# A Matrix Formulation for the Moment Distribution Method

Authors: Arlindo Pires Lopes; Adriana Alencar Santos; Rogério Coelho Lopes
Journal: Advances in Civil Engineering
 (2011)

---

## Abstract

The Moment Distribution Method is a quite powerful hand method of structural analysis, in which the solution is obtained iteratively without even formulating the equations for the unknowns. It was formulated by Professor Cross in an era where computer facilities were not available to solve frame problems that normally require the solution of simultaneous algebraic equations. Its relevance today, in the era of personal computers, is in its insight on how a structure reacts to applied loads by rotating its nodes and thus distributing the loads in the form of member-end moments. Such an insight is the foundation of the modern displacement method. This work has a main objective to present an exact solution for the Moment Distribution Method through a matrix formulation using only one equation. The initial moments at the ends of the members and the distribution and carry-over factors are calculated from the elementary procedures of structural analysis. Four continuous beams are investigated to illustrate the applicability and accuracy of the proposed formulation. The use of a matrix formulation yields excellent results when compared with those in the literature or with a commercial structural program.


---

## Body

## 1. Introduction

It is observed in practice that the greater the complexity of a structure, the greater the number of unknowns and therefore the greater the number of simultaneous equations requiring solution. Hand methods of analysis then become extremely tedious if not impracticable, so that alternatives are desirable. One obvious alternative is to employ computer-based techniques, but another quite powerful hand method is an iterative procedure known as the moment distribution method (MDM).The MDM for the analysis of rigid-jointed structures was introduced by Professor Hardy Cross over seventy five years ago. In the past, when structural engineers had to compute the forces and displacements in a statically indeterminate structure, they inevitably turned to what was generally known as the Hardy Cross method. In this method, the fixed-end moments in the framing members are gradually distributed to adjacent members in a number of steps, such that the system eventually reaches its natural equilibrium configuration.The method is still taught in every undergraduate civil engineering course on the analysis and design of engineering structures, and although the Hardy Cross Method has been superseded by more powerful procedures such as the finite element method, the MDM made possible the efficient and safe design of many reinforced concrete buildings during an entire generation.The original paper written by Cross [1] explaining the method was followed afterwards by a famous textbook written by Cross and Morgan [2]. In the former one, he wrote the following: “The essential idea which the writer wishes to present involves no mathematical relations except the simplest arithmetic.” The MDM depends on the solution of three problems for beam constants: the determination of fixed-end moments, of the stiffness at each end of a beam, and of the carry-over factor at each end for every member of the structure under consideration, which will be explained afterwards.Lightfoot [3] believes that the simpler possibilities of the MDM have now all been discovered and that its economical use for the analysis and design of complex structures must now be assessed in relation to available electronic computing programs and facilities.The purpose of this paper is to enhance and improve the MDM used in solving continuous beams. In this study, the original MDM has been formulated in a closed form through a matrix formulation. Numerical results are presented to demonstrate the accuracy and efficiency of the proposed formulation.


