A Beginner's Guide to Structural Mechanics/Analysis

Continuous Beam Analysis

(c) 2007, 2008 T. Bartlett Quimby

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Section CB.2.1

Moment Distribution

Last Revised: 08/02/2008

Moment distribution is a great method for quickly computing end moments on continuous beams.  Over the years, several variations of the method have been presented.  These methods take advantage of various observations made about the process.  While this method can be applied to a variety of indeterminate structures, the discussion here is limited to continuous beams that may or may not be continuous with supporting columns.

The moment distribution method begins by determining the relative flexural stiffness, in the plane of loading, of all the elements rigidly connected to each joint.  Rotational stiffness of a member is proportional to material and geometric stiffness:

Relative Rotational Stiffness = EI/L

Where:

  • E is the modulus of elasticity of the material,
  • I is the moment of inertia about the axis of bending in the plane of the frame, and
  • L is the length of the member.

The basic premise of the moment distribution method is that any unbalanced moment on a joint is redistributed to the members rigidly attached to that joint in proportion to contributions each element makes to the total rotational stiffness of the joint.  Consequently, distribution factors are computed at each joint, one for each element attached to the joint that provides rotational stiffness in the considered plane.  The equation for computing the distribution factor of the ith member at a joint with n members is:

Note that if any of the variables is constant for all members of the joint, then the constant variable will cancel out of the equation.

The distribution factors are computed joint by joint for the ends of each member connected to the joint.

The moment distribution method assumes that all joints are "fixed" at the start of the problem.  This means that when loads are applied to each span, fixed end moments are developed at each end of the loaded span.  The fixed end moments are a function of the nature and location of the applied loads on the span.  If there is more than one load source on a span, super position can be used to determine the total fixed end moment at the end of each span.

Concerning notation, in this method the end moments are considered to be reactions.  As a result, the right hand rule is used to determine the sign of the moments.  Do not use beam internal force notation.

After computing the fixed end moments (FEM) for each span, you will observe that there is an unbalanced moment at each joint.  In other words, the sum of moments at the joint does not equal zero, a necessary requirement for equilibrium.  To rectify this problem, a joint is "released" and the unbalanced moment (i.e. the difference of the FEMs) is divided among the members attached to joint in proportion to their contributions to the joint rotational stiffness using the distribution factors.  Once the distributed moments are added to each preexisting FEM at the joint, the resulting FEMs are "balanced".  In other words, the sum of the moments at the joint equal zero and equilibrium is satisfied at that joint (for now!).

The addition of moment to each element attached to the joint induces a moment on the opposite end of each member.  This is called "carry over" (CO).  The carry over moment is added to the preexisting FEM at that joint.  The CO moment tends to unbalance the joints adjacent to newly balanced joint.  These joints are, in turn released, balanced and send back CO moments to unbalance the adjacent joint. 

Joints are successively released and balanced until the CO moments get small enough to ignore.  This means that this is an iterative process that looks for convergence.

An example may help to illustrate the process.

Consider the beam shown in Figure CB.2.1.1.  This beam will have a constant E and I for all three spans, so the relative stiffness of each can be computed as 1/L.

Figure CB.2.1.1
Beam Problem Definition

Compute the Distribution Factors

For Joint "A":

Two items contribute to the rotational stiffness at A.  One is the beam AB the other is the infinitely stiff support.  The distribution factor at the left end of beam AB then is:

DFAB,L = (1/7.5) / (infinity + 1/7.5) = 0

The distribution factor for the support at A is:

   DFA = infinity / (infinity + 1/7.5) = 1

For Joint "B":

The two items contributing to rotation stiffness at B are the beams that come in from either side.  The support is a roller so it does not contribute to the rotational support of the joint.  The two distribution factors are:

DFAB,R = (1/7.5) / ( (1/7.5) + (1/5) ) = 0.400

DFBC,L = (1/5) / ( (1/7.5) + (1/5) ) = 0.600

For Joint "C":

The two items contributing to rotation stiffness at C are the beams that come in from either side.  The support is a roller so it does not contribute to the rotational support of the joint.  The two distribution factors are:

DFBC,R = (1/5) / ( (1/6.25) + (1/5) ) = 0.556

DFCD,L = (1/6.25) / ( (1/6.25) + (1/5) ) = 0.444

For Joint "D":

There is only one item contributing to rotation stiffness at D: Beam CD.  The support is a roller so it does not contribute to the rotational support of the joint.  The two distribution factors are:

DFCD,R = (1/6.25) / ( (1/6.25) + (0) ) = 1.000

DFD = (0) / ( (1/6.25) + (0) ) = 0.000

Table CB.2.1.1 summarizes the progress to this point.

