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VTU – EDUSAT Programme – 7, , Class: B.E. V Sem (Civil Engineering) Sub: Structural Analysis – II (CV51), Session on 08.10.2007, , KANI’S METHOD FOR ANALYSIS OF INDETERMINATE, STRUCTURES:, By A.B.Harwalkar, P.D.A.College of Engg Gulbarga, This method was developed by Dr. Gasper Kani of Germany in 1947. This, method offers an iterative scheme for applying slope deflection method. We shall now, see the application of Kani’s method for different cases., I., , Beams with no translation of joints:, , Let AB represent a beam in a frame, or a continuous structure under transverse, loading, as show in fig. 1 (a) let the M AB & MBA be the end moment at ends A & B, respectively., Sign convention used will be: clockwise moment +ve and anticlockwise moment –ve., The end moments in member AB may be thought of as moments developed due, to a superposition of the following three components of deformation.

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1. The member ‘AB’ is regarded as completely fixed. (Fig. 1 b). The fixed end, moments for this condition are written as MFAB & MFBA, at ends A & B respectively., 2. The end A only is rotated through an angle A by a moment 2 M 'AB inducing a, moment M 'AB at fixed end B., 3. Next rotating the end B only through an angle B by moment 2M 'BA while keeping, end ‘A’ as fixed. This induces a moment M 'BA at end A., Thus the final moment MAB & MBA can be expressed as super position of three, moments, MAB = MFAB + 2M 'AB  M 'BA, MBA = MFBA + 2M 'BA + M 'AB, , ………(1), , For member AB we refer end ‘A’ as near end and end ‘B’ as far end. Similarly, when we refer to moment MBA, B is referred as near end and end A as far end., Hence above equations can be stated as follows. The moment at the near end of a, member is the algebraic sum of (a) fixed end moment at near end. (b) Twice the rotation, moment of the near end (c) rotation moment of the far end., , Rotation factors:, Fig. 2 shows a multistoried frame., , Consider various members meeting at joint A. If no translations of joints occur,, applying equation (1), for the end moments at A for the various members meeting at A, are given by:

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MAB = MFAB + 2M 'AB  M 'BA, MAC = MFAC + 2M 'AC + M 'CA, MAD = MFAD + 2 M 'AD + M 'DA, MAE = MFAE + 2M 'AE + M 'EA, For equilibrium of joint A, MA = 0, MFAB+ 2 M 'AB +  M 'BA = 0, ………………….(2), where ,, MFAB =Algebraic sum of fixed end moments at A of all members meeting at A.,  M 'AB = Algebraic sum of rotation moments at A of all member meeting at A.,  M 'BA = Algebraic sum of rotation moments of far ends of the members meeting at A., from equation (2), 1, M FAB   M 'BA ……………… (3),  M 'AB = , , 2, 4EI AB, We know that 2M 'AB =, A = 4EKAB A, L AB, I, Where KAB = AB , relative stiffness of member AB, L AB, , , , , , …………….(4), M 'AB = 2E KAB A,   M 'AB = 2EA KAB …………….(5) (At rigid joint A all the members undergo same, rotation A ), Dividing Equation (4)/(5) gives, , K AB, M 'AB, =, M 'AB K AB, K AB, M 'AB,  M 'AB =, K AB, , ……………(5), , Substituting value of  M 'AB from (3) in (5),  1  K AB, M 'AB    , M FAB  M 'BA,  2  K AB, , , , , , , , , , = UAB M FAB  M 'BA ………………(6), 1 K AB, where UAB = is called as rotation factor for member AB at joint A., 2 K AB

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Analysis Method:, In equation (6) MFAB is a known quantity. To start with the far end rotation, moments M 'BA are not known and hence they may be taken as zero. By a similar, approximation the rotation moments at other joints are also determined. With the, approximate values of rotation moments computed, it is possible to again determine a, more correct value of the rotation moment at A from member AB using equation (6)., The process is carried out for sufficient number of cycles until the desired degree, of accuracy is achieved., The final end moments are calculated using equation (1).

