An isotropic elastic titanium alloy component has Young's modulus $E = 110\text{ GPa}$ and Poisson's ratio $\nu = 0.30$. Under a hydrostatic compressive pressure of $p = 100\text{ MPa}$, the volumetric strain $\epsilon_v = \frac{\Delta V}{V}$ experienced by the component is closest to:

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### 1. Strategy & Theoretical Principles Under hydrostatic pressure $p$, the state of stress is triaxial with $\sigma_{xx} = \sigma_{yy} = \sigma_{zz} = -p$. The Bulk Modulus $K$ relates hydrostatic pressure to volumetric strain $\epsilon_v$: $$K = \frac{E}{3(1 - 2\nu)}$$ $$\epsilon_v = -\frac{p}{K} = -\frac{3(1 - 2\nu) p}{E}$$ ### 2. Analytical Formulation & Calculation Given: - $E = 110\text{ GPa} = 110 \times 10^9\text{ Pa}$ - $\nu = 0.30 \implies 1 - 2\nu = 1 - 0.60 = 0.40$ - $p = 100\text{ MPa} = 100 \times 10^6\text{ Pa}$ 1. Bulk Modulus $K$: $$K = \frac{110 \times 10^9}{3 \times 0.40} = \frac{110 \times 10^9}{1.20} \approx 91.667 \times 10^9\text{ Pa} = 91.67\text{ GPa}$$ 2. Volumetric strain $\epsilon_v$: $$\epsilon_v = -\frac{100 \times 10^6}{91.667 \times 10^9} = -\frac{100}{91667} \approx -0.0010909 \approx -0.00109$$ Or expressed in scientific notation $-1.09 \times 10^{-3}$. ### 3. Distractor Forensics - Using linear strain $\epsilon = -p(1-2\nu)/E$ gives $-0.000363$, which is one-third of the volumetric strain $\epsilon_v = 3\epsilon$. ### 4. Exam Trap & Standard Reference - **Trap Alert:** Remember $\epsilon_v = \epsilon_x + \epsilon_y + \epsilon_z = 3\epsilon$ for isotropic materials. - **Standard Reference:** Megson, T.H.G., *Aircraft Structures for Engineering Students*, Chapter 1.

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