Finding the force exerted on a truss

  • Thread starter Patdon10
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  • #1
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Homework Statement


Find the Force exerted on member ABCD by the pin at B in figure P4.162 (figure is attached)


The Attempt at a Solution


This is also attached. As you can see I solved for everything I could. If I could solve for just one more unknown, I could solve the problem. I just don't see where I could possibly solve for any of them. Can anyone nudge me in the right direction?
 

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Answers and Replies

  • #2
nvn
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Patdon10: Hint: Let your free body be bars BF and CE. The unknown external forces on this free body occur at points B and C. This free body has three equations, and three unknowns. You know the force in bar CE is axial.
 
  • #3
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What are the 3 unknowns? B_x, B_y, and then just C?
 
  • #4
nvn
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Yes.
 
  • #5
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I'm having trouble getting a grasp on this, how do I know what direction C points in?
 
  • #6
nvn
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You don't (or you do not need to know). However, you do know C is an axial force, like I said. Regarding whether it is tension or compression, you just assume a direction. Then if its numeric value, in the solution, comes out negative, you know you drew it backwards.
 
  • #7
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But what direction would it be? What are my options? because the pin connects to truss AD, does the direction have to go along that truss? or is it along it's own truss - meaning it would simply be up or down.
 
  • #8
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F_x = B_x - 1000 = 0
B_x = 1000N

Summation of Fy = B_y - C = 0
B_y = C

M_B = (-0.25)C + (0.73)(1000) = 0
C = B_y = 2920 N

Am I understanding this correctly?
 
  • #9
nvn
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"The force in bar CE is axial" means the force in bar CE is along (parallel to) bar CE. It is also explained in post https://www.physicsforums.com/showthread.php?t=484381#post3211259".

Your solution in post 8 looks good, except you rounded too much. Generally always maintain at least four significant digits throughout all your intermediate calculations, then round only the final answer to three significant digits.

Keep in mind, you found the force bar AD exerts on bar BF. The question in post 1 asks for the force bar BF exerts on bar AD, which is the opposite.
 
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