# Moment Distribution Method for Continuous Beams

## Introduction

Moment Distribution is another method for analyzing indeterminate structures. Like the Slope Deflection Method, it relies on equilibrium conditions to solve for the forces in the structure. Unlike the Slope Deflection Method of solving simultaneous equations, the Moment Distribution method is an iterative approach of distributing moments caused by external forces until all the joints in the structure are in equilibrium. [1] In structures where 3 or more equations are required to solve the system with the slope deflection method, the Moment Distribution Method is often quicker and simpler to use.

The Moment Distribution Method was first developed in 1924 by Hardy Cross, but was not published till 1930 at which point it became very popular.[1] [2]

## Terms & Concepts

"Figure 1: Positive Sign Convention"

In order to understand this method a few concepts and terms must first be defined:

Sign Convention: Sign convention is the same as for the Slope-Deflection Method for Continuous Beams[1]. As seen in Figure 1: up, right and counterclockwise are always positive.
Bending Stiffness: A member's bending stiffness describes a beams resistance to bending[1].The stiffer a beam the more energy it can absorb before it deflects. When a loaded member has a change in bending stiffness, the stiffer section of the beam will carry more of the load[1].
Unbalanced moment:When a joint that is normally free to rotate in a structure is temporarily clamped or restrained from rotation, an 'Unbalanced Moment' develops at this joint due to applied load on the beams connected to the joint.[3]
Fixed End Moment: The moment that would develop at the ends of a beam if their ends were fixed[1].
Distribution Factor: Distribution factors show how the Unbalanced Moment on a joint is distributed amongst the beams framed into the joint[3]. They are a function of the member's bending stiffness and their sum must equal one[3].
Carry Over Factor: Determines how much of a moment at a joint is transferred to the next joint in the beam.[1]

## Theory

"Figure 2 : Adapted from [1] "

As previously stated the Moment Distribution Method applies the concept of equilibrium. In a continuous beam such as the one seen in Figure 2, the value of the internal moment at B is continuous[1] (ie: the moment just to the left and just to the right are equal). Furthermore, rotation at B is also continuous, therefore any rotation to the left of B

($\theta_B_A$) must equal the rotation to the right of B ($\theta_B_C$)[1].

### Unbalanced Moments and Fixed End Moments

If joint B was clamped or restrained from rotation, Unbalanced Moments (UM) would develop at B due to the external loading on the beam[3]. These Unbalanced Moments (UM) are the sum of the Fixed End Moments (FEM) at B from beam BA and beam BC[3]. These Fixed End Moments can be calculated by applying the slope deflection equation in it's FEM form [1]:

$FEM_A_B={\frac{2}{L^2}}(2g_B-g_A)$ [1] FEM at A due to loading on member AB
$FEM_B_A={\frac{2}{L^2}}(2g_A-g_B)$ [1] FEM at B due to loading on member AB

$g_A$ and $g_B$ terms represent the moments of the area of the M/EI diagram for a simply supported beam with the same loading pattern[1]. These terms can be solved for using the Moment Area Theorem[1], however for ease of calculation common FEM conditions and Equations are summarized in a FEM Tables such as the one found here [4] and at the back of the Kassimali Structural Analysis Textbook [1] or the Hibbeler Structural Analysis Textbook [2].

From equilibrium we know that at support B the moment on each side must be equal ($\theta_B_A=\theta_B_C$). When the imaginary clamp on joint B is released, the joint will want to reach equilibrium again. It does this by distributing or 'sharing' the Unbalance Moment across the beams framed into joint B.

