Solution to Problem 251 Statically Indeterminate

$\Sigma M_{pin \,\, support} = 0$

$4P_A + 8P_B = 10(6600)$

$P_A + 2P_B = 16500$       → Equation (1)
 

By ratio and proportion
$\dfrac{\delta_A}{4} = \dfrac{\delta_B}{8}$

$\delta_A = 0.5 \delta_B$

$\left( \dfrac{PL}{AE} \right)_A = 0.5 \left( \dfrac{PL}{AE} \right)_B$

$\dfrac{P_A (4)}{AE} = \dfrac{0.5 P_B (6)}{AE}$

$P_A = 0.75P_B$
 

From Equation (1)
$0.75P_B + 2P_B = 16500$

$P_B = 6000 \, \text{ lb}$           answer
 

$P_A = 0.75(6000)$

$P_A = 4500 \, \text{ lb}$           answer