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                             Indeterminate Structure- solution by Moment Distribution method  

Problem 8-1

Use Moment distribution method to find the resultant end moments for the continuous beam shown in figure 8-1(a). Take EI as constant for the beam.

continuous beam

Figure 8-1(a)

Solution:

Step 1: The given continuous beam has three spans. Consider each span (AB, BC, and CD) with both ends fixed and calculate the fixed-end moments as follows;

MAB = {wa2 (6L2 - 8aL + 3a2 )}/(12L2 ) = -17.1 kNm

MBA = -{wa3 (4L - 3a}/(12L2 ) = +9.9 kNm

MBC = wL2 /12 = -6.67 kNm

MCB = -wL2 /12 = +6.67 kNm

MCD = PL/8 = -10 kNm

MDC = -PL/8 = +10 kNm

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Step 2: Stiffness Coefficients

kAB = 4EI/L = 4EI/5 =0.8EI

kBA = 3EI/L = 3EI/5 =0.6EI

kBC = 4EI/L = 4EI/4 = EI

kCB = 4EI/L = 4EI/4 = EI

kCD = 4EI/L = 4EI/4 = EI

kDC = 4EI/L = 4EI/4 = EI

Step 3: Distribution factors

DAB = 1

DBA = 0.6EI/(0.6EI+EI) = 0.375

DBC = EI/(0.6EI+EI) = 0.625

DCB = EI/(EI+EI) =0.5

DCD = EI/(EI+EI) = 0.5

DDC = 0

Calculation for moment distribution is shown in a systematic way in a tabular form. Each joint is balanced by distributing the unbalanced moment into the connected members  in proportion of distribution factors. These calculation steps for balancing of joints are shown with grey colour. The carry-over moments (half of the balancing moment at near end) are applied on the far end, e.g. carry-over half of the balancing moment from BC to CB.  Resultant bending moment diagram in figure 8-1(b) is plotted by superposing the span moment diagram and support moment diagram.

Note: Please use bending  moment calculator to get the values of span moments

continuous beam

   bending moment diagram of continuous beam

                          (all values in kNm)

   (span moment in blue colour, support moment in red colour)

                            Figure 8-1(b)    

Table for Moment Distribution
Joints

A

B

C

D
Members AB

BA

BC

CB

CD DC
Distribution Factors 1 0.375 0.625 0.5 0.5 0
Initial Fixed-end Moments -17.1 +9.9 -6.67 +6.67 -10 +10
Balancing +17.1 -3.23x0.375

-1.21

-3.23x0.625

-2.02

+3.33x0.5

1.665

+3.33x0.5

1.665

 
Carry-over   8.505 0.833 -1.01   0.833
Balancing   -9.34x0.375

-3.5

-9.34x0.625

-5.84

1.01x0.5

0.505

1.01x0.5

0.505

 
Carry-over     0.253 -2.92   0.253
Balancing   -0.253x0.375

-0.095

-0.253x0.625

-0.158

2.92x0.5

1.46

2.92x0.5

1.46

 
Carry-over     0.73 -0.079   0.73
Balancing   -0.73x0.375

-0.274

-0.73x0.625

-0.456

0.079x0.5

0.04

0.079x0.5

0.04

 
Carry-over     0.02 -0.23   0.02
Balancing   -0.02x0.375

-0.007

-0.02x0.625

-0.012

0.23x0.5

0.12

0.23x0.5

0.12

 
Carry-over     0.06 -0.006   0.06
Final End-Moments 0 13.28 -13.27 6.2 -6.2 11.95

Problem 8-2 for solving indeterminate frame by moment distribution method
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Last updated on Sunday December 02, 2012