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Created: 2024/06/05 22:25
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# Experiment no. 06

# Experiment name: Test of closely coiled helical spring

## Objective

1. To find the stiffness (k) of the spring.
2. To compare theoretical stiffness with observed stiffness.
3. To find out max shear stress and max torque.

## Significance

The stiffness of a spring is a fundamental property that describes how much load is needed to stretch or compress the spring by a unit distance. Understanding and accurately determining the stiffness of springs is crucial in systems in automotive industry such as suspension and brakes, aerospace industry such as landing gears, as well as consumer products such as mattresses.

## Apparatus

1. Slide calipers
2. Vernier scale
3. Universal testing machine

## Data sheet

![](../../Resources/IMG_20240605_233058.jpg)

## Calculation and graph

### Modulus of rigidity

- Modulus of elasticity, $E = 210\times 10^3\space MPa$ 
- Poisson ratio, $\mu = 0.27$

Therefore, modulus of rigidity,
$$
G = \frac{E}{2(1 + \mu)} = 82.677\space GPa
$$

### Theoretical stiffness

- Wire diameter, $d = 1\space mm$
- Spring radius, $R = \frac{13 - 1}{2} = 6\space mm$
- Number of coils, $N = 10$

Therefore, theoretical stiffness,

$$
K_{theo} = \frac{Gd^4}{64R^3N} = \frac{82.677\times 10^3\times 1^4}{64\times 6^3\times 10} = 0.5981\space Nmm^{-1}
$$

### Actual stiffness for individual deflection

1. For load = $1\space N$, deflection = $1.885\space mm$; stiffness = $\frac{1}{1.885} = 0.5305\space Nmm^{-1}$
2. For load = $2\space N$, deflection = $3.705\space mm$; stiffness = $\frac{2}{3.705} = 0.5398\space Nmm^{-1}$
3. For load = $3\space N$, deflection = $5.280\space mm$; stiffness = $\frac{3}{5.280} = 0.5681\space Nmm^{-1}$
4. For load = $4\space N$, deflection = $7.225\space mm$; stiffness = $\frac{4}{7.225} = 0.5536\space Nmm^{-1}$
5. For load = $5\space N$, deflection = $8.980\space mm$; stiffness = $\frac{5}{8.980} = 0.5567\space Nmm^{-1}$
6. For load = $6\space N$, deflection = $10.680\space mm$; stiffness = $\frac{6}{10.680} = 0.5617\space Nmm^{-1}$

Therefore, average actual stiffness,

$$
K_{actual} = \frac{0.5305 + 0.5398 + 0.5681 + 0.5536 + 0.5567 + 0.5617}{6} = 0.5517\space Nmm^{-1}
$$

### Actual stiffness from load v. deflection graph

Actual stiffness,

$$
K = \frac{\Delta load}{\Delta deflection} = \frac{1.11\space N}{2\space mm} = 0.5550\space Nmm^{-1}
$$

![](../../Resources/IMG_20240606_000924.jpg)

## Result

The actual stiffness of the helical spring was found to be $0.5550\space Nmm^{-1}$ while the calculated average actual stiffness and theoretical stiffness was found to be $0.5517\space Nmm^{-1}$ and $0.5981\space Nmm^{-1}$ respectively.

## Discussion

In this experiment, the stiffness (k) of a helical spring was determined. The spring was subjected to various loads, and the corresponding displacements were measured. The theoretical stiffness of the spring was calculated using the formula derived from Hooke's Law and the spring's material and geometric properties. The average actual stiffness was determined from the average of the displacements corresponding to each applied load and also graphically by plotting the load versus displacement and finding the slope of the best-fit line.

The theoretical stiffness of the spring was calculated to be 0.5981 N/mm. This value is derived assuming ideal conditions without any imperfections.

The average actual stiffness, calculated from the displacements observed during the application of the six loads, was 0.5517 N/mm. This value is slightly lower than the theoretical stiffness.

The actual stiffness obtained from the graph of load versus displacement was 0.555 N/mm, which is also lower than the theoretical value but very close to the average experimental stiffness.

The experimentally determined stiffness values (both average and graphical) are slightly lower than the theoretical value. However, the discrepancy is within an acceptable range, suggesting that the theoretical model provides a good approximation of the spring's behavior.