Moment Distribution Method for Frames without Sidesway

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The moment-distribution method for frames without sidesway is used to analyze statically indeterminate rigid frames that are properly restrained against sidesway. This method is similar to the moment-distribution method for continuous beams, except there can be more than two members meeting in a frame at a joint [1]. Therefore, one must be aware of the calculated values as to avoid mistakes [1]. The unbalanced moment at a joint is the sum of fixed-end beam moments, or carry over moments, and is then distributed to the connecting members according to their distribution factors.


Considerations for Frames without Sidesway

Frames without sidesway do not consider lateral displacements at the joints of frames. The equation to determine the "sidesway degree of freedom, ss" is as follows [1]:

$ ss = 2j - [2(f+h) + r +m] $

where:

  • j is equal to the number of joints in the frame
  • f is equal to the number of fixed-ends of the frame
  • h is equal to the number of hinged ends in the frame
  • r is the number of roller ends in the frame
  • m is the number of members in the frame


The equation of sidesway degree of freedom must be equal to zero. Therefore two times the number of joints in the member must equal the summation of two times the number of fixed-ends, two times the number of hinged ends, the number of roller ends, and the number of members in the frame.


If "ss" does not equal zero then the frame must consider sidesway to solve the moment distribution. See Moment Distribution Method for Frames with Sidesway for more details.


Exceptions

There is one condition in which a frame may be analysed using the moment distribution method for frames without sidesway despite having a degree of freedom not equal to zero. This occurs when the frame is completely symmetric about an axis (usually a vertical axis). The frame must be symmetric in all aspects including, physical shape, structural properties such as the moment of inertia "I" and Young's modulus "E", and loading. The loads must be symmetric in location, type, and magnitude.

Important Terms

Definitions:

Bending Stiffness.
"Equal to the product of the elastic modulus E and the area moment of inertia I of the beam cross-section about the axis of interest." [2]
Deflection
"The degree to which a structural element is displaced under a load. It may refer to an angle or a distance." [3]
Fixed End Moment
"Reaction moment developed in a beam member under certain load conditions with both ends fixed." [4]
Rigid Frame
"The load-resisting skeleton constructed with straight or curved members interconnected by mostly rigid connections, which resist movements induced at the joints of members. Its member can take bending moment,shear and axial loads." [5]
Statically Indeterminate
"The static equilibrium equations are insufficient for determining the internal forces and reactions on that structure." [6]
Young's Modulus
"A measure of the stiffness of an elastic material and is a quantity used to characterize materials. It is defined as the ratio of the stress along an axis over the strain along that axis in the range of stress in which Hooke's law holds." [7]


Assumptions

The following assumptions were adapted from the Kassimali Structural Analysis textbook. [1] For simplicity:

  • Assume that the frames that will be analyzed will all have fixed-end supports.
  • Assume that the beams and frames are continuous.
  • Assume the counter clockwise direction to be positive and the clockwise direction to be negative.
  • Assume Young's modulus of elasticity (E) is constant.


Derivations

The following derivations were adapted from the Kassimali Structural Analysis textbook. [1]

Bending Stiffness

For a fixed-end member the moment can be denoted as:


$ M=\left(\begin{array}{c}\frac{4EI}{L}\end{array}\right)\theta $

Where:

$ M $ is the moment

$ E $ is Young's modulus of elasticity

$ I $ is the inertia

$ L $ is the length of the member (column or beam)

$ \theta $ is the angle of deflection


However, bending stiffness will cause a unit rotation and thus can be represented the same as moment. The angle of the deflectioncan be set to 1 radian.