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VTU – EDUSAT Programme – 7, , Class: B.E. V Sem (Civil Engineering) Sub: Structural Analysis – II (CV51), Session on 09.10.2007, , KANI’S METHOD FOR ANALYSIS OF INDETERMINATE, STRUCTURES (CONTD.), By A.B.Harwalkar, P.D.A.College of Engg Gulbarga, , Kani’s method for beams without translation of joints, is illustrated in following, examples:, Ex: 1 Analyze the beam show in fig 3 (a) by Kani’s method and draw bending, moment diagram, , Solution:, a) Fixed end moments:,  10  5 2, MFAB =, = - 20.83 kNm, 12, MFBA= + 20.83 kNm,  25  3 12, MFBC =, = - 4.69 kNm, 42, 25  3 2 1, MFCB =, = 14.06 kNm, 42,  40 x 5, MFCD =, =  25 kNm, 8, MFDC = 25 kNm, b) Rotation Factors:, Jt., , Member, , Relative, stiffness (K), , K, , Rotation Factor

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K, 1, =- x, 2 K, B, , C, , BA, , I/5 = 0.2I, , BC, , 2I/4 = 0.5I, , CB, , 2I/4 = 0.5I, , CD, , I/5 = 0.2I, , - 0.14, 0.7I, , -0.36, - 0.36, , 0.7I, , -0.14, , c) Sum fixed end moment at joints:, MFB = 20.83 – 4.69 = 16.14 kNm, MFC = 14.06 – 25 = 10.94 kNm, The scheme for proceeding with method of rotation contribution is shown in, figure 3 (b). The FEM, rotation factors and sum of fixed end moments are entered in, appropriate places as shown in figure 3 (b)., , d) Iteration Process:, Rotation contribution values at fixed ends A &D are zero. Rotation contributions, at joints B & C are initially assumed as zero arbitrarily. These values will be improved in, iteration cycles until desired degree of accuracy is achieved., The calculations for two iteration cycles have been shown in following table. The, remaining iteration cycle values for rotation contributions along with these two have been, shown directly in figure 3 (c)., , Jt, , B, , C, , Rotation, Contributio, , M 'CB, , M 'BA, , M 'CD, , M 'BC, , n, Iteration 1, , -0.14 (16.14 + -0.36 (16.14 + 0), 0) = -2.26 = -5.81, , -0.36 (-10.94 – -0.14 (-10.94 – 5.81), 5.81) = 2.35

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= 6.03, Iteration 2, , -0.14 (16.14 + -0.36 (16.14 + 6.03), 6.03) = -3.1 = -7.98, , Fig.3(c), , -0.36 (-10.94 – -0.14 (-10.94 – 7.98), 7.98) = 2.65, = 6.81

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Iterations are done up to four cycles yielding practically the same value of rotation, contributions., , e) Final moments:, MAB = - 20.83 + 0 – 3.22 = -24.05 kNm, MBA = 20.83 + 2  (-3.22) + 0 = 14.39 kNm, MBC = 4.69 + 2  (-8.3) + 6.93 =  14.36 kNm, MCB = 14.06 + (2  6.93) – 8.3 = 19.62 kNm, MCD = 25 + (2  2.69) + 0 = -19.62 kNm, MDC = 25 + 0 + 2.69 = 27.69 kNm, Bending moment diagram is shown in fig.3 (d), , Fig.3 (d), , Ex 2: Analyze the continuous beam shown in fig. 4 (a), , Solution:, a) Fixed end moments:, b 3a  l , 2.5 3 1.5  4 , MFAB=, Mo ,  24 = 1.88 kNm, 2, l, 42

