Axially loaded member problem

  • Thread starter whynot314
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  • #1
whynot314
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[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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  • #2
whynot314
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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)([itex]\frac{A(al)}{A(st)}[/itex])([itex]\frac{E(al)}{E(st)}[/itex]). Using this I was able to get an answer but it was wrong.
 
  • #3
whynot314
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Note equation of deformation is [itex]\delta[/itex]=[itex]\frac{PL}{AE}[/itex].
 
  • #4
PhanthomJay
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Did you solve for the first part correctly (Question 4-41)? The top alum piece, and the bottom steel piece, and the bottom alum piece, all have the same axial deformations. Using your deformation equations for each, and Newton's 1st law, gives you 4 equations with 4 unknowns, which you can solve for the forces and deformation.

In the next question 4-42, you can solve for the applied force and internal forces necessary to deform all 3 pieces equally at 5 mm each, using the same approach, but considering the bottom al-steel end free to move without deformation. Now if the applied force turns out less than 400 N to give this 5 mm deformation, then the remaing force (400 - the applied force you calculated) should be used similar to the first question to get the additional forces in the sections.
 

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