Table CB.2.1.1
Distribution Factors

 

Span AB

 

Span BC

 

Span CD

 

Units

Span

7.5   5   6.25

 

m

1/L

0.133333   0.2   0.16

 

 

Joint

A B B C C D

 

DF

0.000 0.400 0.600 0.556 0.444 1.000

 

Compute the Fixed End Moments

The fixed end moments are computed span by span using the given loading for each span and considering the ends of the beam to be fixed.

Figure CB.2.1.2
Fixed End Moments

Any number of sources can provide equations for fixed end moments.  Maybe, we'll eventually add a table of them here.  For now, you can use the beam tables found in the Steel Construction Manual, the Fundamentals of Engineering (FE) exam reference manual, or derive your own.

Figure CB.2.1.2 shows each of the spans, their loading, and the definition of the FEMs.  Note that each of the right end FEMs are right hand rule negative moments.  It is important to keep the sign.

For Span AB, the FEMS are:

FEMAB,L = + PL / 8 = + (10 kN) (7.5 m) / 8 = + 9.375 kN/m

FEMAB,R = - PL / 8 = - (10 kN) (7.5 m) / 8 = - 9.375 kN/m

The two FEMs have the same magnitude since the load is centered on the span.  They would be different otherwise.

For Span BC, the FEMS are:

FEMBC,L = + wL2 / 12 = + (2 kN/m) (5 m)2 / 12 = + 4.167 kN/m

FEMBC,L = + wL2 / 12 = - (2 kN/m) (5 m)2 / 12 = - 4.167 kN/m

For Span CD, the FEMS are:

FEMCD,L = + wL2 / 12 = + (1.5 kN/m) (6.25 m)2 / 12 = + 4.883 kN/m

FEMCD,L = + wL2 / 12 = - (1.5 kN/m) (6.25 m)2 / 12 = - 4.883 kN/m

Table CB.2.1.2 show our spreadsheet table expanded to include the initial FEMs.

Table CB.2.1.2
Distribution Factors

  Span AB   Span BC   Span CD   Units
Span 7.5   5   6.25   m
1/L 0.133333   0.2   0.16    
Load 10 kN 2 kN/m 1.5 kN/m  
Joint A B B C C D  
DF 0.000 0.400 0.600 0.556 0.444 1.000  
FEM 9.375 -9.375 4.167 -4.167 4.883 -4.883 kN-m

Balancing the Joints and Doing the Carry Over

At this time, all joints are said to be fixed.  In other words, it is as if an invisible lock has taken the imbalance at each joint.  This explains why equilibrium is not satisfied at a joint.  Lets take each joint and do a balance.

Joint A:

The table that we are developing does not show the support side of the joint.  This is common.  Joint A is always balanced because the fixed support moment matches (with opposite sign) the beam end moment.  This joint does not require balancing.

Joint B:

The unbalanced moment a this joint equals (-9.375 kN-m + 4.167 kN-m) = - 5.208 kN-m.  To balance this joint we need to add + 5.208 kN-m to the joint.  This is done by adding to each beam end in proportion to  its contribution to the rotational stiffness of the joint.  This is what the distribution factors are for.

The new FEMs are:

FEMAB,R =  - 9.375 kN/m + DFAB,R (+5.208 kN-m)

FEMAB,R =  - 9.375 kN/m + 0.400 (+5.208 kN-m) = -9.375 kN-m + 2.083 kN-m

FEMAB,R =  - 7.292 kN/m

FEMBC,L = + 4.167 kN/m  + DFBC,L (+5.208 kN-m)

FEMBC,L = + 4.167 kN/m  + (0.600) (+5.208 kN-m) = + 4.167 kN/m  + 3.125 kN-m

FEMBC,L = + 7.292 kN/m 

Note that the sum of the two FEMs is zero.  The joint is balanced.

