Moment Distribution Method for Frames with Sidesway
Contents
Introduction
The moment distribution method is a structural analysis method used to analyze continuous beams and frames. There are multiple methods available to assist with the analysis of structures and the moment distribution method is classified as part of the displacement method family. The moment distribution method resembles very similarly the slopedeflection method but still contains its own traits. The moment distribution method differentiates itself by solving equilibrium of moments of joints in an iterative style instead of the traditional simultaneous method.^{[1]}
The moment distribution method is described in detail here. In this section we will take a more detailed look at the moment distribution method for frames with sidesway. Frames with sidesway are frames whose joints may undergo both rotation and translation when subjected to external loads.^{[1]}
In this section we will begin by covering key terms and concepts required to the understanding of the moment distribution method. A detailed stepbystep process required for the analysis of frames with sidesway will be provided. Lastly, we will analyse a sample frame with sideway using the previously described steps.
Important Terms
 Statics^{[2]}
 Statics refer to a body at rest under equilibrium conditions and all the forces that act on it. It also deals with bodies that are moving at a constant velocity under equilibrium conditions.
 Sign Convention
 In order to understand momentdistribution method a sign convention must be adopted for positive and negative moment and shear forces.^{[1]} The positive sign convention we will use for momentdistribution method will consist of shear acting upwards and counterclockwise moments. The negative sign convention we will use will consist of shear acting downwards and clockwise moments. Consider an arbitrarily loaded beam such as the one in Fig 1.1. When a beam is loaded downwards and cut at any point along the length of the beam, the load on the beam will result in a positive internal moment on the left side of the cut, a negative internal moment on the right side of the cut, a positive end moment on the right end of the beam and a negative end moment on the left end of the beam as shown in figure 1.2.^{[3]}In this case, the shear will be positive on the right side of the cut and negative on the left side of the cut.
 End Moments
 End moments refer to the moment at the end of a beam or portion of a frame. Our sign convention states that end moments are positive when they act in a counterclockwise direction. End moments are usually caused by a loading, translation, rotation or point moment.
 Fixed End Moments^{[1]}
 Fixed end moments are the moments at the ends of a beam that is fixed at both ends. Fixedend moments are caused by a loading, a point moment, a rotation, θ, of one of the ends or a small translation, Δ, at the end of a beam relative to the other end of the beam. The equation for fixed end moments depends on the loading conditions of the beam. A few common conditions are shown below in Figure 2.
 Sidesway
 Sidesway is any sideward translation of the top of a column or beam relative to its base. This translation can be linked to translation in the horizontal or vertical direction. One equation for calculating if sidesway is possible in a frame is given by:
 $ ss = 2j  [2 (f + h) + r + m] $^{[8]}
 Where:
 $ ss $ = number of sidesway degrees of freedom
 $ j $ = number of joints
 $ f $ = number of fixed supports
 $ h $ = number of hinged supports
 $ r $ = number of roller supports
 $ m $ = number of members
 If $ ss $ is equal to zero, sidesway is not possible. However, a $ ss $ value greater than zero may mean a frame is without sidesway if the frame and loading are symmetric.^{[8]}
 Member Stiffness^{[1]}
 Bending stiffness is the bending moment required to rotate the end of a beam by 1 radian. The bending stiffness at the end of a member is equal to:
$ \[ K = \left\{ \begin{array}{l l} \frac{4EI}{L} & \qquad \text{For Member Fixed at Other End}\\ \frac{3EI}{L} & \qquad \text{For Member Hinged at Other End} \end{array} \right.\] $
 When the elastic modulus of a beam is constant, relative bending stiffness at the end of a member can be used and is taken as follows:
$ \[ \bar{K} = \left\{ \begin{array}{l l} \frac{I}{L} & \qquad \text{For Member Fixed at Other End}\\ \frac{3}{4} & \qquad \text{For Member Hinged at Other End} \end{array} \right.\] $
 Carryover Moment^{[1]}
 Carryover moment is the moment created at the end of a beam when another point moment is applied at the opposing end. Carryover moment at the end of a beam is equal to:
$ \[ M = \left\{ \begin{array}{l l} \frac{M}{2} & \qquad \text{For Member Fixed at Other End}\\ \0 & \qquad \text{For Member Hinged at Other End} \end{array} \right.\] $
 Carryover Factor (COF)^{[1]}
 The carryover factor is the relationship between the carryover moment and the applied moment. The carryover factor is given as follows:
$ \[ COF = \left\{ \begin{array}{l l} \frac{1}{2} & \qquad \text{For Member Fixed at Other End}\\ \0 & \qquad \text{For Member Hinged at Other End} \end{array} \right.\] $
 Moment Area Method
 See main article: Moment Area Theorem
 Moment Area Method is a method of calculating slopes and deflections of a beam based on the beam’s curvature (M/EI) diagram. The moment Area Method consists of two theorems.
 The first moment area theorem states that the area under the curvature diagram between two points on a prismatic beam with a uniform modulus of elasticity is equal to the change in tangential slope between those two points. ^{[9]} This is modeled by the equation:
 $ \Theta_{BA} = \Theta_B  \Theta_A = \int_A^B \frac{M}{EI} dx $^{[1]}
 The second moment area theorem states that the moment of the area under the curvature diagram between two points is equal to the tangential deviation of the first point, A, from the tangent at the second point, B.^{[9]} This is modeled by the equation:
 $ \Delta_{BA} = \int_A^B \frac{M}{EI}\,dx $^{[1]}
 Moment Area Theorem can be used as an alternate way to calculate bending moment and carryover moment.^{[1]}
 Prismatic Beams
 A prismatic beam is a beam that has a uniform moment of inertia. When a beam is nonprismatic, it must be split into multiple prismatic beams in order to perform the moment distribution method.^{[1]}
 Distribution Factors
 Distribution factors are used to distribute the stress from an applied moment at a joint to the members connected to that joint. The distribution factor for the end of a member is equivalent to the ratio the relative bending stiffness of a member to the sum of relative bending stiffnesses of all members connected to the joint where the moment is applied.^{[1]} This is represented by the following equation:
 $ DF = \frac{\bar{K}}{\Sigma \bar{K}} $
 Distribution factors are essential in the application of the moment distribution method.
 Balanced Moments
 When using the moment distribution method, balanced moments must be used after removing the original support conditions in order to properly analyze a beam.
 Balancing a moment requires multiplying the fixed end moment of a member by the distribution factor.
 Superposition
 Superposition is a method of calculating the deflection of a beam under multiple loadings by calculating the deflection caused by the individual loads and adding the individual results. ^{[1]} Superposition can be used in the application of the moment distribution method.
Procedure

Determine if the frame is subjected to sidesway (Determine the Degrees of Freedom of the frame). The following equation is mentioned in the previous section of Moment Distribution Method for beams without sidesway.
$ s_{s}=2j[2(f+h)+r +m] $ ^{[1]}
 Superimpose the frame into separate systems: One system (a) will be externally loaded with the forces on the primary beam, but with a restraint added at the site of the Degree of Freedom to restrain sidesway. The second system (b) will not be restrained, but will have a force R acting opposite to the restraint in the previous system. System (b) will not be loaded externally
 Draw Free Body Diagrams of system (a), separating the frame into individual beams. Draw the beams as if they had fixedfixed end conditions and then determine their FixedEnd Moments.
To determine how the FEM's will be distributed across the frame, the bending stiffness for each segment of the frame must be known. If the moment of inertia of the frame is constant, a *relative* bending stiffness can be used. The case of bending stiffness can be generalized into 2 cases, the farend member being fixed or the farend member being hinged. Following are the general equations for calculating bending stiffness or relative bending stiffness of a member.
