Description

QUESTION 1

[3 Marks]

Answer the following questions:

(a) Why long center-to-center distances are not recommended for V belts? What is the

recommended range for center-to- center distance?

(1 mark)

(b) What are the advantages of having endless V-belt types?

(1 mark)

(c) Draw a free body diagram of a pulley and part of a belt showing the effective arc and the

idle arc. What is the difference between the two arcs?

(1 mark)

Final Exam, ME 414, Sem 372, Version A, JUC Male Branch

Page 1 of 12

QUESTION 2

[4 Marks]

A helical compression spring is wound using 1 mm-diameter music wire. The outside coil

diameter of the spring is 11 mm. The ends are squared and there are 12 total turns.

(a)

(b)

(c)

(d)

(e)

Estimate the torsional yield strength of the wire.

Estimate the static load corresponding to the yield strength.

Estimate the scale of the spring.

Estimate the solid length of the spring.

Is there any possibility for buckling?

Final Exam, ME 414, Sem 372, Version A, JUC Male Branch

(1 mark)

(1 mark)

(1 mark)

(0.5 mark)

(0.5 mark)

Page 2 of 12

Final Exam, ME 414, Sem 372, Version A, JUC Male Branch

Page 3 of 12

QUESTION 3

[4 Marks]

An 02-series single-row deep-groove ball bearing with a 65-mm bore is loaded with a 3-kN axial

load and a 7-kN radial load. The outer ring rotates at 500 rev/min.

(a) Determine the equivalent radial load that will be experienced by this particular bearing.

(2 marks)

(b) Determine whether this bearing should be expected to carry this load with a 95 percent

reliability for 10 kh.

(2 marks)

(The Weibull parameters are x0 = 0.02, (ÃŽÂ¸ Ã¢Ë†â€™ x0) = 4.439, and b = 1.483.)

Final Exam, ME 414, Sem 372, Version A, JUC Male Branch

Page 4 of 12

Final Exam, ME 414, Sem 372, Version A, JUC Male Branch

Page 5 of 12

QUESTION 4

[6 Marks]

A compound gearbox is needed to provide an exact 45:1 increase in speed, while minimizing the

overall gearbox size. With a 20o pressure angle,

(a) Specify appropriate numbers of teeth to minimize the gearbox size while avoiding the

interference problem in the teeth.

(4 marks)

(b) If the module of all gears is 3 mm, find the center distance between the input and the

output shafts.

(2 marks)

Final Exam, ME 414, Sem 372, Version A, JUC Male Branch

Page 6 of 12

Final Exam, ME 414, Sem 372, Version A, JUC Male Branch

Page 7 of 12

QUESTION 5

[6 Marks]

Figure 1 shows a steel spur pinion that has a module of 1 mm and 16 teeth cut on the 20Ã¢â€”Â¦ fulldepth system. If the pinion is to carry 0.15 kW at 400 rev/min determine:

(a) A suitable face width based on an allowable bending stress of 150 MPa. (4 marks)

(b) The force and torque exerted by shaft a against pinion 2.

(2 marks)

Final Exam, ME 414, Sem 372, Version A, JUC Male Branch

Page 8 of 12

Final Exam, ME 414, Sem 372, Version A, JUC Male Branch

Page 9 of 12

QUESTION 6

[7 Marks]

A 150-mm-wide polyamide F-1 flat belt is used to connect a 50-mm-diameter pulley to drive a

larger pulley with an angular velocity ratio of 0.5. The center-to-center distance is 2.7 m. The

angular speed of the small pulley is 1750 rev/min as it delivers 1.5 kW. The service is such that a

service factor Ks of 1.25 is appropriate.

(a) Estimate the centrifugal tension Fc and the torque T.

(b) Estimate the allowable F1, F2, and Fi.

(c) Check the friction development.

Final Exam, ME 414, Sem 372, Version A, JUC Male Branch

(3 marks)

(3 marks)

(1 marks)

Page 10 of 12

Final Exam, ME 414, Sem 372, Version A, JUC Male Branch

Page 11 of 12

– END OF EXAM PAPER Ã¢â‚¬â€œ

Final Exam, ME 414, Sem 372, Version A, JUC Male Branch

Page 12 of 12

QUESTION 1

[2 Marks]

(a) Compare between life, rating life and median life of bearings. (1.5 marks)

(b) What is Equivalent radial load. (0.5 Mark)

Mid Term Exam, ME 414, Sem 421, JUC Male Branch

Page 1 of 7

QUESTION 2

[6 Marks]

A helical compression spring is to be made of oil-tempered wire of with a spring index of C =

10. The spring is to operate inside a hole, so buckling is not a problem and the ends can be left

plain. The free length of the spring should be 80 mm. A force of 50 N should deflect the spring

15 mm. If the force corresponding to a solid length is 180 N and the factor of safety is 1.3, find:

(a)

The spring rate (stiffness).

(b) The wire diameter.

(c) The mean coil diameter.

(d)

The total number of coils.

(0.5 Mark)

(4 Marks)

(0.5 Marks)

(1 Mark)

Mid Term Exam, ME 414, Sem 421, JUC Male Branch

Page 2 of 7

Mid Term Exam, ME 414, Sem 421, JUC Male Branch

Page 3 of 7

QUESTION 3

[6 Marks]

An angular-contact, inner ring rotating, 02-series ball bearing is required for an application in

which the life requirement is 20 kh at 480 rev/min. The design radial load is 3 kN. The

application factor is 1.2. The reliability goal is 0.90.

(a) Find the multiple of rating life xD required and the catalog rating C10.

(b) Choose a bearing and estimate the existing reliability in service.

(3.5 Marks)

(2.5 Marks)

(The Weibull parameters are x0 = 0.02, (ÃŽÂ¸ Ã¢Ë†â€™ x0) = 4.439, and b = 1.483)

Mid Term Exam, ME 414, Sem 421, JUC Male Branch

Page 4 of 7

Mid Term Exam, ME 414, Sem 421, JUC Male Branch

Page 5 of 7

QUESTION 4

[6 Marks]

A 02-series single-row deep-groove ball bearing with a 30 mm bore is loaded with a 2-kN axial

load and a 5-kN radial load. The inner ring rotates at 400 rev/min.

(a) Determine the equivalent radial load that will be experienced by this particular bearing.

(4 Marks)

(b) Determine the predicted life (in hours) that this bearing could be expected to give in this

application with a 99 percent reliability. Assume the application factor is 1.

(2 Marks)

(The Weibull parameters are x0 = 0.02, (ÃŽÂ¸ Ã¢Ë†â€™ x0) = 4.439, and b = 1.483.)

Mid Term Exam, ME 414, Sem 421, JUC Male Branch

Page 6 of 7

– END OF EXAM PAPER Ã¢â‚¬â€œ

Mid Term Exam, ME 414, Sem 421, JUC Male Branch

Page 7 of 7

Machine Design II

ME 414

Mechanical Springs

Lect. 1.1

Dr. Mir Md Maruf Morshed

Assessment

Assessment

Task

%

1

Assignment 1

5

2

Quiz 1

5

3

Mid Term

20

4

Assignment 2

5

5

Quiz 2

5

6

Final Exam.

30

7

Practical

30

Syllabus

Unit1. Mechanical Springs

Unit2. Bearings (Journal & Anti-Friction)

Unit3. Gear Design

Unit4. Clutches and Brakes

Unit5. Flexible Mechanical Elements

Mechanical Springs

Exert Force

Ã¯â€šâ€” Provide flexibility

Ã¯â€šâ€” Store or absorb energy

Ã¯â€šâ€”

Type of Springs

In general, springs may be classified as either

wire springs,

flat springs,

or special-shaped springs,

and there are variations within these divisions.

Types of Springs

Wire springs include helical springs of

round or square wire that are;

cylindrical or conical in shape and are made to resist:

tensile, compressive, or torsional loads.

Helical Extension

Helical compression

Helical Torsion

Helical Spring

Ã¯â€šâ€”

Among the various springs helical or coil compression springs

are the widely used ones and hence discussions will be confined

to the helical (coil) compression springs.

The basic nomenclature of helical springs

are illustrated below:

d

Ã¯â€šâ€” D

Ã¯â€šâ€” J

Ã¯â€šâ€” Lo

Ã¯â€šâ€” p

Ã¯â€šâ€”

Wire diameter

Mean coil diameter

Polar Moment of Inertia

Free length

Pitch

Helical Spring

d Wire diameter

Ã¯â€šâ€” D Mean coil diameter

Ã¯â€šâ€” Di Inside coil diameter

Ã¯â€šâ€” Do Outside coil diameter

Ã¯â€šâ€” J Polar Moment of Inertia

Ã¯â€šâ€” Lo (or Lf) Free length

Ã¯â€šâ€” Ls Solid length

Ã¯â€šâ€” Nt Number of coils (turns)

Ã¯â€šâ€”

Helical Spring

Ã¯â€šâ€”

Ã¯â€šâ€”

Helical coil spring with round wire

Equilibrium forces at cut section anywhere in the body of the

spring indicates direct shear and torsion

Fig. 10Ã¢â‚¬â€œ1

Ends of Compression Springs

https://www.youtube.com/watch?v=wFhyZ

UvZSDA

Fig. 10Ã¢â‚¬â€œ2

Formulas for Compression Springs With Different Ends

Table 10Ã¢â‚¬â€œ1

Na is the number of active coils

Some Common Spring Steels

Ã¯â€šâ€”

Ã¯â€šâ€”

Ã¯â€šâ€”

Hard-drawn wire (0.60-0.70C)

Ã¢â€”Â¦ Cheapest general-purpose

Ã¢â€”Â¦ Use only where life, accuracy, and deflection are not too important

Oil-tempered wire (0.60-0.70C)

Ã¢â€”Â¦ General-purpose

Ã¢â€”Â¦ Heat treated for greater strength and uniformity of properties

Ã¢â€”Â¦ Often used for larger diameter spring wire

Music wire (0.80-0.95C)

Ã¢â€”Â¦ Higher carbon for higher strength

Ã¢â€”Â¦ Best, toughest, and most widely used for small springs

Ã¢â€”Â¦ Good for fatigue

Ã¯Æ’Ëœ The unified numbering system (UNS) is an alloy designation system

widely accepted in North America.

Ã¯Æ’Ëœ American Iron and Steel Institute (AISI)

Ã¯Æ’Ëœ American Society for Testing and Materials (ASTM)

Some Common Spring Steels

Ã¯â€šâ€”

Chrome-vanadium

Ã¢â€”Â¦ Popular alloy spring steel

Ã¢â€”Â¦ Higher strengths than plain carbon steels

Ã¢â€”Â¦ Good for fatigue, shock, and impact

Ã¯â€šâ€”

Chrome-silicon

Ã¢â€”Â¦ Good for high stresses, long fatigue life, and shock

Machine Design II

Mechanical Springs

Lect. 1.2

Dr. Mir Md Maruf Morshed

Helical Spring

Ã¯â€šâ€”

Ã¯â€šâ€”

Helical coil spring with round wire

Equilibrium forces at cut section anywhere in the body of the

spring indicates direct shear and torsion

Fig. 10Ã¢â‚¬â€œ1

Stresses in Helical Springs

Ã¯â€šâ€”

Torsional shear and direct shear

Additive (maximum) on inside fiber of

cross-section

Ã¯â€šâ€”

Substitute terms

Ã¯â€šâ€”

Fig. 10Ã¢â‚¬â€œ1b

Stresses in Helical Springs

Factor out the torsional stress

d Ã¯Æ’Â¶Ã¯Æ’Â¦ 8FD Ã¯Æ’Â¶

Ã¯Æ’Â¦

Ã¯ÂÂ´ Ã¯â‚¬Â½ Ã¯Æ’Â§1 Ã¯â‚¬Â«

Ã¯Æ’Â·Ã¯Æ’Â§

3 Ã¯Æ’Â·

2

D

Ã¯ÂÂ°

d

Ã¯Æ’Â¨

Ã¯Æ’Â¸Ã¯Æ’Â¨

Ã¯Æ’Â¸

Define Spring Index

Define Shear Stress Correction Factor

1

2C Ã¯â‚¬Â« 1

Ks Ã¯â‚¬Â½ 1 Ã¯â‚¬Â«

Ã¯â‚¬Â½

2C

2C

Maximum shear stress for helical spring

Curvature Effect

Ã¯â€šâ€”

Can account for effect by replacing Ks with Wahl factor or

BergstrÃƒÂ¤sser factor which account for both direct shear and

curvature effect

Deflection of Helical Springs

Use CastiglianoÃ¢â‚¬â„¢s method to relate force and deflection

Fig. 10Ã¢â‚¬â€œ1a

Critical Deflection for Stability

Ã¯â€šâ€”

Buckling type of instability can occur in compression springs

when the deflection exceeds the critical deflection ycr

Ã¯â€šâ€”

ÃŽÂ»eff is the effective slenderness ratio

The spring slenderness ratio is the proportion of the spring’s mean diameter to its length. If

a spring’s mean diameter is more than 3 times its length, it will tend to buckle.

Ã¯â€šâ€”

a is the end-condition constant, defined on the next slide

Ã¯â€šâ€”

C’1 and C’2 are elastic constants

End-Condition Constant

Ã¯â€šâ€”

Ã¯â€šâ€”

Ã¯â€šâ€”

The a term in Eq. (10Ã¢â‚¬â€œ11) is the end-condition constant.

It accounts for the way in which the ends of the spring are

supported.

Values are given in Table 10Ã¢â‚¬â€œ2.

