Simple Forces Problem

  • Thread starter nikkelm
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  • #1
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I am studying for the FE exam and came to this problem:

A circular rod of 1in. cross-section area is subjected to several forces. Determine the displacement relative to C (in the attachment). Assume E =29x10^6 psi.

my solution is:

del=1/29x10^6(Fa-b(10in)/1in+Fc-b(8in)/1in)

yet I don't know how to get the right forces Fa-b and Fc-b

signed,

Mental Block
 

Answers and Replies

  • #2
tiny-tim
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Welcome to PF!

A circular rod of 1in. cross-section area is subjected to several forces. Determine the displacement relative to C (in the attachment) …

Hi nikkelm! Welcome to PF! :smile:

erm … no attachment … ! :cry:

Actually, it takes ages to get an attachment approved.

It'll be much quicker if you just describe the picture to us in words! :smile:
 
  • #3
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Oh... I guess that's for copy right infringement stuff... anyways, the cylinder axis is oriented horizontally and it is 18 in long from A to C and 10 inches long from A to B. A is at the left C at the far right with B in the middle, of course. The force at A is 2kip to the left. The force at B is 4kip to the right. The force at C is 4kip to the left again.
 
  • #4
PhanthomJay
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Before you go too far, I think you mean the force at C is 2K to the left, or else you're not in equilibrium.
 
  • #5
nvn
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nikkelm: Once you resolve the equilibrium issue, you could perform a section cut anywhere you wish, then solve for the unknown force acting on your free-body diagram, to obtain the force in each segment.
 
  • #6
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That explains things.. there is an error in the problem because it definitely shows a 4kip force
 
  • #7
nvn
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Is there any support on the rod? We would need to know about any support(s) on the rod, since we can't see your diagram. If it has no supports, then the rod is accelerating to the left. If accelerating, you might be able to use d'Alembert's principle to obtain the force on any section cut.
 
  • #8
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no, it isn't a kinematics question I am pretty sure it is a mistake. it is a multiple choice and the forces are supposed to be 2kip ea. which makes sense if they meant for it to be 2kip to the left at C.
 

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