Solution to Problem 212 Axial Deformation

Based on maximum stress of steel rod:
$\Sigma M_A = 0$

$5P = 2P_{st}$

$P = 0.4P_{st}$

$P = 0.4 \sigma_{at} \, A_{st}$

$P = 0.4 \, [ \, 30(0.50) \, ]$

$P = 6 \, \text{kips}$
 

 

Based on movement at C:
$\dfrac{\delta_{st}}{2} = \dfrac{0.1}{5}$

$\delta_{st} = 0.04 \, \text{in}$

$\dfrac{P_{st} \, L}{AE} = 0.04$

$\dfrac{P_{st} \, (4 \times 12)}{0.50(29 \times 10^6)} = 0.04$

$P_{st} = 12 083.33 \, \text{lb}$
 

$\Sigma M_A = 0$

$5P = 2P_{st}$

$P = 0.4P_{st}$

$P = 0.4(12\,083.33)$

$P = 4833.33 \, \text{lb} = 4.83 \, \text{kips}$
 

Use the smaller value, P = 4.83 kips