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                                Solved example on Moment Distribution method for non-sway Frame  

Problem 8-2

Use Moment distribution method to find the resultant end moments for the non-sway frame shown in figure 8-2(a). Take EI as constant for all the members of the frame.

Figure 8-2(a)

Solution:

Step 1: The given rigid frame is non-sway because of lateral support at D. Consider each span (AB, BC, CD and CE) with both ends fixed and calculate the fixed-end moments as follows;

MAB = 0

MBA = 0

MBC = wL2 /12 = -6.67 kNm

MCB = -wL2 /12 = +6.67 kNm

MCD = PL/8 = -10 kNm

MDC = -PL/8 = +10 kNm

MCE = 0

MEC = 0

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Step 2: Stiffness Coefficients

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

kBA = 4EI/L = 4EI/5 =0.8EI

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

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

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

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

kCE = 4EI/L = 4EI/4 = EI

kEC = 4EI/L = 4EI/4 = EI

Step 3: Distribution factors

DAB = 0

DBA = 0.8EI/(0.8EI+EI) = 0.44

DBC = EI/(0.8EI+EI) = 0.56

DCB = EI/(EI+EI+EI) =0.33

DCD = EI/(EI+EI+EI) = 0.33

DCE = EI/(EI+EI+EI) = 0.33

DDC = 0

DEC = 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 8-2(b) is plotted by superposing the span moment diagram and support moment diagram.

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

      bending moment diagram for frame

                               (all values in kNm)

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

                                 Figure 8-2(b)     

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Table for Moment Distribution
Joints

A

B

C

D E
Members AB

BA

BC

CB

CE

CD

DC EC
Distribution Factors 0 0.44 0.56 0.33 0.33 0.33 0 0
Initial Fixed-end Moments 0 0 -6.67 +6.67 0 -10 +10 0
Balancing   6.67x0.44

2.94

6.67x0.56

3.74

+3.33x0.33

1.11

+3.33x0.33

1.11

+3.33x0.33

1.11

   
Carry-over 1.47   0.555 1.87     0.555 0.555
Balancing   -0.555x0.44

-0.242

-0.555x0.56

-0.312

-1.87x0.33

-0.62

-1.87x0.33

-0.62

-1.87x.33

-0.62

   
Carry-over -0.121   -0.31 -0.156     -0.31 -0.31
Balancing   0.31x0.44

0.14

0.31x0.56

0.17

0.156x0.33

0.052

0.l56x0.33

0.052

0.156x0.33

0.052

   
Carry-over 0.07   0.026 0.09     0.026 0.026
Balancing   -0.026x0.44

-0.011

-0.026x0.56

-0.015

-0.09x0.33

-0.03

-0.09x0.33

-0.03

-0.09x0.33

-0.03

   
Carry-over -0.006   -0.015 -0.007     -0.015 -0.015
Final End-Moments 1.41 2.83 -2.83 8.98 0.51 -9.49 10.26 0.26
 

Problem 8-1 for solving indeterminate beam by moment distribution method
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Last updated on Sunday December 02, 2012