Thus, $ \overline{K} =\left(\begin{array}{c}\frac{4EI}{L}\end{array}\right) $

Where:

$ \overline{K} $ is the bending stiffness


It can be assumed that Young's modulus of elasticity is constant and can simplify the bending stiffness to :


$ {K} =\left(\begin{array}{c}\frac{I}{L}\end{array}\right) $


This is the simplified form of each bending stiffness for a fixed-end. However, if the end of a member is a hinge, the bending stiffness is multiplied by a factor of 0.75 because it is reduced by a factor of 0.25 from that of a fixed-end. [1]


$ {K} =\left(\begin{array}{c}\frac{3I}{4L}\end{array}\right) $

Distribution Factor

The distribution factor indicates the areas of the applied moment on a specific section of a frame. The distribution factor is typically analyzed at a joint. It is represented as a percentage and thus the values of each distribution from each section of a joint should add up to 1 (or 100%).[8]


The equation for the distribution factor is as follows:

$ DF=\frac{K}{\sum K} $

Where:

$ DF $ is the distribution factor

$ K $ is the bending stiffness

Carryover

Carryover moment is a moment that evens out an unbalanced moment (sum of all moments must equal zero). It acts similar to that of a fixed-end moment. The moment created from an x-distance force has a moment at the opposite end (point B). This is called the carryover moment.[9]

The Carryover Factor (COF) is the percentage of moment applied at one end of a beam that affects the opposite end of the beam. A beam with a fixed-end and a hinged end has a COF of 0.5, which is that of the moment applied at the hinged end. There is no COF for a beam with two hinged ends. [1]

Fixed-End Moment

To find more information about fixed-end moment (FEM), please refer to Unbalanced Members and Fixed End Moments.


Fixed-end moment equations for frames or beams with a:


  • Point load between two fixed-ends [9]

$ FEM = \frac{Pab^{2}}{L^{2}} $

Where:

$ FEM $ is the fixed end moment

$ P $ is the point load

$ a $ is the distance from the joint to the point load

$ b $ is the distance from the point load to the following joint


  • Uniform distributed load between two fixed-ends [9]

$ FEM = \frac{wL^{2}}{12} $

Where:

$ w $ is the distributed load applied

Explanation of the Procedure

This method was adapted from the video Explaining the Moment Distribution Method. [10]

1) Determine whether sidesway must be considered for this frame.

2) Fix all the joints

If it is assumed that all joints are fixed, there is no rotation and, therefore, a fixed-end moment is created at both ends of all the members.

3) Calculate Fixed-End Moments

Draw each member separately and draw the FEM in the positive direction at both ends of each member. Use FEM charts and equations to calculate the Moments.

4) Balance each joint using equilibrium

Usually the moments do not add up to 0, therefore use equilibrium by adding a Balancing Moment in the direction that makes the total moment at the joint equal 0.

5) Distribute Balancing Moment to connecting members based on Distribution Factor

The Distribution Factor is a ratio of the stiffness of the member to the stiffness of the joint. Essentially, a portion of the Balancing Moment - calculated using the Distribution Factor - is applied to the moment at the end of each member at the joint.

6) Carry over moments to opposite end of members using Carry Over Factor and equilibrium

The Carry Over Factor (COF) is a portion of the distributed moment that gets carried over to the other end of the member. The process of carrying over the distributed moment is an iterative process, meaning that the COF is reapplied to the balancing moment repeatedly until the balancing moment approaches 0. Once the balancing moment is near 0, that joint is ‘locked’ in place; no rotation can happen because the joint is in equilibrium.

Example

Moment Distribution for Frames without Sidesway

Step 1: Confirm that the frame does not need to consider sidesway

Use the equation to determine the sidesway degree of freedom, ss, to confirm that the frame does not need to consider sidesway when calculating the moment using the moment distribution method.


$ ss = 2j-[2(f+h)+r+m] $

where:

  • $ j = 5 $
  • $ f = 2 $
  • $ h = 1 $
  • $ r = 0 $
  • $ m = 4 $


$ ss = 2(5) - [2(2+1)+0+4] $

$ ss = 10 - (6+4) $

$ ss = 0 $


Therefore this beam has no degrees of freedom and sidesway does not need to be considered.

Step 2: Fix all the joints

Since supports A and D are already fixed-ends, we can look over this step for these joints. Since support E is a pin, we assume it is a fixed support.

