Solution to Problem 250 Statically Indeterminate

$P_{st} = P_{br}$

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

$\frac{3}{4} \sigma_{st} = 1.5 \, \sigma_{br}$

$\sigma_{st} = 2 \sigma_{br}$
 

For one turn of the nut:
$\delta_{st} + \delta_{br} = \frac{1}{32}$

$\left( \dfrac{\sigma \, L}{E} \right)_{st} + \left( \dfrac{\sigma \, L}{E} \right)_{br} = \dfrac{1}{32}$

$\dfrac{\sigma_{st} (40)}{29 \times 10^6} + \dfrac{\sigma_{br} (40)}{12 \times 10^6} = \dfrac{1}{32}$

$\sigma_{st} + \frac{29}{12} \sigma_{br} = 22\,656.25$

$2\sigma_{br} + \frac{29}{12} \sigma_{br} = 22\,656.25$

$\sigma_{br} = 5129.72 \, \text{psi}$

$\sigma_{st} = 2(5129.72) = 10 259.43 \, \text{psi}$
 

Initial stresses:
$\sigma_{br} = 4000 \, \text{psi}$

$\sigma_{st} = 2(4000) = 8000 \, \text{psi}$
 

Final stresses:
$\sigma_{br} = 4000 + 5129.72 = 9\,129.72 \, \text{ psi}$           answer

$\sigma_{st} = 2(9129.72) = 18\,259.4 \, \text{ psi}$           answer
 

Required number of turns to reduce σ_br to zero:
$n = \dfrac{9129.72}{5129.72} = 1.78 \, \text{ turns }$

The nut must be turned back by 1.78 turns