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Analysis of Indeterminate Structure Session 05-22 Matakuliah: S0725 – Analisa Struktur Tahun: 2009.

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Presentation on theme: "Analysis of Indeterminate Structure Session 05-22 Matakuliah: S0725 – Analisa Struktur Tahun: 2009."— Presentation transcript:

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2 Analysis of Indeterminate Structure Session Matakuliah: S0725 – Analisa Struktur Tahun: 2009

3 Bina Nusantara University 3 Contents 3 Equations Method Flexibility Method Slope Deflection Method Cross Method/ Moment Distribution Method Matrix Analysis

4 Bina Nusantara University 4 3 Equation Method This method will be used for analizing the indeterminate structure ( support reaction, internal loads )

5 Bina Nusantara University 5 3 Equation Method Degree of indeterminacy  i = r – 3 i = 7 – 3 = 4 A BCD E P1q2P2q1 Gambar 1.1. Struktur statis tak tentu

6 Bina Nusantara University 6 3 Equation Method Pada bagian konstruksi diantara 2 perletakan yang berdekatan diberikan kebebasan untuk berputar sudut satu sama lain. Sebagai akibatnya akan timbul CELAH pada balok di tempat tumpuan ( perletakan ) sebagai akibat dari adanya beban luar. ( Gambar 1.2.a ) Principles : l k P1P2q1 celah  o kr  o kl  o kr  o kl lklk k - 1k + 1 Gambar 1.2.a. Putran sudut akibat beban luar

7 Bina Nusantara University 7 3 Equation Method Pada hakekatnya balok ini adalah menerus, utuh dan tidak boleh ada celah, maka harus ada MOMEN pada tumpuan / perletakan antara yang berfungsi mengembalikan celah tadi menjadi utuh kembali. Sebagai gaya statis kelebihan harus dipilih berupa momen-momen pada perletakan antara, umumnya momen – momen yang bekerja adalah MOMEN NEGATIF. ( Gambar 1.2.b ) Princiles : l k  ' kr  ' kl M k lklk k - 1k + 1 M k-1 M k+1 Gambar 1.2.b. Momen lentur negatif

8 Bina Nusantara University 8 3 Equation Method Compatibility eq : Sebagai suatu syarat kompatibilitas, joint ‘k’ merupakan rotasi yang kompatibel, sehingga persamaan kompatibilitasnya menjadi  o kl +  o kr =  ’ kl +  ’ kr Dimana :  o dan  ’  diambil nilai harga mutlaknya. Sehingga harga Mk yang positif berarti bekerja sebagai momen lentur negatif, dimana ini berarti bahwa kita harus MERUBAH tanda gaya statis kelebihan yang didapatkan.

9 Bina Nusantara University 9 3 Equation Method PERSYARATAN KOMPATIBILITAS : Jika tidak ingin merubah tanda tersebut maka harus dimasukkan anggapan – anggapan bahwa momen peralihan merupakan momen lentur positif, sehingga persamaan kompatibilitas menjadi : (  o kl +  o kr ) + (  ’ kl +  ’ kr ) = 0 Sehingga hasil-hasil momen peralihan sudah langsung berikut tandanya menyatakan MOMEN LENTUR SEBENARNYA, jika hasilnya negatif, berarti bekerja sebagai momen lentur negatif.

10 Bina Nusantara University 10 3 Equation Method Karena  nilai  ’ kl tergantung pada M k-1 dan M k nilai  ’ kr tergantung pada M k dan M k+1 m a k a... dari persamaan kompatibilitas di atas akan selalu didapatkan maksimum sebanyak 3 momen tumpuan antara yang terlibat dalam persamaan kompatibilitas, sehingga metoda ini disebut juga  PERSAMAAN TIGA MOMEN

