Tài liệu Handbook of Mechanical Engineering Calculations P23 pdf

SECTION 23 SPRING SELECTION ANDDetermining Spring Dimensions Before and After Load Applications 23.5 Analysis and Design of Flat Metal Springs 23.6 Sizing Torsional Leaf Springs 23.7 Des

SECTION 23 SPRING SELECTION AND

Determining Spring Dimensions Before

and After Load Applications 23.5

Analysis and Design of Flat Metal

Springs 23.6

Sizing Torsional Leaf Springs 23.7

Designing a Spring to Fire a Projectile

Spring Wire Length and Weight 23.15

Helical Compression and Tension Spring

Analysis 23.16

Selection of Helical Compression and

Tension Springs 23.17

Sizing Helical Springs for Optimum

Dimensions and Weight 23.19

Selection of Square- and Wire Helical Springs 23.20

Rectangular-Curved Spring Design Analysis 23.21 Round- and Square-Wire Helical Torsion- Spring Selection 23.23

Torsion-Bar Spring Analysis 23.26 Multirate Helical Spring Analysis 23.27 Belleville Spring Analysis for Smallest Diameter 23.28

Belleville Spring Computations for Disk Deflection, Load, and Number 23.30 Ring-Spring Design Analysis 23.32 Liquid-Spring Selection 23.34 Selection of Air-Snubber Dashpot Dimensions 23.37

Design Analysis of Flat Plastic Springs 23.39

Reinforced-Life of Cyclically Loaded Mechanical Springs 23.42

Shock-Mount Deflection and Spring Rate 23.43

PROPORTIONING HELICAL SPRINGS BY

MINIMUM WEIGHT

The detent helical spring in Fig 1 is to be designed so that force P1for the extendedcondition is to be 20 lb (88.9 N) After an additional compression of 0.625 in (1.59cm) the shearing stress in the spring is to be 75,000 lb / in2(517 MPa) The spring

index, c, is approximately 8, based on past experience, and the modulus of elasticity

in shear is 11,500,000 lb / in2(79.2 GPa) Find the diameter of the wire, helix radius,and number of active coils for the smallest amount of material in the spring

Calculation Procedure:

1. Find the diameter of the spring wire

A spring which will contain the smallest possible amount of material will havedesign parameters selected so that the maximum force, stress, and deflection are

FIGURE 1 Spring-loaded

de-tent (Product Engineering.)

exactly twice the minimum force, stress, and deflection, respectively With suchparameters, the spring wire diameter,

16P c1

d⫽冪␲(S ) s max where d⫽spring wire diameter, in (cm); P1⫽force in the extended condition, lb

(N); c ⫽ spring index, dimensionless; ␲ ⫽ 3.1416; (S s)max ⫽ shear stress in thespring wire, lb / in2(kPa) Substituting, d⫽(16⫻20⫻8 / 75,000␲)0.5⫽0.1042 in

(0.265 cm) Referring to a spring wire table, choose No 12 wire with d⫽0.1055

in (0.268 cm) as the nearest commercially available wire size

2. Compute the mean radius of the spring helix

The helix mean radius,

3. Determine the number of active coils in the spring

Use the relation, N⫽number of active coils in the spring,

(␦ ⫺ ␦2 1) dG

N⫽ 2␲R (S )2 s maxwhere,␦2⫽maximum compression of spring, in (cm);␦1⫽minimum compression

of the spring, in (cm); G⫽modulus of elasticity in shear; other symbols as before

Substituting, N⫽(0.625⫻0.1055⫻11,500,000) / (2␲ ⫻0.43232⫻75,000⫽8.61active coils

4. Find the volume of the spring

Volume of a helical spring is given by V ⫽ 0.5 ⫻ ␲2 ⫻ d2 ⫻ RN, where the symbols are as given earlier Substituting, V ⫽ 0.5⫻ 3.142 ⫻ 0.10552 ⫻0.4323

⫻8.61⫽ 2.044 in3(30.0335 dm3)

Related Calculations. Because of the many variables involved, the ing of a helical spring is usually done by trial and error An unlimited number ofsprings can be obtained that will satisfy a given set of equations for stress anddeflection However, all springs found this way will contain more material thannecessary unless the conditions for the mathematical minimum are fulfilled.The functioning of a helical spring usually occurs when the spring is extended,and is least able to exert a force For example, in the indexing mechanism shown

proportion-in Fig 1, when the ball is proportion-in the detent, the holdproportion-ing power of the mechanismdepends on the force exerted by the spring After relative motion between the partsoccurs, the ball is out of the detent and the additional force from the increasedcompression serves no useful purpose and may be a disadvantage In this conditionthe shearing stress in the wire is usually the governing factor

Many similar applications arise in which the design is controlled by (a) the spring force in the extended condition and (b) the shearing stress for the most

compressed condition It has been proved for this instance, that the spring willcontain the smallest possible amount of material when the design parameters areselected so that the maximum force, stress, and deflection are exactly twice theminimum force, stress, and deflection, respectively

Although not directly specified, the designer can usually estimate a suitable value

for the spring index, c Taking the maximum load as twice the minimum load, the wire-diameter, d, is computed using the equation given in step 1, above Next the helix radius, R, is computed using the d value from step 1 Then the number of

active coils is computed, step 3 Lastly, the minimum volume of the spring isdetermined, as in step 4

This procedure is the work of M F Spotts, Northwestern Technological Institute,

as reported in Product Engineering magazine.

