To solve this problem, let's break it into smaller steps:

Given Data:

1. Outside diameter ($D_{outer}$) = $80 \, \text{mm}$
2. Maximum axial load ($F_{max}$) = $1 \, \text{kN} = 1000 \, \text{N}$
3. Number of coils ($n$) = $10$
4. Shear modulus ($G$) = $0.89 \times 10^5 \, \text{kN}/\text{mm}^2 = 89 \times 10^9 \, \text{N}/\text{m}^2$
5. Material: C50 steel
6. Ends: Ground flat
7. Find:
   • Maximum shear stress ($\tau_{max}$)
   • Total deflection ($\delta$)
   • Factor of safety (FoS)

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Step 1: Determine the mean diameter and wire diameter

The mean diameter ($D$) is related to the outside diameter and wire diameter ($d$): $D = D_{outer} - d$

We'll need $d$ to proceed further. For now, we'll proceed symbolically, solving general equations, and you can either supply the wire diameter, or I can estimate it based on typical proportions.

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Step 2: Shear Stress in the Spring

The maximum shear stress ($\tau_{max}$) in a helical spring is given by:

$\tau_{max} = \frac{16 F_{max} R}{\pi d^3} \cdot K$

Where:

• $R = \frac{D}{2}$ is the spring radius.
• $K = \frac{4C - 1}{4C - 4} + \frac{0.615}{C}$ is the Wahl correction factor.
• $C = \frac{D}{d}$ is the spring index.

Step 3: Deflection of the Spring

The total deflection ($\delta$) of the spring under axial load is given by:

$\delta = \frac{8 F_{max} R^3 n}{G d^4}$

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Step 4: Factor of Safety

The factor of safety (FoS) is calculated using:

$\text{FoS} = \frac{\tau_{yield}}{\tau_{max}}$

For C50 steel, the typical yield strength ($\tau_{yield}$) is about $600 \, \text{MPa}$.

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Would you like me to proceed with a wire diameter value you supply, or shall I assume one (e.g., $d = 10 \, \text{mm}$)?