odule 3 Analysis of Statically Indeterminate Structures by the Displacement ethod Version CE IIT, Kharagpur
Lesson The ultistory Frames with Sidesway Version CE IIT, Kharagpur
Instructional Objectives After reading this chapter the student will be able to. Identify the number of independent rotational degrees of freedom of a rigid frame.. Write appropriate number of equilibrium equations to solve rigid frame having more than one rotational degree of freedom. 3. Draw free-body diagram of multistory frames. 4. Analyse multistory frames with sidesway by the slope-deflection method. 5. Analyse multistory frames with sidesway by the moment-distribution method.. Introduction In lessons 7 and, rigid frames having single independent member rotational Δ ( ψ ) degree of freedom (or joint translation Δ ) is solved using slopedeflection and moment-distribution method respectively. However multistory h frames usually have more than one independent rotational degree of freedom. Such frames can also be analysed by slope-deflection and moment-distribution methods. Usually number of independent member rotations can be evaluated by inspection. However if the structure is complex the following method may be adopted. Consider the structure shown in Fig..a. Temporarily replace all rigid joints of the frame by pinned joint and fixed supports by hinged supports as shown in Fig..b. Now inspect the stability of the modified structure. If one or more joints are free to translate without any resistance then the structure is geometrically unstable. Now introduce forces in appropriate directions to the structure so as to make it stable. The number of such externally applied forces represents the number of independent member rotations in the structure. Version CE IIT, Kharagpur
Version CE IIT, Kharagpur
In the modified structure Fig..b, two forces are required to be applied at level CD and level F for stability of the structure. Hence there are two independent member rotations ( ψ ) that need to be considered apart from joint rotations in the analysis. The number of independent rotations to be considered for the frame shown in Fig..a is three and is clear from the modified structure shown in Fig..b. From the above procedure it is clear that the frame shown in Fig..3a has three independent member rotations and frame shown in Fig.4a has two independent member rotations. Version CE IIT, Kharagpur
For the gable frame shown in Fig..4a, the possible displacements at each joint are also shown. Horizontal displacement is denoted by u and vertical displacement is denoted by v. Recall that in the analysis, we are not considering Version CE IIT, Kharagpur
the axial deformation. Hence at and D only horizontal deformation is possible and joint C can have both horizontal and vertical deformation. The displacements u, uc, u D and ud should be such that the lengths C and CD must not change as the axial deformation is not considered. Hence we can have only two independent translations. In the next section slope-deflection method as applied to multistoried frame is discussed.. Slope-deflection method For the two story frame shown in Fig..5, there are four joint rotations θ,θ,θ and θ and two independent joint translations (sidesway) at the ( C D E ) level of CD and Δ at the level of E. Δ Six simultaneous equations are required to evaluate the six unknowns (four rotations and two translations). For each of the member one could write two slope-deflection equations relating beam end moments to (i) externally applied loads and (ii) displacements (rotations and translations). Four of the required six equations are obtained by considering the moment equilibrium of joint, C, D and E respectively. For example, Version CE IIT, Kharagpur