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1.5 3  2.5  4 , a 3b  l , Mo ,  24  7.88 kNm, 2, l, 42, M, 32, MFBC = o =, = 8 kNm, 4, 4, Mo, MFCB =, = 8 kNm, 4, 36 1 2 2, MFCD = , = -16 kNm, 32, 36 12  2, MFDC =, = 8 kNm, 32, , MFBA =, , b) Modification in fixed end moments:, Actually end ‘D’ is a simply supported. Hence moment at D should be zero. To, make moment at D as zero apply –8 kNm at D and perform the corresponding carry over, to CD., Modified MFDC = 8 – 8 = 0, 1, Modified MFCD= 16 + (-8) = 20 kNm, 2, Now joint D will not enter the iteration process., , c) Rotation Factors:, K, , Rotation Factor, K, 1, U=- x, 2 K, , 0.5 I, ,  0.25, , Joint, , Member, , Relative, stiffness (K), , B, , BA, , I/4 = 0.25I, , BC, , I/4 = 0.25I, , 0.25, , CB, , I/4 = 0.25I, , 0.25, , CD, , 3 I,  = 0.25I, 4 3, , C, , d) Sum of fixed end moments at joints:, MFB = 7.88 + 8 = 15.88 kNm, MFC = 8  20 = 12kNm, , 0.5I, , 0.25

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e) Iteration process, Joint, , B, , Rotation, Contribution, , M, , ', BA, , C, M 'BC, , Rotation factor, , 0.25 0.25, , Iteration 1, , 0.25  (15.88 + 0.25  (15.88 + 0, , started at B, , 0 + 0) = -3.97 + 0) = -3.97, , M 'CB, , M 'CD, , 0.25 0.25, 0.25  (-12 –3.97 0.25  (-12 –3.97 +, + 0) = 3.97 0) = 3.97, , assuming M 'CB, = 0 & taking, , M 'AB = 0, M 'DC = 0., , Iteration 2, , 0.25  (15.88 +, 0 + 3.97) = , , 0.25  (15.88 + 0, + 3.97) =  4.96, , 4.96, Iteration 3, , 0.25 (15.88 + 0 0.25 (15.88 + 0 +, + 4.24) = 5.03 4.24) = 5.03, , Iteration 4, , 0.25 (15.88 + 0 0.25 (15.88 + 0 +, + 4.26) = 5.04 4.26) = 5.04, , 0.25  (-12 4.96 0.25 (12 4.96 +, + 0) = 4.24 0), = 4.24, 0.25  (-12 0.25 (12 5.03 +, 5.03+ 0) = 4.26 0), = 4.26, 0.25  (-12 0.25 (12 5.03 +, 5.03+ 0) = 4.26 0), = 4.26

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Iteration process has been stopped after 4th cycle since rotation contribution values are, becoming almost constant. Values of fixed end moments, sum of fixed end moments,, rotation factors along with rotation contribution values after end of each cycle in, appropriate places has been shown in fig. 4 (b).

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f) Final moments, Final moments,  M Fij  2M ij'  M 'ji, , , , Member, (ij), , FEM MFij (kNm), , AB, , 1.88, , 0, , -5.04, , (kNm), -3.16, , BA, , 7.88, , 2 (-5.04) = 10.08, , 0, , -2.2, , BC, , 8, , 2 (-504) = 10.08, , 4.26, , +2.2, , CB, , 8, , 2  4.26 = 8.52, , -5.04, , 11.48, , CD, , -20, , 2 x 4.26 = 8.52, , 0, , -11.48, , BMD is shown below:, , ', ij, , 2M (kNm), , 1, ji, , M (kNm), , 

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c) Rotation factors:, , Jt., , Member, , Relative stiffness, (K), , B, , BA, , I/4 = 0.25I, , BC, , 3 1 .5 I, = 0. 375I, , 4 3, , CB, , 1.5I/3 = 0.5I, , CD, , 0, , C, , K, , Rotation Factor, K, 1, U=- x, 2 K,  0.2, , 0.625I, , 0.3, 0.5, , 0.5I, , 0, , d) Sum of fixed end moments at joints:, MFB = 6.67 –3.13 = 3.54 kNm, e) Iteration Process, Joint, , B, , Rotation Contribution, , M 'BA (kNm), , M 'BC (kNm), ,  0.2 0.3, , Rotation factor, Iteration 1 started at B, taking M 'AB = 0, , 0.2  (3.54 + 0 + 0) = 0.71  0.3  (3.54 + 0 + 0) = 1.06, , & M 'CB = 0, , Since ‘B’ is the only joint needing rotation correction, the iteration process will, stop after first iteration. Value of FEMs, sum of FEM at joint, rotation factors along with, rotation contribution values in appropriate places is shown in fig. 5 (b), , Fig.5(b)

