structural anaylsis

7/28/2019 structural anaylsis

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UNIT 9 KAN19SMETHODStructure

9.1 IntroductionObjectives

9.2 ApplicationY .2.1 Analysis ofFrame.-,9.2.2 Sway Correction9.2.3 Analysis forVertical Lmds. 9.2.4 Analysis for Horimntal L d s

9.3 Examples9.4 Summary9.5 Key Words9.6 Answers to SAQ s

9.1 INTRODUCTIONKani's Method was derived by Dr. Kani. This method is suitable to work out appro ximatemoments for the whole fram e, beam or statically indeterminate structure, due to dead load,live load or wind load.When a structure, whethe r a beam, a rigid frame or a truss, is statically indeterminate, i.e.the force response cannot be determined by the laws of statics alone, some of the unknownreactions or member forces, equal in number to the degree of indeterminacy, can beregarded as unknown forces acting on a partly determinate structure.Their magnitudes canbe obtained. at the very beginning, from the conditions of consistent deformation. Inestablishing them, the conditions of geometry, q m b e r properties such as modulus ofelasticity, stiffness constant etc. are required. Compatibility and adaptability to recentsoftware packages are the major advan tages of this method.ObjectivesAfter studying this unit, you should be able to

analyse the framed structure whether determinate or indeterminate,work out the mom ents for various frame work, considering differentcategories of loads, i.e. dead load, live load, wind load etc., andana ly~eramed structures having unequal spans or storeys.

9.2 APPLICATION9.2.1 Analysis of FramekIn thismeth od, for various members, first suitable stiffnesses are assumed. Moments due todead load, live load'and wind loads, are worked out fo r the whole frame.me sections aredesigned for the bending moment, shear force and axial fi#ce. If there is a markeddifference between actual values of stiffness and the assumed values, modified values areassumed and the analysis is done again.Let us consider member AB of a frame given in Figure 9.1 where is the slope at A and8 is the slope at B. Considering AB as a fixed beam, moment at A will consist of fixedend momentM, nd moment caused by rotation 8, at A and rotation 0, at B.ReferringFigure 9.1, we get -Moment at A caused by rotation 8, will be as follows :

= 2 x Ek x (28,)

where k is the stiffnessoT member AB and m~ =21WA


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Moment at A caused by rotation 8, at B will be= 2BiOB= mBA

Hence, total moment for ABMM =M,+ 2m, + mBA

But, sum of the moments at a joint should be zero.Hence,

At ioint A, moments of various memberswill be proportional to their stiffnesses.

where, k = is the stiffness of AB and is the sum of stiffnesses of all members1meeting at A. Further,- is called the rototionfocrorof AB at A. This is expressed as

2 c kyAB' For each member at each joint, the rotation factorsakworked out. The unbalancedmoment at a ioint is the sum of the fixed end moments calculated and written inside thecircle at each oint.Hence, unbalanced moment at A,MA = M,

>Considering the first joint and assuming that, at any other joint, there is no otation,m = y MA and m = y MAwhereMA isunbalanced moment a t joint A.Coming next to joint B, rotation effect of end A is considered.Thus, meA= YBA( m ~

\For all other joints, similar procedure is adopted and the fm t approximation is completed.Again for joint A, the rotational effect at B and E will be considered to find themodifiedvalue of mm This procedure is followed for the other joints also and the second cycle iscompleQd. The moments can be found by repeating a few cycles.9.2.2 Sway CorrectionHorizontalshear will be produced in each column due to moments in the top and bottomcolumns. The horizontal forces acting on the frame above each storey shatl be equal to thetotal shear in that storey.9.2.3 Analysis for Vertical LoadsIf only vertical loads are acting, the total shear in each storey should be zero. If the sum ofthe moments at the top and the bottom of all columns in the same storey is not zero, for the


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\1

same he~ gh t f colunms, the moment is balanced in propportions of the stiffnesses of the Kani's Methodcolum1s.For a column AB, if S e the displacement of the frame, the moment m', caused by thisdisplacement will be,

6EI6 C ~ A B- - - - - Im ' ~ h2 h where C= 6E6 and stiffness kAE=;

