Influence Lines for Continuous Beams. a Direct Method

7/27/2019 Influence Lines for Continuous Beams. a Direct Method

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526 T h et r u c t u r a ln g i n e e r ,o v e m b e r 1945INFLUENCE LINES FOR CONTINUOUSBEAMS

A DIRECT METHOD%yJ. W. H* ing, MSc., Assuc.M.Inst.C.E., A.M.1.Struct.E.ABSTRACT

Thepaper presents a directmethod of obtainingordinates forplotting the Influence Lines for a Support Reaction, or Bending Momentor Shear Force at a point, using any of the current methods of analysisof Continuous Beams, but 'avoiding on the one hand the necessity ofcalculating such ordinates from numerous solutions with the unit pointload disposed at different pointson hebeam, and on the other theinaccuracy inherent in the laboratory method using splines or the like,unless considerable care is taken, and accurate methods of measurementare used.In brief, the method is to calculate the Moment effect of unit dis-placement, or unit rotation at a support, or at two supports, and thenceto calculate the end rotations of the various spans, and so the deflectioncurve,which will be he nfluenceLine equired. In the examplesgiven, Moment Distribution is used to obtain the Moment effect of theunit movements, from which the end rotations are then derived.

As an alternative,agraphicalmethod of solution, using Charac-teristic Points, is given, wherein the end rotations are obtained directlywithout reference to the end moments in the spans.Examples are given forcases of Influence Lines for Moment at asupport, or within a span, Reaction at a support, or Shear Force withina span.The Paper is accompanied by 8 diagrams, appertaining to thesolutions for a particular Continuous Bcam of four sparis, having spansof difkrent lengths and stiffncsses, and one end built in.

PAPERThe presentcurrentmethods of determiningInfluence Lines forBending Moment, Shear Force, Support Reactions, etc., for ContinuousBeams of more than two simply supportedpans, involve eitherlaborious plotting from numerousdiagrams drawn for a number ofdispositions of theunit point oad,or a laboratorymethod, which

whilst simple in theory, is not easy to carry out accurately in practice,at short notice, except in a laboratory.The methodhere ubmitted needs one hortcalculation to getthe whole Influence Line for a support reaction or a support moment.The Influence Line for moment or shear force at a point in a span isdetermined from notmore than two support moments, or reactions


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Influence Lines f o r Cont inuous Beam8 527at all supports to one ide of the point in question, the former in generalbeing the more convenient.

In generalprinciple, hemethod embodies the idea used in helaboratory method, wherein unit otation applied to the two endsof a beam broken at a point will give the influence line for moment atthat point, aqd unitvertical movement of a support will give theinfluence line for reaction at that support, but replaces experiment bycalculation.The application of the above unitmotions at a support,has a momenteffect which can be calculated for the whole span, very simply, by anyof the tandard methods, MomentDistribution, or CharacteristicPoints being the most convenient as a rule, according toone's preferencefor arithmetical or graphical methods. The former method has beenwidely published of late, and proof of the latter very simple method, asused later in the text, has been given by the author elsewhere.l Fromthese moments, the deflection curve for the beam is simply obtained, thisbeing the required influence line. The method of obtaining the deflection,curve from end moments is via the end rotations. These are connectedby the well known formulae, which give for an unloadedspan AB,having +ve (hogging) moments at the ends of Ma and Mb, and norelative settlement of the ends,

8 is +ve downwards from left to right.Where here is no relativedeflection of or applied moment at a support, Ob followsfrom 8a asb ( 2 M a +M b ) =- a (2Mb +Ma) . This can then be used to find &,8d, etc., in the same way. This is particularly usefulfor the MomentInfluence Lines. The deflection curve for any span follows simply fromthe known fact that if unit rotation is applied to one end of a built-inbeam, the resulting curve is the influence line for moment at that point,whence it follows that the deflection of any point K in a span AB whereAK =a, KB =b, and AB =L, will be & X b2a/L due to slope at A,and & X - 2b/L due to slope at B. If the spanL is divided into say s ixequal parts, the deflections due to 8a at points 1/6, 1/3, 1/2, 2/3, and 5/6 Lfrom A will thus beL .Ba (25, 32, 27, 16, and 5)/216. The resulting curveis as Fig. 1. Since the above figures are standard and apply for all spans,they can be worked out once for all at whatever intervals are deemedconvenient. Where a support has been deflected, this effect must beadded to give the complete deflection curve, as, is done in the examplebelow on a support reaction (Example 1 ) . In such cases a discontinuitywill be noticed in the slopes on either side of a support, as worked outfrom the moments alone.After this preamble, themethodcan be mosteasily explained bycarrying out the working of several examples, on a given beam. The

