Let's solve **Problem 1.7-9** step by step. --- ## **Given Data** - **Loads:** \( P_1 = 26.5 \) kips \( P_2 = 22. \) kips - **Diameters and Thicknesses:** \( d_{AB} = 1.25 \) in, \( t_{AB} = .5 \) in \( d_{BC} = 2.25 \) in, \( t_{BC} = .375 \) in - **Modulus of Elasticity:** \( E = 14,000 \) ksi - **Increase in wall thickness of BC:** \( \Delta t_{BC} = 200 \times 10^{-6} \) in --- ## **Part (a): Increase in the Inner Diameter of Pipe Segment BC** ### **Step 1: Find Original Inner Diameter of BC** \[ d_{i,BC} = d_{BC} - 2t_{BC} \] \[ d_{i,BC} = 2.25 - 2 \times .375 = 2.25 - .75 = 1.50 \text{ in} \] ### **Step 2: Find Increase in Inner Diameter** The increase in wall thickness means the outer diameter remains constant, so the inner diameter must decrease by twice the increase in thickness: \[ \Delta d_{i,BC} = -2 \Delta t_{BC} \] \[ \Delta d_{i,BC} = -2 \times 200 \times 10^{-6} = -400 \times 10^{-6} = -.0004 \text{ in} \] **Final Answer:** \[ \boxed{\text{The inner diameter of segment BC decreases by } .0004 \text{ in}} \] --- ## **Part (b): Find Poisson's Ratio for the Brass** ### **Step 1: Axial Strain Calculation** Axial strain (from mechanics of materials): \[ \varepsilon = \frac{\Delta t_{BC}}{t_{BC}} \] \[ \varepsilon = \frac{200 \times 10^{-6}}{.375} = 533.33 \times 10^{-6} \] ### **Step 2: Axial Stress in Segment BC** Total load on BC: \( P_1 + P_2 \) \[ P = P_1 + P_2 = 26.5 + 22. = 48.5 \text{ kips} = 48,500 \text{ lb} \] Cross-sectional area: \[ A = \pi \left[ \frac{(d_{BC})^2 - (d_{i,BC})^2}{4} \right] \] \[ A = \pi \left[ \frac{(2.25)^2 - (1.50)^2}{4} \right] = \pi \left[ \frac{5.0625 - 2.25}{4} \right] = \pi \left[ \frac{2.8125}{4} \right] = \pi \times .7031 = 2.209 \text{ in}^2 \] Axial stress: \[ \sigma = \frac{P}{A} = \frac{48,500}{2.209} = 21,956 \text{ psi} = 21.96 \text{ ksi} \] ### **Step 3: Axial Strain from Stress and Modulus** \[ \varepsilon_{axial} = \frac{\sigma}{E} = \frac{21,956}{14,000 \times 100} = .001568 \] ### **Step 4: Poisson's Ratio** Lateral strain: \[ \varepsilon_{lateral} = \frac{\Delta t_{BC}}{t_{BC}} = 533.33 \times 10^{-6} \] Poisson's ratio, \( \nu \): \[ \varepsilon_{lateral} = -\nu \varepsilon_{axial} \implies \nu = -\frac{\varepsilon_{lateral}}{\varepsilon_{axial}} \] \[ \nu = -\frac{533.33 \times 10^{-6}}{.001568} = -.340 \] (Take absolute value since Poisson's ratio is positive.) **Final Answer:** \[ \boxed{\nu = .34} \] --- ## **Part (c): Increase in Wall Thickness and Inner Diameter of Segment AB** ### **Step 1: Find Axial Stress in Segment AB** - **Load on AB:** Only \( P_1 \) passes through AB. \[ A_{AB} = \pi \left[ \frac{(d_{AB})^2 - (d_{i,AB})^2}{4} \right] \] \[ d_{i,AB} = d_{AB} - 2t_{AB} = 1.25 - 2 \times .5 = .25 \text{ in} \] \[ A_{AB} = \pi \left[ \frac{(1.25)^2 - (.25)^2}{4} \right] = \pi \left[ \frac{1.5625 - .0625}{4} \right] = \pi \left[ \frac{1.5}{4} \right] = \pi \times .375 = 1.178 \text{ in}^2 \] \[ \sigma_{AB} = \frac{P_1}{A_{AB}} = \frac{26,500}{1.178} = 22,494 \text{ psi} = 22.49 \text{ ksi} \] ### **Step 2: Axial Strain in AB** \[ \varepsilon_{axial,AB} = \frac{22,494}{14,000 \times 100} = .001607 \] ### **Step 3: Lateral Strain (Thickness Increase)** \[ \varepsilon_{lateral,AB} = \nu \varepsilon_{axial,AB} = .34 \times .001607 = .000546 \] \[ \Delta t_{AB} = t_{AB} \times \varepsilon_{lateral,AB} = .5 \times .000546 = .000273 \text{ in} \] ### **Step 4: Change in Inner Diameter of AB** \[ \Delta d_{i,AB} = -2 \Delta t_{AB} = -2 \times .000273 = -.000546 \text{ in} \] --- ## **Final Answers** ### **(a)** \[ \boxed{\text{The inner diameter of pipe segment BC decreases by } .0004 \text{ in}} \] ### **(b)** \[ \boxed{\nu = .34} \] ### **(c)** - **Increase in wall thickness of AB:** \(\boxed{.000273 \text{ in}}\) - **Decrease in inner diameter of AB:** \(\boxed{.000546 \text{ in}}\)