Let's solve ** 2** by step: ## Given Data- Wire diameter \( d = 5.080) mm - Mean coil diameter, \( D =50.800) mm - Number of coils, \( N = 12 \) - Free length, \( L_{free} = 127.000 \) mm - Squared ends (assume closed and ground for calculations) --- ## (a) Find the Solid Length **Solid length** is the length of the spring when all coils are fully compressed and touching each other. \[ L_{solid} = N \times d \] **Substitute the values:** \[ L_{solid} = 12 \times 5.080 = 60.96 \text{ mm} \] --- ## (b) Force Necessary to Deflect Spring to Solid Length To find the force (\( F \)) required to compress the spring to its solid length, we need the spring rate (\( k \)): \[ k = \frac{G d^4}{8 D^3 N} \] Where: - \( G \) = Modulus of rigidity for oil-tempered wire (assume \( G = 79,300 \) MPa = \( 79,300 \) N/mm²) - \( d = 5.080 \) mm - \( D = 50.800 \) mm - \( N = 12 \) ### Step 1: Calculate Spring Rate (\( k \)) \[ k = \frac{79,300 \times (5.080)^4}{8 \times (50.800)^3 \times 12} \] Calculate numerator: \[ (5.080)^4 = 668.5 \] \[ 79,300 \times 668.5 = 53,008,050 \] Calculate denominator: \[ (50.800)^3 = 131,048 \] \[ 8 \times 131,048 \times 12 = 12,579,008 \] So, \[ k = \frac{53,008,050}{12,579,008} \approx 4.214 \text{ N/mm} \] ### Step 2: Maximum Deflection (\( \delta_{max} \)) \[ \delta_{max} = L_{free} - L_{solid} = 127.000 - 60.96 = 66.04 \text{ mm} \] ### Step 3: Force to Solid Length \[ F = k \times \delta_{max} = 4.214 \times 66.04 = 278.4 \text{ N} \] --- ## (c) Factor of Safety Against Yielding ### Step 1: Maximum Shear Stress at Solid Length \[ \tau_{max} = \frac{8 F D}{\pi d^3} \cdot K_w \] Where \( K_w \) is the Wahl correction factor: \[ C = \frac{D}{d} = \frac{50.800}{5.080} = 10 \] \[ K_w = \frac{4C - 1}{4C - 4} + \frac{.615}{C} \] \[ K_w = \frac{4 \times 10 - 1}{4 \times 10 - 4} + \frac{.615}{10} = \frac{39}{36} + .0615 \approx 1.083 + .0615 = 1.1445 \] \[ \tau_{max} = \frac{8 \times 278.4 \times 50.8}{\pi \times (5.08)^3} \times 1.1445 \] \[ (5.08)^3 = 131.13 \] \[ \pi \times 131.13 = 411.79 \] \[ 8 \times 278.4 \times 50.8 = 113,314.56 \] \[ \frac{113,314.56}{411.79} = 275.1 \] \[ \tau_{max} = 275.1 \times 1.1445 = 314.9 \text{ MPa} \] ### Step 2: Yield Strength of Material Assume yield strength for oil-tempered wire, \( S_{sy} \approx 800 \) MPa. ### Step 3: Factor of Safety \[ n = \frac{S_{sy}}{\tau_{max}} = \frac{800}{314.9} = 2.54 \] --- ## **Final Answers** ### (a) Solid Length \[ \boxed{60.96 \text{ mm}} \] ### (b) Force for Solid Length \[ \boxed{278.4 \text{ N}} \] ### (c) Factor of Safety \[ \boxed{2.54} \] --- If you need further explanation or clarification for any step, let me know!