Step 3: Calculate Fixed-End Member

Fixed-end moments (FEMs) are the moments that are applied at fixed-ends due to external loading.


The fixed-end moments for each of the members in the frame are as follows:


$ FEM_{AB}= \frac{(50 kN)\cdot (2m)\cdot (3m)^{2}}{(5m)^{2}}=+36kN \cdot m $


$ FEM_{BA}= \frac{(50 kN)\cdot (3m)\cdot (2m)^{2}}{(5m)^{2}}=-24kN \cdot m $


$ FEM_{BC}= \frac{(20 kN/m)\cdot (6m)^{2}}{12}=+60kN\cdot m $


$ FEM_{CB}= \frac{(20 kN/m)\cdot (6m)^{2}}{12}=-60kN\cdot m $


$ FEM_{CD}= 0kN\cdot m $


$ FEM_{DC}= 0kN\cdot m $


$ FEM_{CE}= 0kN\cdot m $


$ FEM_{EC}= 0kN\cdot m $


At joint C, there is no fixed-end moment from the member CD; there is only a fixed-end moment coming from the member CB.

Step 4: Balancing Moments

To balance the joints, the previous moments at each joint are analyzed.

At joint B, there is a moment of + 60 kNm from member BC and a moment of - 24 kNm from member BA. There is a difference of - 36 kNm that needs to be added to balance out the moment equilibrium. The unbalanced moment must be then distributed across each member accordingly. Therefore the Distribution Factor must be calculated.

At joint C, the unbalanced moment is equal and opposite in direction to the fixed-end moment of member CD. The unbalanced moment is then equal to + 60 kNm. Again, this moment must be distributed across each of the members of the frame from joint C.

Step 5: Calculate Distribution Factor

The distribution factor is analyzed at each joint.


Joint B

$ DF_{BA}=\frac{\frac{200}{5}}{\frac{200}{5}+\frac{400}{6}}=0.375 $


$ DF_{BC}=\frac{\frac{400}{6}}{\frac{200}{5}+\frac{400}{6}}=0.625 $


Joint C

$ DF_{CB}=\frac{\frac{400}{6}}{\frac{200}{5}+\frac{400}{6}+\frac{400}{2}}=0.217 $


$ DF_{CD}=\frac{\frac{200}{5}}{\frac{200}{5}+\frac{400}{6}+\frac{400}{2}}=0.130 $


$ DF_{CE}=\frac{\frac{400}{2}}{\frac{200}{5}+\frac{400}{6}+\frac{400}{2}}=0.652 $


Joint E

$ DF_{EC} = \frac{\frac{400}{2}}{\frac{400}{2}} = 1.0 $

At joint E, 100% of the distribution is from member EC.


Balancing Joints

NOTE: Only joint B and joint C need to be analysed because their distribution factor is not 1.0.


Joint B

If joint B is analyzed, then sections BA and BC will be inspected. The difference in moments at joint B is equal to:

$ - [60 - 24 ] kN\cdot m = -36 kN\cdot m $ .


Since the section BA carries 37.5% of the load, the balanced joint will be:

$ - 36 kN\cdot m \times 0.375 = -13.5 kN\cdot m $ .


Section BC carries 62.5% of the load. Therefore, the balanced joint will be:

$ - 36 kN\cdot m \times 0.625 = -22.5 kN\cdot m $ .


Joint C

Joint C is analyzed the same way as joint B; sections CB and CD represent the moments at joint C. However, the difference in moments between joint CB and CD is:

$ - [ -60 - 0 ] kN\cdot m = + 60 kN\cdot m $ .

This is because there is no additional moment for section CD (there is no external focus acting on this column).


Since the section CB carries 21.7% of the load, the balanced joint will be:

$ + 60 kN\cdot m \times 0.217 = +13.02 kN\cdot m $ .