11 Bina Nusantara University 11 3 Equation Method Dengan mennjau nilai putaran sudut akibat momen – momen peralihan persamaan kompatibilitas dapat dituliskan sebagai   ’ kl = M k ( l l / 3EI ) + M k-1 ( l l / 6EI )  ’ kr = M k ( l r / 3EI ) + M k+1 ( l r / 6EI ) +  ’ kl +  ’ kr = M k-1 ( l l / 6EI )+M k {( l l / 3EI )+( l r / 3EI )}+ M k+1 ( l r / 6EI )

12 Bina Nusantara University 12 3 Equation Method Sehingga persamaan 3 momen pada tumpuan k... M k-1 ( l l / 6EI )+M k {( l l / 3EI )+( l r / 3EI )}+ M k+1 ( l r / 6EI ) + (  o kl +  o kr ) = 0 Agar persamaan ini dapat digunakan secara efisien dan tepat maka diperlukan rumus – rumus dari berbagai type beban, baik dari beban luar maupun momen peralihan ( momen pada tumpuan antara ).

13 Bina Nusantara University 13 3 Equation Method C a t a t a n... Jika pada perletakan ujung – ujung adalah SENDI & ROL, maka i  jumlah perletakan antara. Sedangkan jika perletakan ujung adalah JEPIT, maka perletakan jepit ini diperlakukan sebagai perletakan antara, dengan MENGUBAH menjadi SENDI dan Momen Jepit sebagai gaya kelebihan. ( Gambar 1.3 )

14 Bina Nusantara University 14 3 Equation Method M Gambar 1.3. Struktur dengan perletakan Jepit yang diubah  Sendi + Momen jepit

15 Bina Nusantara University 15 While analyzing indeterminate structures, it is necessary to satisfy (force) equilibrium, (displacement) compatibility and force-displacement relationships Method Analysis

16 Bina Nusantara University 16 (a) Force equilibrium is satisfied when the reactive forces hold the structure in stable equilibrium, as the structure is subjected to external loads (b) Displacement compatibility is satisfied when the various segments of the structure fit together without intentional breaks, or overlaps (c) Force-displacement requirements depend on the manner the material of the structure responds to the applied loads, which can be linear/nonlinear/viscous and elastic/inelastic; for our study the behavior is assumed to be linear and elastic Method Analysis

17 Bina Nusantara University 17 (a) Force equilibrium is satisfied when the reactive forces hold the structure in stable equilibrium, as the structure is subjected to external loads Method Analysis

18 Bina Nusantara University 18 (b) Displacement compatibility is satisfied when the various segments of the structure fit together without intentional breaks, or overlaps Method Analysis

19 Bina Nusantara University 19 (c) Force-displacement requirements depend on the manner the material of the structure responds to the applied loads, which can be linear/nonlinear/viscous and elastic/inelastic; for our study the behavior is assumed to be linear and elastic Method Analysis

20 Bina Nusantara University 20 Two methods are available to analyze indeterminate structures, depending on whether we satisfy force equilibrium or displacement compatibility conditions – They are: Force method and Displacement Method Method Analysis

21 Bina Nusantara University 21 Force Method satisfies displacement compatibility and force- displacement relationships; it treats the forces as unknowns – Two methods which we will be studying are Method of Consistent Deformation and (Iterative Method of) Moment Distribution Method Analysis

22 Bina Nusantara University 22 Displacement Method satisfies force equilibrium and force- displacement relationships; it treats the displacements as unknowns – Two available methods are Slope Deflection Method and Stiffness (Matrix) method Method Analysis

23 Bina Nusantara University 23 Method Analysis MethodUnknownsEquations used for solution Coefficient of the unknowns Forced Method ForcesCompatibility and force displacement Flexibility Coefficients Displacem ent Method Displaceme nts Equilibrium and force displacement Stiffness Coefficients

