# Step-by-Step Solution ## **Given Data** **Geometric Properties:** - Diameter (\(d\)) = 6 in - Thickness (\(t\)) = .06 in **Material Properties:** - Young's modulus (\(E\)) = 10 Msi - Poisson's ratio (\(\nu\)) = .3 - Yield strength (\(Y\)) = 20 ksi - Coefficient of thermal expansion (\(\alpha\)) = \(16 \times 10^{-6}\) (in/in)/°F - Thermal conductivity (\(K\)) = 80 Btu/hr ft² °F/ft --- ## a) **Find the value of internal pressure \(p\) when the tube begins to yield (Tresca Criterion)** ### **Step 1: Stresses in a Thin-Walled Cylinder** For a thin-walled cylinder under internal pressure \(p\): - **Hoop (circumferential) stress:** \[ \sigma_h = \frac{p r}{t} \] - **Longitudinal (axial) stress:** \[ \sigma_a = \frac{p r}{2t} \] Where \(r = \frac{d}{2}\) is the inner radius. ### **Step 2: Tresca Yield Criterion** The Tresca criterion says yielding starts when the maximum shear stress reaches half the yield strength: \[ \max |\sigma_i - \sigma_j| = Y \] For this cylinder: - Principal stresses: \(\sigma_1 = \sigma_h\), \(\sigma_2 = \sigma_a\), \(\sigma_3 = \) (assuming open ends) So, the differences are: - \(|\sigma_h - \sigma_a|\) - \(|\sigma_h - |\) - \(|\sigma_a - |\) The largest is \(|\sigma_h - | = \sigma_h\). ### **Step 3: Set Up the Yield Condition** At yield, \[ \sigma_h = Y \] Plug in the formula for hoop stress: \[ \frac{p r}{t} = Y \] \[ p = \frac{Y t}{r} \] --- ### **Step 4: Plug in the Numbers** - \(Y = 20\,\text{ksi} = 20,000\,\text{psi}\) - \(t = .06\,\text{in}\) - \(d = 6\,\text{in} \implies r = 3\,\text{in}\) Now, \[ p = \frac{20,000 \times .06}{3} \] \[ p = \frac{1,200}{3} = 400\,\text{psi} \] --- ### **Final Answer (Part a):** \[ \boxed{p = 400\,\text{psi}} \] --- ## b) **Dilatational and Deviatoric Stresses at Onset of Yielding** ### **Step 1: Stress Components at Yield** At yield: - \(\sigma_h = Y = 20,000\,\text{psi}\) - \(\sigma_a = \frac{Y}{2} = 10,000\,\text{psi}\) - \(\sigma_r = \) (thin wall assumption, neglect radial stress) --- ### **Step 2: Mean (Dilatational) Stress** Mean (hydrostatic or dilatational) stress: \[ \sigma_m = \frac{\sigma_1 + \sigma_2 + \sigma_3}{3} \] \[ \sigma_m = \frac{20,000 + 10,000 + }{3} = \frac{30,000}{3} = 10,000\,\text{psi} \] --- ### **Step 3: Deviatoric Stresses** Deviatoric stress for each principal direction: \[ s_i = \sigma_i - \sigma_m \] - \(s_1 = 20,000 - 10,000 = 10,000\,\text{psi}\) - \(s_2 = 10,000 - 10,000 = \,\text{psi}\) - \(s_3 = - 10,000 = -10,000\,\text{psi}\) --- ### **Final Answers (Part b):** - **Dilatational (mean) stress:** \(\boxed{10,000\,\text{psi}}\) - **Deviatoric stresses:** \(\boxed{10,000\,\text{psi},\, \,\text{psi},\, -10,000\,\text{psi}}\) --- ## **Summary Table** | Quantity | Value | |------------------------------|--------------------------| | Pressure at yield (Tresca) | \(400\,\text{psi}\) | | Dilatational Stress | \(10,000\,\text{psi}\) | | Deviatoric Stresses | \(10,000, , -10,000\) psi | If you have any questions or need a more detailed explanation of any step, let me know!