unit-6 theorem of three moments

7/28/2019 Unit-6 Theorem of Three Moments

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UNIT 6 THEOREM OF THREE MOMENTSStructure

6.1 IntroductionOb j d v e s

6.2 Deriva tion of Three Mom ents Equation6.3 Application of E quation for Exterior Fixed End6.4 Sinking of Su pport6.5 Summary6.6 Key Words6.7 Answers to SAQ s

6.1 INTRODUCTIONIn Unit 5, the various types of indeterminate structures are described. The continuousbeam can be analysed by the method described in this unit. The theorem of threemoments was developed by the F rench engineer Clapeyron in 1 857. This equation is arelationship that exists between the bending mom ents at three adjacent supports in acontinuous member. It is particularly helpful in determining the bending moments at thesupports of statically indeterminate beams. As mom ents (any category of actions) ate theunknown s, it is one of the force methods. The prerequisite for this is to know theprinciples of statics and a knowledge of shear force and bend ing moment diagrams forstatically determinate beams. Moreover, the knowledge of moment area method or anyother method to calculate the slope and deflection will be useful in understanding thederivation of this theorem.ObjectivesAfter studying this unit, you should be able to

compute the unknown moments in the indeterminate beams subjected to- external load(s ) with sink ing of support, or- external load(s) without sinking of support,find reactions f or indeterminate h e a m and to draw sh ear force diagram, andsketch the bending moment diagram.

Sign C onventionsSagging bending mom ent is considered positive. In the final bending m omentdiagram (BM D), the ordinates are drawn oil that side of heam where tei~sion dueto bending) is developed. Th us, in the case of a simply-supported beam positiveordinates are drawn downw ards. For the case of hogging (negative) bendingmom ent, the lenrio n is developed at the top fibres of the ba r n . Hence, tileordinates are drawn upwards (neg ative sign).

6.2.DERIVATION OF THREE MOMENTS EQUATION .This is applicable to any two adjoining spans of a continuous beam and can be repeatedlyused for other two adjoining spans. For n numbers of spans (n- ) sin~ultaneousequations may be w ritten which give solu tio~ i f redundant mo ments at the supports alongwith the conditions of statics.Consider two adjoining spans AB and BC of a continuous beam ABC as shown inFigure 6.1 (a). Span AB and BC are L1 and L2 respectively. Mom ents of inertia for AB andBC are 1, and I2 respectively.Figure 6.1 (b) shows bending m oment diagram due to external loads as well as due tounknown m oments at supports. Sagg mg mom ent is considered positive. The lower part ofbendine m om ent d i ae ra~ nn F i u re 6.1 is due to unknown t iloments


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Hcre, MA ,MB ,M(- are assunled as hogging (neg ative). The upp er pait shows BMdiagram due lo loads Tor span AB and BC . Thus, these diagrams are drawn consideringeach heam as'separarc sinlply supp orted heam subjected to given external loads.

C 1 ,l a )

Figure 6.1Let Al an d A2 be area of bending moment diagram due to loads on siniply supportedspans AB and B C respectively. XI and X2 are distances of their centres of gravity from Aand C respectively. It may be noted Ulat fo r spa n AB -BC, B is the intermediate supportwhereas A and C are left and right supports respectively with respect to B.In order to solve for unknown m oments, we need additional equation. It can be seen fromFigure 6.1 ( c) Lhat slope at B whether left or right of it, is sam e, i.e.

ORA = OBC (6.1)The slopes at B for span BA and BC respectively are same due to continuity at B . This isalso known as a compatibility condition.As slopes are very small, we can write

where ~ A Bs vertical dev iation of A with respect to tangent at B, andt c ~s vertical deviation o f C with respect to tangent at B.

Here, ~ A B nd tea can be obtaiiled by using moment area method. From m oment areaM Mtheorem, ~ A B s moment of- rea between A and B about A an d t c ~s moment of-l Elarea between B an d C about C.Thu s, here as given in Figure 6.1 (c),

M .ta u = Z (Area of Qagram between A and B) x X IIandHere, t A ~s upward (+ve) and t r ~s down ward (-ve) with respect to tangent at B.


