Solution to Problem 213 Axial Deformation

Free body diagram:
 

 

For aluminum:
$\Sigma M_B = 0$

$6P_{al} = 2.5(50)$

$P_{al} = 20.83 \, \text{kN}$
 

$\delta = \dfrac{PL}{AE}$

$\delta_{al} = \dfrac{20.83(3)1000^2}{500(70\,000)}$

$\delta_{al} = 1.78 \, \text{mm}$
 

For steel:
$\Sigma M_A = 0$

$6P_{st} = 3.5(50)$

$P_{st} = 29.17 \, \text{kN}$
 

$\delta = \dfrac{PL}{AE}$

$\delta_{st} = \dfrac{29.17(4)1000^2}{300(200\,000)}$

$\delta_{st} = 1.94 \, \text{mm}$
 

Movement diagram:
 

 

$\dfrac{y}{3.5} = \dfrac{1.94 - 1.78}{6}$

$y = 0.09 \, \text{mm}$

$\delta_B = \, \text{vertical movement of } P$

$\delta_B = 1.78 + y = 1.78 + 0.09$

$\delta_B = 1.87 \, \text{mm}$           answer