I

EPprurc9.2

If h is same for all columns, CtdAB = C,kABwhere C, = -hWhile ?rrtM is the moment due to displacement at A, total moment at A for column AB willbe,

,MAB= + 2% + ?tlBA+mrMand MBA= + n l A ~ 2 ~ 1 ~mfMFor vertical loads,

- - = 0 for columns.To td nioment for all columns should be zero.Hence, C(MAB +ME,,) 0

=> (2mAB+mm +mkE+m, + 2mE,,+mtM) = O=> 3C (m, + m,) = - 2 C mtAB=> 3Cmt,= --2 C m, +%#A)~ A BBut (m',) = -. m', where, stiffness of column AB is kM and sum of stiffhessof a l lC kcolumns isD.

Hence, *m'M = ---~ (mu + rnm): [-2Ck kM is known asdisplacementtconsidered.Sway correction is applied only after the completion of the first cycle of distribution. Afterthat, seuxidXqcle is started. In the second cycle, effect of displacement contribution isNote : In the above derivation, heightsof all columns are assumed to be same. If thecolumns have unequal heights, the above formula will have to be -ed..9.2.4 'Analysisfor Horizontal LoadsIf P is the horizontal force above the storeyefor which horizontal shear is being considered,

~ ( M M MBA)h + P = 0 for equal heights of columns. 7


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Hence. z > ) I , +m g, +mgM+hBAmm) + ~h = o=> z m b+nlgBA) -Ph -3z (nrM+ nlBA)AS. ??I,, = mgBA,

PhHere. -5- is called storey moment denoted by M.In this case, analysis is exactly the sam e as in the first case, but only ef fect of storeyPhmoment -is taken into consideration when applying sway correction.3CdlpnnswithUnequalHelghts

Let us consider a frame having columns of unequal heights (Figure 9.3). If P is thewind force acting upto the bottom of the storey a d M is thehorizon& force at thebase of column AB,

Likewise,

But, total shear in a storey must be zero.fience, +P =0where. is total horizontal shear in all columns.

Now, or any reference height hRtaken as the maximum height of column,


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MC D +M~~ i l ~ M~~+= X - = Ccn and so onh, hc, h,h ~ .where, C, =- ccD=*,nd so on.~ A B h w

But ~ + P = o

Now, the moment produced at the end of the member due to sway,6EIAB6 k~~~ =- c . , (where C = 6E6)~AB ' ~ A B

C .h2,=> m - - x k A B x -A B K - hR h h ~

Summing for all members, we get

It givesSubstitutingEq. 9.3) n Eq. (9.2), we have

SubstitutingEy. 9.1) in Eq. (9.4), we have

3 ~ABCABThus, isplacement factor 6, = -- x C(kABcm2)PARand storey moment =-'If no horizontal force is acting, we get

If heights of all the columns are same, i.e. C,, = 1,

Kani'sMethod


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9.3 EXAMPLESExample 9.1

Analyse the frame shown in Figure 9.4 by Kani's method.

Elgore 9.4Solution

Fixed end moments

Rotation contributions at different joints are as follows :At Joint B,

1 2U6Ym = -1 + 3V6)1 = - 0.2

1 3V6Yar = -? [(2UL) )+h~6)]- 0.3At joint C,

.1*Ice = - -x (3V6)2 [(3V6) (m)l= - 0.31YcD = +-x (ID)2 [(3V6) (m)1 - 0.2

Considering the reference height hRas 6m, we get

Displacement factors,

In Figure 9.5, the distribution process is shown. At each joint, two concentric circlesare drawn. The fixed end moments are written above the beams, outside the circles.The ro tation factors are written in the space between inner circles and outer circles,towards each side of the member. Rotation contributions are written below the line


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for horizontal member and towards left side of vertical line for vertical members.Unbalanced moment at a joint is written in the inner circle. Displacementcontribution for each column are written horizontally at centre of column.

