Solution to Problem 215 Axial Deformation

$6P_{st} = 2W$

$P_{st} = \frac{1}{3} W$
 

$\Sigma M_{st} = 0$

$6P_{al} = 4W$

$P_{al} = \frac{2}{3} W$
 

$\delta_{st} = \delta_{al}$

$\left[ \dfrac{PL}{AE} \right]_{st} = \left[ \dfrac{PL}{AE} \right]_{al}$

$\dfrac{\frac{1}{3}W (6 \times 12)}{A_{st} (29 \times 10^6)} = \dfrac{\frac{2}{3}W (4 \times 12)}{A_{al} (10 \times 10^6)}$

$\dfrac{A_{al}}{A_{st}} = \dfrac{\frac{2}{3}W (4 \times 12)(29 \times 10^6)}{\frac{1}{3}W (6 \times 12)(10 \times 10^6)}$

$\dfrac{A_{al}}{A_{st}} = 3.867$           answer