Tài liệu Handbook of Mechanical Engineering Calculations P20 pptx

SECTION 20 GEAR DESIGN AND APPLICATIONAnalyzing Gears for Dynamic Loads 20.1 Helical-Gear Layout Analysis 20.12 Analyzing Shaft Speed in Epicyclic Gear Trains 20.14 Speeds of Gears and G

SECTION 20 GEAR DESIGN AND APPLICATION

Analyzing Gears for Dynamic Loads

20.1

Helical-Gear Layout Analysis 20.12

Analyzing Shaft Speed in Epicyclic Gear

Trains 20.14

Speeds of Gears and Gear Trains 20.17

Selection of Gear Size and Type 20.18

Gear Selection for Light Loads 20.21

Selection of Gear Dimensions 20.25

Horsepower Rating of Gears 20.26

Moment of Inertia of a Gear Drive

20.28

Bearing Loads in Geared Drives 20.29 Force Ratio of Geared Drives 20.30 Determination of Gear Bore Diameter 20.31

Transmission Gear Ratio for a Geared Drive 20.32

Epicyclic Gear Train Speeds 20.33 Planetary-Gear-System Speed Ratio 20.34

ANALYZING GEARS FOR DYNAMIC LOADS

A two-stage, step-up gearbox drives a compressor and has a lubrication pumpmounted on one of the gear shafts, Fig 1 Tables 1, 2, and 3 show the spur-geardata, tolerances for tooth errors, and polar moments of inertia for the masses in thecompressor drive All gears in the drive have 20⬚ pressure angles The gears aremade of steel; the compressor is made of cast iron A 50-hp (37.3-kW) motor drivesthe gearbox at 3550 rpm What are the dynamic loads on this gearbox?

Calculation Procedure:

1. Determine the pitchline velocity, V, applied load, W, at each mesh, and shaft speed, n

In the equations that follow, subscripts identify the shaft, gear, or mesh under

con-sideration For example, the subscript A denotes Shafts A1and A2; B, Shaft B; and

C, Shaft C Also, subscripts 1, 2, 3, 4 refer respectively to Gears 1, 2, 3, and 4 in the gearbox The subscripts a and b denote the mesh of Gears 1 and 2 and the mesh of Gears 3 and 4, respectively Finally, subscripts r and n refer to the driver

gear and driven gear Full nomenclature appears at the end of this procedure

For Shaft A1and A2,

FIGURE 1 Two-stage step-up gearbox driving a compressor (Machine Design.)

TABLE 1 Spur Gear Data

TABLE 2 Tolerances for Tooth Errors

Tolerance for tooth profile

Tabulated values do not include the polar moment of inertia for the lubrication pump.

A material mass factor B⫽ 0.00087 lb
s 2 / in 3
ft was assumed for steel; B⫽ 0.00080 lb
sec 2 / in 3
ft (0.000073 kg
s 2 / cm 3
m) cast iron.

The polar moment of inertia for cylindrical masses was calcualted as, I o⫽BD4L.

Therefore, for Shafts B and C, the following values for speed, torque, load and

pitchline velocity are obtained

Find the Total Effective Mass at Mesh a and Mesh b

Total effective mass m t must be determined at each gear mesh The parameter m t

can be calculated from

1 / m t⫽1 / m r⫹1 / m n

where m r is the effective mass at the driver gear and m nis the effective mass at thedriven gear

For rotating components such as gears, an effective mass m at the pitch radius

R can be calculated from

m⫽ 兺I / R o

The gears in the compressor drive mesh at two points, for which the total

ef-fective mass m tmust be determined

At Mesh a and Mesh b, m tis

Consequently, the equations for m tbecome

3. Compute the acceleration force for the spur gears

For spur gears, the acceleration force ƒ1is

2

ƒ1⫽Hm V t

where H⫽A1(1 / R r⫹1 / R n ) The calculation factor A1⫽0.00086 for 14.5⬚teeth,0.00120 for 20⬚teeth, and 0.00153 for 25⬚teeth The parameters R r and R nare thepitch radii of driver and driven gears, respectively

Therefore at Mesh a,

4. Calculate the deflection force

The deflection force ƒ2is

1 / ƒ2⫽1 / C⫹(1 / C1⫹1 / C2⫹


1 / C ) x where C accounts for deflection from bending and compressive loads in the gear teeth, and C x accounts for deflection from torsional loads in shafts, flexible cou-

plings, and other components C is the load required to deflect gear teeth by an amount equal to the error in action C x is the load required at the pitch radius todeflect a shaft or coupling by an amount equal to the error in action Previously,

calculations of ƒ2only considered the parameter C With the addition of C xto the

equation for ƒ2, the value of ƒ2will always be less than the smallest value of C,

C1, C2 C x

The parameter C is

C⫽W⫹1000 eFA where A is the load required to deflect teeth by 0.001 in (0.0254 mm).