The process is started at Joint B. At first the rotation contribution of end C isassumed as zero. The rotation contribution of end A is zero as it is fixed en d Thus,the unbalanced moment of -3 is distributed for BA and BC in proportion of rotationfactors- .2 and - 0.3 respectively. This gives, m,, =+ 0.6 and mBc=+0.9. At jointC, net distribution is carried out. Rotation contribution of end D is zero as it is a fixedend and rotation contribution of end B is now t e e n as+0.9. The unbalancedmoment of (+ 3 + 0.9)=+ 3.9 is distributed for CB and CD proportionally with therotation factors - 0.3 and - .2 giving mm =- 1.17,mcD=- 0.78.Thus, the firstcycle is completed.In the second cycle, at joint B, rotation contribution of end A is zero and that of C isnow taken as - 1.17.Unbalanced moment of (- 3- 1.17)=- 4.17 is distributed forBA and BC, proportionally with the rotation factors. Their cycles are completed inthis way and then sway correction is applied. The disttibution is shown in Table 9.1and Table 9.2.

Table9.1

After third cycle,(m, +m,)C, = (0.855+0 ) x 1+ (- 0.856 +0 ) x 2 = - .857

m', = - 0.3 x (- 0.857) = +0.257

-'a Method


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A:. "12- i:! :FL ccmnns, these moments are written honzonlally.Now, the fourth cycle is started at B. Moment to be distributed will be sum ofunbalanced fixed end moment, -3, rotation contribution of C, -1.285, anddisplacement contribution of column BA, +0.257. Thus, M = (- 3- 1.285+ 0.257).Thishas been taken as - 4.005 which is not correct. This is distributed at B for BCand BA proportionally with the rotation contributions.Considering the displacement contributions of the columns, fourth, fifth and sixthcycles are completed. In further distribution, the error of the fourth cycle is rectified.Distribution of rotation contributions is given in Table 9.1 and distribution ofdisplacement contributions is given in Table 9.2.

Table 9.2 : way Correction

Final moments are,MM = 2 M , +m , +m',Hence, M , = +0.825+0.342=+ 1 I67

M,, = +0.825 +0.825+0.342= + 1.992MBc = - 3 -0.238 + 1.237=- 2.001McB = +3- 0.238 - 1.475= + 1.287McD = -0.983 - 0.988+0.684=- 1.282M Dc = -0.983 +0.684 =- 0.299

Errors at joints are,M BA+MBc = 1.992- 2.001 = -0.009

Actually, the sum of moments at joints should be zero. However, the error calculatedabove is neglible.Example 9.2

A frame is shown in Figure 9.6. Analyse the same by Kani's method.Solution

The fixed end moments are all zero because the loading is on joints only,Rotation factors at joint C,

After3rd CydeX (m M + mE A)C M= 0.855 X 1+ (- 0.856)X 2=0.855- 1 .712=- 0.857

m'm =- 0.3 x (- 0.857)=+0.257

m ' c ~ -0.6 x (- 0.857)==+ .514

Similarly, at joint D,

After 5th CycleZ(mAB +~ M ) C A B= 0.818 x 1+ (- 0.976)X 2~ 0 . 8 1 8 - .952=- 1.134

m'm = - 0.3 x (- 1.134)=+ 0.34

m ' c ~ - 0.6x (- 1.134)=+0.68

After4thCydeZ ( ~ A B~ B A ) C M= 0.805 x 1+ (- 0.944)X 2=0.805 - 1.888=- 1.083

m'm = - 0.3 x (- 1.083)=+0.325

m'cn =-0.6x (- 1.083)=+0.65

After 6th CycleZ ( ~ A BmR4)CM= 0.825 X I+ (- 0.983) X 2=0.825 - 1.966=- 1.141

m'~g - 0.3 x (- 1.141)=+0.342

m'c~ - 0.6 x (- 1.141)=+0.684


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Similarly, at joint B,

Similarly, at joint E,

Displacement FactorsFor upper storey, all the vertical mem bers are of equal heights.Hence,

For lower storey. if hR= 8 m, then C, = 1, C,,= = 2(7Hence,3aAB= - - X (4IA) 1Thus, we get, =- 0.32 [(4US)< 1 + (21/4) x 231

8Storey moment for upper storey = 6 x - = 16kNm38Storey moment for upper storey = (6+ 12)x - = 48 kNm3

All fixed end mom ents are zero in this case. Hence, unbalanced m omen ts at all jointsare zero. The process is started w ith distribution of storey momen ts. In the ratio ofdisplacem ent factors, storey m omen t for upper storey is distributed in upper storeycolumns and storey moment for lower storey is distributed in lower storey columns.

