Module 3. Analysis of Statically Indeterminate Structures by the Displacement Method. Version 2 CE IIT, Kharagpur
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1 odule 3 Analysis of Statically Indeterminate Structures by the Displacement ethod
2 Lesson 21 The oment- Distribution ethod: rames with Sidesway
3 Instructional Objectives After reading this chapter the student will be able to 1. Extend moment-distribution method for frames undergoing sidesway. 2. Draw free-body diagrams of plane frame. 3. Analyse plane frames undergoing sidesway by the moment-distribution method. 4. Draw shear force and bending moment diagrams. 5. Sketch deflected shape of the plane frame not restrained against sidesway Introduction In the previous lesson, rigid frames restrained against sidesway are analyzed using moment-distribution method. It has been pointed in lesson 17, that frames which are unsymmetrical or frames which are loaded unsymmetrically usually get displaced either to the right or to the left. In other words, in such frames apart from evaluating joint rotations, one also needs to evaluate joint translations (sidesway). or example in frame shown in ig 21.1, the loading is symmetrical but the geometry of frame is unsymmetrical and hence sidesway needs to be considered in the analysis. The number of unknowns is this case are: joint rotations θ B and θ C and member rotationψ. Joint B and C get translated by the same amount as axial deformations are not considered and hence only one independent member rotation need to be considered. The procedure to analyze rigid frames undergoing lateral displacement using moment-distribution method is explained in section 21.2 using an example.
4 21.2 Procedure A special procedure is required to analyze frames with sidesway using momentdistribution method. In the first step, identify the number of independent rotations (ψ ) in the structure. The procedure to calculate independent rotations is explained in lesson 22. or analyzing frames with sidesway, the method of superposition is used. The structure shown in ig. 21.2a is expressed as the sum of two systems: ig. 21.2b and ig. 21.2c. The systems shown in figures 21.2b and 21.2c are analyzed separately and superposed to obtain the final answer. In system 21.2b, sidesway is prevented by artificial support atc. Apply all the external loads on frame shown in ig. 21.2b. Since for the frame, sidesway is prevented, moment-distribution method as discussed in the previous lesson is applied and beam end moments are calculated. ' ' ' ' ' ' Let,,, CB, and DC be the balanced moments obtained by distributing fixed end moments due to applied loads while allowing only joint rotations ( θ B and θ C ) and preventing sidesway. Now, calculate reactions and (ref. ig 21.3a).they are, H A1 H D 1
5 H + ' ' A 1 + h2 Pa h 2 H D1 ' ' + DC (21.1) h 1 again, R P H A + H ) (21.2) ( 1 D1
6 In ig 21.2c apply a horizontal force in the opposite direction of R. Now k R, then the superposition of beam end moments of system (b) and k times (c) gives the results for the original structure. However, there is no way one could analyze the frame for horizontal force, by moment-distribution method as sway comes in to picture. Instead of applying, apply arbitrary known displacement / sidesway Δ' as shown in the figure. Calculate the fixed end beam moments in the column and for the imposed horizontal displacement. Since joint displacement is known beforehand, one could use moment-distribution method to analyse this frame. In this case, member rotations ψ are related to joint '' '' '' '' '' '' translation which is known. Let,,, and are the, CB balanced moment obtained by distributing the fixed end moments due to assumed sidesway Δ' at joints B and C. Now, from statics calculate horizontal force due to arbitrary sidesway Δ '. DC
7 H A2 '' + h 2 '' H D2 '' '' + DC (21.3) h 1 H A + H ) (21.4) ( 2 D2 In ig 21.2, by method of superposition k R or k R / Substituting the values of R and from equations (21.2) and (21.4), P ( H A1 + H D1) k (21.5) ( H + H ) A2 Now substituting the values of,, and in 21.5, D2 H A1 H A2 H D1 H D 2 k ' + ' ' + ' Pa P + + h2 h2 h1 '' + '' '' + '' DC + h2 h1 DC (21.6) Hence, beam end moment in the original structure is obtained as, + k original system( b) system( c) If there is more than one independent member rotation, then the above procedure needs to be modified and is discussed in the next lesson.