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~ A B ~ C BFor 0~~ = OBC, we have - --L1 L2'On simplifying,

Theorem of Three(6.3) Moments

This is the equation of three moments. We observe that MA ,Mu and Mc are threeunknown mom ents. The terms on right hand side are obtained from bending m omentdiagram considering simply supported spans subjected to external loads. When there aren numbers of unknown mom ents for multispan continuous beams, n number ofcompatibility conditions are required and the continuity of slope at various supports isconsidered.If a bean1 has sam e a o s s section in all the spans, i.e. for prismatic beam, 1, = 12 = I . Thethree moment equation simplifies to

Example 6.1Analyse continuous beam ABC given in Figure 6.2 (a). Find the reactions anddraw shear force and bending mo ment diagrams.

Figure 6.2


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Indetenniwte SolutionStludWC~1 The equation of three moments, [Eq. (6.4)] is as follows :

M ~ L l- - 6A1XI 6A2X2- -- --lI l hI21Here, in this case, since A and C are simple supports, moments MA= MC = 0,L 1 = 6 m , h = 8 m , I I = I and 12=21 .For simply supported beam AB-subjected to uniform aly distributed load (w ),

w Lbe ing moment diagram is a parab ola and malrimum BM s -. For span AB,ld 8AI is area of parabolic BM iagram and X I is distance of C .G. of BM iagramfrom left su pport A.12

Lland XI = - s m easured from A.2Hence,

w ( ~ i ) \ x 6 ~ 5 x 2 1 6 - 2 7 0LI 4 4 - 4For span BC, At is area of parabolic BM iagram and X2 is distanc e of C.G. ofBM iagram from right support C.

6A2x2 w (~ 2) ' 8 x 83 - 1024- 4 -L2 4

Thus, the three moment equation for spans AB -BC is as follows :

We get, 20MB = - 782: MB =- 39.1 kN m (hogging)

e

To find reactions, the reactions due ta load [Figu re 6.2 (b)] and those due to endmoment MB [Figure 6.2 (c )] are superimpose d. Thus, we get,RA= 15 - 6.517 = 8.483 kNRBA= 15 + 6.517 = 21.517 kNRBc= 32 + 4.888 = 36.888 kNRB' BA+ RBC= 58.4045 kNRc= 32-4.888 = 27.112 kNcan be stated as algebraic su m of vertical

(8 x 8) + (5 x 6) - 8.483 + 21.517 + 36.888 +27.1 12)= 0 (OK)The shear force and bending mom ent diagrains are shown in Figure 6.2 (d) and6.2 (e)respectively. The bending m oment at B is hogging. Therefore, tension is attop and hence, negative BM iagram is drawn at top and the positive BMdiagram considering simply supported spans AB and CD is superimposed. Theresulting bending mom ent diagram is redrawn in Figu re 6.2 ( f) on a horizontalabscissa.Maximum positive bending momeilt will w cu r w here SF is equal to 0.


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From SF diagram taking similar triangles,

Shear is zero in spanAB atxl fromA.

Similarly,x2 = 3.389 m in span BCHence,Mmdc = 32x2- x2- 4 . 8 8 8 ~ ~45.94 kNm (sagging)

SAQ 1Analyse the corilinuous beam shown in Figure 6.3.Also draw bending momentdiagram.

Figure6.3

Example 6.2If aqy span is subjected to

(a) point load as shown in Figure 6.4 (a), and(b) partial u.d.1. as shown in Figure 6.5 (a),

w1 6AX2 if the span is on left or on right with respectalculate the terms- and- Lto intermediate support.Note : This example is aimed to W e ou understand the computation ofterms on right ha d side of three moment equation.This depends ontypes of loads and their disposition.

