M2l8

511 views

Published on

this is useful for structural students

Published in: Design
0 Comments
0 Likes
Statistics
Notes
  • Be the first to comment

  • Be the first to like this

No Downloads
Views
Total views
511
On SlideShare
0
From Embeds
0
Number of Embeds
1
Actions
Shares
0
Downloads
15
Comments
0
Likes
0
Embeds 0
No embeds

No notes for slide

M2l8

  1. 1. Module 2Analysis of Statically Indeterminate Structures by theMatrix Force Method Version 2 CE IIT, Kharagpur
  2. 2. Lesson 8The Force Method of Analysis: Beams Version 2 CE IIT, Kharagpur
  3. 3. Instructional ObjectivesAfter reading this chapter the student will be able to1. Solve statically indeterminate beams of degree more than one.2. To solve the problem in matrix notation.3. To compute reactions at all the supports.4. To compute internal resisting bending moment at any section of the continuous beam.8.1 IntroductionIn the last lesson, a general introduction to the force method of analysis is given.Only, beams, which are statically indeterminate to first degree, were considered.If the structure is statically indeterminate to a degree more than one, then theapproach presented in the previous example needs to be organized properly. Inthe present lesson, a general procedure for analyzing statically indeterminatebeams is discussed.8.2 Formalization of ProcedureTowards this end, consider a two-span continuous beam as shown in Fig. 8.1a.The flexural rigidity of this continuous beam is assumed to be constant and istaken as EI . Since, the beam is statically indeterminate to second degree, it isrequired to identify two redundant reaction components, which need be releasedto make the beam statically determinate. Version 2 CE IIT, Kharagpur
  4. 4. Version 2 CE IIT, Kharagpur
  5. 5. The redundant reactions at A and B are denoted by R1 and R2 respectively.The released structure (statically determinate structure) with applied loading isshown in Fig. 8.1b. The deflection of primary structure at B and C due toapplied loading is denoted by (Δ L )1 and (Δ L )2 respectively. Throughout thismodule (Δ L )i notation is used to denote deflection at i th redundant due toapplied loads on the determinate structure. 4 7 PL3 (Δ L )1 = − wL − (8.1a) 8 EI 12 EI 4 27 PL3 (Δ L )2 = − 7 wL − (8.1b) 24 EI 16 EIIn fact, the subscript 1 and 2 represent, locations of redundant reactionsreleased. In the present case R A (= R1 ) and RB (= R2 ) respectively. In the presentand subsequent lessons of this module, the deflections and the reactions aretaken to be positive in the upward direction. However, it should be kept in mindthat the positive sense of the redundant can be chosen arbitrarily. The deflectionof the point of application of the redundant should likewise be considered positivewhen acting in the same sense.For writing compatibility equations at B and C , it is required to know deflection ofthe released structure at B and C due to external loading and due to redundants.The deflection at B and C due to external loading can be computed easily. Sinceredundants R1 and R2 are not known, in the first step apply a unit load in thedirection of R1 and compute deflection, a11 at B , and deflection, a 21 at C , asshown in Fig.8.1c. Now deflections at B and C of the given released structuredue to redundant R1 are, (Δ R )11 = a11 R1 (8.2a) (Δ R )21 = a 21 R1 (8.2b)In the second step, apply unit load in the direction of redundant R2 and computedeflection at B (point 1), a12 and deflection at C , a 22 as shown in Fig 8.1d. It maybe recalled that the flexibility coefficient a ij is the deflection at i due to unit valueof force applied at j . Now deflections of the primary structure (releasedstructure) at B and C due to redundant R2 is (Δ R )12 = a12 R2 (8.3a) (Δ R )22 = a 22 R2 (8.3b) Version 2 CE IIT, Kharagpur
