Chapter TEN Moment-Distribution Method 10.1 Introduction The moment-distribution method was originally presented by Prof. Hardy cross in 1930 and is considered one of the most important contributions ever made to the structural analysis of continuous beams and rigid frames. The moment-distribution is a method of successive approximations that may be carried out to any desired degree of accuracy. Essentially, the method begins by assuming each joint of a structure is fixed. Then, by unlocking and locking each joint in succession, the internal moments at the joints are distributed and balanced until the joints have rotated to their final or nearly final positions. 243 Chapter Ten 10.2 General Principles and Definitions Before developing the procedures of the moment-distribution method, it is necessary to adopt a sign convention and define the various terms used in the analysis. Sign Convention In applying the moment-distribution method, we will adopt the same sign convention as used previously for the slopedeflection method: Clockwise moments that act on member ends are considered positive, whereas counterclockwise moment are negative. Fixed-End Moments (F.E.Ms) The moments at the fixed end of a loaded member are called fixed-end moments. These moments can be determined from Table (9.1) of the preceding chapter, depending upon the type of loading on the member. Member Stiffness Factor (kij) Consider a prismatic beam AB, which is hinged at end A and fixed at end B, as shown in Fig. (10.1-a). If we apply a moment M at the end A, the beam rotates by an angle θ at the hinged end A and develops a moment MBA at the fixed end B, as shown in the figure. In Chapter 9 we related M to θ using the conjugate beam method. This resulted in Eq. (9.6), that is, 242 Moment-Distribution Method MBA=carryover moment M=applied moment θ (a) B A L EI=constant M=applied moment A B θ (b) L EI=constant Real beam M/EI (c) A' RA' B' L Conjugate beam RB' Figure (10.1) The term is referred to as the stiffness factor at A and can be defined as the moment that must be applied at the end A of the member to cause a unit rotation ( ) at A. 244 Chapter Ten Now, suppose that the far end B of the beam is hinged, as shown in Fig. (10.1-b). The relationship between the applied moment M and the rotation θ of the end A of the beam can be determined by using the conjugate beam method, as illustrated in Fig. (10.1-b), that is This expression indicates that the stiffness factor of the member at A for this case is Thus it may summarized that the stiffness factor of a member ij at the end i is given by if far end j of member is fixed (10.1) if far end j of member is hinged (10.2) Carry-Over Moment (C.O.M.) Let us considered again the hinged-fixed beam in Fig. (10.1a). When a moment M is applied at the hinged end A of the beam, a moment MBA develops at the fixed end B, as shown in the figure. The moment MBA is termed the carryover moment. It was shown in Chapter 9 that: 243 Moment-Distribution Method (Eq. 9.6) and (Eq. 9.7) or (10.3) This equation indicates that, when a moment of magnitude M is applied at one end of a beam, one-half of the applied moment is carried over to the far end, provided that the far end is fixed. Note that the direction of the carryover moment, MBA, is the same as the applied moment, M. When the far end of the beam is hinged, as shown in Fig. (10.1-b), the carry over moment MBA is zero. Thus we can express the carryover moment as { (10.4) The ratio of the carryover moment to the applied ⁄ moment ( is called the carryover factor of the member. It represents the fraction of the applied moment M that is carried over to the far end of the member. Thus we can express the carryover factor (COF) as { (10.5) 243 Chapter Ten Distribution Factor (D.F) When an external moment is applied to a joint of a structure where two or more members meet, an important question that arises is how to distribute this moment among the various members connected at that joint. Consider the joint B show in Fig. (10.2-a) at which three members meet, and suppose that a moment M is applied to this joint, causing it to rotate by an angle θ. To determine what fraction of the applied moment M is resisted by each of the three members connected to the joint, we draw the free-body diagram of joint B as show in Fig. (10.2-b). By considering the moment equilibrium of joint B (that is, ∑ ), we have C M A θ θ B M θ MBC MBA B MBD (a) (b) D Figure (10.2) 243 Moment-Distribution Method or (10.6) The moments at the ends B of the members can be expressed in terms of the joint rotation θ and the stiffness factor k of the members, Eqs. (10.1) and (10.2), as (10.7) (10.8) (10.9) Substituting these equations into Eq. (10.6), we obtain ( or ∑ where ∑ (10.10) and represents the sum of the stiffnesses of all the members connected to joint B. From Eq. (10.10) we can write (10.11) ∑ The above equations, Eqs. (10.7) to (10.9), can now be written as ∑ ∑ 243 Chapter Ten ∑ ⁄∑ , The ratios ⁄∑ , and ⁄∑ indicate the portions of the moment M that are resisted by members BA, BC, and BD, respectively. The ratio ⁄∑ for a member ij is termed the distribution factor (D.F) of the member and it will be given the symbol d, that is ∑ (10.12) The distribution factor for a member is thus equal to the stiffness of the member divided by the sum of the stiffnesses of all members meeting at the joint. 10.3 Procedure for Analysis The procedure for the analysis of structures by the momentdistribution method can be summarized as follows: 1. The joints on the structure should be identified. 2. Compute the fixed-end moments. Assuming that all the free joints are clamped against rotation, evaluate for each member, the fixed-end moments due to the external loads and support settlements (if any) by using the fixed-end moment expressions given in Table (9.1) of Chapter 9. 3. Calculate the distribution factors. The stiffness factors for each member of the structure at the joints should be 243 Moment-Distribution Method calculated. Using these values the distribution factors can be determined from Eq. (10.12). Remembering that the distribution factor is zero for a fixed end and one for an end roller or hinged support. The sum of all the distribution factors at a joint must equal 1. 4. Balance the moments at all the joints that are free to rotate by applying the moment-distribution process as follows: a. At each joint, evaluate the unbalanced moment and distribute it to the members connected to the joint. The distributed moment at each member end rigidly connected to the joint is obtained by multiplying the negative of the unbalanced moment by the distribution factor for the member end. b. Carryover one-half of each distributed moment to the opposite (far) end of the member. c. Repeat steps (4-a) and (4-b) until either all the free joints are balanced or the unbalanced moments at these joints are negligibly small. 5. Determine the final member end moments by algebraically summing the fixed-end moment and all the distributed and carryover moments at each member end. If the moment distribution has been carried out correctly, then the final moments must satisfy the 233 Chapter Ten equations of moment equilibrium at all joints of the structure that are free to rotate. 6. Compute member end shearing forces and axial forces by considering the equilibrium of the members of the structure. 7. Determine support reactions by considering the equilibrium of the joints of the structure. 