Solution to Problem 211 Axial Deformation

Based on allowable stresses:

Steel:

$P_{st} = \sigma_{st} \, A_{st}$

$P = 140(480) = 67\,200 \, \text{N}$

$P = 67.2 \, \text{kN}$

 

Bronze:

$P_{br} = \sigma_{br} \, A_{br}$

$2P = 120(650) = 78\,000 \, \text{N}$

$P = 39\,000 \, \text{N} = 39 \, \text{kN}$

 

Aluminum:

$P_{al} = \sigma_{al} \, A_{al}$

$2P = 80(320) = 25\,600 \, \text{N}$

$P = 12\,800 \, \text{N} = 12.8 \, \text{kN}$

Based on allowable deformation:
(steel and aluminum lengthens, bronze shortens)
$\delta = \delta_{st} - \delta_{br} + \delta_{al}$

$3 = \dfrac{P(1000)}{480(200\,000)} - \dfrac{2P(2000)}{650(83\,000)} + \dfrac{2P(1500)}{320(70\,000)}$

$3 = \left( \frac{1}{96\,000} - \frac{2}{26\,975} + \frac{3}{22\,400} \right) P$

$P = 42 \, 733.52 \, \text{N} = 42.73 \, \text{kN}$
 

Use the smallest value of P, P = 12.8 kN