Section CD carries 13.0% of the load. Resultantly, the balanced joint will be:

$ + 60 kN\cdot m \times 0.130 = + 7.80 kN\cdot m $ .


Section CE carries 65.2% of the load, making the balance joint:

$ + 60 kN\cdot m \times 0.652 = + 39.12 kN\cdot m $ .


Below is a table to demonstrate the work described above.

Moment Ends AB BA BC CB CD CE DC EC
Distribution Factor 0.375 0.625 0.217 0.130 0.652 1.0
Fixed-End Moments + 36 - 24 + 60 - 60
Balance Joints - 13.5 - 22.5 + 13.02 + 7.80 + 39.12

Step 6: Carry Over Factor

Since this example is a frame with two fixed-ends and a hinged end, the carryover factor (COF) is equal to half; the COF for the fixed ends is going to be 0.5 and and the COF for the pinned end will be zero (0). Half of the remaining moment for fixed ends will be distributed to the other portion of the joint. Therefore, as an example, half the moment in member BA will be distributed to member AB. This process is repeated continuously until the carryover moments/ moments from balancing the joints approaches zero. [1]


Half of the moment from BA carries over to member AB:

$ - 13.5 kN\cdot m \times 0.5 = - 6.75 kN\cdot m $


Half of the moment from BC that carries over to member CB:

$ - 22.5 kN\cdot m \times 0.5 = - 11.25 kN\cdot m $


Half of the moment moment from CB that carries over to member BC:

$ + 13.02 kN\cdot m \times 0.5 = + 6.51 kN\cdot m $


Half of the moment from CD that carries over to member DC:

$ + 7.80 kN\cdot m \times 0.5 = + 3.90 kN\cdot m $


None of the moment from CE will carry over to member EC because it is a pinned end; the moment at any pinned end will be zero:

$ + 39.12 kN\cdot m \times 0 = 0.0 kN\cdot m $


A sample of the one carryover has been shown in the table below:

Moment Ends AB BA BC CB CD CE DC EC
Distribution Factor 0.375 0.625 0.217 0.130 0.652 1.0
Fixed-End Moments + 36 - 24 + 60 - 60
Balance Joints - 13.5 - 22.5 + 13.02 + 7.80 + 39.12
Carryover - 6.75 + 6.51 - 11.25 + 3.90


Above, we see that the moment after the carryover procedure is not close to zero; therefore the system is not yet in equilibrium. The above process must be carried out until the moments in the carryover process/balancing the joints are negligible. For this particular example, joint B and joint C are being inspected because they have a distribution factor (more than one member connects at the joint). For point A and point D there is only one member that is connected here. Therefore the distribution factor at A is 100%; all moments applied at point A is distributed across member AB and all moments applied at point D is distributed across DC. Each joint is analysed. Therefore the difference in moments at joint B is -6.51 kNm because column BA did not have a carryover moment from column AB. The difference in moment is then divided according to the distribution factor for each member (BA and BC). The same procedure is followed through for joint C. As shown, the difference in moment at joint C is + 11.25 kNm. There is no carryover moment for joint CD or CE, therefore only joint CB has a moment. This remaining moment is also distributed to each of the members streaming from joint C according to the distribution factor. This process is carried out until the moments become negligible (approach zero).

Once the moments approach zero, all of the moment adjustments are summed and as a result, this will give the final moments of each section of a frame without sidesway.

The process is presented in a table below.

Table of Data

Moment Ends AB BA BC CB CD CE DC EC
Distribution Factor 0.375 0.625 0.217 0.130 0.652 1.0
Fixed-End Moments + 36 - 24 + 60 - 60 0.0
Balance Joints - 13.5 - 22.5 + 13.02 + 7.80 + 39.12 0.0
Carryover - 6.75 + 6.51 - 11.25 + 3.90
Balance Joints - 2.44 - 4.07 + 2.44 + 1.46 + 7.34
Carryover - 1.22 + 1.22 - 2.04 + 0.73
Balance Joints - 0.46 - 0.76 + 0.44 + 0.27 + 1.33
Carryover - 0.23 + 0.22 - 0.38 + 0.14
Balance Joints - 0.08 - 0.14 + 0.08 + 0.05 + 0.25
Carryover - 0.04 + 0.04 - 0.07 + 0.025
Final Moments + 27.4 - 40.5 + 40.4 - 57.8 + 9.6 + 48.0 + 4.8 0.0