8 Example 21.1 Analyse the rigid frame shown in ig 21.4a. Assume EI to be constant for all members. Also sketch elastic curve. Solution In the given problem, joint C can also rotate and also translate by an unknown amount Δ. This problem has to be solved in two steps. In the first step, evaluate the beam-end moment by preventing the sidesway. In the second step calculate beam end moments by moment-distribution method for known translation (see ig 21.4b). By appropriately superposing the two results, the beam end moment of the original structure is obtained. a) Calculate stiffness and distribution factors K EI ; K 0. 25EI ; K CB 0. 25EI ; K EI Joint B : K EI D 0.571; D Joint C : K EI
9 D ; D (1) b) Calculate fixed end moment due to applied loading. CB 0 ; 0 kn.m +10 kn.m ; CB 10 kn.m 0 kn.m ; 0 kn.m. (2) DC Now the frame is prevented from sidesway by providing a support at C as shown in ig 21.4b (ii). The moment-distribution for this frame is shown in ig 21.4c. Let ', ', ' and ' be the balanced end moments. Now calculate DC horizontal reactions at A and D from equations of statics. H A1 ' + ' KN ( ) H D kn( ). 3
10 R 10 ( ) 10 kn( ) (3) d) oment-distribution for arbitrary known sidesway Δ '. Since Δ ' is arbitrary, Choose any convenient value. Let 150 Δ ' EI Now calculate fixed end beam moments for this arbitrary sidesway. 6EIψ 6 EI 150 ( ) 100 kn.m L 3 3EI 100 kn.m +100 kn.m (4) DC
11 The moment-distribution for this case is shown in ig 24.4d. Now calculate horizontal reactions and. H A2 H D H A kn( ) H D kn( ) kn( )
12 Let k be a factor by which the solution of case ( iii ) needs to be multiplied. Now actual moments in the frame is obtained by superposing the solution ( ii ) on the solution obtained by multiplying case ( iii ) by k. Thus k cancel out the holding force R such that final result is for the frame without holding force. Thus, k R. 10 k (5) Now the actual end moments in the frame are, ' + k '' ( ) kn.m ( ) kn.m ( 52.98) kn.m ( 52.97) kn.m CB ( ) kn.m ( ) kn.m DC The actual sway is computed as, 150 Δ kδ' EI EI The joint rotations can be calculated using slope-deflection equations. 2EI where ψ L + [ 2θ A + θ B 3ψ ] Δ L 2EI + L [ 2θ + θ 3ψ ] B A
13 In the above equation, except θ A and θ B all other quantities are known. Solving for θ A and θ B, 9.55 θ A 0 ; θ B. EI The elastic curve is shown in ig. 21.4e.
14 Example 21.2 Analyse the rigid frame shown in ig. 21.5a by moment-distribution method. The moment of inertia of all the members is shown in the figure. Neglect axial deformations. Solution: In this frame joint rotations B and evaluated. C and translation of joint B and C need to be a) Calculate stiffness and distribution factors. At joint B : K 0.333EI ; K 0. 25EI K CB 0.25EI ; K EI K EI D ; D At joint C : K EI
15 D CB ; D b) Calculate fixed end moments due to applied loading kn.m ; 9.0 kn.m kn.m; 0 kn.m CB 0 kn.m; 0 kn.m DC c) Prevent sidesway by providing artificial support at C. Carry out momentdistribution ( i.e. Case shown in ig. 21.5c. A in ig. 21.5b). The moment-distribution for this case is
16 Now calculate horizontal reaction at A and D from equations of statics H A kn ( ) H D kn ( ) 3 R 12 ( ) 5.23 kn( ) d) oment-distribution for arbitrary sidesway Δ '(case B, ig. 21.5c) Calculate fixed end moments for the arbitrary sidesway of 150 Δ '. EI 6 E(2 I) 12EI 150 ψ ( ) + 50 kn.m ; + 50 kn.m ; L 6 6EI
17 6 EI ( ) 6EI 150 ψ ( ) kn.m ; DC kn.m ; L 3 3EI The moment-distribution for this case is shown in ig. 21.5d. Using equations of H A2 H D 2 static equilibrium, calculate reactions and kn( kn( 3 H A 2 H D 2 ) ) ( ) kn ( ) e) inal results Now, the shear condition for the frame is (vide ig. 21.5b)
18 ( H A1 + H D1) + k( H A2 + H D2 ) 12 ( ) + k( ) 12 k Now the actual end moments in the frame are, ' + k '' ( ) kn.m ( ) kn.m ( ) kn.m ( ) kn.m CB ( ) kn.m ( ) kn.m DC The actual sway 150 Δ k Δ' EI EI The joint rotations can be calculated using slope-deflection equations. or 2E(2I ) + L [ 2θ + θ 3ψ ] A L 12EIψ L 12EIψ [ 2 θ + θ ] + A B 4EI L B 4EI L L 12EIψ L 12EIψ [ 2 θ + θ ] + B A 4EI L 4EI L kn.m
19 0.629 kn.m ( ) (50) kn.m ( ) (50) 2.55 kn.m 1 change in near end + - change in far end 2 θ A 3EI L 1 ( ) + ( ) EI θ B EI Example 21.3 Analyse the rigid frame shown in ig. 21.6a. The moment of inertia of all the members are shown in the figure.