Kani's Method


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At joint B, first cycle is started. Ullbalanced molllent at this joint will be Ulr sunl ordisplacement contribution of upper and lower storeys, i.e. (- 12- 14.2)=- 26.2. Inproportion of rotation contributions, this moment is distributed in members BA, BEand BC. Afterwards, at joint C, unbalanced moment will be rotation contributioil(+2.62) of joint B and displacement contribution (-12) of upper storey. The suill ofthese moments, i.e. - .38 is distributed proportionally with the rotation factors. Thisprocess is carried out at joints D and E. Now, the first cycle is completed., d

i"' IConsidering the rotation contributions of first cycle , displacement con tributions arecalculated. Then second cycle is started. This process is continued till the d ifferencein d isplacement contributions and rotation contributions is negligible.Final moments for horizontal members are,MM = 2mM +m , +M,

McD= + 8.76+ 5.1 = + 13.86kN mM , = + 8.76 + 3.66 = + 12.42 kN mMBE = -k 18.4 + 6.2 = + 24.6 kN mMEB = + 18.4 + 12.2 = + 30.6 kN mFinal moments for vertical members are, MBA= 2m, +mBA+m',M , = + 6.2 - 23.58 = - 17.38 kNmM,, = +6 .2+6 .2 -23 .58 = - 1 1 . 1 8 k N m

By carrying out one m ore cycle, better result can be obtained. The entire process isshown in Table 9.3, Table 9.4 and Figure 9.7.


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T a b l e 9.3

T a b l e 9.4 ( a ) : w a y CorrectionFor U p p e r S to re y

1stCycleJoint BM= - 14.2 - 12

=- 26.2mgc=+ 2.62r n g ~ + 5.24rrrm =+ 5.24P Point CM=-20.06- 18.8+ 1 .55 + 6.82=- 9.38m c ~ = + 3 . 1 2m c ~ = 1.56

JointDM =- 1 2 +3 .1 2=- 8.88m ~ c .+ 2.96

m c ~+ 1.48

JointEM =- 1 2 -2 8 .8+5.24 + 1.48

=- 4.08m m = + 3 . 4 1mEB = + 6.82~ E F+ 6.82

Table9.4 (b): w a y CorrectionFor Lower S t o r e y

2nd CycleJointBM = - 20.06 - 18.8

+ 1.55 + 6.82=- 30.48m gc =+ 3.05m g ~ = +.1

m a = + 6 1JointCM = - 8.8+3.05

+ 2.96=- 12.79111c.n + 4.26me-E=+2.13

JointDM = - 8.8+4.26

+3.41= - 11.24

rrlne =+ 3.7 1I I I D E=+ 1.86Joint EM = - 18.8-40.12+6.10+ 1.86= - 50.96n ~ ~ n = + S . lmEB =+ 10.2nlEF =+ 10.2

Before 1stCycleM = 6~ ' B E = - O . ~ ~ X6=,T 12 + 6.41 = 25.07 +5.10 = 28.14~ D E = - 12 ~ ' C D m'nc

=- 0.75 X 25.07 =- 0.75 X 28.14=- 18.8

After 3rd CycleM = 16+3.11+2.38+5.77

+ 1.87 = 29.13m ' c ~ ~ ' D C=- 0.75 x 29.13=-21.85

3 r d C yc le .Joint BM =- 2.35 - 21.1+2.13+10.20=-31.12 "mgc.=+3.11nlBE =+ 6.22n t h =+ 6.22Joint CM=-21.1 +3.11