8. Draw the shearing force, axial force, and bending moment diagrams. Example 10.1: Draw the bending moment diagram for the beam shown in Fig. (10.3) by using the moment-distribution method. Solution: First of all the joints are numbered. We have three joints, 1, 2, and 3, which divide the beam into two members. Fixed-End Moments: By assuming that joints 1, 2, and 3 are fixed (or locked), we calculate the fixed-end moments at the ends of each member due to the external loads, Fig. (10.3-b). By using Table (9.1) in Chapter 9, we obtain for member 1-2 kN.m 233 Moment-Distribution Method 38 kN 18 kN/m (a) 1 I 2 3 2m 2m 2I 6m 38 kN 18 kN/m (b) 1 2 6m 3 38 kN 18 kN/m (c) 4m 2 2 1 3 +35 kN.m (d) 2 Unbalanced moment=+35 kN.m (e) 1 3 2 Carryover moment Distributed moments Carryover moment M12=-54-10=-64 kN.m M21=+54-20=+34 kN.m M32=+19-7.5=+11.5 kN.m 38 kN 18 kN/m (f) 1 2 M23=-19-15=-34 kN.m Figure (10.3) 3 233 Chapter Ten -64 + _ -34 + _ (g) - -11.5 (+B.M. on compression side -B.M. on tension side) Figure (10.3) Continued kN.m and for member 2-3 kN.m kN.m Distribution Factors: We must determine the distribution factor at the two ends of each member. Using Eq. (10.1) the stiffness factors at these ends are ( ( ( ( 232 Moment-Distribution Method The distribution factor can then be determined for the ends of member by using Eq. (10.12). At joint 1 and 3, the distribution factor depends on the member stiffness factor and the stiffness factor of the fixed support (wall). Since in theory it would take an infinite size moment to rotate the wall one radian, the wall stiffness factor is infinite. Thus for joints 1 and 3 we have and for member ends connected at joint 2 we have ⁄( ⁄( ) ) Note that the sum of the distribution factors at each joint must always equal 1. Balancing Joint 2: There is a kN.m fixed-end moment at end 2 of member (1-2), whereas the end 2 of member (2-3) is subjected to a kN.m fixed-end moment. Thus the unbalanced moment UM2 at joint 2 is, Figs. (10.3-c) and (10.3-d) kN.m 234 Chapter Ten To balance joint 2, we will apply an equal, but opposite moment of kN.m to the joint and allow it to rotate freely. As a result, portions of this moment are distributed in members (2-1) and (2-3) in accordance with the distribution factors of these members at the joint. Specifically, the distributed moment D.M21 in member (2-1) is ( kN.m and that D.M23 in member (2-3) is ( kN.m The distributed moment at end 2 of member (2-1) induces a carryover moment at the far end 1 (Fig. 10.3-e), which can be determined by multiplying the distributed moment by the carryover factor of the member. Since joint 1 is fixed, the carryover factor is . Thus the C.O.M at end 1 of member (1-2) is ( kN.m These results are shown in Fig. (10.3-e). In this particular case only one cycle of moment distribution is necessary, since no further joints have to be balanced (or unlocked) to satisfy joint equilibrium. The final member end moments are obtained by algebraically summing the fixed-end moment and all the 233 Moment-Distribution Method distributed and carryover moments at each member end. The results are depicted in Fig. (10.3-f) and the bending moment diagram is shown in Fig. (10.3-g). Since the process of moment distribution, as explained above, is both long and cumbersome, it is usually convenient to carry out the analysis in tabular form as shown in Table (10.1). Table (10.1) Joints 1 Members connected to joint Distribution factors Members 12 21 23 32 D.Fs. 0 4/7 3/7 0 Fixed-end moments F.E.Ms. -54 +54 -19 Distributed moments D.Ms. -20 Carryover moments C.O.Ms. -10 Final End Moments -64 +34 -34 +11.5 2 3 +19 -15 -7.5 Example 10.2: Draw the bending moment diagram for the beam shown in Fig. (10.4) by using the moment-distribution method. Solution: Fixed-End Moments: From Table (9.1) in Chapter 9 233 