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Need help with a Statically Indeterminate System!

  • Thread starter TA1068
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  • #1
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Homework Statement



Three steel bars are pin-connected to a rigid member K. Determine the force developed in each bar. Determine the load carried be each of the tension members and the elongation of each member

http://img3.imageshack.us/img3/6850/problemdiagram.jpg [Broken]

Known:
[tex]A_A_B[/tex] = 0.10 in^2
[tex]E_A_B[/tex] = 30E6 psi
[tex]A_C_D[/tex] = 0.20 in^2
[tex]E_C_D[/tex] = 15E6 psi
[tex]A_F_H[/tex] = 0.30 in^2
[tex]E_F_H[/tex] = 10E6 psi


Homework Equations



[tex]\delta[/tex] = (PL) / (AE)


The Attempt at a Solution



Finding the moment about B:
10[tex]P_C_D[/tex] + 20[tex]P_F_H[/tex] = 15(15000)
or
[tex]P_F_H[/tex] = 7500 - (1/3)[tex]P_C_D[/tex]

The equation [tex]\delta[/tex] = (PL) / (AE) yields:
[tex]P_A_B[/tex] = 150,000[tex]\delta_A_B[/tex]
[tex]P_C_D[/tex] = 200,000[tex]\delta_C_D[/tex]
[tex]P_F_H[/tex] = 300,000[tex]\delta_F_H[/tex]

And the sum of forces in the Y direction gives:
[tex]P_A_B[/tex] + [tex]P_C_D[/tex] + [tex]P_F_H[/tex] = 15000


This is where I'm stuck. If any point along K was fixed it would be easy; K is rigid, so then the distance from the fixed point can be turned into a ratio to find the other [tex]\delta[/tex] values. I think all 3 points (B, D, and H) are pulled downward, but I'm not sure what there relation is to each other. Any clues?
 
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Answers and Replies

  • #2
13
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Yikes, I think my TEX tags are all messed up. Not sure how to fix it, let me know if you have any questions!
 
  • #3
13
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Now if it was something like this:
http://img194.imageshack.us/img194/8167/fixedpoint.jpg [Broken]
I would say sigma_CD = 10(theta) and sigma_FH = 30(theta) and all would be good.


But since it's like this:
http://img199.imageshack.us/img199/9023/nonfixedpoint.jpg [Broken]
I have another variable in there with x. Now sigma_AB = (x)(theta), sigma_CD = (10+x)(theta), and sigma_FH = (30+x)(theta)

Hmm...
 
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