24 Bina Nusantara University 24 Two methods are available to analyze indeterminate structures, depending on whether we satisfy force equilibrium or displacement compatibility conditions - They are: Force method and Displacement Method Force Method satisfies displacement compatibility and force- displacement relationships; it treats the forces as unknowns - Two methods which we will be studying are Method of Consistent Deformation and (Iterative Method of) Moment Distribution Displacement Method satisfies force equilibrium and force- displacement relationships; it treats the displacements as unknowns - Two available methods are Slope Deflection Method and Stiffness (Matrix) method Flexibility Method

25 Bina Nusantara University 25 FORCED METHOD This method is also known as flexibility method or compatibility method. In this method the degree of static indeterminac y of the structure is determined and the redundants are identified. Flexibility Method VTU Programme

26 Bina Nusantara University 26 FORCED METHOD A coordinate is assigned to each redundant. Thus,P1, P Pn are the redundants at the coordinates 1,2, n.If all the redundants are removed, the resulting structure known as released structure, is statically determinate Flexibility Method VTU Programme

27 Bina Nusantara University 27 FORCED METHOD This released structure is also known as basic determinate structure. From the principle of super position the net displacement at any point in statically indeterminate structure is some of the displacements in the basic structure due to the applied loads and the redundants. This is known as the compatibility condition and may be expressed by the equation; Flexibility Method VTU Programme

28 Bina Nusantara University 28 FORCED METHOD ∆ 1 = ∆ 1 L + ∆ 1 R Where ∆ ∆n = Displ. At Co-ord.at 1,2 - -n ∆ 2 = ∆ 2 L + ∆ 2 R ∆ 1 L ---- ∆ n L = Displ.At Co-ord.at 1, n | | | Due to aplied loads | | | ∆ 1 R ----∆ n R = Displ.At Co-ord.at 1, n ∆ n = ∆ n L + ∆ n R Due to Redudants Flexibility Method VTU Programme

29 Bina Nusantara University 29 FORCED METHOD The above equations may be return as [∆] = [∆ L ] + [∆ R ] (1) ∆ 1 = ∆ 1 L + δ 11 P 1 + δ 12 P δ 1n P n ∆ 2 = ∆ 2 L + δ 21 P 1 + δ 22 P δ 2n P n |||| | |||| | (2) ∆ n = ∆ n L + δ n1 P 1 + δ n2 P δ nn P n Flexibility Method VTU Programme

30 Bina Nusantara University 30 FORCED METHOD ∆ = [∆ L] + [δ] [P] (3)  [P]= [δ] -1 { [∆] – [∆ L] } (4) If the net displacements at the redundants are zero then ∆1, ∆ ∆n =0, Then  [P] = - [δ] -1 [∆ L] (5) The redundants P1,P2, Pn are Thus determined Flexibility Method VTU Programme

31 Bina Nusantara University 31 Slope Deflection Slope deflection equations The slope deflection equations express the member end moments in terms of rotations angles. The slope deflection equations of member ab of flexural rigidity EI ab and length L ab are:

32 Bina Nusantara University 32 Slope Deflection Slope deflection equations where θ a, θ b are the slope angles of ends a and b respectively, Δ is the relative lateral displacement of ends a and b. The absence of cross-sectional area of the member in these equations implies that the slope deflection method neglects the effect of shear and axial deformations.

33 Bina Nusantara University 33 Slope Deflection Slope deflection equations The slope deflection equations can also be written using the stiffness factor. and the chord rotation

34 Bina Nusantara University 34 Slope Deflection

35 Bina Nusantara University 35 Slope Deflection Slope deflection equations When a simple beam of length Lab and flexural rigidity EIab is loaded at each end with clockwise moments M ab and M ba, member end rotations occur in the same direction. These rotation angles can be calculated using the unit dummy force method or the moment-area theorem

36 Bina Nusantara University 36 Slope Deflection

37 Bina Nusantara University 37 Slope Deflection Slope deflection equations Joint equilibrium conditions imply that each joint with a degree of freedom should have no unbalanced moments i.e. be in equilibrium. Therefore,