The result of the balancing creates moments at the far ends of Beams AB and BC equal to half the balancing moments.  Table CB.2.1.3 shows the balancing moments at joint B and the resulting carry over moments at joints A and C.  The summation line shows the sum of the FEM, balance, and CO lines to show the current end moments on each span.  Note that only joint B is "balanced" at this point.

Table CB.2.1.3
Joint B Balance and Carry Over

  Span AB   Span BC   Span CD   Units
Span 7.5   5   6.25   m
1/L 0.133333   0.2   0.16    
Load 10 kN 2 kN/m 1.5 kN/m  
Joint A B B C C D  
DF 0.000 0.400 0.600 0.556 0.444 1.000  
FEM 9.375 -9.375 4.167 -4.167 4.883 -4.883 kN-m
Balance  0.000 2.083 3.125       kN-m
CO 1.042  0.000   1.563     kN-m
Sum 10.417 -7.292 7.292 -2.604 4.883 -4.883 kN-m

Joint C:

Ignoring the carry over moment from the prior joint, the unbalanced moment at this joint equals (-4.167 kN-m + 4.883 kN-m) = + 0.716 kN-m.  To balance this joint we need to add - 0.716 kN-m to the joint.  This is done by adding to each beam end in proportion to its contribution to the rotational stiffness of the joint.  This is what the distribution factors are for.

The new FEMs are:

FEMBC,R =  - 4.167 kN/m + DFBC,R (-0.716 kN-m)

FEMBC,R =  - 4.167 kN/m + 0.556 (-0.716 kN-m) = -4.167 kN-m - 0.398 kN-m

FEMBC,R =  - 4.565 kN/m

FEMCD,L = + 4.883 kN/m  + DFCD,L (-0.716 kN-m)

FEMCD,L = + 4.883 kN/m  + (0.444) (-0.716 kN-m) = + 4.883 kN/m  - 0.318 kN-m

FEMBC,L = + 4.565 kN/m 

Note that the sum of the two FEMs is zero.  The joint is balanced, if you continue to ignore the carryover from joint B.  Table CB.2.1.4 shows the balancing moments and carry over moments (highlighted in yellow) resulting from this operation.

Table CB.2.1.4
Joint B Balance and Carry Over

  Span AB   Span BC   Span CD   Units
Span 7.5   5   6.25   m
1/L 0.133333   0.2   0.16    
Load 10 kN 2 kN/m 1.5 kN/m  
Joint A B B C C D  
DF 0.000 0.400 0.600 0.556 0.444 1.000  
FEM 9.375 -9.375 4.167 -4.167 4.883 -4.883 kN-m
Balance 0.000 2.083 3.125 -0.398 -0.318   kN-m
CO 1.042  0.000 -0.199 1.563   -0.159 kN-m
Sum 10.417 -7.292 7.093 -3.002 4.565 -5.042 kN-m

 

Joint D:

As we did for Joint C, we ignore the carry over moment from the prior joint.  As we did at joint A, we did not include support moment in the table.  For this case the joint is a pin, so no moment can be there when we end.  The unbalanced moment a this joint equals -4.883 kN-m.  To balance this joint we need to add  +4.883 kN-m to the joint.  This is done by adding to each beam end in proportion to  its contribution to the rotational stiffness of the joint.  This is what the distribution factors are for.

The new FEM is:

FEMCD,R =  - 4.883 kN/m + DFCD,R (+4.883 kN-m)

FEMCD,R =  - 4.883 kN/m + 1.000 (+4.883 kN-m) = -4.167 kN-m + 4.883 kN-m

FEMCD,R =  0 kN/m

Note that the FEM is zero.  The joint is balanced, if you continue to ignore the carry over from joint C.  Table CB.2.1.5 shows the balancing moment and carry over moment (highlighted in yellow) resulting from this operation.

