Module 4. Analysis of Statically Indeterminate Structures by the Direct Stiffness Method. Version 2 CE IIT, Kharagpur

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1 Module Analysis of Statically Indeterminate Structures by the Direct Stiffness Method Version CE IIT, Kharagur

2 Lesson 9 The Direct Stiffness Method: Beams (Continued) Version CE IIT, Kharagur

3 Instructional Objectives After reading this chater the student will be able to. Comute moments develoed in the continuous beam due to suort settlements.. Comute moments develoed in statically indeterminate beams due to temerature changes.. Analyse continuous beam subjected to temerature changes and suort settlements. 9. Introduction In the last two lessons, the analysis of continuous beam by direct stiffness matrix method is discussed. It is assumed in the analysis that the suorts are unyielding and the temerature is maintained constant. However, suort settlements can never be revented altogether and hence it is necessary to make rovisions in design for future unequal vertical settlements of suorts and robable rotations of fixed suorts. The effect of temerature changes and suort settlements can easily be incororated in the direct stiffness method and is discussed in this lesson. Both temerature changes and suort settlements induce fixed end actions in the restrained beams. These fixed end forces are handled in the same way as those due to loads on the members in the analysis. In other words, the global load vector is formulated by considering fixed end actions due to both suort settlements and external loads. At the end, a few roblems are solved to illustrate the rocedure. 9. Suort settlements Consider continuous beam ABC as shown in ig. 9.a. Assume that the flexural rigidity of the continuous beam is constant throughout. Let the suort B settles by an amount Δ as shown in the figure. The fixed end actions due to loads are shown in ig. 9.b. The suort settlements also induce fixed end actions and are shown in ig. 9.c. In ig. 9.d, the equivalent joint loads are shown. Since the beam is restrained against dislacement in ig. 9.b and ig. 9.c, the dislacements roduced in the beam by the joint loads in ig. 9.d must be equal to the dislacement roduced in the beam by the actual loads in ig. 9.a. Thus to incororate the effect of suort settlement in the analysis it is required to modify the load vector by considering the negative of the fixed end actions acting on the restrained beam. Version CE IIT, Kharagur

4 Version CE IIT, Kharagur

5 9. Effect of temerature change The effect of temerature on the statically indeterminate beams has already been discussed in lesson 9 of module in connection with the flexibility matrix method. Consider the continuous beam ABC as shown in ig. 9.a, in which san BC is subjected to a differential temerature T at to and T at the bottom of the beam. Let temerature in san AB be constant. Let d be the deth of beam and EI be the flexural rigidity. As the cross section of the member remains lane after bending, the relative angle of rotation dθ between two cross sections at a distance dx aart is given by ( T T ) dx dθ α (9.) d where α is the co-efficient of the thermal exansion of the material. When beam is restrained, the temerature change induces fixed end moments in the beam as shown in ig. 9.b. The fixed end moments develoed are, M ( T T ) T M α EI (9.) d T Corresonding to the above fixed end moments; the equivalent joint loads can easily be constructed. Also due to differential temeratures there will not be any vertical forces/reactions in the beam. Version CE IIT, Kharagur

6 Examle 9. Calculate suort reactions in the continuous beam ABC (vide ig. 9.a) having constant flexural rigidity EI, throughout due to vertical settlement of suort B, by mm as shown in the figure. Assume E GPa and I m. Version CE IIT, Kharagur

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8 The continuous beam considered is divided into two beam elements. The numbering of the joints and members are shown in ig. 9.b. The ossible global degrees of freedom are also shown in the figure. A tyical beam element with two degrees of freedom at each node is also shown in the figure. or this roblem, the unconstrained degrees of freedom are u and u. The fixed end actions due to suort settlement are, M AB EIΔ 9 kn.m; M 9 kn.m BA L M BC 9 kn.m ; M 9 kn.m () CB The fixed-end moments due to suort settlements are shown in ig. 9.c. The equivalent joint loads due to suort settlement are shown in ig. 9.d. In the next ste, let us construct member stiffness matrix for each member. Member : L m, node oints -. Global d. o. f [ k' ] EI () Member : L m, node oints -. Global d. o. f [ k ] EI () On the member stiffness matrix, the corresonding global degrees of freedom are indicated to facilitate assembling. The assembled global stiffness matrix is of order K is given by,. Assembled stiffness matrix [ ] Version CE IIT, Kharagur