+ 3.7 1= - 14.28

I I I ~ - I >=+ 4.76nrcg =+ 2.38Joint I)M=-21.1 +4.76

+ 5.10=- 11.24

rrrnc-=+ 3.75rllnE=+ 1.87Joint EM=-44.7-21.1

+ 1.87 + 6.22=- 57.71

I I I E ~+ 5.77IIlEB =+ 1 1.54111EF=+ 1 1.54

Before 1stCycleM = 48~ ' A B- .3 x 48 =- 14.4m 'm= - .6 x 48 =- 8.8

Af'ter 3rd CycleM =.48 + 6.22+ (11.54 x 2 )

= 77.3~ ' A B - 0.3 X 77.3 =- 3.2m ' p =- .6 X 77.3=- 6.4

Af ter 4th CycleM = 6+2.50+1.85+3.11

+6.02 = 29.48~ ' C D ~ ' D C

=- 0.75 X 29.48=-22.11

4thCycleJointBM =- 23.2 - 2.38-21.85+11.54=-31.13me(-=+3.11r n g ~ + 6.22m u =+ 6.22JointCM=-21.85+3.11

+ 3.75=- 14.99ntc-n=+ 5.0

I I I ~ - E+ 2.5Joint DM=-21.85+6.0

+ 5.77= - 1.03

r~lnc'= 3.691 1 1 ~ ~+ 1.85Joint EM=-46.1 -21.85

+ 1.85 + 6.22=-60.18I I I U , =+ 6.02IrlEB =+ 12.04IIlEF =+ 6.82

After 5th CycleM = 16+2.55i 1.83+3.11+6.10 = 29.59~ ' C D m ' ~ c

=- .75 X 29.59=- 22.19

After 1st CycleM = 48 +5:24+ (2x6 .82)

= 66.68~ ' A B - 0.3 x 66.88 =- 20.06m ' p =- 0.6 x 66.88 =+ 40.12

After 4th CycleM = 48 + 6.22 + (2 X 12.04)

= 78.3~ ' A B - 0.3 x 78.3 =- 23.49m%=- 0.6 x 78.3 =- 46.98

5thCycleJoint BM =- 23.5 - 22.1 1+2.5+12.04=-31.06m ~ c = + 3 . 1 1BE=+ 6.2m m = +6 .2 -JointCM=-22.11 +3.11

+3.69=- 15.31m c ~ = + S . lm c ~+ 2.55

JointDM=-22.11+5.10+ 6.02

=- 10.99nine.=+3.66n l n ~+ 1.83JointEM=-46.98-22 .1

+ 1.85 +6.20=+ 61.04

I ? I E ~+ 6.1film =+ 12.20n l f f=+ 12.20I

-After 2nd Cycle

M = 48+6.10+ ( 2 ~0.20)= 74.5m ' ~ g - .3 x 74.5 =- 2.35m ' p = - 0.6 x 74.5 =- 4.7

After 5th CycleM = 43 + 6.20 + ( 2 x 12.20)

= 78.60rnu=- .3 x 78.60 =- 3.58 'm ' a = - 0.6 x 78.60= - 7.16


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h d&d d e SAQ 1strpcbme-II . A continuousbeam is shown in Figure 9.8. Analyse the same by Kani' Method.

SAQ 2For vertical loads, analyse the frame shown in Figure 9.9. The live load and deadload areassumed as 5kN/m2 and 3.5 kN11n~espectively. Spacing of frames is4 m.

9.4 SUMMARYIn this unit,you have 'beenntroducedto the Kani's Method for analysis of frames.Youhave 'been given guidelines to work out moments for various types of ftames, suchasframeswith unequal heights, multi-storey frame works andcorresponding diffezntmoments, such as moments resulting due to sway, horizontal forces, vertical forces anddead loads.9.5 m Y ORDSRotationFactor : Rotation factor of a member is the negative ratio of

stiffness of all individual,members, to the sumof stiffnessof all the members connected to that joint, multiplied by0.5.Dlsplacement Factor : 'Ihis actor comes intoplay when the sum of moments attop and bottomof a l l the columns in the storey is not zero.'Ihe value of this factor differsaccording ta the type of theproblem. To sum up, here are the cases :


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(a) Assume the vertical members to be AB and CD ofsame height.