If the Lewis form factor y is known, A can be calculated from

Summary of Calculations for Deflection Load

Previously, the error in action e had to be assumed for a given class of gears,

based on recommendations tabulated in handbooks However, this error can beapproximated as

e⫽ 兹(e pr⫹e ) sr ⫹(e pn⫹e ) sn

where for the driver and driven gears, e p is the tooth profile error and e sis the tooth

spacing error If actual measurements of e p and e sare not available, then the

tol-erances for allowable e p and e scan be used for calculations instead Therefore, for

both Mesh a and Mesh b, the approximate error in action e is

4 2

C x⫽1,080,000eD / R L s s

where D s is the shaft diameter, and L s is the shaft length Generally, for other

components such as flexible couplings, C xhas to be determined experimentally orfrom information supplied by the component manufacturer

In the gear box, the portion of Shaft A2between the motor and Gear 1 deflects

interdependently with Shaft A1 On the other hand, the portion of Shaft A2betweenthe lubrication pump and Gear 1 deflects independently of the shaft elements to

the right of Gear 1 The table below summarizes the calculations of C xfor all shaftelements affecting each mesh

Summary of Calculations for Shaft Deflection Load

Shaft

elements

for Mesh a

Shaft diameter,

Shaft length,

Pitch radius,

Shaft deflection

For example, Shaft A2 has a different diameter than Shaft A1 has Consequently,

the effective value for C xbetween the motor and pump is

C x⫽C4⫽C3⫹1 / (1C2⫹1 / C )1

⫽910⫹1 / (1 / 546⫹1 / 73)

⫽974 lb (442.2 kg)

Note that values of C x for each shaft element in the table are differentiated from

each other by the use of a subscript x, enumerated as x ⫽ 1, 2, 3, 4 Thissubscript should not be confused with numerical subscripts used to denote Gears

At Mesh a and Mesh b, W d / W ⫽ 3.49 and 4.13 respectively Because the

W d / W ratio for both meshes is greater than two, there will be free impact between

the gear teeth Also, the drive will be noisy and may wear rapidly The dynamic

load W dcan de decreased by placing a flexible coupling between the motor and

Shaft A1 This will effectively isolate the motor mass Also, the lubrication pump

can be isolated by driving it with a quill shaft Finally, the diameter of Shaft B can

be minimized to meet the required torque capacity while increasing shaft resilience

to lower the magnitude of W d

Related Calculations. Gears in mesh never operate under a smooth, continuousload Factors such as manufacturing errors in tooth profile and spacing, tooth de-flections under load, and imbalance all interact to create a dynamic load on gearteeth The resulting action is similar to that of a variable load superimposed on asteady load

Consider, for example, how tooth loads can fluctuate from manufacturing errors.The maximum, instantaneous tooth load occurs at the maximum error in action.The average load is the applied load on the teeth at the pitchline of the gears

As each pair of teeth moves through its duration of contact, errors in actioncreate periods of sudden acceleration that momentarily separate mating gear teeth

Impact occurs as each pair of teeth returns to mesh to complete its contact duration.The impact loads can be significantly greater than the applied load at the pitchline.The maximum magnitude of the dynamic load depends on gear and pinion masses,connected component masses, operating speed, and material elasticity Elastic de-flections in gear teeth and drive components, such as shafts and flexible couplings,help reduce dynamic loads.

