Solution to Problem 209 Axial Deformation

P₁ = 6000 lb tension
P₂ = 1000 lb compression
P₃ = 4000 lb tension
 

 

$\delta = \dfrac{PL}{AE}$
 

$\delta = \delta_1 - \delta_2 + \delta_3$

$\delta = \dfrac{6000(3 \times 12)}{0.5(10 \times 10^6)} - \dfrac{1000(5 \times 12)}{0.5(10 \times 10^6)} + \dfrac{4000(4 \times 12)}{0.5(10 \times 10^6)}$

$\delta = 0.0696 \, \text{ in. (lengthening)}$           answer