Table 10Ã¢â‚¬â€œ2

Absolute Stability

Ã¯â€šâ€”

Absolute stability occurs when, in Eq. (10Ã¢â‚¬â€œ10),

2

C2Ã¯â€šÂ¢ / Ã¯ÂÂ¬eff

Ã¯â‚¬Â¾1

Ã¯â€šâ€”

This results in the condition for absolute stability

Ã¯â€šâ€”

For steels, this turns out to be

Strength of Spring Materials

Ã¯â€šâ€”

Ã¯â€šâ€”

Ã¯â€šâ€”

Ã¯â€šâ€”

With small wire diameters, strength is a function of diameter.

A graph of tensile strength vs. wire diameter is almost a straight

line on log-log scale.

The equation of this line is

where A is the intercept and m is the slope.

Values of A and m for common spring steels are given in Table

10Ã¢â‚¬â€œ4.

Constants for Estimating Tensile Strength

Table 10Ã¢â‚¬â€œ4

Estimating Torsional Yield Strength

Since helical springs experience shear stress, shear yield strength

is needed.

Ã¯â€šâ€” If actual data is not available, estimate from tensile strength

Ã¯â€šâ€” Assume tensile yield strength is between 60-90% of tensile

strength

0.6Sut Ã¯â€šÂ£ Ssy Ã¯â€šÂ£ 0.9Sut

Ã¯â€šâ€”

Ã¯â€šâ€”

Ã¯â€šâ€”

Assume the distortion energy theory can be employed to relate

the shear strength to the normal strength.

Ssy = 0.577Sy

This results in

Mechanical Properties of Some Spring Wires (Table 10Ã¢â‚¬â€œ5)

Diameter

d, mm

3

3

Maximum Allowable Torsional Stresses

Set Removal

Set removal or presetting is a process used in manufacturing a

spring to induce useful residual stresses.

Ã¯â€šâ€” The spring is made longer than needed, then compressed to solid

height, intentionally exceeding the torsional yield strength.

Ã¯â€šâ€” This operation sets the spring to the required final free length.

Ã¯â€šâ€” Yielding induces residual stresses opposite in direction to those

induced in service.

Ã¯â€šâ€” 10 to 30 percent of the initial free length should be removed.

Ã¯â€šâ€” Set removal increases the strength of the spring and is useful for

energy-storage purposes but it is not recommended when springs

are subject to fatigue.

Ã¯â€šâ€”

Example 10Ã¢â‚¬â€œ1

11 mm.

Example 10Ã¢â‚¬â€œ1

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Example 10Ã¢â‚¬â€œ1

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Machine Design II

Mechanical Springs

Lect. 1.3

Dr. Mir Md Maruf Morshed

Helical Compression Spring Design for Static Service

Limit the design solution space by setting some practical limits

Ã¯â€šâ€” Preferred range for spring index

Ã¯â€šâ€”

Ã¯â€šâ€”

Preferred range for number of active coils

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Helical Compression Spring Design for Static Service

Ã¯â€šâ€”

To achieve best linearity of spring constant, preferred to limit

operating force to the central 75% of the force-deflection curve

between F = 0 and F = Fs.

This limits the maximum operating force to Fmax Ã¢â€°Â¤ 7/8 Fs

Define fractional overrun to closure as ÃŽÂ¾ where

Ã¯â€šâ€”

This leads to

Ã¯â€šâ€”

Solving the outer equality for ÃŽÂ¾,

Ã¯â€šâ€”

Thus, it is recommended that

Ã¯â€šâ€”

Ã¯â€šâ€”

ÃŽÂ¾ = 1/7 = 0.143 Ã¯Ââ€š0.15

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Summary of Recommended Design Conditions

Ã¯â€šâ€”

The following design conditions are recommended for helical

compression spring design for static service

where ns is the factor of safety at closure (solid height).

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Figure of Merit for High Volume Production

For high volume production, the figure of merit (fom) may be the

cost of the wire.

Ã¯â€šâ€” The fom would be proportional to the relative material cost,

weight density, and volume

Ã¯â€šâ€”

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Design Flowchart for Static Loading

Choose d

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Design Flowchart for Static Loading

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Example 10Ã¢â‚¬â€œ2

an 89 N

50.8 mm. Because of

25.4 mm and the free

101.6 mm. Design the spring.

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Example 10Ã¢â‚¬â€œ2

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Practice Problems

Example (Problem -10-5)

Exercise (Problem- 27, 28, 30)

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Lecture Slides

Rolling-Contact Bearings

Lect. 2.1

The McGraw-Hill Companies Ã‚Â© 2012

Bearings

Why bearings?

https://www.youtube.com/watch?v=lgGd44

VRneY

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Bearings

Type of bearings

– Rolling contact bearing (Ball Bearing & Roller Bearing)

is used to transfer the main load through elements in rolling contact

rather than in sliding contact.

– Sliding-contact bearing (Journal bearing)

In a rolling bearing the starting friction is about twice the running

friction, but still it is negligible in comparison with the starting

friction of a sleeve (journal) bearing.

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Bearings

What affect the frictional characteristics of a bearing?

– Load,

– speed,

– and the operating viscosity of the lubricant

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Ball Bearings

Bearings are manufactured to take

– pure radial loads,

– pure thrust loads,

– or a combination of the two kinds of loads.

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Bearings

The nomenclature of a ball bearing is

illustrated in Fig. 11Ã¢â‚¬â€œ1, which also

shows the four essential parts of a

bearing. These are

– the outer ring,

– the inner ring,

– the balls or rolling elements,

– and the separator.

In low-priced bearings, the separator

is sometimes omitted, but it has the

important function of separating the

elements so that rubbing contact will

not occur.

Fig. 11Ã¢â‚¬â€œ1

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Types of Ball Bearings

Some of the various types of standardized bearings that are manufactured

are shown in Fig. 11Ã¢â‚¬â€œ2.

The single-row deep-groove bearing (Fig. 11Ã¢â‚¬â€œ2a) will take radial load as

well as some thrust load..

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Types of Ball Bearings

The use of a filling notch (Fig. 11Ã¢â‚¬â€œ2b) in the inner and outer rings enables

a greater number of balls to be inserted, thus increasing the load capacity.

The thrust capacity is decreased, however, because of the bumping of the

balls against the edge of the notch when thrust loads are present.

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Types of Ball Bearings

The angular-contact bearing (Fig. 11Ã¢â‚¬â€œ2c) provides a greater thrust

capacity.

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Types of Ball Bearings

All these bearings may be obtained with shields (Fig. 11Ã¢â‚¬â€œ2d) on one or

both sides. The shields are not a complete closure but do offer a measure

of protection against dirt.

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Types of Ball Bearings

It has seals (Fig. 11Ã¢â‚¬â€œ2e) that prevent lubricant from going out of the

bearing and dirt or any liquid from entering inside the bearing.

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Types of Ball Bearings

Single-row bearings will withstand a small amount of shaft misalignment of

deflection, but where this is severe, self-aligning bearings (Fig. 11Ã¢â‚¬â€œ2f and

h) may be used.

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Types of Ball Bearings

Double-row bearings (Fig. 11Ã¢â‚¬â€œ2g) are made in a variety of types and

sizes to carry heavier radial and thrust loads. Sometimes two single-row

bearings are used together for the same reason, although a double-row

bearing will generally require fewer parts and occupy less space.

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Types of Ball Bearings

The one-way ball thrust bearings (Fig. 11Ã¢â‚¬â€œ2i) are made in many types

and sizes.

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Types of Roller Bearings

Some of the large variety of standard roller bearings available are

illustrated in Fig. 11Ã¢â‚¬â€œ3.

Straight roller bearings (Fig. 11Ã¢â‚¬â€œ3a) will carry a greater radial load

than ball bearings of the same size because of the greater contact area.

However, they have the disadvantage of. requiring almost perfect

geometry of the raceways and rollers

A slight misalignment will cause the rollers to skew and get out of

line. For this reason, the retainer must be heavy. Straight roller bearings

will not, of course, take thrust loads.

.

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Types of Roller Bearings

The straight-roller bearing (Fig. 11Ã¢â‚¬â€œ3a) takes high radial load, but does

not take thrust load.

(Fig. 11Ã¢â‚¬â€œ3)

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Types of Roller Bearings

The spherical-roller thrust bearing (Fig. 11Ã¢â‚¬â€œ3b) is useful where heavy

loads and misalignment occur. It takes both radial and thrust loads.

The spherical elements have the advantage of increasing their contact

area as the load is increased.

(Fig. 11Ã¢â‚¬â€œ3)

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Types of Roller Bearings

Needle bearings (Fig. 11Ã¢â‚¬â€œ3d) are very useful where radial space is

limited.

They have a high load capacity when separators are used, but may be

obtained without separators.

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Types of Roller Bearings

Tapered roller bearings (Fig. 11Ã¢â‚¬â€œ3e, f ) combine the advantages of ball

and straight roller bearings, since they can take either radial or thrust

loads or any combination of the two, and in addition, they have the high

load-carrying capacity of straight roller bearings.

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Lecture Slides

Lubrication and

Journal Bearings

Lect. 2.4

The McGraw-Hill Companies Ã‚Â© 2012

Lubrication

The object of lubrication:

to reduce friction, wear, and heating of machine parts that

move relative to each other.

A lubricant:

is any substance that, when inserted between the moving

surfaces, accomplishes these purposes.

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Lubrication

In a sleeve bearing, a shaft, or journal, rotates or oscillates within a

sleeve, or bushing, and the relative motion is sliding.

In an antifriction bearing, the main relative motion is rolling.

A follower may either roll or slide on the cam.

Gear teeth mate with each other by a combination of rolling and

sliding.

Pistons slide within their cylinders.

All these applications require lubrication to reduce friction,

wear, and heating

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Types of Lubrication

Hydrodynamic

Ã¯â€šâ€” Hydrostatic

Ã¯â€šâ€” Elastohydrodynamic

Ã¯â€šâ€” Boundary

Ã¯â€šâ€” Solid film

Ã¯â€šâ€”

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Types of Lubrication

Ã¯â€šâ€”

Hydrodynamic

Hydrodynamic lubrication means that the load-carrying surfaces of the bearing are separated

by a relatively thick film of lubricant, so as to prevent metal-to-metal contact.

Hydrodynamic lubrication does not depend upon the introduction of the lubricant

under pressure, but it does require the existence of an adequate supply at all times. The film

pressure is created by the moving surface itself pulling the lubricant into a wedge-shaped

zone at a velocity sufficiently high to create the pressure necessary to separate the surfaces

against the load on the bearing.

Hydrodynamic lubrication is also called full-film, or fluid, lubrication.

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Types of Lubrication

Hydrostatic

Hydrostatic lubrication is obtained by introducing the lubricant,

which is sometimes air or water, into the load-bearing area at a

pressure high enough to separate the surfaces with a relatively thick

film of lubricant. So, unlike hydrodynamic lubrication, this kind of

lubrication does not require motion of one surface relative to

another.

Ã¯â€šâ€”

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Types of Lubrication

Elastohydrodynamic

Elastohydrodynamic lubrication is the

phenomenon that occurs when a lubricant is

introduced between surfaces that are in

rolling contact, such as mating gears or

rolling bearings.

Boundary Lubricant

Insufficient surface area, a drop in the velocity of the moving surface, a lessening in the

quantity of lubricant delivered to a bearing, an increase in the bearing load, or an

increase in lubricant temperature resulting in a decrease in viscosityÃ¢â‚¬â€ any one of these

Ã¢â‚¬â€may prevent the buildup of a film thick enough for full-film lubrication. When this

happens, the highest asperities may be separated by lubricant films only several

molecular dimensions in thickness. This is called boundary lubrication.

Solid film

Solid film lubricants are paint-like coatings of very

fine particles of lubricating pigment blended with a binder and other additives. The

lubricant is applied to a substrate by spray, dip or brush methods and, once cured,

creates a solid film which repels water, reduces friction and increases the wear life of the

substrate to which it has been applied. Certain film lubricants also offer additional

properties such as corrosion inhibition. Solid film lubricants are used in the automotive,

transportation and aerospace industries.

