m must be differentfrom I. Any variation in which both y t and ym are kept equal to zero willnot disturb either of the equalities which have been associated with thesevariables, and therefore corresponds to movement within the sub-spacecommon to two of the hyper-planes bounding R.Each repetition of the analysis results in one more equality being set

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268 R. K. LIVESLEYup among the weighted moments Xa, and one more restriction beingimposed on-variation in the variables x4. As each new variable yt, ym,...is introduced, so the number of non-zero gradient components in equation(24) is reduced, until finally only one of the original variables remains, and(r 1) equalities similar to (14) have been set up. This implies that a linejoining two vertices has been reached, and the next application of theprocess produces a movement along the line to one of these vertices. Ther equalities similar to (14) now determine the coordinates of the vertexcompletely, and in fact the solution has already been obtained in the formof the vector X%+1) generated by repeated application of (17).5.2. Finding the minimum vertex

The series of transformations described in 5.1 achieves two results. Inthe first place it alters the initial trial solution X 1* to a solutionassociated with some vertex of R. The components of the vectorsatisfy r equalities of the type |X5| = \Xa\ (a ^ a), where a indicates theset appropriate to it. In the example shown in Fig. 2, for instance, twocycles of the analysis result in the vertex C being reached from the initialtrial point A, via the path ABC shown. In the second place the originalmatrix A^ appearing in (2) is transformed into a new matrix A'at, asso-ciated with a new set of variables yo so that each such variable is associatedwith relaxation of only one of the equalities which together define thevertex.

If the vertex which has been reached is the one for which 0 attains itsminimum, then any further movement will inorease 0. If the minimumhas not been attained, however, there must be at least one line connectingthe vertex to another of less weight. If such a line can be found, the analysisalready developed can be used to determine a new vertex. If several linesexist, the best results will be achieved by choosing the line on which 0decreases most rapidly. Thu is not essential, however as the process willalways terminate if any of the lines on which 0 decreases is chosen at eachstage.

In attaining the first vertex the path lies either in it or on its boundaries.The set a therefore remains the same as it was at the original trial pointmW. In considering further movement it is convenient to distinguishbetween variations which lie along the boundaries of R and those whichdo not.5.2.1. Variation along a boundary of R. The original equation (8) hasnow been transformed into

^K (25)


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THE AUTOMATIC DESIGN OF STRUCTURAL FRAMES 269Consider the variable y, associated with the equality X^ = Xr , whereXp is the dominant quantity of its group in R. If only yt is varied, (25) be-C m e 8 Xa = A' |Xr| is that which has the same sign asB^. This variation implies no change in the 5-set, and represents move-ment along the line in the direction in which it forms a boundary of R.

It is apparent that the column of coefficients A'^ in (26) plays exactlythe same role as the column Pa in equation (13). Equation (26) gives

and henoe sgn Bur = [Y A' sgnX+1)]sgn Buv, (28)tyi la J

where, as before, a represents the set appropriate to R. Equation (28) givesthe differential coefficient of 0 with respect to a variation y,BgnB llv , i.e.the direction for which the line in question forms a bounding line of R.If this expression is negative, then a movement from the vertex to a trialpoint where y^ f l sgn i ^ , (6> 0)and the other variables remain zero will result in a decrease in 0. Multiply-ing the column A' by sgn.8^ results in a situation in which positivevariation in y, will reduce G.

5.2.2. Variation away from R. If the minimum vertex is not a vertexof R, then at some stage a point will be reached where a decrease in G canonly be achieved by a movement along a line leading away from R. Sucha movement involves a change in the position of a moment of maximummodulus and hence in the set a. Two cases must be distinguished.

(o) No other equality within the group. If the equality X^ = Xr asso-ciated with the variable y, is the only one in its group, then a variation inwhich Xr becomes the dominant quantity will not affect any other equali-ties, since these do not contain X^. In such circumstances a variation inwhich |Z r | > |XM| represents movement along a line in the opposite

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270 R. K. LIVE SLE Ydirection to that discussed in 5.2.1, i.e. a movement in which y, has thesame sign as B^. The differential coefficient of 0 in this direction is

sgn-B^, (29)where the summation giving dQ\byx now includes the term A'viBgnX lJ'+1) ,associated with the new dominant quantity X r, in place of the termA'fl lBgaX^+1K Expression (29) may be written

