1. The problem statement, all variables and given/known data 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 2. Relevant equations [tex]\delta[/tex] = (PL) / (AE) 3. 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?