## 2. The Moment Distribution Method

The MDM relies on a series of calculations that are repeated until every cycle comes closer to the final situation. In this way, it is possible to avoid solving simultaneous algebraic equations. This method is a unique method of structural analysis, in which the solution is obtained iteratively without even formulating the equations for the unknowns. The following beam will be used to illustrate the method.In the first stage of Figure1, rotation is possible at both B and C. The second stage is that rotation at B and C is prevented, and the load is applied obtaining the initial moments. Afterwards, in the third stage, allow B to rotate until moment equilibrium is reached and then rotation at B will induce a moment at C. Finally, in the fourth stage of Figure 1, allow C to rotate until moment equilibrium is reached. The rotation of C will induce a moment at B. This process is repeated until moment equilibrium is reached at all nodes.Figure 1
Continuous beam subjected to static loads.Assume that the sum of the initial moments at the nodeB is equal to M0. Rotation will take place until moment equilibrium is attained, that is, ∑MB=0 (Figure 2).Figure 2
Free-body diagram of nodeB.Therefore,(1)MBAD+MBCD+M0=0,
where MBAD and MBCD are the moments as a result of the rotation at B and are called the distribution moments. Remember that all the other rotations and sway are fixed. Using the equations from the slope-deflection method [4, 5], it is easy to demonstrate that(2)MBAD=2EIABLAB(2θB)=4EIABLABθB,MBCD=2EIBCLBC(2θB)=4EIBCLBCθB,
where EI is the cross-section rigidity, L is the member length, and θB is the rotation at B.Applying (2) into (1) and solving for θB, the rotation at B can be written as(3)θB=-M04EIAB/LAB+4EIBC/LBC.Finally, the distribution moments are obtained, as follows:(4)MBAD=-(4EIAB/LAB)M04EIAB/LAB+4EIBC/LBC=-kBAM0kBA+kBC=-kBAM0∑kB,MBCD=-(4EIBC/LBC)M04EIAB/LAB+4EIBC/LBC=-kBCM0kBA+kBC=-kBCM0∑kB,
where kBA is the stiffness of the member BA at node B. It is also the moment that would be induced if a unit rotation was applied at B in the member BA and the rotation at A was zero. If B rotates, a bending moment will be induced at A and C. Assuming a rotation θB, the moment at A can be written as follows:(5)MABD=-(2EIAB/LAB)M04EIAB/LAB+4EIBC/LBC.For this case, the distributed bending moment is half the value of the distributed bending moment atB. This is called the carry-over factor, CBA=0,5. It should be mentioned that stiffness and carry-over factors may vary with several parameters, depending on the shape of the structural member, the restraint conditions, and the structural effects considered.In summary, the MDM still remains an important tool for analysis of structures in practice and can be programmed for complicated problems without the use of modern structural analysis software. At the first stage of the analysis, it is assumed that the rigid joints of the structural members are initially fixed against rotations as in the displacement method. The reactive moments produced by external loads are computed. These moments are unbalanced at the joints of the original nonrestrained structure. In order to equilibrate the joints the moments are distributed proportionally to the corresponding member stiffness. These distributed moments are associated with the so-called carry-over moments at the opposite ends of structural members. They are considered to be new incremental unbalanced moments, and the procedure repeats until the unbalanced moments become negligible. The final moments at the ends of all members are the sum of all distributed moment increments. This procedure assumes the joint rotations only.According to Volokh [6], the MDM is nothing but Jacobi iterative scheme in disguise. It is shown that the method is the incremental form of Jacobi iterative scheme applied to the classical displacement formulation of the problem.


## 3. Sign Convention

The MDM convention that the final moments at a joint or a support are considered to be acting from the joint or support to the member is used. The convention adopted here is easier for moments and just as easy for forces, so its use is recommended. The positive direction for final moments is taken as clockwise.The use of a rotational sense to determine the sign of final moments automatically arranges that the algebraic sum of such moments equates to zero, at any joint in equilibrium. The clockwise direction also implies the use of the fourth quadrant for the definition of positive linear directions. Thus, forces, distances, and linear displacements are taken as positive when measured from left to right or vertically downwards. This is in accordance with the positive directions used in beam theory [7].This sign convention is convenient when considering joint equilibrium in the slope-deflection and moment distribution methods. For a horizontal beam subjected to downward loading, the MDM convention gives negative and positive fixed-end moment at the left-hand and right-hand supports, respectively.


