[URL=http://s1341.photobucket.com/user/nebula-314/media/20130909_125150_zpse5d4c881.jpg.html][PLAIN]http://i1341.photobucket.com/albums/o745/nebula-314/20130909_125150_zpse5d4c881.jpg[/URL][/PLAIN]

I am not sure how to go beyond this point or even if I have the compatibility problem correct, I am assuming that the bar does hit the other side of the wall. But when I go to solve this I have to many unknowns.

[URL=http://s1341.photobucket.com/user/nebula-314/media/20130909_123600_zps3c277e60.jpg.html][PLAIN]http://i1341.photobucket.com/albums/o745/nebula-314/20130909_123600_zps3c277e60.jpg[/URL][/PLAIN]

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Also because the steel rod is bonded to the aluminum rod i figured that member BC would have the same displacement, thus giving me F(al)=F(st)($\frac{A(al)}{A(st)}$)($\frac{E(al)}{E(st)}$). Using this I was able to get an answer but it was wrong.

Note equation of deformation is $\delta$=$\frac{PL}{AE}$.

PhanthomJay