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Viscosity

Ã¯â€šâ€”

Shear stress in a fluid is proportional to the rate of change of

velocity with respect to y

Ã¯â€šâ€”

m is absolute viscosity, also called dynamic viscosity

Fig. 12Ã¢â‚¬â€œ1

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Viscosity

Ã¯â€šâ€”

For most lubricating fluids, the rate of shear is constant, thus

Ã¯â€šâ€”

Fluids exhibiting this characteristic are called Newtonian fluids

Fig. 12Ã¢â‚¬â€œ1

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Units of Viscosity

Ã¯â€šâ€”

Ã¯â€šâ€”

Ã¯â€šâ€”

Ã¯â€šâ€”

Units of absolute viscosity

Ã¢â€”Â¦ ips units: reyn = lbfÃ‚Â·s/in2

Ã¢â€”Â¦ SI units: PaÃ‚Â·s = NÃ‚Â·s/m2

Ã¢â€”Â¦ cgs units: Poise =dynÃ‚Â·s/cm2

cgs units are discouraged, but common historically in lubrication

Viscosity in cgs is often expressed in centipoise (cP), designated

by Z

Conversion from cgs to SI and ips:

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Measurement of Viscosity

Saybolt Universal Viscosimeter used to measure viscosity

Ã¯â€šâ€” Measures time in seconds for 60 mL of lubricant at specified

temperature to run through a tube 17.6 mm in diameter and 12.25

mm long

Ã¯â€šâ€” Result is kinematic viscosity

Ã¯â€šâ€” Unit is stoke = cm2/s

Ã¯â€šâ€” Using Hagen-Poiseuille law kinematic viscosity based on seconds

Saybolt, also called Saybolt Universal viscosity (SUV) in seconds

is

Ã¯â€šâ€”

where Zk is in centistokes (cSt) and t is the number of seconds

Saybolt

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Measurement of Viscosity

Ã¯â€šâ€”

In SI, kinematic viscosity n has units of m2/s

Conversion is

Eq. (12Ã¢â‚¬â€œ3) in SI units,

Ã¯â€šâ€”

To convert to dynamic viscosity, multiply v by density in SI units

Ã¯â€šâ€”

Ã¯â€šâ€”

where r is in kg/m3 and m is in pascal-seconds

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Comparison of Absolute Viscosities of Various Fluids

Fig. 12Ã¢â‚¬â€œ2

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

PetroffÃ¢â‚¬â„¢s Lightly Loaded Journal Bearing

Ã¯â€šâ€”

Fig. 12Ã¢â‚¬â€œ3

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

PetroffÃ¢â‚¬â„¢s Equation

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Important Dimensionless Parameters

Ã¯â€šâ€”

Some important dimensionless parameters used in lubrication

Ã¢â€”Â¦ r/c radial clearance ratio

Ã¢â€”Â¦ mN/P

Ã¢â€”Â¦ Sommerfeld number or bearing characteristic number

Ã¯â€šâ€”

Interesting relation

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Stable Lubrication

To the right of AB, changes in

conditions are self-correcting

and results in stable lubrication

Ã¯â€šâ€” To the left of AB, changes in

conditions tend to get worse

and results in unstable

lubrication

Ã¯â€šâ€” Point C represents the

approximate transition between

metal-to-metal contact and

thick film separation of the

parts

Ã¯â€šâ€” Common design constraint for

point B,

Ã¯â€šâ€”

Fig. 12Ã¢â‚¬â€œ4

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Thick Film Lubrication

Ã¯â€šâ€”

Formation of a film

Fig. 12Ã¢â‚¬â€œ5

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Machine Design II

ME 414

Mechanical Springs

Lect. 1.1

Dr. Mir Md Maruf Morshed

Assessment

Assessment

Task

%

1

Assignment 1

5

2

Quiz 1

5

3

Mid Term

20

4

Assignment 2

5

5

Quiz 2

5

6

Final Exam.

30

7

Practical

30

Syllabus

Unit1. Mechanical Springs

Unit2. Bearings (Journal & Anti-Friction)

Unit3. Gear Design

Unit4. Clutches and Brakes

Unit5. Flexible Mechanical Elements

Mechanical Springs

Exert Force

Ã¯â€šâ€” Provide flexibility

Ã¯â€šâ€” Store or absorb energy

Ã¯â€šâ€”

Type of Springs

In general, springs may be classified as either

wire springs,

flat springs,

or special-shaped springs,

and there are variations within these divisions.

Types of Springs

Wire springs include helical springs of

round or square wire that are;

cylindrical or conical in shape and are made to resist:

tensile, compressive, or torsional loads.

Helical Extension

Helical compression

Helical Torsion

Helical Spring

Ã¯â€šâ€”

Among the various springs helical or coil compression springs

are the widely used ones and hence discussions will be confined

to the helical (coil) compression springs.

The basic nomenclature of helical springs

are illustrated below:

d

Ã¯â€šâ€” D

Ã¯â€šâ€” J

Ã¯â€šâ€” Lo

Ã¯â€šâ€” p

Ã¯â€šâ€”

Wire diameter

Mean coil diameter

Polar Moment of Inertia

Free length

Pitch

Helical Spring

d Wire diameter

Ã¯â€šâ€” D Mean coil diameter

Ã¯â€šâ€” Di Inside coil diameter

Ã¯â€šâ€” Do Outside coil diameter

Ã¯â€šâ€” J Polar Moment of Inertia

Ã¯â€šâ€” Lo (or Lf) Free length

Ã¯â€šâ€” Ls Solid length

Ã¯â€šâ€” Nt Number of coils (turns)

Ã¯â€šâ€”

Helical Spring

Ã¯â€šâ€”

Ã¯â€šâ€”

Helical coil spring with round wire

Equilibrium forces at cut section anywhere in the body of the

spring indicates direct shear and torsion

Fig. 10Ã¢â‚¬â€œ1

Ends of Compression Springs

https://www.youtube.com/watch?v=wFhyZ

UvZSDA

Fig. 10Ã¢â‚¬â€œ2

Formulas for Compression Springs With Different Ends

Table 10Ã¢â‚¬â€œ1

Na is the number of active coils

Some Common Spring Steels

Ã¯â€šâ€”

Ã¯â€šâ€”

Ã¯â€šâ€”

Hard-drawn wire (0.60-0.70C)

Ã¢â€”Â¦ Cheapest general-purpose

Ã¢â€”Â¦ Use only where life, accuracy, and deflection are not too important

Oil-tempered wire (0.60-0.70C)

Ã¢â€”Â¦ General-purpose

Ã¢â€”Â¦ Heat treated for greater strength and uniformity of properties

Ã¢â€”Â¦ Often used for larger diameter spring wire

Music wire (0.80-0.95C)

Ã¢â€”Â¦ Higher carbon for higher strength

Ã¢â€”Â¦ Best, toughest, and most widely used for small springs

Ã¢â€”Â¦ Good for fatigue

Ã¯Æ’Ëœ The unified numbering system (UNS) is an alloy designation system

widely accepted in North America.

Ã¯Æ’Ëœ American Iron and Steel Institute (AISI)

Ã¯Æ’Ëœ American Society for Testing and Materials (ASTM)

Some Common Spring Steels

Ã¯â€šâ€”

Chrome-vanadium

Ã¢â€”Â¦ Popular alloy spring steel

Ã¢â€”Â¦ Higher strengths than plain carbon steels

Ã¢â€”Â¦ Good for fatigue, shock, and impact

Ã¯â€šâ€”

Chrome-silicon

Ã¢â€”Â¦ Good for high stresses, long fatigue life, and shock

Machine Design II

Mechanical Springs

Lect. 1.2

Dr. Mir Md Maruf Morshed

Helical Spring

Ã¯â€šâ€”

Ã¯â€šâ€”

Helical coil spring with round wire

Equilibrium forces at cut section anywhere in the body of the

spring indicates direct shear and torsion

Fig. 10Ã¢â‚¬â€œ1

Stresses in Helical Springs

Ã¯â€šâ€”

Torsional shear and direct shear

Additive (maximum) on inside fiber of

cross-section

Ã¯â€šâ€”

Substitute terms

Ã¯â€šâ€”

Fig. 10Ã¢â‚¬â€œ1b

Stresses in Helical Springs

Factor out the torsional stress

d Ã¯Æ’Â¶Ã¯Æ’Â¦ 8FD Ã¯Æ’Â¶

Ã¯Æ’Â¦

Ã¯ÂÂ´ Ã¯â‚¬Â½ Ã¯Æ’Â§1 Ã¯â‚¬Â«

Ã¯Æ’Â·Ã¯Æ’Â§

3 Ã¯Æ’Â·

2

D

Ã¯ÂÂ°

d

Ã¯Æ’Â¨

Ã¯Æ’Â¸Ã¯Æ’Â¨

Ã¯Æ’Â¸

Define Spring Index

Define Shear Stress Correction Factor

1

2C Ã¯â‚¬Â« 1

Ks Ã¯â‚¬Â½ 1 Ã¯â‚¬Â«

Ã¯â‚¬Â½

2C

2C

Maximum shear stress for helical spring

Curvature Effect

Ã¯â€šâ€”

Can account for effect by replacing Ks with Wahl factor or

BergstrÃƒÂ¤sser factor which account for both direct shear and

curvature effect

Deflection of Helical Springs

Use CastiglianoÃ¢â‚¬â„¢s method to relate force and deflection

Fig. 10Ã¢â‚¬â€œ1a

Critical Deflection for Stability

Ã¯â€šâ€”

Buckling type of instability can occur in compression springs

when the deflection exceeds the critical deflection ycr

Ã¯â€šâ€”

ÃŽÂ»eff is the effective slenderness ratio

The spring slenderness ratio is the proportion of the spring’s mean diameter to its length. If

a spring’s mean diameter is more than 3 times its length, it will tend to buckle.

Ã¯â€šâ€”

a is the end-condition constant, defined on the next slide

Ã¯â€šâ€”

C’1 and C’2 are elastic constants

End-Condition Constant

Ã¯â€šâ€”

Ã¯â€šâ€”

Ã¯â€šâ€”

The a term in Eq. (10Ã¢â‚¬â€œ11) is the end-condition constant.

It accounts for the way in which the ends of the spring are

supported.

Values are given in Table 10Ã¢â‚¬â€œ2.

Table 10Ã¢â‚¬â€œ2

Absolute Stability

Ã¯â€šâ€”

Absolute stability occurs when, in Eq. (10Ã¢â‚¬â€œ10),

2

C2Ã¯â€šÂ¢ / Ã¯ÂÂ¬eff

Ã¯â‚¬Â¾1

Ã¯â€šâ€”

This results in the condition for absolute stability

Ã¯â€šâ€”

For steels, this turns out to be

Strength of Spring Materials

Ã¯â€šâ€”

Ã¯â€šâ€”

Ã¯â€šâ€”

Ã¯â€šâ€”

With small wire diameters, strength is a function of diameter.

A graph of tensile strength vs. wire diameter is almost a straight

line on log-log scale.

The equation of this line is

where A is the intercept and m is the slope.

Values of A and m for common spring steels are given in Table

10Ã¢â‚¬â€œ4.

Constants for Estimating Tensile Strength

Table 10Ã¢â‚¬â€œ4

Estimating Torsional Yield Strength

Since helical springs experience shear stress, shear yield strength

is needed.

Ã¯â€šâ€” If actual data is not available, estimate from tensile strength

Ã¯â€šâ€” Assume tensile yield strength is between 60-90% of tensile

strength

0.6Sut Ã¯â€šÂ£ Ssy Ã¯â€šÂ£ 0.9Sut

Ã¯â€šâ€”

Ã¯â€šâ€”

Ã¯â€šâ€”

Assume the distortion energy theory can be employed to relate

the shear strength to the normal strength.

Ssy = 0.577Sy

This results in

Mechanical Properties of Some Spring Wires (Table 10Ã¢â‚¬â€œ5)

Diameter

d, mm

3

3

Maximum Allowable Torsional Stresses

Set Removal

Set removal or presetting is a process used in manufacturing a

spring to induce useful residual stresses.

Ã¯â€šâ€” The spring is made longer than needed, then compressed to solid

height, intentionally exceeding the torsional yield strength.

Ã¯â€šâ€” This operation sets the spring to the required final free length.

Ã¯â€šâ€” Yielding induces residual stresses opposite in direction to those

induced in service.

Ã¯â€šâ€” 10 to 30 percent of the initial free length should be removed.

Ã¯â€šâ€” Set removal increases the strength of the spring and is useful for

energy-storage purposes but it is not recommended when springs

are subject to fatigue.

Ã¯â€šâ€”

Example 10Ã¢â‚¬â€œ1

11 mm.

Example 10Ã¢â‚¬â€œ1

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Example 10Ã¢â‚¬â€œ1

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Machine Design II

Mechanical Springs

Lect. 1.3

Dr. Mir Md Maruf Morshed

Helical Compression Spring Design for Static Service

Limit the design solution space by setting some practical limits

Ã¯â€šâ€” Preferred range for spring index

Ã¯â€šâ€”

Ã¯â€šâ€”

Preferred range for number of active coils

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Helical Compression Spring Design for Static Service

Ã¯â€šâ€”

To achieve best linearity of spring constant, preferred to limit

operating force to the central 75% of the force-deflection curve

between F = 0 and F = Fs.

This limits the maximum operating force to Fmax Ã¢â€°Â¤ 7/8 Fs

Define fractional overrun to closure as ÃŽÂ¾ where

Ã¯â€šâ€”

This leads to

Ã¯â€šâ€”

Solving the outer equality for ÃŽÂ¾,

Ã¯â€šâ€”

Thus, it is recommended that

Ã¯â€šâ€”

Ã¯â€šâ€”

ÃŽÂ¾ = 1/7 = 0.143 Ã¯Ââ€š0.15

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Summary of Recommended Design Conditions

Ã¯â€šâ€”

The following design conditions are recommended for helical

compression spring design for static service

where ns is the factor of safety at closure (solid height).

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Figure of Merit for High Volume Production

For high volume production, the figure of merit (fom) may be the

cost of the wire.

Ã¯â€šâ€” The fom would be proportional to the relative material cost,

weight density, and volume

Ã¯â€šâ€”

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Design Flowchart for Static Loading

Choose d

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Design Flowchart for Static Loading

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Example 10Ã¢â‚¬â€œ2

an 89 N

50.8 mm. Because of

25.4 mm and the free

101.6 mm. Design the spring.

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Example 10Ã¢â‚¬â€œ2

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Practice Problems

Example (Problem -10-5)

Exercise (Problem- 27, 28, 30)

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Lecture Slides

Rolling-Contact Bearings

Lect. 2.1

The McGraw-Hill Companies Ã‚Â© 2012

Bearings

Why bearings?

https://www.youtube.com/watch?v=lgGd44

VRneY

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Bearings

Type of bearings

– Rolling contact bearing (Ball Bearing & Roller Bearing)

is used to transfer the main load through elements in rolling contact

rather than in sliding contact.

