Analysis Using Moment Distribution

Introduction

With the continuous advancement of computer-based analysis within the field of structural engineering, the tools used for analysis by hand are becoming less relevant. Many will, however, argue that these tools are more important today if we are to fully appreciate the outputs of computer programs. Hence, this post describes one of the most powerful and fastest methods of analysis available: Moment Distribution.

Moment distribution is a very fast method of analyzing statically indeterminate structures elastically. It is based on the relative stiffness of members consisting of a structure. An out of balance moment is determine and distributed according to the relative stiffness until all joints within the structure are balanced.

Analysis Principles

The principle of moment distribution is based on creating fixed end moments at joints in a structure and then releasing them sequentially in order to derive the bending moments within it. This is done via an iterative process that relies on achieving equilibrium as the joints in the structure are released.

Terminologies of Moment Distribution

Fixed End Moment (FEM)

Fixed end moments are moments produced at member ends by assuming them to be fully fixed. Table 1 gives the fixed end moment for some common loading conditions within structures.

Table 1: Fixed End Moment
Stiffness

This is the ratio of the flexural modulus and the length of members defined as (EI/L). The fixed end moment is distributed in proportion to the relative members in the structure.

k=\frac { 4EI }{ L } far-end\quad is\quad fixed
k=\frac { 3EI }{ L } far-end\quad is\quad hinged
Distribution Factors

The distribution factor can be defined as the proportion of unbalanced moment carried by members connected to a joint. It is a ratio of relative stiffness across a joint. In mathematical terms, the distribution factor of a member ‘k’ framed at a joint ‘u’ is defined as

{ D }_{ uk }=\frac { \frac { { E }_{ k }{ I }_{ k } }{ L } }{ \sum _{ i }^{ n }{ \frac { { E }_{ i }{ I }_{ i } }{ L } } }
Carry-over Factors

When a joint is released, balancing moment occurs to counterbalance the unbalanced moment. The balancing moment is initially the same as the fixed-end moment. This balancing moment is then carried over to the member’s other end. A ratio of the carried-over moment at the other end to the fixed-end moment of the initial end is the carryover factor.

  • For Continuous Support C.O = 1/2
  • Hinged End C.O = 1.0
  • Fixed End C.O = 0
Sign Conventions

Counter-clockwise moments are taken as negative while clockwise moments are positive. Figure 1 further describes the sign convention. The sign convention on the left hans side is used in drawing bending moment diagrams while that on the right handside is used for determining fixed end moments

Fig 1: Moment Distribution Sign Convention
Support Settlement

There are instances when the support of a structure is subjected to vertical movement giving rise to additional moments within the support. This condition can be modelled in moment distribution by defining the vertical deflection and calculating the corresponding fixed end moment to that effect. Table 2 shows two typical instances where the support has dropped a distance ‘Δ’ and the resulting fixed end bending moments as shown

Table 2: Fixed End Moment of Displaced Supports

Procedure for Analysis

  1. Restrain all displacement and calculate relative stiffness of the elements consisting of the structure.
  2. Calculate the distribution factors of each joint and determine carry-over factors
  3. Evaluate the fixed end moment using Table 1 & 2 for the appropriate loading and support condition
  4. Carry-out successive distribution and carry-over until the joints become balanced.
  5. Determine the bending moment and shear-force at critical sections and draw diagrams.

Worked Example

Figure 2 shows a multi-span beam having constant flexural modulus EI at all section. Determine the bending moments and shear in the structure using moment distribution method. Assuming support B settles by 12mm downwards. E=210×103N/mm2 and I= 9.0×108mm4

Figure 2: Worked Example

The first thing for us to do is to calculate the relative stiffness of each segment of the beam, and from the figure, we can see that the far ends are fixed.

Relative Stiffness
{ k }_{ AB }={ k }_{ BC }=\frac { 4EI }{ l } =\frac { 4EI }{ 12 } =0.33EI\quad ;{ k }_{ CD }=\frac { 4EI }{ l } =\frac { 4EI }{ 8 } =0.5EI
Distribution Factors

Our distribution factors at support A and D is usually taken as 0 for fixed ends, because the supports are assumed to be infinitely rigid compared to the member.

{ D }_{ AB }={ D }_{ DC }=0
{ D }_{ BA }={ D }_{ BC }=\frac { { k }_{ AB } }{ { k }_{ AB }+{ k }_{ BC } } =\frac { 0.33EI }{ 0.33EI+0.33EI } =0.5
{ D }_{ CB }=\frac { { k }_{ BC } }{ { k }_{ BC }+{ k }_{ CD } } =\frac { 0.33EI }{ 0.33EI+0.5EI } =0.398
{ D }_{ CB }=\frac { { k }_{ CD } }{ { k }_{ BC }+{ k }_{ CD } } =\frac { 0.5EI }{ 0.33EI+0.5EI } =0.602
FEM

We use table 1 to determine the fixed end moments for each load conditions N:B The fixed end moment due to sinking of support B exerts a counterclockwise moment on member AB and a clockwise moment on member BC, therefore, its effect would be Negative in AB and Positive in BC.

