Let's break down and solve the problem step-by-step. --- ### **Given Data:** - **Spring wire:** No. 4 music wire - **Spring index (\(C\))**: 6 - **Factor of safety (\(FS\))**: 1.5 - **Average load (\(F_{avg}\))**: 120 lb We need to **find the permissible values for the maximum and minimum loads.** --- ### **Step 1: Find the Wire Diameter (\(d\))** From the notes: - For No. 4 music wire, \(d = 0.207\) in \( = 5.2578\) mm (from tables). --- ### **Step 2: Find the Ultimate Shear Strength (\(S_{ut}\))** From the notes: - \(S_{ut} = 241,000\) psi \( = 1661.6\) MPa --- ### **Step 3: Find the Endurance Limit (\(S_{se}\))** From the notes: - \(S_{se} = 0.23 \times S_{ut}\) - \(S_{se} = 0.23 \times 241,000\) psi \( = 55,430\) psi --- ### **Step 4: Find Shear Yield Strength (\(S_{sy}\))** From the notes: - \(S_{sy} = 0.40 \times S_{ut}\) - \(S_{sy} = 0.40 \times 241,000\) psi \( = 96,400\) psi --- ### **Step 5: Calculate the Permissible Stresses** Since the spring carries a fluctuating load, the allowable stress will be based on both mean and alternating stresses. The factor of safety is applied to both. #### **Goodman Criterion for Springs:** \[ \frac{\tau_m}{S_{sy}/FS} + \frac{\tau_a}{S_{se}/FS} \leq 1 \] where: - \(\tau_m\) = mean shear stress - \(\tau_a\) = alternating shear stress - \(S_{sy}, S_{se}\) = yield and endurance strengths in shear - \(FS\) = factor of safety --- ### **Step 6: Calculate Shear Stress in the Spring** Shear stress in a helical spring: \[ \tau = \frac{8 F D}{\pi d^3} K_w \] Where: - \(F\) = Force/load (lb) - \(D\) = Mean coil diameter = \(C \cdot d = 6 \cdot 0.207 = 1.242\) in - \(d\) = wire diameter = 0.207 in - \(K_w\) = Wahl factor (accounts for curvature of the wire) #### **Wahl Factor:** \[ K_w = \frac{4C-1}{4C-4} + \frac{0.615}{C} \] For \(C = 6\): \[ K_w = \frac{24-1}{24-4} + \frac{0.615}{6} = \frac{23}{20} + 0.1025 = 1.15 + 0.1025 = 1.2525 \] --- ### **Step 7: Express the Mean and Alternating Load** Let: - \(F_{max}\) = maximum permissible load - \(F_{min}\) = minimum permissible load Given: - \(F_{avg} = \frac{F_{max} + F_{min}}{2} = 120\) lb - \(F_{a} = \frac{F_{max} - F_{min}}{2}\) (alternating component) --- ### **Step 8: Write Goodman Criterion in Terms of Load** \[ \tau_m = \frac{8 F_{avg} D}{\pi d^3} K_w \] \[ \tau_a = \frac{8 F_{a} D}{\pi d^3} K_w \] Plug into Goodman equation: \[ \frac{\tau_m}{S_{sy}/FS} + \frac{\tau_a}{S_{se}/FS} \leq 1 \] \[ \frac{\frac{8 F_{avg} D}{\pi d^3} K_w}{S_{sy}/FS} + \frac{\frac{8 F_a D}{\pi d^3} K_w}{S_{se}/FS} \leq 1 \] Let’s substitute all known values: - \(F_{avg} = 120\) lb - \(D = 1.242\) in - \(d = 0.207\) in - \(K_w = 1.2525\) - \(S_{sy}/FS = 96,400 / 1.5 = 64,267\) psi - \(S_{se}/FS = 55,430 / 1.5 = 36,953\) psi \[ \frac{8 \times 120 \times 1.242 \times 1.2525}{\pi \times (0.207)^3 \times 64,267} + \frac{8 \times F_{a} \times 1.242 \times 1.2525}{\pi \times (0.207)^3 \times 36,953} = 1 \] Calculate denominator: \[ \pi \times (0.207)^3 = \pi \times 0.00887 = 0.02787 \] Calculate numerator for mean load: \[ 8 \times 120 \times 1.242 \times 1.2525 = 8 \times 120 \times 1.5565 = 8 \times 186.78 = 1,494.24 \] So, \[ \frac{1,494.24}{0.02787 \times 64,267} + \frac{8 \times F_{a} \times 1.5565}{0.02787 \times 36,953} = 1 \] Calculate the denominators: - \(0.02787 \times 64,267 = 1,792.5\) - \(0.02787 \times 36,953 = 1,029.2\) So, \[ \frac{1,494.24}{1,792.5} + \frac{12.452 \times F_{a}}{1,029.2} = 1 \] \[ 0.834 + \frac{12.452 F_{a}}{1,029.2} = 1 \] \[ \frac{12.452 F_{a}}{1,029.2} = 1 - 0.834 = 0.166 \] \[ 12.452 F_{a} = 0.166 \times 1,029.2 = 170.85 \] \[ F_{a} = \frac{170.85}{12.452} = 13.73 \text{ lb} \] --- ### **Step 9: Find Maximum and Minimum Loads** Recall: - \(F_{avg} = \frac{F_{max} + F_{min}}{2} = 120\) - \(F_{a} = \frac{F_{max} - F_{min}}{2} = 13.73\) So: \[ F_{max} = F_{avg} + F_{a} = 120 + 13.73 = 133.73 \text{ lb} \] \[ F_{min} = F_{avg} - F_{a} = 120 - 13.73 = 106.27 \text{ lb} \] --- ## **Final Answer** ### **Permissible Maximum and Minimum Loads:** \[ \boxed{ F_{max} = 134 \text{ lb (rounded)}, \quad F_{min} = 106 \text{ lb (rounded)} } \] **Explanation:** The maximum and minimum permissible loads are calculated using the Goodman fatigue criterion, considering the fluctuating nature of the applied load, material properties, spring geometry, and factor of safety. If you need the calculations shown more explicitly, let me know!