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Analysis of Indeterminate Structure Session 05-22

Analysis of Indeterminate Structure Session 05-22. Matakuliah : S0725 – Analisa Struktur Tahun : 2009. Contents. 3 Equations Method Flexibility Method Slope Deflection Method Cross Method/ Moment Distribution Method Matrix Analysis. 3 Equation Method.

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Analysis of Indeterminate Structure Session 05-22

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

  2. Contents 3 Equations Method Flexibility Method Slope Deflection Method Cross Method/ Moment Distribution Method Matrix Analysis

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

  4. P1 q1 P2 q2 A E B C D Gambar 1.1 . Struktur statis tak tentu 3 Equation Method Degree of indeterminacy i = r – 3 i = 7 – 3 = 4

  5. jokl jokr celah P1 q1 P2 jokr jokl k - 1 k + 1 ll k lk Gambar 1.2.a. Putran sudut akibat beban luar 3 Equation Method Principles : 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 )

  6. Gambar 1.2.b. Momen lentur negatif Mk Mk-1 Mk+1 j'kl j'kr k - 1 k + 1 ll k lk 3 Equation Method Princiles : 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 )

  7. 3 Equation Method Compatibility eq : Sebagai suatu syarat kompatibilitas , joint ‘k’ merupakan rotasi yang kompatibel, sehingga persamaan kompatibilitasnya menjadi jokl + jokr = j’kl + j’kr Dimana : jo danj’ 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.

  8. 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 : ( jokl + jokr) + (j’kl + j’kr ) = 0 Sehingga hasil-hasil momen peralihan sudah langsung berikut tandanya menyatakan MOMEN LENTUR SEBENARNYA , jika hasilnya negatif , berarti bekerja sebagai momen lentur negatif.

  9. 3 Equation Method Karena nilaij’kltergantung padaMk-1danMk nilaij’krtergantung padaMkdanMk+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

  10. 3 Equation Method Dengan mennjau nilai putaran sudut akibat momen – momen peralihan persamaan kompatibilitas dapat dituliskan sebagai  j’kl= Mk ( ll / 3EI ) + Mk-1 ( ll / 6EI ) j’kr= Mk ( lr / 3EI ) + Mk+1 (lr / 6EI ) + j’kl + j’kr = Mk-1 ( ll / 6EI )+Mk {( ll / 3EI )+( lr / 3EI )}+ Mk+1 (lr / 6EI )

  11. 3 Equation Method Sehingga persamaan 3 momen pada tumpuan k ... Mk-1 ( ll / 6EI )+Mk {( ll / 3EI )+( lr / 3EI )}+ Mk+1 (lr / 6EI ) + ( jokl + jokr) = 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 ).

  12. 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 )

  13. M 3 Equation Method Gambar 1.3. Struktur dengan perletakan Jepit yang diubah  Sendi + Momen jepit

  14. Method Analysis While analyzing indeterminate structures, it is necessary to satisfy (force) equilibrium, (displacement) compatibility and force-displacement relationships

  15. Method Analysis (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

  16. Method Analysis (a) Force equilibrium is satisfied when the reactive forces hold the structure in stable equilibrium, as the structure is subjected to external loads

  17. Method Analysis (b) Displacement compatibility is satisfied when the various segments of the structure fit together without intentional breaks, or overlaps

  18. Method Analysis (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

  19. Method Analysis 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

  20. Method Analysis Force Methodsatisfies 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

  21. Method Analysis 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

  22. Method Analysis

  23. Flexibility Method 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 Methodsatisfies 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

  24. Flexibility Method FORCED METHOD This method is also known as flexibility method or compatibility method. In this method the degree of static indeterminacy of the structure is determined and the redundants are identified. VTU Programme

  25. Flexibility Method FORCED METHOD A coordinate is assigned to each redundant. Thus,P1, P2 - - - - - -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 VTU Programme

  26. Flexibility Method 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; VTU Programme

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

  28. Flexibility Method FORCED METHOD The above equations may be return as [∆] = [∆L] + [∆R] - - - - (1) ∆1 = ∆1L + δ11 P1 + δ12 P2 + - - - - - δ1nPn ∆2 = ∆2L + δ21 P1 + δ22 P2 + - - - - - δ2nPn | | | | | | | | | | - - - - - (2) ∆n = ∆nL + δn1 P1 + δn2 P2 + - - - - - δnnPn VTU Programme

  29. Flexibility Method FORCED METHOD ∆ = [∆ L] + [δ] [P] - - - - - - (3) • [P]= [δ]-1{[∆] – [∆ L]} - - - - - - (4) If the net displacements at the redundants are zero then ∆1, ∆2 - - - - ∆n =0, Then [P] = - [δ] -1 [∆ L] - - - - - -(5) The redundants P1,P2, - - - - - Pn are Thus determined VTU Programme

  30. 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 EIab and length Lab are:

  31. Slope Deflection • Slope deflection equations where θa, θb 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.

  32. Slope Deflection • Slope deflection equations The slope deflection equations can also be written using the stiffness factor . and the chord rotation

  33. Slope Deflection

  34. Slope Deflection • Slope deflection equations When a simple beam of length Lab and flexural rigidity EIab is loaded at each end with clockwise moments Mab and Mba, 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

  35. Slope Deflection

  36. 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,

  37. Slope Deflection Method FORCED METHOD ∆ = [∆ L] + [δ] [P] - - - - - - (3) • [P]= [δ]-1{[∆] – [∆ L]} - - - - - - (4) If the net displacements at the redundants are zero then ∆1, ∆2 - - - - ∆n =0, Then [P] = - [δ] -1 [∆ L] - - - - - -(5) The redundants P1,P2, - - - - - Pn are Thus determined VTU Programme

  38. Slope Deflection Method 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.

  39. A A B  = B  = rigid body motion = /L  (i) Due to externally applied loads FEMBA FEMAB + (ii) Due to rotation A at support A MBA=(2EIA)/L A MAB=(4EIA)/L Slope Deflection Method

  40. + B A B (iii) Due to rotation B at support B MBA=(4EIB)/L MAB=(2EIB)/L L + MAB=(-6EI)/L2 MBA=(-6EI)/L2 A B (iv) Due to differential settlement of  (between A and B)  L Slope Deflection Method

  41. Slope Deflection Method

  42. A  = B (FEM)BA (FEM)AB + (FEM)BA (FEM)BA/2 Slope Deflection Method

  43. + A MAB=(3EIA)/L +  MAB=(3EI)/L2  = PL3/(3EI), M = PL = (3EI/L3)(L) = 3EI/L2 MAB = [(FEM)AB - (FEM)BA/2]+(3EIA)/L -(3EI)/L2 Modified FEM at end A Slope Deflection Method

  44. Slope Deflection Method

  45. Slope Deflection Method

  46. Slope Deflection Method

  47. Slope Deflection Method 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 θAθB.

  48. Slope Deflection Method

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

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