Chapter Ten 80 kN (a) 24 kN/m 2 2I 6m 8m 1 3 3I I 4 6m 6m +120 + -28.295 (b) -93.437 +108 kN.m _ + -18.850 _ _ (+B.M. on compression side -B.M. on tension side) _ -9.528 Figure (10.4) kN.m kN.m kN.m kN.m there is no load on member (3-4) Stiffness Factors: From Eq. (10.1), 233 Moment-Distribution Method Distribution Factors: From Eq. (10.12), fixed end ⁄( ⁄( ⁄( ) ⁄( ) fixed end Starting with the fixed-end moment, the moments at joints 2 and 3 are balanced and distributed simultaneously. These moments are then carried over simultaneously to the respective ends of each member. The resulting moments are again simultaneously distributed and carried over. The process is continued until the resulting moments are diminished an appropriate amount. The resulting moments are found by summation. The complete analysis is shown in Table (10.2). The bending moment diagram is shown in Fig. (10.4-b) 233 Chapter Ten Table (10.2) Joints 1 Members 12 21 23 32 34 43 D.Fs. 0 1/3 2/3 3/4 1/4 0 F.E.Ms. -30 +90 -72 +72 0 0 D.Ms. 0 -6 -12 -54 -18 0 C.O.Ms. -3 0 -27 -6 0 -9 D.Ms. 0 +9 +18 +4.5 +1.5 0 C.O.Ms. +4.5 0 +2.25 +9 0 +0.75 D.Ms. 0 -0.75 -1.5 -6.75 +2.25 0 C.O.Ms. -0.375 0 -3.375 -0.75 0 -1.125 D.Ms. 0 +1.125 +2.25 +0.562 +0.188 0 C.O.Ms. +0.562 0 +0.281 +1.125 0 +0.094 D.Ms. 0 -0.094 -0.187 -0.844 -0.281 0 C.O.Ms. -0.047 0 -0.422 -0.094 0 -0.241 D.Ms. 0 +0.141 +0.281 +0.070 +0.024 0 C.O.Ms. +0.071 0 +0.035 +0.145 0 +0.012 D.Ms. 0 -0.012 -0.023 -0.108 -0.036 0 C.O.Ms. -0.006 0 -0.054 -0.012 0 -0.018 Final End Moments -28.295 +93.410 -93.464 +18.844 -18.855 -9.528 Average 2 3 93.437 4 18.85 10.4 Hinged or Simple Supports at Ends The analysis of structures, that are hinged or simply supported at one or more ends, can be considerably simplified by using Eq. (10.2) to calculate the stiffness 233 Moment-Distribution Method factors for members adjacent to the hinged or simple end supports. Also, the joints at the hinged or simple end supports are balanced only once during the moment distribution process, after which they are left unclamped so that no moment can be carried to them as the interior joints of the structure are balanced. Example 10.3: Draw the bending moment diagram for the beam shown in Fig. (10.5). By using the moment-distribution method. constant. 80 kN 24 kN/m (a) 1 5m 3 2 1.25 m 1.25 m +75 kN.m -52.5 + _ -45 _ (b) +50 + 0 (+B.M. on compression side -B.M. on tension side) Figure (10.5) Solution: Fixed-End Moments: From Table (9.1), kN.m 233 Chapter Ten kN.m kN.m kN.m Stiffness Factors: From Eq. (10.1), From Eq. (10.2), From Eq. (10.1), Distribution Factors: fixed end ⁄( ) ⁄( ) hinged end The process of moment distribution is conducted in Table (10.3), and the bending moment diagram is shown in Fig. (10.5-b). The final end moments on the last line of Table 233 Moment-Distribution Method (10.3) are obtained by algebraically summing the modified fixed-end moment and all the distributed and carryover moments at each member end (that is each column). Table (10.3) Joints 1 2 3 Members 12 21 23 32 D.Fs. 0 2/5 3/5 1 F.E.Ms. -50 +50 -25 +25 Balance joint (3) -25 C.O.Ms. Modified F.E.Ms. -12.5 -50 D.Ms. C.O.Ms Final End Moments +50 -37.5 -5 -7.5 -2.5 -52.5 0 0 +45.0 -45.0 0 10.5 Structures with Cantilever Overhangs Sometimes, a structure may have a cantilever overhang at one or more ends, Fig. (10.6). In such a case, the bending moment at end A of the overhanging portion will be due to the load over this portion and will remain constant, irrespective of the moments on the other portions of the structure. Since the overhanging portion AB does not contribute to the rotational stiffness of joint A, the distribution factor for its end A is zero. Thus, joint A in Fig. (10.6-b) can be treated as a simple end support in the 233 Chapter