Generally, gears should be designed so that their bending and wear capacitiesare equal to or greater than the maximum, instantaneous dynamic load However,the exact magnitude of the maximum dynamic load is seldom known Althoughmany gear studies have been conducted, there is no full agreement on the singlebest method for determining dynamic loads

One of the most widely accepted methods for calculating dynamic loads

con-siders the maximum dynamic load W d, resulting from elastic impact, to be

W d⫽W⫹W i The term W is the applied load at the pitchline of a gear For gears, the incremental load W iis

W i⫽ 兹ƒ (2ƒ a 2⫺ƒ ) a

In the above equation, ƒ ais the resultant force required to accelerate masses in

a system of elastic bodies The resultant acceleration force ƒ ais

1 / ƒ a⫽1 / ƒ1⫹1 / ƒ2The term ƒ1is the force required to accelerate masses in a system of rigid bodies

The term ƒ2is the force required to deflect the system elastically by the amount oferror in action As mentioned previously, free impact loads occur in gears whenthe teeth separate and return to mesh suddenly during the contact interval For free

impact, the value of the ratio W d / W will always be greater than two.

If forces ƒ1, ƒ2, and ƒ a are plotted as functions of pitchline velocity, force ƒ2is

seen to be an asymptote of the incremental load W i The equations that define ƒ1and ƒ2depend on the type of gears being analyzed for dynamic loads The example

presented in this procedure provides equations defining ƒ1 and ƒ2 for spur gearapplications

Sometimes this method for calculating W d gives values for dynamic load thatare conservative These high dynamic load estimates can lead to overdesigned gears

Conservative calculations of W d can be minimized if ƒ2 can be calcualted more

accurately A less conservative calculation of ƒ2effectively reduces W d because ƒ2

is an asymptote for the incremental load W i

The method presented here has been refined to give more accurate values for

ƒ2 Previously, only the elastic deflection in the gear teeth was considered in the

calculation of ƒ2 Now, elastic deflections in other mechanical components, such as

flexible couplings and shafts, are also considered for their effects on ƒ2through

inclusion of a parameter C x

Also, the calculation of ƒ2is further refined by a more accurate approximation

of the error in action Previously, the error in action was assumed for a given class

of gears, based on recommendations commonly tabulated in gear design handbooks.Now the error in action can be calculated from tooth profile errors and tooth-to-tooth spacing errors The equation for approximating error in action results fromtests where gears were measured for error in action, profile error, spacing error, andrunout error

This procedure shows how the refinements in the dynamic load analysis apply

in spur gear applications However, the principles can be applied to other forms ofgearing as well Furthermore, the example describes how resilience in mechanicalcomponents, such as shafts, can be applied to reduce dynamic loads on gears.Reduced dynamic loads result in lower operating noise and longer component life

In all new applications where there is no design experience, gear drives should

be analyzed for the possibility of dynamic loading Also, all gear drives operating

at peripheral speeds of about 1000 ft / min (304.8 m / min) and higher should bechecked

At low operating speeds, usually less than 1000 ft / min (304.8 m / min), geardrives follow the dynamics of rigid bodies Consequently, elastic deflections incomponents have little effect on dynamic loads Inertial forces generated by mo-mentum variations appear to be directly proportional to mass magnitude and thesquare of mass velocity At intermediate operating speeds, elastic deflections helpreduce instantaneous dynamic loads In this range, inertial forces appear to be di-rectly proportional to mass velocity and the square root of mass magnitude Athigher operating speeds, elastic deflections have a pronounced influence on dynamicloads The deflections tend to limit inertial forces to an asymptotic value Conse-quently, inertial forces appear to be independent of mass magnitude and mass ve-locity

Some applications must be carefully analyzed for dynamic loading Examplesinclude:

• Drives where large masses are directly connected, or connected with short shafts,

to gear drives;

• Gear drives connected with solid couplings;

• Drives with fluid pumps connected directly to gears or gear shafts;

• Long gear trains, such as those in printing presses, paper machines, and processmachinery;

• Auxiliary drives with small gears connected to high power systems;

• Drives with multiple power inputs;

• High-speed drives that include servodrives;

• Any gear drive where gears shown signs of distress or operate with high noisefor no apparent reason

This procedure is the work of Eliot K Buckingham, President, Buckingham

As-sociates, Inc., as reported in Machine Design magazine.

