Module 3. Analysis of Statically Indeterminate Structures by the Displacement Method. Version 2 CE IIT, Kharagpur
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1 odule Analysis of Statically Indeterminate Structures by te Displacement etod Version E IIT, Karagpur
2 Lesson 17 Te Slope-Deflection etod: rames wit Sidesway Version E IIT, Karagpur
3 Instructional Objectives After reading tis capter te student will be able to 1. Derive slope-deflection equations for te frames undergoing sidesway.. Analyse plane frames undergoing sidesway., Draw sear force and bending moment diagrams. 4. Sketc deflected sape of te plane frame not restrained against sidesway Introduction In tis lesson, slope-deflection equations are applied to analyse statically indeterminate frames undergoing sidesway. As stated earlier, te axial deformation of beams and columns are small and are neglected in te analysis. In te previous lesson, it was observed tat sidesway in a frame will not occur if 1. Tey are restrained against sidesway.. If te frame geometry and te loading are symmetrical. In general loading will never be symmetrical. Hence one could not avoid sidesway in frames. or example, consider te frame of ig In tis case te frame is symmetrical but not te loading. Due to unsymmetrical loading te beam end moments and are not equal. If b is greater tan a, ten >. In Version E IIT, Karagpur
4 suc a case joint and are displaced toward rigt as sown in te figure by an unknown amount Δ. Hence we ave tree unknown displacements in tis frame: rotations θ, and te linear displacement Δ. Te unknown joint rotations θ θ and θ are related to joint moments by te moment equilibrium equations. Similarly, wen unknown linear displacement occurs, one needs to consider force-equilibrium equations. Wile applying slope-deflection equation to columns Δ in te above frame, one must consider te column rotation ψ as unknowns. It is observed tat in te column, te end undergoes a linear displacement Δ wit respect to end A. Hence te slope-deflection equation for column is similar to te one for beam undergoing support settlement. However, in tis case Δ is unknown. or eac of te members we can write te following slope-deflection equations. EI + [ θ A + θ ψ ] were ψ Δ ψ is assumed to be negative as te cord to te elastic curve rotates in te clockwise directions. A D D EI A + [ θ + θ A ψ ] EI + [ θ + θ ] EI + [ θ + θ ] EI Δ D + [ θ + θ D ψ D ] ψ D EI D + [ θ D + θ ψ D ] (17.1) As tere are tree unknowns ( θ, θ and Δ ), tree equations are required to evaluate tem. Two equations are obtained by considering te moment equilibrium of joint and respectively (17.a) A (17.b) Now consider free body diagram of te frame as sown in ig Te orizontal sear force acting at A and of te column is given by D Version E IIT, Karagpur
5 H1 (17.a) A + Similarly for memberd, te sear force is given by H H + D D (17.b) Now, te required tird equation is obtained by considering te equilibrium of member, X H1+ H 0 0 A + + D + D 0 (17.4) Substituting te values of beam end moments from equation (17.1) in equations (17.a), (17.b) and (17.4), we get tree simultaneous equations in tree unknowns θ, θ and Δ, solving wic joint rotations and translations are evaluated. Version E IIT, Karagpur
6 Knowing joint rotations and translations, beam end moments are calculated from slope-deflection equations. Te complete procedure is explained wit a few numerical examples. Example 17.1 Analyse te rigid frame as sown in ig. 17.a. Assume EI to be constant for all members. Draw bending moment diagram and sketc qualitative elastic curve. Solution In te given problem, joints and rotate and also translate by an amount Δ. Hence, in tis problem we ave tree unknown displacements (two rotations and one translation) to be evaluated. onsidering te kinematically determinate structure, fixed end moments are evaluated. Tus, 0 ; 0 ; + 10 kn. m ; 10kN. m ; 0 ; 0. (1) A Te ends A and D are fixed. Hence, θ A θ D 0. Joints and translate by te same amount Δ. Hence, cord to te elastic curve ' and D' rotates by an amount (see ig. 17.b) Δ ψ ψ D () ords of te elastic curve ' and D' rotate in te clockwise direction; enceψ and ψ D are taken as negative. D D Version E IIT, Karagpur
7 Now, writing te slope-deflection equations for te six beam end moments, EI + [ θ + θ ψ ] A 0 ; θ A 0 ; ψ Δ. A D EIθ + EIΔ 4 EIθ + EIΔ 10 + EIθ EIθ EIθ + EIθ 4 EIθ + EIΔ Version E IIT, Karagpur
8 D EIθ + EIΔ () Now, consider te joint equilibrium of and (vide ig. 17.c) (4) A D (5) Te required tird equation is written considering te orizontal equilibrium of te entire frame i.e. 0 (vide ig. 17.d). X H H 0 H + H 10. (6) 1 Version E IIT, Karagpur
9 onsidering te equilibrium of te column andd, yields H 1 A + and H D + D (7) Te equation (6) may be written as, (8) A + D D Substituting te beam end moments from equation () in equations (4), (5) and (6).EIθ + 0.5EIθ EIΔ 10 (9).EIθ + 0.5EIθ EIΔ 10 (10) Version E IIT, Karagpur
