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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.
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 |
(all values in kNm)
(span moment in
blue colour, support moment in
red colour)
Figure 8-1(b)
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