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An analysis of the common types of engineering springs

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Close-coiled Helical Springs.

$D = Mean\;coil\;diameter.$

(1)

$d = Wire\;diameter$

(2)

$n = Number\;of\;coils.$

(3)

Helical Spring under axial load

As the angle of the helix is small , the action on any cross section is approximately a pure torque and the effects bending and shear can be neglected. The value of the torque is given by:

$Torque\;=W\times \frac{D}{2}$

(4)

The wire can therefore be considered to being twisted like a shaft and if $\theta$ is the total angle of twist along the wire and if x is the deflection of W along the axis of the coil,

$Then\;\;\;\;\;x = \left(\frac{D}{2} \right)\theta$

(5)

approximately

$and\;\;\;\;\;l = \pi \,D\,n$

(6)

Applying the formula for torsion of shafts and making the above substitution

$\frac{W\frac{D}{2}}{\frac{\pi d^4}{32}} = \frac{2\,\hat{s}}{d}\,= \frac{C\times \frac{2\,x}{D}}{\pi D\,n}$

(7)

$or\;\;\;\;\frac{8\,W\,D}{\pi \,d^4} = \frac{2\hat{s}}{d} = \frac{C\,x}{\pi \,D^2\,n}$

(8)

$The\;spring\;stiffness\;\mu = \frac{W}{x} = \frac{C\,d^4}{8D^3\,n}$

(9)

$The \;strain \;energy\;\;\;U = \frac{1}{2}W\,x$

(10)

Which by substitution in terms of $\hat{s}$ from equation (7) can be reduced to:

$U = \left(\frac{\hat{s}^2}{4C} \right)\times \;volume$

(11)

Helical Spring under an axial torque T

This will produce approximately a pure bending moment of magnitude T at all cross sections. the total strain energy is therefore given by:

$U = \frac{T^2\,l}{2E\,I} = \frac{T^2\,\pi Dn}{2E\times \frac{\pi d^4}{64}} = \frac{32T^2\,Dn}{Ed^4}$

(12)

But if T causes a rotation of one end of the spring through an angle $\phi$ about the axis relative to the other end then:

$U = \frac{1}{2}T\phi$

(13)

Substituting this in equation (11)

$\phi = \frac{T\,l}{EI} = \frac{64TDn}{Ed^4}$

(14)

$Maximum\;bending\;stress = \frac{T\times \frac{d}{2}}{\frac{\pi d^4}{64}} = \frac{32T}{\pi d^3}$

(15)

Example 1 [imperial]

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. $C = 6\times 10^6lb.in^{-2}$ .

Workings

We know that,

$Stiffness\;\mu = \frac{Cd^4}{8D^3n}$

(16)

$i.e.\;\;\;\;\;5 = \frac{6\times 10^6\times d^4}{8D^3n}$

(17)

$or\;\;\;\;\;d^4 = \frac{2D^3n}{3\times 10^5}$

(18)

Again,

$Maximum\;stress\;\hat{s} = \frac{8WD}{\pi d^3}$

(19)

$\therefore\;\;\;\;D = 250\pi d^3$

(20)

$But\;the\;solid\;length\;=1.8in.\;=nd$

(21)

Substituting from equations(5) and (6) in equation (3)

$d^4 = \frac{2}{3\times 10^5}\left(250\pi \right)^3d^9\;\frac{1.8}{d}$

(22)

$Giving \;\;\;\;\;\;d = 0.1145in.$

(23)

Substituting this value in equation (3)

$D = 1.12in.$

i.e. mean coil radius = $0.56 in$

And from equation (5) the number of coils, $n = 15.7$

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 $\alpha$ be the angle of the helix, then the length of the wire $l = \frac{\pi Dn}{cos\alpha }$

In the diagram OX is the polar axis ( axis of twisting) at any normal cross section and is inclined at an angle $\alpha$ to the vertical OV. All the axis 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

$= \left( \frac{WD}{2} \right)cos\alpha + T\: sin\alpha$

(35)

and the combined bending couple

$= T\;cos\alpha - \left(\frac{WD}{2} \right)sin\alpha$

(36)

both of which tend to increase the curvature.