## 4. Matrix Formulation

This section reports some of the basic relations of the proposed formulation for the MDM based on a matrix notation. Let the distribution and carry-over factors and the initial moments be written as follows:(6){M}={M0}+[β]{M0}+[α][β]{M0}+[β][α][β]{M0}+[α][β][α][β]{M0}+[β][α][β][α][β]{M0}+⋯,
where {M} is a vector which contains the final moments at the ends of the members and dimension (m×1), {M0} is also a column vector which contains the fixed-ends moments of the members and dimension (m×1), [β] is a square matrix which contains the distribution factors and dimension (m×m), [α] is a square matrix which contains the carry-over factors and dimension (m×m), and, finally, m is the number of the structural members multiplied by two.Note that the first term of the series represented by (6) is the vector which contains the fixed-ends moments of the members. The second term is a product between the distribution factors and the fixed-ends moments, obtaining a new set of moments when all nodes of the structure are released simultaneously. The third term of the series is the product between the distribution factors and the set of moments obtained previously, and the series continues for n cycles. Equation (6) can be rewritten as(7){M}=([I]+[β]+[α][β]+[β][α][β]+[α][β][α][β]+[β][α][β][α][β]+⋯){M0},
where [I] is a simplest nontrivial diagonal matrix known as identity matrix. After n cycles of the method and using some properties of the identity matrix, the following equation is obtained:(8){M}={[I]+[β]}{[I]+[α][β]+([α][β])2+([α][β])3+⋯+([α][β])n-1}{M0}
or(9){M}={[I]+[β]}{Sn}{M0},
where {Sn}={[I]+[α][β]+([α][β])2+([α][β])3+⋯+([α][β])n-1}. Since a geometric progression series is a sum of terms in which two successive terms always have the same ratio, {Sn} is a geometric progression with a common ratio of r={[α][β]}.The sum of the firstn terms of a geometric progression is given by(10){Sn}=a1(1-rn)(1-r),
where a1 is the first term and r is the ratio of the series.Substituting[I] as the first term and [α][β] as the ratio of the series into (10) gives(11){Sn}=[I]{[I]-([α][β])n}{[I]-[α][β]}.Note that the terms in matrix[β] are less than the unity. Therefore, when n in the term ([α][β])n tends to the infinite, {Sn} can be written as(12){Sn}=[I]{[I]-[α][β]}or{Sn}={[I]-[α][β]}-1.Substituting (12) into (9) gives(13){M}={[I]+[β]}{[I]-[α][β]}-1{M0},
which represents the closed form of the MDM through a matrix formulation.


## 5. Reference Coordinates

The analysis of a continuous beam by the closed form of the MDM can be achieved, as previously demonstrated, by solving (13) which is expressed in a matrix form. The development of the matrix elements of the latter equation must be described with respect to the reference coordinates. It is necessary to identify the ends of the members as shown in Figure 3.Figure 3
Reference coordinates of the structure.In the case of matrix[β] which represents the distribution factors, it is observed that in the MDM the distribution factor at the end of member corresponding to reference coordinate 2 will multiply the bending moments associated to reference coordinates 2 and 3. Thus, the distribution factor has to be repeated two times in line 2 of the matrix and located in columns 2 and 3. The distribution factor corresponding to reference coordinate 5 will multiply the bending moments associated to reference coordinates 4 and 5. This distribution factor will be repeated two times in line 5 of the matrix, and located in columns 4 and 5. In other words, the reference coordinate of each distribution factor defines the line of the matrix and its value is repeated as many times according to the number of the end members in a node. The location in the columns is defined by the reference coordinates of the end members, while the other values are zero.In this way, the matrix[β] for the continuous beam in Figure 3 is given by(14)[β]=[-β1000000-β2-β20000-β3-β3000000-β4-β40000-β5-β5000000-β6]6×6.In order to evaluate the matrix[α] in Figure 3, which represents the carry-over factors, consider, for instance, the case at the end of the member corresponding to reference coordinate 2. The carry-over factor will multiply the bending moments associated to reference coordinate 2, but the result will be added to the bending moments associated to reference coordinate 1. Therefore, this carry-over factor will be located in column 2 and line 1 in matrix [α]. Similarly, the carry-over factor corresponding to reference coordinate 5 will multiply the bending moments associated to reference coordinates 5, but the result will be added to the bending moments associated to reference coordinate 6. This carry-over factor will be located in column 5 and line 6 in matrix [α]. To summarize the procedure, in the case of the carry-over factors, the member’s reference coordinate identifies the column in which the correspondent value will be located and the opposite reference coordinate identifies the line of the matrix, while the other values are zero.Applying this procedure in the matrix of the carry-over factors leads to(15)[α]=[0α20000α100000000α40000α300000000α60000α50]6×6.After all of the required matrices have been obtained, some of them with respect to the chosen reference coordinates, the closed form of the MDM can be done.