DETERMINING SAFE TORSIONAL STRESS FOR A

HELICAL SPRING

The load on a helical spring is 1600 lb (726.4 kg) and the corresponding deflection

is to be 4 in (10.2 cm) Rigidity modulus is 11⫻ 106lb / in2 (75.8 GPa) and themaximum intensity of safe torsional stress is 60,000 lb / in2(413.4 MPa) Designthe spring for the total number of turns if the wire is circular in cross section with

a diameter of 0.625 in (1.5875 cm) and a centerline radius of 1.5 in (3.81 cm)

Calculation Procedure:

1. Find the number of active coils in the spring

Use the relation y⫽N(64)( P)(r3) / G(d4), where y⫽total deflection of the spring,

in (cm); N⫽ number of active coils in the spring; P⫽ load on spring, lb (N); r

⫽radius as axis to centerline of wires, in (cm); G⫽modulus of elasticity of springmaterial in shear, lb / in2(kPa); d ⫽spring wire diameter, in (cm) Solving for the

number of active coils, N ⫽ y(G)(d4) / 64( P)(r3) Or N ⫽ 4(11 ⫻ 116)(0.6254) /64(1600)(1.53)⫽19.4 active coils

2. Check for the safe limit of torsional stress in the spring

Use the relation S, ⫽ 16( P)(r) /␲(d3) Solving, S⫽ 16(1600)(1.5) /␲(0.6253) ⫽

50,066 lb / in2(344.95 MPa) This torsional stress is less than the safe limit of 60,000

lb / in2(413.4 MPa) chosen for this spring Hence, the spring is acceptable

The total number spring turns for two inactive coils (one at each end) is N⫹2

when under load when G⫽12⫻106lb / in2(82.7 GPa) and mean spring diameter(outside diameter⫺ wire diameter) is to be 0.779 in (1.98 cm)

Calculation Procedure:

1. Find the safe shearing stress in the spring

Use the relation, S t⫽ 8PD /␲d3, where S⫽ fiber stress in shear, lb / in2(kPa); P

⫽axial load on spring, lb (N); D⫽mean diameter of spring as defined above, in

(cm); d ⫽diameter of spring wire, in (cm) Substituting, S⫽9(140)(0.779)(2) /␲

(0.1923)⫽78,475 lb / in2(540.7 MPa)

2. Determine the deflection of the spring

Use the relation, y⫽4(␲)(N )(r2)(S) / Gd, where y⫽total deflection, in (cm); N⫽

number of active coils; G ⫽rigidity modulus of the spring material in shear, lb /

in2 (kPa); other symbols as defined earlier Substituting, y ⫽ 4(␲)(7)(0.7792)(78,475) / 12⫻ 106(0.192)⫽1.818 in (4.6 cm), or about 113⁄16in

3. Compute the total number of coils in the spring

For the ends of a spring to be closed and ground smooth, 1.5 coils should be taken

as inactive In compression springs the number of active coils depends on the style

of ends, as follows: open ends, not ground—all coils are active; open ends, ground—0.5 coil inactive; closed ends, not ground—1 coil inactive; closed ends, ground—1.5 coils inactive; squared ends, ground—2 coils inactive Since this

spring is to have ends that are closed and round; i.e., closed and ground, use 1.5inactive coils Hence, the total number of coils will be 7⫹1.5⫽8.5, say 9 coils

4. Find the length of the spring when under load

The solid height of the spring when it is entirely compressed is 9 coils ⫻ 0.192

DETERMINING SPRING DIMENSIONS BEFORE

AND AFTER LOAD APPLICATION

A cylindrical helical spring of circular cross-section wire is to be designed to safelycarry an axial compressive load of 1200 lb (5338 N) at a maximum stress of110,000 lb / in2(757.9 MPa) The spring is to have a deflection scale of approxi-mately 150 lb / in (262 N / cm) Spring proportions are to be: Mean diameter ofcoil / diameter of wire⫽6 to 8; spring length when closed / mean diameter of coil

⫽1.7 to 2.3 Determine (a) mean diameter of coil; (b) diameter of wire; (c) length

of coil when closed, and (d ) length of coil before load application Use G⫽rigiditymodulus of steel in shear⫽11.5 ⫻106(79,235 MPa)

Calculation Procedure:

1. Find the trial wire size

In this trial design we can assume the ratio of D / d to be 6:8, as given Then d⫽

D / 7, using the midpoint ratio, where D⫽mean diameter of spring coil / diameter

of spring wire Use the relation, D⫽S(␲)(d3) / 8( P), where S⫽spring stress, lb /

in2(kPa); D⫽mean diameter of the spring ⫽outside diameter ⫺wire diameter,

in (cm); d ⫽ diameter of spring wire, in (cm) Substituting, D ⫽

(110,000)(3.1416)[(D / 7)3] / 8(1200); D⫽4.45 in (2.02 cm)

The wire diameter is d⫽4.45 / 7⫽0.635 in (1.6 cm) The nearest wire standard

wire size is 0.6666 in (1.69 cm) Solving for D / d⫽4.45 / 0.6666⫽6.68 This lies

within the limits of 6 to 8 set for this spring for D / d.

2. Determine the spring scale

Use the relation P / y⫽ G(d4) / N(64)(r3), where y⫽ total deflection of the spring,

in (cm); N ⫽ number of coils; other symbols as before For this spring, P / y ⫽

150 Substituting, 150 ⫽ (11.5 ⫻ 106)(0.6254) / N(64)(2.383) Solving, N ⫽ 13.6active coils

Total turns for this spring, assuming closed ends ground⫽13.6⫹1.5⫽15.1

3. Evaluate the length closed/mean diameter ratio

This ratio must be between 1.7 and 2.3 if the spring is to meet its design

require-ments Using the relation, (d )(total turns) / D, we have (15.5)(0.6666) / 4.45⫽2.26,which is within the required limits Then, the length of the closed coil ⫽15.1 ⫻

0.6666⫽10.065 in (25.56 cm)

4. Find the coil length before load application

To find the length of the spring coil before the load is applied we add the closedlength of the coil to the total deflection of the spring To find the total deflection,use the relation, y ⫽ N(64)( P)(2.383) / 11.5 ⫻ 106(0.66664); or y ⫽