7. W. H. King, "Bending Moments in Continuous Beams," Conmete and ConstluctwnaiEnginckring, vol. 34, p . 265 (MY,939).


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528 T h e Structural E n g i n e e r , November 1945chosen beam ABCDE, is freely supported at A, B, C, and D, and builtin at E. Details of the spans are as follows :

Span AB Length L, Stiffness K,Span BC Length L, =3L, Stiffness K, =2K,/3Span CD Length L, =3L, Stiffness K, =2K,/3Span DE Length L, =2L, Stiffness K , = 3K,/4

Moment Distribution willbeused or kxamples to ind Reactions atA and B ( R a and Rb), and Moments at B and C (Mb and MC), ndthese results will be used to obtain the influence lines for shear force ata point P in span AB and a point Q in span BC, and also the influenceiine for moment at Q. The Characteristic Points method will then beused for R b and Mb, the construction being given in detail.Examble I . To F i n d the Influence Li ne for R h .

For convenience,as is usual with the Moment Distribution Method, themoment convention will be altered to one calling all clockwise moments-he. Using this convention we then get ea =(2M - Mb)/6K,and Ob =(2Mb - Ma)/6K1, etc., instead of those quotedearlier.The end moments induced in spans AB and BC, due to unit upwardmovement of B, rotation being prevented at B and C, are, at A =0,at B = 3K,/L, and - K,/L,, and at C - K,/L,. The two last areequal to - 4K1/3L,. The work then proceeds as follows :


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Influence U n e e f o r Continuous Beams 329

e -

A0.000 3.000-0.883

- 228

- 027

'T- 003

0-000 1 869

-*310620

8 IC1--le333 -1.333 0.000 0.0;-0.784 - 392

431 863 862 431

-- 203 - 101 - 01- 03051 101 101 051- 024 - 012 - 12 - 24

006 012 012 006~~- 003 - 001 - 01 - 03

001 001-1.859 - 862862258- 714 -034 ~ 3 6 7so86

90-000 0 ~ 0 0 0

- 28 - 14- 17 --- 014~-- 03 -- 001

x K ,L,+ L,The various values of 8 are worked out from the moments as follows :- -859K1 - 310 2 X 1.859 ,620.Span AB. 8a = - -- : 8b =6K1. I;, - L1 CiL, L1--(- 2 x 1 4 5 9 +462j K 2.856 -714 .Span BC, O b = x L= - - . - - *6 X 2K1/3 Ll 4L1 L,

- 1.724 +1'859 0.135 '034 .ec = .4Ll ---4L1 - - etc., he fullL1list of values being given in the above table.If theordinates of the deflection curve are nowworked out forintervals of one-sixth of each span, using the coefficients given above,and referring to the slope values as th at the left hand end of each span,and O a t the right hand end, we get the following table.Defln. at 5AB/6AB/3 AB/2B/3B/6 BC/6

-667S338336675003 3 367ue to Ab-007 -002072092-077046014rom&e317248007023039o46036rom81BC/3----p______

~ _ _ _Total -217425616782912.083991

-_ L _~-_ _ _ _ _ _ _ _ _ _~~~~~

BC/2-268

5BC/6BC/3

U 167333 -500-012015013-049159---p----

DE/32DE/3a022 -013-000 -000-022013~-~~~~ 5DE/6004000

044


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530 T he Struct1tra.l E n g h e e r , Novembe r 1945The deflections at supports A, C, D, and E are 0, and a t B, 1.000.These deflections are plotted in Fig. 2, and give at once the InfluenceLine for Reaction at B. I t will be noticed that the curve is continuouswhen plotted, the effect of the deflection at B having wiped out hediscontinuity evident at B and C from a study of the 8 values alone.

Example 2. To Find the Influence Linefor MbWhat is required eventually for this influence line is the introductionof unit rotation between the ends of spans AB and BC a t B. If this isintroduced before theMoment Distribution is carried out, the resulteventually is to alter its value, so that it is not convenient to carry outthe work in this way. The method adopted is to introduce unit momentclockwise at B, the portion allocated to each span being proportionalto the distribution factor for that span, i.e., as $Kl to K, with the neces-sary carry over moment to C. This last is balanced out, and the result,as seen in the table elow is to give the effect on thebeam of unit momentat B. If the beam is now broken at B and equal moments applied tothe two portions, of such a value as to cause unit relative rotation oftheir ends at B, it is clear that these rotations will be inversely as thenewly calculated moments at B. Hence since the two moments sumto unity, we can say at once that the required value of & in span AB isthe Mb moment in span BC, and the value of & in span BC is minusthe Mb moment im pa n AB. All other values of 8 are derived from thesetwo, since the deflected curve will be continuous over all supports save B.