3DFcn = -- C (k, + kc,)(b) Consider the vertical members to be AB and CD of

different heights and AB >CD; D to be hinged,

MA, =11, Me, = 1 x 0.75When the joint is hinged, i.e. D in this case, McD ismultiplied by 0.75 or else it remains to be 1.Heightis multiplied by 1.5.Hence, "RCc = (he, x 1.5)where, h, = Reference height.

Storey Moment : This takes place in case of analysis of horizontal forceabove the storey . Its value is found out by drawing freebody diagram of the column in which the force acts andthen finding out the force 'P' y applying = 0

PhHence, storey moment = -_*l3Then, this storey moment is considered into the cycle.

9.6 ANSWERS TO SAOsSAQ 1

% = - iLEG2 =12G = -= 2 + 1 2 m m- ( 4 x 3 2 = -+2.35 +0.75) - 3 W rnM~~ - - (4 x 4) (4 x 4)M,, = + 3 3 m- ( 4 = - 2 m mG - - 8- = + + m mM ~ ~ =8

Kani's Method


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1 kRotation Factor,y = -- -kI For Joint B,

For Joint C,

The process of distribution is shown in Figure 9.10. Now, the fixed end moments androtation factors are calculated. At each joint, two concentric circles are drawn. Thefixed end moments have been written outside the circles, above the beams. Rotationcontributions have been written below below the beams. Rotation factors have beenwritten in between the inner circles and the outer circles, towards each side of themember. Unbalanced moment at any joint have been written in the innefcircle. Theprocess is started from joint B. At the begining, for joint B, rotation contributions ofend C is assumed as zero. The rotation contribution of end A is also zero as it is afixed end. Thus, the unbalanced moment of + 9 at joint B is distributed for BA andBC proportionally with the rotation factors- .2 and- .3 respectively.Hence, m , = - 1.8 and mBc = - .7At joint C, next distribution is carried out. Rotation contribution of D is assumed aszero and that of end B isknown as- 2.7. The unbalanced moment of(- 2.7 + 1 ) =- 1.7 is distributed for CB and CD proportionally with the rotationfactors- 0.25 and- .25 respectively.Hence, mcB = m , = + 0.425Next for joint D, taking rotation contribution of C as +0.425,Unbalanced moment will be (+ 2 + 0.425) = + 2.425Distribution for DC will be (- 0.5 x 2.425) = - 1.213The first cycle is completed in this way.In the second cycle, at joint B, rotation contribution of end A is zero. Rotationcontribution of C is taken as+0.425. Unbalanced moment of (+ 9 + 0.425)=+ 9.425is distributed for BA and BC proportionally with the rotation factors. 'Ihe process iscontinued until1 very negligible variation in rotation contributions is achieved. Thecalculations for four cycles are &tailed in the Table 9.5.Final moments at a joint are

'Ihe values of (m , +maA)for member are written at the centre after the final cycle.So, final moment at a joint is this value plus fixed end moment plus rotationcontribution of the end.M, = - 12- 1.965 = - 13.965kN m


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Table 9.5 K d ' r Method

Errors

JointB

c

D

Total moment at joint B = MBA+ M,, = + 8.07- 8.055

1s tCycle 2ndCycle 3rd Cycle 4th CycleM = + 9 M = + 9 + 0 .4 2 5 M = + 9 + 0 .7 6 M = + 9 + 0 .8 2 7m ~ ~ = - 0 . 2 X 9 =+9.425 =+ 9.76 =+ 9.827=- 1.8 l t t ~ - 0.2 X 9.425 1 1 1 ~ ~- 0.2 x 9.76 ~ A B- 0.2 X 9.827m ~ c = - 0 . 3 X9 =- 1.885 = - 1.925 = - 1.965= - 2.7 rnsc= - 0.3 x 9.425 met= - 0.3 x 9.76 m sc = - 0.3~9.827=- 2.827 = - 2.928 =- .948M = + l - 2 . 7 M = + l - 2 . 8 2 7 - 1 .2 1 3 M = + I - 2 . 9 2 8 - 1 . 38 M = + l - 2 . 9 4 8 - 1 .4 1 4=- 1.7 =-3.04 =- 3.308 = - 3.362m c ~ = - 0 . 2 5 X (- 1.7) mce =- 0.25 x (- 3.04) rtlcs=- 0.25 x (- 3.308) mcs = - 0.25 x (- 3.?52)=+0.425 =+ 0.76 =+ 0.827 =+0.841m a = - 0.25 x (- 1.7) mca= - 0.25 X (- 3.04) mcn =- 0.25 x t- 3.308) mcn = - 0.25 x (- 3.362)=+0.425 =+0.76 =+ 0.827 =+ 0.841M=+2+0.425 M=+ 2+0.76 M =+2+0.827 M=+ 2+0.8415 +2.42-5 =+2.76 = + 2.827 =+2.841mnc- - .5 x 2.425 mnc= - .5 x 2.76 rnr,c= -0.5 x 2.827 m m = - .5 x 2.841=- 1.213 = - 1.38 = - 1.414 = - 1.421