Theorem of ~h re eMorncntv


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Solution(a) Figure 6.4 (b) shows the bending moment diagram considering simplysupported beam carrying concentrated p i n t load P, the area

PabA = [+Ix?Ix=Distance of C.G. f area from left end = Xi (L +a)BDistance of C.G. f area from right end =X2 = (L +b)B

MIThus, - -WELA romA.L L- .Similarly,Note :When point load is at centre of s@, we get

(b) Figure 6.5 (b) shows bending moment diagram of the simply supported beamloaded as shown in Figure 6.5 (a).

Due .toreac t ~ o n

To find-X1 for parabolic potti&, i.e. between C and D,he expression faLbending moment between x =4 to 6 from A is as follows :The moment of area of bending moment under parabola between C and Dwith respect toA is


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(-Insubstituting the values, we get6

Similarly, the expression for bending moment between x = 2 to 4 from B

Momen t of area under parabola (between D to C with respect to B

AX I for left triangle with respect to left support = lG.?!l (y)AX2 for left triangle with respect to right suppo rt ="r" (4'4)2AX I for right triangle with respect to left support = 2AX2 for right triangle with respect to right support = 2 \AXI due to partial u.d.1. w ith respect to left support,

-(6+$) = 1 5 2 0 k ~ m 'Therefore,%L for partial u.d.1. w ith respect to le ft supp art is

Similarly,AX2 du e to partial u.d.1. with respect to right support.- ( 72 x4 ) 4 + - + 4 4 8 +2-( :]

Therefore,6AX2

L- or partial u.d.1. with respect to right sup port is1296= 6x- = 9 7 2 k ~ m '8

Theorem of Thm cMoments


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A lte rn ate ~ e t h o d ( M e t h o d f Superposition)Consider Figure 6.5 (a) and'assume end B as tixed, RA= 18 kN is deemed tobe a load, revised bending mo ment diagram w ill be as per F igure 6.5 (c).

AX2 with respect to right end = moment of area about B

Similarly,AXl with respect to left end =moment of area about A

which are the same as those by integration method.SAQ 2

An intermediate span PQ is subjected to(a) unit'ornlly distributed load w = 30 kN/m as shown in Figure 6.6 (a) , and(b) external m oment M as shown in Figure 6.6 (b).

6AX1 6AX2Calculate-and-with respect to left suppo rt and right support.

Figure 6.6

SAQ 3A~ialysche co~~t i l luouseam shown in Figure 6.7 and draw bending mom ent 'and

- . shear force cliagr;uns

Figure 6.7


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6.3 APPLICATION O F EQUATION FOR EXTERIORFIXED ENDTheorem orThree

Moment3

For a continuous beam with fixed ends, end moment is developed at the fixed support.The rotation at fixed end does not take place and the s u p w ~ t oment is required to becalculated. In order to a nalyse such a beam by the theorem of three moments, anadditional equation is required. In such a case whenever the exterior support is fixed, anadditional imaginary sp an of z ero length without any load is assumed.Example 6.3

hla lyse the propped cantilever beani given in Figure 6.8 (a ) and draw bending~ n o n i e i ~ tiagrnin. Take EI as constant.

SolutionConsidering equilibrium of AB and of joint B, equivalent bending momen t -PL(hoi ging ) and vertical load P can be taken at roller B as shown in Figure 6.8 (b).

MB = -PL (hogging)Considering spans BC- CC' with no load on span BC and CC', the threemoment equation is as follows :MB x 2 L + 2 M c ( 2 L + 0 ) + 0 = 0

-MBC)n simplification, this gives M c =-Thus, we get MC = (sagging)The bending mom ent diagram is shown in Figure 6.8 (c).

SAQ 4 A i ~ x c t l e am ot cp;lli L I S subjected to ccccnrnc pcunt load a\ \ I I ( \W I I 111Flgurc (1.9.Calculate tJ~e lxctl cnd nlonlentc ill,, ' u ~ dZtfiP


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SAQ 5Analyse the fixed beam with ii l ter~lal inge as shown in Figure 6.10.

Example6.4Analyse the prismatic beam given in.Figwe 6.11 (a) and draw bending momentdiagram.

