Let's solve part (a) (i)–(iii) **step-by-step** with clear explanations and include a diagrammatic summary for clarity. --- # Part (a): Helical Springs ## **Given Data** | Spring | Length (mm) | No. of coils | Coil dia. (mm) Wire dia. (mm) | |--------|-------------|--------------|----------------|----------------| | A | 200 | 12 | 100 | 10 | | B | 175 | 15 | 75 | 6 | - **Shear modulus, G = 80 GPa = 80,000 N/mm²** - **Compression in system = 12.5 mm** (from part ii) --- ### **Step 1: Calculate Spring Stiffness (k) for Each Spring** The **stiffness** of a close-coiled helical spring: \[ k = \frac{G d^4}{8 n D^3} \] Where: - \( G \) = shear modulus (N/mm²) - \( d \) = wire diameter (mm) - \( n \) = number of coils - \( D \) = mean coil diameter (mm) #### **Spring A:** - \( d_A = 10 \) mm, \( n_A = 12 \), \( D_A = 100 \) mm \[ k_A = \frac{80,000 \times 10^4}{8 \times 12 \times 100^3} \] \[ k_A = \frac{80,000 \times 10,000}{8 \times 12 \times 1,000,000} \] \[ k_A = \frac{800,000,000}{96,000,000} = 8.33 \text{ N/mm} \] #### **Spring B:** - \( d_B = 6 \) mm, \( n_B = 15 \), \( D_B = 75 \) mm \[ k_B = \frac{80,000 \times 6^4}{8 \times 15 \times 75^3} \] \[ k_B = \frac{80,000 \times 1,296}{8 \times 15 \times 421,875} \] \[ k_B = \frac{103,680,000}{50,625,000} = 2.05 \text{ N/mm} \] --- ### **Step 2: System Calculation** Both springs are in **series** (from the diagram): \[ \frac{1}{k_{eq}} = \frac{1}{k_A} + \frac{1}{k_B} \] \[ \frac{1}{k_{eq}} = \frac{1}{8.33} + \frac{1}{2.05} = .12 + .488 = .608 \] \[ k_{eq} = \frac{1}{.608} = 1.645 \text{ N/mm} \] --- ## **(i) Force Exerted on Shoulder D by Disk C** For a compression \( x \): \[ F = k_{eq} \cdot x \] Given: \( x = 12.5 \) mm \[ F = 1.645 \times 12.5 = 20.56 \text{ N} \] --- ## **(ii) Force to Hold the Disk 12.5 mm from Shoulder** This is the **same as above** since you are compressing the springs by 12.5 mm: \[ F = 20.56 \text{ N} \] --- ## **(iii) Maximum Shear Stress in Each Spring** Shear stress in a spring wire: \[ \tau_{max} = \frac{8 F D}{\pi d^3} \] ### **Spring A:** \[ \tau_{A} = \frac{8 \times F_A \times D_A}{\pi d_A^3} \] But for springs in series, **force in each spring is equal**: \[ F_A = F_B = F = 20.56 \text{ N} \] \[ \tau_A = \frac{8 \times 20.56 \times 100}{\pi \times 10^3} = \frac{16,448}{3141.59} = 5.24 \text{ N/mm}^2 \] ### **Spring B:** \[ \tau_B = \frac{8 \times 20.56 \times 75}{\pi \times 6^3} = \frac{12,336}{678.58} = 18.18 \text{ N/mm}^2 \] --- ## **Summary Table** | Parameter | Spring A | Spring B | |--------------|-------------|-------------| | Stiffness (N/mm) | 8.33 | 2.05 | | Max Shear Stress (N/mm²) | 5.24 | 18.18 | --- ## **Graphical Representation** ```plaintext Force vs. Deflection for Series Springs |\ | \ | \ | \ | \ |_____\ x ``` *The slope represents the equivalent spring stiffness, which is much lower than individual stiffnesses.* --- ## **Final Answers** 1. **Force exerted by disk C:** **20.56 N** 2. **Force to hold disk 12.5 mm from shoulder:** **20.56 N** 3. **Maximum shear stress in springs:** - **Spring A:** 5.24 N/mm² - **Spring B:** 18.18 N/mm² --- ### **Let me know if you want part (b) solved as well!**