Theory of Structures II Moment Distribution Frames with Sideway Reference: Structural Analysis 10th Edition in SI Units by R.C. Hibbeler

1. CED 426
Structural Theory II
Lecture 19
Moment Distribution:
Frames with Sideways
Mary Joanne C. Aniñon
Instructor

2. Moment Distribution for Frames: Sideway
• Frames that are nonsymmetrical or subjected to
nonsymmetrical loadings have a tendency to
sidesway.
• An example of one such case is shown in Fig.11–16a.
• Here the applied loading P will create unequal
moments at joints B and C such that the frame will
sidesway an amount ∆ to the right.
• To determine this deflection and the internal
moments at the joints using moment distribution, we
will use the principle of superposition.

3. Moment Distribution for Frames: Sideway
• The frame is first considered held
from sidesway by applying an
artificial joint support at C, Fig. 11–
16b.
• Using moment distribution and
statics the restraining force R is
determined.
• The equal, but opposite, restraining
force is then applied to the frame,
Fig. 11–16c, so that the moments
in the frame can be calculated

4. Moment Distribution for Frames: Sideway
• One method for doing this is to assume a numerical value for one of the
internal moments, say M′BA.
• Using moment distribution and statics, the deflection ∆′ and external force
R′ corresponding to this assumed value of M′BA are calculated.
• Since the force R′ develops moments in the frame that are proportional to
those developed by R, then the moment at B developed by R will be MBA =
MBA′ (R>R′).
• Finally, addition of the joint moments for both cases, Figs. 11–16b and 11–
16c, will yield the actual moments in the frame, Fig. 11–16a.

5. Procedure For Analysis
• The following procedure provides a general method for determining
the end moments on the beam spans using moment distribution.
1. Determine the Distribution Factors and Fixed-End Moments
2. Perform Moment-Distribution Process

6. Distribution Factors and Fixed-End Moments
1.a. Identify the joints and spans on the beam.
1.b. Calculate member and joint stiffness factors for each span.
1.c. Determine the distribution factors (DF). Remember that DF = 0 for
fixed end and DF = 1 for an end pin or end roller support
1.d. Determine the fixed-end moments

7. Moment-Distribution Process
Assume all joints are initially locked. Then:
2.a. Determine the moment that is needed to put each joint in
equilibrium
2.b. Release or unlock the joints and distribute the counterbalancing
moments into the members at each joint
2.c. Carry these moments over to the other end of the member by
multiplying each moment by the carry-over factor +
1
2

8. Example 1
• Determine the moments at
each joint of the frame show
in Fig. 11-18a

9. Example 1 | Solution:
• Step 1.a. First, we consider the frame held from sidesway as shown in Fig.
11-18b.
=
The resultant moment in the frame (a) is equal to the sum of those calculated for the
frame in (b) plus the proportionate amount of those for the frame in (c).
𝑀 = 𝑀𝑤𝑖𝑡ℎ 𝑟𝑒𝑠𝑡𝑟𝑎𝑖𝑛𝑖𝑛𝑔 𝑓𝑜𝑟𝑐𝑒 + (
𝑅
𝑅′
)𝑀𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑅

10. Example 1 | Solution:
Step 1.b. Identify the joints and
spans in the beam: A, B, C, D and
span AB, BC, CD

11. Example 1 | Solution:
Step 1.b. Identify the joints and
spans in the beam: A, B, C, D and
span AB, BC, CD

12. Example 1 | Solution:
Step 1.c. Calculate member and joint
stiffness factors for each span.
• The stiffness factor of each span is
calculated using 4EI/L.
𝐾𝐴𝐵 =
4𝐸𝐼
5
𝐾𝐵𝐶 =
4𝐸𝐼
5
𝐾𝐷𝐶 =
4𝐸𝐼
5

13. Example 1 | Solution:
• Step 1.d. Determine the
distribution factors (DF).
𝐷𝐹 =
𝐾
𝐾
𝐷𝐹𝐴𝐵 = 0 (fixed)
𝐷𝐹𝐵𝐴 = 𝐷𝐹𝐵𝐶 =
4𝐸𝐼/5
4𝐸𝐼
5
+
4𝐸𝐼
5
= 0.5
𝐷𝐹𝐶𝐵 = 𝐷𝐹𝐶𝐷 =
4𝐸𝐼/5
4𝐸𝐼
5
+
4𝐸𝐼
5
= 0.5
𝐷𝐹𝐷𝐶 = 0 (fixed)

14. Example 1 | Solution:
1.e. Determine the fixed-end
moments

15. Example 1 | Solution:
Step 2:
Moment Distribution Table

16. Example 1 | Solution:
Step 3.a: Solve the reactions.
• Using the results, the equations of
equilibrium are applied to the free-body
diagrams of the columns in order to
determine 𝐴′𝑥 and 𝐷′𝑥 in Fig. 11-18e.

17. Example 1 | Solution:
Step 3.b: Solve the restraining force, R.
• With 𝐴′𝑥 and 𝐷′𝑥 determined, we can obtain
the joint restraint R from the free-body
diagram of the entire frame
A’x = 1.73 kN
D’x = 0.81 kN

18. Example 1 | Solution:
Step 3.c: The value of opposite R.
• An equal but opposite value of R = 0.92kN
must now be applied to the frame at C
and the internal moments at the joints
must be calculated.

19. Example 1 | Solution:
Step 4: Assume the FEM due to deflection
• To begin, R′ causes the frame to deflect ∆′ as shown
in Fig.11–18f. Here the joints at B and C are
temporarily restrained from rotating, and as a result
the fixed-end moments at the ends of the columns
are determined from the formula for deflection
found on the inside back cover, that is,

20. Example 1 | Solution:
Step 4: Assume the FEM due to deflection
• Since both B and C happen to be displaced the
same amount , and AB and DC have the same E, I,
and L, the FEM in AB will be the same as that in DC.
• To find R’, first assume a certain value for the fixed-
end moments:

21. Example 1 | Solution:
Step 5: Moment Distribution Table
.
𝑀𝑡𝑟𝑢𝑒
𝑅
=
𝑀𝑎𝑠𝑠𝑢𝑚𝑒
𝑅′
𝑀𝑡𝑟𝑢𝑒 = 𝑀𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑅 =
𝑅
𝑅′
× 𝑀𝑎𝑠𝑠𝑢𝑚𝑒

22. Example 1 | Solution:
Step 6.a: Solve the reactions.
• From equilibrium, the
horizontal reactions at A and D
are calculated
Step 6.b: Solve the R’.

23. Example 1 | Solution:
Step 7: Solve the resultant moment.
𝑀 = 𝑀𝑤𝑖𝑡ℎ 𝑟𝑒𝑠𝑡𝑟𝑎𝑖𝑛𝑖𝑛𝑔 𝑓𝑜𝑟𝑐𝑒 + (
𝑅
𝑅′
)𝑀𝑎𝑠𝑠𝑢𝑚𝑒
𝑀 = 𝑀𝑤𝑖𝑡ℎ 𝑟𝑒𝑠𝑡𝑟𝑎𝑖𝑛𝑖𝑛𝑔 𝑓𝑜𝑟𝑐𝑒 + 𝑀𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑅