Everything You Need to Know About Helical Springs

Impact and One-Dimensional Wave Propagation

H.R. Harrison , T. Nettleton , in Advanced Engineering Dynamics, 1997

six.xiv Waves in a helical spring

The helical spring volition exist treated as a thin wire, that is plane cross-sections remain aeroplane. This assumption has been shown to be acceptable by mechanical testing. An element of the wire has six degrees of freedom, 3 displacements and three rotations, so the possibility exists for six modes of propagation. If the helix angle is small these modes separate into two groups each having three degrees of freedom, one set consisting of the in-plane motion and the other the out-of-aeroplane motility. The effect of helix angle will be discussed later but here nosotros shall develop the theory for out-of-plane motion for a spring with cypher helix angle. This, as we shall encounter, is associated with axial motion of the spring, that is with the bound being in its compression or tensile mode.

Figure six.37 defines the co-ordinate organisation to be used. The unit vector i is tangent to the centrality of the wire, j is along the radius of curvature directed towards the middle and k completes the right-handed triad. For naught helix bending k is parallel to the axis of the spring. The radius of curvature is R and s is the distance measured along the wire. θ is the angle through which the radius turns. Thus

Fig. vi.37.

First nosotros shall consider the differentiation of an arbitrary vector V with respect to southward

where the prime signifies differentiation with respect to non-rotating axes and Ω is the rate of rotation of the axes with distance s. Thus

and using equation (6.137)

From equation (vi.138) the components of the derivative are

or in matrix form

where

(six.144) $[T\mathrm{one}]\equiv \left[\begin{array}{ccc}p& -\mathrm{one}/R& 0\\ 1/R& p& 0\\ 0& 0& p\end{array}\right]$

and

A cross-section has a displacement u and a rotation ø from its equilibrium position. The spatial charge per unit of change of u is due to stretching and shearing of the chemical element of wire and also to rigid body rotation, and so the strain

(6.146) $\begin{array}{l}(\varepsilon )=[T_{\mathrm{i}}](u)-{[ø]}^{\text{x}}(\text{d}\mathrm{southward})/\text{d}s\\ =[T_{\mathrm{one}}](u)+{[\text{d}s/\text{d}s]}^{\text{x}}(ø)\end{array}$

Now (ds) = (ds 0 0) and therefore

(6.147) $\frac{{[\text{d}s]}^{\text{x}}}{\text{d}s}=\left[\begin{array}{ccc}0& 0& 0\\ 0& 0& -\mathrm{i}\\ 0& 1& 0\end{array}\right]\equiv [T_2]$

(see Appendix 1)

The components of strain are, therefore,

(six.148) $\begin{array}{cc}\text{axial\hspace{0.17em}strain}& {\varepsilon }_i=p{u}_i-u_j/R\end{array}$

(6.149) $\begin{array}{cc}\text{shear\hspace{0.17em}strain}& {\varepsilon }_j=p{u}_j+u_i/R-{\varnothing }_k\end{array}$

(6.150) $\begin{array}{cc}\text{shear\hspace{0.17em}strain}& {\varepsilon }_k=p{u}_k+{ø}_j\end{array}$

These are related to the elastic constants by

where q is a factor to permit for the shear stress distribution not being uniform. A typical value for circular cross-sections is 0.9.

The relationship between the elastic constants, Young'southward modulus East, shear modulus G and Poisson'due south ratio u is E = twoG(1 + v), so let

Combining equations (6.148) to (6.154)

(six.155) $P_i=\mathrm{Thou}A(\mathrm{ii}mp{u}_i-\mathrm{ii}\mathrm{thou}{u}_j/R)$

(6.156) $P_j=GA(qp{u}_j+q{u}_i/R-q{ø}_k)$

For bending nosotros use the usual technology relationships for bending and torsion of shafts. If the shape of the cross-section has point symmetry then with J being the polar second moment of area

Too E = Thousand2m then that EI_j = EI_k = mGJ, and therefore

(half dozen.158) ${\mathrm{1000}}_i=G{I}_i\frac{\text{d}{ø}_i}{\text{d}s}=GJ(p{ø}_i-{ø}_j/R)$

(6.159) $M_j=\mathrm{Due\; east}{I}_j\frac{\text{d}{ø}_j}{\text{d}\mathrm{due\; south}}=GJk(p{ø}_j-{ø}_i/R)$

(6.160) $M_k=E{I}_k\frac{\text{d}{ø}_m}{\text{d}\mathrm{due\; south}}=GJm(p{ø}_k)$

The equations of motion tin can exist derived with reference to Fig. 6.38. Resolving forces acting on the element, neglecting any external forces,

Fig. 6.38.

$\left((P)+\frac{\text{d}(P)}{\text{d}{\mathrm{southward}}_{t=\text{constant}}}\text{d}s\right)-(P)=\rho A\text{d}\mathrm{south}\frac{{\partial }^2(u)}{\partial t^2}$

or, letting $D=\frac{\partial }{\partial t}$

or

The component equations are

Now considering moments about the centre of mass of the element

(6.165) $\left((K)+\frac{\text{d}(M)}{\text{d}{\mathrm{southward}}_{t=\text{abiding}}}\text{d}s\right)-(M)+{(\text{d}\mathrm{southward})}^{\text{10}}(P)=\frac{\partial (\mathrm{Fifty})}{\partial t}$

where (50) is the moment of momentum. For small rotations

(6.166) $(\mathrm{Fifty})=\left[\begin{array}{c}\rho I_i\text{d}\mathrm{south}D{ø}_i\\ \rho I_j\text{d}sD{ø}_j\\ \rho I_k\text{d}\mathrm{due\; south}D{ø}_k\end{array}\right]$

and

(6.167) $\frac{\partial (L)}{\partial t}=\rho J{D}^2\left[\begin{array}{l}{ø}_i\\ {ø}_j/\mathrm{ii}\\ {ø}_{\mathrm{grand}}/2\end{array}\right]\text{d}s=\frac{\text{d}(M)}{\text{d}{\mathrm{south}}_{t=\text{constant}}}\text{d}s+{(\text{d}s)}^{\text{x}}(P)$

The iii component equations are, after dividing by ds,

(vi.169) $p{\mathrm{Grand}}_j+K_i/R-P_k=\rho J{D}^2{ø}_j/\mathrm{two}$

Substituting the six equations of state ((6.155) to (6.160)) into the equations of motion ((half dozen.162–six.164) and (6.168–half dozen.170)) will yield vi equations in the vi co-ordinates and these will separate into 2 groups of 3. So, substituting equations (6.157), (6.158) and (half dozen.159) into equations (vi.164), (half-dozen.168) and (six.169) leads to

(half dozen.171) $GJ(p^2{ø}_i-p{ø}_j/R)-\mathrm{Thousand}J\frac{\mathrm{grand}}{R}(p{ø}_j+{ø}_i/R)=\rho J{D}^2{ø}_i$

(half-dozen.172) $GJm(p^2{ø}_j+p{ø}_i/R)+\frac{GJ}{R}(p{ø}_i-{ø}_j/R)-q\mathrm{Yard}A(p{u}_{\mathrm{chiliad}}+{ø}_j)=\rho J{D}^2{ø}_j/2$

(6.173) $q\mathrm{Thousand}A(p^{\mathrm{ii}}{u}_k+p{ø}_j)=\rho A{D}^2{u}_{\mathrm{yard}}$

which comprise only the three out-of-plane co-ordinates.

It is convenient for discussion purposes to put the to a higher place equations into non-dimensional class. To this end we ascertain the following terms

Thus equations (half-dozen.171) to (6.173) may be written in matrix form as

(vi.177) $\left[\begin{array}{ccc}({\tilde{p}}^{\mathrm{ii}}-m)& -\tilde{p}(1+m)& 0\\ \tilde{p}(1+g)& (m{\tilde{p}}^2-1-{\alpha }^{2}q)& -q{\alpha }^{\mathrm{two}}\tilde{p}\\ 0& q\tilde{p}& q{\tilde{p}}^2\end{array}\right]\left[\begin{array}{c}{ø}_i\\ {ø}_j\\ U_k\end{array}\right]={\tilde{D}}^2\left[\begin{array}{l}{ø}_i\hfill \\ {ø}_j/\mathrm{ii}\hfill \\ U_{\mathrm{thou}}\hfill \end{array}\right]$

(Note that this matrix equation can exist written in symmetrical form; however, we can discuss the manner of wave propagation simply likewise in the current form.)

In a fashion similar to previous cases we shall assume a wave travelling along the centrality of the wire. Thus

Now

and

with similar expressions for the other two co-ordinates.

Allow us ascertain the non-dimensional wavenumber

and the non-dimensional frequency

from which we have

the non-dimensional phase velocity.

From equations (6.181) to (6.184) we write

so equation (6.177) can now be written as

(vi.188) $\left[\begin{array}{ccc}({\mathrm{West}}^2-\mathrm{yard}-M^2)& \text{j}K(1+\mathrm{one\; thousand})& 0\\ -\text{j}\mathrm{1000}(\mathrm{ane}+\mathrm{1000})& \left(\frac{1}{2}{W}^2-1-{\alpha }^{2}q\right)& -q\text{j}G{\alpha }^{\mathrm{two}}\\ 0& -q\text{j}\mathrm{Thousand}& (W^2-q{K}^2)\end{array}\right]\left[\begin{array}{c}{ø}_i\\ {ø}_j\\ U_k\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right]$

For a not-trivial solution the determinant of the square matrix must equate to zero. Thus

(six.189) $\left|\begin{array}{ccc}(W^2-m-K^2)& \text{j}\mathrm{One\; thousand}(1+m)& 0\\ -\text{j}\mathrm{Yard}(1+m)& (½W^2-\mathrm{i}-{\alpha }^{\mathrm{two}}q)& -q\text{j}M{\alpha }^2\\ 0& -q\text{j}\mathrm{Chiliad}& ({\mathrm{Due\; west}}^2-q{K}^2)\end{array}\right|=0$

This leads to a cubic in W ² for whatever given value of K. In that location are, therefore, 3 branches to the dispersion diagram and these are shown in Fig. six.39. The fourth root of ω is plotted as ordinate in gild to compress the scale. The two highest modes have cut-off frequencies and therefore sinusoidal wave propagation only exists in these modes at loftier frequencies. The of import everyman mode is shown in particular in Fig. vi.40.

Fig. 6.39. Dispersion diagram for helical spring

Fig. 6.twoscore.

The Due west—K diagram for the lowest mode exhibits a zero frequency when Thou = 0 and also when M = 1 or λ = iiπR. At this wavelength particles having maximum positive velocity are one plow apart and the maximum negative velocity particles are diametrically opposite.

For low values of wavenumber equation (6.189) reduces to

Now α² =2R ²/r ² for solid round cantankerous-section wire. The spring index (R/r) is unlikely to be less than 3 and so the minimum value of α² is about 18; a more typical index of 5 gives αⁱⁱ = 50. Since q is of the order unity equation (6.190) is, to a shut approximation,

Returning to the dimensional form

from which the phase velocity and the group velocity are given by

This approximation is quite reasonable for wavelengths longer than 5 turns. Also shown on Fig. half dozen.40 are the amplitude ratios and it is interesting to note that although the strain associated with long wavelengths is torsional in nature there is very little rotation almost the wire axis (ø _i → 0). This is true for the static case, represented here by cypher frequency and infinitely long wavelength.

The dispersion diagram shown is a plot of Wα versus K for α = 10 but on the scale used no departure is seen for α ranging from iii to xxx.

The issue of the helix angle being greater than null is to couple the in-aeroplane and out-of-aeroplane co-ordinates, but for small helix angle and depression wavenumber the essential nature of the curves does non alter. The more than noticeable event is around K = 1 where the curve is more rounded for the everyman longitudinal way and the curve for the torsional fashion no longer goes to zero. The two lowest dispersion curves are shown in Fig. 6.41 which likewise shows the results of mechanical steady-state vibration tests. Impact tests were as well carried out from which the inflow times of various frequency components were measured and compared with the theory; some results are shown in Fig. 6.42.

Fig. 6.41. Dispersion curve for helical leap (data from Ph.D thesis, H.R. Harrison 1971)

Fig. 6.42.

In department half dozen.12 the dispersion diagram for a periodic mass–spring system was developed and shown in Fig. six.36. The similarity with the lowest mode for the spring as shown on Fig. six.xl is quite noticeable. The numerical similarity is strong if in the lumped parameter model the mass and the stiffness of the components are those of a single turn of the spring. This model gives skillful agreement for wavelengths as short as ane turn of the helix.

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Discretization of the Domain

Singiresu Southward. Rao , in The Finite Element Method in Engineering (Fifth Edition), 2011

2.3.1 Blazon of Elements

Oftentimes, the type of elements to be used will exist evident from the physical problem. For instance, if the trouble involves the analysis of a truss construction under a given set of load weather condition (Figure 2.8(a)), the blazon of elements to exist used for idealization is obviously the "bar or line elements" as shown in Figure 2.8(b). Similarly, in the case of stress analysis of the short axle shown in Figure 2.nine(a), the finite element idealization can be done using 3-dimensional solid elements every bit shown in Figure ii.9(b). However, the type of elements to exist used for idealization may not be apparent, and in such cases one has to choose the type of elements judicially. As an example, consider the trouble of analysis of the thin-walled shell shown in Figure 2.10(a). In this case, the shell can exist idealized by several types of elements as shown in Figure 2.10(b). Hither, the number of dof needed, the expected accurateness, the ease with which the necessary equations can exist derived, and the caste to which the physical structure can exist modeled without approximation volition dictate the choice of the element type to exist used for idealization. In certain problems, the given body cannot be represented every bit an assemblage of only i type of elements. In such cases, we may have to use two or more types of elements for idealization. An example of this would exist the analysis of an shipping wing. Since the wing consists of top and bottom covers, stiffening webs, and flanges, 3 types of elements—namely, triangular plate elements (for covers), rectangular shear panels (for webs), and frame elements (for flanges)—have been used in the idealization shown in Figure 2.11.

Figure 2.8. A Truss Construction.

Figure ii.9. A Brusk Beam.

Figure ii.ten. A Thin-Walled Beat out under Pressure level.

Figure ii.11. Idealization of an Shipping Fly Using Different Types of Elements.

Example 2.1

A helical jump is subjected to a compressive load equally shown in Figure 2.12(a). Suggest different methods of modeling the jump using one-dimensional elements.

Figure two.12. Modeling of a Helical Spring.

Solution

Arroyo: Use diverse one-dimensional or line elements.

The helical bound (in the form of curved wire) can exist divided into several line or one-dimensional segments. These segments can be straight or curved. Each of the straight line segments (or elements) can be assumed to exist a spatial truss element with each of its endpoints (or nodes) having three displacement dof (parallel to the x, y, and z axes) as shown in Figure 2.12(b). Since this element has only translational degrees of liberty (with no rotational degrees of freedom), it will not be able to carry any moment. As such, the element may not be able to represent the behavior of the helical spring accurately.

Alternately, each of the straight line segments (or elements) tin he assumed to be a spatial frame element with each of its endpoints (or nodes) having three displacement dof (parallel to the x, y, and z axes) and 3 rotational dof (about the x, y, and z axes) every bit shown in Figure 2.12(c). In the example of the curved line segments (elements), each element tin can be treated equally a curved frame element with 3 displacement dof (parallel to the x, y, and z axes) and three rotational dof (near the x, y, and z axes) at each end every bit shown in Effigy 2.12(d). Because of the inclusion of rotational degrees of freedom, the models shown in Figures 2.12(c) and (d) volition exist able to simulate the behavior of the helical spring more accurately.

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Definitions and abbreviations

In Handbook of Valves and Actuators, 2007

COGUANOR

Comisian Guatemalteca de Normas, the National Standards Authorisation of Guatemala.

Coil bound

The condition of helical springs when compressed until all the coils touch. The spring length when coil spring is chosen the solid length. Some spring designs are not suitable for compression to solid length and permanent plastic deformation occurs rendering the spring useless for its intended purpose. Encounter also Permanent set up.

Cold differential ready pressure level

When a spring-loaded valve which will commonly operate hot is preset on a cold test rig, the cold set pressure must be increased slightly and then the valve volition operate at the right force per unit area when hot. Typical corrections are shown in Table i.seven.

Tabular array 1.7. Cold differential set pressures

Operating temperature	% age increase in cold set pressure level
−xviii to 120°C	0
121 to 315°C	i
316 to 425°C	2
426 to 530°C	3

Cold drawing

A technique for improving the surface finish and correcting the sizes of pipes, tubes, bars, shapes or wire. The semi-processed material is drawn through a series of dies which polish the surface and reduce the bore/shape slightly to the correct size. The material is work-hardened and the enhanced mechanical properties tin can exist utilised provided information technology is not welded or heated.

Cold pull

Tension applied to a piping system during erection when common cold. Thermal expansion, on heating, relieves the initial stresses.

Common cold-setting chemical compound

An adhesive, liquid gasket, thread seal or impregnation compound which hardens or thickens at ambient temperature without the application of external heat.

Cold water finish see Mirror finish

Common cold working

Altering the shape or size of a metallic by plastic deformation. Processes include rolling, cartoon, pressing, spinning, extruding and heading. It is carried out below the recrystallisation bespeak commonly at room temperature. Hardness and tensile strength are increased with the caste of common cold work whilst ductility and impact values are lowered. The cold rolling and cold drawing of steel significantly improves surface cease.

Cavalcade separation

When a non-return valve closes, the liquid travelling towards the valve may create water hammer. The inertia of the liquid travelling away from the valve may create column separation; that is the liquid may divide and create a vapour pocket. Dissolved gas tin also be released in these situations. If, when the inertia energy is prodigal, the liquid column reverses and returns to the non-return valve, a classic cavitation status is created.

Compliance

Requirements in specifications and standards must be verifiable in order to establish compliance. Any requirement which cannot be verified is worthless. CEN editorial policy dictates that requirements must exist verifiable at the manufacturer's works prior to acceleration. A simple example may clarify the situation.

An EN standard may state that cast steel earth valves shall exist painted with a total paint thickness of 100 pm. A tertiary-party inspector tin audit a finished valve, measure the paint thickness, and state categorically that the valve does or does non comply with the painting requirements. An EN standard cannot state valve packing shall take a minimum life of ii years and the leakage rate shall not increase by more than a cistron of 2. This requirement is not verifiable in the manufacturer'due south works, before shipment, and cannot, sensibly, exist considered as part of a buy understanding.

The requirement is too awarding specific and not suitable for inclusion in a standard. Standards can contain notes which are explanatory or draw policy. A note could exist every bit follows, "Valves complying with standard are intended to exist heavy-duty and packing should be suitable for a 2 year working life." This would be a policy statement, not an enforceable, verifiable requirement.

Compressibility factor Z

A factor used in gas applications to convert the ideal gas formula to existent gas. The basic gas equation, pV = mRT, becomes pV = ZmRT. Values for Z are obtained from charts depending upon the operating pressure and temperature.

Pinch coupling

A pipe fitting which relies on the compression of a ferrule to seal the pipe and locate the pipe axially. Compression fittings with one or two metallic ferrules are limited to 1″ or 2″ od Imperial tube but are capable of loftier to very high pressures; fittings for thin wall copper tube are available up to 108 mm. Non-metallic ferrules tin can be used for lower pressure applications with tubes up to 125 mm od.

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SPRINGS

E.J. HEARN Ph.D., B.Sc. (Eng.) Hons., C.Eng., F.I.Mech.E., F.I.Prod.E., F.I.Diag.Due east. , in Mechanics of Materials ane (Third Edition), 1997

Case 12.i

A shut-coiled helical leap is required to blot ii.25 × 10³ joules of free energy. Determine the bore of the wire, the hateful bore of the spring and the number of coils necessary if:

(a)

the maximum stress is non to exceed 400 MN/m²;

(b)

the maximum compression of the bound is limited to 250 mm;

(c)

the mean bore of the bound can exist causeless to be eight times that of the wire.

How would the answers change if appropriate Wahl factors are introduced? For the leap material G = 70 GN/m².

Solution

The spring is required to absorb 2.25 × 10³ joules or 2.25 kN yard of free energy.

$\therefore \text{piece of work\hspace{0.17em}done}=\frac{1}{2}W\delta =2.25\times {\mathrm{x}}^3$

But δ is limited to 250 mm.

∴

$\begin{array}{l}\frac{1}{2}W\times 250\times {10}^{-3}=\mathrm{ii.25}\times {10}^{\mathrm{three}}\\ \mathrm{Westward}=\frac{2.25\times {\mathrm{ten}}^3\times 2}{250\times {\mathrm{ten}}^{-\mathrm{iii}}}=\mathrm{xviii}gN\end{array}$

Thus the maximum load which can be carried by the spring is xviii kN.

Now the maximum stress is non to exceed 400 MN/mⁱⁱ; therefore from eqn. (12.one),

$\frac{\mathrm{ii}\mathrm{West}R}{\pi r^{\mathrm{three}}}=400\times {\mathrm{x}}^6$

Simply R = viiir∴

$\begin{array}{l}\frac{2\times 18\times {\mathrm{x}}^{\mathrm{three}}\times 8r}{\pi r^{\mathrm{iii}}}=400\times {10}^6\\ r^2=\frac{2\times 18\times {\mathrm{ten}}^3\times 8}{\pi \times 400\times {10}^6}=229\times {\mathrm{ten}}^{-6}\\ r=\mathrm{15.i}\times {\mathrm{x}}^{-3}=\mathrm{15.ane}m\mathrm{thou}\end{array}$

The required diameter of the wire, for practical convenience, is, therefore,

2 × 15 = xxx mm

and, since R = eightr, the required mean diameter of the coils is

8 × 30 = 240 mm

At present full deflection

$\begin{array}{l}\delta =\frac{4\mathrm{Due\; west}{R}^3\mathrm{northward}}{K{r}^4}=250m\mathrm{one\; thousand}\\ n=\frac{250\times {10}^{-\mathrm{iii}}\times 70\times {\mathrm{ten}}^{\mathrm{ix}}\times {\left(15\times {10}^{-3}\right)}^4}{4\times \mathrm{eighteen}\times {10}^{\mathrm{three}}\times {\left(120\times {\mathrm{x}}^{-\mathrm{iii}}\right)}^3}\\ =7.12\end{array}$

Once again from applied considerations, the number of consummate coils necessary = 7. (If 8 coils were chosen the maximum deflection would exceed 250 mm.)

The result of introducing Wahl correction factors is determined equally follows: From the given information C = D/d = 8 ∴ From Fig. 12.half dozen K = 1.184.

Now

$\begin{array}{l}{\tau }_{\max }=\mathrm{Chiliad}\left[\frac{\mathrm{viii}\mathrm{Due\; west}D}{\pi d^3}\right]=G\left[\frac{2\mathrm{Due\; west}R}{\pi r^3}\right]=400\times {10}^{\mathrm{six}}\\ \therefore 400\times {10}^6=\frac{1.184\times \mathrm{two}\times 18\times {\mathrm{ten}}^3\times 8r}{\pi r^3}\\ \therefore r^2=\frac{1.184\times \mathrm{two}\times 18\times {10}^3\times 8}{\pi \times 400\times {10}^6}=271.35\times {10}^{-6}\\ \therefore r=16.47\times {10}^{-3}=16.47mm\end{array}$

i.e. for practical convenience d = 2 × 16.5 = 33 mm, and since D = 8d, D = viii × 33 = 264 mm.

Total deflection

$\delta =\frac{4\mathrm{Westward}{R}^{\mathrm{iii}}\mathrm{northward}}{G{r}^4}=250\mathrm{thou}\mathrm{1000}$

$\begin{array}{l}\therefore n=\frac{250\times {10}^{-3}\times \mathrm{lxx}\times {10}^{\mathrm{nine}}\times {\left(16.5\times {10}^{-\mathrm{iii}}\right)}^4}{\mathrm{four}\times \mathrm{eighteen}\times {10}^3\times {\left(132\times {10}^{-\mathrm{iii}}\right)}^3}\\ =\mathrm{vii.83}\end{array}$

Although this is considerably greater than the value obtained before, the number of complete coils required remains at seven if maximum deflection is strictly limited to 250 mm.

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Mechanical Systems I

Nicolae Lobontiu , in Arrangement Dynamics for Technology Students, 2010

2.1.2 Spring Elements

Springs serving equally elastic supports for translatory and rotary motion are studied in this section in relation to their lumped stiffness (or leap abiding), denoted past k. Springs are mechanical elements that generate elastic forces in translatory move and elastic torques in rotary motion that oppose the spring deformation; these rubberband reactions are proportional to the leap deformation (linear or athwart displacement). Figure 2.4 sketches a helical bound and gives the stiffness equations respective to either axial (translatory) motility or to torsion-generated (rotary) motility. The parameters defining the helical leap are the radius R, the wire diameter d, the number of active turns n, and the textile shear and Immature'southward modulii, G and E (more details on these quantities are given in Appendix D).

Figure 2.iv. Helical Leap and Symbols for Translatory or Rotary Motion: Translatory Stiffness, k_t = Gd ^four/(64nR ⁱⁱⁱ); Rotary Stiffness, k_r = Ed ^four/(64nR).

For a spring whose terminate points undergo the displacements x ₁ and 10 ₂ (as shown in Figure 2.4), the rubberband forcefulness developed in the spring is proportional to the leap deformation, which is the difference betwixt the two end point displacements, and can be expressed as

(2.fourteen) $f_e\text{(}t\text{)}=\mathrm{thou}\text{Δ}\mathrm{10}\text{(}t\text{)}=k\left[x_{\text{one}}\text{(}t\text{)}-x_{\text{ii}}\text{(}t\text{)}\right]$

Similarly, an rubberband torque is generated past a spring in rotation whose cease points undergo the rotations θ_ane and θ₂, the elastic torque being expressed equally

(2.xv) $m_e\text{(}t\text{)}=g\text{Δθ(}t\text{)}=k[{\theta }_{\text{1}}\text{(}t\text{)}-{\theta }_{\text{2}}\text{(}t\text{)}]$

These equations assume the springs are linear; therefore, the stiffness is abiding. Stiffness equations are given in the companion website Affiliate 2 for other translatory and rotary springs. A spiral leap, for instance, such equally the ane sketched in Effigy 2.v and whose full length is fifty, is used in rotary move applications.

FIGURE 2.5. Screw Torsion Spring for Rotary Movement; Spring Stiffness Is k = πEd ⁴/(64l).

For a translatory spring, the elastic energy stored corresponding to a deformation Δx is

(2.16) $U_e=\frac{\text{1}}{\text{2}}\mathrm{thou}[\text{Δ}x{\text{(}t\text{)}]}^{\text{two}}$

Similarly, for a rotary spring, the elastic energy relative to an angular deformation Δθ is

(two.17) $U_e=\frac{\text{1}}{\text{ii}}k[\text{Δθ}{\text{(}t\text{)}]}^{\text{2}}$

Springs tin can exist combined in series or in parallel, as sketched in Figure 2.6, where information technology has been assumed that the serial and parallel chains are clamped at one terminate.

Effigy two.half dozen. Translatory Spring Combinations: (a) Series; (b) Parallel.

For serial springs, the force is the aforementioned in each component and is equal to f in Figure 2.6(a), whereas the total deformation is the sum of private deformations. Conversely, for parallel spring combinations, the displacements are identical for all springs, whereas the sum of individual spring forces equals the externally applied force f at equilibrium. The equivalent series stiffness thousand_s and the parallel stiffness k_p corresponding to north bound elements are derived in the companion website Chapter 2; their equations are

(2.18) $\{\begin{array}{l}\frac{\text{i}}{{\mathrm{thou}}_s}=\frac{\text{1}}{{\mathrm{one\; thousand}}_{\text{1}}}+\frac{\text{1}}{{\mathrm{thou}}_{\text{2}}}+\cdots +\frac{\text{ane}}{k_n}\hfill \\ k_p=k_{\text{1}}+k_{\text{2}}+\cdots +{\mathrm{one\; thousand}}_n\hfill \end{array}$

Case two.2

Four identical translatory helical springs are combined in 2 arrangements such that, in each of the two combinations, there are both series and parallel connections. When the same strength is applied separately to each spring arrangement at the free terminate (the other one beingness fixed), it is determined that the ratio of the complimentary-end displacements is 25/iv. Identify the two spring combinations and calculate the equivalent stiffness for each when d = 1 mm, R = 6 mm, n = 10, and G = 160 GPa.

Solution

The largest displacement is obtained when all four springs are coupled in series, considering the stiffness is minimal, run into first Eq. (ii.eighteen). Conversely, the smallest deportation corresponds to a full parallel jump connection when the stiffness is maximal, every bit indicated by the second Eq. (two.18). Yet, these connections are not allowed in this example. 2 combinations are sketched in Effigy 2.7, which are candidates satisfying this example's requirement to mix series and parallel pairs.