### Distribution Factors

To determine how much of the UM is transferred to each beam framed into joint B, each beam's Distribution Factor (DF), must be determined. A beam's Distribution Factor is a function of the bending stiffness's of all the beams connected to the joint in consideration[3]. In Figure 2, the Distribution Factors for beam BA and BC connected to joint B are calculated as follows:

$DF_B_A={\frac{K_B_A}{\Sigma K}}$ [3]

And for beam BC:

$DF_B_C={\frac{K_B_C}{\Sigma K}}$ [3]

#### Bending Stiffness

"Figure 3 : Adapted from [1] "
Bending stiffness $\bar{K}$ is defined as "the moment that must be applied at an end of the member to cause a unit rotation of that end" [1].
$\bar{K}={\frac{M}{\theta}}$[1]
Consider the two beams in Figure 3, one with two hinge supports and the other with a hinge and a fixed support. If a unit rotation is applied to the left end of the hinged-fixed beam the bending stiffness can be determined by applying the slope deflection equation and substituting $M_n_f=M, \theta_n=\theta, and \theta_f=\psi=FEM_n_f=0$.[1]
$M={\frac{4EI}{L}}\theta \Rightarrow {\frac{M}{\theta}}={\frac{4EI}{L}}$ [1]
Therefore : $\bar{K}={\frac{4EI}{L}}$ , Fixed End
For the hinged-hinged beam with a unit rotation, the bending stiffness for a hinged end can be determined by applying the slope deflection equation modified for a hinged end support by substituting in $M_r_h=M, \theta_r=\theta, and \psi=FEM_r_h=FEM_h_r=0$.[1]
$M={\frac{3EI}{L}}\theta \Rightarrow {\frac{M}{\theta}}={\frac{3EI}{L}}$ [1]
Therefore : $\bar{K}={\frac{3EI}{L}}$ , Hinged End
It is also interesting to note that these bending stiffness relationships can also be derived using alternative methods of analysis such as the conjugate beam method.[2].

Once all the bending stiffness are calculated, they can be used to calculate the required distribution factors. A distribution factor is needed for each beam that frames into a joint with an unbalanced moment. For example in Figure 2, the Bending Stiffness and Distribution Factors for member AB and BC are required.

After all the bending stiffness are calculated and used to determine the distribution factors, the appropriate portion of the Unbalanced Moment at joint B can be distributed to member AB and BC. From here, the influence the distributed moments have on the far end of their beam is determined by a Carry Over Coefficient.

### Carry Over Coefficient

A Carry Over Coefficient describes how much of an Unbalanced Moment is transferred from the joint where the UM occurs to adjacent joints. Similarly to member bending stiffness, COFs are dependent on a beams end condition:

$COF={\frac{1}{2}}$, Fixed End [3]
$COF=0$, Hinged End [3]

These COFs are derived from the slope deflection equations for fixed and hinged end supports respectively[1]. However, for a hinged end, COF=0 intuitively makes sense. A hinged support can not resist any moment, therefore no moment can be carried through the beam to this support.

To mathematically derive the COF for a fixed end, once again consider the Hinged-Fixed beams in Figure 3. The slope deflection equation can be used to solve for $M_B_A$, which is the moment at B due to external loads at A.[1]

$M_n_f={\frac{2EI}{L}}(2\theta_n+\theta_f-3\psi)+FEM_n_f$[1]
Substitute in $M_n_f=M_B_A, \theta_f=\theta$ and $\theta_n=\psi=FEM_n_f=0$.[1] Therefore:
$M_B_A={\frac{2EI}{L}}\theta$[1]
Recall from calculating bending stiffness above for a fixed end $M={\frac{4EI}{L}}\theta$[1], rearrange for $\theta$ and sub it in to get:
$M_B_A={\frac{2EI}{L}}({\frac{ML}{4EI}}) \Rightarrow M_B_A={\frac{M}{2}}$[1]

## Steps of Analysis

Modified from [1] and [3]

The following steps of Analysis and a Moment-Distribution Table can help make calculations quick and easy. A Moment Distribution Table helps organize all FEM, Unbalanced Moments and carry over moments occurring at the end of each beam in a structure. Therefore a Moment Distribution Table needs a column for each member end in the structure. Note: beams are cut at supports creating two members, therefore two columns are required for each joint in the middle of a beam.

