Module 2. Analysis of Statically Indeterminate Structures by the Matrix Force Method. Version 2 CE IIT, Kharagpur

Transcription

1 Module Anlysis of Stticlly Indeterminte Structures by the Mtrix Force Method Version CE IIT, Khrgpur

2 esson 8 The Force Method of Anlysis: Bems Version CE IIT, Khrgpur

3 Instructionl Objectives After reding this chpter the student will be ble to. Solve stticlly indeterminte bems of degree more thn one.. To solve the problem in mtrix nottion.. To compute rections t ll the supports.. To compute internl resisting bending moment t ny section of the continuous bem. 8. Introduction In the lst lesson, generl introduction to the force method of nlysis is given. Only, bems, which re stticlly indeterminte to first degree, were considered. If the structure is stticlly indeterminte to degree more thn one, then the pproch presented in the previous exmple needs to be orgnized properly. In the present lesson, generl procedure for nlyzing stticlly indeterminte bems is discussed. 8. Formliztion of Procedure Towrds this end, consider two-spn continuous bem s shown in Fig. 8.. The flexurl rigidity of this continuous bem is ssumed to be constnt nd is tken s EI. Since, the bem is stticlly indeterminte to second degree, it is required to identify two redundnt rection components, which need be relesed to mke the bem stticlly determinte. Version CE IIT, Khrgpur

4 Version CE IIT, Khrgpur

5 The redundnt rections t A nd B re denoted by nd respectively. The relesed structure (stticlly determinte structure) with pplied loding is shown in Fig. 8.b. The deflection of primry structure t B nd C due to respectively. Throughout this pplied loding is denoted by ( ) ( ) i nd ( ) th module nottion is used to denote deflection t i redundnt due to pplied lods on the determinte structure. w 7P (8.) 8EI EI ( ) 7w 7P (8.b) EI ( ) In fct, the subscript nd represent, loctions of redundnt rections relesed. In the present cse A ( ) nd B ( ) respectively. In the present nd subsequent lessons of this module, the deflections nd the rections re tken to be positive in the upwrd direction. However, it should be kept in mind tht the positive sense of the redundnt cn be chosen rbitrrily. The deflection of the point of ppliction of the redundnt should likewise be considered positive when cting in the sme sense. For writing comptibility equtions t B nd C, it is required to know deflection of the relesed structure t B ndc due to externl loding nd due to redundnts. The deflection t B ndc due to externl loding cn be computed esily. Since redundnts nd re not known, in the first step pply unit lod in the direction of nd compute deflection, t B, nd deflection, t C, s shown in Fig.8.c. Now deflections t B nd C of the given relesed structure due to redundnt re, ( ) (8.) ( ) (8.b) In the second step, pply unit lod in the direction of redundnt nd compute deflection t B (point ), nd deflection t C, s shown in Fig 8.d. It my be reclled tht the flexibility coefficient is the deflection t i due to unit vlue of force pplied t j. Now deflections of the primry structure (relesed structure) t B nd C due to redundnt is ( ) ij (8.) ( ) (8.b) Version CE IIT, Khrgpur

6 It is observed tht, in the ctul structure, the deflections t joints B nd C is zero. Now the totl deflections t B nd C of the primry structure due to pplied externl loding nd redundnts nd is, ( ) + + (8.) ( ) + + (8.b) The eqution (8.) represents the totl displcement t B nd is obtined by superposition of three terms: ) Deflection t B due to ctul lod cting on the stticlly determinte structure, ) Displcement t B due to the redundnt rection cting in the positive direction t B (point ) nd ) Displcement t B due to the redundnt rection cting in the positive direction t C. The second eqution (8.b) similrly represents the totl deflection t C. From the physics of the problem, the comptibility condition cn be written s, ( ) (8.5) ( ) (8.5b) The eqution (8.5) nd (8.5b) my be written in mtrix nottion s follows, ( ) 0 + ( ) 0 (8.6) {( ) } + [ A]{ } { 0} (8.6b) In which, {( ) } ( ) ( ) ; [ A] nd { } Solving the bove set of lgebric equtions, one could obtin the vlues of redundnts, nd. { } [ A] { } (8.7) Version CE IIT, Khrgpur