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(f) Final moments:, Final moments,  M Fij  2M ij'  M 'ji, , , , Member, (ij), , FEM MFij (kNm), , AB, ,  6.67, , 0, , 0.71, , (kNm), 7.38, , BA, , 6.67, , 2  (-0.71) =, , 0, , 5.25, , BC, , 3.13, , 2  (1.06), , 0, , 5.25, , ', ij, , 2M (kNm), , 1, ji, , M (kNm), , CB, CD, DC, , , , +5, -, , -, , -, , 5, 0, , FBD of each span along with reaction values which have been calculated from statics are, shown below:

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BMD and SFD are shown below, , II., , Kani’s method for members with translatory joints:, Fig. 6 shows a member AB in a frame which has undergone lateral displacement, at A & B so that the relative displacement is  = B  A, If ends A & B are restrained from rotation FEM corresponding to this displacement are, , ', ', M 'AB,  M 'BA, , , 6EI , ………………(7), L2, , When translation of joints occurs along with rotations the true end moments are given by, ', MAB = MFAB + 2M 'AB + M 'BA + M 'AB

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', MBA = MFBA + 2M 'BA + M 'AB + M 'BA, , If ‘A’ happens to be a joint where two or more members meet then from, equilibrium of joint we have,  MAB = 0, ', MFAB + 2 M 'AB +  M 'BA +  M 'AB, =0, 1, ',   M 'AB = , ……………(8), M FAB  M 'BA  M 'AB, 2, we know from equation (5), K AB, M 'AB, M 'AB =, K AB, 1 K AB, ', M FAB  M 'BA  M 'AB, Using equation (8) M 'AB = , 2 K AB, , , , , , , , Similarly, , M 'BA, , , , , , , , , , ', = U AB M FA  M 'BA  M 'AB, ', = UBA M FB  M 'AB  M 'BA, ……………..(9), , Using the above relationships rotation contributions can be determined by, iterative procedure. If lateral displacements are known the displacement moments can be, determined from equation (7). If lateral displacements are unknown then additional, equations have to be developed for analyzing the member., , Ex 4: In a continuous beam shown in fig. 7 (a). The support ‘B’ sinks by 10mm., Determine the moments by Kani’s method & draw BMD., , Take I = 1.2 x 104 mm4 & E = 2 x 105 N/mm², Solution:, (a) Calculation of FEM:, 20  6 2 6  2 10 5 1.2 10 4 1012 10, , MFAB = , 12, 60002  10 6, = 60 40, = 100 kNm

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MFBA = +60 40= 20 kNm, 50  3  2 2 6  2 10 5 1.2 10 4 1012 10, , MFBC = , 52, 50002  10 6, = 24 + 57.6, = 33.6 kNm, 50  3 2  2 6  2 10 5 1.2 10 4 1012 10, , MFCB = +, 52, 50002 10 6, = 36 + 57.6, = 93.6 kNm, C & D are at same level, 20  4 2,  MFCD = ,   26.67 kNm, 12, MFDC = + 26.67 kNm, b) Modification in fixed end moments:, Since end ‘D’ is a simply supported, moment at ‘D’ is zero. To make moment at, D as zero apply a moment of  26.67 kNm at end D and perform the corresponding carry, over to CD.,  Modified MFDC = + 26.67 26.67 = 0, 1, Modified MFCD =  26.67 +, (26.67), 2, = 40 kNm, Other FEMs will be same as calculated earlier. Now joint ‘D’ will not enter the iteration, process., c) Rotation factors:, , Joint, , Member, , Relative stiffness, (K), , B, , BA, , I/6 = 0.17 I, , C, , K, , 0.37 I, , Rotation Factor, K, 1, U=- x, 2 K, 0.23, , BC, , I/5 = 0.2 I, , 0.27, , CB, , I/5 = 0.2I, , 0.26, , CD, , 3, x I/4 = 0.19 I, 4, , 0.39I, , 0.24