38 Bina Nusantara University 38 FORCED METHOD ∆ = [∆ L] + [δ] [P] (3)  [P]= [δ] -1 { [∆] – [∆ L] } (4) If the net displacements at the redundants are zero then ∆1, ∆ ∆n =0, Then  [P] = - [δ] -1 [∆ L] (5) The redundants P1,P2, Pn are Thus determined Slope Deflection Method VTU Programme

39 Bina Nusantara University 39 Consider portion AB of a continuous beam, shown below, subjected to a distributed load w(x) per unit length and a support settlement of  at B; EI of the beam is constant. Slope Deflection Method

40 Bina Nusantara University 40 Slope Deflection Method

41 Bina Nusantara University 41 M  BA =(4EI  B )/L BB M  AB =(2EI  B )/L + A B (iii) Due to rotation  B at support B L M  AB =(-6EI  )/L 2 A B L M  BA =(-6EI  )/L 2  + (iv) Due to differential settlement of  (between A and B) Slope Deflection Method

42 Bina Nusantara University 42 Slope Deflection Method

43 Bina Nusantara University 43 AA BB  = (FEM) AB (FEM) BA (FEM) BA /2 ( FEM) BA + Slope Deflection Method

44 Bina Nusantara University 44 + AA M  AB =(3EI  A )/L +  M  AB =(3EI  )/L 2 M AB = [(FEM) AB - (FEM) BA /2]+(3EI  A )/L -(3EI  )/L 2 Modified FEM at end A  = PL 3 /(3EI), M = PL = (3EI  /L 3 )(L) = 3EI  /L 2 Slope Deflection Method

45 Bina Nusantara University 45 Slope Deflection Method

46 Bina Nusantara University 46 Slope Deflection Method

47 Bina Nusantara University 47 Slope Deflection Method

48 Bina Nusantara University 48 This method is at the core of the moment distribution method, and is also very powerful. Consider a beam of length L, subjected to end moments (clockwise positive), and downward transverse loads either distributed or concentrated The end slopes are θ A θ B. Slope Deflection Method

49 Bina Nusantara University 49 Slope Deflection Method

50 Bina Nusantara University 50 are the numerical values of the fixed-end moments, e.g. wL2/8, PL/8, Pab2/L2, etc… Slope Deflection Method

51 Bina Nusantara University 51 Slope Deflection Method

52 Bina Nusantara University 52 Here, Slope Deflection Method

53 Bina Nusantara University 53 (Method developed by Prof. Hardy Cross in 1932) The method solves for the joint moments in continuous beams and rigid frames by successive approximation. Moment Distribution Method

54 Bina Nusantara University 54 Thus the Moment Distribution Method (also known as the Cross Method) became the preferred calculation technique for reinforced concrete structures. Moment Distribution Method

55 Bina Nusantara University 55 The description of the moment distribution method by Hardy Cross is a little masterpiece. He wrote: "Moment Distribution. The method of moment distribution is this: Imagine all joints in the structure held so that they cannot rotate and compute the moments at the ends of the members for this condition; at each joint distribute the unbalanced fixed-end moment among the connecting members in proportion to the constant for each member defined as " stiffness "; Moment Distribution Method

56 Bina Nusantara University 56 multiply the moment distributed to each member at a joint by the carry-over factor at the end of the member and set this product at the other end of the member; distribute these moments just "carried over"; Moment Distribution Method

57 Bina Nusantara University 57 repeat the process until the moments to be carried over are small enough to be neglected; and add all moments - fixed-end moments, distributed moments, moments carried over - at each end of each member to obtain the true moment at the end." [Cross 1949:2] Moment Distribution Method

58 Bina Nusantara University Restrain all possible displacements. 2. Calculate Distribution Factors: The distribution factor DFi of a member connected to any joint J is where S is the rotational stiffness, and is given by Moment Distribution Method