  6. 6. It is observed that, in the actual structure, the deflections at joints B and C iszero. Now the total deflections at B and C of the primary structure due to appliedexternal loading and redundants R1 and R2 is, Δ 1 = (Δ L )1 + a11 R1 + a12 R2 (8.4a) Δ 2 = (Δ L )2 + a 21 R1 + a 22 R2 (8.4b)The equation (8.4a) represents the total displacement at B and is obtained bysuperposition of three terms: 1) Deflection at B due to actual load acting on the statically determinate structure, 2) Displacement at B due to the redundant reaction R1 acting in the positive direction at B (point 1) and 3) Displacement at B due to the redundant reaction R2 acting in the positive direction at C .The second equation (8.4b) similarly represents the total deflection at C . Fromthe physics of the problem, the compatibility condition can be written as, Δ 1 = (Δ L )1 + a11 R1 + a12 R2 = 0 (8.5a) Δ 2 = (Δ L )2 + a 21 R1 + a 22 R2 = 0 (8.5b)The equation (8.5a) and (8.5b) may be written in matrix notation as follows, ⎧ (Δ L )1 ⎫ ⎡ a11 a12 ⎤ ⎧ R1 ⎫ ⎧0⎫ ⎨ ⎬+ ⎨ ⎬=⎨ ⎬ (Δ L )2 ⎭ ⎢a 21 a 22 ⎥ ⎩ R2 ⎭ ⎩0⎭ (8.6a) ⎩ ⎣ ⎦ {(Δ L )1 } + [A]{R} = {0} (8.6b)In which, ⎧( Δ ) ⎫ ⎡a a12 ⎤ ⎧ R1 ⎫ {( Δ ) } = ⎪( Δ ) ⎪ ; [ A] = ⎢a L 1 ⎨ L 1 ⎬ 11 ⎥ and {R} = ⎨ R ⎬ ⎪ ⎩ L 2 ⎪ ⎭ ⎣ 21 a22 ⎦ ⎩ 2⎭Solving the above set of algebraic equations, one could obtain the values ofredundants, R1 and R2 . {R} = −[A]−1 {Δ L } (8.7) Version 2 CE IIT, Kharagpur
  7. 7. In the above equation the vectors {Δ L } contains the displacement values of theprimary structure at point 1 and 2, [A] is the flexibility matrix and {R} is columnvector of redundants required to be evaluated. In equation (8.7) the inverse of theflexibility matrix is denoted by [A] . In the above example, the structure is −1indeterminate to second degree and the size of flexibility matrix is 2 × 2 . Ingeneral, if the structure is redundant to a degree n , then the flexibility matrix is ofthe order n × n . To demonstrate the procedure to evaluate deflection, considerthe problem given in Fig. 8.1a, with loading as given below w=w; P = wL (8.8a)Now, the deflection (Δ L )1 and (Δ L )2 of the released structure can be evaluatedfrom the equations (8.1a) and (8.1b) respectively. Then, 4 7 wL4 17 wL4 (Δ L )1 = − wL − =− (8.8b) 8 EI 12 EI 24 EI 7 wL4 27 wL4 95wL4 (Δ L )2 =− − =− (8.8c) 24 EI 16 EI 48EIThe negative sign indicates that both deflections are downwards. Hence thevector {Δ L } is given by wL4 ⎧34⎫ {Δ L } = − ⎨ ⎬ (8.8d) 48 EI ⎩95⎭The flexibility matrix is determined from referring to figures 8.1c and 8.1d. Thus,when the unit load corresponding to R1 is acting at B , the deflections are, L3 5 L3 a11 = , a 21 = (8.8e) 3EI 6 EISimilarly when the unit load is acting at C , 5L3 8L3 a12 = , a 22 = (8.8f) 6 EI 3EIThe flexibility matrix can be written as, L3 ⎡2 5 ⎤ [A] = ⎢5 16⎥ (8.8g) 6 EI ⎣ ⎦ Version 2 CE IIT, Kharagpur
  8. 8. The inverse of the flexibility matrix can be evaluated by any of the standardmethod. Thus, 16 − 5⎤ [A]−1 = 6 EI ⎡ 3 ⎢ ⎥ (8.8h) 7 L ⎣− 5 2 ⎦Now using equation (8.7) the redundants are evaluated. Thus, ⎧ R1 ⎫ 6 EI wL4 ⎡ 16 − 5⎤ ⎧34⎫ ⎨ ⎬= 3 × ⎢− 5 2 ⎥ ⎨95⎬ ⎩ R2 ⎭ 7 L 48 EI ⎣ ⎦⎩ ⎭ 69 20Hence, R1 = wL and R2 = wL (8.8i) 56 56Once the redundants are evaluated, the other reaction components can beevaluated by static equations of equilibrium.Example 8.1Calculate the support reactions in the continuous beam ABC due to loading asshown in Fig. 8.2a. Assume EI to be constant throughout. Version 2 CE IIT, Kharagpur