I d C

where (80ldy t)R indicates the summation previously carried out by usingthe a-set appropriate to R. If expression (30) is negative, then, as before,multiplication of the column A'^ by s g n f i ^ produces a situation inwhich positive variation of y, reduces 0.(b) Several equ alities in one grou p. If there is only one equality in agroup , the n bot h positive an d negative va riation s in th e associated va riableare permissible. In geom etrical term s, th e line associated with the varia tionextends on both sides of the vertex, as at point C in Fig. 2. W hen there ismore th an one equality in a single group, however, each line will end a t th evertex instea d of continuing throu gh it. Th e reason may be seen from aconsideration of the n ature of th e equalities. Le t there be s equalities

|Z M | = | Z r | , |Z ^ | = |Z A | , |X M | = |X p |associated with the variables y lt ym, yn,.. , where X^ is the dominantweighted moment in R for a certain gro up. I t is possible to va ry y t insuch a way that |X^| > \X \ without affecting the other equalities, butmovement in the opposite direction renders these equalities meaningless,since they do not now contain the new dominant quantity X r. Thissituation occurs at points * and E in Fig. 2. The equations of the a lineslying on the boundary of R are

\X V\ < |X J = |XA| = \X p\ = ...I X J = \x y\ = \x p\ =...

These equations imply an additional line lying away from R\X p\ = .... (32)

The lines defined by (31) have already been considered in 5.2.1, while theimplied line (32) can be dealt with as follows.

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TH E AUTOMATIC DESIGN OF STRUCTURAL FRAMES 271Consider first the equalities \X \ = \X,\, \Xp\ = \X^\, associated withthe variables y t, ym. If a new variable y'm is defined by

y l> ( 3 3 )where the sign depends on the sign of the implied equality X v =substituting for ym in (25) gives

(34)I t can easily be verified th a t if A"^ represents the new column of coefficientsof y, appearing in (34) thenAU = A, (35)according to th e sign of th e implied equa lity. Th us if y, is varied keepingall other variables including y'm constant, the equality |X K | = \X\\ is notaffected.In exactly the same way, by subtracting a suitable multiple of thecolumn A'an from the column A"^ a new column A^ can be derived in whicha variation of y t does not upset the equalities \X V\ = \X^\ = \Xp\. Theprocess may be continued until all the equalities occurring in the group

have been dealt with, and the column finally obtained may be treated forvariation along the line given by (32) in a similar manner to that describedunder (a) above.5.2.3. Determining the new vertex. Once a line has been found alongwhich positive variation of the appropriate variable produces a decreasein G, the analysis developed in 5.1 can be applied. Apar t from a multi-plying factor the column A'^ (or the combined column in case (b ) above)associated with y t is equivalent to the column P a associated with thevariable 6 in equation (13). Provided that the solution 9=0 (associatedwith the vertex already found) is discarded, the smallest positive value of6 given by (16) defines a new equality, which can be associated with thevariable y t in the same way as before. Thu s a new vertex is reached a ndthe matrix A'ai modified by the column A'^ in the usual way. W hen avertex is reached a t which 0 does no t decrease on any line, tha t vertex mustbe the required solution, and (4) then defines the associated critical design6. A fully autom atic p rogr am m eSince the steps comprising the analysis presented in 5 are all simpleoperations of linear algebra, the development of a computer programmeto carry out these steps automatically presents few intrinsic difficulties.The programme developed by the author for the Manchester Computer has

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272 R. K. LIVESLE Ya capa city of 32 mo ment points an d 16 redu nda nts, bu t there is no reasonwhy this could no t be extended.As in the simple iterative scheme described previously, the programmecommences by reading the m atrix A^, the vector X (\ and the coordinatesof the initial trial point mjx). The computer then evaluates and prints theinitial trial solution X^1', and records the 5-set appropriate to the initialregion R. Subsequent operations follow exactly the analysis set out in 5,the machine finally printing the solution associated with the minimumvertex, together w ith the value of 0. The solutions associated with inter-m ediate vertices m ay also be prin ted if required.

The programm e was conceived as a 's tan da rd ' p rogramm e, which couldbe applied to any frame within its capacity by practising structuraldesigners. In a recent pap er (Livesley and Charlton, 8) the au tho rsdescribed a similar programme for elastic structural analysis, and dis-cussed the general principles underlying the use of electronic computersin solving rou tine engineering problem s. Since a stan da rd prog ram memay be used by many different machine operators, it is essential that itshould work correctly when applied to any conoeivable problem within itsrange. This means th at it m ust be constructed and tested extremely care-fully, and that all 'special cases', however unlikely, must be allowed for.In the programme at present under discussion, a large part of the develop-ment time was in fact spent on making the programme function correctlyin all situation s. Much of this wo rk was merely a m atte r of organizing th erather complicated record-keeping required, but certain points seem ofsufficient computational interest to be worth recording here.