= +0.015kNmTotal Joint at joint C = McB +McD =+0.734 - 0.739

= -0.005kNmMoment at supportD will be - 0.001 kN mActual sum of the moments atB, (3 and D is very small and hence, may beconsidered as zero.

SAQ 2The stiffnesses for various members are assumed is shown in Figure 9.11. For totalload on span, the fixed end moments are calculated. Self weight of beams IJ, JK andKL are assumed as 5 kNim, 3 kN/m and 4 kN/m respectively.

Dead load from slab per metre run of girder = 3.5 x 4 = 14 kNLive load per metre run of girder = 5 x 4 = 20Fixed end moments for dead load and total load for girders LT,JK and KL are shownin Table 9.6,Table 9.7 and Table 9.8.

1 kRotation Factor, y = ---DHence at Joint A,


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IndeterminateStructures - I1

Similarly, the rotation factors at others joints are calculated.3 kDisplacement Factor=---2C k

Hence, displacement factor for each storey will be as follows :

Table9.6

BC and FG 27.75

Cycles of distribution of moments are shown in Figure 9.12. After second cycleonwards,the sway correction is applied. This gives results which converge easily. InTables9.7 and 9.8, exact calculations for moments at each joint are shown.


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Table9.7

Table9.8


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In Figure 9.12, rotation contiibution are written below the beams and fixed endmoments above the beams. Displacement contributions for each column are writtenhorizontally at the centre of the column. Rotation contributions for columns arewritten towards the left sides.Final moments in horizontal members are as follows : ,

MAE= -208+111.56-26.52 = -69.92WmM , = + 208- 85.04 + 26.52 = + 149.48 W mMBc = - 27.75- 42.52 - 20.44 = - 90.71 kN mMCB = + 27.75 + 21.58 - 20.44 = + 28.39 kN mMCD = - 79.17 + 32.37 - 5.94 = - 52.74 W mM , = + 79.17 - 31.37 - 5.94 = + 34.52 kN mM , = - 208+ 68.04 + 16.44 = - 123.52W mMFE= + 208- 51.6+ 16.44 = + 172.84kN mMFG = -27.75 - 25.8 - 14.24 = - 67.79 W mM,, = + 27.75+ 11.58- 14.24 = +25.07 kNmMGH= -79.17 + 17.34- 2.03 = - 63.86kN mMHG= +79.17 - 19.37 - 2.03 = + 57.77W m

Finalmoments n vertical members are as follows :M , = + 27.89 + 44.90 - 3.03 = + 62.76 W mM , = + 17.01+44.9 - 3.03 = + 58.88kN mMBF= - 21.26 - 34.16- 3.03 = - 58.45W mMFB= - 12.90- 34.16- 3.03 = - 50.09W mMcG = + 10.79'+ 16.57- 3.03 = + 24.33 W mM , = + 5.78 + 16.57- 3.03 = + 19.32 W mM,, = - 12.77- 19.23-3.03 = - 35.03 kN mMHD= - 6.46 - 19.23- 3.03 = - 28.72 W mME, = + (34.02x 2) - 3.83 = + 64.21kN mME = + 34.02- 3.83 = + 30.19W mMFI = - 25.8 x 2) - 3.83 = - 55.43 kN mM f l = - 25.8 - 3.83 = -55.43 kN m

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structural anaylsis