59 Bina Nusantara University Determine carry-over factors The carry-over factor to a fixed end is always 0.5, otherwise it is Calculate Fixed End Moments. These could be due to in-span loads, temperature variation and/or relative displacement between the ends of a member. Moment Distribution Method

60 Bina Nusantara University Do distribution cycles for all joints simultaneously Each cycle consists of two steps: 1. Distribution of out of balance moments Mo, 2.Calculation of the carry over moment at the far end of each member. The procedure is stopped when, at all joints, the out of balance moment is a negligible value. In this case, the joints should be balanced and no carry-over moments are calculated. 6. Calculate the final moment at either end of each member. This is the sum of all moments (including FEM) computed during the distribution cycles. Moment Distribution Method

61 Bina Nusantara University 61 Consider the continuous beam ABCD, subjected to the given loads, as shown in Figure below. Assume that only rotation of joints occur at B, C and D, and that no support displacements occur at B, C and D. Due to the applied loads in spans AB, BC and CD, rotations occur at B, C and D. Moment Distribution Method

62 Bina Nusantara University 62 Moment Distribution Method 15 kN/m10 kN/m 8 m6 m8 m A BC D III 3 m PROBLEMS

63 Bina Nusantara University 63 Step I The joints B, C and D are locked in position before any load is applied on the beam ABCD; then given loads are applied on the beam. Since the joints of beam ABCD are locked in position, beams AB, BC and CD acts as individual and separate fixed beams, subjected to the applied loads; these loads develop fixed end moments. 8 m -80 kN.m 15 kN/m A B 6 m kN.m kN.m B C 8 m kN.m 10 kN/m C D 150 kN kN.m 3 m Moment Distribution Method

64 Bina Nusantara University 64 Step I The joints B, C and D are locked in position before any load is applied on the beam ABCD; then given loads are applied on the beam. Since the joints of beam ABCD are locked in position, beams AB, BC and CD acts as individual and separate fixed beams, subjected to the applied loads; these loads develop fixed end moments. Moment Distribution Method

65 Bina Nusantara University 65 8 m -80 kN.m 15 kN/m A B 6 m kN.m kN.m B C 8 m kN.m 10 kN/m C D 150 kN kN.m 3 m Moment Distribution Method

66 Bina Nusantara University 66 In beam AB Fixed end moment at A = -wl 2 /12 = - (15)(8)(8)/12 = - 80 kN.m Fixed end moment at B = +wl 2 /12 = +(15)(8)(8)/12 = + 80 kN.m In beam BC Fixed end moment at B = - (Pab 2 )/l 2 = - (150)(3)(3) 2 /6 2 = kN.m Fixed end moment at C = + (Pab 2 )/l 2 = + (150)(3)(3) 2 /6 2 = kN.m In beam AB Fixed end moment at C = -wl 2 /12 = - (10)(8)(8)/12 = kN.m Fixed end moment at D = +wl 2 /12 = +(10)(8)(8)/12 = kN.m Moment Distribution Method

67 Bina Nusantara University 67 Step II Since the joints B, C and D were fixed artificially (to compute the the fixed-end moments), now the joints B, C and D are released and allowed to rotate. Due to the joint release, the joints rotate maintaining the continuous nature of the beam. Due to the joint release, the fixed end moments on either side of joints B, C and D act in the opposite direction now, and cause a net unbalanced moment to occur at the joint Moment Distribution Method

68 Bina Nusantara University 68 Moment Distribution Method

69 Bina Nusantara University 69 Step III These unbalanced moments act at the joints and modify the joint moments at B, C and D, according to their relative stiffnesses at the respective joints. The joint moments are distributed to either side of the joint B, C or D, according to their relative stiffnesses. These distributed moments also modify the moments at the opposite side of the beam span, viz., at joint A in span AB, at joints B and C in span BC and at joints C and D in span CD. This modification is dependent on the carry-over factor (which is equal to 0.5 in this case); when this carry over is made, the joints on opposite side are assumed to be fixed. Moment Distribution Method

