Let's break down the problem and solve it step by step: --- ## Given - Stress-strain relation: \(\bar{\sigma} = 22,000 \bar{\epsilon}^{.25}\) Where: - \(\bar{\sigma}\) = true (axial) stress (psi) - \(\bar{\epsilon}\) = true (axial) strain A thin-walled tube with closed ends is internally pressurized. --- ## (a) **At what value of \(\bar{\epsilon}\) will instability occur with respect to pressure?** ### **Step 1: Instability Condition** For thin-walled tubes (cylinders) under internal pressure, **pressure instability** (maximum internal pressure) occurs at the **maximum point of the pressure-strain curve**, i.e., when: \[ \frac{dP}{d\bar{\epsilon}} = \] For a thin-walled tube, the **hoop strain** (circumferential) is related to the **axial strain** for closed ends, but for instability with respect to pressure, we use the following: The **internal pressure** can be expressed (thin wall) as: \[ P = \frac{\sigma_{\theta} \, t}{r} \] where: - \(P\): internal pressure - \(\sigma_{\theta}\): hoop (circumferential) stress - \(t\): wall thickness - \(r\): radius For closed ends, the **axial stress** is half the hoop stress: \[ \sigma_{z} = \frac{\sigma_{\theta}}{2} \] But the instability criterion comes from the **maximum pressure**, which in turn is linked to the **maximum of \(P\) as a function of strain**. ### **Step 2: Express Pressure in Terms of \(\bar{\epsilon}\)** Let’s relate pressure to strain. For thin-walled cylinder: \[ \sigma_{\theta} = \frac{P r}{t} \] \[ \sigma_{z} = \frac{P r}{2t} \] Given the constitutive relation for the material: \[ \sigma_{z} = 22{,}000 \bar{\epsilon}^{.25} \] So, \[ \frac{P r}{2t} = 22{,}000 \bar{\epsilon}^{.25} \] \[ P = \frac{2t}{r} \cdot 22{,}000 \bar{\epsilon}^{.25} \] But as the tube stretches, radius and thickness change: - \(r = r_ (1 + \bar{\epsilon}_{\theta})\) - \(t = t_/(1+\bar{\epsilon}_{\theta})\) - For thin walls, the **hoop strain** relates to the **axial strain** for closed ends by incompressibility (volume constancy), but for instability, what matters is how \(P\) depends on \(\bar{\epsilon}\). ### **Step 3: Find Maximum \(P\) with Respect to \(\bar{\epsilon}\)** The instability occurs at the **maximum pressure**, i.e., when: \[ \frac{dP}{d\bar{\epsilon}} = \] From above, \[ P = K \bar{\epsilon}^{.25} \] where \(K = \frac{2t}{r} \cdot 22{,}000\) (but \(t\) and \(r\) are functions of strain). #### However, for power-law solids with closed-end thin-walled cylinders, the instability (maximum pressure) occurs at: \[ \frac{d\sigma_{z}}{d\bar{\epsilon}} = \sigma_{z} \] Plug in the given constitutive relation: \[ \sigma_{z} = 22,000 \bar{\epsilon}^{.25} \] \[ \frac{d\sigma_{z}}{d\bar{\epsilon}} = 22,000 \cdot .25 \bar{\epsilon}^{-.75} \] Set: \[ \frac{d\sigma_{z}}{d\bar{\epsilon}} = \sigma_{z} \] So, \[ 22,000 \cdot .25 \bar{\epsilon}^{-.75} = 22,000 \bar{\epsilon}^{.25} \] \[ .25 \bar{\epsilon}^{-.75} = \bar{\epsilon}^{.25} \] Divide both sides by \(\bar{\epsilon}^{-.75}\): \[ .25 = \bar{\epsilon}^{.25 + .75} = \bar{\epsilon}^{1} \] \[ \bar{\epsilon} = .25 \] --- ### **Final Answer (a):** \[ \boxed{\bar{\epsilon} = .25} \] --- ## (b) **If the tube had an initial diameter of 4 in. and wall thickness \(1/16\) in., what is the pressure at instability?** ### **Step 1: Initial Geometry** - Initial diameter, \(d_ = 4\) in. ⇒ \(r_ = 2\) in. - Wall thickness, \(t_ = 1/16\) in. ### **Step 2: Calculate \(\sigma_{z}\) at Instability** From (a), instability at \(\bar{\epsilon} = .25\): \[ \sigma_{z} = 22,000 \times (.25)^{.25} \] Calculate \(.25^{.25}\): \[ .25^{.25} = e^{.25 \ln .25} = e^{.25 \times (-1.3863)} = e^{-.3466} \approx .7071 \] So, \[ \sigma_{z} = 22,000 \times .7071 \approx 15,556 \text{ psi} \] ### **Step 3: Relate Pressure to Axial Stress** \[ \sigma_{z} = \frac{P r}{2t} \implies P = \frac{2 t \sigma_{z}}{r} \] But after strain, the final radius and thickness change: - Axial strain \(\bar{\epsilon}\) = .25 - In thin-walled approximation and small strains, we can use initial geometry for a rough estimate. \[ P = \frac{2 t_ \sigma_{z}}{r_} \] Plug in the numbers: - \(t_ = \frac{1}{16}\) in. = .0625 in. - \(r_ = 2\) in. - \(\sigma_{z} \approx 15,556\) psi \[ P = \frac{2 \times .0625 \times 15,556}{2} = .0625 \times 15,556 \approx 972 \text{ psi} \] --- ### **Final Answer (b):** \[ \boxed{P \approx 972 \text{ psi}} \] --- ## **Summary Table** | Part | Answer | |------|--------| | (a) | \(\bar{\epsilon} = .25\) | | (b) | \(P \approx 972 \text{ psi}\) | --- Let me know if you'd like to see more detailed derivations or explanations for any step!