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d) Sum of fixed end moments:, MFB = 20 + 33.6 = 53.6 kNm, MFC= 93.6 40 = 53.6 kNm, e) Iteration process:, , Joint, , B, , Rotation, Contribution, Rotation factor, Iteration 1, (Started at B by, taking M 'AB = 0, and assuming, M 'CB = 0, Iteration 2, , Iteration 3, , M, , ', BA, , (kNm), , M 'BC (kNm), , 0.23 0.27, 0.23  (53.6 + 0.27 x (53.6 +, 0 + 0) = 12.33 0+0)= 14.47, , 0.23 (53.6  0.27 (53.6 10.17), , M 'CB (kNm), , M 'CD (kNm), , 0.26 0.24, 0.26(53.6 – 0.24 (53.6  14.47), 14.47+ 0) = 10.17 10.96= 9.39, , 0.26 (53.6 11.73) 0.24 (53.6 11.73), , 10.17) = 10.00 = 11.73, , = 10.89 =10.05, , 0.23, , 0.26 (53.6 11.53) 0.24 (53.6 11.53), , (53.610.89), = 9.82, Iteration 4, , C, , 0.27 (53.6 10.89), =11.53, , 0.23 (53.6, 10.94), = 9.81, , = 10.94 =10.10, 0.26 (53.6 11.52) 0.24 (53.6 11.52), , 0.27 (53.6 10.94), = 11.52, , = 10.94 = 10.1, , Iteration process has been stopped after fourth cycle since rotation contribution values are, becoming almost constant. Values of FEMs, sum of fixed end moments, rotation factors, along with rotation contribution values after end of each cycle in appropriate places has, been shown in Fig. 7 (b).

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f) Final moments:, Final moments,  M Fij  2M ij'  M 'ji, , , , Member, (ij), , FEM MFij, (kNm), , 2M (kNm), , AB, ,  100, , 0, , 9.81, , (kNm), 109.81, , BA, , 20, , 2  (-9.81)= -19.62, , 0, , +0.38, , BC, , 33.6, , 2  11.52) = 23.04, , 10.94, , 0.38, , CB, , 93.6, , 2  (10.94) = 21.88, , 11.52, , 60.2, , CD, , 40, , 2  (10.1) = 20.2, , 0, , 60.2, , DC, , 0, , 0, , 0, , 0, , g), , ', ij, , 1, ji, , M (kNm), , BMD is shown below:, , 109.81, , 60.2, , 0.38, , 20x6² / 8 = 90, KNM, , 50x3x2/5 = 60, KNM, , 20x4²/8 = 40KNM, , 

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VTU – EDUSAT Programme – 7, Class: B.E. V SEM (Civil Engineering) SUB: Structural Analysis – II (CV51), Session on 15.10.2007, KANI’S METHOD FOR ANALYSIS OF INDETERMINATE STRUCTURES, (CONTD.), BY A.B.HARWALKAR, P.D.A.College of Engg Gulbarga, III), , Analysis of frames with no translation of joints:, The frames, in which lateral translations are prevented, are analyzed in the same, way as continuous beams. The lateral sway is prevented either due to symmetry of frame, and loading or due to support conditions. The procedure is illustrated in following, example., Example-5. Analyze the frame shown in Figure 8 (a) by Kani’s method. Draw BMD., (VTU Jan 2005 Exam)., , Fig-8(a), Solution:, (a), Fixed end moments:, MFAB = MFBA = MFCD = MFDC = 0, - 40 x 6 2, MFBC =, = -120kNm., 12, MFCB = +120kNm., (b), , Rotation factors:, , Joint, , Member, , Relative Stiffness (k) k, , B, , BC, BA, CB, CD, , 3I/6 = 0.5I, I/3 = 0.33I, 3I/6 = 0.5I, I/3 = 0.33I, , C, , 0.83I, 0.83I, , Rotation factor, = -½k/k, -0.3, -0.2, -0.3, -0.2