The total strain energy due to bending and twisting is given by:

$U = \frac{\left[\left(\frac{WD}{2}\right)cos\,\alpha + T\,sin\;\alpha \right]^2l}{2CJ} + \frac{\left[T\,cos\;\alpha - \left(\frac{WD}{2} \right)sin\,\alpha \right]^2l}{2EI}$

(37)

Using Castiliano's Theorem, the axial deflection $x = \frac{\delta U}{\delta W}$ and the axial rotation $\phi = \frac{\delta U}{\delta T}$ . 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.

1) Axial load only

$x = \left(\frac{\delta U}{\delta W} \right)_{T\,=\,0}$

(38)

$= \frac{2\left[\frac{WD}{2} \right]cos\,\alpha \left(\frac{D}{2} \right)cos\,\alpha \; .l}{2CJ} + \frac{2\left[- \left(\frac{WD}{2} \right)sin\,\alpha \right]\left[-\left(\frac{D}{2} \right) sin\,\alpha \right]l}{2EI}$

(39)

$= \left(\frac{WD^2l}{4} \right)\left(\frac{cos^2\,\alpha }{CJ} + \frac{sin^2\,\alpha}{EI} \right)$

(40)

$= \left(\frac{8WD^3\: n\: cos\,\alpha }{d^4} \right)\left(\frac{cos^2\,\alpha }{C} + \frac{2\: sin^2\,\alpha}{E} \right)$

(41)

$Note\;\;\;\;\left( J = \frac{\pi d^4}{32}\;\;\;\;and\;\;\;\;I = \frac{\pi d^4}{64} \right)$

(42)

$\phi = \left(\frac{\delta U}{\delta T} \right)_{T=0}$

(43)

$= \frac{2\left[\left(\frac{WD}{2} \right)cos\,\alpha \right]sin\,\alpha .\;l}{2CJ} + \frac{2\left[-\left(\frac{WD}{2} \right)sin\,\alpha \right]cos\: \alpha .\;l}{2EI}$

(44)

$= \left(\frac{WD\,l}{2} \right)sin\,\alpha \,.\,cos\,\alpha \left(\frac{1}{CJ} - \frac{1}{EI} \right)$

(45)

$= \left(\frac{16\,WD^2n\,sin\alpha }{d^4} \right)\left(\frac{1}{C} - \frac{1}{E} \right)$

(46)

2) Axial load only

$\phi = \left(\frac{\delta U}{\delta T} \right)_{W=0}$

(47)

$= \frac{2(T\,sin\,\alpha )sin\,\alpha \;.l}{2CJ} + \frac{2(T\;cos\: \alpha )cos\;\alpha \;.l}{2EI}$

(48)

$= T\,l\left(\frac{sin^2\,\alpha }{CJ} + \frac{cos^2\,\alpha }{EI} \right)$

(49)

$= \left(\frac{32TDn\,cos\,\alpha }{d^4} \right)\left(\frac{sin^2\,\alpha }{C} + \frac{2cos^2\alpha }{E} \right)$

(50)

$x = \left(\frac{\delta U}{\delta W} \right)_{W=0}$

(51)

$= \frac{2(T\: sin\,\alpha)(\frac{D}{2})cos\,\alpha \;.l }{2CJ} + \frac{2(Tcos\,\alpha )[-(\frac{D}{2})sin\,\alpha ]l}{2EI}$

(52)

$= \left(\frac{TD\;l\,sin\,\alpha .cos\,\alpha }{2} \right)\left(\frac{1}{CJ} - \frac{1}{EI} \right)$

(53)

$= \left(\frac{16\,TD^2\,n\,sin\alpha }{d^4} \right)\left(\frac{1}{C} - \frac{2}{E} \right)$

(54)

Example 1 [imperial]

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.