70 Bina Nusantara University 70 Step IV The carry-over moment becomes the unbalanced moment at the joints to which they are carried over. Steps 3 and 4 are repeated till the carry-over or distributed moment becomes small. Moment Distribution Method

71 Bina Nusantara University 71 Step V Sum up all the moments at each of the joint to obtain the joint moments. Moment Distribution Method

72 Bina Nusantara University 72 Stiffness and Carry-over Factors Stiffness = Resistance offered by member to a unit displacement or rotation at a point, for given support constraint conditions  A MAMA MBMB A B A RARA RBRB L E, I – Member properties A clockwise moment M A is applied at A to produce a +ve bending in beam AB. Find  A and M B. Moment Distribution Method

73 Bina Nusantara University 73 Using method of consistent deformations L AA A MAMA B L f AA A B 1 Moment Distribution Method

74 Bina Nusantara University 74 Applying the principle of consistent deformation, Stiffness factor = k  = 4EI/L Moment Distribution Method

75 Bina Nusantara University 75 Moment Distribution Method Considering moment M B, M B + M A + R A L = 0  M B = M A /2= (1/2)M A Carry - over Factor = 1/2

76 Bina Nusantara University 76 Distribution Factor Distribution factor is the ratio according to which an externally applied unbalanced moment M at a joint is apportioned to the various members mating at the joint Moment Distribution Method

77 Bina Nusantara University 77 Distribution Factor moment M A C B D I1L1I1L1 I3L3I3L3 I2L2I2L2 A D B C M BA M BC M BD At joint B M - M BA -M BC -M BD = 0 M Moment Distribution Method

78 Bina Nusantara University 78 Distribution Factor Moment Distribution Method i.e.,M = M BA + M BC + M BD

79 Bina Nusantara University 79 Moment Distribution Method Modified Stiffness Factor The stiffness factor changes when the far end of the beam is simply- supported. AA MAMA AB RARA RBRB L

80 Bina Nusantara University 80 Moment Distribution Method Modified Stiffness Factor As per earlier equations for deformation, given in Mechanics of Solids text-books.

81 Bina Nusantara University 81 Moment Distribution Method Solution of that problems above Fixed end moments

82 Bina Nusantara University 82 Moment Distribution Method Solution of that problems above Stiffness Factors (Unmodified Stiffness)

83 Bina Nusantara University 83 Moment Distribution Method Solution of that problems above Distribution Factors

84 Bina Nusantara University 84 Moment Distribution Method Solution of that problems above Moment Distribution Table

85 Bina Nusantara University 85 Moment Distribution Method Solution of that problems above Computation of Shear Forces 8 m 3 m 8 m III 15 kN/m 10 kN/m 150 kN A BC D

86 Bina Nusantara University 86 Moment Distribution Method Solution of that problems above Computation of Shear Forces

87 Bina Nusantara University 87 Moment Distribution Method Solution of that problems above Shear Force and Bending Moment Diagrams m m S. F. D.

88 Bina Nusantara University 88 Moment Distribution Method Solution of that problems above Shear Force and Bending Moment Diagrams M max = kN.m Max= kN.m 3.74 m m B. M. D

89 Bina Nusantara University 89 Moment Distribution Method Solution of that problems above Simply-supported bending moments at center of span M center in AB = (15)(8) 2 /8 = +120 kN.m M center in BC = (150)(6)/4 = +225 kN.m M center in AB = (10)(8) 2 /8 = +80 kN.m

90 Bina Nusantara University 90 The section will discuss moment distribution method to analyze beams and frames composed of non prismatic members. Moment Distribution Method MOMENT DISTRIBUTION METHOD FOR NONPRISMATIC MEMBER

91 Bina Nusantara University 91 First the procedure to obtain the necessary carry- over factors, stiffness factors and fixed-end moments will be outlined. Then the use of values given in design tables will be illustrated. Finally the analysis of statically indeterminate structures using the moment distribution method will be outlined Moment Distribution Method MOMENT DISTRIBUTION METHOD FOR NONPRISMATIC MEMBER