## 6. Numerical Examples

To compare the performance of the proposed formulation with the traditional one previously reported in the literature, four numerical examples are presented.The first example is a continuous beam shown in Figure4 given by Süssekind [8]. This beam is selected as a benchmark problem in the present study. In this particular, example certain features should be noted. First, the support at A is a fixed one and the support at E is an outside pinned support, so that the final moment at E must be zero. The loading and the span lengths are given in Figure 4. As seen before, it is necessary to identify the ends of the members, with respect to reference coordinates, as shown in Figure 5.Figure 4
Continuous beam for the first example.Figure 5
Reference coordinates for the first example.The fixed-end moments of the members and the distribution and carry-over factors are calculated from the elementary procedures of structural analysis. These values are given in Table1 as follows.Table 1
Initial moments at the ends of the members and the distribution and carry-over factors for the first example.
NodeABCDEMembersABBABCCBCDDCDEEDFixed-end moments−44−44−44−7.50Distribution factors00.670.330.50.50.570.431Carry-over factors0.50.50.50.50.50.50.50Using the values from Table1 and putting into a matrix form, the matrices [β] and [α] for the continuous beam in Figure 4 are given, respectively, by
(16)[β]=[000000000-0.67-0.67000000-0.33-0.3300000000-0.5-0.5000000-0.5-0.500000000-0.57-0.57000000-0.43-0.4300000000-1]8×8,[α]=[00.50000000.500000000000.50000000.500000000000.50000000.500000000000.500000000]8×8.Using (13), which represents the closed form of the MDM through a matrix formulation, leads to the following final moments at the ends of the members:(17)M1=-3.91kN⋅m,M2=4.19kN⋅m,M3=-4.19kN⋅m,M4=3.48kN⋅m,M5=-3.48kN⋅m;M6=5.87kN⋅m,M7=-5.87kN⋅m,M8=0kN⋅m.The numerical results obtained with the matrix formulation were compared with the results obtained in the literature to illustrate the applicability and accuracy of the proposed formulation. All the final moments at the ends of the members were exactly the same as given by Süssekind [8].In the second example, the behavior of the matrix formulation was tested in a continuous nonprismatic beam shown in Figure6 given by R. C. Lopes and A. P. Lopes [9]. This beam has four supports at A, B, C, and D. The loading and the span lengths are also given in Figure 6, and the reference coordinates are given in Figure 7.Figure 6
Continuous beam for the second example.Figure 7
Reference coordinates for the second example.The initial moments at the ends of the members and the distribution and carry-over factors are given in Table2.Table 2
Initial moments at the ends of the members and the distribution and carry-over factors for the second example.
NodeABCDMembersABBABCCBCDDCFixed-end moments−215.1178.17−218.40358.60−78.17215.11Distribution factors10.8750.1250.1250.8751Carry-over factors0.3720.6800.6370.6370.6800.372Using the values from Table2 and putting into a matrix form, the matrices [β] and [α] for the continuous beam in Figure 6 are given, respectively, by(18)[β]=[-1000000-0.875-0.8750000-0.125-0.125000000-0.125-0.1250000-0.875-0.875000000-1]6×6,[α]=[00.68000000.372000000000.63700000.637000000000.37200000.6800]6×6.Using (13), which represents the closed form of the MDM through a matrix formulation, leads to the following final moments at the ends of the members:(19)M1=0kN⋅m,M2=226.65kN⋅m,M3=-226.65kN⋅m,M4=333.43kN⋅m,M5=-333.43kN⋅m,M6=0kN⋅m.The numerical results obtained with the matrix formulation were compared with a commercial structural program known as SAP2000 v. 14.1.0. The matrix formulation yields excellent results.In the third example, the proposed formulation was tested in a symmetric continuous beam shown in Figure8 given by Leet et al. [10]. This beam has five supports at A, B, C, D, and E. Supports A and B are fixed, and the flexural stiffness is assumed to be constant. The loading and the span lengths are also given in Figure 8, and the reference coordinates are given in Figure 9.Figure 8
Continuous beam for the third example.Figure 9
Reference coordinates for the third example.The fixed-end moments of the members and the distribution and carry-over factors are calculated from the elementary procedures of structural analysis. These values are given in Table3 as follows.Table 3
Initial moments at the ends of the members and the distribution and carry-over factors for the third example.
