Act like a helpful tutor and exlain me :
Design an infinite-life helical coil extension spring with full end loops and generous loop-bend radii for a minimum load Fmin of 7 lbf and a maximum load Fma

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Act like a helpful tutor and exlain me : 
Design an infinite-life helical coil extension spring with full end loops and generous loop-bend radii for a minimum load Fmin of 7 lbf and a maximum load Fmax of 14 lbf, with an accompanying stretch of 1/4 in. The spring is for a food-service equipment and must be stainless steel. The outside diameter of the coil cannot exceed 1 in, and the free length cannot exceed 2.5 in. Using a fatigue design factor of nf = 2, complete the design. Use the Gerber criterion with Table 10-8.

VIPSolutions · Accepted Answer

Let's break down your **helical extension spring design problem** step by step, using mechanical design principles and the constraints you provided.

---

## **1. Problem Restatement & Given Data**
- **Spring type:** Infinite-life helical extension spring, full loops, generous bend radii.
- **Material:** Stainless steel, for food-service (likely 302, 304, or 316; let’s use ASTM A313 Type 302/304 as typical).
- **Loads:**  
  - Minimum load, $ F_{min} = 7 $ lbf  
  - Maximum load, $ F_{max} = 14 $ lbf  
  - Corresponding total extension, $ \Delta L = 0.25 $ in
- **Geometry constraints:**  
  - Max coil outside diameter $ D_{o,max} = 1 $ in  
  - Max free length $ L_{f,max} = 2.5 $ in
- **Fatigue design factor:** $ n_f = 2 $
- **Fatigue criterion:** Gerber, with reference to Table 10-8 (Shigley 10th ed.)
---

## **2. Select Material Properties**

Assume **ASTM A313 Type 302/304 Stainless Steel** (from Table 10-8, Shigley):

| Property                | Value (typical)         |
|-------------------------|-------------------------|
| Ultimate tensile ($S_{ut}$) | 200 ksi                |
| Shear yield ($S_{sy}$)      | ~90 ksi                 |
| Endurance limit ($S_{se}$)  | ~45 ksi (use for infinite life) |

---

## **3. Calculate Required Spring Rate**

The **spring rate** is:
$$
k = \frac{F_{max} - F_{min}}{\Delta L} = \frac{14 - 7}{0.25} = \frac{7}{0.25} = 28\ \text{lbf/in}
$$

---

## **4. Estimate Coil Diameter and Wire Size**

Let’s denote:
- $ d $: wire diameter (in)
- $ D $: mean coil diameter (in)
- $ D_o $: outside diameter = $ D + d \leq 1 $ in

**Spring rate for extension/close-wound spring**:
$$
k = \frac{G d^4}{8 D^3 n_a}
$$
where:
- $ G $: modulus of rigidity, stainless steel $\approx 11.2 \times 10^6 $ psi
- $ n_a $: number of active coils

Let’s **assume** (to be checked) that 1.5 coils are taken by the full loops, so:
$$
n_a = n_t - 1.5
$$
where $ n_t $ = total coils.

---

## **5. Initial Wire Diameter Guess**

Let’s try $ d = 0.09 $ in (music wire’s common size and available for SS).

Then mean coil diameter:
$$
D = D_o - d = 1 - 0.09 = 0.91\ \text{in}
$$

Let’s estimate active coils needed for the desired rate:
$$
k = \frac{G d^4}{8 D^3 n_a}
\implies n_a = \frac{G d^4}{8 D^3 k}
$$

Plug in numbers:
$$
n_a = \frac{11.2 \times 10^6 \times (0.09)^4}{8 \times (0.91)^3 \times 28}
$$

Calculate numerator:
$$
(0.09)^4 = 0.0000656
$$
$$
11.2 \times 10^6 \times 0.0000656 \approx 734.72
$$

Denominator:
$$
8 \times (0.91)^3 \times 28 = 8 \times 0.753 \times 28 \approx 168.7
$$

So,
$$
n_a = \frac{734.72}{168.7} \approx 4.35
$$

So, **total coils**:
$$
n_t = n_a + 1.5 \approx 5.85
$$

---

## **6. Check Free Length**

**Free length:**
$$
L_f = n_t d + \text{loops length}
$$
For full loops, **loop lengths are usually** $ \approx 1 d $ per loop, so for 2 loops: $ 2d $.

So:
$$
L_f \approx n_t d + 2d = (5.85 + 2) \times 0.09 = 7.85 \times 0.09 \approx 0.71\ \text{in}
$$

This is **well below** the 2.5 in max—so you could possibly use a larger wire or more coils.

---

## **7. Check Stresses (Fatigue)**

### **Calculate Stresses**

The **maximum shear stress** in the coil (Wahl’s equation):
$$
\tau = \frac{8 F D}{\pi d^3} K_w
$$
Where $ K_w $ is the Wahl factor:
$$
C = \frac{D}{d} = \frac{0.91}{0.09} = 10.1
$$
$$
K_w = \frac{4C-1}{4C-4} + \frac{0.615}{C} = \frac{40.4-1}{40.4-4} + \frac{0.615}{10.1}
= \frac{39.4}{36.4} + 0.061 = 1.082 + 0.061 \approx 1.14
$$

**Max force:** $ F_{max} = 14 $ lbf

$$
\tau_{max} = \frac{8 \times 14 \times 0.91}{\pi \times (0.09)^3} \times 1.14
$$

Calculate denominator:
$$
\pi \times (0.09)^3 = 3.1416 \times 0.000729 = 0.00229
$$

Numerator:
$$
8 \times 14 \times 0.91 = 101.92
$$

So:
$$
\tau_{max} = \frac{101.92}{0.00229} \times 1.14 \approx 44,519 \times 1.14 \approx 50,751 \text{ psi}
$$

**Min force:** $ F_{min} = 7 $ lbf

$$
\tau_{min} = \frac{8 \times 7 \times 0.91}{0.00229} \times 1.14 = \frac{50.96}{0.00229} \times 1.14 \approx 22,263 \times 1.14 \approx 25,382 \text{ psi}
$$

### **Alternating and Mean Stresses**

$$
\tau_m = \frac{\tau_{max} + \tau_{min}}{2} = \frac{50,751 + 25,382}{2} = 38,067 \text{ psi}
$$
$$
\tau_a = \frac{\tau_{max} - \tau_{min}}{2} = \frac{50,751 - 25,382}{2} = 12,684 \text{ psi}
$$

---

## **8. Fatigue Check (Gerber, Table 10-8)**

**For infinite life:**
$$
\left( \frac{\tau_m}{S_{se}} \right)^2 + \frac{\tau_a}{S_{sy}} \leq \frac{1}{n_f}
$$

From Table 10-8 for stainless steel:
- **Endurance limit ($ S_{se} $)**: 45,000 psi
- **Shear yield ($ S_{sy} $)**: 90,000 psi
- **Design factor ($ n_f $)**: 2

Plug in values:
$$
\left( \frac{38,067}{45,000} \right)^2 + \frac{12,684}{90,000} \leq 0.5
$$
$$
\left(0.846\right)^2 + 0.141 \leq 0.5
$$
$$
0.716 + 0.141 = 0.857 > 0.5
$$

**NOT SAFE**. The stresses are higher than allowed for infinite life with $ n_f = 2 $.

---

## **9. Adjust Design (Increase Wire Size)**

Let’s try **d = 0.12 in**.

- $ D = 1 - 0.12 = 0.88 $ in
- \( C = 0.88/0.12 = 7.

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