92 Bina Nusantara University 92 Moment Distribution Method Stiffness and Carry-over Factors Use moment-area method to find the stiffness and carry -over factors of the non-prismatic beam.  A B PAPA MBMB MAMA AA C AB = Carry-over factor of moment M A from A to B

93 Bina Nusantara University 93 Moment Distribution Method Stiffness and Carry-over Factors A B M A =C BA M B =C BA K B MB(KB)MB(KB) M B =C AB M A =C AB K A MA(KA)MA(KA)  A (= 1.0) MAMA  B (= 1.0) A B (a) (b) MBMB

94 Bina Nusantara University 94 Use of Betti-Maxwell’s reciprocal theorem requires that the work done by loads in case (a) acting through displacements in case (b) is equal to work done by loads in case (b) acting through displacements in case (a) Moment Distribution Method

95 Bina Nusantara University 95

96 Bina Nusantara University 96

97 Bina Nusantara University 97 Moment Distribution Method

98 Bina Nusantara University 98 Moment Distribution Method

99 Bina Nusantara University 99 Moment Distribution Method

100 Bina Nusantara University 100 Moment Distribution Method

101 Bina Nusantara University 101 Moment Distribution Method

102 Bina Nusantara University 102 Moment Distribution Method

103 Bina Nusantara University 103 Moment Distribution Method

104 Bina Nusantara University 104 Moment Distribution Method

105 Bina Nusantara University 105 Stiffness-Factor Modification Moment Distribution Method

106 Bina Nusantara University 106 Moment Distribution Method

107 Bina Nusantara University 107 Moment Distribution Method

108 Bina Nusantara University 108 Moment Distribution Method

109 Bina Nusantara University 109 Moment Distribution Method

110 Bina Nusantara University 110 Moment Distribution Method

111 Bina Nusantara University 111 Moment Distribution Method

112 Bina Nusantara University 112 Moment Distribution Method

113 Bina Nusantara University 113 Moment Distribution Method

114 Bina Nusantara University 114 Moment Distribution Method

115 Bina Nusantara University 115 Moment Distribution Method

116 Bina Nusantara University 116 Moment Distribution Method

117 Bina Nusantara University 117 Moment Distribution Method

118 Bina Nusantara University 118 Moment Distribution Method

119 Bina Nusantara University 119 Moment Distribution Method

120 Bina Nusantara University 120 Moment Distribution Method

121 Bina Nusantara University 121 Moment Distribution Method

122 Bina Nusantara University 122 Moment Distribution Method

123 Bina Nusantara University 123 Moment Distribution Method

124 Bina Nusantara University 124 Moment Distribution Method

125 Bina Nusantara University 125 Moment Distribution Method

126 Bina Nusantara University 126 Symmetric Beam and Loading Moment Distribution Method

127 Bina Nusantara University 127 Symmetric Beam with Antisymmetric Loading Moment Distribution Method

128 Bina Nusantara University 128 Moment Distribution Method

129 Bina Nusantara University 129 Moment Distribution Method

130 Bina Nusantara University 130 Moment Distribution Method

131 Bina Nusantara University 131 Moment Distribution for frames: No sidesway Moment Distribution Method

132 Bina Nusantara University 132 Moment Distribution Method

133 Bina Nusantara University 133 Moment Distribution Method