Ten analysis. The moment at end A of the cantilever, however does affect the unbalanced moment at joint A, Figs. (10.6-a and -b) and must be included along with the other fixed-end moments in the analysis. P w P w A B A (a) B (b) Figure (10.6) Example 10.4: Draw the bending moment diagram for the beam shown in Fig. (10.7) by using the moment-distribution method. constant. Solution: Fixed-End Moments: From Table (9.1), there is no load on member (1-2) kN.m kN.m kN.m 232 Moment-Distribution Method 30 kN 10 kN/m (a) 2 1 3 4m 9m 6m 4 (a) Continuous beam M34=-30*4=-120 kN.m (b) 30 kN 3 4 30 kN (b) Statically determinate cantilever portion -27.5 +13.75 (c) +27.5 +120 10 kN/m -120 30 kN 3 4 2 1 (c) Final end moments +101.25 kN.m + -120 _ -27.5 _ (d) + +13.75 (d) Bending moment diagram (+B.M. on compression side -B.M. on tension side) Figure (10.7) Stiffness Factors: From Eq. (10.1), From Eq. (10.2), _ 234 Chapter Ten From Eq. (10.1), Distribution Factors: From Eq. (10.12), fixed end ⁄( ) ⁄( ) hinged end Table (10.4) shows the moment distribution process. Table (10.4) Joints 1 Members 12 21 23 32 34 D.Fs. 0 2/3 1/3 1 0 F.E.Ms. 0 0 -67.5 +67.5 -120 +52.5 0 +120 -120 2 3 Balance joint (3) C.O.Ms. Modified F.E.Ms. +26.25 0 D.Ms. C.O.Ms 0 -41.25 +27.5 +13.75 +13.75 Final End Moments +13.75 0 +27.5 -27.5 +120 -120 233 Moment-Distribution Method 10.6 Moment Distribution Sidesway for Frames: No The procedure for the analysis of frames without sidesway is similar to that for the analysis of continuous beams presented in the preceding sections. However, unlike the continuous beams, more than two members may be connected to a joint of a frame. In such cases, care must be taken to record the computation in such a manner that mistakes are avoided. Example 10.5: Draw the bending moment diagram for the frame shown in Fig. (10.8) by using the moment-distribution method. Solution: Fixed-End Moments: From Table (9.1), kN.m kN.m there is no load on member (2-4) kN.m kN.m kN.m 233 Chapter Ten 2 kN/m 3 2I 4 5 2I 10 m I I 1 2 8000.99 40 kN 10 m (a) 30 m +225 + 30 m -205.7 -186.2 _ _ +225 -116 -115.9 _ -19.3 _ +200 + (b) -92.1 _ - -9.6 (+B.M. on compression side -B.M. on tension side) Figure (10.8) kN.m Stiffness Factors: From Eq. (10.1), From Eq. (10.2), + 233 Moment-Distribution Method From Eq. (10.1), Distribution Factors: From Eq. (10.12), fixed end ⁄( ) ⁄( ) fixed end ⁄( ⁄( ⁄( ) ) ) hinged end Moment Distribution: The moment distribution process is carried out in Table (10.5). The table, which is similar in form to those used previously for the analysis of continuous beams, contains one column for each member end of the frame. Note that the columns for all member ends, which are connected to the same joint, are grouped together, so that any unbalanced 233 Chapter Ten moment at the joint can be conveniently distributed among the members connected to it. Care must be taken when carrying over moments from one end of the member to the other. In this frame, no moment can be carried over from end 1 to end 3 of member (1-3) and from end 2 to end 4 of member (2-4), because joints 1 and 2, which are at fixed supports, will not be released during the moment distribution process. Table (10.5) Joints 1 Members 13 31 34 43 45 D.Fs. 0 0.429 0.571 0.4 F.E.Ms. -100 +100 -150 +150 3 4 2 5 42 24 54 0.3 0.3 0 1 -150 0 0 +150 Balance joint 5 -150 C.O.Ms. Modified F.E.Ms. -75 -100 D.Ms. C.O.Ms. +150 -225 0 +21.45 +28.55 +30 +22.5 +22.5 +15 +14.275 -8.565 -5.71 -2.855 -4.283 +1.630 +1.713 +0.857 +0.815 -0.489 -0.326 -0.163 -0.244 -116.0 +186.2 -6.435 -3.218 D.Ms. C.O.Ms. -150 +10.725 D.Ms. C.O.Ms. +100 +1.225 +0.613 D.Ms. -0.368 C.O.Ms. -0.184 Final End Moments -92.1 +115.9 0 0 +11.25 -4.283 -4.283 -2.142 +1.285 +1.285 +0.643 -0.244 -0.244 -0.122 -205.7 +19.3 +9.6 0 233 Moment-Distribution Method The bending moment diagram is shown in Fig. (10.8-b) Example 10.6: Draw the bending moment diagram for the frame shown in Fig. (10.9) by using the moment-distribution method. 