20 Solution: a) Calculate stiffness and distribution factors 2EI K 0.392EI ; K 0. 50EI 5.1 K CB 0.50EI ; K EI At joint B : K EI D ; D At joint C : K EI D ; D (1) CB b) Calculate fixed end moments due to applied loading. 0 kn.m DC 2.50 kn.m 2.50 kn.m (2) CB c) Prevent sidesway by providing artificial support atc. Carry out momentdistribution for this case as shown in ig. 21.6b.
21 Now calculate reactions from free body diagram shown in ig. 21.5d.
22 Column 0 5H V 0 A A1 1 5H + V 2.29 (3) A1 1 Column H V D D H V D1 2 (4) Beam C 0 2V V kn ( ) 1 V kn ( ) (5) Thus from (3) H A kn( ) from (4) H D kn( ) (6)
23 0 H + H + R 50 X A1 D1 ( ) R kn (7) d) oment-distribution for arbitrary sidesway Δ '. Calculate fixed end beam moments for arbitrary sidesway of Δ ' EI The member rotations for this arbitrary sidesway is shown in ig. 21.6e.
24 BB" Δ1 Δ ' 5.1 Δ ' ψ ; Δ 1 L L cosα 5 2 Δ ' Δ Δ ' 5 Δ ' Δ ' ψ ( clockwise ) ; ψ ( clockwise ) 5 5 ψ Δ2 2 Δ 'tan α Δ ' ( counterclockwise ) EI 6 E(2 I) ψ kn.m L 5.1 5EI kn.m 6EI 6 EI ( ) ψ 7.65 kn.m L 2 5EI 7.65 kn.m CB 6EI 6 E(2 I) ψ kn.m L 5.1 5EI kn.m DC The moment-distribution for the arbitrary sway is shown in ig. 21.6f. Now reactions can be calculated from statics.
25 Column 0 5H V 0 A A1 1 A2 1 5H + V (3) Column 0 5H V 0 D D1 2 D2 2 5H V (4) Beam C 0 2V V kn ( ) V kn( ) ; + (5) Thus from 3 H A kn( ) from 4 H D kn( ) (6)
26 7.766 kn( ) (7) e) inal results k R k Now the actual end moments in the frame are, ' + k '' ( ) kn.m ( ) kn.m ( 6.567) kn.m ( 6.567) kn.m CB (6.567) kn.m (6.283) kn.m DC The actual sway Δ k Δ' EI EI Summary In this lesson, the frames which are not restrained against sidesway are identified and solved by the moment-distribution method. The moment-distribution method is applied in two steps: in the first step, the frame prevented from sidesway but subjected to external loads is analysed and subsequently, the frame which is undergoing an arbitrary but known sidesway is analysed. Using shear equation for the frame, the moments in the frame is obtained. The numerical examples are explained with the help of free-body diagrams. The deflected shape of the frame is sketched to understand its deformation under external loads.
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