$ \[ K = \left\{ \begin{array}{l l}\frac{4EI}{L} & \qquad \text{Far Member Fixed}\\\frac{3EI}{L} & \qquad \text{Far Member Hinged} \end{array} \right.\] $^{[10]}
$ \[ \widehat{K} = \left\{ \begin{array}{l l}\frac{I}{L} & \qquad \text{Far Member Fixed}\\\frac{3}{4}\frac{I}{L} & \qquad \text{Far Member Hinged}\end{array} \right.\] $^{[10]}
The distribution factor for when these end conditions are released must be calculated to know how the moments are distributed over the member.
$ DF = \frac{\bar{K}}{\Sigma \bar{K}} $
^{[10]}The next important step is to account for Carry Over Moments that are distributed to the far end of a member. The Carry Over Factor can be generalized again for two cases, the far end being hinged or fixed.
$ \[ COF = \left\{ \begin{array}{l l}\frac{1}{2} & \qquad \text{Far Member Fixed}\\\0 & \qquad \text{Far Member Hinged} \end{array} \right.\] $^{[10]}
 At the point which the actual moments are found, basic static equilibrium can be used to find the support reactions of the frame.
 Draw a Free Body Diagram of system (b) and be sure to include a force R in opposite direction of the added restraint (effectively counteracting it). For this Force R, we cannot calculate it using the Moment Distribution Method, so we assume the force R causes an arbitrary, known displacement (D) at the joint.
 Express the FEMs of the frame in terms of E, I, and displacement D.
 Analyze this system using the MomentDistribution Method to find the support moments. Since D is of our choice, we can use our knowledge of internal forces to solve for shear at the frame's joints, then use equilibrium to solve for the force Q being applied.
Once Q is known, the final moments of the original frame can be calculated using superposition. Using the equation below, the final step of this method is completed.
$ M_A= M_{AO}+\frac{R}{Q}*M_{AQ} $^{[10]}
^{[10]}This equation is interpreted as; The true moment of joint A is equal to the moment of joint A is system (a) plus the moment of joint A of system (b) times the restraint reaction over force
^{[1]}
Example
The following example draws upon methods used in the Kassimali textbook.^{[1]}
Question
Find the reaction moments at A and D. Draw the moment distribution diagram for the frame. Note that the I of section BC is twice that of AB and CD.
Solution
Part 1: Solving restrained system (System with no sidesway)
First we solve our restrained system in figure 4 using moment distribution method. Our first order of business if to find the FEMs, Distribution Factors (DFs), and the carry over factors (COFs).
FEM of different members
We first need to figure out what our fixed moment area is for each shape. To do this, we determine how we want to subdivide our frame into members. Generally, at any corner and any point where I changes would be good to add a member division. For our problem, we have selected 3 members, as shown in the figures
The idea behind fixed end moments is taking each end of a section and fixing it against moment, or "clamping". To do this, we put each of our members into a fixedfixed support situation, as the figures above show. Using equations predetermined equations, we calculate what the end moments for each fixedfixed member. This is our FEM.
 $ FEM_{AB}=FEM_{BA}=\frac{PL}{8}=\frac{8*3}{8}=3 $
 $ FEM_{BC}=FEM_{CB}=\frac{w*L^2}{12}=\frac{4*3^2}{8}=3 $
 $ FEM_{CD}=FEM_{DC}=0 $
Note that $ FEM_{CD} $ and $ FEM_{DC} $ are zero because there is no load applied to member CD.
Distribution Factors
As explained in the procedure, these distribution factors will tell you how the fixed end moment will distribute when we "unclamp" each fix joint. We find 2 distribution factors per connecting joint
Joint B
 $ DF_{BA}= \frac{k}{\sum k}=\frac{k_{BA}}{k_{BA}+k_{BC}}=\frac{\frac{I}{3}}{\frac{2*I}{3}+\frac{I}{3}}=\frac{1}{3} $
 $ DF_{BC}=\frac{\frac{2*I}{3}}{\frac{2*I}{3}+\frac{I}{3}}=\frac{2}{3} $
Joint C
 $ DF_{CB}=\frac{k_{CB}}{k_{CB}+k_{BC}}=\frac{\frac{2*I}{3}}{\frac{2*I}{3}+\frac{I}{3}}=\frac{2}{3} $
 $ DF_{CD}=\frac{\frac{I}{3}}{\frac{2*I}{3}+\frac{I}{3}}=\frac{1}{3} $
CHECK: The sum of all distribution factors at one joint should be 1. This won't confirm if your DFs are correct, but it will help you check your arithmetic.
