6.4 A cylindrica specimen of a titantum alloy having an elastic modulus of 107 GPa (15.5times 10^6psi) and an original diameter of 3.8 mm (0.15 in.)will experience only elastic deformation when a tensile load of 2000 N (450 Iby) is applied.Compute the maximum length of the specimen before deformation if the maximum allowable clongation is 0.42 min (0.0165 in).

To solve this problem, we need to use the formula for elastic deformation:<br /><br />\[ \epsilon = \frac{\sigma}{E} \]<br /><br />where:<br />- \(\epsilon\) is the strain (deformation),<br />- \(\sigma\) is the stress,<br />- \(E\) is the elastic modulus.<br /><br />Given:<br />- Elastic modulus, \(E = 107 \text{ GPa} = 15.5 \times 10^6 \text{ psi}\)<br />- Original diameter, \(d = 3.8 \text{ mm} = 0.15 \text{ in.}\)<br />- Tensile load, \(F = 2000 \text{ N} = 450 \text{ lbf}\)<br />- Maximum allowable elongation, \(\Delta L = 0.42 \text{ mm} = 0.0165 \text{ m}\)<br /><br />First, we need to calculate the original cross-sectional area \(A\) of the specimen:<br /><br />\[ A = \frac{\pi d^2}{4} \]<br /><br />Substituting \(d = 3.8 \text{ mm}\):<br /><br />\[ A = \frac{\pi (3.8 \text{ mm})^2}{4} = \frac{\pi \times 14.44 \text{ mm}^2}{4} = 11.36 \text{ mm}^2 \]<br /><br />Next, we calculate the maximum allowable stress \(\sigma_{\text{max}}\):<br /><br />\[ \sigma_{\text{max}} = \frac{F}{A} \]<br /><br />Substituting \(F = 2000 \text{ N}\) and \(A = 11.36 \text{ mm}^2\):<br /><br />\[ \sigma_{\text{max}} = \frac{2000 \text{ N}}{11.36 \text{ mm}^2} = 175.88 \text{ MPa} \]<br /><br />Now, we can find the maximum allowable strain \(\epsilon_{\text{max}}\):<br /><br />\[ \epsilon_{\text{max}} = \frac{\sigma_{\text{max}}}{E} \]<br /><br />Substituting \(\sigma_{\text{max}} = 175.88 \text{ MPa}\) and \(E = 107 \text{ GPa}\):<br /><br />\[ \epsilon_{\text{max}} = \frac{175.88 \text{ MPa}}{107 \text{ GPa}} = 0.00164 \]<br /><br />Finally, we can find the maximum length \(L_{\text{max}}\) before deformation:<br /><br />\[ \Delta L = \epsilon_{\text{max}} \times L_{\text{max}} \]<br /><br />Solving for \(L_{\text{max}}\):<br /><br />\[ L_{\text{max}} = \frac{\Delta L}{\epsilon_{\text{max}}} \]<br /><br />Substituting \(\Delta L = 0.42 \text{ mm}\) and \(\epsilon_{\text{max}} = 0.00164\):<br /><br />\[ L_{\text{max}} = \frac{0.42 \text{ mm}}{0.00164} = 256.10 \text{ mm} \]<br /><br />Therefore, the maximum length of the specimen before deformation is approximately 256.10 mm.