Sure! Let's break down **Problem 3** step-by-step: --- # **Given Data** - **Hole diameter** = 25.4 mm - **Maximum outside spring diameter** = 22.86 mm **Spring index** (C) = 8 - **Free length** = 76.2 mm - **Solid length** = 25.4 mm - **Ends:** Squared and peened - **Material:** Music wire - **Spring use:** Compressed from free length to solid length ### **Material Properties for Music Wire** - **Ultimate tensile strength** (\(S_{ut}\)): For music wire, \( S_{ut} \approx 230 \) MPa (for \( d \approx 2 \) mm, varies with diameter) - **Modulus of elasticity** (\(E\)): 207 GPa - **Shear modulus** (\(G\)): 79.3 GPa --- # **(a) Determine a Suitable Wire Diameter** ## **Step 1: Use Spring Index Relation** Spring index: \[ C = \frac{D}{d} \] Where: - \(C\) = Spring index = 8 - \(D\) = Mean coil diameter - \(d\) = Wire diameter ## **Step 2: Relate Mean and Outside Diameter** \[ D_{o} = D + d \leq 22.86\,\text{mm} \] \[ D = 8d \] So, \[ D_o = 8d + d = 9d \leq 22.86 \] \[ d \leq \frac{22.86}{9} = 2.54\,\text{mm} \] Let's select a standard wire diameter **just below 2.54 mm** (from music wire tables): - Standard size: **2.5 mm** (choose next standard size below limit) **Selected wire diameter:** \[ \boxed{d = 2.5\,\text{mm}} \] --- # **(b) Determine the Total Number of Coils** ## **Step 1: Solid Length Formula** Solid length (\(L_s\)) for squared ends: \[ L_s = N_t \cdot d \] Where \(N_t\) is **total number of coils**. Given: \[ L_s = 25.4\,\text{mm} \] \[ d = 2.5\,\text{mm} \] \[ N_t = \frac{L_s}{d} = \frac{25.4}{2.5} = 10.16 \] **Round down to the nearest whole number:** \[ \boxed{N_t = 10} \] --- # **(c) Determine the Spring Constant (\(k\))** ## **Step 1: Mean Coil Diameter** \[ D = C \cdot d = 8 \times 2.5 = 20\,\text{mm} \] ## **Step 2: Number of Active Coils (\(N_a\))** For squared and ground ends: \[ N_a = N_t - 2 = 10 - 2 = 8 \] ## **Step 3: Spring Constant Formula** \[ k = \frac{G d^4}{8 D^3 N_a} \] Where: - \(G = 79,300\,\text{MPa} = 79,300\,\text{N/mm}^2\) - \(d = 2.5\,\text{mm}\) - \(D = 20\,\text{mm}\) - \(N_a = 8\) Plug in values: \[ k = \frac{79,300 \times (2.5)^4}{8 \times (20)^3 \times 8} \] \[ (2.5)^4 = 39.0625 \] \[ (20)^3 = 8,000 \] \[ k = \frac{79,300 \times 39.0625}{8 \times 8,000 \times 8} \] \[ = \frac{3,099,218.75}{512,000} \] \[ = 6.05\,\text{N/mm} \] \[ \boxed{k = 6.05\,\text{N/mm}} \] --- # **(d) Static Factor of Safety (FOS) when Compressed to Solid Length** ### **Step 1: Max Load at Solid Length** Maximum **deflection**: \[ \delta_{max} = L_{free} - L_{solid} = 76.2 - 25.4 = 50.8\,\text{mm} \] \[ F_{max} = k \times \delta_{max} = 6.05 \times 50.8 = 307.34\,\text{N} \] ### **Step 2: Maximum Shear Stress** \[ \tau_{max} = \frac{8 F_{max} D}{\pi d^3} K_w \] Where \( K_w \) is the **Wahl factor**: \[ K_w = \frac{4C-1}{4C-4} + \frac{.615}{C} \] \[ = \frac{4 \times 8 - 1}{4 \times 8 - 4} + \frac{.615}{8} = \frac{31}{28} + .0769 = 1.107 + .0769 = 1.1839 \] Now, \[ \tau_{max} = \frac{8 \times 307.34 \times 20}{\pi \times (2.5)^3} \times 1.1839 \] \[ (2.5)^3 = 15.625 \] \[ \tau_{max} = \frac{49,174.4}{49.087} \times 1.1839 \] \[ = 1,002.24 \times 1.1839 = 1,187.2\,\text{MPa} \] ### **Step 3: Factor of Safety** Music wire's yield strength (\(S_{sy}\)) is about .45 of \(S_{ut}\) (for music wire): \[ S_{sy} \approx .45 \times 230 = 1,035\,\text{MPa} \] \[ \text{Static FOS} = \frac{S_{sy}}{\tau_{max}} = \frac{1,035}{1,187.2} = .87 \] **This suggests the spring yields before reaching solid length!** You need to use a larger wire or fewer coils for a safe design. --- # **(e) Fatigue Factor of Safety (Gerber-Zimmerli Criterion)** ### **Step 1: Amplitude and Mean Stresses** For full range cycling (free to solid): \[ \tau_{max} = 1,187.2\,\text{MPa} \] \[ \tau_{min} = \,\text{MPa} \] \[ \tau_m = \frac{\tau_{max} + \tau_{min}}{2} = \frac{1,187.2}{2} = 593.6\,\text{MPa} \] \[ \tau_a = \frac{\tau_{max} - \tau_{min}}{2} = 593.6\,\text{MPa} \] ### **Step 2: Endurance Limit for Torsion** For music wire: \[ S_{se} = .22 \times S_{ut} = .22 \times 2,300 = 506\,\text{MPa} \] ### **Step 3: Gerber-Zimmerli Criterion** The fatigue safety factor \(n_f\) is: \[ \left( \frac{\tau_m}{S_{sy}} \right)^2 + \frac{\tau_a}{S_{se}} = \frac{1}{n_f} \] So, \[ \left(\frac{593.6}{1035}\right)^2 + \frac{593.6}{506} = \frac{1}{n_f} \] \[ (.574)^2 + 1.174 = .329 + 1.174 = 1.503 \] \[ n_f = \frac{1}{1.503} = .67 \] --- # **Summary of Answers** | Part | Answer | |------|--------| | (a) | Wire diameter: **2.5 mm** | | (b) | Total coils: **10** | | (c) | Spring constant: **6.05 N/mm** | | (d) | Static FOS: **.87** (**Unsafe!**) | | (e) | Fatigue FOS: **.67** (**Unsafe!**) | --- ## **Tutor's Note:** Both static and fatigue factors of safety are **below 1**, meaning the spring **will fail** under the prescribed use. To obtain a safe design, you must **increase the wire diameter**, **reduce the number of coils**, or **reduce the required deflection**. If you want, I can recalculate for a larger wire diameter. Let me know!