– Sliding-contact bearing (Journal bearing)

In a rolling bearing the starting friction is about twice the running

friction, but still it is negligible in comparison with the starting

friction of a sleeve (journal) bearing.

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Bearings

What affect the frictional characteristics of a bearing?

– Load,

– speed,

– and the operating viscosity of the lubricant

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Ball Bearings

Bearings are manufactured to take

– pure radial loads,

– pure thrust loads,

– or a combination of the two kinds of loads.

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Bearings

The nomenclature of a ball bearing is

illustrated in Fig. 11Ã¢â‚¬â€œ1, which also

shows the four essential parts of a

bearing. These are

– the outer ring,

– the inner ring,

– the balls or rolling elements,

– and the separator.

In low-priced bearings, the separator

is sometimes omitted, but it has the

important function of separating the

elements so that rubbing contact will

not occur.

Fig. 11Ã¢â‚¬â€œ1

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Types of Ball Bearings

Some of the various types of standardized bearings that are manufactured

are shown in Fig. 11Ã¢â‚¬â€œ2.

The single-row deep-groove bearing (Fig. 11Ã¢â‚¬â€œ2a) will take radial load as

well as some thrust load..

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Types of Ball Bearings

The use of a filling notch (Fig. 11Ã¢â‚¬â€œ2b) in the inner and outer rings enables

a greater number of balls to be inserted, thus increasing the load capacity.

The thrust capacity is decreased, however, because of the bumping of the

balls against the edge of the notch when thrust loads are present.

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Types of Ball Bearings

The angular-contact bearing (Fig. 11Ã¢â‚¬â€œ2c) provides a greater thrust

capacity.

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Types of Ball Bearings

All these bearings may be obtained with shields (Fig. 11Ã¢â‚¬â€œ2d) on one or

both sides. The shields are not a complete closure but do offer a measure

of protection against dirt.

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Types of Ball Bearings

It has seals (Fig. 11Ã¢â‚¬â€œ2e) that prevent lubricant from going out of the

bearing and dirt or any liquid from entering inside the bearing.

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Types of Ball Bearings

Single-row bearings will withstand a small amount of shaft misalignment of

deflection, but where this is severe, self-aligning bearings (Fig. 11Ã¢â‚¬â€œ2f and

h) may be used.

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Types of Ball Bearings

Double-row bearings (Fig. 11Ã¢â‚¬â€œ2g) are made in a variety of types and

sizes to carry heavier radial and thrust loads. Sometimes two single-row

bearings are used together for the same reason, although a double-row

bearing will generally require fewer parts and occupy less space.

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Types of Ball Bearings

The one-way ball thrust bearings (Fig. 11Ã¢â‚¬â€œ2i) are made in many types

and sizes.

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Types of Roller Bearings

Some of the large variety of standard roller bearings available are

illustrated in Fig. 11Ã¢â‚¬â€œ3.

Straight roller bearings (Fig. 11Ã¢â‚¬â€œ3a) will carry a greater radial load

than ball bearings of the same size because of the greater contact area.

However, they have the disadvantage of. requiring almost perfect

geometry of the raceways and rollers

A slight misalignment will cause the rollers to skew and get out of

line. For this reason, the retainer must be heavy. Straight roller bearings

will not, of course, take thrust loads.

.

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Types of Roller Bearings

The straight-roller bearing (Fig. 11Ã¢â‚¬â€œ3a) takes high radial load, but does

not take thrust load.

(Fig. 11Ã¢â‚¬â€œ3)

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Types of Roller Bearings

The spherical-roller thrust bearing (Fig. 11Ã¢â‚¬â€œ3b) is useful where heavy

loads and misalignment occur. It takes both radial and thrust loads.

The spherical elements have the advantage of increasing their contact

area as the load is increased.

(Fig. 11Ã¢â‚¬â€œ3)

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Types of Roller Bearings

Needle bearings (Fig. 11Ã¢â‚¬â€œ3d) are very useful where radial space is

limited.

They have a high load capacity when separators are used, but may be

obtained without separators.

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Types of Roller Bearings

Tapered roller bearings (Fig. 11Ã¢â‚¬â€œ3e, f ) combine the advantages of ball

and straight roller bearings, since they can take either radial or thrust

loads or any combination of the two, and in addition, they have the high

load-carrying capacity of straight roller bearings.

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Lecture Slides

Rolling-Contact Bearings

Lect 2.2

The McGraw-Hill Companies Ã‚Â© 2012

Bearing Life Definitions

Ã¯â€šâ€”

Ã¯â€šâ€”

When the ball or roller of rolling-contact bearings rolls, contact

stresses occur on the inner ring, the rolling element, and on the

outer ring.

If a bearing is

Ã¢â€”Â¦ clean and properly lubricated,

Ã¢â€”Â¦ mounted and sealed against the entrance of dust and dirt,

Ã¢â€”Â¦ is maintained in this condition, and is operated at reasonable

temperatures,

then metal fatigue will be the only cause of failure.

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Bearing Life Definitions

Common life measures are

Ã¢â‚¬Â¢ Number of revolutions until the first tangible evidence

of fatigue

Ã¢â‚¬Â¢ Number of hours of use at a standard angular speed until

the first tangible evidence of fatigue

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Bearing Life Definitions

Ã¯â€šâ€”

Bearing Failure: Spalling or pitting of an area of 0.01 in2

Ã¯â€šâ€”

Life: Number of revolutions (or hours @ given speed) required for

failure.

Ã¢â€”Â¦ For one bearing

Ã¯â€šâ€”

Rating Life: Life required for 10% of sample to fail.

Ã¢â€”Â¦ For a group of bearings

Ã¢â€”Â¦ Also called Minimum Life or L10 Life

Ã¯â€šâ€”

Median Life: Average life required for 50% of sample to fail.

Ã¢â€”Â¦ For many groups of bearings

Ã¢â€”Â¦ Also called Average Life or Average Median Life

Ã¢â€”Â¦ Median Life is typically 4 or 5 times the L10 Life

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Load Rating Definitions

Catalog Load Rating, C10:

Constant radial load that causes 10% of a group of bearings to fail at

the bearing manufacturerÃ¢â‚¬â„¢s rating life.

Ã¢â€”Â¦ Depends on type, geometry, accuracy of fabrication, and

material of bearing

Ã¢â€”Â¦ Also called Basic Dynamic Load Rating, and Basic Dynamic

Capacity

Ã¯â€šâ€” Basic Load Rating, C:

A catalog load rating based on a rating life of 106 revolutions of the

inner ring.

Ã¢â€”Â¦ The radial load that would be necessary to cause failure at such a

low life is unrealistically high.

Ã¯â€šâ€”

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Load-Life Relationship

Ã¯â€šâ€”

Ã¯â€šâ€”

Ã¯â€šâ€”

Nominally identical groups of bearings are tested to the life-failure

criterion at different loads.

A plot of load vs. life on log-log scale is approximately linear.

Using a regression equation

to represent the line,

Ã¢â€”Â¦ a = 3 for ball bearings

Ã¢â€”Â¦ a = 10/3 for roller bearings

(cylindrical and tapered

roller)

Fig. 11Ã¢â‚¬â€œ4

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Load-Life Relationship

Ã¯â€šâ€”

Applying Eq. (11Ã¢â‚¬â€œ1) to two load-life conditions,

Ã¯â€šâ€”

Denoting condition 1 with R for catalog rating conditions, and

condition 2 with D for the desired design conditions,

Ã¯â€šâ€”

The units of L are revolutions. If life is given in hours at a given

speed n in rev/min, applying a conversion of 60 min/h,

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Load-Life Relationship

Solving Eq. (a) for FR, which is just another notation for the catalog load

rating,

Ã¯â€šâ€”

Ã¯â€šâ€”

Ã¯â€šâ€”

The desired design load FD and life LD come from the problem

statement.

The rated life LR will be stated by the specific bearing

manufacturer. Many catalogs rate at LR = 106 revolutions.

The catalog load rating C10 is used to find a suitable bearing in the

catalog.

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Load-Life Relationship

Ã¯â€šâ€”

It is often convenient to define a dimensionless multiple of rating

life

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Example 11Ã¢â‚¬â€œ1

2 kN

2

16.2 kN

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Reliability vs. Life

At constant load, the life measure distribution is right skewed.

Ã¯â€šâ€” The Weibull distribution is a good candidate.

Ã¯â€šâ€” Defining the life measure in dimensionless form as x = L/L10, the

reliability is expressed with a Weibull distribution as

Ã¯â€šâ€”

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Reliability vs. Life

Ã¯â€šâ€”

An explicit expression for the cumulative distribution function is

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Relating Load, Life, and Reliability

Ã¯â€šâ€”

Ã¯â€šâ€”

Ã¯â€šâ€”

Catalog information is at point A, at coordinates C10 and

x10=L10/L10=1, on the 0.90 reliability contour.

The design information is at point D, at coordinates FD and xD,

on the R=RD reliability contour.

The designer must move from point D to point A via point B.

Fig. 11Ã¢â‚¬â€œ5

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Relating Load, Life, and Reliability

Ã¯â€šâ€”

Along a constant reliability contour (BD), Eq. (11Ã¢â‚¬â€œ2) applies:

Fig. 11Ã¢â‚¬â€œ5

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Relating Load, Life, and Reliability

Ã¯â€šâ€”

Along a constant load line (AB), Eq. (11Ã¢â‚¬â€œ4) applies:

Ã¯â€šâ€”

Solving for xB,

Fig. 11Ã¢â‚¬â€œ5

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Relating Load, Life, and Reliability

Ã¯â€šâ€”

Substituting xB into Eq. (a),

Ã¯â€šâ€”

Noting that FB = C10, and including an application factor af,

Ã¯â€šâ€”

Note that when RD = 0.90, the denominator equals one and the

equation reduces to Eq. (11Ã¢â‚¬â€œ3).

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Weibull Parameters

The Weibull parameters x0, q, and b are usually provided by the

catalog.

Ã¯â€šâ€” Typical values of Weibull parameters are given on p. 608 at the

beginning of the end-of-chapter problems, and shown below.

Ã¯â€šâ€” Manufacturer 1 parameters are common for tapered roller

bearings

Ã¯â€šâ€” Manufacturer 2 parameters are common for ball and straight

roller bearings

Ã¯â€šâ€”

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Relating Load, Life, and Reliability

Ã¯â€šâ€”

Ã¯â€šâ€”

Eq. (11Ã¢â‚¬â€œ6) can be simplified slightly for calculator entry. Note

that

where pf is the probability for failure

Thus Eq. (11Ã¢â‚¬â€œ6) can be approximated by

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Example 11Ã¢â‚¬â€œ3

1840 N

(1.2) (1.84)

29.7 kN

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Lecture Slides

Rolling-Contact Bearings

Lect.2.3

The McGraw-Hill Companies Ã‚Â© 2012

Dimension-Series Code

Ã¯â€šâ€”

Ã¯â€šâ€”

Ã¯â€šâ€”

Ã¯â€šâ€”

Ã¯â€šâ€”

ABMA standardized dimension-series code represents the relative

size of the boundary dimensions of the bearing cross section for

metric bearings.

Two digit series number

First digit designates the width series

Second digit designates the diameter series

Specific dimensions are tabulated in catalogs under a specific

series

Fig. 11Ã¢â‚¬â€œ7

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Representative Catalog Data for Ball Bearings (Table 11Ã¢â‚¬â€œ2)

Representative Catalog Data for Cylindrical Roller Bearings

(Table 11Ã¢â‚¬â€œ3)

Combined Radial and Thrust Loading

Ã¯â€šâ€”

Equivalent Radial Load, Fe:

Is the load that does the same damage as the combined radial

and thrust loads together.

Ã¯â€šâ€”

Basic Static Load Rating, Co:

is the load that will produce a total permanent deformation

in the raceway and rolling element at any contact point of 0.0001

times the diameter of the rolling element (0.0001d).

Ã¢â€”Â¦ d = diameter of roller

Ã¢â€”Â¦ Used to check for permanent deformation

Ã¢â€”Â¦ Used in combining radial and thrust loads into an equivalent

radial load

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Combined Radial and Thrust Loading

Ã¯â€šâ€”

Ã¯â€šâ€”

Ã¯â€šâ€”

When ball bearings carry both an

axial thrust load Fa and a radial load

Fr, an equivalent radial load Fe that

does the same damage is used.

A plot of Fe/VFr vs. Fa/VFr is

obtained experimentally.

V is a rotation factor to account for

the difference in ball rotations for

outer ring rotation vs. inner ring

rotation.

Ã¢â€”Â¦ V = 1 for inner ring rotation

Ã¢â€”Â¦ V = 1.2 for outer ring rotation

Fig. 11Ã¢â‚¬â€œ6

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Combined Radial and Thrust Loading

Ã¯â€šâ€”

The data can be approximated by

two straight lines

X is the ordinate intercept and Y is

the slope

Ã¯â€šâ€” Basically indicates that Fe equals Fr

for smaller ratios of Fa/Fr, then

begins to rise when Fa/Fr exceeds

some amount e

Ã¯â€šâ€”

Fig. 11Ã¢â‚¬â€œ6

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Combined Radial and Thrust Loading

Ã¯â€šâ€”

It is common to express the two

equations as a single equation

where

i = 1 when Fa/VFr Ã¢â€°Â¤ e

i = 2 when Fa/VFr > e

Ã¯â€šâ€”

X and Y factors depend on geometry

and construction of the specific

bearing.