1) Calculate the distribution factors at every joint that is free to rotate. This is done by calculating each member's stiffness and then plugging them into the Distribution Factor Equation (Note: the sum of the distribution factors at any joint must equal one). Place these Distribution Factors in the Moment-Distribution Table at the top of the column corresponding to the correct member end.
2) Using the Fixed End Moment equations provided in the table here [4] or in the back cover of [1] or [2], calculate the fixed-end moments assuming that all the joints that are free to rotate are clamped. Add these numbers in the next row of the Moment-Distribution table.
3) Start the iterative distribution process. This process will likely be repeated a few times:
a) Choose a joint to begin with and remove the imaginary clamp. Sum the fixed end moments on the joint. (Note: ensure you are accounting for the direction change.)
b) Using the distribution factors, find how much of the moment is distributed to each side of the joint.
c) Use the carry-over moments to calculate how much of the distributed moment is carried over to adjacent joints and enter these values in the next row of the table.
d) Check all the values are filled into the Moment-Distribution table.
e) Repeat step 3 until all joints are balanced (no changes to moment values after carry-over is applied).
4) Find the final moments at each joint by summing ALL moments and carry-over moments that have been calculated for that joint. The moment-distribution table makes this step much easier to complete. Simply sum each column to find the final moment value. If done correctly the moments will satisfy the moment equilibrium equation (ie: the moments on either side of a joint should be equal and opposite values).
5) Calculate the member end shears by looking at the equilibrium of the members of the structure and applying $\Sigma M$ and $\Sigma F$ equations.
6) Calculate the support reactions by looking at the equilibrium of the joints in the structure $\Sigma F$ equations.
7) Construct the shear and bending moment diagrams.

## Example Problems

### Problem 1

Let's look at a relatively simple example using moment distribution method. In this example we have a beam that is fixed on both ends with a roller support 5 ft from the left end. There is also a distributed load of 10 k/ft over the first 5 ft and a point load of 20 k at 4 ft to the right of the roller.

"Example 1: Indeterminate structure with point load and UDL"

Step 1: Calculate the bending stiffness.

$K_A_B=4E{\frac{I_A_B}{L_A_B}=4E{\frac{150in^4}{5 ft\times12 in/ft}=10E in^4$

$K_B_C=4E{\frac{I_B_C}{L_B_C}=4E{\frac{600in^4}{8 ft\times12 in/ft}=25E in^4$

$K_{B}=K_A_B+K_B_C=35E in^{4}$

Step 2: Calculate the distribution factors.

$DF_B_A=\frac{K_B_A}{K_B}=\frac{10E}{35E}=0.286$

$DF_B_C=\frac{K_B_C}{K_B}=\frac{25E}{35E}=0.714$

Step 3: Calculate the fixed-end moments (see table at [4] for equations).

$FEM_A_B=\frac{wL^{2}}{12}=\frac{-10k/ft\times (5ft)^{2}}{12}=-20.83 k.ft$

$FEM_B_A=20.83$

$FEM_B_C=\frac{PL}{8}=\frac{-20k\times8ft}{8}=-20 k.ft$

$FEM_C_B=20k.ft$

Step 4: Release clamp at B and solve the moment-distribution table.

"Example 1: Final moments and forces acting on the beam"
Members AB BA BC CB
DF 0 0.286 0.714 0
FEM -20.83 20.83 -20 20
Distributing 0 -0.237 -0.593 0
Carry Over -0.1185 0 0 -.2965
Sum -20.95 20.59 -20.59 19.70

Step 5: Solve the shears.

Member AB:

$\sum M_A_B = 0 \rightarrow -20.95 - (10k/ft \times 5ft)\times2.5ft + V_B_A(5ft)+20.59=0$

$V_B_A=25.1k$

$\sum F=0\rightarrow V_A_B-(10)(5)+25.1=0$

$V_A_B=24.9$

Member BC:

$\sum M_B_C=0\rightarrow -20.59+19.7-20k(4ft)+V_C_B(8ft)$

$V_C_B=10.11k$

$\sum F=0\rightarrow V_B_C-20k-V_C_B=0$

$V_B_C=9.89k$

Step 6: From here the support reactions, shear and bending-moment diagrams can be created.

### Problem 2

Let's look at another example in SI unit and solve it by using the moment distribution method.

"Example 2:Indeterminate structure with 5 dofs"

Step 1: Determine the distribution factors at joint B,C,E,F.