  9. 9. Select two reactions viz, at B(R1 ) and C (R2 ) as redundants, since the given beamis statically indeterminate to second degree. In this case the primary structure isa cantilever beam AC . The primary structure with a given loading is shown in Fig.8.2b.In the present case, the deflections (Δ L )1 , and (Δ L )2 of the released structure at B and C can be readily calculated by moment-area method. Thus, (Δ L )1 = − 819.16 EIand (Δ L )2 =− 2311.875 (1) EIFor the present problem the flexibility matrix is, 125 625 a11 = a 21 = 3EI 6 EI 625 1000 a12 = a 22 = (2) 6 EI 3EIIn the actual problem the displacements at B and C are zero. Thus thecompatibility conditions for the problem may be written as, a11 R1 + a12 R2 + (Δ L )1 = 0 (3) a 21 R1 + a 22 R2 + (Δ L )2 = 0 ⎧ R1 ⎫ 3EI ⎡ 1000 − 312.5⎤ 1 ⎧ 819.16 ⎫ ⎨ ⎬= ⎢ × ⎨ ⎬ 125 ⎥ EI (5) ⎩ R2 ⎭ 27343.75 ⎣− 312.5 ⎦ ⎩2311.875⎭Substituting the value of E and I in the above equation, R1 = 10.609kN and R2 = 3.620 kNUsing equations of static equilibrium, R3 = 0.771 kN and R4 = −0.755 kN.m Version 2 CE IIT, Kharagpur
  10. 10. Example 8.2A clamped beam AB of constant flexural rigidity is shown in Fig. 8.3a. The beamis subjected to a uniform distributed load of w kN/m and a central concentratedmoment M = wL2 kN.m . Draw shear force and bending moment diagrams byforce method.Select vertical reaction (R1 ) and the support moment (R2 ) at B as theredundants. The primary structure in this case is a cantilever beam which couldbe obtained by releasing the redundants R1 and R2 . The R1 is assumed to bepositive in the upward direction and R2 is assumed to be positive in thecounterclockwise direction. Now, calculate deflection at B due to only appliedloading. Let (Δ L )1 be the transverse deflection at B and (Δ L )2 be the slope at Bdue to external loading. The positive directions of the selected redundants areshown in Fig. 8.3b. Version 2 CE IIT, Kharagpur
  11. 11. Version 2 CE IIT, Kharagpur
  12. 12. The deflection (Δ L )1 and (Δ L )2 of the released structure can be evaluated fromunit load method. Thus, 4 3wL4 wL4 (Δ L )1 = − wL − =− (1) 8EI 8EI 2 EI 3 wL3 2 wL3and (Δ L )2 = − wL − =− (2) 6 EI 2 EI 3EIThe negative sign indicates that (Δ L )1 is downwards and rotation (Δ L )2 isclockwise. Hence the vector {Δ L } is given by wL3 ⎧3L ⎫ {Δ L } = − ⎨ ⎬ (3) 6 EI ⎩4⎭The flexibility matrix is evaluated by first applying unit load along redundant R1and determining the deflections a11 and a 21 corresponding to redundants R1 andR2 respectively (see Fig. 8.3d). Thus, L3 L2 a11 = and a 21 = (4) 3EI 2 EISimilarly, applying unit load in the direction of redundant R2 , one could evaluateflexibility coefficients a12 and a 22 as shown in Fig. 8.3c. L2 L a12 = and a 22 = (5) 2 EI EINow the flexibility matrix is formulated as, L ⎡2 L2 3L ⎤ [A] = ⎢ ⎥ (6) 6 EI ⎣ 3L 6⎦The inverse of flexibility matrix is formulated as, Version 2 CE IIT, Kharagpur
  13. 13. − 3L ⎤ [A]−1 = 6 EI ⎡ 6 3 ⎢ ⎥ 3L ⎣− 3L 2 L2 ⎦The redundants are evaluated from equation (8.7). Hence, ⎧ R1 ⎫ 6 EI ⎡ 6 − 3L ⎤ ⎛ wL3 ⎞⎧3L ⎫ ⎨ ⎬ =− 3 ⎢ ⎥ ×⎜− ⎟⎨ ⎬ ⎩ R2 ⎭ 3L ⎣− 3L 2 L2 ⎦ ⎜ 6 EI ⎟⎩ 4 ⎭ ⎝ ⎠ w ⎧ 6L ⎫ = ⎨ ⎬ 3 ⎩− L2 ⎭ wL2 R1 = 2wL and R2 = − (7) 3The other two reactions ( R3 and R4 ) can be evaluated by equations of statics.Thus, wL2 R4 = M A = − and R1 = R A = − wL (8) 6The bending moment and shear force diagrams are shown in Fig. 8.3g andFig.8.3h respectively.SummaryIn this lesson, statically indeterminate beams of degree more than one is solvedsystematically using flexibility matrix method. Towards this end matrix notation isadopted. Few illustrative examples are solved to illustrate the procedure. Afteranalyzing the continuous beam, reactions are calculated and bending momentdiagrams are drawn. Version 2 CE IIT, Kharagpur

×