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(c), , Sum of FEM:, MFB = -120+0 = -120, MFC = 120+0 = +120, (d), Iteration process:, Joint, B, , C, , Rotation, Contribution, , M’BA, , M’BC, , M’CB, , M’CD, , Rotation, Factor, , -0.2, , -0.3, , -0.3, , -0.2, , Iteration, 1 -0.2(-120+0), stated, with =24, end B taking, M’AB=0 and, assuming, M’CB=0, , -0.3(-120+0), =36, , -0.2(120+36+0), = -46.8, , -0.2(120+36+0), = -31.2, , Iteration 2, , -0.2(-120-46.8), =33.6, , -0.3(-120-46.8), =50.04, , -0.3(120+50.04), = -51.01, , -0.2(120+50.04), = -34.01, , Iteration 3, , -0.2(-120-51.01), =34.2, , -0.3(-120-51.01), =51.3, , -0.3(120+51.3), = -51.39, , -0.2 (120+51.3), = -34.26, , Iteration 4, , -0.2(-120-51.39), =34.28, , -0.3(-120-51.39), =51.42, , -0.3(120+51.42), = -51.43, , -0.2 (120+51.42), = -34.28, , The fixed end moments, sum of fixed and moments, rotation factors along with, rotation contribution values at the end of each cycle in appropriate places is shown in, figure 8(b)., , Fig-8(b)

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(e), , Final moments:, , Member, (ij), , MFij, , 2M’ij (kNm), , M’ji (kNm), , (kNm) Final moment =, MFij + 2M’ij + M’ji, , AB, , 0, , 0, , 34.28, , 34.28, , BA, , 0, , 2 x 34.28, , 0, , 68.56, , BC, , -120, , 2 x 51.42, , -51.43, , -68.59, , CB, , 120, , 2 x (-51.43), , 51.42, , 68.56, , CD, , 0, , 2 x (-34.28), , 0, , -68.56, , DC, , 0, , 0, , -34.28, , -34.28, , BMD is shown below in figure-8 (c), , Fig-8 (c), , IV), , Analysis of symmetrical frames under symmetrical loading:

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Considerable calculation work can be saved if we make use of symmetry of, frames and loading especially when analysis is done manually. Two cases of symmetry, arise, namely, frames in which the axis of symmetry passes through the centerline of the, beams and frames with the axis of symmetry passing through column line., Case-1: (Axis of symmetry passes through center of beams):, Let AB be a horizontal member of the frame through whose center, axis of, symmetry passes. Let Mab and Mba be the end moments. Due to symmetry of, deformation Mab and Mba are numerically equal but are opposite in their sense., , A, , =, =, , Slope due to Mab + slope due to Mba, M ab l M ba l M ab l, +, =, 3EI 6EI 2EI, , Let this member be replaced by member AB’ whose end A will undergo the, rotation A due to moment Mab applied at A, the end B’ being fixed., , Mab l',  A =, 4EI', Hence for equality of rotations between original member AB and the substitute, member AB’, Mab l Mab l', , 2EI 4EI', I 2I', =, l l'

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K = 2K’,  K’ =, , K, 2, , Thus if K is the relative stiffness of original member AB, this member can be, K, replaced by substitute member AB’ having relative stiffness, . With this substitute, 2, member, the analysis need to be carried out for only, one half of the frame considering, line of symmetry as fixed.