The chief difficulty, which appears to occur quite generally in linearprogramming work, is due to the fact that the course of the calculationdepends at certain stages on the equality or disparity of certain numbers.Th e section of the analysis^ devoted to determ ining th e ne xt equa lityencou ntered is particularly sensitive to small differences in num bers whichare supposed to be exactly equal, and a malfunctioning of this part of theprogramme which results in the same equality being selected twice has adisastrou s effect on th e subsequent calculations. If the equality IZ^I = \X r\is the first to be selected, for instance, then from equation (17),

| X | = \X\. ( 3 6 )On the second cycle of analysis, when th e vector P^ = A'^ Miscalculated,equations (22) and (24) imply 1^1 = \K\, (37)so that equation (16) gives an indeterminate value for 8 when the equality(14) is considered on all sub seq uent occasions. I t is straigh tforw ard to

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THE AUTOMATIC DESIGN OFS TR U C TU R A L F R A M ES 273arrange that the programme ignores this value, but if rounding errorscause equalities (36) and (37) to be only approx im ate, the progra mm e m aymistake the equality |X M | = \Xr\ for one no t already attaine d. Fo r thisreason the routines are arranged so that equalities of the type representedby (22), (36), and (36) are always digitally exa ct. In th e case of th e m atrixalteration specified by (20), this exact equality is achieved by calculatingthe columns of the new matrix according to the equations

instead of the algebraically equivalent expression appearing in (20). Asimilar procedure is used in calculating the column A'^ in (34).As has already been mentioned, the occurrence of both positive andnegative equalities similar to (15) must be considered when determiningth e ne xt equality encountered. Fo r machines which hav e a relatively slowdivision time a tes t is useful to find which eq ua lity is relevant. It is straight-forward to show that the sign of the expressionP.Xgi-PuXP, (38)

is the same as the sign of the equality (15) which gives the smaller valueof 6. Thus if (33) is positive, the relevant value of 0 will be that asso-ciated with the equality X& = Xa, i.e.nFTrr

while if it is negative the only possible equality is the negative oneXa = -Xa.

In some cases, of course, neither equality will give a positive value of 6.The section of the programme which tests the lines meeting at a vertexis arranged so t h a t it follows the first line it comes to on which 0 hasnegative gradient. The variables y t are scanned in turn and the approp riatecolumns of the m atrix A'a{ tested as described in 5.2. I t would doubtlessbe better from the point of view of convergence to test all the lines andchoose the one for which the gradient has largest negative value, butsuccessful functioning of the programme on a number of test problemsindicates that this is not absolutely necessary.7. Exam ples of problems solved by the program m e

Th e following exam ples were among those used during th e developm entand testing of the computer programmes described in this paper. They&0M.S5 T

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274 R. K. LIVESLEYwere originally chosen to assist in the eradication of various logical faultsand program ming e rrors, b u t th ey give some idea of th e possibilities of themachine in structural design work.Example 1The first exam ple is the continuous beam shown in Fig . 1, whose weightfunction is depicted in Fig. 2. As a typica l case, startin g from the point A(m B = 2-76, m o = 4-25) the computer traced th e p at h ABODE, asBhown in Fig. 2. Th e solution tim e was as follows:

Inp ut of problem tapeCalculation time .Prin ting of solution12 sec.8 10 30 These figures do not include the time required to feed the programmetap e into the machine . This extr a tim e (about 45 seconds) will, of course,be the same for all problems.

100

10100 3

-+ IO tXt-i

0

F I G . 3.Example 2

This struc ture is shown in Fig. 3. There are three redundancies a nd theequations of equilibrium are as follows:



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THE AUTOMATIC DESIGN OF STRUCTURAL FRAMES 275202025

15 50

000

- 1 3 * -00

- 1 0- 2 025

30 20

0

000000

1030

M 0017,50010,0000

010,0000

where = Mf, m , =and th e division into g roups is indicated by square brack ets. This problemproved extremely difficult to solve with the iterative program me describedin 4, for reasons mentioned in tha t section. Using th e second program m e,however, no difficulty was encountered.In the case where the initial trial point was chosen to be the origin theresults printed by the computer were as follows:

xaxhx cxdx.x,x.xkx33338,796

x a = - x b

It will be noticed that movement from the first to the minimum vertextakes place along an ' implied line', as described in 5.2.2(6). The t imetaken by the machine to produce the above solution was 40 seconds, ofwhich approximately 12 seconds was devoted to actual calculation, theremainder being spent on input and printing.Example 3The largest frame so far t reated by the programme is shown in Fig. 4.There are 30 mo men t points and 12 redundancies. The programme success-fully found the same value for the minimum weight from a wide range ofinitial trial points, although it was found that there were several vertices

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276 R. K. LIVESLEYpossessing the same (minimum) weight. Star ting from th e origin m t = 0tb.3 solution time was 7J minutes, this being divided as follows:

Input and initial printingCalculation of first vertexSelection of a minimum vertexPrinting of solution

. 45 sec.. 3 min. 5 ,,. 3 25 . . 20 7 35

W ith th e above starting-po int 14 vertices were tested before the minim um8000

t/rs})