Fig. 11Ã¢â‚¬â€œ6

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Equivalent Radial Load Factors for Ball Bearings

X and Y for specific bearing obtained from bearing catalog.

Ã¯â€šâ€” Table 11Ã¢â‚¬â€œ1 gives representative values in a manner common to

many catalogs.

Table 11Ã¢â‚¬â€œ1

Ã¯â€šâ€”

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Equivalent Radial Load Factors for Ball Bearings

Table 11Ã¢â‚¬â€œ1

Ã¯â€šâ€”

Ã¯â€šâ€”

X and Y are functions of e, which is a function of Fa/C0.

C0 is the basic static load rating, which is tabulated in the catalog.

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Bearing Life Recommendations (Table 11Ã¢â‚¬â€œ4)

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Recommended Load Application Factors (Table 11Ã¢â‚¬â€œ5)

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Example 11Ã¢â‚¬â€œ4

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Example 11Ã¢â‚¬â€œ4

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Bearing Lubrication

Ã¯â€šâ€”

The purposes of bearing lubrication

Ã¢â€”Â¦ To provide a film of lubricant between the sliding and rolling

surfaces

Ã¢â€”Â¦ To help distribute and dissipate heat

Ã¢â€”Â¦ To prevent corrosion of the bearing surfaces

Ã¢â€”Â¦ To protect the parts from the entrance of foreign matter

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Bearing Lubrication

Ã¯â€šâ€”

Either oil or grease may be used, with each having advantages in

certain situations.

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Realized Bearing Reliability

Ã¯â€šâ€”

Eq. (11Ã¢â‚¬â€œ6) was previously derived to determine a suitable catalog

rated load for a given design situation and reliability goal.

Ã¯â€šâ€”

An actual bearing is selected from a catalog with a rating greater

than C10.

Sometimes it is desirable to determine the realized reliability from

the actual bearing (that was slightly higher capacity than needed).

Solving Eq. (11Ã¢â‚¬â€œ6) for the reliability,

Ã¯â€šâ€”

Ã¯â€šâ€”

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Realized Bearing Reliability

Ã¯â€šâ€”

Similarly for the alternate approximate equation, Eq. (11Ã¢â‚¬â€œ7),

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Group Example (11.2)

An angular-contact, inner ring rotating, 02-series ball bearing is required

for an application in which the life requirement is 50 kh at 480 rev/min.

The design radial load is 2745 N. The application factor is 1.4. The

reliability goal is 0.90. Find the multiple of rating life xD required and the

catalog rating C10 with which to enter Table 11Ã¢â‚¬â€œ2. Choose a bearing and

estimate the existing reliability in service.

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Group Example (11.20)

An 02-series single-row deep-groove ball bearing with a 65-mm bore

(see Tables 11Ã¢â‚¬â€œ1 and 11Ã¢â‚¬â€œ2 for specifications) is loaded with a 3-kN axial

load and a 7-kN radial load. The outer ring rotates at 500 rev/min.

(a) Determine the equivalent radial load that will be experienced by this

particular bearing.

(b) Determine whether this bearing should be expected to carry this

load with a 95 percent reliability for 10 kh.

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Machine Design II

Gears Ã¢â‚¬â€œ General

Lect. 3.1

Dr. Mir Md Maruf Morshed

Gears

Gears are toothed members which transmit power /

motion between two shafts by meshing without any slip

Advantages & Disadvantages of Gear Drive

Advantages

a.

It transmits the exact velocity ratio,

b. It may be used to transmit large power

c.

It has reliable service,

d.

It has high efficiency,

e.

It has compact layout.

Disadvantages

a.

The manufacture of gear require special tools and equipment,

b.

The error in cutting teeth may cause vibrations and noise during

operation.

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Types of Gears

Spur

Bevel

Helical

Figs. 13Ã¢â‚¬â€œ1 to 13Ã¢â‚¬â€œ4

Worm

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Types of Gears

Spur gears

Spur gears have teeth parallel to the axis of rotation and are used to transmit

motion from one shaft to another parallel shaft.

Of all types, the spur gear is the simplest.

Spur

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Types of Gears

Helical gears,

Helical gears have teeth inclined to the axis of rotation.

Can be used for the same applications as spur gears.

Are not noisy, because of the more gradual engagement of the teeth during

meshing.

The inclined tooth also develops thrust loads and bending couples, which

are not present with spur gearing.

Helical

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Types of Gears

Bevel gears

Bevel gears have teeth formed on conical surfaces and are used mostly for

transmitting motion between intersecting shafts.

The figure actually illustrates straight-tooth bevel gears.

Spiral bevel gears are cut so the tooth is no longer straight, but forms a

circular arc.

Hypoid gears are quite similar to spiral bevel gears except that the shafts are

offset and nonintersecting.

Spiral bevel gear

Bevel

Straight-tooth

bevel gear

Hypoid bevel gear

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Types of Gears

Worm gears

Worms and worm gears represent the fourth basic gear type.

The worm resembles a screw.

The direction of rotation of the worm gear, also called the worm wheel,

depends upon the direction of rotation of the worm and upon whether the

worm teeth are cut right-hand or left-hand.

Worm gear sets are mostly used when the speed ratios of the two shafts are

quite high, say, 3 or more.

Worm

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Nomenclature of Spur-Gear Teeth

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Nomenclature of Spur-Gear Teeth

The pitch circle

is a theoretical circle upon which all calculations are usually based; its

diameter is the pitch diameter.

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Nomenclature of Spur-Gear Teeth

The pitch circle

The pitch circles of a pair of mating gears are tangent to each other.

A pinion is the smaller of two mating gears.

The larger is often called the gear.

pinion

gear

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Nomenclature of Spur-Gear Teeth

The circular pitch p

is the distance, measured on the pitch circle, from a point on one tooth to a

corresponding point on an adjacent tooth.

Thus the circular pitch is equal to the sum of the tooth thickness and the width

of space.

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Nomenclature of Spur-Gear Teeth

The module m

is the ratio of the pitch diameter to the number of teeth. The customary

unit of length used is the millimeter. The module is the index of tooth size in SI.

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Nomenclature of Spur-Gear Teeth

The diametral pitch P

is the ratio of the number of teeth on the gear to the pitch diameter.

Thus, it is the reciprocal of the module.

Since diametral pitch is used only with U.S. units, it is expressed as teeth per

inch.

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Nomenclature of Spur-Gear Teeth

The addendum a

is the radial distance between the top land and the pitch circle.

The dedendum b

is the radial distance from the bottom land to the pitch circle.

The whole depth ht is the sum of the addendum and the dedendum.

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Nomenclature of Spur-Gear Teeth

The clearance circle

is a circle that is tangent to the addendum circle of the matinggear.

The clearance c

is the amount by which the dedendum in a given gear exceeds the

addendum of its mating gear.

The backlash

is the amount by which the width of a tooth space exceeds the

thickness of the engaging tooth measured on the pitch circles.

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Tooth Size

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Machine Design II

Gears Ã¢â‚¬â€œ General

Lect. 3.2

Dr. Mir Md Maruf Morshed

Conjugate Action

Ã¯â€šâ€”

When surfaces roll/slide against

each other and produce constant

angular velocity ratio, they are

said to have conjugate action.

Fig. 13Ã¢â‚¬â€œ6

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Conjugate Action

Ã¯â€šâ€”

Forces are transmitted on

line of action which is

normal to the contacting

surfaces.

Ã¢â‚¬Â¢ When one curved surface pushes

against another, as seen in the figure

13-6, the point of contact occurs where

the two surfaces are tangent to each

other (Point c) and the forces will be

directed along the common normal (line

ab) which is also called the Ã¢â‚¬Å“line of

actionÃ¢â‚¬Â or the Ã¢â‚¬Å“pressure lineÃ¢â‚¬Â.

Fig. 13Ã¢â‚¬â€œ6

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Conjugate Action

Ã¯â€šâ€”

Angular velocity ratio is

inversely proportional to the

radii to point P, the pitch

point.

Ã¯â€šâ€”

Circles drawn through P

from each fixed pivot are

pitch circles, each with a

pitch radius.

Fig. 13Ã¢â‚¬â€œ6

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Involute Profile

Ã¯â€šâ€”

The most common conjugate profile is the involute profile.

Fig. 13Ã¢â‚¬â€œ8

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Involute Profile

The most common conjugate profile is the involute profile.

Ã¯â€šâ€” Can be generated by unwrapping a string from a cylinder, keeping

the string taut and tangent to the cylinder.

https://www.tec-science.com/wpÃ¯â€šâ€” Circle is called base circle.

content/uploads/2018/10/en-involute-gearÃ¯â€šâ€”

involute-circle-construction.mp4

Fig. 13Ã¢â‚¬â€œ8

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Circles of a Gear Layout

Fig. 13Ã¢â‚¬â€œ9

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Sequence of Gear Layout (Fundamentals)

Ã¢â‚¬Â¢ Pitch circles in contact

Ã¢â‚¬Â¢ Pressure line at desired

pressure angle

Ã¢â‚¬Â¢ Base circles tangent to

pressure line

Ã¢â‚¬Â¢ Involute profile from

base circle

Ã¢â‚¬Â¢ Cap teeth at addendum

circle at 1/P from pitch

circle

Ã¢â‚¬Â¢ Root of teeth at

dedendum

circle at 1.25/P from

pitch circle

Ã¢â‚¬Â¢ Tooth spacing from

circular pitch, p = p / P

Fig. 13Ã¢â‚¬â€œ9

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Relation of Base Circle to Pressure Angle

Fig. 13Ã¢â‚¬â€œ10

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Tooth Action

First point of

contact at a

where flank of

pinion touches

tip of gear

Ã¯â€šâ€” Last point of

contact at b

where tip of

pinion touches

flank of gear

Ã¯â€šâ€” Line ab is line of

action

Ã¯â€šâ€” Angle of action

is sum of angle

of approach and

angle of recess

Ã¯â€šâ€”

Fig. 13Ã¢â‚¬â€œ12

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Tooth Action

Pitch circles

Fig. 13Ã¢â‚¬â€œ12

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Contact Ratio

Arc of action qt is the sum of the arc of approach qa and the arc of

recess qr., that is qt = qa + qr

Ã¯â€šâ€” The contact ratio mc is the ratio of the arc of action and the circular

pitch.

Ã¯â€šâ€”

Ã¯â€šâ€”

The contact ratio is the average number of pairs of teeth in contact.

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Contact Ratio

Ã¯â€šâ€”

Contact ratio can also be found from the length of the line of action

Ã¯â€šâ€”

The contact ratio should be at least 1.2

Watch this following Video for Gear Nomenclatures:

Fig. 13Ã¢â‚¬â€œ15

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Forming of Gear Teeth

Ã¯â€šâ€”

Common ways of forming gear teeth

Ã¢â€”Â¦ Sand casting

Ã¢â€”Â¦ Shell molding

Ã¢â€”Â¦ Investment casting

Ã¢â€”Â¦ Permanent-mold casting

Ã¢â€”Â¦ Die casting

Ã¢â€”Â¦ Centrifugal casting

Ã¢â€”Â¦ Powder-metallurgy

Ã¢â€”Â¦ Extrusion

Ã¢â€”Â¦ Injection molding (for thermoplastics)

Ã¢â€”Â¦ Cold forming

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Cutting of Gear Teeth

Ã¯â€šâ€”

Common ways of cutting gear teeth

Ã¢â€”Â¦ Milling

Ã¢â€”Â¦ Shaping

Ã¢â€”Â¦ Hobbing

Ã¢â€”Â¦ Milling a tooth gear Video:

https://www.youtube.com/watch?v=27qaZz

i3ZCU

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Shaping with Pinion Cutter

Fig. 13Ã¢â‚¬â€œ17

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Shaping with a Rack

Fig. 13Ã¢â‚¬â€œ18

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Hobbing a Worm Gear

Fig. 13Ã¢â‚¬â€œ19

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Standard and Commonly Used Tooth Systems for Spur Gears

Table 13Ã¢â‚¬â€œ1

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Tooth Sizes in General Use

Table 13Ã¢â‚¬â€œ2

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Machine Design II

Gears Ã¢â‚¬â€œ General

Lect. 3.3

Dr. Mir Md Maruf Morshed

Interference

Ã¯â€šâ€”

Ã¯â€šâ€”

Ã¯â€šâ€”

Contact of portions of

tooth profiles that are not

conjugate is called

interference.

Occurs when contact

occurs below the base

circle

Now notice that the points

of tangency of the

pressure line with the base

circles C and D are

located inside of points A

and B. Interference is

present.

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Interference

Ã¯â€šâ€”

Occurs when contact

occurs below the base

circle (on the noninvolute

portion of the flank).

If teeth were produced by

generating process, then

the generating process

removes the interfering

portion; known as

undercutting.

– the undercut tooth is

considerably weakened.

Ã¯â€šâ€”

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Interference of Spur Gears

Ã¯â€šâ€”

On spur and gear with one-to-one gear ratio, smallest number of

teeth which will not have interference is

Ã¯â€šâ€”

k =1 for full depth teeth. k = 0.8 for stub teeth

Ã¯â€šâ€”

On spur meshed with larger gear with gear ratio mG = NG/NP = m,

the smallest number of teeth which will not have interference is

Ã¯â€šâ€”

Largest gear with a specified pinion that is interference-free is

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Interference of Spur Gears

Ã¯â€šâ€”

k =1 for full depth teeth. k = 0.8 for stub teeth

Ã¯â€šâ€”

Depth: Standard full-depth teeth have working depths of 2/P. If the teeth have

equal addenda (as in standard interchangeable gears), the addendum is 1/P. Stub

teeth have a working depth usually 20% less than full-depth teeth. Full-depth

teeth have a larger contact ratio than stub teeth.