Carry Over Factors
You carry over Factor tells you how much of the distributed moment after you unclamp the joint will transfer over to the other end.
$ FixedFixed=0.5 $
$ FixedHinged=0 $
Moment Distribution
So this is the moment we've been waiting for. To help us with book keeping, we will be using a table, much like how the Kassimali textbook shows.^{[1]}
Joint  A  B  B  C  C  D 

Distribution Factors  1/3  2/3  2/3  1/3  
Fixed Moment Areas  3  3  3  3  
Unclamp C  1  << 2  1 >>  1/2  
Unclamp B  1/6  << 1/3  2/3 >>  1/3  
Unclamp C  1/9  << 2/9  1/9 >>  1/18  
Unclamp B  1/54  << 1/27  2/27 >>  1/27  
Unclamp C  1/81  << 2/81  1/81 >>  1/162  
Unclamp B  1/486  << 1/243  2/243 >>  1/243  
Unclamp C  1/729  << 2/729  1/729 >>  1/1458  
Unclamp B  1/4374  << 1/2187  2/2187 >>  1/2187  
Sum  2.81  3.37  3.37  1.13  1.13  0.56 
CHECK: Notice how moments at joint B and joint C cancel each other out. If you have distributed the moments correctly, and the last term is small enough, this should be the case. To quickly check the moments, you can make a qualitative deflection diagram and see if the moments you solved make sense.
Moments at A and D without sidesway
$ M_{AO}=2.81[kN*m] $
$ M_{DO}=0.56 [kN*m] $
Part II: Sidesway
Up to this point we have solved a system without sidesway. In order to do this, we applied a restraint which delivered a reactive force that kept the frame from swaying. We have chosen to call this force R. Using the principle of superimposition, we find an equivalent system with a negative R applied to it instead of R. In other words, we want to find the effect that R had on our system, so we can "subtract" it from the nonsway system to get the sway system, or in other words, add it's opposite effect.^{[1]}
Finding R
First we need to find R. We use the equilibrium of the frame in the x direction (the direction of sway and the force R) to find R. We first need to find the shear at each end of the members, $ V_A $ and $ V_D $.
Member AB
Fix FBD numbers
$ \sum M_B=0 $
$ 0=2.813.37+1.5*8+3*V_A $
$ V_A=3.81 kN $
Member CD
$ \sum M_C=0 $
$ 0=1.69+3*V_D $
$ V_D=0.56 kN $
Force R
$ \sum F_x=0 $
$ 0=R+V_A+V_D $
$ R=3.63 kN $
Since the R we are looking for is the opposite force required to restrain the frame from sidesway, the R we want is:
$ R=3.63 kN $
Solve for Q & D'
The moment distribution method is a versatile method, but one of it's shortcomings is that it can not solve end moments using displacements.^{[1]} So to solve our system, we use a parallel system with force Q causing D' displacement.^{[1]} This Q system is an arbitrary system, meaning we can assign one of it's parameters to anything we want, and then solve for the other parameters. In our case, we will choose what one of our FEM will be, and then solve the other FEM in terms of our chosen value. To do so, we first develop FEM equations in terms of E, I, L, and D'.
$ FEM_{AB}=FEM_{BA}=\frac{6*EI*D'}{L^2}= \frac{2*EI*D'}{3} $
$ FEM_{BC}=FEM_{CB}=\frac{2*E(2I)*0}{3}=0 $
$ FEM_{DC}=FEM_{CD}= \frac{2*EI*D'}{3} $
We set our $ FEM_{AB} $ to 3, which is a convenient value given out distribution factors. Since our $ FEM_{DC} $, $ FEM_{CD} $ and $ FEM_{BA} $ are all the same as $ FEM_{AB} $, our solution simply is that:
$ FEM_{AB}=FEM_{BA}=3kN $
$ FEM_{CD}=FEM_{DC}=3kN $
Notice how moments at both end of a beam are positive. This is different from the previously when we solved FEM with forces because deformations usually result in moments at both ends of the member to have the same sign and value.
