# Springs

**Contents**

## Introduction

Springs are load bearing elastic objects that are used to store and transfer mechanical energy. They are usually made from a low alloy, medium or high carbon steel. Their relatively high yield strength allows them to return to their original shape and size after a temporary deformation. There are several types of springs, including helical springs, flat springs and torsion springs among others. Each of these types is suited to its own specific application, however there are all constructed from a pliable / resilient material that can withstand a certain degree of deformation without fracture or failure. In engineering applications springs are often used for their compression or load-bearing ability, as well as for dampening out the effects of shock or impact loads.##### MISSING IMAGE!

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### 1 - Under Axial Load, W

- As the angle of the helix is small, the action on any cross section is approximately a pure torque and the effects of bending and shear can be neglected. The value of the torque is given by: The wire can therefore be considered to being twisted like a shaft. If is the total angle of twist along the wire, and x is the deflection of
##### MISSING IMAGE!

**23287/Springs-001.png**cannot be found in /users/23287/Springs-001.png. Please contact the submission author.*W*along the axis of the coil, then and Applying the formula for torsion of shafts and making the above substitution or The spring stiffness The strain energy Which by substitution in terms of from equation (7) can be reduced to:

### 2- Under Axial Torque Load, T

- This will produce approximately a pure bending moment of magnitude
*T*at all cross sections. The total strain energy is therefore given by: But if*T*causes a rotation of one end of the spring through an angle about the axis relative to the other end, then: Substituting this in equation (9)

##### Example - Helical Spring Under An Axial Torque T

^{2}. The solid length of the spring(i.e. coils touching) is 1.8in. Find the wire diameter; mean coil radius and the number of coils. .

## Open-coiled Helical Spring.

Let be the angle of the helix, then the length of the wire##### MISSING IMAGE!

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*O*. If now an axial load W and an axial torque

*T*are applied to the spring, the latter tending to increase the curvature, the actions at O are couples WD/2 about

*H*and

*T*about OV (the effect of the shearing force

*W*may be neglected). Resolving these couples about the axis OX and OY the combined twisting couple and the combined bending couple both of which tend to increase the curvature. The total strain energy due to bending and twisting is given by: Using Castiliano's Theorem, the axial deflection and the axial rotation . The general case can be derived from the above expressions but usually the loading is either

*W*only or

*T*only and the solution to these cases is given below.

##### Example - Axial extension of spring subjected to a load

## Leaf Springs

This type of spring was universally used on cars, lorries, and railway trucks. Whilst the introduction of independent suspension has reduced the automotive use, leaf springs are still in common use. The spring is made up of a number of leaves of equal width but varying length, placed in laminations and loaded as a beam. There are two main types. The "Semi-elliptic" is simply supported at both ends and loaded at it's centre whilst the quarter-elliptic is arranged as a cantilever. Semi-Elliptical Type In order to develop a simplified theory, it is assumed that the ends of each leaf (where they extend beyond their neighbour)are tapered uniformly to a point. It is also assumed that the "pack" is complete and that the shortest leaf is diamond shaped. These assumptions are not realised in practice. The main leaf must by necessity retain it's full width where it is supported. These slight departures from design do not seriously affect the the theory.##### MISSING IMAGE!

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*l*= span ( assumed constant)*b*= width of leaves*t*= thickness of leaves*W*= central load*y*= rise of crown above the level of the ends*n*= The number of leaves in the spring

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*R*in the strained case must be the same for all leaves and contact continues to be through the ends only. Friction between the leaves is ignored and it is assumed that each leaf is free to slide over it's neighbour. And since they all maintain the same radius of curvature, they can be imagined to be arranged side by side to form a curved beam of constant depth and varying width (as shown)

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### Quarter-elliptic Type

- The analysis is similar to to that used above. In this case the equivalent plan section varies from zero to nb at the fixed end, and the other values at this end are: Substituting in equation (34) or and
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##### Example - Leaf Springs

^{2}.. Plates are available in multiples of for thickness and for width. Determine suitable values for width; thickness; number of plates and calculate the radius to which the plates should be formed. Assume that the width is twelve times thickness and that

Now,

Dividing (3) by (7) and From equation (7) The actual deflection under the proof load of 0.75 tons. since the spring is now straight, the initial radius of curvature is given by:-