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Interference

Ã¯â€šâ€”

For 20Ã‚Âº pressure angle, the most useful values from Eqs. (13Ã¢â‚¬â€œ11)

and (13Ã¢â‚¬â€œ12) are calculated and shown in the table below.

Minimum NP

Max NG

Integer Max NG

Max Gear Ratio

mG= NG/NP

13

14

15

16

17

16.45

26.12

45.49

101.07

1309.86

16

26

45

101

1309

1.23

1.86

3

6.31

77

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Interference

Ã¯â€šâ€”

Increasing the pressure angle to 25Ã‚Âº allows smaller numbers of

teeth

Minimum NP

Max NG

Integer Max NG

Max Gear Ratio

mG= NG/NP

9

10

11

13.33

32.39

249.23

13

32

249

1.44

3.2

22.64

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Interference

Ã¯â€šâ€”

Ã¯â€šâ€”

Ã¯â€šâ€”

Ã¯â€šâ€”

Ã¯â€šâ€”

Interference can be eliminated by using more teeth on the pinion.

However, if tooth size (that is diametral pitch P) is to be

maintained, then an increase in teeth means an increase in

diameter, since P = N/d.

Interference can also be eliminated by using a larger pressure

angle. This results in a smaller base circle, so more of the tooth

profile is involute.

This is the primary reason for larger pressure angle.

Note that the disadvantage of a larger pressure angle is an increase

in radial force for the same amount of transmitted force.

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Standard and Commonly Used Tooth Systems for Spur Gears

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Tooth Sizes in General Use

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Machine Design II

Gears Ã¢â‚¬â€œ General

Lect. 3.4

Dr. Mir Md Maruf Morshed

Gear Trains

Ã¯â€šâ€”

The combination of two or more gears transmitting power from

one shaft to another is called gear train.

Ã¯â€šâ€”

Types of Gear Trains

Ã¢â€”Â¦

Ã¢â€”Â¦

Ã¢â€”Â¦

Ã¢â€”Â¦

Simple Gear Train

Compound Gear Train

Reverted Gear Train

Epicyclic Gear Train

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Simple Gear Train

Ã¢â€”Â¦ When there is one gear on each shaft, the arrangement is called simple gear

train

(a)

Ã¯â€šâ€”

(b)

(c)

The speed ratio of gear train is the ratio of the speed of the driver to the speed

of the driven or follower:

(a) Speed ratio = n1 / n2 =N2/N1

(b) Speed ratio = n1 / n3 = N3/N1

Speed ratio = Speed of the driver / Speed of the driven

= No. of teeth on driven / No. of teeth on driver

Train value = Speed of the driven / Speed of the driver

= No. of teeth on driver / No. of teeth on driven

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Train Value

L: last F: First

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Compound Gear Train

Ã¯â€šâ€”

Ã¯â€šâ€”

Ã¯â€šâ€”

A practical limit on train value for one pair of gears is 10 to 1

To obtain more, compound two gears onto the same shaft

A two-stage compound gear train, such as shown in Fig. below,

can obtain a train value of up to 100 to 1.

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Compound Gear Train

The design of gear trains to accomplish a specific train value is

straightforward.

Ã¯â€šâ€” The numbers of teeth on gears must be integers (determine them

first then obtain pitch diameters).

Ã¯â€šâ€” Determine the number of stages necessary to obtain the overall

ratio, then divide the overall ratio into portions to be accomplished

in each stage.

Ã¯â€šâ€” Keep the portions as evenly divided between the stages as

possible. In cases where the overall train value need only be

approximated, each stage can be identical.

For example, in a two-stage compound gear train, assign the

square root of the overall train value to each stage.

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Compound Gear Train

If an exact train value is needed, attempt to factor the overall train

value into integer components for each stage.

Ã¯â€šâ€” Then assign the smallest gear(s) to the minimum number of teeth

allowed for the specific ratio of each stage, in order to avoid

interference (see Sec. 13Ã¢â‚¬â€œ7).

Ã¯â€šâ€” Finally, applying the ratio for each stage, determine the necessary

number of teeth for the mating gears. Round to the nearest integer

and check that the resulting overall ratio is within acceptable

tolerance.

Ã¯â€šâ€”

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Example 13Ã¢â‚¬â€œ3

Substituting m = 5.4772 and ÃÂ¤ = 20

o

15.848

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Example 13Ã¢â‚¬â€œ3

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Example 13Ã¢â‚¬â€œ4

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Example 13Ã¢â‚¬â€œ4

Substituting m = 6 and ÃÂ¤ = 20

15.96

N3 = 16

Substituting m = 5 and ÃÂ¤ = 20

15.76

o

o

N5 = 16

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Compound Reverted Gear Train

Ã¯â€šâ€”

Ã¯â€šâ€”

A compound gear train with input and output shafts in-line

Geometry condition must be satisfied

Fig. 13Ã¢â‚¬â€œ29

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Example 13Ã¢â‚¬â€œ5

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Example 13Ã¢â‚¬â€œ5

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Example 13Ã¢â‚¬â€œ5

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Example 13Ã¢â‚¬â€œ5

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Planetary Gear Train

Ã¯â€šâ€”

Ã¯â€šâ€”

Ã¯â€šâ€”

Ã¯â€šâ€”

Ã¯â€šâ€”

Planetary, or epicyclic

gear trains allow the axis

of some of the gears to

move relative to the other

axes

Sun gear has fixed center

axis

Planet gear has moving

center axis

Planet carrier or arm

carries planet axis relative

to sun axis

Allow for two degrees of

freedom (i.e. two inputs)

Fig. 13Ã¢â‚¬â€œ30

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Planetary Gear Trains

Ã¯â€šâ€”

Train value is relative to arm

Fig. 13Ã¢â‚¬â€œ31

Fig. 13Ã¢â‚¬â€œ30

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Example 13Ã¢â‚¬â€œ6

Fig. 13Ã¢â‚¬â€œ30

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Example 13Ã¢â‚¬â€œ6

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Example 13Ã¢â‚¬â€œ6

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Machine Design II

Gears Ã¢â‚¬â€œ General

Lect. 3.5

Dr. Mir Md Maruf Morshed

Force Analysis Ã¢â‚¬â€œ Spur Gearing

Ã¯â€šâ€”

In the Figure shown below

Ã¢â€”Â¦ Gear 2 (the pinion) is the driver

Ã¢â€”Â¦ Gear 3 is the driven

Ã¢â€”Â¦ The shafts are a and b for the pinion

and the gear, respectively.

Ã¢â€”Â¦ The pinion mounted on shaft a

rotating clockwise at n2 rev/min and

driving a gear on shaft b at n3 rev/min

Fig. 13Ã¢â‚¬â€œ32

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Force Analysis Ã¢â‚¬â€œ Spur Gearing

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Force Analysis Ã¢â‚¬â€œ Spur Gearing

Ã¯â€šâ€”

Ã¯â€šâ€”

Ã¯â€šâ€”

Ã¯â€šâ€”

Ã¯â€šâ€”

The force exerted by gear 2 against gear 3

is F23.

The force of gear 2 against a shaft a is F2a .

We can also write Fa2 to mean the force of

a shaft a against gear 2.

The reactions between the mating teeth

occur along the pressure line.

In Fig. b the pinion has been separated

from the gear and the shaft, and their

effects have been replaced by forces. Fa2

and Ta2 are the force and torque,

respectively, exerted by shaft a against

pinion 2. F32 is the force exerted by gear 3

against the pinion.

Using a similar approach, we obtain the

free-body diagram of the gear shown in

Fig. c.

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Force Analysis Ã¢â‚¬â€œ Spur Gearing

Ã¯â€šâ€”

This figure shows the

FBD of the pinion

Ã¯â€šâ€”

Transmitted load Wt is

the tangential load

Ã¯â€šâ€”

It is the useful component

of force, transmitting the

torque

Fig. 13Ã¢â‚¬â€œ33

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Power in Spur Gearing

Ã¯â€šâ€”

Transmitted power H

Ã¯â€šâ€”

Pitch-line velocity is the linear velocity of a point on the gear at the

radius of the pitch circle. It is a common term in tabulating gear

data.

/ 60

m/s

m

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Power in Spur Gearing

Ã¯â€šâ€”

Useful power relation in customary units,

Ã¯â€šâ€”

In SI units,

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Example 13Ã¢â‚¬â€œ7

Fig. 13Ã¢â‚¬â€œ34

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Example 13Ã¢â‚¬â€œ7

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

Example 13Ã¢â‚¬â€œ7

ShigleyÃ¢â‚¬â„¢s Mechanical Engineering Design

ISBN: 0073529281

Author: Budynas / Nisbett

Title: ShigleyÃ¢â‚¬â„¢s Mechanical

Engineering Design, 8e

Front endsheets

Color: 2C (Black & PMS 540 U)

Pages: 2,3

ISBN: 0073529281

Author: Budynas / Nisbett

Title: ShigleyÃ¢â‚¬â„¢s Mechanical

Engineering Design, 8e

Front endsheets

Color: 2C (Black & PMS 540 U)

Pages: 2,3

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10

Mechanical Springs

Chapter Outline

10Ã¢â‚¬â€œ1

Stresses in Helical Springs

10Ã¢â‚¬â€œ2

The Curvature Effect

10Ã¢â‚¬â€œ3

Deflection of Helical Springs

10Ã¢â‚¬â€œ4

Compression Springs

10Ã¢â‚¬â€œ5

Stability

10Ã¢â‚¬â€œ6

Spring Materials

10Ã¢â‚¬â€œ7

Helical Compression Spring Design for Static Service

10Ã¢â‚¬â€œ8

Critical Frequency of Helical Springs

10Ã¢â‚¬â€œ9

Fatigue Loading of Helical Compression Springs

518

519

520

520

522

523

528

534

536

10Ã¢â‚¬â€œ10

Helical Compression Spring Design for Fatigue Loading

10Ã¢â‚¬â€œ11

Extension Springs

10Ã¢â‚¬â€œ12

Helical Coil Torsion Springs

10Ã¢â‚¬â€œ13

Belleville Springs

10Ã¢â‚¬â€œ14

Miscellaneous Springs

10Ã¢â‚¬â€œ15

Summary

539

542

550

557

558

560

517

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Mechanical Engineering Design

When a designer wants rigidity, negligible deflection is an acceptable approximation

as long as it does not compromise function. Flexibility is sometimes needed and is

often provided by metal bodies with cleverly controlled geometry. These bodies can

exhibit flexibility to the degree the designer seeks. Such flexibility can be linear or

nonlinear in relating deflection to load. These devices allow controlled application of

force or torque; the storing and release of energy can be another purpose. Flexibility

allows temporary distortion for access and the immediate restoration of function.

Because of machineryÃ¢â‚¬â„¢s value to designers, springs have been intensively studied;

moreover, they are mass-produced (and therefore low cost), and ingenious configurations have been found for a variety of desired applications. In this chapter we will

discuss the more frequently used types of springs, their necessary parametric relationships, and their design.

In general, springs may be classified as wire springs, flat springs, or specialshaped springs, and there are variations within these divisions. Wire springs include

helical springs of round or square wire, made to resist and deflect under tensile, compressive, or torsional loads. Flat springs include cantilever and elliptical types, wound

motor- or clock-type power springs, and flat spring washers, usually called Belleville

springs.

10Ã¢â‚¬â€œ1

Stresses in Helical Springs

Figure 10Ã¢â‚¬â€œ1a shows a round-wire helical compression spring loaded by the axial force F.

We designate D as the mean coil diameter and d as the wire diameter. Now imagine

that the spring is cut at some point (Fig. 10Ã¢â‚¬â€œ1b), a portion of it removed, and the effect

of the removed portion replaced by the net internal reactions. Then, as shown in the

figure, from equilibrium the cut portion would contain a direct shear force F and a torsion T = F D/2.

The maximum stress in the wire may be computed by superposition of the direct

shear stress given by Eq. (3Ã¢â‚¬â€œ23), p. 89, with V = F and the torsional shear stress given

by Eq. (3Ã¢â‚¬â€œ37), p. 101. The result is

Ãâ€žmax =

Tr

F

+

J

A

F

F

Figure 10Ã¢â‚¬â€œ1

(a) Axially loaded helical

spring; (b) free-body diagram

showing that the wire is

subjected to a direct shear and

a torsional shear.

d

T = FDÃ¥â€¦Â¾2

F

(b)

F

D

(a)

(a)

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Mechanical Springs

519

at the inside fiber of the spring. Substitution of Ãâ€žmax = Ãâ€ž , T = F D/2, r = d/2, J =

Ãâ‚¬d 4/32, and A = Ãâ‚¬d 2 /4 gives

Ãâ€ž=

8F D

4F

+

3

Ãâ‚¬d

Ãâ‚¬d 2

(b)

Now we define the spring index

C=

D

d

(10Ã¢â‚¬â€œ1)

which is a measure of coil curvature. The preferred value of C ranges from 4 to 12.1

With this relation, Eq. (b) can be rearranged to give

Ãâ€ž = Ks

8F D

Ãâ‚¬d 3

(10Ã¢â‚¬â€œ2)

where K s is a shear stress-correction factor and is defined by the equation

Ks =

2C + 1

2C

(10Ã¢â‚¬â€œ3)

The use of square or rectangular wire is not recommended for springs unless

space limitations make it necessary. Springs of special wire shapes are not made in

large quantities, unlike those of round wire; they have not had the benefit of refining

development and hence may not be as strong as springs made from round wire. When

space is severely limited, the use of nested round-wire springs should always be considered. They may have an economical advantage over the special-section springs, as

well as a strength advantage.