Nomenclature

A⫽ Load which deflects gear teeth by 0.001 in, lb (0.0254 mm, kg)

A1⫽ Calculation factor for H

B⫽ Material mass factor, lb
s 2 / in 3
ft (kg
s 2 / cm 3 -m)

C⫽ Load which deflects gear teeth by an amount equal to the error in action, lb (kg)

C1,2,3 r⫽ Load at the pitch radius which deflects a shaft by an amount equal to the error in

e s⫽ Tooth spacing error, in (mm)

ƒ⫽ Face width, in (mm)

ƒ a⫽ Resultant force required to accelerate masses in a system of elastic bodies, lb (kg)

ƒ1⫽ Force required to accelerate masses in a system of rigid bodies, lb (kg)

ƒ2⫽ Force required to deflect elastically the system by the amount of error in action, lb (kg)

H ⫽ Calculation factor for ƒ1

I a⫽ Polar moment of inertia, in 2
lb
s 2 / ft (cm 2
kg
s 2 / m)

L⫽ Cylinder length, in (mm)

L s⫽ Shaft length, in (mm)

m⫽ Effective mass, lb
s 2 / ft (kg
s 2 / m)

m n⫽ Effective mass at the driven gear, lb
s 2 / ft (kg
s 2 / m)

m r⫽ Effective mass at the driver gear, lb
s 2 / ft (kg
s 2 / m)

m t⫽ Total effective mass at the mesh, lb
s 2 / ft (kg
s 2 / m)

N⫽ Number of gear teeth

n⫽ Shaft speed, rpm

P⫽ Power, hp (kW)

R⫽ Pitch radius, in (mm)

T⫽ Torque, lb
in (N
m)

V⫽ Pitchline velocity, ft/min (m/min)

W⫽ Applied load at the pitch radius, lb (kg)

W d⫽ Dynamic load, lb (kg)

W i⫽ Incremental load, lb (kg)

y⫽ Lewis form factor

Z ⫽ Calculation factor for C

Subscripts

1, 2, 3, 4 ⫽ Gear 1, 2, 3, and 4

a, b ⫽ Mesh a and b

n, r⫽ Driven gear, driver gear

HELICAL-GEAR LAYOUT ANALYSIS

Helical gears are to be designed for shafts 4-in (10.16-cm) apart and at right angles,

Fig 2 The normal dimeteral pitch, P n⫽ 20; the number of teeth, N, in the gear

or worm wheel ⫽ 64; the number of teeth, n, in the pinion or threads in the

worm⫽32 Determine the suitable helix angle, or angles,␣, of the gear or worm

Calculation Procedure:

1. Determine if the selected variables for this gearset are suitable

In the preliminary layout of gear drives, the designer must determine if the meteral pitch, gear ratio, and center distance are suitable for the proposed layout.With this settled, the designer can proceed to calculate the specific helix angle

dia-To check a proposed layout, use the relation,

FIGURE 2 Helical-gear layout.

P n⫽normal diameteral pitch

N⫽number of teeth in gear or worm wheel

n ⫽number of teeth in pinion or threads in worm

R⫽ratio, n / N

a ⫽helix angle of gear or worm wheel, deg

Find the value of the lefthand portion of this equation for this gear layout from:2(4)(20) / 64⫽2.5 Then, R⫽32 / 64⫽0.5

2. Find the suitable helix angle or angles

Knowing the value of the lefthand side of the equation and the value of R we can

assume suitable values for alpha and solve the equation If the chosen helix angle

is correct, the two sides of the equation will be equal

As a first choice, assume a helix angle of 20⬚ Substituting, using values fromstandard trigonometric tables, we have 2.5⫽(0.5 / 0.3584)⫹(1 / 0.9336)⫽2.466.This helix angle is not suitable because the two sides of the equation are not equal.Try 20⬚30⬘ Or, 2.5⫽(0.5 / 0.3486)⫹(1 / 0.9373)⫽2.4968 This is much closerthan the first try Another trial calculation shows that 20⬚25⬘ is the correct helixangle for this layout

Another angle, namely 58⬚30⬘is also suitable, based on the equation above Forgreater efficiency and less wear, the gear and pinion helix angles should be nearlyequal Thus, the gear angle chosen could be 58⬚30⬘and the pinion angle becomes

This procedure is the work of Wayne A Ring, Barber-Colman Company, as

reported in Product Engineering magazine.

FIGURE 3 Epicyclic gear with drive and driven shafts rotating in same direction (Product Engineering.)

ANALYZING SHAFT SPEED IN EPICYCLIC GEAR

TRAINS

Analyze the input and output shaft rpm for the six epicyclic gear trains shown inFigs 3 through 8 The gear train in Fig 3 has arm A integral with the righthandshaft Gears C and D are keyed to a short length of shaft which is mounted in abearing in arm A Gear C meshes with internal gear E which is keyed to the lefthandshaft

Calculation Procedure:

1. Find the ratio of the shaft speeds

To find the ratio of the speed of shaft E to the speed of shaft A, proceed thus: Let

N b be the number of teeth in gear B, N cthe number in gear C, and so on Let arm

A, which was originally in a vertical position, be given an angular ment,␪

displace-In giving arm A the angular displacement, gear C will traverse through arc ab

on gear B Since angles are inversely proportion to radii, or to the number of teeth,gears C and D with turn through angle (␪)N b / N c

While the foregoing was taking place, gears D and E were rotating on each other

through the equal arcs ed and ef Gear E will have been turned in the reverse

direction through angle (␪)(N b / N c )(N d / N e)

The net effect of these two operations is to move the point of gear E, which

was originally vertical at g, over to location ƒ Gear E has thus rotated through

FIGURE 4 Drive and driven shafts rotate in opposite directions (Product Engineering.)

FIGURE 5 Drive and driven shafts rotate in same direction (Product Engineering.)

FIGURE 6 Drive and driven shafts rotate in same direction (Product Engineering.)

FIGURE 7 Drive and driven shafts rotate in opposite directions (Product Engineering.)

angle (1⫺ [N b / N d ] / [N c / N e])(␪) This latter value, when divided by␪, the angularmovement of shaft A, gives the ratio of the rotations of shafts E and A, respectively

2. Extend the analysis to other gear arrangements

The above analysis can be applied to the gear arrangements shown in Fig 4 through

8 When this is done the speed ratios for each arrangement is as given in theillustrations

FIGURE 8 Drive and driven shafts rotate in same direction (Product Engineering.)

Related Calculations. This procedure is a valuable analysis for epicyclic geartrains of all types The resulting speed equations can be used to determine bothinput and output speeds of the gearset

E F Spotts, Northwestern Technological Institute, prepared this procedure which

was reported in Product Engineering magazine.

SPEEDS OF GEARS AND GEAR TRAINS

A gear having 60 teeth is driven by a 12-tooth gear turning at 800 r / min What isthe speed of the driven gear? What would be the speed of the driven gear if a 24-tooth idler gear were placed between the driving and driven gear? What would bethe speed of the driven gear if two 24-tooth idlers where used? What is the direction

of rotation of the driven gear when one and two idlers are used? A 24-tooth drivinggear turning at 600 r / min meshes with a 48-tooth compound gear The second gear

of the compound gear has 72 teeth and drives a 96-tooth gear What are the speedand direction of rotation of the 96-tooth gear?

Calculation Procedure:

1. Compute the speed of the driven gear

For any two meshing gears, the speed ratio R D / R d⫽ N d / N D , where R D⫽ rpm of

driving gear; R d⫽rpm of driven gear; N d⫽number of teeth in driven gear; N D⫽

number of teeth in driving gear By substituting the given values, R D / R d⫽N d / N D,

or 800 / R d⫽60 / 12; R d⫽160 r / min

2. Determine the effect of one idler gear

An idler gear has no effect on the speed of the driving or driven gear Thus, the

speed of each gear would remain the same, regardless of the number of teeth in

the idler gear An idler gear is generally used to reduce the required diameter ofthe driving and driven gears on two widely separated shafts.

3. Determine the effect of two idler gears

The effect of more than one idler is the same as that of a single idler—i.e., thespeed of the driving and driven gears remains the same, regardless of the number

of idlers used

4. Determine the direction of rotation of the gears

Where an odd number of gears are used in a gear train, the first and last gears turn

in the same direction Thus, with one idler, one driver, and one driven gear, the driver and driven gear turn in the same direction because there are three gears (i.e.,

an odd number) in the gear train

Where an even number of gears is used in a gear train, the first and last gears

turn in the opposite direction Thus, with two idlers, one driver, and one driven gear, the driver and driven gear turn in the opposite direction because there are four

gears (i.e., an even number) in the gear train

5. Determine the compound-gear output speed

A compound gear has two gears keyed to the same shaft One of the gears is driven

by another gear; the second gear of the compound set drives another gear In acompound gear train, the product of the number of teeth of the driving gears andthe rpm of the first driver equals the product of the number of teeth of the drivengears and the rpm of the last driven gear