Table CB.2.1.5
Joint B Balance and Carry Over

  Span AB   Span BC   Span CD   Units
Span 7.5   5   6.25   m
1/L 0.133333   0.2   0.16    
Load 10 kN 2 kN/m 1.5 kN/m  
Joint A B B C C D  
DF 0.000 0.400 0.600 0.556 0.444 1.000  
FEM 9.375 -9.375 4.167 -4.167 4.883 -4.883 kN-m
Balance 0.000 2.083 3.125 -0.398 -0.318 4.883 kN-m
CO 1.042 0.000 -0.199 1.563 2.441 -0.159 kN-m
Sum 10.417 -7.292 7.093 -3.002 7.006 -0.159 kN-m

Adding the original FEMs to the corresponding balancing moments and carry over moments we arrive at a new FEM state.  Notice that the imbalance at each joint is significantly less than what the original FEMs provided.  We are starting to converge on the solution.

At this point we repeat the balance, carry over and sum steps until the imbalances become small enough for our purposes.  Generally, this occurs when the unbalanced moments are about 1% of the end moments.  With a spreadsheet, repeating the steps is simply a matter of copying the balance, carry over and sum steps once the formulas are all set up.  Table CB.2.1.6 shows the results of nine balancing cycles.

Table CB.2.1.6
The Full Solution

  Span AB   Span BC   Span CD   Units
Span 7.5   5   6.25   m
1/L 0.133333   0.2   0.16    
Load 10 kN 2 kN/m 1.5 kN/m  
Joint A B B C C D  
DF 0.000 0.400 0.600 0.556 0.444 1.000  
FEM 9.375 -9.375 4.167 -4.167 4.883 -4.883 kN-m
Balance 0.000 2.083 3.125 -0.398 -0.318 4.883 kN-m
CO 1.042 0.000 -0.199 1.563 2.441 -0.159 kN-m
Sum 10.417 -7.292 7.093 -3.002 7.006 -0.159 kN-m
Balance 0.000 0.080 0.119 -2.224 -1.780 0.159 kN-m
CO 0.040 0.000 -1.112 0.060 0.080 -0.890 kN-m
Sum 10.456 -7.212 6.100 -5.167 5.306 -0.890 kN-m
Balance 0.000 0.445 0.667 -0.077 -0.062 0.890 kN-m
CO 0.222 0.000 -0.039 0.334 0.445 -0.031 kN-m
Sum 10.679 -6.767 6.729 -4.910 5.689 -0.031 kN-m
Balance 0.000 0.015 0.023 -0.433 -0.346 0.031 kN-m
CO 0.008 0.000 -0.216 0.012 0.015 -0.173 kN-m
Sum 10.687 -6.752 6.535 -5.331 5.358 -0.173 kN-m
Balance 0.000 0.087 0.130 -0.015 -0.012 0.173 kN-m
CO 0.043 0.000 -0.008 0.065 0.087 -0.006 kN-m
Sum 10.730 -6.665 6.658 -5.282 5.433 -0.006 kN-m
Balance 0.000 0.003 0.005 -0.084 -0.067 0.006 kN-m
CO 0.002 0.000 -0.042 0.002 0.003 -0.034 kN-m
Sum 10.731 -6.662 6.620 -5.363 5.369 -0.034 kN-m
Balance 0.000 0.017 0.025 -0.003 -0.002 0.034 kN-m
CO 0.008 0.000 -0.001 0.013 0.017 -0.001 kN-m
Sum 10.740 -6.645 6.644 -5.354 5.383 -0.001 kN-m
Balance 0.000 0.001 0.001 -0.016 -0.013 0.001 kN-m
CO 0.000 0.000 -0.008 0.000 0.001 -0.007 kN-m
Sum 10.740 -6.645 6.637 -5.370 5.371 -0.007 kN-m
Balance 0.000 0.003 0.005 -0.001 0.000 0.007 kN-m
CO 0.002 0.000 0.000 0.002 0.003 0.000 kN-m
Sum 10.742 -6.642 6.641 -5.368 5.373 0.000 kN-m

Notice that the changes to the end moments have become "small".  It is time to stop.

With these end moments, the remaining shears and moments can be found using equilibrium equations.