10 8 EIθ + EIθ + EIΔ 0 (11) Equations (9), (10) and (11) indicate symmetry and tis fact may be noted. Tis may be used as te ceck in deriving tese equations. Solving equations (9), (10) and (11), EIθ 9.57 ; EIθ 1.55 and EI Δ Substituting te values of EIθ, EIθ and EIΔ in te slope-deflection equation (), one could calculate beam end moments. Tus, 5. kn.m (counterclockwise) 1.14 kn.m(clockwise) A 1.10 kn.m kn.m kn.m D kn.m. D Te bending moment diagram for te frame is sown in ig. 17. e. And te elastic curve is sown in ig 17. f. te bending moment diagram is drawn on te compression side. Also note tat te vertical atcing is used to represent bending moment diagram for te orizontal members (beams). Version E IIT, Karagpur
11 Version E IIT, Karagpur
12 Example 17. Analyse te rigid frame as sown in ig. 17.4a and draw te bending moment diagram. Te moment of inertia for all te members is sown in te figure. Neglect axial deformations. Version E IIT, Karagpur
13 Solution: In tis problem rotations and translations at joints and need to be evaluated. Hence, in tis problem we ave tree unknown displacements: two rotations and one translation. ixed end moments are kn. m ; A 9 kn. m 6 0 ; 0 ; 0 ; 0. D D ; (1) Te joints and translate by te same amount Δ. Hence, te cord to te elastic curve rotates in te clockwise direction as sown in ig. 17.b. and ψ D Δ 6 Δ ψ () Now, writing te slope-deflection equations for six beam end moments, (EI ) θ Δ EIθ + 0. EIΔ A EIθ + 0. EIΔ Version E IIT, Karagpur
14 EIθ EIθ 0. 5EIθ + EIθ D 1.EIθ EIΔ D 0.667EIθ EIΔ () Now, consider te joint equilibrium of and (4) A (5) Te required tird equation is written considering te orizontal equilibrium of te entire frame. onsidering te free body diagram of te member (vide ig. 17.4c), D H + H 0. 1 (6) Version E IIT, Karagpur
15 Te forces H1 and H are calculated from te free body diagram of column andd. Tus, and H 1 H 6 A D + D (7) Substituting te values of and H into equation (6) yields, H (8) A + D D Substituting te beam end moments from equation () in equations (4), (5) and (8), yields.eiθ + 0.5EIθ + 0.EIΔ 9.EIθ + 0.5EIθ EIΔ 0 EIθ + 4EIθ +.EIΔ 6 (9) Solving equations (9), (10) and (11), EIθ.76 ; EIθ 4.88 and EI Δ Substituting te values of EIθ, EIθ and EIΔ in te slope-deflection equation (), one could calculate beam end moments. Tus, kn.m (counterclockwise) 0.5 kn.m(clockwise) A 0. kn.m.50 kn.m.50 kn.m D 6.75 kn.m. D Te bending moment diagram for te frame is sown in ig d. Version E IIT, Karagpur
16 Version E IIT, Karagpur
17 Example 17. Analyse te rigid frame sown in ig a. oment of inertia of all te members are sown in te figure. Draw bending moment diagram. Under te action of external forces, te frame gets deformed as sown in ig. 17.5b. In tis figure, cord to te elastic curve are sown by dotted line. ' is perpendicular to and " is perpendicular to D. Te cords to te elastic Version E IIT, Karagpur
18 curve " rotates by an angle ψ D as sown in figure. Due to symmetry, D figure, ψ, "" rotates by ψ and D rotates by ψ ψ. rom te geometry of te ψ " Δ 1 L L ut Tus, Δ ψ 1 Δ cosα L Δ Δ cosα 5 ψ D Δ 5 Δ Δ tanα ψ Δ tan Δ α (1) 5 We ave tree independent unknowns for tis problem A and D are fixed. Hence, θ θ 0. ixed end moments are, A D θ, θ and Δ. Te ends 0 ; 0 ; +.50 kn. m ;.50kN. m ; 0 ; 0. A Now, writing te slope-deflection equations for te six beam end moments, D D E(I) 5.1 [ θ ψ ] A 0.784EIθ EIΔ A 1.568EIθ EIΔ.5 + EIθ + EIθ 0. EIΔ EIθ + EIθ 0. 6 EIΔ D 1.568EIθ EIΔ D 0.784EIθ EIΔ () Now, considering te joint equilibrium of and, yields Version E IIT, Karagpur
19 0 + 0 A.568EIθ + EIθ 0.19EIΔ.5 () D.568EIθ + EIθ 0.19EIΔ.5 (4) Sear equation for olumn olumn D 5H1 1 A + (1) V 0 (5) 5H D D + (1) V 0 (6) eam 0 V (7) Version E IIT, Karagpur
20 X 0 H 1 + H 5 (8) Y V 1 V 10 0 (9) 0 rom equation (7), V rom equation (8), H1 5 H rom equation (9), 1 10 V V 10 Substituting te values of V, H 1 1and V in equations (5) and (6), 60 10H (10) A H (11) Eliminating H in equation (10) and (11), D D (1) + A D D Substituting te values of equation. Tus,,,, A D D in (1) we get te required tird Simplifying, 0.784EIθ EIΔ EIθ EIΔ EIθ EIΔ EIθ EIΔ -(.5 + EIθ + EIθ 0. EIΔ )- (.5 + EIθ + EIθ 0. EIΔ ) EIθ 0.648EIθ +.084EIΔ 5 (1) Solving simultaneously equations () (4) and (1), yields EIθ ; EIθ 1.05 and EI Δ Substituting te values of EIθ, EIθ and EIΔ in te slope-deflection equation (), one could calculate beam end moments. Tus,.8 kn.m Version E IIT, Karagpur
21 A.70 kn.m.70 kn.m 5.75 kn.m 5.75 kn.m D 4.81 kn.m. (14) D Te bending moment diagram for te frame is sown in ig d. Summary In tis lesson, slope-deflection equations are derived for te plane frame undergoing sidesway. Using tese equations, plane frames wit sidesway are analysed. Te reactions are calculated from static equilibrium equations. A couple of problems are solved to make tings clear. In eac numerical example, te bending moment diagram is drawn and deflected sape is sketced for te plane frame. Version E IIT, Karagpur
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