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Springs
An analysis of the common types of engineering 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.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 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:
[imperial]
Example - Helical Spring Under An Axial Torque T
Problem
A close-coiled helical spring is to have a stiffness of 5lb./in.in compression, with a maximum load of 9 lb. and a maximum shear stress of 18,000lb./in^{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. .
Workings
We know that,
Again,
Substituting from equations(5) and (6) in equation (3)
Substituting this value in equation (3)
i.e. mean coil radius =
And from equation (5) the number of coils,
Solution
Wire diameter, d = 0.1145 in
Mean coil radius = 0.56 in
The number of coils, n = 15.7
Open-coiled Helical Spring.
Let be the angle of the helix, then the length of the wire In the diagram OX is the polar axis (axis of twisting) at any normal cross section, and is inclined at an angle to the vertical OV. All the axes OX,OY,OH,and OV are in the vertical plane, which is tangential to the helix at 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:
[imperial]
Example - Axial extension of spring subjected to a load
Problem
An open coiled spring is made having ten turns wound to a mean diameter of 4.5 in. The wire diameter is 3/8 in. and the coils make an angle of 30 degrees with a plane perpendicular to the axis of the coil.
Find the axial extension when subjected to a load of 20 lb. and find the angle through which the free end will turn with this load if free to rotate.
Workings
Axial extension
Angle of rotation of free end.
= \frac{8\times 20\times 4.5^3\times 10}{\left(\frac{3}{8} \right)^4\;cos\,30}\left(\frac{cos^2\,30}{12\times 10^6}\;+
\[\frac{2\,sin^2\,30}{30\times 10^6} \right)
= \frac{8^5\times 4\times 4.5^3}{3\times \sqrt{3}\times 10^4}\left(\frac{1}{16} + \frac{1}{60}} \right) = 0.672\,in.
\phi = \frac{16WD^2\,n\;sin\,\alpha }{d^4}\,\left(\frac{1}{C} - \frac{2}{E} \right)}
Solution
Axial extension = 0.672 in
Angle of rotation of free end =
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. Let,- 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
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
Example:
[imperial]
Example - Leaf Springs
Problem
A laminated steel spring , simply supported at the ends and centrally loaded, with a span of 30 in., is required to carry a proof load of 0.75 tons whilst the central deflection is not to exceed 2 in. The bending stress must not exceed 25tons/in^{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
Workings
We know
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:-Solution
Width,
Thickness,
Number of plates = 11
The radius to which the plates should be formed,