Solution to Problem 255 Statically Indeterminate

$\delta_B = \delta_C + \delta_2$

$\delta_2 = \delta_B - \delta_C$

$\dfrac{\delta_1}{6} = \dfrac{\delta_2}{2}$

$\delta_1 = 3 \delta_2$
 

$\delta_A = \delta_C + \delta_1$

$\delta_A = \delta_C + 3 \delta_2$

$\delta_A = \delta_C + 3(\delta_B - \delta_C)$

$\delta_A = 3 \delta_B - 2 \delta_C$

$\left( \dfrac{PL}{AE} \right)_A = 3 \left( \dfrac{PL}{AE} \right)_B - 2 \left( \dfrac{PL}{AE} \right)_C$

$\dfrac{P_A (5)}{AE} = \dfrac{3 P_B (6)}{AE} - \dfrac{2 P_C (6)}{AE}$

$P_A = 3.6P_B - 2.4P_C$       → Equation (1)
 

$\Sigma F_V = 0$

$P_A + P_B + P_C = 600$

$(3.6P_B - 2.4P_C) + P_B + P_C = 600$

$4.6P_B - 1.4P_C = 600$       → Equation (2)
 

$\Sigma M_A = 0$

$4P_B + 6P_C = 3(600)$

$P_B = 450 - 1.5P_C$       → Equation (3)
 

Substitute P_B = 450 - 1.5 P_C to Equation (2)
$4.6(450 - 1.5P_C) - 1.4P_C = 600$

$8.3P_C = 1470$

$P_C = 177.11 \, \text{ kN}$           answer
 

From Equation (3)
$P_B = 450 - 1.5(177.11)$

$P_B = 184.34 \, \text{ kN}$           answer
 

From Equation (1)
$P_A = 3.6(184.34) - 2.4(177.11)$

$P_A = 238.56 \, \text{ kN}$           answer