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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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