7 In the bove eqution the vectors { } contins the displcement vlues of the primry structure t point nd, [ A ] is the flexibility mtrix nd { } is column vector of redundnts required to be evluted. In eqution (8.7) the inverse of the flexibility mtrix is denoted by [ A ]. In the bove exmple, the structure is indeterminte to second degree nd the size of flexibility mtrix is. In generl, if the structure is redundnt to degree n, then the flexibility mtrix is of the order n n. To demonstrte the procedure to evlute deflection, consider the problem given in Fig. 8., with loding s given below w w ; P w (8.8) ( ) Now, the deflection nd ( ) of the relesed structure cn be evluted from the equtions (8.) nd (8.b) respectively. Then, w 7w 7w (8.8b) 8EI EI EI ( ) 7w 7w 95w (8.8c) EI 8EI ( ) The negtive sign indictes tht both deflections re downwrds. Hence the vector { } is given by w (8.8d) 8EI 95 { } The flexibility mtrix is determined from referring to figures 8.c nd 8.d. Thus, when the unit lod corresponding to is cting t B, the deflections re, 5, (8.8e) EI Similrly when the unit lod is cting t C, 5 8, (8.8f) EI The flexibility mtrix cn be written s, 5 A (8.8g) 5 6 [ ] Version CE IIT, Khrgpur

8 The inverse of the flexibility mtrix cn be evluted by ny of the stndrd method. Thus, 6 5 A (8.8h) 7 5 [ ] Now using eqution (8.7) the redundnts re evluted. Thus, w 7 8EI Hence, 69 w nd w (8.8i) Once the redundnts re evluted, the other rection components cn be evluted by sttic equtions of equilibrium. Exmple 8. Clculte the support rections in the continuous bem ABC shown in Fig. 8.. Assume EI to be constnt throughout. due to loding s Version CE IIT, Khrgpur

9 B ( ) ( ) Select two rections viz, t nd C s redundnts, since the given bem is stticlly indeterminte to second degree. In this cse the primry structure is cntilever bem AC. The primry structure with given loding is shown in Fig. 8.b. In the present cse, the deflections ( ), nd ( ) of the relesed structure t B nd C cn be redily clculted by moment-re method. Thus, 89.6 EI.875 () EI ( ) nd ( ) For the present problem the flexibility mtrix is, 5 EI 65 6 EI EI () In the ctul problem the displcements t B ndc re zero. Thus the comptibility conditions for the problem my be written s, ( ) ( ) () EI EI.875 (5) Substituting the vlue of E nd I in the bove eqution, Using equtions of sttic equilibrium, 0.609kN nd.60 kn 0.77 kn nd kn.m Version CE IIT, Khrgpur

10 Exmple 8. A clmped bem AB of constnt flexurl rigidity is shown in Fig. 8.. The bem is subjected to uniform distributed lod of w kn/m nd centrl concentrted moment M w force method. kn.m. Drw sher force nd bending moment digrms by ( ) ( ) Select verticl rection nd the support moment t B s the redundnts. The primry structure in this cse is cntilever bem which could be obtined by relesing the redundnts nd. The is ssumed to be positive in the upwrd direction nd is ssumed to be positive in the counterclockwise direction. Now, clculte deflection t B due to only pplied loding. et be the trnsverse deflection t be the slope t B ( ) B nd ( ) due to externl loding. The positive directions of the selected redundnts re shown in Fig. 8.b. Version CE IIT, Khrgpur

11 Version CE IIT, Khrgpur

12 ( ) ( ) The deflection nd of the relesed structure cn be evluted from unit lod method. Thus, w w w () 8EI 8EI EI ( ) nd ( ) w w w () EI EI The negtive sign indictes tht ( ) clockwise. Hence the vector { } { } is given by ( ) is downwrds nd rottion is w () The flexibility mtrix is evluted by first pplying unit lod long redundnt nd determining the deflections nd corresponding to redundnts nd respectively (see Fig. 8.d). Thus, nd EI () EI Similrly, pplying unit lod in the direction of redundnt flexibility coefficients nd s shown in Fig. 8.c., one could evlute nd EI (5) EI Now the flexibility mtrix is formulted s, [ ] The inverse of flexibility mtrix is formulted s, A (6) 6 Version CE IIT, Khrgpur

13 6 [ A ] The redundnts re evluted from eqution (8.7). Hence, 6 w w 6 w w nd (7) The other two rections ( nd ) cn be evluted by equtions of sttics. Thus, w M A nd A w (8) 6 The bending moment nd sher force digrms re shown in Fig. 8.g nd Fig.8.h respectively. Summry In this lesson, stticlly indeterminte bems of degree more thn one is solved systemticlly using flexibility mtrix method. Towrds this end mtrix nottion is dopted. Few illustrtive exmples re solved to illustrte the procedure. After nlyzing the continuous bem, rections re clculted nd bending moment digrms re drwn. Version CE IIT, Khrgpur