I I 1 Jw w 7

Fio. 4.was attained . The average time of 15 seconds per vertex was divided asfollows: S e l e c t i o n o f n e w l i n e . . . . . 3 s e c .C a l c u l a t i o n of 9 2 M a t r i x a l t e r a t i o n . . . . . . 1 0 ,,One interesting feature of this problem w as th at during t he solution severalcases of 'coincident vertices' appeared. A t such a poin t more than requalities are satisfied, but this does not affect the operation of the pro-gram m e. The vertex is treated as two separate points joined by a linewhich, although of zero length, has a definite gradient associated with it,an d th is line is tested in the normal way. I t is also interesting to note th a tan elastic analysis of the same frame took the com puter app roxim ately th esame t im e (9).8. The calculation of collapse loads

So far the problem considered has been the choice of a suitable set offully plastic m om ents to support a given system of loading. The analysiswhich has been developed is, however, equally suitable for solving therelated problem, in which the fully plastic moments are given and it isnecessary to find the factor A by whioh the loading must be increased tocause collapse. F ro m the linear natu re of th e equation s of equilibrium itis app arent th at t he latter problem is equ ivalent to finding the largest factor

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TH E A U TO M A TIC D E S I G N O F S TR U C TU R A L F R A M ES 277A by w hich th e known fully plastic mom ents of a struc ture can be dividedbefore the structure collapses under a given load.To solve this problem by the analysis developed in 5 i t is nece ssaryto re-define the variables Xa. These will now be used to represent thequantities MJJKa>where ^fo is the fully p lastic mom ent of the m ember a tth e point a and is assumed to be know n. Since the factor A will affect eachpoint equally, it is natural to include all the points a in the same group.Th e symbol a th us indica tes simply the v alue of a for which \X a\ is greatest .Dividing the equations represented by (1) by the app ropriate values of ufagives a set of equ ations whioh m ay be represented b y (2), prov ided th a t th ecoefficients A^ are suita bly reinterpreted.With these slight modifications in notation the analysis proceeds exactlyas before. Since the re is now only one group , th e function 0 becomes

T ha t is to say, 0 represents the largest va lue of the ra tio \Ma\lUfa. For anygiven moment system, therefore, the largest factor A by which the fullyplastic moments -# o can be divided without violating the conditions \M a\ is given by

Clearly, the ma xim um possible value of A will be attaine d when th e function0 is minimized. As before, this m inimum w ill norm ally occur a t a ve rte x,where r equalities of the type \X a\ = \X a\ are satisfied. A t such a ve rte xthe re w ill be (r-f 1) values of Xa of equal maxim um modu lus, correspondingto the (r+1) plastic hinges which are in general necessary for collapse.How ever, if AJJ^ is attain ed at more th an one vertex, a collapse mode willexist which has less th an (r-f-1) p lastic hinges. This will be associated w ithlocal failure of a certain part of the structure.Although the method described will determine the correct load factoran d final collapse mode in all cases, th e p at h tra ce d o ut in m J-space will notin general correspond to the tru e behaviour of the m om ents in the str uc tur eas th e plastic hinges develop. An investigation of this beha viour requiresknowledge of the elastic characteristics of the structure, as well as theequations of equilibrium.9. Conclusions

The analytical method described in this paper gives an exact solution ofthe general mathematical problem considered, while the digital computerprogra m m e enables rapid solutions to be obtained in individual cases. I t

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278 TH E AUTOMATIC DESIGN OF STRUCTUR AL FRAM ESis of course true that there are many factors which affect the cost of astructure besides its weight, and in a practical design several differentloading systems m ust often be considered. Th e computer prog ramm e,however, provides a method of solution so rapid and au tomatio th at it m aybe of value in giving th e engineer a roug h guide in the in itial stages of hisdesign work.

R E F E R E N C E S1. J . F . B A X B B , J. Inst. Struct. Eng. 27 (1949), 397.2. J . Fouuoss , Quart. J. Applied Math. 10 (1963), 347.3. Proc. Roy. Soc. A, 223 (1954), 482 .4. A. jE to m ra s, M .Sc. Thesis (Manohester U niversity, 1954).5. J . H B Y K A N , Quart. J. Applied Math. 8 (1951), 373.6. H. J. GOBKNBEBG and W. PBAQBB, Proo. Amer. Soc. Civ. Eng. 77 (1951), 59.7. A . CHABNBS, W. W. COOPKB, aad A. HENDBBSOIT, An Introdu ction to LinearProgramming (New York, 1953).8. R . K. LIVKSLHY and T. M. CHABLTON, Trans. N.E. Coast Inat. Eng. Shipbuilders,71 (1954), 67.9. Engineering, 176 (1953), 230 and 277.



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