Note:A lot of you are probably thinking that was too easy, and unfortunately, you're right. Generally, after setting one of the FEMs to a certain value, you solve for the other FEMs interms of your D' for the FEM you set. So be sure to go through the whole process and not just assume all the FEMs are the same.
Since I, E, and L are the same for both systems, our distribution factor is the same. Next we use the table method to solve the reaction moments at A and D.
Moment Distribution
Joint  A  B  B  C  C  D 

Distribution Factors  1/3  2/3  2/3  1/3  
Fixed Moment Areas  3  3  3  3  
Unclamp C  1  << 2  1 >>  1/2  
Unclamp B  1/3  << 2/3  4/3 >>  2/3  
Unclamp C  2/9  << 4/9  2/9 >>  1/9  
Unclamp B  1/27  << 2/27  4/27 >>  2/27  
Unclamp C  2/81  << 4/81  2/81 >>  1/81  
Unclamp B  1/243  << 2/243  4/243 >>  2/243  
Unclamp C  2/729  << 4/729  2/729 >>  1/729  
Unclamp B  1/2187  << 2/2187  4/2187 >>  2/2187  
Unclamp C  2/6561  << 4/6561  2/6561 >>  1/6561  
Sum  2.63  2.25  2.25  2.25  2.25  2.63 
$ M_{AQ}=2.63 kN*m $
$ M_{DQ}=2.63 kN*m $
Lastly, we need to find Q.
Shear at D
$ M_{C}=0 $
$ 0=2.63+2.25+3*V_D $
$ V_D=1.63kN*m $
Shear at A
$ M_{B}=0 $
$ 0=2.63+2.25+3*V_A $
$ V_A=1.63kN*m $
Force Q
$ \sum F_x=0 $
$ 0=Q+V_A+V_D $
$ Q=3.26 KN $
Finding moment of sidesway system
$ M_A= M_{AO}+\frac{R}{Q}*M_{AQ} $
$ = 2.81+\frac{3.63}{3.26}*2.63 $
$ =5.74 kN*m $
$ M_D= M_{DO}+\frac{R}{Q}*M_{DQ} $
$ = 0.56+\frac{3.63}{3.26}*(2.63)} $
$ =3.49 kN*m $
Moment Distribution Diagram
Our first order of business is to find the member end moments and shears. To find the moments, we take our moment distibuted moments and apply the same equations we used to find $ M_{A} $ and $ M_{D} $.
$ M_{AB}= 5.74 kN*m $
$ M_{BA}= 3.37+frac{3.63}{3.26}*(2.25) kN*m $
$ M_{BA}= 3.37+frac{3.63}{3.26}*(2.25) kN*m $
$ M_{BA}= 0.86 kN*m $
$ M_{BC}= 3.37+frac{3.63}{3.26}*(2.25) kN*m $
$ M_{BC}= 0.86 kN*m $
$ M_{CB}= 1.13+frac{3.63}{3.26}*(2.25) kN*m $
$ M_{CB}= 3.64 kN*m $
$ M_{CD}= 1.13+frac{3.63}{3.26}*(2.25) kN*m $
$ M_{CD}= 3.64 kN*m $
$ M_{DC}= 3.49 kN*m $
Finding Shears
Using equilibrium and the moments we just found, we find the shears of the ends of each member.