$E = 30\times 10^6 lb.in^{-2}\;\;\;\;and\;\;\;\;C = 12\times 10^6\,lb.in^{-2}$

Workings

Axial extension

$x = \frac{8WD^3n}{d^4\,cos\,\alpha }\left(\frac{cos^2\,\alpha }{C} + \frac{2\,sin^2\,\alpha }{E} \right)$

(55)

$= \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)$

(56)

$= \frac{8^5\times 4\times 4.5^3}{3\times \sqrt{3}\times 10^4}\left(\frac{1}{16} + \frac{1}{60}} \right) = 0.672\,in.$

(57)

Angle of rotation of free end.

$\phi = \frac{16WD^2\,n\;sin\,\alpha }{d^4}\,\left(\frac{1}{C} - \frac{2}{E} \right)}$

(58)

$= \frac{16\times 20\times 4.5^2\times 10\,sin\,30}{\left(\frac{3}{8} \right)^4}\left(\frac{1}{12\times 10^6} - \frac{2} {30\times 10^6}\right)$

(59)

$= \frac{8^4\times 16\times 4.5^2}{3^4\times 10^4}\left(\frac{1}{12} - \frac{1}{15} \right)$

(60)

$= 0.027\;radians = 1.55^0$

(61)

Solution

Axial extension = 0.672 in

Angle of rotation of free end = $1.55^0$

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 length 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 assume 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

If the leaves are initially curved to circular arcs of the same radius $R_0$ , contact between the leaves will only take place at their ends and consequently the loading of any one leaf will be as shown in the following diagram.

Over the central portion both M and I are constant whilst over the end section both M and I are proportional to the distance from the end. Consequently over the whole leaf M/I is constant.

$but\;\;\;\;\;\;\frac{M}{EI} = \frac{1}{R} - \frac{1}{R_0}\;\;\;\;(see.\; "Bending\;of\;curved\;bars")$

(62)

Since $R_0$ is assumed to remain constant, the radius of curvature R in the strained case must be the same for all leaves and contact continues to be through the eds 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)

As the bending moment for the equivalent section is directly proportional to the distance from either end and I also varies uniformly, it can be seen that the spring is equivalent to a beam of uniform strength ( i.e. the beam has the same maximum strength at all sections)

Now consider any convenient cross section. In the following analysis the central section has been used.

$M\;= - \frac{Wl}{4}\;\;\;\;\;\;\;(tending\;to\;decrease\;the\;curvature)$

(63)

$I = \frac{nbt^3}{12}$

(64)

Using equations of a circle

$y(2R - y) = \left(\frac{l}{2} \right)\left(\frac{l}{2} \right)$

(65)

and treating y as small compared to R

$\frac{1}{R} = \frac{8\,y}{l^2}$

(66)

Rewriting equation (65)

$\frac{- \frac{Wl}{4}}{E\frac{nbt^3}{12}} = \frac{8}{l^2}\left(y - y_0 \right)$

(67)

The deflection $\delta$ is given by:

$\delta = y_0 - y = \frac{3W\,t^3}{8nbt^3\,E}$

(68)

The load required to straighten the load is called the "Proof Load" and is given by

$\frac{8nbt^3Ey_0}{3l^3}$

(69)

The maximum bending stress is given by:

$\hat{f} = \left(\frac{M}{I} \right)\left(\frac{t}{2} \right)$

(70)

$= \left(\frac{Wl}{4} \right)\left(\frac{t}{2} \right)\div \frac{nbt^3}{12}$

(71)

$= \frac{3Wl}{2nbt^2}$

(72)

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:

$M\;= - Wl$

(73)

$I = \frac{nbt^3}{12}$

(74)

$\frac{1}{R} = \frac{2y}{l^2}$

(75)

Substituting in equation 65

$\frac{M}{EI} = \frac{1}{R} - \frac{1}{R_0}$

(76)

$or\;\;\;\;\;\delta = y_0 - y = \frac{6Wl}{nbt^3E}$

(77)

$and\;\;\;\;\;\hat{f} = \left(\frac{M}{I} \right)\left(\frac{t}{2} \right)$

(78)