+ + 0 (.) 0 A C E The other two equations are obtained by considering the force equilibrium of the members. Thus, the shear at the base of all columns for any story must be equal to applied load. Thus 0 at the base of top story gives (ref. Fig..6) F X P H C H D 0 (.) Similarly F X 0 at the base of frame results in P P H A H 0 (.3) + F Thus we get six equations in six unknowns. Solving the above six equations all the unknowns are evaluated. The above procedure is explained in example.. Example. Analyse the two story rigid frame shown in Fig..7a by the slope-deflection method. Assume EI to be constant for all members. Version CE IIT, Kharagpur
In this case all the fixed end moments are zero. The members A and Δ EF undergo rotations ψ (negative as it is clockwise) and member C and 5 Δ ED undergo rotationsψ. Now writing slope-deflection equations for 5 beam end moments. EI A 0 + [ θ A + θ 3ψ ] θ A 0; 5 0.4EI + 0.4EIΔ A θ ψ Δ 5 0.8EI + 0.4EIΔ A θ C 0.8EIθ + 0.4EIθ + 0.4EIΔ C C 0.8EIθ + 0.4EIθ + 0.4EIΔ C Version CE IIT, Kharagpur
E E CD 0.8EIθ + 0. 4EIθ 0.8EIθ + 0. 4EIθ E C E 0.8EIθ + 0. 4EIθ D DC 0.8EIθ + 0. 4EIθ D C DE ED 0.8EIθ + 0.4EIθ + 0.4EIΔ D 0.8EIθ + 0.4EIθ + 0.4EIΔ E E D 0.8EI + 0.4EIΔ EF θ E 0.4EI + 0.4EIΔ FE θ E () Version CE IIT, Kharagpur
oment equilibrium of joint, C, D and E requires that (vide Fig..7c). + C + E 0 0 A C + CD DC + DE 0 E + ED + EF 0 () The required two more equations are written considering the horizontal equilibrium at each story level. i e. F 0 (vide., Fig..7d). Thus,. X H C + H D 0 H H 60 (3) A + F Considering the equilibrium of column A, EF, C and ED, we get (vide.7c) Version CE IIT, Kharagpur
H C C + 5 C H D DE + 5 ED H A A + 5 A H F EF + FE (4) 5 Using equation (4), equation (3) may be written as, C + C + DE + ED 00 + + 300 (5) A + A EF FE Substituting the beam end moments from equation () in () and (5) the required equations are obtained. Thus,.4θ + 0.4θ C + 0.4θ E + 0.4Δ + 0.4Δ.6θ 0.4θ + 0.4θ + 0.4Δ 0 C + D.6θ 0.4θ + 0.4θ + 0.4Δ 0 D + C E.4θ E + 0.4θ + 0.4θ D + 0.4Δ + 0.4Δ.θ.θ +.θ +.θ + 0.96Δ 00 + C D E.θ.θ + + 0.96Δ 300 (6) + E Solving above equations, yields 0 0 65.909 7.73 7.73 θ ; θ C ; θ E ; θ D EI EI EI 337. 477.7 Δ ; Δ EI EI 65.909 ; EI (7) Version CE IIT, Kharagpur
Substituting the above values of rotations and translations in equation () beam end moments are evaluated. They are, A C E CD DE EF 88.8 kn.m ; 6.8 kn.m 7.7 kn.m ; 3.7. kn.m A C 79.09 kn.m ; 79.09 kn.m 3.7 kn.m ; 3.7 kn.m E DC 3.7 kn.m ; 7.7 kn.m ED 6.8 kn.m ; 88.8 kn.m FE.3 oment-distribution method The two-story frame shown in Fig..8a has two independent sidesways or member rotations. Invoking the method of superposition, the structure shown in Fig..8a is expressed as the sum of three systems; ) The system shown in Fig..8b, where in the sidesway is completely prevented by introducing two supports at E and D. All external loads are applied on this frame. ) System shown in Fig..8c, wherein the support E is locked against sidesway and joint C and D are allowed to displace horizontally. Apply arbitrary sidesway Δ' and calculate fixed end moments in column C and DE. Using moment-distribution method, calculate beam end moments. 3) Structure shown in Fig..8d, the support D is locked against sidesway and joints and E are allowed to displace horizontally by removing the support at E. Calculate fixed end moments in column A and EF for an arbitrary sidesway Δ' as shown the in figure. Since joint displacement as known beforehand, one could use the moment-distribution method to analyse the frame. Version CE IIT, Kharagpur
All three systems are analysed separately and superposed to obtain the final answer. Since structures.8c and.8d are analysed for arbitrary sidesway Δ' and Δ' respectively, the end moments and the displacements of these two analyses are to be multiplied by constants kand k before superposing with the results obtained in Fig..8b. The constants and k must be such that k Version CE IIT, Kharagpur