Figure ii.7. Translatory Spring Combinations: (a) Three Parallel Branches; (b) Iii Series Branches.

The equivalent stiffnesses of the leap connections shown in Figure two.vii are

(2.19) $\{\begin{array}{l}{\mathrm{1000}}_{\text{one}}=\frac{\mathrm{thou}}{\text{2}}+k+\mathrm{one\; thousand}+\frac{\text{5}}{\text{2}}\mathrm{thou}\hfill \\ \frac{\text{ane}}{m_{\text{2}}}=\frac{\text{1}}{\mathrm{thou}}+\frac{\text{1}}{\mathrm{thousand}}+\frac{\text{1}}{\text{ii}\mathrm{one\; thousand}}=\frac{\text{5}}{\text{2}\mathrm{chiliad}}\hfill \end{array}$

Considering f = 1000 _one x ₁ = k _two x _ii, Eqs. (2.xix) yield

(2.twenty) $\frac{k_{\mathrm{i}}}{k_2}=\frac{{\mathrm{ten}}_2}{{\mathrm{ten}}_{\mathrm{one}}}=\frac{25}{\mathrm{four}}$

which is indeed the displacement ratio given in this instance; as a event, the spring arrangements are the ones shown in Figure 2.7. By using the equation of a translatory helical leap, as given in the explanation of Figure two.4, and the specified numerical values, the following results are obtained: k = 1157.4 Northward/m, grand _one = fivethousand/2 = 2893.5 N/grand, and g ₂ = 2grand/5 = 462.96 North/chiliad.

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Loftier Force per unit area Vessels

Dennis R. Moss , Michael Basic , in Pressure level Vessel Design Manual (Fourth Edition), 2013

Helicoflex

Helicoflex o-rings are metal gaskets with a helical spring on the inside of the o-ring. This is a variation of the other types of o-rings based upon the plastic deformation of a jacket of greater ductility than the flange material. The close wound helical spring is selected to have a specific compression resistance. During compression, the pressure level forces the jacket to yield and fill the flange imperfections. Each coil of the helical spring acts independently and allows the seal to adapt to flange imperfections. These o-rings can be used from cryogenic applications upwardly to 1800°F and pressures to 50,000 PSI and higher for special applications.

Jacket materials can exist made of aluminum, silver, copper, soft iron, balmy steel, nickel, monel, tantalum, stainless steel, inconel or titanium.

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Assembly

Corrado Poli , in Pattern for Manufacturing, 2001

Tangling and Nesting

Parts that nest or tangle (vending cups, helical springs, etc.; run into Figure 12.6) are difficult to grasp singly and dispense with one hand; they nowadays obvious handling difficulties. In add-on, parts that are gluey (a part coated with grease or an adhesive), sharp (razor blade), delicate (glass), glace (ball bearing coated with lite oil), or flexible (belts, gaskets, etc.; see Figure 12.7) are as well difficult to grasp and manipulate with one hand, and thus should besides exist avoided by designers when possible.

Figure 12.6. Some examples of parts that tangle.

FIGURE 12.seven. An example of flexible parts where two easily are needed to maintain orientation prior to insertion.

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The valve spring

Hiroshi Yamagata , in The Science and Technology of Materials in Automotive Engines, 2005

vii.i Functions

Figure 7.one shows a valve spring. The valve spring is a helical spring used to close the poppet valve and maintain an air-tight seal past forcing the valve to the valve seat. A spring accumulates kinetic energy during wrinkle and the energy is prodigal upon expansion. In that location are many types, shapes and sizes of steel springs.

Fig 7.1. Valve spring. More often than not, coil springs of a wire diameter below five   mm ϕ are cold-formed at room temperature, while wires above 11   mm are normally hot-formed. Compression valve springs are provided with the ends plain and ground.

The valve train consists mainly of valves, valve springs and camshafts. At low camshaft revolutions, the valve jump tin follow the valve lift hands so that the valve moves regularly. Past contrast, at high revolutions, information technology is more than difficult for the valve and valve spring to follow the cam. Valve float is the term given to unwanted movements of the valve and valve spring due to their inertial weights. To avert this, the load of the valve spring should be fix high. The load applied at the longest length is called the set load, and the valve leap is e'er set to accept a loftier compressive stress above gear up load weather.Figure 7.ii shows double springs, which are used to raise the prepare load while minimizing the increase in summit.

Fig vii.ii. Double springs installed in a bucket blazon valve lifter.