Member AB:
$ \sum M_{B}= 0 $
$ 0=8*1.5+5.740.86+3*V_A $
$ V_A=5.63 kN $
$ \sum F_x=0 $
$ 0=85.63+V_B $
$ V_B=2.37 kN $
Member BC:
$ \sum M_{B}= 0 $
$ 0=0.863.643*4*1.5+3*V_C $
$ V_C=6.93 kN $
$ \sum F_y=0 $
$ 0=6.934*3+V_B $
$ V_B=5.07 kN $
Member CD:
$ \sum M_{C}= 0 $
$ 0=3.64+3.49+3*V_D $
$ V_D=2.38 kN $
$ \sum F_y=0 $
$ 0=V_C4.18 $
$ V_B=2.38 kN $
Note: When we performed our moment distribution, we took values up the accuracy of the hundredth decimal. Notice how $ P $, $ V_D $, and $ V_A $ don't quite add up. This is because we did not do moment distribution far enough and are value are not precise enough. In the FBD, we have changed the values to reflect equilibrium.
Developing Moment Distribution Diagrams
Next, we take the shear and moment values and use them to construct moment distribution diagrams.
References
 ↑ ^{1.00} ^{1.01} ^{1.02} ^{1.03} ^{1.04} ^{1.05} ^{1.06} ^{1.07} ^{1.08} ^{1.09} ^{1.10} ^{1.11} ^{1.12} ^{1.13} ^{1.14} ^{1.15} ^{1.16} ^{1.17} ^{1.18} ^{1.19} Kassimali, A. (2011). Structural Analysis: SI Edition (4th ed.). Stamford, CT: Cengage Learning.
 ↑ Hibbeler, R. C. Engineering Mechanics: Statics & Dynamics. 12th ed. Upper Saddle River, NJ: PrenticeHall, 2010. Print.
 ↑ University of Sidney. "Chapter 5 Bending Moments and Shear Force Diagrams for Beams."Mechanics of Solids 1. Aerospace, Mechanical & Mechatronic Engineering, University of Sydney, 2009. Web. 21 Nov. 2013. <http://web.aeromech.usyd.edu.au/AMME2301/Documents/mos/Chapter05.pdf>.
 ↑ Sjhan81. Fem1.png. Digital image. Wikimedia Commons. The Wikimedia Foundation, 3 Sept. 2007. Web. 21 Nov. 2013. <http://commons.wikimedia.org/wiki/File:Fem1.png>.
 ↑ Sjhan81. Fem2.png. Digital image. Wikimedia Commons. The Wikimedia Foundation, 3 Sept. 2007. Web. 21 Nov. 2013. <http://commons.wikimedia.org/wiki/File:Fem2.png>.
 ↑ Sjhan81. Fem3.png. Digital image. Wikimedia Commons. The Wikimedia Foundation, 3 Sept. 2007. Web. 21 Nov. 2013. <http://commons.wikimedia.org/wiki/File:Fem3.png>.
 ↑ Sjhan81. Fem4.png. Digital image. Wikimedia Commons. The Wikimedia Foundation, 3 Sept. 2007. Web. 21 Nov. 2013. <http://commons.wikimedia.org/wiki/File:Fem4.png>.
 ↑ ^{8.0} ^{8.1} Erochko, Jeffrey. "Lecture #19 SlopeDeflection Method Continued." CIVE 3203  Intro to Structural Analysis. Canada, Ottawa. 15 Nov. 2013. Lecture.
 ↑ ^{9.0} ^{9.1} Zhen, Yubao. "Lecture 18: Bending (VI) — Methods of MomentArea and Superposition." Lecture. Harbin Institute of Technology, 16 Nov. 2012. Web. 20 Nov. 2013. <http://am.hit.edu.cn/courses/mechmat2012/Courseware_files/18_masupp_presentation.pdf>.
 ↑ ^{10.0} ^{10.1} ^{10.2} ^{10.3} ^{10.4} ^{10.5} Bera, Padmlochan."Lesson 21The Moment Distribution Method: Frames with Sidesway": Indian Institute of Technology, Web 18 Nov. 2013 <http://www.nptel.iitm.ac.in/courses/Webcoursecontents/IIT%20Kharagpur/Structural%20Analysis/pdf/m3l21.pdf>.