{ M’ }_{ BA }=-\frac { 6EI\triangle }{ { l }^{ 2 } } =-\frac { 6\times 210\times { 10 }^{ 3 }\times 9.0\times { 10 }^{ 8 }\times 12 }{ { 12000 }^{ 2 } } \times { 10 }^{ -6 }=-94.5kN.m
{ M’ }_{ BA }=-\frac { 6EI\triangle }{ { l }^{ 2 } } =-\frac { 6\times 210\times { 10 }^{ 3 }\times 9.0\times { 10 }^{ 8 }\times 12 }{ { 12000 }^{ 2 } } \times { 10 }^{ -6 }=-94.5kN.m
{ M’ }_{ BC }=-\frac { { wl }^{ 2 } }{ 12 } +\frac { 6EI\triangle }{ { l }^{ 2 } } =-\frac { 20\times { 12 }^{ 2 } }{ 12 } +\frac { 6\times 210\times { 10 }^{ 3 }\times 9.0\times { 10 }^{ 8 }\times 12 }{ { 12000 }^{ 2 } } \times { 10 }^{ -6 }=-145.5kN.m
{ M’ }_{ CB }=\frac { { wl }^{ 2 } }{ 12 } +\frac { 6EI\triangle }{ { l }^{ 2 } } =-\frac { 20\times { 12 }^{ 2 } }{ 12 } +\frac { 6\times 210\times { 10 }^{ 3 }\times 9.0\times { 10 }^{ 8 }\times 12 }{ { 12000 }^{ 2 } } \times { 10 }^{ -6 }=334.5kN.m
{ M’ }_{ CD }=-{ M }’_{ DC }=-\frac { { Pl }^{ 2 } }{ 8 } =\frac { -250\times { 8 } }{ 8 } =-250kN.m
Distribution

Table 3 presents the distribution table for the beam. The process here is very simple. First, we determine the out of balance moment of a joint and distribute the negative of this moment according to the distribution factor. Secondly, carry over half of the distributed moment to the other end of the continuous support. This process is then repeated until the remaining distributed moments are less than 1% of the initially distributed moments.

For example, joint B is initially out of balance by (-94.5-145.5) = -240kN.m, so in order to balance the joint +240kN.m is distributed according to the distribution factors of the members connected to the joint. for member BA =0.5×240=120kN.m, similar for member BC and so forth. Half of this distributed moment is carried over to the other side, which is 120/2 =60 in the case of AB

N:B The moments at fixed ends were not carried over because the carry-over factor for fixed ends is zero.

Bending Moment & Shear Force Diagrams

In-order to draw the bending moment and shear diagrams, we must determine the values of the moment at the spans and shear force at the supports. This is done by analyzing each segment of the beam discretely using simple rules of statics using figure 3.

Figure 3: Statics of Beam
Shears at Support
{ V }_{ AB }=-{ V }_{ BA }=\left[ \frac { { M }_{ BA }-{ M }_{ AB } }{ L } \right] =\left[ \frac { 40.6-27.09 }{ 12 } \right] =1.125kN
{ V }_{ BC }=\left[ \frac { { M }_{ BC }-{ M }_{ CB } }{ l } \right] +\frac { w{ l }^{ 2 } }{ 2l }
{ V }_{ BC }=\left[ \frac { 40.6-341.45 }{ 12 } \right] +\frac { 20\times { 12 }^{ 2 } }{ 2\times 12 } =\quad 94.9kN
{ V }_{ CB }=wl-{ V }_{ BC }=(20\times 12)-94.9=145.1kN
{ V }_{ CD }=\left[ \frac { { M }_{ CD }-{ M }_{ DC } }{ l } \right] +\frac { Pl }{ 2l }
{ V }_{ CD }=\left[ \frac { 341.45-204.37 }{ 8 } \right] +\frac { 250\times 8 }{ 2\times 8 } =142.14kN
{ V }_{ DC }=P-{ V }_{ CD }=250-142.14=107.86kN
Bending Moment at Spans

The bending moment at span AB = 0k.Nm

Bending moment at span BC

{ M }_{ max }={ V }_{ BA }x-\frac { w{ x }^{ 2 } }{ 2 } -{ M }_{ BC }
{ M }_{ max\quad }occurs\quad when\frac { dM }{ dx } =0
V_{ BC }-wx=0
x=\frac { { V }_{ BC } }{ w } =\frac { 94.9 }{ 20 } = 4.75m
{ M }_{ max }=(94.9\times 4.75)-\frac { 20\times 4.75^{ 2 } }{ 2 } -40.6=184.6kN.m

Bending Moment at span CD

{ M }_{ max }={ V }_{ CD }x-{ M }_{ CD }
x=4m
{ M }_{ max }=(142.14\times 4)-341.45=227.11kN.m
Figure 4: Shear Force and Bending Moment Diagrams

Further Reading & References

  • Brohn, D.M.: (2005) Understanding Structural Analysis 3rd Ed. New Paradigm Solutions
  • Hibbeler, R.C: (2015) Structural Analysis 9th Ed. Pearson.

THANKYOU FOR READING!!!

Omotoriogun Victor
About Omotoriogun Victor 66 Articles
A dedicated, passion-driven and highly skilled engineer with extensive knowledge in research, construction and structural design of civil engineering structures to several codes of practices

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