Some other resistance phenomenon that occurs at high revolutions is surging, due to resonance. Surging occurs when each turn of the coil spring vibrates upwards and down at high frequency, independently of the motion of the entire leap. Information technology takes place when the natural frequency of the valve spring coincides with the particular rotational speed of the engine. Generally, surging occurs at high revolutions, and the surging stress generated is superimposed on the normal stress. The full stress is likely to exceed the allowable fatigue limit of the leap material and tin intermission the spring. A variable pitch spring reduces the gamble of surging. This spring has two portions along the length, a roughly coiled portion and a densely coiled portion, which ensures that the natural frequency of the jump is not constant and therefore not susceptible to resonance.

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VIBRATION Transmission

Due south.I. Hayek , in Encyclopedia of Vibration, 2001

Ii Semi-Space Beams Joined past a Hinge and Helical Jump

Consider two semi-infinite beams (plates) joined by a hinge and connected through a helical spring of rotational stiffness γ at x=0. Here the displacements, moments and shears are continuous at x=0 and the moment is given by the helical spring stiffness times the difference of the slopes of the two beams at x=0:

[24] ${\alpha }_{\text{nr}}=-\frac{\mathrm{i}+\text{i}}{1-\text{i}\left(1+4\overline{m}\overline{\alpha }\right)}$

[25] ${\alpha }_{\text{nt}}=-\frac{2\text{i}\left(1+\mathrm{two}\overline{k}\overline{\alpha }\right)}{\mathrm{ane}-\text{i}\left(1+4\overline{k}\overline{\alpha }\right)}$

[26] ${\alpha }_{\text{fr}}=-\frac{1}{1-\text{i}\left(1+\mathrm{iv}\overline{\mathrm{grand}}\overline{\alpha }\right)}$

[27] ${\alpha }_{\text{ft}}=-\frac{\text{i}\left(1+4\overline{k}\overline{\alpha }\right)}{\mathrm{i}-\text{i}\left(\mathrm{one}+4\overline{k}\overline{\alpha }\right)}$

[28] ${\tau }_{\text{r}}=-\frac{1}{1+{\left(1+4\overline{\mathrm{one\; thousand}}\overline{\alpha }\right)}^{\mathrm{ii}}}$

[29] ${\tau }_{\text{t}}=\frac{{\left(1+\mathrm{iv}\overline{k}\overline{\alpha }\right)}^2}{\mathrm{ane}+{\left(\mathrm{one}+4\overline{\mathrm{one\; thousand}}\overline{\alpha }\right)}^2}$

Plots of $\left|{\alpha }_{\text{fr}}\right|,\left|{\alpha }_{\text{ft}}\right|,\left|{\alpha }_{\text{nr}}\right|,\text{and}\left|{\alpha }_{\text{nt}}\right|$ in dB calibration are shown in Figures 1–4, respectively, vs non-dimensional moving ridge number g for diverse values of the nondimensional helical spring stiffness $\overline{\alpha }=0.1,1,10,\text{and}100$ .

Effigy 1. Far-field reflection coefficient for swivel with helical jump.

Figure 2. Far-field transmission coefficient for hinge with helical spring.

Figure 3. Near-field reflection coefficient for hinge with helical jump.

Figure 4. Near-field transmission coefficient for hinge with helical bound.

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Car Components

DAN B. MARGHITU , ... NICOLAE CRACIUNOIU , in Mechanical Engineer's Handbook, 2001

3.4.three DEFLECTION, δ

The deflection-strength relations are obtained using Castigliano'southward theorem. The total strain energy for a helical leap is

(three.13) $U=U_t+U_s=\frac{T^{\mathrm{ii}}l}{2GJ}+\frac{F^{\mathrm{two}}\mathrm{50}}{2AG},$

where

(3.14) $U_t=\frac{T^{\mathrm{ii}}\mathrm{50}}{2GJ}$

is the torsional component of the energy, and

(3.15) $U_s=\frac{F^{\mathrm{two}}l}{2A\mathrm{1000}}$

is the shear component of the energy. The spring load is F, the torsion torque is T, the length of the wire is l, the 2nd moment of inertia is J, the cross-exclusive surface area of the wire is A, and the modulus of rigidity is G.

Substituting T = FD/two, l = πDN, J = πd⁴/32, and A = πd ² /4 in Eq. (three.13), one may obtain the full strain free energy as

(3.16) $U=\frac{\mathrm{iv}{F}^2{D}^3\mathrm{Due\; north}}{d^{4}G}+\frac{2F^{2}D\mathrm{North}}{d^2\mathrm{Chiliad}},$

where N = N_a is the number of active coils.

When Castigliano's theorem is applied, the deflection of the helical jump is

(3.17) $\delta =\frac{\partial U}{\partial F}=\frac{8F{D}^{3}N}{d^{\mathrm{iv}}G}+\frac{4FDN}{d^2\mathrm{Thousand}}.$

If we use the bound index C = D/d, the deflection becomes

(three.18) $\delta =\frac{8F{D}^{3}N}{d^{4}G}\left(1+\frac{\mathrm{one}}{\mathrm{two}{C}^2}\right)\approx \frac{8F{D}^{\mathrm{three}}N}{d^{4}G}.$

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Everything You Need to Know About Helical Springs

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