9 K EI () [ ] Thus the global load vector corresonding to unconstrained degrees of freedom is, k () 9 { } Thus the load dislacement relation for the entire continuous beam is, 9 EI u u.9 u u. u.9 u () Since, u u u u due to suort conditions. We get, 9 EI....8 u u Thus solving for unknowns uand u, u u. EI EI u.9 radians; u.7 radians (7) Now, unknown joint loads are calculated by, Version CE IIT, Kharagur

10 EI..... EI.8 7. (8) Now the actual suort reactions,, and must include the fixed end suort reactions. Thus, (9).88 kn;.7 kn; 8.9 kn.m;.7 kn () Examle 9. A continuous beam ABCD is carrying a uniformly distributed load of kn / m as shown in ig. 9.a. Comute reactions due to following suort settlements. Suort B Suort C.m vertically downwards..m vertically downwards. Assume E GPa and I m. Version CE IIT, Kharagur

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12 Solution The node and member numbering are shown in ig. 9.(b), wherein the continuous beam is divided into three beam elements. It is observed from the figure that the unconstrained degrees of freedom are u and u. The fixed end actions due to suort settlements are shown in ig. 9.(c). and fixed end moments due to external loads are shown in ig. 9.(d). The equivalent joint loads due to suort settlement and external loading are shown in ig. 9.(e). The fixed end actions due to suort settlement are, M A EI ( ψ ) where ψ is the chord rotation and is taken + ve if the L rotation is counterclockwise. Substituting the aroriate values in the above equation, M A 9. 9 kn.m. M kn.m. B M kn.m. C M 9 kn.m. () D The vertical reactions are calculated from equations of equilibrium. The fixed end actions due to external loading are, w L M A. kn.m. M.. kn.m. B M C M. kn.m. () D In the next ste, construct member stiffness matrix for each member. Member, L m, node oints -. Version CE IIT, Kharagur

13 Global d. o. f [ k' ] EI () Member, L m, node oints -. Global d. o. f [ k ] EI () Member, L m, node oints -. Global d. o. f [ k ] EI () On the member stiffness matrix, the corresonding global degrees of freedom are indicated to facilitate assembling. The assembled global stiffness matrix is of the order 8 K is, 8. Assembled stiffness matrix [ ] Version CE IIT, Kharagur

14 [ K ] EI () The global load vector corresonding to unconstrained degree of freedom is, 9 k (7) 9 { } Writing the load dislacement relation for the entire continuous beam, EI u. u u.9 u u u. u.9 u 7 8 (8) We know thatu u u u u7 u8. Thus solving for unknowns dislacements uand u from equation, 9.. u EI (9) 9.. u Version CE IIT, Kharagur

15 u u.(8. ) () u.8 radians; u. radians () The unknown joint loads are calculated as, 7 8 ( 8 ) () Now the actual suort reactions,,,, 7 and 8 must include the fixed end suort reactions. Thus, Version CE IIT, Kharagur

16 Version CE IIT, Kharagur () 8.8 kn.m;. kn;. kn; 8 kn. kn.m;. kn; 8 7. () Summary The effect of temerature changes and suort settlements can easily be incororated in the direct stiffness method and is discussed in the resent lesson. Both temerature changes and suort settlements induce fixed end actions in the restrained beams. These fixed end forces are handled in the same way as those due to loads on the members in the analysis. In other words, the global load vector is formulated by considering fixed end actions due to both suort settlements and external loads. At the end, a few roblems are solved to illustrate the rocedure.

MODULE 3 ANALYSIS OF STATICALLY INDETERMINATE STRUCTURES BY THE DISPLACEMENT METHOD

MODULE 3 ANALYSIS OF STATICALLY INDETERMINATE STRUCTURES BY THE DISPLACEMENT METHOD ODULE 3 ANALYI O TATICALLY INDETERINATE TRUCTURE BY THE DIPLACEENT ETHOD LEON 19 THE OENT- DITRIBUTION ETHOD: TATICALLY INDETERINATE BEA WITH UPPORT ETTLEENT Instructional Objectives After reading this