In this gearset, the first driver has 24 teeth and the second driver has 72 teeth.The rpm of the first driver is 600 The driven gears have 48 and 96 teeth, respec-tively Speed of the final gear is unknown Applying the above rule gives

(24)(72)(600)2(48)(96)(R d ); R d⫽215 r / min

Apply the rule in step 4 to determine the direction of rotation of the final gear.Since the gearset has an even number of gears, four, the final gear revolves in theopposite direction from the first driving gear

Related Calculations. Use the general procedure given here for gears and geartrains having spur, bevel, helical, spiral, worm, or hypoid gears Be certain to de-termine the correct number of teeth and the gear rpm before substituting values inthe given equations

SELECTION OF GEAR SIZE AND TYPE

Select the type and size of gears to use for a 100-ft3/ min (0.047-m3/ s) reciprocatingair compressor driven by a 50-hp (37.3-kW) electric motor The compressor andmotor shafts are on parallel axes 21 in (53.3 cm) apart The motor shaft turns at

1800 r / min while the compressor shaft turns at 300 r / min Is the distance betweenthe shafts sufficient for the gears chosen?

Calculation Procedure:

1. Choose the type of gears to use

Table 4 lists the kinds of gears in common use for shafts having parallel, secting, and nonintersecting axes Thus, Table 14 shows that for shafts having par-allel axes, spur or helical, external or internal, gears are commonly chosen Since

inter-TABLE 4 Types of Gears in Common Use*

external gears are simpler to apply than internal gears, the external type is chosenwherever possible Internal gears are the planetary type and are popular for appli-cations where limited space is available Space is not a consideration in this appli-cation; hence, an external spur gearset will be used

Table 5 lists factors to consider in selecting gears by the characteristics of theapplication As with Table 4, the data in Table 5 indicate that spur gears are suitablefor this drive Table 6, based on the convenience of the user, also indicates thatspur gears are suitable

2. Compute the pitch diameter of each gear

The distance between the driving and driven shafts is 21 in (53.3 cm) This distance

is approximately equal to the sum of the driving gear pitch radius r D in and the

driven gear pitch radius r d in Or d D⫹r d⫽ 21 in (53.3 cm)

In this installation the driving gear is mounted on the motor shaft and turns

at 1800 r / min The driven gear is mounted on the compressor shaft and turns at

300 r / min Thus, the speed ratio of the gears (R D , driver rpm / R d, driven rpm)⫽

1800 / 300⫽6 For a spur gear, R D / R d⫽r d / r D, or 6⫽r d / r D , and r d⫽6r D Hence,

substituting in r D⫹r d⫽21, r D⫹6r D⫽ 21; r D⫽3 in (7.6 cm) Then 3⫹r d⫽

21, r d⫽18 in (45.7 cm) The respective pitch diameters of the gears are d D⫽2⫻

3⫽6.0 in (15.2 cm); d d⫽2⫻18 ⫽36.0 in (91.4 cm)

3. Determine the number of teeth in each gear

The number of teeth in a spur gearset, N D and N d, can be approximated from the

ratio R D / R d ⫽N d / N D, or 1800 / 300⫽ N d / N D ; N d⫽ 6N D Hence, the driven gearwill have approximately six times as many teeth as the driving gear

As a trial, assume that N d⫽72 teeth; then N D⫽N d/ 6⫽72 / 6⫽12 teeth Thisassumption must now be checked to determine whether the gears will give the

desired output speed Since R D / R d⫽N d / N D, or 1800 / 300⫽72 / 12; 6⫽6 Thus,the gears will provide the desired speed change

The distance between the shafts is 21 in (53.3 cm)⫽r D⫹r d This means thatthere is no clearance when the gears are meshed Since all gears require someclearance, the shafts will have to be moved apart slightly to provide this clearance

If the shafts cannot be moved apart, the gear diameter must be reduced In thisinstallation, however, the electric-motor driver can probably be moved a fraction of

an inch to provide the desired clearance

4. Choose the final gear size

Refer to a catalog of stock gears From this catalog choose a driving and a drivengear having the required number of teeth and the required pitch diameter If gears

of the exact size required are not available, pick the nearest suitable stock sizes