NodeABCDEMembersABBABCCBCDDCDEEDFixed-end moments−206.225206.225−75.62575.625−75.62575.625−206.225206.225Distribution factors00.430.570.50.50.570.430Carry-over factors0.50.50.50.50.50.50.50.5Using the values from Table3 and putting into a matrix form, the matrices [β] and [α] for the continuous beam in Figure 8 are given, respectively, by(20)[β]=[000000000-0.43-0.43000000-0.57-0.5700000000-0.5-0.5000000-0.5-0.500000000-0.57-0.57000000-0.43-0.43000000000]8×8,[α]=[00.50000000.500000000000.50000000.500000000000.50000000.500000000000.50000000.50]8×8.Using (13), which represents the closed form of the MDM through a matrix formulation, leads to the following final moments at the ends of the members:(21)M1=-234.28kN⋅m,M2=150.11kN⋅m,M3=-150.11kN⋅m,M4=38.38kN⋅m,M5=-38.38kN⋅m,M6=150.11kN⋅m,M7=-150.11kN⋅m,M8=234.28kN⋅m.The numerical results obtained with the matrix formulation were compared with a commercial structural program known as SAP2000 v. 14.1.0. The bending moment diagram is plotted in Figure10. The matrix formulation yields excellent results, and the minor differences less than 1% are due to shear deformations included in the analysis via SAP2000.Figure 10
Bending moment diagram for the third example.In the fourth example, the proposed formulation was tested in an asymmetric continuous beam shown in Figure8 given by Leet et al. [10]. This beam has four supports at A, B, C, and D. Supports A and B are fixed and the cross-section of the beam is a C-Shape (C10 × 25) given by the AISC steel construction manual [11]. The loading and the span lengths are also given in Figure 11, the section properties are given in Figure 12, and the reference coordinates are given in Figure 13.Figure 11
Continuous beam for the fourth example.Figure 12
Section properties for the fourth example.Figure 13
Reference coordinates for the fourth example.Figure 14
Bending moment diagram for the fourth example.The initial moments at the ends of the members and the distribution and carry-over factors are given in Table4.Table 4
Initial moments at the ends of the members and the distribution and carry-over factors for the fourth example.
NodeABCDMembersABBABCCBCDDCFixed-end moments−63.87563.875−133.225133.22500Distribution factors00.50.50.330.670Carry-over factors0.50.50.50.50.50.5Using the values from Table4 and putting into a matrix form, the matrices [β] and [α] for the continuous beam in Figure 6 are given, respectively, by(22)[β]=[0000000-0.5-0.50000-0.5-0.5000000-0.33-0.330000-0.67-0.670000000]6×6,[α]=[00.500000.5000000000.500000.5000000000.500000.50]6×6.Using (13), which represents the closed form of the MDM through a matrix formulation, leads to the following final moments at the ends of the members:(23)M1=-40.06kN⋅m,M2=111.51kN⋅m,M3=-111.51kN⋅m,M4=105.22kN⋅m,M5=-105.22kN⋅m,M6=-52.61kN⋅m.The numerical results obtained with the matrix formulation were compared with a commercial structural program known as SAP2000 v. 14.1.0. The bending moment diagram is plotted in Figure13. The matrix formulation yields excellent results, and the minor differences less than 2% are due to shear deformations included in the analysis via SAP2000.


## 7. Conclusions

A new formulation for the MDM based on a matrix formulation has been developed. The main advantage of the proposed scheme is the use of only one equation to solve the problem without iteration. The results obtained here show that accurate values for the final moments at the ends of the members have been achieved compared to previous examples given in the literature or using a commercial structural program. It should be mentioned that the matrix formulation can be applied to framed structures too. Nowadays, it seems that the simpler possibilities of the MDM have all been discovered and that its economical use for the analysis and design of complex structures must now be assessed in relation to available electronic computing programs and facilities. From this point of view, the matrix formulation of the MDM emerges as an option to solve structural problems. It is apparent, therefore, that with research developments here and abroad the MDM still has many exciting possibilities to add to its considerable achievements. The next step of this research is to extend the proposed formulation to three-dimension problems and also the analysis of structures considering shear deformations.



---

*Source: 725395-2011-05-18.xml*