Moment-Distribution Method

Transcription

1 Moment-istribution Method Structural nalysis y slam Kassimali FRMES Theory of Structures-II M Shahid Mehmood epartment of ivil Engineering Swedish ollege of Engineering & Technology, Wah antt

2 nalysis of Frames Without Sidesway The procedure for the analysis of frames without sidesway is similar to that for the analysis of continuous beam. Unlike the continuous beams, more than two members may be connected to a joint of a frame. 2

3 Example 1 etermine the member end moments and reactions for the frame shown by the moment-distribution method. 2 k/ft 10 ft 10 ft 40 k I = 800 in 4 I = 1,600 in 4 I = 1,600 in 4 I = 800 in 4 E E = 29,000 ksi 30 ft 30 ft 3

4 Solution 1.istribution Factors istribution Factors at Joint, F F F F hecks 4

5 istribution Factors at Joint, E F F F hecks E F F F

6 istribution Factors at Joint E, F E 1 2.Fixed-End Moments (s) 100 k - ft 100 k - ft 0 E 150 k - ft E 150 k - ft 3.Moment istribution 4.Final Moments 6

7 arryover Member Ends istribution Factors 1.Fixed-end Moments 2.alance Joints 3.arryover alance Joints 5.arryover 6.alance Joints arryover 8.alance Joints 9.arryover 10.alance Joints E E Final Moments arryover alance Joints

8 2 k/ft 2 k/ft E k

9 nalysis of Frames With Sidesway onsider the rectangular frame shown in Figure. Δ w Δ P EI = constant qualitative deflected shape of the frame for an arbitrary loading is shown in the figure using an exaggerated scale. 9

10 While the fixed joints and of the frame are completely restrained against rotation as well as translation, the joints and are free to rotate and translate. Δ w Δ P ctual Frame Since the members of the frame are assumed to be inextensible and the deformations are assumed to be small, the joints and displace by the same amount Δ, in the horizontal direction only. 10

11 The M 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. w P Frame with sidesway prevented External loads are then applied to this frame, and MEM are computed by applying the M process in the usual manner. 11

12 With the MEM known, the restraining force (reaction) R that develops at the imaginary support is evaluated by applying the equations of equilibrium. w R P Imaginary roller Frame with sidesway prevented 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 the next slide. 12

13 Δ Δ R Frame subjected to R 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 moments in the actual frame. If M, M O, and M R denote, respectively, the MEM in the actual frame, the frame with sidesway prevented, and the frame subjected to R, then we can write M M O M R 13

14 Δ w Δ M M O M R P M Moments w = R Δ Δ R P + M O Moments M R Moments 14

15 n important question that arises in the second part is how to determine the member end moments M R that develop when the frame undergoes sidesway under the action of R. The MM 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 an unload load Q acting at the location and in the direction of R. From the known joint translation, Δ, we determine the relative translation between the ends of each member, and we calculate the member s in the same manner as done previously in the case of support settlements. 15

16 Δ Δ Q Frame subjected to an arbitrary Translation Δ M Q Moments 16

17 The s thus obtained are distributed by the M process to determine the MEMs M Q caused by the yet-unknown load Q. Once the MEMs M Q have been determined, the magnitude of Q can be evaluated by the application of equilibrium equations. With the load Q and the corresponding moments M Q known, the desired moments M R due to the lateral load R can now be determined easily by multiplying M Q by the ratio R/Q; that is M R Q R M Q y substituting this Equation into the last Equation, w can express the MEMs in the actual frame as M M R Q O M Q 17

18 Example 2 etermine the member end moments and reactions for the frame shown by the moment-distribution method. 40 kn 7 m 5 m EI = constant 3 m 4 m 18

19 Solution 40 kn istribution Factors t joint F F I I 7 7 m EI = constant 5 m t joint F F I I 7 I I 7 I 7 I m 4 m F F hecks 19

20 Part 1: Sidesway Prevented The sidesway of frame is prevented by adding an imaginary roller at joint. R 40 kn Imaginary Roller Frame with sidesway prevented ssuming that joint and of this frame are clamped against rotation, we calculate the s due to the external loads to be 39.2 kn. m 29.4 kn. m 0 20

21 The M of these s is then performed, as shown on the M Table to determine the MEMs M O in the frame with sidesway prevented Member End Moments for Frame with Sidesway Prevented - M O 21

22 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 and by considering the moment equilibrium of the free bodies of thecolumns m 5 m

23 Next, by considering the equilibrium of the horizontal forces acting on the entire frame, we determine the restraining force R to be R = F x 0 R R 2.06 kn 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. 23

24 Part 2: Sidesway Permitted Since the actual frame is not supported by a roller at joint, we neutralize the effect of the restraining force by applying a lateral load R = 2.06 kn in the opposite direction to the frame. R = 2.06 kn Frame subjected to R = 2.06 kn M R Moments M method cannot be used directly to compute MEMs M R due to the lateral load R = 2.06 kn, we use an indirect approach in which the frame is subjected to an arbitrary known joint translation Δ caused by an unknown load Q acting at the location in the direction of R. 24

