Let's break down and solve **Problem 2** step-by-step. I'll explain each part clearly. --- ## **Given Data** - **Wire material:** Music wire - **Wire diameter (\( d \)):** .148 inch - **Rod diameter (core, \( d_{rod} \)):** ⅜ inch = .375 inch - **Annular clearance:** ⅛ inch = .125 inch - **Number of turns (\( N \)):** 11 total turns - **Ends:** Squared and ground --- ### **Step 1: Estimate the Torsional Yield Strength of the Wire** Music wire (ASTM A228) has: - **Ultimate tensile strength (\( S_{ut} \)):** ~230–250 ksi for \( d = .148" \) - **Torsional yield strength (\( S_{sy} \)):** \[ S_{sy} \approx .577 \times S_{ut} \] Let's use \( S_{ut} = 230 \) ksi. \[ S_{sy} = .577 \times 230 = 132.7 \text{ ksi} \] **Final answer:** \(\boxed{133\ \text{ksi}}\) (rounded). --- ### **Step 2: Estimate the Static Load Corresponding to the Yield Strength** #### **Find Mean Coil Diameter (\( D \)):** \[ D = d_{rod} + \text{annular clearance} + d \] \[ D = .375 + .125 + .148 = .648\ \text{in} \] #### **Spring Index (\( C \)):** \[ C = \frac{D}{d} = \frac{.648}{.148} = 4.38 \] #### **Shear Stress in the Wire:** \[ \tau_{max} = \frac{8 F D}{\pi d^3} K_w \] Where \( K_w \) is the Wahl factor: \[ K_w = \frac{4C - 1}{4C - 4} + \frac{.615}{C} \] \[ K_w = \frac{4 \times 4.38 - 1}{4 \times 4.38 - 4} + \frac{.615}{4.38} \] \[ K_w = \frac{17.52 - 1}{17.52 - 4} + \frac{.615}{4.38} = \frac{16.52}{13.52} + .140 = 1.222 + .140 = 1.362 \] Set \(\tau_{max} = S_{sy}\) to solve for \(F\): \[ S_{sy} = \frac{8 F D}{\pi d^3} K_w \] \[ F = \frac{S_{sy} \cdot \pi \cdot d^3}{8 D K_w} \] Plug in the values (\(S_{sy} = 133,000\) psi, \(d = .148\) in, \(D = .648\) in, \(K_w = 1.36\)): \[ F = \frac{133,000 \times \pi \times (.148)^3}{8 \times .648 \times 1.36} \] \[ (.148)^3 = .00324 \] \[ \text{Numerator: } 133,000 \times 3.1416 \times .00324 = 1,353 \] \[ \text{Denominator: } 8 \times .648 \times 1.36 = 7.05 \] \[ F = \frac{1,353}{7.05} = 192 \text{ lbf} \] **Final answer:** \(\boxed{192\ \text{lbf}}\) --- ### **Step 3: Estimate the Scale of the Spring** Scale = mean coil diameter : wire diameter \[ \text{Scale} = \frac{D}{d} = \frac{.648}{.148} = 4.38 \] **Final answer:** \(\boxed{4.4} \) (rounded) --- ### **Step 4: Estimate the Deflection Caused by the Load in (2)** The spring deflection: \[ \delta = \frac{8 F D^3 N}{G d^4} \] Where \(G\) (modulus of rigidity for music wire) = 11.5 Mpsi = 11,500,000 psi Plug in: - \(F = 192\) lbf - \(D = .648\) in - \(N = 9\) (Active coils: For squared and ground, subtract 2 from total turns: \(N_a = N-2 = 9\)) - \(G = 11,500,000\) psi - \(d = .148\) in \[ \delta = \frac{8 \times 192 \times (.648)^3 \times 9}{11,500,000 \times (.148)^4} \] \[ (.648)^3 = .272 \] \[ (.148)^4 = .000479 \] \[ \text{Numerator: } 8 \times 192 \times .272 \times 9 = 3758 \] \[ \text{Denominator: } 11,500,000 \times .000479 = 5,509 \] \[ \delta = \frac{3758}{5509} = .682 \text{ in} \] **Final answer:** \(\boxed{.68\ \text{in}}\) --- ### **Step 5: Estimate the Solid Length of the Spring** Solid length = total coils \(\times\) wire diameter \[ L_{solid} = N \times d = 11 \times .148 = 1.63\ \text{in} \] **Final answer:** \(\boxed{1.63\ \text{in}} \) --- ### **Step 6: Checking Buckling** A spring will buckle if its free length \(L_f\) exceeds a certain multiple of the mean coil diameter. \[ \frac{L_f}{D} < 2.63 \text{ (for both ends fixed)} \] Let’s estimate free length (\(L_f\)): \[ L_f = \text{solid length} + \text{deflection} = 1.63 + .68 = 2.31\ \text{in} \] \[ \frac{L_f}{D} = \frac{2.31}{.648} = 3.57 \] Since \(3.57 > 2.63\), the spring **may be prone to buckling**. **Final answer:** **May be prone to buckling** --- ### **Step 7: Find the Pitch** Pitch is the distance between centers of adjacent coils (for squared and ground ends): \[ \text{Pitch} = \frac{L_{free} - d}{N-1} \] \[ \text{Pitch} = \frac{2.31 - .148}{10} = \frac{2.162}{10} = .216\ \text{in} \] **Final answer:** \(\boxed{.216\ \text{in}}\) --- ## **Summary Table** | Part | Quantity | Answer | |------|--------------------------------------|------------------| | a | Torsional yield strength | 133 ksi | | b | Static load for yield | 192 lbf | | c | Scale (D/d) | 4.4 | | d | Deflection at load | .68 in | | e | Solid length | 1.63 in | | f | Buckling check | May buckle | | g | Pitch | .216 in | --- If you need any step explained in more detail, ask!