,B, ,c, 8%The working out is then as follows :il --.000 0.529 0.000 0.000.000 O * O O O.471 0-235_.__-- i059 .- * l 1 7 - * l 1 8 --a069--- _ _ _ _* 031 *031 e016014 a028028 a014- 007- 014 -e014 -.007_ _ .- ---* 004

l -003 -002

!0.000 I.150 -0-036-0.0360.564 0.1500.218 0.436=0.036 0.017 10.118 4 . 0 3 5.436 0-118-000 0.564

!-004 002 102 -003~ _ _ - - o02 - * 002--- 1~ _ _ _- l

_ _ _


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Inf luence Linea f o r Continuoue Beam8 531I t will be seen that the values of 8 a t B are the moments reversed,with a minus sign given to the one in span BC. The other values of 8

a r e then derived as for the following case of &.(2 X . l l S . ) - 436 * 200(2 X -436)- 118 754

ec = X ( 4 5 6 4 ) =- ,564 =,150, andso on.multiplied by L,, the length of the first span.The deflection table is thenas follows, all values requiring to be

Defln a t I AB/6 I AB/3 AB12-027

2AB/3

-081082-065055-016~---

AB/6 11 BC16 1 BC/3 1 BC12 ]2BC/3 lSBC/O

IDefln. a t DE/3E/6CD/6CD/3D/2D13D/6-___-______Due to 81

-011008-.055--.076--.070--.049--.023otalI . - ----.003--.008--.014--.016--.013ue to er. o i loo8- .052--.067--.056--.033--.010---~-~-~-_E12 a002C05009 5DE/6DE/3---- _ - __ _

-009 1 -005 I *002The curve is plotted in Fig. 3.Example 3. T o Find Inflluence Lihe for MC

This example contains no new principle, but its solution is necessaryfor finding shear force and moment influence lines when required fora point Q within BC.


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532 T h e StructuTaE Engineer, November 1945

l

Resulting deflections, which need to be multiplied by L,, as before,are tabulated below, the working not being shown.Defln. at hB/6AB/3AB/2AB/3 5AB/6 BC/6 BC/3 BC/2 2BC/3 5BC/6Total 1 - . 0 1 ~ - . 0 1 8 ~ - . 0 2 ~ - . 0 ~ ~ ~ - . 0 1 ~ ~076 1 -163 1 -232 1 -248 I - 1 82

.

These deflections are plotted in Fig. 4, which is the Influence Linefor Moment at C.Example 4. To-find heInfluence Linefor R a

This may be found directly as for Rb, but for the case where A is afree support, and Mb has already been found, it is evident that R a is=[l-z/LJ - Mb/L,, where z is the distance of the unit load from A,and the terms in hesquare bracket are ignored after he first span.The Mb Influence Line can therefore be modified as in Fig. 3, by takingFBE as the baseline.Example 5 . T o i n d the Influence Line for SF a t P. a point in span AB

Since this is equal to 1 - a when the load is to the left of P, and-Ra when the load is to the right of P, Fig. 3 can again be modified b,-taking as baseline AGHBE.

A E


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ltcjluence Linea for Continuoue B e a m s 533Example 6. Tof i n d the Influence Line f o r SF a t Q , point in span BC

This can be found from the reactions Ra and Rb, s 1- Ra+Rb),for loads to the left of Q, and - R a +Rb) or loads to the right of Q.In general, for interior spans, it is more convenient to derive it from theend moments and the free span SF influence line. For this case it will bethe sum of (MC - Mb)/L,, and the free influence line. If this last is addedgraphically, by reversing it and treating it as the baseline,we get adiagram as inFig. 5 , which clearly needs very little modification to servefbr any. other point in span BC. The ordinates of the moment part oft h e curve, obtained by taking one-third of the ordinate difference M C--Mh, since L2=3L,, are as follows, for the same points as have been takenpreviously, it being unnecessary to multiply by L, in this case.