13.6(64)(1200)(2.383) / 11.5 ⫻ 106(0.66664) ⫽ 6.2 in(15.75 cm) Then, the length

of the coil before load application⫽ 10⫹6.2⫽16.2 in (41.1 cm)

Related Calculations. The steps in this procedure are useful when designing

a spring to meet certain preset requirements Using the given steps you can develop

a spring meeting any of several dimensional or stress parameters

ANALYSIS AND DESIGN OF FLAT METAL

SPRINGS

Determine the width and thickness of the leaves of a six-leaf steel cantilever spring13-in (33-cm) long to carry a load of 375 lb (1668 N) with a deflection of 1.25 in(3.2 cm) The maximum stress allowed in this spring is 50,000 lb / in2(344.5 MPa);the modulus of elasticity of the steel spring material is 30⫻ 106lb / in2(206,700MPa)

Calculation Procedure:

1. Find the leaf thickness for this steel spring

See the list of equations below Knowing the deflection, we can use the equation,

F⫽S(l2) / E(t ), where F⫽deflection, in (cm); S⫽safe tensile stress in the springmetal, lb / in2(kPa); l⫽spring length, in (cm); E⫽modulus of elasticity as given

above; t ⫽ spring leaf thickness in (cm) substituting, 1.25 ⫽ 50,000(132) / 30 ⫻

106(t ); t⫽0.225 in (0.57 cm)

W⫽save load or pull, lb (N)

F⫽deflection at point of application, in (cm)

S⫽safe tensile stress of material, lb / in2(kPa)

E⫽modulus of elasticity, 30 ⫻106for steel (kPa)

Other symbols as shown above

2. Determine the width of each leaf in the spring

Use the relation, W ⫽ S(N )(b)(t2) / 6( l ), where the symbols are as defined above and W⫽safe load or pull, lb (N) Substituting, 375⫽ 50,000(6)(b)(.2252) / 6(13);

b⫽1.93 in (4.9 cm)

Related Calculations. Use the spring layout and equations above to design any

of the three types of metal springs shown Flat metal springs find wide use in avariety of mechanical applications The three types shown above—flat parallel, flattriangular, and flat leaf—are the most popular today Using the data given here,engineers can design a multitude of flat springs for many different uses

SIZING TORSIONAL LEAF SPRINGS

Size a torsional leaf spring, Fig 2, for a 38-in⫻38-in (96.5-cm⫻96.5-cm) workplatform weighing 145 lb (65.8 kg) which requires an assist spring to reduce thelifting force that must be exerted by a human being A summation of the momentsabout a line through both platform pivots, Fig 3, shows that a lifting force of only

23 lb (102.3 N) is needed if a torsional spring provides an assist moment of M t⫽

1880 lb䡠in (212.4 N䡠m) An assist moment is required to raise the platform fromeither the stowed or the operating position As the platform is rotated from oneposition to the other, the assist spring undergoes a load reversal The required springwind-up angle,␪ ⫽ ␲/ 2 radians; the spring length⫽36 in (91.4 cm) AISI 4130alloy steel heat-treated to 31 Rc minimum with an allowable shear stress of 68,000

lb / in2(468.5 MPa) and a shear modulus of G⫽ 11.5⫻106lb / in2(79,235 MPa)

is to be used for this spring Size the spring, determining the required dimensionsfor blade thickness, number of blades, aspect ratio, and blade width

Calculation Procedure:

1. Estimate the blade thickness

Use the equation

b ␶

␤ ⫽ ⫽ ,

l 2G␪

FIGURE 2 Torsional leaf springs normally are twisted to a specific wind-up angle This

two-blade spring is wound at right angles (Machine Design.)

FIGURE 3 Work platform requires an assist spring to reduce required lifting force (Machine Design.)

where the symbols are as given below Solving for the blade thickness using the

given data, b⫽(68.000)(36) / 2(11.5⫻106)(␲/ 2)⫽0.068 in (0.172 cm) Note that

this is the blade half-thickness.

To keep manufacturing costs low, a stock plate thickness would be chosen The

nearest stock plate thickness, 2bƒ, is 0.125 in (0.3175 cm) So bƒ⫽ 0.125 / 2 ⫽

0.0625 in (0.159 cm), which is less than b⫽ 0.068 in (0.172 cm) the computedrequired blade half-thickness

Nomenclature

a⫽ blade half-width, in (cm)

aƒ⫽ selected blade half-width, in (cm)

a r⫽ required blade half-width, in (cm)

b⫽ blade half-thickness, in (cm)

bƒ⫽ selected blade half-thickness, in (cm)

F⫽ torsional stress factor, 1 / in3(1 / cm3)

ƒ⫽ normalized torsional stress factor

G⫽ shear modulus, lb/in 2 (kPa) R0 ⫽ estimated aspect ratio

K⫽ torsional stiffness factor, in 4 (cm 4 ) R r⫽ required aspect ratio

k⫽ normalized torsional stiffness s⫽ normalized shear stress

l⫽ spring length, in (cm) ␤ ⫽ normalized half-thickness

M t⫽ applied moment, lb 䡠 in (N 䡠 m) ␪ ⫽ spring wind-up angle, rad

m t⫽ normalized applied moment ␶ ⫽ torsional stress, lb/in 2 (kPa)

n⫽ number of blades ␶a⫽ allowable torsional stress, lb/in 2 (kPa)

n f⫽ selected number of blades

R⫽ aspect ratio

2. Select the number of blades for this torsion leaf spring

Use the equation,

3. Choose the aspect ratio

The required aspect ratio, R r, is established by employing the computed value for

R0in step 2 as an initial estimate in the equation

l M t

冉 冊冉 冊G ␪

R i⫹1⫽

16 3.36 14

n bƒ ƒ冋3 ⫺ R i 冉1⫺12R4i冊册

where i⫽0, 1, 2,

Usually, after three or so iterations, computed values R i⫹1converge to a single value

which is the required aspect ratio, R r Substituting, using R0⫽9.2, as computed in

4. Determine the blade width

Given R r⫽9.83 and 2bƒ⫽0.125 in (0.3175 cm), the required blade width, 2a, is found from 2a r ⫽ R r (2bƒ) Or, 2a r ⫽ (9.83)(0.125) ⫽ 0.123 in (0.31 cm) Formanufacturing simplicity, the blades will be cut to a 1.25-in (3.18-cm) width from

stock 0.125-in (3.18-cm) thick plates Therefore, let 2aƒ⫽1.25 in (3.18 cm)

5. Check the assist moment and stress

The assist spring has been sized to have five 36-in (91.4-cm)-long blades with a0.125-in (3.18-cm) wide by 0.125-in (3.18-cm) thick cross section The assist mo-ment provided by such a spring stack is determined from

65(11.5⫻10 ) 3

be less than 23 lb (102.3 N)

6. Find the shear stress in each blade

Use the equation,

Related Calculations. Leaf springs, in most applications where they transmitforces, are subjected primarily to bending loads Design of such springs normally

is based upon cantilever beam theory, which has been well-documented in theliterature on structures and machine elements

It is not widely known that leaf springs also can be designed to twist When soused, leaf springs can generate high torques while occupying a small space Suchtorsional leaf springs are most commonly used as assist devices on hatches, doors,and folding platforms or walkways Typical applications are on a variety of vehicles

in which space is restricted The procedure presented here applies to blades havingidentical dimensions, the usual configuration

A manual available from the Society of Automotive Engineers provides designtechniques for leaf springs for vehicular applications The SAE manual describes adesign procedure for leaf springs consisting of blades with either the same or dif-ferent widths and thicknesses The procedure given here is said, by its developer(see below), to be a more straightforward method

Stresses in torsional leaf springs can be reduced significantly by increasing eitherthe number of blades, the aspect ratio, or both Single torsional springs made ofsquare, round, or rectangular bar materials have higher stresses than do torsionalleaf springs, given the same assist moment and angular deflection

SI Values

8 in (20.3 cm)

26 in (66 cm)

FIGURE 4 Spring for firing a projectile.

According to Timoshenko and Goddier, a local irregularity in the stress bution occurs near the attachment points when the aspect ratio of the spring islarge Because no specific information is generally available, experience and goodjudgment must be used to determine how large the aspect ratio can be before localattachment stresses become significant Generally, the aspect ratio should be greaterthan 4 but less than 15 Also, the number of blades should be limited so that thespring stack height does not exceed the spring width

distri-This procedure is the work of Russel Lilliston, Staff Engineer, Martin Marietta

Aerospace, as reported in Machine Design magazine.

DESIGNING A SPRING TO FIRE A PROJECTILE

A 22-coil squared and ground spring is designed to fire a 10-lb (4.54-kg) projectileinto the air The spring has a 6-in (15.24-cm) diameter coil with 0.75-in (1.9-cm)diameter wire Free length of the spring, Fig 4, is 26 in (66 cm); it is compressed

to 8 in when loaded or set The shear elastic limit for the spring material is 85,000lb/in2 (585.65 MPa) The spring constant Wahl factor is 1.18 Shear modulus ofelasticity of the spring material is 12⫻106lb / in2(82,680 MPa) Determine (a) the height to which the projectile will be fired; (b) the safety factor for this spring.

Calculation Procedure:

1. Find the actual stress in the spring

Use the relation, actual stress, S a⫽ (L c )(G)(d )(W ) / (4␲)(n)(r2), where S c⫽actualstress, lb / in2(kPa); L⫽compressed length of spring, in (cm); G⫽shear modulus

of elasticity, lb / in2(kPa); d ⫽spring-wire diameter, in (cm); W⫽Wahl factor; n

⫽ number of coils; r ⫽ spring radius ⫽ diameter / 2, in (cm) Substituting, S c ⫽

(8)(12⫻106)(0.75)(1.18) / (4␲)(22)(32)⫽ 34,146 lb / in2(235.2 MPa)

2. Compute the force to which the spring is stressed when loaded

Use the relation, P⫽0.1963(d3)(S c ) / (r)(W ), where P⫽loaded force, lb (N); other

symbols as given earlier Substituting, P ⫽ (0.1963)(0.753)(34,146) / (3)(1.18) ⫽

798.8 lb (3553 N)

3. Calculate the distance the projectile will be fired

Use the relation (weight, lb)(distance fired, s) ⫽ ( P)(deflection) / 2 Or s ⫽

(798.8)(8) / (2)(10)⫽319.5 in (811.6 cm)

4. Determine the factor of safety for the spring

The factor of safety⫽ (stress at elastic limit) / (actual stress)⫽85,000 / 34,146 ⫽

Calculation Procedure:

1. Set up the frequency and amplitude equations for this machine

The natural frequency of free vibration of this system is given by

1 kg

ƒ⫽2␲冪W where k⫽spring modulus, lb / in (kg / cm); g ⫽gravitational constant, in / s2(cm /

s2); W⫽weight of each spring, lb (kg)

The amplitude of the forced vibration of this machine, A, is given by

2. Set up the magnification factor

The ratio of ƒ1 / ƒis called the frequency ratio, and the ratio of 1 / (1⫺[ ƒ ]1/ƒ2) may

be interpreted as the magnification factor Now, because of proportionality factors,

For the reduction in the disturbing force which we seek, the ratio A / x sis ative Solving for ƒ in the above equation gives ƒ⫽4.339 cps.

neg-3. Find the spring modulus

Substitute the value found for ƒ in the first equation above for natural frequencyand we find ƒ⫽4.339⫽(1 / 6.28)(k⫻386.4 / [4000 / 6] ); k0.5 ⫽128.1 lb / in (2243

N / cm)

Related Calculations. Isolating the vibration caused by machinery is an portant design challenge, especially in heavily populated structures Springs arewidely used to isolate the vibrations produced by reciprocating engines and com-pressors, pumps, presses, and similar machinery This general procedure gives auseful approach to analyzing springs for such applications

im-SPRING SELECTION FOR A KNOWN LOAD AND

DEFLECTION

Give the steps in choosing a spring for a known load and an allowable deflection.Show how the type and size of spring are determined

Calculation Procedure:

1. Determine the load that must be handled

A spring may be required to absorb the force produced by a falling load or therecoil of a mass, to mitigate a mechanical shock load, to apply a force or torque,

to isolate vibration, to support moving masses, or to indicate or control a load ortorque Analyze the load to determine the magnitude of the force that is acting andthe distance through which it acts

Once the magnitude of the force is known, determine how it might beabsorbed—by compression or extension (tension) of a spring In some applications,either compression or extension of the spring is acceptable

2. Determine the distance through which the load acts

The load member usually moves when it applies a force to the spring This ment can be in a vertical, horizontal, or angular direction, or it may be a rotation

move-With the first type of movement, a compression, or tension, spring is generally chosen With a torsional movement, a torsion-type spring is usually selected Note

that the movement in either case may be negligible (i.e., the spring applies a largerestraining force), or the movement may be large, with the spring exerting only anominal force compared with the load

3. Make a tentative choice of spring type

Refer to Table 1, entering at the type of load Based on the information knownabout the load, make a tentative choice of the type of spring to use

4. Compute the spring size and stress

Use the methods given in the following calculation procedures to determine thespring dimensions, stress, and deflection

TABLE 1 Metal Spring Selection Guide

5. Check the suitability of the spring

Determine (a) whether the spring will fit in the allowable space, (b) the probable spring life, (c) the spring cost, and (d ) the spring reliability Based on these findings,

use the spring chosen, if it is satisfactory If the spring is unsatisfactory, chooseanother type of spring from Table 1 and repeat the study

SPRING WIRE LENGTH AND WEIGHT

How long a wire is needed to make a helical spring having a mean coil diameter

of 0.820 in (20.8 mm) if there are five coils in the spring? What will this springweigh if it is made of oil-tempered spring steel 0.055 in (1.40 mm) in diameter?

Calculation Procedure:

1. Compute the spring wire length

Find the spring length from l ⫽␲nd m , where l⫽wire length, in; n⫽number of

coils in the spring, in Thus, for this spring, l⫽␲(5)(0.820)⫽12.9 in (32.8 cm)

2. Compute the weight of the spring

Find the spring weight from w ⫽ 0.224ld2, where w ⫽ spring weight, lb; d ⫽

spring wire diameter, in For this spring, w ⫽ 0.224(12.9)(0.055)2 ⫽ 0.0087 lb(0.0387 kg)

Related Calculations. The weight equation in step 2 is valid for springs made

of oil-tempered steel, chrome vanadium steel, silica-manganese steel, and chromium steel For stainless steels, use a constant of 0.228, in place of 0.224, inthe equation The relation given in this procedure is valid for any spring having acontinuous coil—helical, spiral, etc Where a number of springs are to be made,simply multiply the length and weight of each by the number to be made to de-termine the total wire length required and the weight of the wire

silicon-HELICAL COMPRESSION AND TENSION SPRING

ANALYSIS

Determine the dimensions of a helical compression spring to carry a 5000-lb(22,241.1-N) load if it is made of hard-drawn steel wire having an allowable shearstress of 65,000 lb / in2 (448,175.0 kPa) The spring must fit in a 2-in (5.1-cm)diameter hole What is the deflection of the spring? The spring operates at atmos-pheric temperature, and the shear modulus of elasticity is 5 ⫻106lb / in2(34.5⫻

109Pa)

Calculation Procedure:

1. Choose the tentative dimensions of the spring

Since the spring must fit inside a 2-in (5.1-cm) diameter hole, the mean diameter

of the coil should not exceed about 1.75 in (4.5 cm) Use this as a trial mean

diameter, and compute the wire diameter from d ⫽ [8Ld m k/(␲s s)]1 / 3, where d ⫽

spring wire diameter, in; L⫽load on spring, lb; D m⫽mean diameter of coil, in;

k⫽spring curvature correction factor⫽(4c⫺1) / (4c⫺4)⫹0.615 / c for heavily coiled springs, where c⫽2r m / d⫽d m / d; s s⫽allowable shear stress material, lb /

in2 For lightly coiled springs, k⫽1.0 Thus, d⫽[8⫻5000⫻1.75⫻1.0 / (␲ ⫻

65,000)]1 / 3⫽0.70 in (1.8 cm) So the outside diameter d o of the spring will be d o

⫽d m⫹2(d / 2)⫽d m⫹d⫽1.75⫹0.70⫽2.45 in (6.2 cm) But the spring must

fit a 2-in (5.1-cm) diameter hole Hence, a smaller value of d mmust be tried

Using d m⫽ 1.5 in (3.8 cm) and following the same procedure, we find d⫽[8

Thus, c ⫽2r m / d ⫽2(1.25 / 2) / 0.625⫽ 2.0 Note that r m⫽ d m/ 2⫽ 1.25 / 2 in

this calculation for the value of c Then k ⫽ [(4 ⫻ 2) ⫺ 1] / [(4 ⫻ 2) ⫺ 4] ⫹

0.615 / 2⫽ 2.0575 Hence, the assumed value of k ⫽ 1.0 was inaccurate for this

spring Recalculating, d⫽[8⫻5000⫻1.25⫻2.0575 / (␲ ⫻65,000)]1 / 3⫽0.796

in (2.0 cm) Now, 1.25⫹ 0.796⫽2.046 in (5.2 cm), which is still too large

Using d m⫽1.20 in (3.1 cm) and assuming k⫽2.0575, then d⫽[8⫻5000⫻

It is now obvious that a practical trade-off must be utilized so that a spring can

be designed to carry a 5000-lb (22,241.1-N) load and fit in a 2-in (5.1-cm) diameter

hole, d h, with suitable clearance Such a trade-off is to use hard-drawn steel wire

of greater strength at higher cost This trade-off would not be necessary if d hcould

be increased to, say, 2.13 in (5.4 cm)