SolutlonSpan BC is subjected to two types of load s, hence the method of superp osition isapplied . For simplicity, partial u.d.1 and poin t load are conside red separately.SpanAB-BC

M2X2 for point load =art A of- Pab (L +b )Lz Lpart B of-A2X2 tor partial u.d.1. about ck!

a2x2 = 972 kN m2 N f e r Example 6.2 (b)]' ~ ' ~ a r t 3 ,

Applying three moment theorem , putting MA=02 ( 6 + 8 ) M B + 8 M c = -1296-900-972 = -3168


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SpanBC-CC 'L 1 = B C = 6 m , k = C C '= OPartAof- Pal ( L+a)" X' due to point load =L1

6 41x1PartB of- L1 due to partial u.d.1. B[y= 1140 kNm2 [refer Example 6.2 (b)]Part B

641x1Total- 1260 + 1140 = 2400 kN m2LApplying moment area theorem, L1 = 8 m, Lz =0

8 M B + 1 6 M c = -2400M B + 2 M c = - 3 0 0Solution of simultaneous Eqs. (6.5) and (6.6) yields,

MB =- 82 kNm (hogging);Me =- 109 kN m (hogging)Figure 6.1 1 (b) shows the bending moment diagram (BMD).

6.4 SINKING OF SUPPORTIn the previous section, we have considered continuous beam resting on rigid supportswhich do not yield. However, if any of the supports sinks, it causes change in bendingmoments of the beam. Let us now consider a case ofcontinuous beam AB C in whichsupportsB and C sink. For continuous beamABC as shown in Figure 6.1 2 (a), supportAdoes not deflect whereas supportB sinks by P1 downward with respect toA apd C inksby (6, - 62)with respect toA. Hence, vertical deviation of intermediate supportB is 61with respect toA andhwith respect to C.This is considered as positive sign convention,Figure 6.12 (b) gives deflected shape of original beam and oonesponding bendingmoment diagram is as shown in Figure 6.12 (c).


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The compatibility condition at R is~ B A ~ B c

From Figure 6.12 (b),

On substituting the values, we get

The equation becomes

When there is sinking of support in case of prismatic beam, we have 11 = 12 = I and theequation becomes

Note :While writing signs of S1 and & we have to take care. For span AB, S1 is positive ifintermediate support B is at lower level by 6 1 han left support (A ). For span BC, & spositive, if intermediate support B is lower by 62 with resp ect to right supp ort C.Example 6.5

For a continuous beam ABC as shown in Figure 6.13, if support B sinks by 5 mrnand support C inks by 2 mm with respect to unyielded support A, find bendingmoment at B. Take E l = 36 x lo 3 kN m 2 as constant for the beam.

0 Figure6.13Solution

6 1 with respect to B =0.005 m (positive)while 62 wi& respect to B = 0.005 - 0.002 = 0.003 m (positive)Applying three moment theorem,

28MB = 182 .= Me = 6.5 kN m (sagging)


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SAQ 6Attempt Example 6.5, if supporl B sinks by 2 mrn and support C sirlks by 5 irilnwith respect to A and EI for span BC = 72 x 1 0 % ~ 2.

Example 6.6For the loaded beam sh own in F igure 6.11 of Exam ple 6.4, if B settles by 8 mm,analyse the beam. Take EI = 72 x lo3kN m 2 for all spans.

SolutionSpanAB-BC

Here, LI =AB = 6 m, & =BC = 8 m and intermediate support sinks.Therefore,61= 62 = 0.008 mApplying three moment theorem ,

For MA= 0 and refering Example 6.4, AX lL can be directly written down toform equations

where - 3168 is cartied forward from Example 6.4; refer Eq. (6.5)..'. 7MB+ ?MC = - 540SpanBC-CC '

B is exterior support and L1= 8 m6EZ6S1 = -0.008m and- - 4 3 2 k N mL1

where - 2400 is carried forward from Example 6.4; refer Eq. 6.6)MB+ 2Mc = - 354 (6.8)

On solving Eqs. (6.7) and (6.8), we getMB =- 31kN m (hogging)M c =- 161.5 kN m (hogging)

6.5 SUMMARY

Theoreni of ThreeMoments

In this unit, you have learnt the three moment equatlon which gives relationshipbetween three bending m omen ts at three consecutive supports in two adjoining spans of acontinuous beam. In this method, the u n ho w n s are redundant moments at the supports.1 1 e number of unknow ns n is equal to the degree of static indeterminacy. Here, ncomp atibility equations are obtained by writing three mclment equations for each grou p ofadjacent spans one by one.