25 Q Δ Δ Frame subjected to an rbitrary Translation Δ M Q Moments ssuming that the joints and of the frame are clamped against rotation as shown in figure on next slide, s due to the translation Δ are given by 6EI' EI' 5 2 6EI' 49 6EI' 25 25

26 In which negative sign have been assign to the s for the columns, because these moments must act in the clockwise direction, as shown. Δ Δ s due to Known Translation Δ Instead of arbitrarily assuming a numerical value for Δ to compute the s, it is usually more convenient to assume a numerical value for one of the s, evaluate Δ from the expression of that, and use the value of Δ to compute the remaining s. 26

27 Thus, we arbitrarily assume the to be -50 kn.m by solving for Δ, we obtain ' 6EI' EI 50 kn.m by substituting this value of Δ into the expressions for and, we determine the consistent values of these moments to be kN.m These s are then distributed by the usual M process, to determine the MEMs M Q caused by the yet unknown load Q. 27

28 Member End Moments ue to Known Translation Δ - M Q 28

29 To evaluate the magnitude of Q that corresponds to these MEMs, we first calculate shears at the lower ends of the columns by considering their moment equilibrium and then apply the equation of equilibrium in the horizontal direction to the entire frame Q = m m F x 0 Q Q 34.41kN 29

30 which indicates that the moments M Q computed are caused by a lateral load Q = kn. Since the moments are linearly proportional to the magnitude of the load, the desired moment M R due to the lateral load R = 2.06 kn must be equal to the moment M Q multiplied by the ratio R/Q = 2.06/ ctual Member End Moments The actual MEMs, M, can now be determined by algebraically summing the MEMs M O and 2.06/34.41 times the MEMs M Q. 30

31 M M M M M M kn. m 26.1kN. m 26 kn. m 21.3 kn. m 21.3 kn. m NS NS NS NS NS kn. m NS 31

32 40 kn ctual Member End Moments (kn. m) 32

33 Example 3 etermine the member end moments and reactions for the frame shown by the moment-distribution method. 30 k 16 ft 12 ft 12 ft 20 ft EI = constant 8 ft 33

34 Solution istribution Factors t joint I F F 20 I k 16 ft 12 ft 12 ft 20 ft 8 ft EI = constant 34

35 Solution istribution Factors t joint I 3 I F 0.49 F I 3 I I 3 I k 16 ft 12 ft 12 ft 20 ft 8 ft EI = constant 35

36 MEMs due to an rbitrary Sidesway Δ Since no external loads are applied to the members of the frame, the MEMs M O in the frame restrained against sidesway will be zero. To determine the MEMs M due to the 30-k lateral load, we subject the frame to an arbitrary known horizontal translation Δ at joint. Figure on the next slide shows a qualitative deflected shape of the frame with all joints clamped against rotation and subjected to the horizontal displacement Δ at joint. 36

37 k Δ Δ ' 3 3 ' 4 s due to an rbitrary Translation Δ 37

38 MEMs due to an rbitrary Sidesway Δ Note that, since the frame members are assumed to be inextensible and deformations are assumed to be small, an end of a member can translate only in a direction perpendicular to the member. From this figure, we can see that the relative translation Δ between the ends of members in the direction perpendicular to the member can be expressed in terms of the joint translation Δ as ' 5 4 ' 1.25' 38

39 MEMs due to an rbitrary Sidesway Δ Similarly, the relative translation for members and are given by ' 1 ' 2 3 ' ' 1.417' ' 1.202' The s due to the relative translation are 6EI 6 6EI 1.25' 2 20 EI ' ' 39

40 MEMs due to an rbitrary Sidesway Δ in which the s for members and are W (positive), whereas those for member are W (negative). If we arbitrarily assume that then 6EI 1.202' k -ft EI' and therefore 54.1k - ft 61.3 k - ft 40

41 The s are distributed by the M process to determine the MEMs M Q. To determine the magnitude of the load Q that corresponds to the MEMs M Q we first calculate the shears at the ends of the girder by considering the moment equilibrium of the free body of the girder as shown Evaluation of Q

42 M TLE Member End Moments ue to Known Translation Δ - M Q 42

43 The girder shears (5.58 k) thus obtained are then applied to the free bodies of the inclined members and. Next, we apply the equations of moment equilibrium to members and to calculate the horizontal forces at the lower ends of these members Evaluation of Q 43

44 The magnitude of Q can now be determined by considering the equilibrium of horizontal forces acting on the entire frame as shown below F x 0 Q Q k Q = k Evaluation of Q 44

45 ctual MEMs The actual MEMs, M, due to the 30-k lateral load can now be evaluated by multiplying the moments M Q computed in Table by theratio 30/Q=30/19.49: M M M M M M k - ft 87 k - ft 86.8 k - ft 85 k - ft 85 k - ft NS NS NS NS NS NS 45

46 Member End Forces k Evaluation of Q 46

47 Support Reactions 30 k 17.2 k 12.8 k 85.1 k-ft 8.59 k 8.59 k 47