36 kN 64.8 kN/m 2 1 (a) 2I 3 5m I 4 5m 1.5 m +202.5 kN.m + _ -162 -81 -54 _ _ (b) -27 _ (+B.M. on compression side -B.M. on tension side) - -13.5 Figure (10.9) Solution: Fixed-End Moments: 233 Chapter Ten From Table (9.1), there is no load on member (2-4) kN.m kN.m kN.m Stiffness Factors: From Eq. (10.1), Distribution Factors: From Eq. (10.12), ⁄( ) ⁄( ) fixed end fixed end Using these data, the moment distribution is carried out in Table (10.6). The bending moment diagram is shown in Fig. (10.9-b). 233 Moment-Distribution Method Table (10.6) Joints 2 3 4 Members 21 24 23 32 42 D.Fs. 0 1/3 2/3 0 0 F.E.Ms. +54 0 -135 +135 0 D.Ms. 0 +27 +54 +27 +13.5 +162 +13.5 C.O.Ms Final End Moments +54 +27 -81 10.7 Analysis of Frames with Sidesway Frames that are nonsymmetrical or subjected to nonsymmetrical loading have a tendency to sidesway, or deflect horizontally. An example of one such case is shown in Fig. (10.10-a). Here the applied load P will create unequal moments at joints B and C such that the frame will deflect an amount Δ to the right. Since the deformations are assumed to be small, the joints B and C displace by the same amount, Δ, as shown in Fig. (10.10-a). The moment distribution analysis of such a frame, with sidesway, is carried out in two parts. In the first part, the sidesway of the frame is prevented by adding an imaginary roller to the structure, as shown in Fig. (10.10-b). External loads are then applied to this frame, and member end moments are computed by applying the moment-distribution 233 Chapter Ten process in the usual manner. With the member end moments known, the restraining force (reaction) R that develops at the imaginary support is evaluated by applying the equations of equilibrium. Δ P P Δ B C A D (a) Actual frame M moments R R B C B C A D A D (b) Frame with sideway (c) Frame subjected prevented to R M0 moments MR moments δ δ Q (d) Frame subjected to an arbitrary translation δ (MQ moments) Figure (10.10) In the second part of the analysis, the frame is subjected to the force R, which is applied in the opposite direction, as shown in Fig. (10.10-c). The moments that develop at the member ends are determined and superimposed on the moments computed in the first part to obtain the member end 232 Moment-Distribution Method moments in the actual frame. If M, M0, and MR denote, respectively, the member end moments in the actual frame, the frame with sidesway prevented, and the frame subjected to R, then we can write (10.13) A question that arises in the second part of the analysis is how to determine the member end moments MR that develop when the frame undergoes sidesway under the action of R. Since the moment-distribution method cannot be used directly to compute the moments due to the known lateral load R, we employ an indirect approach in which the frame is subjected to an arbitrary known joint translation δ caused by unknown load Q acting at the location and in the direction of R, as shown in Fig. (10.10-d). From the known joint translation, δ, we determine the relative translation between the ends of each member, and we calculate the member fixed-end moments. These moments are distributed by the moment-distribution process to determine the member end moments MQ caused by the yet unknown load Q. Once the member end moments MQ have been determine, the magnitude of Q can be evaluated by the application of equilibrium equations. With the load Q and the corresponding moments MQ known, the desired moments MR due to the lateral load R can 234 Chapter Ten now be determined easily