$ \sum AB = (+36 - 6.75 - 1.22 - 0.23 - 0.04) = +27.4 kN\cdot m $

$ \sum BA = (-24 - 13.5 - 2.44 - 0.46 - 0.08) = -40.5 kN\cdot m $

$ \sum BC = (+60 - 22.5 + 6.51 - 4.07 + 1.22 - 0.76 + 0.22 - 0.14 + 0.04) = +40.4 kN\cdot m $

$ \sum CB = (-60 + 13.02 - 11.25 +2.44 - 2.04 + 0.44 - 0.38 +0.08 - 0.07) = - 57.8 kN\cdot m $

$ \sum CD = (+7.80 + 1.46 + 0.28 + 0.05) = + 9.6 kN\cdot m $

$ \sum CE = (+ 39.12 + 7.34 + 1.39 = 0.25) = + 48.0 kN\cdot m $

$ \sum DC = (+ 3.9 + 0.73 + 0.14 + 0.025) = + 4.8 kN\cdot m $

$ \sum EC = 0.0 kN\cdot m $


Therefore, the moment at A is 27.4 kNm in the counter clockwise direction. The moment at at joint B along member BA is 40.5 kNm in the clockwise direction. The moment at joint B along member BC is 40.4 kNm in the counter clockwise direction. The moment at joint C along member CB is 57.8 kNm in the clockwise direction. The moment at joint C along member CD is 9.6 kNm in the counter clockwise direction. The moment at joint C along member CE is 48.0 kNm in the counter clockwise direction. The moment at joint D is 4.8 kNm in the clockwise direction. The moment at joint E is equal to 0 kNm. Point E is supported with a pin, therefore the moment must be equal to zero.


Final Member End Moments


As a final check for this process, all the moments at each joint should sum to zero. Since this is an approximate method and due to some rounding errors, the summation of moments is reasonably close.

Want More Information?

Here are some YouTube Videos to further explain this method.


Related Topics

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References

  1. 1.0 1.1 1.2 1.3 1.4 1.5 1.6 1.7 Kassimali, A. (2011). Structural Analysis: SI Edition (4th ed.). Stamford, CT: Cengage Learning.
  2. Bending Stiffness. In Wikipedia. Retrieved November 11, 2013, from http://en.wikipedia.org/wiki/Bending_stiffness
  3. Deflection. In Wikipedia. Retrieved November 11, 2013, from http://en.wikipedia.org/wiki/Deflection_(engineering)
  4. Fixed End Moment. In Wikipedia. Retrieved November 11, 2013, from http://en.wikipedia.org/wiki/Fixed_end_moment
  5. Rigid Frame. In Wikipedia. Retrieved November 11, 2013, from http://en.wikipedia.org/wiki/Rigid_frame
  6. Statically Indeterminate. In Wikipedia. Retrieved November 11, 2013, from http://en.wikipedia.org/wiki/Statically_indeterminate
  7. Young's Modulus. In Wikipedia. Retrieved November 11, 2013, from http://en.wikipedia.org/wiki/Young's_modulus
  8. Thandavamoorthy, T.S. (2011). Structural Analysis / Thandavamoorthy. New Delhi. Oxford University Press.
  9. 9.0 9.1 9.2 Leet, Kenneth. (2011). Fundamentals of Structural Analysis. New York, NY. McGraw-Hill.
  10. Structurefree. “Explaining the Moment Distribution Method – Structural Analysis.” On-line video. YouTube. YouTube, Sept. 25, 2013/ Web. Nov. 10, 2013.