Using d m⫽ 1.25 in (3.2 cm) and a clearance of, say, 0.08 in (0.20 cm), then d

⫽d h⫺ d m⫺0.08 ⫽2.00⫺ 1.25⫺0.08 ⫽0.67 in (1.7 cm) Also, c⫽ d m / d⫽

1.25 / 0.67⫽ 1.866 and k⫽[(4⫻1.866) ⫺1] / [(4⫻1.866)⫺4]⫹0.615 / 1.866

⫽2.196 The new allowable shear stress can now be found from s s⫽8Ld m k /␲d3

⫽8⫻5000⫻1.25⫻2.196 / (␲ ⫻0.673)⫽116,206 lb / in2(801,212.1 kPa)

Hard-drawn spring wire (ASTM A-227-47) is available with s s ⫽ 117,000 lb / in2

(806,686.6 kPa) and G⫽11.5⫻106lb / in2(79.29⫻109Pa)

Hence, d⫽[8⫻5000⫻1.25⫻2.196 / (␲ ⫻117,000)]1 / 3⫽0.668 in (1.7 cm)

Thus, d o⫽1.25⫹0.668⫽1.918 in (4.9 cm) Now, the spring is acceptable because

further recalculations would show that d o will remain less than 2 in (5.1 cm),regardless

2. Compute the spring deflection

The deflection of a helical compression spring is given by ƒ ⫽ 64nr L / (d3 4G) ⫽

The number of coils n assumed for this spring is based on past experience with

similar springs However, where past experience does not exist, several trial values

of n can be used until a spring of suitable deflection and length is obtained.

Related Calculations. Use this general procedure to analyze helical coil pression or tension springs As a general guide, the outside diameter of a spring ofthis type is taken as (0.96)(hole diameter) The active solid height of a compression-type spring, i.e., the height of the spring when fully closed by the load, usually is

com-nd, or (0.9) (final height when compressed by the design load).

SELECTION OF HELICAL COMPRESSION AND

TENSION SPRINGS

Choose a helical compression spring to carry a 90-lb (400.3-N) load with a stress

of 50,000 lb / in2 (344,750.0 kPa) and a deflection of about 2.0 in (5.1 cm) Thespring should fit in a 3.375-in (8.6-cm) diameter hole The spring operates at about

70⬚F (21.1⬚C) How many coils will the spring have? What will the free length ofthe spring be?

Calculation Procedure:

1. Determine the spring outside diameter

Using the usual relation between spring outside diameter and hole diameter, we get

d o⫽0.96d h , where d h⫽hole diameter, in Thus, d o⫽0.96(3.375)⫽3.24 in, say3.25 in (8.3 cm)

2. Determine the required wire diameter

he equations in the previous calculation procedure can be used to determine therequired wire diameter, if desired However, the usual practice is to select the wirediameter by using precomputed tabulations of spring properties, charts of springproperties, or a special slide rule available from some spring manufacturers Thetabular solution will be used here because it is one of the most popular methods

TABLE 2 Load and Spring Rates for Helical Compression and Tension

⫽ 180 lb (800.7 N) This means that a 90-lb (400.3-N) load at a 50,000-lb / in2(344,750.0-kPa) stress corresponds to a 180-lb (800.7-N) load at 100,000-lb / in2(689,500-kPa) stress

Enter Table 2 at the spring outside diameter, 3.25 in (8.3 cm), and project tically downward in this column until a load of approximately 180 lb (800.7 N) isintersected At the left read the wire diameter Thus, with a 3.25-in (8.3-cm) outsidediameter and 183-lb (814.0-N) load, the required wire diameter is 0.250 in (0.635cm)

ver-3. Determine the number of coils required

The allowable spring deflection is 2.0 in (5.1 cm), and the spring rate per singlecoil, Table 2, is 208 lb / in (364.3 N / cm) at a tabular stress of 100,000 lb / in2(689,500 kPa) We use the relation, deflection ƒ, in⫽load, lb / desired spring rate,

lb / in, S R; or, 2.0⫽90 / S R ; S R⫽ 90 / 20⫽ 45 lb / in (78.8 N / cm)

4. Compute the number of coils in the spring

The number of active coils in a spring is n⫽ (tabular spring rate, lb / in) / (desired

spring rate, lb / in) For this spring, n⫽208 / 45⫽4.62, say 5 coils

5. Determine the spring free length

Find the approximate length of the spring in its free, expanded condition from l in

⫽ (n⫹i)d ⫹ƒ, where l⫽ approximate free length of spring, in; i⫽ number ofinactive coils in the spring; other symbols as before Assuming two inactive coils

for this spring, we get l⫽(5⫹ 2)(0.25)⫹2 ⫽3.75 in (9.5 cm)

Related Calculations. Similar design tables are available for torsion springs,spiral springs, coned-disk (Belleville) springs, ring springs, and rubber springs.These design tables can be found in engineering handbooks and in spring manu-

facturers’ engineering data Likewise, spring design charts are available from many

of these same sources Spring design slide rules are generally available free ofcharge to design engineers from spring manufacturers

