Solving indeterminate frame by Moment Distribution Method

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Problem 8-2

Use Moment distribution method to find the resultant end moments for the non-sway frame shown in figure 8-2(a). Also draw bending moment diagram. 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;

M_AB = 0

M_BA = 0

M_BC = wL² /12 = -6.67 kNm

M_CB = -wL² /12 = +6.67 kNm

M_CD = PL/8 = -10 kNm

M_DC = -PL/8 = +10 kNm

M_CE = 0

M_EC = 0

Step 2: Stiffness Coefficients

k_AB = 4EI/L = 4EI/5 = 0.8EI

k_BA = 4EI/L = 4EI/5 = 0.8EI

k_BC = 4EI/L = 4EI/4 = EI

k_CB = 4EI/L = 4EI/4 = EI

k_CD = 4EI/L = 4EI/4 = EI

k_DC = 4EI/L = 4EI/4 = EI

k_CE = 4EI/L = 4EI/5 = 0.8EI

k_EC = 4EI/L = 4EI/5 = 0.8EI

Step 3: Distribution factors

D_AB = 0 (fixed support at the end)

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

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

D_CB = EI/(EI+EI+0.8EI) =0.36

D_CD = EI/(EI+EI+0.8EI) = 0.36

D_CE = 0.8EI/(EI+EI+0.8EI) = 0.28

D_DC = 0 (fixed support at the end)

D_EC = 0 (fixed support at the end)

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 to show the illustration. 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 BA to AB, BC to CB, CB to BC, CD to DC and CE to EC etc.

(In case of mobile device please scroll horizontally to view the full width of table)

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	0.36	0.28	0.36	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	+3.33x0.36 1.2	+3.33x0.28 0.93	+3.33x0.36 1.2
Carry-over	1.47		0.6	1.87			0.6	0.47
Balancing		-0.6x0.44 -0.264	-0.6x0.56 -0.336	-1.87x0.36 -0.67	-1.87x0.28 -0.52	-1.87x0.36 -0.67
Carry-over	-0.132		-0.33	-0.168			-0.33	-0.26
Balancing		0.33x0.44 0.145	0.33x0.56 0.185	0.168x0.36 0.06	0.l68x0.28 0.048	0.168x0.36 0.06
Carry-over	0.072		0.03	0.092			0.03	0.024
Balancing		-0.03x0.44 -0.013	-0.03x0.56 -0.017	-0.092x0.36 -0.033	-0.092x0.28 -0.026	-0.092x0.36 -0.033
Carry-over	-0.006		-0.016	-0.008			-0.016	-0.013
Final End-Moments	1.41	2.83	-2.83	8.98	0.51	-9.49	10.26	0.26

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

     

                               (all values in kNm)

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

                                 Figure 8-2(b)     

You can visit the following links of solved examples for Indeterminate Structures

Problem 8-1^New Solution of Indeterminate beam by moment distribution method

Problem 7-4 Solution of Indeterminate Structure by slope-deflection equations

Problem 7-1 Solution of Indeterminate Structure (Propped cantilever) by using compatibility equations

Problem 7-2 Solution of Indeterminate Structure (Continuous Beam) by consistent deformation

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