10Ã¢â‚¬â€œ2

The Curvature Effect

Equation (10Ã¢â‚¬â€œ2) is based on the wire being straight. However, the curvature of the wire

increases the stress on the inside of the spring but decreases it only slightly on the outside. This curvature stress is primarily important in fatigue because the loads are lower

and there is no opportunity for localized yielding. For static loading, these stresses can

normally be neglected because of strain-strengthening with the first application of load.

Unfortunately, it is necessary to find the curvature factor in a roundabout way. The

reason for this is that the published equations also include the effect of the direct shear

stress. Suppose K s in Eq. (10Ã¢â‚¬â€œ2) is replaced by another K factor, which corrects for

both curvature and direct shear. Then this factor is given by either of the equations

KW =

4C Ã¢Ë†â€™ 1 0.615

+

4C Ã¢Ë†â€™ 4

C

(10Ã¢â‚¬â€œ4)

KB =

4C + 2

4C Ã¢Ë†â€™ 3

(10Ã¢â‚¬â€œ5)

The first of these is called the Wahl factor, and the second, the BergstrÃƒÂ¤sser factor.2

Since the results of these two equations differ by the order of 1 percent, Eq. (10Ã¢â‚¬â€œ6)

is preferred. The curvature correction factor can now be obtained by canceling out the

1

Design Handbook: Engineering Guide to Spring Design, Associated Spring-Barnes Group Inc.,

Bristol, CT, 1987.

2

Cyril SamÃƒÂ³nov, Ã¢â‚¬Å“Some Aspects of Design of Helical Compression Springs,Ã¢â‚¬Â Int. Symp. Design and

Synthesis, Tokyo, 1984.

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Mechanical Engineering Design

effect of the direct shear. Thus, using Eq. (10Ã¢â‚¬â€œ5) with Eq. (10Ã¢â‚¬â€œ3), the curvature correction factor is found to be

KB

2C(4C + 2)

Kc =

=

(10Ã¢â‚¬â€œ6)

Ks

(4C Ã¢Ë†â€™ 3)(2C + 1)

Now, K s , K B or K W , and K c are simply stress-correction factors applied multiplicatively to T r/J at the critical location to estimate a particular stress. There is no stressconcentration factor. In this book we will use

Ãâ€ž = KB

to predict the largest shear stress.

10Ã¢â‚¬â€œ3

8F D

Ãâ‚¬d 3

(10Ã¢â‚¬â€œ7)

Deflection of Helical Springs

The deflection-force relations are quite easily obtained by using CastiglianoÃ¢â‚¬â„¢s theorem.

The total strain energy for a helical spring is composed of a torsional component and

a shear component. From Eqs. (4Ã¢â‚¬â€œ18) and (4Ã¢â‚¬â€œ20), p. 162, the strain energy is

U=

T 2l

F 2l

+

2G J

2AG

(a)

Substituting T = F D/2, l = Ãâ‚¬ D N , J = Ãâ‚¬d 4/32, and A = Ãâ‚¬d 2/4 results in

U=

4F 2 D 3 N

2F 2 D N

+

d4G

d2G

(b)

where N = Na = number of active coils. Then using CastiglianoÃ¢â‚¬â„¢s theorem, Eq. (4Ã¢â‚¬â€œ26),

p. 165, to find total deflection y gives

y=

Ã¢Ë†â€šU

8F D 3 N

4F D N

=

+ 2

Ã¢Ë†â€šF

d4G

d G

Since C = D/d, Eq. (c) can be rearranged to yield

8F D 3 N

1

. 8F D 3 N

y=

1

+

=

d4G

2C 2

d4G

(c)

(10Ã¢â‚¬â€œ8)

The spring rate, also called the scale of the spring, is k = F/y, and so

. d4G

k=

8D 3 N

10Ã¢â‚¬â€œ4

(10Ã¢â‚¬â€œ9)

Compression Springs

The four types of ends generally used for compression springs are illustrated in Fig. 10Ã¢â‚¬â€œ2.

A spring with plain ends has a noninterrupted helicoid; the ends are the same as if a

long spring had been cut into sections. A spring with plain ends that are squared or

closed is obtained by deforming the ends to a zero-degree helix angle. Springs should

always be both squared and ground for important applications, because a better transfer

of the load is obtained.

Table 10Ã¢â‚¬â€œ1 shows how the type of end used affects the number of coils and the

spring length.3 Note that the digits 0, 1, 2, and 3 appearing in Table 10Ã¢â‚¬â€œ1 are often

3

For a thorough discussion and development of these relations, see Cyril SamÃƒÂ³nov, Ã¢â‚¬Å“Computer-Aided

Design of Helical Compression Springs,Ã¢â‚¬Â ASME paper No. 80-DET-69, 1980.

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Mechanical Springs

521

Figure 10Ã¢â‚¬â€œ2

Types of ends for compression

springs: (a) both ends plain;

(b) both ends squared; (c) both

ends squared and ground;

(d) both ends plain and ground.

+

+

(a) Plain end, right hand

(c) Squared and ground end,

left hand

+

+

(b) Squared or closed end,

right hand

(d ) Plain end, ground,

left hand

Table 10Ã¢â‚¬â€œ1

Type of Spring Ends

Formulas for the

Dimensional

Characteristics of

Compression-Springs.

(Na = Number of Active

Coils)

Term

Source: From Design

Handbook, 1987, p. 32.

Courtesy of Associated Spring.

Plain

Plain and

Ground

Squared or

Closed

Squared and

Ground

End coils, Ne

0

1

2

2

Total coils, Nt

Na

Na 1

Na 2

Na 2

Free length, L0

pNa d

p(Na 1)

pNa 3d

pNa 2d

Solid length, Ls

d (Nt 1)

dNt

d(Nt 1)

dNt

Pitch, p

(L0 d) Na

L0 (Na 1)

(L0 3d) Na

(L0 2d) Na

used without question. Some of these need closer scrutiny as they may not be integers.

This depends on how a springmaker forms the ends. Forys4 pointed out that squared

and ground ends give a solid length L s of

L s = (Nt Ã¢Ë†â€™ a)d

where a varies, with an average of 0.75, so the entry d Nt in Table 10Ã¢â‚¬â€œ1 may be overstated. The way to check these variations is to take springs from a particular springmaker, close them solid, and measure the solid height. Another way is to look at the

spring and count the wire diameters in the solid stack.

Set removal or presetting is a process used in the manufacture of compression

springs to induce useful residual stresses. It is done by making the spring longer than

needed and then compressing it to its solid height. This operation sets the spring to the

required final free length and, since the torsional yield strength has been exceeded,

induces residual stresses opposite in direction to those induced in service. Springs to

be preset should be designed so that 10 to 30 percent of the initial free length is

removed during the operation. If the stress at the solid height is greater than 1.3 times

the torsional yield strength, distortion may occur. If this stress is much less than 1.1

times, it is difficult to control the resulting free length.

Set removal increases the strength of the spring and so is especially useful when

the spring is used for energy-storage purposes. However, set removal should not be

used when springs are subject to fatigue.

4

Edward L. Forys, Ã¢â‚¬Å“Accurate Spring Heights,Ã¢â‚¬Â Machine Design, vol. 56, no. 2, January 26, 1984.

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Mechanical Engineering Design

10Ã¢â‚¬â€œ5

Stability

In Chap. 4 we learned that a column will buckle when the load becomes too large.

Similarly, compression coil springs may buckle when the deflection becomes too

large. The critical deflection is given by the equation

ycr = L 0 C1

C2 1/2

1Ã¢Ë†â€™ 1Ã¢Ë†â€™ 2

ÃŽÂ»eff

(10Ã¢â‚¬â€œ10)

where ycr is the deflection corresponding to the onset of instability. SamÃƒÂ³nov5 states that

this equation is cited by Wahl6 and verified experimentally by Haringx.7 The quantity

ÃŽÂ»eff in Eq. (10Ã¢â‚¬â€œ10) is the effective slenderness ratio and is given by the equation

ÃŽÂ»eff =

ÃŽÂ±L 0

D

(10Ã¢â‚¬â€œ11)

C1 and C2 are elastic constants defined by the equations

C1 =

E

2(E Ã¢Ë†â€™ G)

C2 =

2Ãâ‚¬ 2 (E Ã¢Ë†â€™ G)

2G + E

Equation (10Ã¢â‚¬â€œ11) contains the end-condition constant ÃŽÂ±. This depends upon how the

ends of the spring are supported. Table 10Ã¢â‚¬â€œ2 gives values of ÃŽÂ± for usual end conditions.

Note how closely these resemble the end conditions for columns.

Absolute stability occurs when, in Eq. (10Ã¢â‚¬â€œ10), the term C2 /ÃŽÂ»2eff is greater than

unity. This means that the condition for absolute stability is that

L0 <
Table 10Ã¢â‚¬â€œ2
Ãâ‚¬ D 2(E Ã¢Ë†â€™ G) 1/2
ÃŽÂ±
2G + E
End Condition
End-Condition
Constants ÃŽÂ± for Helical
Compression Springs*
(10Ã¢â‚¬â€œ12)
Constant
Spring supported between flat parallel surfaces (fixed ends)
0.5
One end supported by flat surface perpendicular to spring axis (fixed);
other end pivoted (hinged)
0.707
Both ends pivoted (hinged)
1
One end clamped; other end free
2
Ã¢Ë†â€” Ends
supported by flat surfaces must be squared and ground.
5
Cyril SamÃƒÂ³nov Ã¢â‚¬Å“Computer-Aided Design,Ã¢â‚¬Â op. cit.
6
A. M. Wahl, Mechanical Springs, 2d ed., McGraw-Hill, New York, 1963.
7
J. A. Haringx, Ã¢â‚¬Å“On Highly Compressible Helical Springs and Rubber Rods and Their Application for
Vibration-Free Mountings,Ã¢â‚¬Â I and II, Philips Res. Rep., vol. 3, December 1948, pp. 401Ã¢â‚¬â€œ 449, and vol. 4,
February 1949, pp. 49Ã¢â‚¬â€œ80.
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Mechanical Springs
523
For steels, this turns out to be
L 0 < 2.63
D
ÃŽÂ±
(10Ã¢â‚¬â€œ13)
For squared and ground ends ÃŽÂ± = 0.5 and L 0 < 5.26D.
10Ã¢â‚¬â€œ6
Spring Materials
Springs are manufactured either by hot- or cold-working processes, depending upon
the size of the material, the spring index, and the properties desired. In general, prehardened wire should not be used if D/d < 4 or if d > 14 in. Winding of the spring

induces residual stresses through bending, but these are normal to the direction of the

torsional working stresses in a coil spring. Quite frequently in spring manufacture,

they are relieved, after winding, by a mild thermal treatment.

A great variety of spring materials are available to the designer, including plain

carbon steels, alloy steels, and corrosion-resisting steels, as well as nonferrous materials such as phosphor bronze, spring brass, beryllium copper, and various nickel alloys.

Descriptions of the most commonly used steels will be found in Table 10Ã¢â‚¬â€œ3. The UNS

steels listed in Appendix A should be used in designing hot-worked, heavy-coil springs,

as well as flat springs, leaf springs, and torsion bars.

Spring materials may be compared by an examination of their tensile strengths;

these vary so much with wire size that they cannot be specified until the wire size is

known. The material and its processing also, of course, have an effect on tensile

strength. It turns out that the graph of tensile strength versus wire diameter is almost

a straight line for some materials when plotted on log-log paper. Writing the equation

of this line as

Sut =

A

dm

(10Ã¢â‚¬â€œ14)

furnishes a good means of estimating minimum tensile strengths when the intercept A

and the slope m of the line are known. Values of these constants have been worked out

from recent data and are given for strengths in units of kpsi and MPa in Table 10Ã¢â‚¬â€œ4.

In Eq. (10Ã¢â‚¬â€œ14) when d is measured in millimeters, then A is in MPa Ã‚Â· mmm and when

d is measured in inches, then A is in kpsi Ã‚Â· inm .

Although the torsional yield strength is needed to design the spring and to analyze

the performance, spring materials customarily are tested only for tensile strengthÃ¢â‚¬â€

perhaps because it is such an easy and economical test to make. A very rough estimate

of the torsional yield strength can be obtained by assuming that the tensile yield strength

is between 60 and 90 percent of the tensile strength. Then the distortion-energy theory

can be employed to obtain the torsional yield strength (Sys = 0.577Sy ). This approach

results in the range

0.35Sut Ã¢â€°Â¤ Ssy Ã¢â€°Â¤ 0.52Sut

(10Ã¢â‚¬â€œ15)

for steels.

For wires listed in Table 10Ã¢â‚¬â€œ5, the maximum allowable shear stress in a spring

can be seen in column 3. Music wire and hard-drawn steel spring wire have a low end

of range Ssy = 0.45Sut . Valve spring wire, Cr-Va, Cr-Si, and other (not shown) hardened and tempered carbon and low-alloy steel wires as a group have Ssy Ã¢â€°Â¥ 0.50Sut .

