# Solution to Problem 250 Statically Indeterminate

**Problem 250**

In the assembly of the bronze tube and steel bolt shown in Fig. P-250, the pitch of the bolt thread is p = 1/32 in.; the cross-sectional area of the bronze tube is 1.5 in.^{2} and of steel bolt is 3/4 in.^{2} The nut is turned until there is a compressive stress of 4000 psi in the bronze tube. Find the stresses if the nut is given one additional turn. How many turns of the nut will reduce these stresses to zero? Use E_{br} = 12 × 10^{6} psi and E_{st} = 29 × 10^{6} psi.

**Solution 250**

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

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

$\sigma_{st} = 2 \sigma_{br}$

**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**