SIZING HELICAL SPRINGS FOR OPTIMUM

DIMENSIONS AND WEIGHT

Determine the dimensions of a helical spring having the minimum material volume

if the initial, suddenly applied, load on the spring is 15 lb (66.7 N), the mean coildiameter is 1.02 in (2.6 cm), the spring stroke is 1.16 in (2.9 cm), the final springstress is 100,000 lb / in2 (689,500 kPa), and the spring modulus of torsion is 11.5

⫻106lb / in2(79.3⫻ 109Pa)

Calculation Procedure:

1. Compute the minimum spring volume

Use the relationv m⫽8ƒLG / s , where2 v m⫽minimum volume of spring, in3; ƒ⫽

spring stroke, in3; L⫽initial load on spring, lb; G⫽modulus of torsion of springmaterial, lb / in2; sƒ ⫽ final stress in spring, lb / in2 For this spring, v m ⫽

8(1.16)(15)(11.5⫻106) / (100,000)2⫽0.16 in3(2.6 cm3) Note: sƒ⫽2s s , where s s

⫽shear stress due to a static, or gradually applied, load

2. Compute the required spring wire diameter

Find the wire diameter from d⫽ [16L d m/ (␲sƒ)]1 / 3, where d⫽wire diameter, in;

d m ⫽mean diameter of spring, in; other symbols as before For this spring, d ⫽

[16⫻15 ⫻1.02 / (␲ ⫻100,000)]1 / 3⫽0.092 in (2.3 mm)

3. Find the number of active coils in the spring

se the relation n⫽4 m/ (␲2d2d m ), where n⫽number of active coils; other symbols

as before Thus, n⫽4(0.16) / [␲2(0.092)2(1.02)]⫽7.5 coils

4. Determine the active solid height of the spring

The solid height H s⫽(n⫹1)d, in, or H s⫽(7.5⫹1)(0.092)⫽0.782 in (2.0 cm).For a practical design, allow 10 percent clearance between the solid height and the

minimum compressed height H c Thus, H c⫽1.1H s⫽1.1(0.782) ⫽0.860 in (2.2

cm) The assembled height H a⫽H c⫹ƒ⫽0.860 ⫹1.16⫽ 2.020 in (5.13 cm)

5. Compute the spring load-deflection rate

The load-deflection rate R ⫽ Gd4/ (8d n), where R3 ⫽ load-deflection rate, lb / in;

m

other symbols as before Thus, R⫽ (11.5 ⫻ 106)(0.092)4/ [8(1.02)3(7.5)] ⫽ 12.9

lb / in (2259.1 N / m)

The initial deflection of the spring is ƒi⫽L / R in, or ƒ i⫽15 / 12.9⫽1.163 in

(3.0 cm) Since the free height of a spring Hƒ⫽ H a⫹ ƒi, the free height of this

spring is Hƒ⫽2.020⫹1.163 ⫽3.183 in (8.1 cm)

Related Calculations. The above procedure for determining the minimumspring volume can be used to find the minimum spring weight by relating the spring

weight W lb to the density of the spring material␳lb / in3in the following manner:

For the required initial load L1 lb, Wmin⫽ ␳(8ƒL1G / sƒ) For the required energy

capacity E in䡠lb, Wmin ⫽ ␳(4ED / sƒ) For the required final load L2lb, Wmin ⫽

␳(2ƒ2L2G / s ).2

The above procedure assumes the spring ends are open and are not ground Forother types of end conditions, the minimum spring volume will be greater by thefollowing amount: For squared (closed) ends, v m⫽ 0.5␲2d2d m For ground ends,

v m⫽ 0.25␲2d2d m The methods presented here were developed by Henry

Swies-kowski and reported in Producted Engineering.

SELECTION OF SQUARE- AND

RECTANGULAR-WIRE HELICAL SPRINGS

Choose a square-wire spring to support a load of 500 lb (2224.1 N) with a deflection

of not more than 1.0 in (2.5 cm) The spring must fit in a 4.25-in (10.8-cm) diameter

hole The modulus of rigidity for the spring material is G ⫽ 11.5 ⫻ 106 lb / in2(79.3 ⫻ 109Pa) What is the shear stress in the spring? Determine the correctedshear stress for this spring

Calculation Procedure:

1. Determine the spring dimensions

Assume that a 4-in (10.2-cm) diameter square-bar spring is used Such a springwill fit the 4.25-in (10.8-cm) hole with a small amount of room to spare

As a trial, assume that the width of the spring wire⫽0.5 in (1.3 cm)⫽a Since

the spring is square, the height of the spring wire⫽ 0.5 in (1.3 cm)⫽b.

With a 4-in (10.2-cm) outside diameter and a spring wire width of 0.5 in (1.3

cm), the mean radius of the spring coil r m⫽ 1.75 in (4.4 cm) This is the radiusfrom the center of the spring to the center of the spring wire coil

2. Compute the spring deflection

The deflection of a square-wire tension spring is ƒ⫽ 45Lr n / (Ga3 4), where ƒ ⫽

m

spring deflection, in; L⫽load on spring, lb; n⫽number of coils in spring; othersymbols as before To solve this equation, the number of coils must be known.Assume, as a trial value, five coils Then ƒ⫽45(500)(1.75)3(5) / [(11.5⫻106)(0.5)4]

⫽ 0.838 in (2.1 cm) Since a deflection of not more than 1.0 in (2.5 cm) is mitted, this spring is probably acceptable

per-3. Compute the shear stress in the spring

Find the shear stress in a square-bar spring from S s ⫽ 4.8Lr m / a3, where S s ⫽

spring shear stress, lb / in2; other symbols as before For this spring, S s ⫽

(4.8)(500)(1.75) / (0.5)3⫽33,600 lb / in2(231,663.8 kPa) This is within the able limits for usual spring steel

allow-4. Determine the corrected shear stress

Find the shear stress in a square-bar spring from S s⫽4.8Lr m / a3, where s s⫽spring

correction factor k⫽ 1 ⫹ 1.2 / c ⫹ 0.56 / c2 ⫹ 0.5 / c3, where c ⫽ 2r m / a For this spring, c⫽(2⫻1.75) / 0.5⫽7.0 Then k⫽1⫹1.2 / 7⫹0.56 / 72⫹0.5 / 73⫽1.184