$= \frac{6Wl}{nbt^3}$

(79)

Example 1 [imperial]

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 $\frac{1}{32}\,in$ for thickness and $\frac{1}{8}\,in.$ 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 $E = 30\times 10^6 lb. in.^{-2}.$

Workings

We know

$\delta = \frac{3Wl^3}{8nbt^3E}$

(80)

$\therefore\;\;\;\;2 = \frac{3\times 0.75\times 2240\times 30^3}{8n(12t)t^3\times 30\times 10^6}$

(81)

$\therefore\;\;\;\;\;nt^4 = 0.02363$

(82)

Now,

$Maximum\;stress,\;\hat{f} = \frac{3Wl}{2nbt^2}$

(83)

$\therefore\;\;25 = \frac{3\times 0.75\times 30}{2\times n(12t)t^2}$

(84)

$\therefore\;\;\;\;nt^3 = 0.1125$

(85)

Dividing (3) by (7)

$t = \frac{nt^4}{nt^3} = 0.21\, in.\;\;\;\;say \;\;\frac{7}{32}\,in.$

(86)

and

$b = 12t = 2\frac{5}{8}\,in.$

(87)

From equation (7)

$n = \frac{\left(0.1125\times 32^3 \right)}{7^3} = 10.75, \;\;\;\;say \;11\;leaves$

(88)

The actual deflection under the proof load of 0.75 tons.

$= \frac{3\times 0.75\times 2240\times 30^3\times 32^3}{8\times 11\times 2.625\times 7^3\times 30\times 10^6}$

(89)

$= \frac{3\times 0.75\times 2240\times 30^3\times 32^3}{8\times 11\times 2.625\times 7^3\times 30\times 10^6}=1.875\,in.$

(90)

since the spring is now straight, the initial radius of curvature is given by:-

$R_0 = \frac{l^2}{8\delta }$

(91)

$\therefore\;\;\;\;R_0 = \frac{30^2}{8\times 1.875} = 60\,in.$

(92)

Solution

Width, $b = 2\frac{5}{8}\,in.$

Thickness, $t = 0.21 in$

Number of plates = 11

The radius to which the plates should be formed, $R_0 = 60 in$

Flat Spiral Springs

This type of spring is used in clockwork mechanisms ans consists of a uniform strip would into a spiral in one plane and pinned at it's outer end. The spring is would up by applying a torque to a spindle attached to the centre of the spiral.

Let T be the torque tending to wind up the spring and X and Y the components of reaction at the outer end of the spring. Taking moments about the spindle axis T = Y R where R is the maximum radius of the spiral. At any point in the spring, defined by coordinates x and y the bending moment $= Y_x - X_y$ tending to increase the curvature.

The Strain energy is given by:

$U = \int \frac{(Y_x - X_y)^2}{2EI}ds$

(93)

$= \int \frac{\left[\left(\frac{T}{R} \right)x - X_y \right]^2}{2EI}ds$

(94)

since O is a fixed point $\frac{\delta U}{\delta x} = 0$ giving

$X = \frac{T}{R}\;\frac{\int xy\,ds}{\int y^2\,ds} = 0\;\;\;\;by\;symmetry$

(95)

$Then\;\;\;\;\theta = \frac{\delta U}{\delta T} = \frac{2T}{R^2}\int x^2\,ds$

(96)

But if we treat the spiral as a uniform disc, then

$\int x^2\,ds\;\approx \;\left(\frac{R^2}{4} + R^2 \right)l$

(97)

The maximum bending moment = Y.2R at the left hand edge = 2T

$Maximum\; Stress,\;\hat{f} = \frac{2T}{Z} = \frac{12T}{bt^2}$

(98)

Example 1 [imperial]

Problem

A flat spiral spring is 0.25 in. wide; 0.01 in. thick and 8 ft. long. Assuming the maximum stress of 120,000 lb./in\^2.to occur at the point of greatest bending moment, calculate the torque, the work stored and the number of turns needed to wind up the spring. $E = 30\times 10^6 \,lb.in.^{-2}$