Example 7 . Tof i n d the InfZuence'Line o r Moment at QFor a point Q in span BC, dividing it into portions of length "a "and "b ", he moment is given by (a/L, X MC) +(b/L, X Mb),+ hemoment at Q considering BC as a simply supported span, and the influ-ence line for Mg will clearly be derived from those for Mb, MC,and Mg

on a free span. The case plotted in Fig. 6 is for a point such that "a "= 2/3.L,, and the free influence line has again been added graphically,although the only advantage in so doing in this case is a slight saving inaddition. The ordinates for the support moment part of the curve willbe found to be L, X the following figures for the same points as before :

whilst the free influence line is a triangle, having its apex above point Qat a height of 2/9.L, =2/3.L,, when it has been inverted so as to servethe baseline in span BC.We will now turn our attention to the graphical method of findingthe values of 8,which is done directly, without first finding the supportmoments.


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534 The Structural E n g i n e e r , Novembe r 1945Graphical Solution f o r 6 by Characteristic Points

This method is rapid, accurate enough for most purposes, and hasthe advantage hat he values of 8 can be derived directly from thediagram, since the slope at a support is given by the interval betweenthe adjacent Characteristic Point, and the Restraining Moment Diagram(such as plrl, p2 r2 ,etc., in Figs. 7 and S) multiplied by 1/2K for theparticular span. Since the slope required is not he true one, but thevalue of 8 relative to the deflected support line in some cases, the valueis found by using the distancepm (with various suffixes, etc.), insteadof the usual pc. I t is necessary-to multiply these values by -1 a t theleft hand end of a span in order to obtain the right sign for 6. Twoexamples only will be given, for Mb and Rb.Construction for 6 values f o r Mb. (See Fig .7 .)First draw a line ABCDE, marked off so that AB represents l /Kl ,BC 1/K2, etc. Trisect each bay in points m, m, m2 m2, etc. At m,and m2 et off perpendiculars m, rl and m2r2= -2K1K2/ (K, +K,) ;rl and r, are the Characteristic Points for support B, the m pointsbeing all others.Starting from a known point on the Restraining Moment Diagram,which will be the final baseline, viz., point A, join to rl, and produceto cut the perpendicular at B in Q. Join Qr2. Join r, r2, cutting BQin k. Mark off on this line a distance r,i =rlk. *JoinAi, and produce


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I n f l uence L in es f o r Cont inuous Beams 535to cut Qrz in S, , which will sthen be a known point on the baseline inspan BC. Proceed from this point to obtain a known point in span CD,viz., Ss, and thence to S,, in span DE. There will now be two knownpoints in span DE, viz., S, and m,', which willdefine the completebaseline de for this span. d andS3 define dc, and so on to A . For rotationsat supports further in than the second, or where the intersection for theS points is bad, the construction should be modified as follows, where astart has been made from he,known point m4' in span ED. The charac-teristic points at D are m, and me', h being the point correspondingto i. Draw any line m,' R cutting the perpendicular t m, in t, and thatat D in R. Join t,h and produce to cut the perpendicular a t ms' in t3'.*JoinRts'. S,' the required known point in span CD is where this inter-sects m,'h. The Characteristic Point heights are set as-2K1K,/(K, +K,), originally, since by so doing the initial set rotations at B are, forspans AB and BC respectively; &/(K, +K,) and -KJ(K, +KZ),summing unity, and being inversely as the span stiffnesses. The subse-quent construction maintains this discontinuity of unity, so that the finalvalues of pm/2K with the sign adjusted for values at the left hand end ofthe span, will be those required for 8. For the case of a built-in end sup-port, such as E, the value given to the end characteristic point heightoriginally is -2 X K for that span. A check on Fig. 7 will indicate thatthe values of 8 derived from the pm heights are sensibly the same asthose obtained in Example 2.Construction for 8 valuesfor Rb. (See Fig. 8 )point heights at m, m,' m, and m,' as follows :In this case, unit upward displacement of B will give characteristicm,r, =-ml'rl =2KJL, :m2r2= -m2'r2' =-2KdL, =-2K1/9L1'

The construction then follows the lines of that for Mb, and a check o nthe values of pm/K will indicate that they arematerially the same as the8 values found in Example1.In conclusion, the author would emphasize that any currentmethodsof analysis may be used to find the 8 values, and that the advantagesclaimed for themethod just enumerated are simplicity and rapidity,using only knowledge which is fundamental to anyone who has mademuch study of Theory of Structures, and requiring no rigging up ofmodels, however simple. I t is hoped that readers may find it useful.

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Influence Lines for Continuous Beams. a Direct Method

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