Hence, the corrected shear stress is S⬘s ⫽ ks s , or S⬘s ⫽ (1.184)(33,600) ⫽ 39,800

lb / in2(274,411.3 kPa) This is still within the limits for usual spring steel

Related Calculations. Use a similar procedure to select rectangular-wiresprings Once the dimensions are selected, compute the spring deflection from

ƒ⫽19.6Lr n / [Gb3 2(a⫺0.566)], where all the symbols are as given earlier in this

m

FIGURE 5 Typical curved spring (Product Engineering.)

calculation procedure Compute the uncorrected shear stress from S s ⫽ Lr m (3a

⫹ 1.8b) / (a2b2) To correct the stress, use the Liesecke correction factor given in

Wahl—Mechanical Springs For most selection purposes, the uncorrected stress is

satisfactory

CURVED SPRING DESIGN ANALYSIS

Find the maximum load P, maximum deflection F, and spring constant C for the

curved rectangular wire spring shown in Fig 5 if the spring variables expressed in

metric units are E⫽14,500 kg / mm2, S b⫽55 kg / mm2, b ⫽1.20 mm, h⫽ 0.30

mm, r1⫽0.65 mm, r2⫽1.75 mm, L⫽9.7 mm, u1⫽1.7 mm, and u2⫽5.6 mm

Calculation Procedure:

1. Divide the spring into analyzable components

Using Fig 6, developed by J Palm and K Thomas of West Germany, as a guide,divide the spring to be analyzed into two or more analyzable components, Fig 5.Thus, the given spring can be divided into two springs—a type D (Fig 6), calledsystem I, and a type A (Fig 6), called system II

2. Compute the spring force

The spring force P ⫽PI⫽ PII Since (u2⫹r2)⬎ (u1⫹r1), the spring in system

II exerts a larger force From Fig 6 for␤ ⫽90⬚, P⫽S␴max/ (u2⫹r2), where S⫽

section modulus, mm3, of the spring wire Since S ⫽ bh2/ 6 for a rectangle, P ⫽

bh2␴max/ [6(u2⫹ r2)] where b ⫽ spring wire width, mm; h⫽ spring wire height,mm; ␴max ⫽ maximum bending stress in the spring, kg / mm3; other symbols as

given in Fig 5 Then P⫽(1.20)(0.30)2⫻(55) / [6(5.6⫹1.75)]⫽0.135 kg

3. Compute the spring deflection

The total deflection of the springs is F⫽ 2FI⫹FII, where F⫽spring deflection,

mm, and the subscripts refer to each spring system Taking the sum of the

deflec-tions as given in Fig 6, we get F⫽ [2P / (3EI )][2K1r (m3 1⫹␤1/ 2)2⫹ (v1⫺ u1)3

⫹K2r (m3 2⫹␤2)3], where E⫽Young’s modulus, kg / mm2; I⫽spring wire moment

of inertia, mm4; K ⫽correction factor for the spring from Fig 7, where the

sub-scripts refer to the radius being considered in the relation u / r ; m⫽u / r ;␤ ⫽angle

FIGURE 6 Deflection, force, and stress relations for curved springs (Product Engineering.)

FIGURE 7 Correction factors for curved springs (Product Engineering.)

FIGURE 8 Typical torsion spring (Product Engineering.)

of spring curvature, rad Where the subscripts 1 and 2 are used in this

equa-tion, they refer to the respective radius identified by this subscript Since I ⫽

bh3/ 12 for a rectangle, or I⫽ (1.20)(0.30)3/ 12 ⫽ 0.0027 mm4, F ⫽ {[2(0.135) /[(3)(14,500)(0.0027)]}[2(0.92)(0.65)(2.62 ⫹ 1.57)3 ⫹ 0 ⫹ 0.94(1.75)3(3.2 ⫹

1.57)3]⫽ 1.34 mm

4. Compute the spring constant

The spring constant C⫽P / F⫽0.135⫽0.135 / 1.34⫽0.101 kg / mm

Related Calculations. The relations given here can also be used for

round-wire springs For accurate results, h / r for flat springs and d o / r for round-wire

springs should be less than 0.6 The various symbols used in this calculation cedure are defined in the text and illustrations Since the equations given hereanalyze the springs and do not contain any empirical constants, the equations can

pro-be used, as presented, for both metric and English units Where a round spring is

analyzed, h⫽b⫽ d o , where d o⫽spring outside diameter, mm or in

ROUND- AND SQUARE-WIRE HELICAL

TORSION-SPRING SELECTION

Choose a round-music-wire torsion spring to handle a moment load of 15.0 lb䡠in(1.7 N䡠m) through a deflection angle of 250⬚ The mean diameter of the springshould be about 1.0 in (2.5 cm) to satisfy the space requirements of the design.Determine the required diameter of the spring wire, the stress in the wire, and thenumber of turns required in the spring What is the maximum moment and angulardeflection the spring can handle? What is the maximum moment and deflectionwithout permanent set?

Calculation Procedure:

1. Select a suitable wire diameter

To reduce the manufacturing cost of a spring, a wire of standard diameter should

be used, whenever possible, for the spring, Fig 8 Usual torsion-spring wire ameters and the side of square-wire springs range from 0.02 to 0.60 in (0.05 to1.52 cm), depending on the moment the spring must carry and the angular deflec-tion

di-Assume a wire diameter of 0.10 in (0.25 cm) and a bending stress of 150,000

lb / in2(1.03⫻109Pa) as trial values for this spring [Typical round-wire and