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In this unit, sagging bending moment is considered positive.Three moment equation for continuous beam as follows :The generalised equation for continu ous beam when su bjected to external loadand support settlement is given by,

When prismatic beam is subjected to external load and there is no supportsettlement, substitute II = 12 = I, then last two terms on R.H.S in the aboveequation are om itted.When supports are yielding and there is no ex ternal load in prismatic beam,II =12 = I then first two terms on R.H.S. in the above equation are omitted.Terms 6AXlIL with respect to left end and 6AX21Lwith respect to right end fortypical types of loads are shown in Table 6.1, treating the beam member assimply supported.

Table 6.1 :Values of the Quantity =for some Important CasesLr - I I

6 AX1 Pab ( L+ a )- - - -L L (hogging)6AXz Pub ( L + bl- -lab

- -L L (hogging)

To w rite three mom ent equation in case of bea ms with fixed support,additional span of zero length without load is assumed.IndeterIhinate beams inc lude- Continuous beams,- Propped cantilevers (w ith or w ithout overhangs),- Fixed beams, and- Fixed Beam with internal hinges.

6.6 KEY WORDSEnd Moments : The mo ments at the ends of m embe rs are called endmoments.Prismatic Beam : Beam having uniform cross section is called prismaticbeam.Continuous Beam : Beams having more than one span and continuous overnumber of suppo rts are called continuou s beams.ThreeMoment Equation : Compatibility condition involving bending moments atthree successive supports is called three moment

equation.


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Yielding of Supports : Vertical settleme nt or rotational slip of the supp orts due toprobable displacementof the footings is defined asyielding of su pports. This iliduces fixed end actions at theends. Settlem ent is also known as sinking of the support.6.7 ANSWERS TO SAQsSAQ 1

The loads and reactions are shown in Figure 6.14 (a). Here,MA and MD arezero and Me and Mc are unknown moments. Thus, two equations are required.Consider spans AB-BC for three mom ent equation.Another equation is formed by conside ring spans BC-CD for three momentequation. Here, L1= 8 m and L2 = 4 rtl and- s zero as span CD is not

L2loaded.The solu tion of these two linear sim ultanec~ us quation s gives the unknow ns MBand Mc as follows :

Shear force and bending mom ent diagrams are given in Figures 6.14 (b) and6.14 (c) respectively.

6 M 1 M [a" 2ubz - b3]b) -- -L L~SAQ 3

ME= 26.4 kN m, Mc =8.3 kN m. BM diagram is given in Figure 6.15.


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SAQ 4 Consid erin g span A 'A-AB and AB-BB ' nd using results of Exanlple 6.2 (a),following equations are obtained :

Pub (L + a )MA+ 2MB = - LOn solving simultaneous equations, we get

purl2MA = - - and MB = -- b t h hogging)L L"SAQ 5

Refer Figure 6.10. Figure 6.16 (a) shows the equivalent beam to he analysedconsitler he;un I)C-CO, and ohlain r1lolllellt.s using Lllrec momenl ll~corcm.

Figure 6.16Mc-= - 3 0 k N m The reaction can be found o ut by superposition. ,RE = 4 0 - 7.5 = 32.5 kN AB is cantilever beamMA= 32.5 x 2 = - 65 kN In (hogging)B.M diagram is shown in Figure 6.16 (b).SAQ 6Here, 6 , = 0.002 m upward with respecl lo B), and

6' = 0.002 - 0.005 = - 0.003 (downward with respect to B).The same procedure as given in Example 6.5 is followed and following value isobtained :

ME = 0.75 kN in (sagging)

unit-6 theorem of three moments

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