Topic4_Moment Distribution with Stiffness Factor Modification.pptx

Theory of Structures II
Moment Distribution
Stiffness Factor Modification
Reference: Structural Analysis 10th Edition in SI Units by R.C. Hibbeler

1. CED 426 Structural Theory II Lecture 18 Moment Distribution: Stiffness-Factor Modification and Frames no Sideways Mary Joanne C. Aniñon Instructor

2. Stiffness-Factor Modification In the previous examples of moment distribution, we have considered each beam span to be constrained by a fixed support at its far end when distributing and carrying over the moments For this reason, we have calculated the stiffness factors, distribution factors, and carry-over factors based on the case shown in Fig. 11-9, where K = 4EI/L and the carry over factor is 1/2

3. Stiffness-Factor Modification In some cases, it is possible to modify the stiffness factor of a span and thereby simplify the process of moment distribution. Three cases where this frequently occurs will be considered: Member Pin Supported at Far End Symmetric Beam and Loading Symmetric Beam with Antisymmetric loading

4. Many beams are supported at their ends by a pin (or roller) as shown in Fig. 11–10a. We can determine the stiffness factor at joint A of this beam by applying a moment M at the joint and relating it to the angle . To do this we must find the shear in the conjugate beam at A’   Member Pin Supported at Far End

5. From the conjugate beam Fig. 11-10b, we have Member Pin Supported at Far End

6. Also, note that the carry-over factor is zero since the pin at B does not support a moment. By comparison, then, if the far end were fixed supported, the stiffness factor K = 4EI/L would have to be modified by 3/4 to model the case of having the far end pin supported. Member Pin Supported at Far End

7. Symmetric Beam and Loading If a beam is symmetric with respect to both its loading and geometry, the bending moment diagram for the beam will also be symmetric. As a result, a modification of the stiffness factor for the center span can be made, so that moments in the beam only have to be distributed through a joint lying on either half of the beam.

8. Symmetric Beam and Loading To develop the appropriate stiffness-factor modification, consider the beam shown in Fig.11-11a. Due to the symmetry, the internal moments at B and C are equal. Assuming this value to be M, the conjugate beam for span BC is shown in Fig.11-11b

9. Symmetric Beam and Loading

10. Symmetric Beam with Antisymmetric loading If the asymmetric beam is subjected to antisymmetric loading, the resulting moment diagram will be antisymmetric. Considering the beam Fig. 11-12a, due to the antisymmetric loading, the internal moment at B is equal but opposite to that of C Assuming the value to be M, the conjugate beam for its center span BC is shown in Fig. 11-12b.

11. Symmetric Beam with Antisymmetric loading

12. Example 1 Determine the internal moments at the supports for the beam shown in Fig. 11-13a

13. Example 1 Step 1.a. Identify the joints and spans in the beam: A, B, C, D and span AB, BC, CD Step 1.b. Calculate member and joint stiffness factors for each span. By inspection, the beam and loading are symmetrical We will apply K=2EI/L to calculate the stiffness factor of the center span BC and therefore use only the left half of the beam for analysis. The analysis can be shortened even further by using K = 3EI/L for calculating the stiffness of segment AB since the far end A is pinned.

14. Example 1 Step 1.b. Calculate member and joint stiffness factors for each span. Step 1.c. Determine the distribution factors (DF).    

15. Example 1 1.d. Determine the fixed-end moments FEM BA FEM AB =0 FEM BC FEM CB FEM CD FEM DC =0

16. Example 1 FEM BA FEM AB =0 FEM BC FEM CB FEM CD FEM DC =0 Joint A B Member AB BA BC DF 1 0.667 0.333 FEM Dist. 60 48.9 -133.33 24.4 ∑M 108.9 -108.9 The data are listed in the table. Calculating the stiffness factors as shown considerably reduces the analysis, since only joint B must be balanced and carry-overs to joint A and C are not necessary.

17. Moment Distribution For Frames: No Sidesway Application of the moment distribution method for frames having no sideways follows the same procedure as that given for beams. To minimize the chance for errors, it is suggested that the analysis be arranged in a tabular form, as in the previous examples The distribution of moments can be shortened if the stiffness factor of a span can be modified.

18. Example 2 Determine the internal moments at the joints of the frame shown in Fig. 11-15a. There is a pin at E and D and a fixed support at A. EI is constant

19. By inspection, the pin at E will prevent the frame from sidesway. Example 2

20. Example 2 Step 1.a. Identify the joints and spans in the beam: A, B, C, D, E and span AB, BC, CD, CE Step 1.b. Calculate member and joint stiffness factors for each span. The stiffness factors of CD and CE can be calculated using K = 3EI/L since the far ends D and E are pinned.  

21. Example 2 Step 1.c. Determine the distribution factors (DF).    

22. Example 2 1.d. Determine the fixed-end moments

23. Example 2 Joint A B C C D E Member AB BA BC CB CE CD DC EC DF 0.545 0.455 0.330 0.372 0.298 1 1 FEM Dist. 73.6 -135 61.4 135 -44.6 -50.2 -40.2 CO Dist. 36.8 12.2 -22.3 10.1 30.7 -10.1 -11.5 -9.1 CO Dist. 6.1 2.8 -5.1 2.3 5.1 -1.7 -1.9 -1.5 CO Dist. 1.4 0.4 -0.8 0.4 1.2 -0.4 -0.4 -0.4 CO Dist . 0.2 0.1 -0.2 0.1 0.2 -0.1 -0.1 ∑M 44.5 89.1 -89.1 115 -64.1 -51.2

24. Moment Diagram for frame + - - +