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VTU – EDUSAT Programme – 7, Class: B.E. V SEM (Civil Engineering) SUB: Structural Analysis – II (CV51), Session on 16.10.2007, , KANI’S METHOD FOR ANALYSIS OF INDETERMINATE, STRUCTURES (CONTD.), BY A.B.HARWALKAR, P.D.A.College of Engg Gulbarga, Example-6: Analyze the frame given in example-5 by using symmetry condition by, Kani’s method., Solution:, Since symmetry axis passes through center of beam only one half of frame as, shown in figure 9 (a) will be considered, , Fig-9(a), ,  Rotation factors, 1, UBA = - x (0.33I / 0.33I + 0.25I) = -0.28, 2, 1, UBC = - x (0.25I / 0.33I + 0.25I) = -0.22, 2, The calculation of rotation contribution values is shown directly in figure-9(b), , Fig-9(b)

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Here we can see that rotation contributions are obtained in the first iteration only., The final moments for half the frame are shown in figure 9(c) and for full frame are, shown in figure 9(d)., , Fig-9(c), , Fig-9(d), Example-7: Analyze the frame shown in figure 10(a) by Kani’s method., , Fig-10(a), , Solution:, Analysis will be carried out taking the advantage of symmetry, (a) Fixed end moments:, MFcd = -[(20 x 1 x 32 / 42) + (20 x 3 x 12 / 42)] = -15kNm, MFbe = -24 x 42 / 12 = -32 kNm.

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The substitute frame is shown in figure 10(b), D’, , 1 2I I, KCD’ = x =, 2 4 4, , Fig-10(b), , Kba =, , 2I I, =, 4 2, , (b) Rotation factors:, Joint, , Member, , Relative Stiffness K, , k, , B, , BA, BE’, , 2I/4, 1 4I, x =I / 2, 2 4, I/4, , 5I/4, , I/4, 1 2I I, x =, 2 4 4, , 2I/4, , BC, C, , CB, CD’, , Rotation factors = –, , 1 K, 2 ΣK, , -1/5, -1/5, 1, 10, -1/4, -1/4, , -, , Rotation contributions calculated by iteration process are directly shown in figure-10(c)., , ', , Fig-10(c), '

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The calculation of final moments for the substitute frame is shown in figure-10(d), , Fig-10(d), , Figure-10(e) shows final end moments for the entire frame., , Fig-10(e)

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VTU – EDUSAT Programme – 7, Class: B.E. V SEM (Civil Engineering) SUB: Structural Analysis – II (CV51), , Session on 19.10.2007, KANI’S METHOD FOR ANALYSIS OF INDETERMINATE, STRUCTURES (CONTD.), BY A.B.HARWALKAR, P.D.A.College of Engg Gulbarga, Case 2: When the axis of symmetry passes through the column:, This case occurs when the number of bays is an even number. Due to symmetry, of the loading and frame, the joints on the axis of symmetry will not rotate. Hence it is, sufficient if half the frame is analyzed. The following example illustrates the procedure., Example-8: Analyze the frame shown in figure-11(a) by Kani’s method, taking advantage, of symmetry and loading., , Fig-11(a), Solution:, Only half frame as shown in figure-11(b) will be considered for the analysis., , D, , Fig-11(b)

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(a) Fixed end moments:, 120 x 6 2, MFBE= = - 360kNm, 12, 120 x 6 2, MFCD = = - 360kNm, 12, (b) Rotation factors:, Joint, , Member, , Relative, Stiffness k, , k, , B, , BA, BE, BC, CB, CD, , I/3, 3I/6 = I/2, I/3, I/3, 3I/6 = I/2, , 7I/6, , C, , 5I/6, , Rotation factors = –, 1 K, 2 ΣK, -1/7, -3/14, -1/7, -1/5, -3/10, , (c) Iteration process:, The iteration process for calculation of rotation contribution values at C & B was, carried up to four cycles and values for each cycle are shown in figure-11(c)., , Fig-11(c)

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Final moments calculations for half the frame are shown in figure-11(d) and final, end moments of all the members of the frame are shown in figure-11(e)., , E, , Fig-4(d), , Fig-11(d), , Fig-11(e)