134 Bina Nusantara University 134 Moment Distribution Method

135 Bina Nusantara University 135 Moment Distribution Method

136 Bina Nusantara University 136 Moment Distribution Method

137 Bina Nusantara University 137 Moment Distribution for frames: sidesway Moment Distribution Method

138 Bina Nusantara University 138 Moment Distribution Method

139 Bina Nusantara University 139 Moment Distribution Method

140 Bina Nusantara University 140 Moment Distribution Method

141 Bina Nusantara University 141 Moment Distribution Method

142 Bina Nusantara University 142 Moment Distribution Method

143 Bina Nusantara University 143 Moment Distribution Method

144 Bina Nusantara University 144 Moment Distribution Method

145 Bina Nusantara University 145 Moment Distribution Method

146 Bina Nusantara University 146 Moment Distribution Method

147 Bina Nusantara University 147 Moment Distribution Method

148 Bina Nusantara University 148 Matrix Analysis D= δ x P D=PL 3 / 3EI = δ x P  δ=L 3 / 3EI δ=Flexibility Coeff. P=K x D P=K x PL 3 / 3EI K=3EI / L 3 K=Stiffness Coeff. FLEXIBILITY AND STIFFNESS MATRICES : SINGLE CO-ORD. VTU Programme

149 Bina Nusantara University 149 Matrix Analysis FLEXIBILITY AND STIFFNESS MATRICES : SINGLE CO-ORD. D= ML / EI D= δ x M=ML / EI  δ=L / EI δ=Flexibility Coeff. M=K x D =K x ML / EI  K=EI / L K=Stiffness Coeff. δ X K= 1 VTU Programme

150 Bina Nusantara University 150 Matrix Analysis FLEXIBILITY AND STIFFNESS MATRICES : TWO CO- ORDINATE SYSTEM VTU Programme

151 Bina Nusantara University 151 Matrix Analysis FLEXIBILITY AND STIFFNESS MATRICES : TWO CO- ORDINATE SYSTEM δ 11 =L 3 / 3EI δ 21 =L 2 / 2EI δ 12 =δ 21 =L 2 / 2EI δ 22 =L / EI Unit Force At Co-ord.(1) Unit Force At Co-ord.(2) VTU Programme

152 Bina Nusantara University 152 STIFFNESS MATRIX Unit Displacement at (1) K 11 =12EI / L 3 K 21 = – 6EI / L 2 Forces at Co-ord.(1) & (2) Unit Displacement at (2) Forces at Co-ord.(1) & (2) K 12 = – 6EI / L 2 K 22 =4EI / L =  P 1 =K 11 D 1 +K 12 D 2 P 2 =K 21 D 1 +K 22 D 2 P1P2P1P2 K 11 K 12 K 21 K 22 D1D2D1D2 K = 12EI / L 3 – 6EI / L 2 4EI / L Matrix Analysis VTU Programme

153 Bina Nusantara University 153 Develop the Flexibility and stiffness matrices for frame ABCD with reference to Coordinates shown The Flexibility matrix can be developed by applying unit force successively at coordinates (1),(2) &(3) and evaluating the displacements at all the coordinates δ ij = ∫ mi mj / EI x ds δ ij =displacement at I due to unit load at j Matrix Analysis VTU Programme

154 Bina Nusantara University 154 Unit Load at (1) Unit Load at (2) Unit Load at (3) PortionDCCBBA II4I OriginDCB Limits m1m1 x55 - x m2m2 0x10 m3m3 - 1 Matrix Analysis VTU Programme

155 Bina Nusantara University 155 δ 11 = ∫ m 1.m 1 dx / EI = 125 / EI δ 21 = δ 12 = ∫ m 1.m 2 dx / EI = 125 / 2EI δ 31 = δ 13 = ∫ m 1.m 3 dx / EI = - 25 / EI δ 22 = ∫ m 2.m 2 dx / EI = 1000 / 3EI δ 23 = δ 32 = ∫ m 2.m 3 dx / EI = - 75 / 2EI δ 33 = ∫ m 3.m 3 dx / EI = 10 / EI  δ = 1 / 6EI Matrix Analysis VTU Programme

156 Bina Nusantara University 156 INVERSING THE FLEXIBILITY MATRIX [ δ ] THE STIFENESS MATRIX [ K ] CAN BE OBTAINED K = EI Matrix Analysis VTU Programme


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