Maximum moments in 2-span beams

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SUMMARY

The discussion focuses on calculating maximum moments in a 2-span continuous beam subjected to different uniformly distributed loads (UDL). Participants confirm that the structure is statically indeterminate and recommend using the force method for analysis. They suggest adapting moment distribution techniques from examples with three spans and emphasize the importance of calculating distribution factors based on span stiffness. The maximum moment at the center support can be approximated by averaging the moments from each span, although this method has limitations for unequal spans and loads.

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  • Understanding of statically indeterminate structures
  • Familiarity with the force method in structural analysis
  • Knowledge of moment distribution techniques
  • Basic principles of bending moment diagrams
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1. How do you calculate the maximum moments in a 2-span continuous beam?

2 span beam with 2 different uniformly distributed loads (UDL) at different spans:

i46.tinypic.com/2d7zpw.jpg

I am a first year civil engineering student. This is apparently a statically indeterminate structure with one redundancy so I can't solve the problem using equilibrium equations.

I have seen a method called Moment Distribution but the examples seem to be available for 3 spans or more. Is there any simple way to find maximum bending moment in the beam?

Thanks in advance.
 
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You mean, a statically indeterminate structure. You should look into the force method for this. Depends on what you need, I think you may also find tables with moment diagrams for such beams.
 
Taking each span as a propped cantilever, the end moment at the centre support with a udl would be q*L*L/8. The difference between the moments at the centre support is then redistributed in proportion to the stiffnesses of the two beams. That is, moment distribution with just one release. You should be able to adapt from an example with three spans.
 
I notice that if you average the end moments (at the centre) in this case, the correct answer is very close (<0.3%) to that, and is what an engineer might use in practice
 
Hi and thanks for responses.

Do you mean average the end moments (at centre support) ie M1=(q*L^2)/8 from the left hand span and M2=(q*L^2)/8 from right hand span?

(M1+M2)/2 = Max bending moment in the whole span

Correct?
 
That is what I meant, although it is not strictly correct for unequal spans and unequal udl's. Then stiffness K1 of left hand span is 1/L1. Hence k2 also. Then distribution factors DF's are k1/(k1+k2) and k2/(k1+k2) (The DF's should add up to 1). The moment at the centre is the maximum moment in the beam, but not as a matter of principle. To get the full bending moment diagram you need to take your final centre moment and use it (with the other loads) to find the reactions at the end supports. Then you can find the M at any section in the beam and draw the M diagram.
 
Thank you!
 

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