k Δ' Δ and Δ' Δ k. (.4) The constants kand k are evaluated by solving shear equations. From Fig..9, it is clear that the horizontal forces developed at the beam level CD in Fig..9c and.9d must be equal and opposite to the restraining force applied at the restraining support at D in Fig..9b. Thus, k ( H C H D ) + k ( H C3 + H D3 ) + (3.5) P From similar reasoning, from Fig..0, one could write, k ( H A H F ) + k ( H A3 + H F 3 ) + (3.6) P Solving the above two equations, and k are calculated. k Version CE IIT, Kharagpur
Example. Analyse the rigid frame of example. by the moment-distribution method. Solution: First calculate stiffness and distribution factors for all the six members. K A 0.0EI ; K C 0.0EI ; K E 0.0EI ; K K C DC 0.0EI ; 0.0EI ; K K CD DE 0.0EI ; 0.0EI ; () K E 0.0EI ; K ED 0.0EI ; K EF 0.0EI Joint : K 0. 60EI DF 0.333; DF 0.333; DF 0.333 A C E Version CE IIT, Kharagpur
JointC : K 0. 40EI DF 0.50; DF 0.50 C CD Joint D : K 0. 40EI DF 0.50; DF 0.50 DC DE Joint E : K 0. 60EI DF 0.333; DF 0.333; DF 0.333 () E ED EF The frame has two independent sidesways: Δ to the right of CD and Δ to the right of E. The given problem may be broken in to three systems as shown in Fig..a. Version CE IIT, Kharagpur
In the first case, when the sidesway is prevented [Fig..0a ( ii )], the only internal forces induced in the structure being 0 kn and 40 kn axial forces in member CD and E respectively. No bending moment is induced in the structure. Thus we need to analyse only (iii) and (iv). Version CE IIT, Kharagpur
Case I : oment-distribution for sidesway Δ' at beam CD [Fig..qa ( iii )]. Let the 5 arbitrary sidesway be Δ '. Thus the fixed end moment in column C and EI DE due to this arbitrary sidesway is F C F ED 6 EIΔ ' 6EI 5 L 5 EI F C + 6.0 kn.m F + 6.0 kn.m (3) DE Now moment-distribution is carried out to obtain the balanced end moments. The whole procedure is shown in Fig..b. Successively joint D, C, and E are released and balanced. Version CE IIT, Kharagpur
From the free body diagram of the column shown in Fig..c, the horizontal forces are calculated. Thus, Version CE IIT, Kharagpur
H C H A 3.53 + 3.7.34 kn; H D.34 kn 5 0.70.4 0.4 kn; H F 0.4 kn (4) 5 Case II : oment-distribution for sidesway Δ' at beam E [Fig..a (iv )]. Let the 5 arbitrary sidesway be Δ ' EI Thus the fixed end moment in column A and EF due to this arbitrary sidesway is F A F FE 6 EIΔ ' 6EI 5 L 5 EI F A + 6.0 kn.m F + 6.0 kn.m (5) EF oment-distribution is carried out to obtain the balanced end moments as shown in Fig..d. The whole procedure is shown in Fig..0b. Successively joint D, C, and E are released and balanced. Version CE IIT, Kharagpur
Version CE IIT, Kharagpur
From the free body diagram of the column shown in Fig..e, the horizontal forces are calculated. Thus, H H.59 0.53 0.4 kn; H 0.4 kn 5 5.+ 4.3.86 kn; HF.86 kn (6) 5 C3 D3 A3 3 For evaluating constants kand k, we could write, (see Fig..a,.c and.d). k k k k ( H C + H D ) + k ( H C3 + H D3 ) 0 ( H A + H F ) + k ( H A3 + H F 3 ) 60 (.34 +.34) + k ( 0.4 0.4) 0 (.4 0.4) + k (.86 +.86) 60 0 Version CE IIT, Kharagpur
k k (.34) + k ( 0.4) 0 (.4) + k (.86) 30 0 Solving which, k 3.47 k 9. 7 (7) Thus the final moments are, A C E CD DE EF 88.5 kn.m ; 6.09 kn.m 7.06 kn.m ; 3.54. kn.m 79.54 kn.m ; 79.54 kn.m A C 3.54 kn.m ; 3.54 kn.m E DC 3.54 kn.m ; 7.06 kn.m ED 6.09 kn.m ; 88.5 kn.m (8) FE Summary A procedure to identify the number of independent rotational degrees of freedom of a rigid frame is given. The slope-deflection method and the momentdistribution method are extended in this lesson to solve rigid multistory frames having more than one independent rotational degrees of freedom. A multistory frames having side sway is analysed by the slope-deflection method and the moment-distribution method. Appropriate number of equilibrium equations is written to evaluate all unknowns. Numerical examples are explained with the help of free-body diagrams. Version CE IIT, Kharagpur