by multiplying MQ by the ratio , that is, ( ) (10.14) By substituting Eq. (10.14) into Eq. (10.13), we can express the member end moments in the actual frame as ( ) (10.15) This method of analysis is illustrated by the following example. Example 10.7: Draw the bending moment diagram for the frame shown in Fig. (10.11) by using the momentdistribution method. constant. Solution: Fixed-End Moments: From Table (9.1), there is no load on member (1-2) kN.m kN.m there is no load on member (3-4) Stiffness Factors: 233 Moment-Distribution Method 40 kN 2 3 5m (a) 7m 4 1 3m 4m 40 kN R 2 3 (b) 4 1 M21=+24 kN.m 2 M34=-24 kN.m 3 (d) (c) 5m 7m M43=-12.2 kN.m 4 M12=+12.1 kN.m 1 Δ R 2 3 R=2.08 kN Δ 3 2 4 4 (e) 1 kN (f) kN 1 Figure (10.11) 233 Chapter Ten M21=+34.5 kN.m 2 (g) M12=+42.3 kN.m 1 Q M34=+45.4 kN.m 3 (h) 7m (i) 5m M43=+71.7 kN.m 4 Figure (10.11) Continued From Eq. (10.1), Distribution Factors: From Eq. (10.12), fixed end ⁄( ) ⁄( ) ⁄( ) ⁄( ) fixed end Moment-Distribution Method 233 Moment-Distribution, Part I: In this part of the analysis, the sidesway of the frame is prevented by adding an imaginary roller at joint 2, as shown in Fig. (10.11-b). The moment-distribution of the fixed-end moments due to the applied external load 40 kN is then performed, as shown in Table (10.7), to determine M0 member end moments. To evaluate the restraining force R that develops at the imaginary roller support, we first calculate the shears at the lower ends of the columns (1-2) and (3-4) by considering the equilibrium of the free bodies of the columns shown in Figs. (10.11-c) and (10.11-d). Next, by considering the equilibrium of the horizontal forces acting on the entire frame (Fig. (10.11-e)), we determine the restraining force R. From Fig. (10.11-c), ∑ + kN From Fig (10.11-d), ∑ + kN From Fig (10.11-e), ∑ + kN 233 Chapter Ten Table (10.7) Joints 1 Members 12 21 23 32 34 43 D.Fs. 0 1/2 1/2 0.417 0.583 0 F.E.Ms. 0 0 -39.2 +29.4 0 0 +19.6 +19.6 -12.3 -17.1 -6.2 +9.8 +3.1 -4.1 -2.1 +1.6 +1.1 -0.7 -0.4 +0.6 +0.2 -0.3 -0.2 +0.1 -24.1 +24.1 D.Ms. C.O.Ms +3.1 +1.6 D.Ms. C.O.Ms +1.1 +0.6 D.Ms. C.O.Ms Final End Moments 3 +9.8 D.Ms. C.O.Ms 2 +0.2 +0.1 +12.1 +24 4 -8.6 -5.7 -2.9 -0.9 -0.5 -0.3 -0.2 -24 -12.2 Note that the restraining force acts to the right, indicating that if the roller would not have been in place, the frame would have swayed to the left. Moment-Distribution, Part II: Since the actual frame is not supported by a roller at joint 2, the frame will be subjected to a lateral load kN at joint 2 in the opposite direction (that is, to the left), as shown in Fig. (10.11-f). Here the joints 2 and 3 are temporarily restrained from rotating, and as a result the fixed-end moments at the ends of the columns are determined by the Eq. (9.12) in Chapter 9, as 233 Moment-Distribution Method Thus, Here Δ is unknown and the method to conduct the analysis is to assume a value for Δ or one of the fixed-end moments. This assumed value will correspond to unknown load Q instead of R. Assuming, kN.m, then from which kN.m The positive sign is necessary since the moment must act clockwise on the column for deflection Δ to the left. The foregoing fixed-end moments are distributed by the usual moment-distribution process, as shown in Table (10.8), to determine the MQ moments caused by unknown load Q. To evaluate the magnitude of Q, we first calculate the shearing forces at the lower ends of the columns by considering their equilibrium and then apply the equation of 233 Chapter Ten equilibrium in the horizontal direction to the entire frame. From Fig. (10.11-g), Table (10.8) Joints 1 Members 12 21 23 32 34 43 D.Fs. 0 1/2 1/2 0.417 0.583 0 F.E.Ms. +50 +50 0 0 +98 +98 -25 -25 -40.9 -57.1 -20.5 -12.5 +10.3 +5.2 +2.6 +5.2 -1.3 -2.2 -1.1 -0.7 +0.6 +0.3 +0.2 +0.3 -0.1 -0.1 -0.05 -0.05 -34.4 -45.5 D.Ms. C.O.Ms -12.5 D.Ms. C.O.Ms +10.3 +5.2 D.Ms. C.O.Ms -1.3 -0.7 D.Ms. C.O.Ms +0.6 +0.3 D.Ms. C.O.Ms Final End Moments ∑ 2 -0.1 -0.05 +42.3 +34.5 + kN From Fig. (10.11-h), ∑ + kN From Fig. (10.11-i), 3 4 -28.6 +7.3 +3.7 -3 -1.5 +0.4 +0.2 -0.2 -0.1 +45.4 +71.7 233 Moment-Distribution Method ∑ + kN which indicates that the moments MQ in Table (10.8) are caused by a lateral load kN. Since the moments are linearly proportional to the magnitude of the load, the desired moments MR due to the lateral load kN must be equal to the moments MQ multiplied by the ratio ⁄ ⁄ . Actual Member End Moments: The actual member end moments, M, can be determined by Eq. (10.15). Thus kN.m kN.m ( kN.m ( kN.m kN.m kN.m 10.8 Analysis of Multistory Frames The forgoing procedure can be extended to the analysis of structures with several independent joint translations. Consider the two-story frame shown in Fig. (10.12-a). This 233 Chapter Ten P3 P3 Δ2 P1 P1 F R2 E E P4 P4 Δ1 P2 R1 P2 D C C A δ1 F C D B (b) Frame with sidesway prevented M0 moments (a) Actual frame M moments E D A B δ2 Q21 F Q12 D C A B (c) Frame subjected to known translation δ1 (MQ1 moments) Q22 E Q11 A F B (d) Frame subjected to known translation δ2 (MQ2 moments) Figure (10.12) structure has two independent joint translations, the sidesway Δ1 of the first story and the displacements Δ2 of the second story. The moment-distribution analysis of this frame is carried out in three parts. In the first part, the sidesway of 232 Moment-Distribution Method both floors of the frame is prevented by adding imaginary rollers at the floor levels, as shown in Fig. (10.12-b). Member end moments M0 that develop in this frame due to the external loads are computed by the moment-distribution process, and the restraining forces R1 and R2 at the imaginary supports are evaluated by applying the equations of equilibrium. In the second part of the analysis, the lower floor of the frame is allowed to displace by a known amount δ1 while the sidesway of the upper floor is prevented, as shown in Fig. (10.12-c). The fixed-end moments caused by this displacement are computed and distributed to obtain the member end moments MQ1. With the member end moments known, the forces Q11 and Q12 at the locations of the roller supports are determined from the equilibrium equations. Similarly, in the third part of the analysis, the upper floor of the frame is allowed to displace by a known amount δ2, as shown in Fig. (10.12-d), and the corresponding member end moments MQ2, and the forces Q12 and Q22, are evaluated. The member end moments M in the actual frame are determined by superposition of the moments computed in the three parts as (10.16) In which c1 and c2 are the constants whose values are obtained by solving the equations of superposition of horizontal forces at the locations of the imaginary supports. 234 Chapter Ten By superimposing the horizontal forces shown in Figs. (10.12-a) to (10.12-d) at joints D and F, respectively, we obtain By solving these equations simultaneously, we obtain the values of the constants c1 and c2, which are then used in Eq. (10.16) to determine the desired member end moments, M. The analysis of multistory frames by the momentdistribution method is quite tedious and time consuming. Therefore, the analysis of such structures is performed on computers using the matrix formulation of the displacement method. Problems 10.1 through 10.4: Determine the reactions and draw the shearing force and bending moment diagrams for the beams shown in Figs. P9.1 to P9.4 by using the moment-distribution method. 10.5 through 10.7: Determine the member end moments for the frames shown in Figs. P9.6, P9.7, P9.9, and P9.10 by using the moment-distribution method.