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The problem was assigned in my Structural Analysis class, and I am having some difficulty with it.

My professor said the problem is "very easy" if we just think about what the fixed end moment on a cantilever is. However, I don't see how this is relevant considering the member ACE is not a cantilever...

The only other way I see to do this is using the moment distribution method, but I'm uncertain as to what happens with the internal hinges; we've never had a problem like this in class, and there is no relevant example in my textbook. Any suggestions?

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  • $\begingroup$ I'm not sure if this qualifies as a duplicate, but this question (and the requisite solution) is very similar to a previous one. $\endgroup$ – Wasabi Dec 5 '15 at 20:09
  • $\begingroup$ It actually seems pretty different. That problem was statically determinate I believe. $\endgroup$ – WGD Dec 5 '15 at 22:13
  • $\begingroup$ @WGD, yes, but the essential trick to solving it can be seen in its answer (which is the same as suggested by CableStay in their answer below). You can "split" the structure into three segments: from the fixed end to the first hinge (which is equivalent to an indeterminate fixed-and-pinned beam with a concentrated load at its end), an isostatic beam between the hinges and a simply supported beam as of the second hinge (which you don't even need to solve in this case). $\endgroup$ – Wasabi Dec 5 '15 at 23:58
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Remember that hinges cannot transfer moment - only shear. Try drawing three separate free body diagrams (member ACD, member DE, and member EFG) knowing that the only reactions at the ends of member DE can be shear.

The free body diagram that you come up with for ACD should look much more manageable than the initial problem - a end loaded cantilever with redundant roller support.

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  • $\begingroup$ Would it be possible for me to use the flexibility method on member ACD as it will still be indeterminate? I have never seen the method used on an isolated segment of a structure before. Also, is there some simple formula to calculate the fixed end moment on a cantilever with a redundant roller support? I feel like this is what my professor was most likely alluding to, but I'm uncertain. $\endgroup$ – WGD Dec 5 '15 at 22:12
  • $\begingroup$ @WGD yes you can use the flexibility method. When you have a Gerber beam (segment between two hinges with no internal supports), you can split the structure and the different segments behave exactly as three different structures which can each he solved by traditional methods. $\endgroup$ – Wasabi Dec 6 '15 at 0:12
  • $\begingroup$ You can use the flexibility method to solve for the roller reaction. Then get the reactions at A. $\endgroup$ – CableStay Dec 6 '15 at 0:12
  • $\begingroup$ @WGD and whatever equation there is for fixed-and-pinned beams will be whatever you get by solving the flexibility method. $\endgroup$ – Wasabi Dec 6 '15 at 0:14

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