$DF_B_A=\frac{K_B_A}{K_B_A+K_B_C}=\frac{\frac{I}{3}}{\frac{I}{3}+\frac{I}{2}}=0.4$

$DF_B_C=\frac{K_B_C}{K_B_A+K_B_C}=\frac{\frac{I}{2}}{\frac{I}{3}+\frac{I}{2}}=0.6$

$DF_C_B=\frac{K_C_B}{K_C_B+K_C_E}=\frac{\frac{I}{2}}{\frac{I}{2}+\frac{I}{2}}=0.5$

$DF_C_E=\frac{K_C_E}{K_C_B+K_C_E}=\frac{\frac{I}{2}}{\frac{I}{2}+\frac{I}{2}}=0.5$

$DF_E_C=\frac{K_E_C}{K_E_C+K_E_F}=\frac{\frac{I}{2}}{\frac{I}{2}+\frac{I}{5}}=0.714$

$DF_E_F=\frac{K_E_F}{K_E_D+K_E_F}=\frac{\frac{I}{5}}{\frac{I}{2}+\frac{I}{5}}=0.286$

Step 2: Determine the fixed-end moments.

$FEM_A_B=FEM_B_A=\frac{wL^{2}}{12}=\frac{10kN/m\times(3m)^{2}}{12}=7.5 kN m$

$FEM_C_E=FEM_E_C=\frac{PL}{8}=\frac{25kN\times(2m)}{8}=6.25 kN m$

$FEM_E_F=FEM_F_E=\frac{wL^{2}}{12}=\frac{5kN/m\times(5m)^{2}}{12}=10.4 kN m$

Step 3: Balance the moments at each member end.

Here are some more steps explaining how to build up a moment-distribution table:

• e.g.
• At line 1, input the fix-end-force at each corresponding moment end.
• At line 2, distribute the unbalanced moment(UM)by the corresponding distribution factor
• e.g. $UM_B=(-7.5)+0=-7.5 kNm$
• Thus, the distributed moment at BA column is $DM_B_A=DF_B_A\times(-UM_B)=0.4\times-(-7.5)=+3 kNm$
• At line 3, distribute the carryover moment to two other members which showing by arrows
• e.g. The carryover moment (COM)for fixed ends is half of the distributed moment $COM_A_B=\frac{1}{2}(DF_B_A)=\frac{1}{2}\times(+3)=+1.5 kNm$
• Repeat the last 3 steps until the unbalanced moment is less than the criteria value (ie: a specified tolerance or significant figures, in this problem it is < 0.1 )
• Determine the final moment end moment by summing the columns

Step 4: Determine the member shear forces by equilibrium equations and show the Free Body Diagram.

therefore,

### Problem 3

For a continue beam with cantilever section at one end, it is a similar procedures as above to solve this question. Let's look at a simple example:

Step 1: Determine the distribution factors at joint B,C,D.

$DF_B_A=\frac{K_B_A}{K_B_A+K_B_C}=\frac{\frac{I}{5}}{\frac{I}{5}+\frac{I}{5}}=0.5$

$DF_B_C=\frac{K_B_C}{K_B_A+K_B_C}=\frac{\frac{I}{5}}{\frac{I}{5}+\frac{I}{5}}=0.5$

$DF_C_D=1$

Step 2: Determine the fixed-end moments.

$FEM_A_B=\frac{PL}{8}=\frac{5kN\times(5m)}{8}=3.125 kN m$

$FEM_C_D=\frac{wL^{2}}{2}=\frac{10kN/m\times(4m)^{2}}{2}=80 kN m$

Step 3: Balance the moments at each member end.

The procedure of making this table is discussed in Steps of Analysis and Problem 2

Step 4: Determine the member shear forces by using equilibrium equations and show the Free Body Diagram.

Therefore, the FBD is

## Moment Distribution Method for Frames

Further explanation on how to use the moment distribution method to analyse indeterminate frames without and with sidesway can be found here Moment Distribution Method for Frames without Sidesway and Moment Distribution Method for Frames with Sidesway