Many nonferrous materials (not shown) as a group have Ssy Ã¢â€°Â¥ 0.35Sut . In view of this,

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Mechanical Engineering Design

Table 10Ã¢â‚¬â€œ3

High-Carbon and Alloy

Spring Steels

Source: From Harold C. R.

Carlson, Ã¢â‚¬Å“Selection and

Application of Spring

Materials,Ã¢â‚¬Â Mechanical

Engineering, vol. 78, 1956,

pp. 331Ã¢â‚¬â€œ334.

Name of

Material

Similar

Specifications

Music wire,

0.80Ã¢â‚¬â€œ0.95C

UNS G10850

AISI 1085

ASTM A228-51

This is the best, toughest, and most

widely used of all spring materials for

small springs. It has the highest tensile

strength and can withstand higher

stresses under repeated loading than

any other spring material. Available in

diameters 0.12 to 3 mm (0.005 to

0.125 in). Do not use above 120Ã‚Â°C

(250Ã‚Â°F) or at subzero temperatures.

Oil-tempered wire,

0.60Ã¢â‚¬â€œ0.70C

UNS G10650

AISI 1065

ASTM 229-41

This general-purpose spring steel is

used for many types of coil springs

where the cost of music wire is

prohibitive and in sizes larger than

available in music wire. Not for shock

or impact loading. Available in

diameters 3 to 12 mm (0.125 to

0.5000 in), but larger and smaller sizes

may be obtained. Not for use above

180Ã‚Â°C (350Ã‚Â°F) or at subzero

temperatures.

Hard-drawn wire,

0.60Ã¢â‚¬â€œ0.70C

UNS G10660

AISI 1066

ASTM A227-47

This is the cheapest general-purpose

spring steel and should be used only

where life, accuracy, and deflection

are not too important. Available in

diameters 0.8 to 12 mm (0.031 to

0.500 in). Not for use above

120Ã‚Â°C (250Ã‚Â°F) or at subzero

temperatures.

Chrome-vanadium

UNS G61500

AISI 6150

ASTM 231-41

This is the most popular alloy spring

steel for conditions involving higher

stresses than can be used with the

high-carbon steels and for use where

fatigue resistance and long endurance

are needed. Also good for shock

and impact loads. Widely used for

aircraft-engine valve springs and for

temperatures to 220Ã‚Â°C (425Ã‚Â°F).

Available in annealed or pretempered

sizes 0.8 to 12 mm (0.031 to 0.500 in)

in diameter.

Chrome-silicon

UNS G92540

AISI 9254

This alloy is an excellent material for

highly stressed springs that require

long life and are subjected to shock

loading. Rockwell hardnesses of C50

to C53 are quite common, and the

material may be used up to 250Ã‚Â°C

(475Ã‚Â°F). Available from 0.8 to 12 mm

(0.031 to 0.500 in) in diameter.

Description

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Mechanical Springs

525

Table 10Ã¢â‚¬â€œ4

Constants A and m of Sut = A/d m for Estimating Minimum Tensile Strength of Common Spring Wires

Source: From Design Handbook, 1987, p. 19. Courtesy of Associated Spring.

ASTM

No.

Exponent

m

Music wire*

A228

0.145

0.004Ã¢â‚¬â€œ0.256

201

Ã¢â‚¬Â

OQ&T wire

A229

0.187

0.020Ã¢â‚¬â€œ0.500

147

Hard-drawn wireÃ¢â‚¬Â¡

A227

0.190

0.028Ã¢â‚¬â€œ0.500

140

Chrome-vanadium wire

A232

0.168

0.032Ã¢â‚¬â€œ0.437

Chrome-silicon wire

A401

0.108

0.063Ã¢â‚¬â€œ0.375

0.146

0.013Ã¢â‚¬â€œ0.10

169

0.263

0.10Ã¢â‚¬â€œ0.20

128

Material

Ã‚Â§

302 Stainless wire

#

A313

Diameter,

in

0.478

Phosphor-bronze wire**

Ã¢Ë†â€” Surface

B159

A,

kpsi inm

Relative

Diameter,

A,

Cost

mm

MPa mmm of Wire

0.10Ã¢â‚¬â€œ6.5

2211

2.6

0.5Ã¢â‚¬â€œ12.7

1855

1.3

0.7Ã¢â‚¬â€œ12.7

1783

1.0

169

0.8Ã¢â‚¬â€œ11.1

2005

3.1

202

1.6Ã¢â‚¬â€œ9.5

1974

4.0

0.3Ã¢â‚¬â€œ2.5

1867

7.6Ã¢â‚¬â€œ11

2.5Ã¢â‚¬â€œ5

2065

0.20Ã¢â‚¬â€œ0.40

90

5Ã¢â‚¬â€œ10

2911

0

0.004Ã¢â‚¬â€œ0.022

145

0.1Ã¢â‚¬â€œ0.6

1000

0.028

0.022Ã¢â‚¬â€œ0.075

121

0.6Ã¢â‚¬â€œ2

0.064

0.075Ã¢â‚¬â€œ0.30

110

2Ã¢â‚¬â€œ7.5

8.0

913

932

is smooth, free of defects, and has a bright, lustrous finish.

Ã¢â‚¬Â

Has a slight heat-treating scale which must be removed before plating.

Ã¢â‚¬Â¡

Surface is smooth and bright with no visible marks.

Ã‚Â§

Aircraft-quality tempered wire, can also be obtained annealed.

Tempered

# Type

to Rockwell C49, but may be obtained untempered.

302 stainless steel.

Ã¢Ë†â€”Ã¢Ë†â€” Temper

CA510.

Joerres8 uses the maximum allowable torsional stress for static application shown in

Table 10Ã¢â‚¬â€œ6. For specific materials for which you have torsional yield information use

this table as a guide. Joerres provides set-removal information in Table 10Ã¢â‚¬â€œ6, that

Ssy Ã¢â€°Â¥ 0.65Sut increases strength through cold work, but at the cost of an additional

operation by the springmaker. Sometimes the additional operation can be done by the

manufacturer during assembly. Some correlations with carbon steel springs show that

the tensile yield strength of spring wire in torsion can be estimated from 0.75Sut . The

corresponding estimate of the yield strength in shear based on distortion energy theory

.

is Ssy = 0.577(0.75)Sut = 0.433Sut = 0.45Sut . SamÃƒÂ³nov discusses the problem of

allowable stress and shows that

Ssy = Ãâ€žall = 0.56Sut

(10Ã¢â‚¬â€œ16)

for high-tensile spring steels, which is close to the value given by Joerres for hardened alloy steels. He points out that this value of allowable stress is specified by Draft

Standard 2089 of the German Federal Republic when Eq. (10Ã¢â‚¬â€œ2) is used without stresscorrection factor.

8

Robert E. Joerres, Ã¢â‚¬Å“Springs,Ã¢â‚¬Â Chap. 6 in Joseph E. Shigley, Charles R. Mischke, and Thomas H. Brown,

Jr. (eds.), Standard Handbook of Machine Design, 3rd ed., McGraw-Hill, New York, 2004.

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Table 10Ã¢â‚¬â€œ5

Mechanical Properties of Some Spring Wires

Material

Elastic Limit,

Percent of Sut

Tension Torsion

Music wire A228

65Ã¢â‚¬â€œ75

HD spring A227

60Ã¢â‚¬â€œ70

45Ã¢â‚¬â€œ60

45Ã¢â‚¬â€œ55

G

E

Diameter

d, in

0.125

28.0

193

11.6

80.0

0.125

Oil tempered A239

85Ã¢â‚¬â€œ90

45Ã¢â‚¬â€œ50

28.5

196.5

11.2

77.2

Valve spring A230

85Ã¢â‚¬â€œ90

50Ã¢â‚¬â€œ60

29.5

203.4

11.2

77.2

Chrome-vanadium A231

88Ã¢â‚¬â€œ93

65Ã¢â‚¬â€œ75

A232

88Ã¢â‚¬â€œ93

Chrome-silicon A401

29.5

203.4

11.2

77.2

29.5

203.4

11.2

77.2

85Ã¢â‚¬â€œ93

65Ã¢â‚¬â€œ75

29.5

203.4

11.2

77.2

A313*

65Ã¢â‚¬â€œ75

45Ã¢â‚¬â€œ55

28

193

10

69.0

17-7PH

75Ã¢â‚¬â€œ80

55Ã¢â‚¬â€œ60

29.5

208.4

11

75.8

414

65Ã¢â‚¬â€œ70

42Ã¢â‚¬â€œ55

29

200

11.2

77.2

420

65Ã¢â‚¬â€œ75

45Ã¢â‚¬â€œ55

29

200

11.2

77.2

431

Stainless steel

72Ã¢â‚¬â€œ76

50Ã¢â‚¬â€œ55

30

206

11.5

79.3

Phosphor-bronze B159

75Ã¢â‚¬â€œ80

45Ã¢â‚¬â€œ50

15

103.4

6

41.4

Beryllium-copper B197

70

50

17

117.2

6.5

44.8

75

50Ã¢â‚¬â€œ55

19

131

7.3

50.3

65Ã¢â‚¬â€œ70

40Ã¢â‚¬â€œ45

31

213.7

11.2

77.2

Inconel alloy X-750

*Also includes 302, 304, and 316.

Note: See Table 10Ã¢â‚¬â€œ6 for allowable torsional stress design values.

Table 10Ã¢â‚¬â€œ6

Maximum Allowable

Torsional Stresses for

Helical Compression

Springs in Static

Applications

Source: Robert E. Joerres,

Ã¢â‚¬Å“Springs,Ã¢â‚¬Â Chap. 6 in Joseph

E. Shigley, Charles R. Mischke,

and Thomas H. Brown, Jr. (eds.),

Standard Handbook of Machine

Design, 3rd ed., McGraw-Hill,

New York, 2004.

526

Maximum Percent of Tensile Strength

Before Set Removed

(includes KW or KB)

After Set Removed

(includes Ks)

Music wire and colddrawn carbon steel

45

60Ã¢â‚¬â€œ70

Hardened and tempered

carbon and low-alloy

steel

50

65Ã¢â‚¬â€œ75

Austenitic stainless

steels

35

55Ã¢â‚¬â€œ65

Nonferrous alloys

35

55Ã¢â‚¬â€œ65

Material

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Mechanical Springs

527

EXAMPLE 10Ã¢â‚¬â€œ1

A helical compression spring is made of no. 16 music wire. The outside coil diam7

eter of the spring is 16

in. The ends are squared and there are 12 12 total turns.

(a) Estimate the torsional yield strength of the wire.

(b) Estimate the static load corresponding to the yield strength.

(c) Estimate the scale of the spring.

(d) Estimate the deflection that would be caused by the load in part (b).

(e) Estimate the solid length of the spring.

( f ) What length should the spring be to ensure that when it is compressed solid and

then released, there will be no permanent change in the free length?

(g) Given the length found in part ( f ), is buckling a possibility?

(h) What is the pitch of the body coil?

Solution

(a) From Table AÃ¢â‚¬â€œ28, the wire diameter is d = 0.037 in. From Table 10Ã¢â‚¬â€œ4, we find

A = 201 kpsi Ã‚Â· inm and m = 0.145. Therefore, from Eq. (10Ã¢â‚¬â€œ14)

Sut =

A

201

=

= 324 kpsi

dm

0.0370.145

Then, from Table 10Ã¢â‚¬â€œ6,

Ssy = 0.45Sut = 0.45(324) = 146 kpsi

Answer

7

Ã¢Ë†â€™ 0.037 = 0.400 in, and so the spring

(b) The mean spring coil diameter is D = 16

index is C = 0.400/0.037 = 10.8. Then, from Eq. (10Ã¢â‚¬â€œ6),

KB =

4C + 2

4 (10.8) + 2

=

= 1.124

4C Ã¢Ë†â€™ 3

4 (10.8) Ã¢Ë†â€™ 3

Now rearrange Eq. (10Ã¢â‚¬â€œ7) replacing Ãâ€ž with Ssy , and solve for F:

F=

Answer

Ãâ‚¬d 3 Ssy

Ãâ‚¬(0.0373 )146(103 )

=

= 6.46 lbf

8K B D

8(1.124) 0.400

(c) From Table 10Ã¢â‚¬â€œ1, Na = 12.5 Ã¢Ë†â€™ 2 = 10.5 turns. In Table 10Ã¢â‚¬â€œ5, G = 11.85 Mpsi,

and the scale of the spring is found to be, from Eq. (10Ã¢â‚¬â€œ9),

k=

Answer

Answer

d4G

0.0374 (11.85)106

=

= 4.13 lbf/in

3

8D Na

8(0.4003 )10.5

y=

(d)

F

6.46

=

= 1.56 in

k

4.13

(e) From Table 10Ã¢â‚¬â€œ1,

L s = (Nt + 1)d = (12.5 + 1)0.037 = 0.500 in

Answer

Answer

(f )

L 0 = y + L s = 1.56 + 0.500 = 2.06 in.

(g) To avoid buckling, Eq. (10Ã¢â‚¬â€œ13) and Table 10Ã¢â‚¬â€œ2 give

L 0 < 2.63
D
0.400
= 2.63
= 2.10 in
ÃŽÂ±
0.5
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Mechanical Engineering Design
Mathematically, a free length of 2.06 in is less than 2.10 in, and buckling is unlikely.
However, the forming of the ends will control how close ÃŽÂ± is to 0.5. This has to be
investigated and an inside rod or exterior tube or hole may be needed.
(h) Finally, from Table 10Ã¢â‚¬â€œ1, the pitch of the body coil is
p=
Answer
10Ã¢â‚¬â€œ7
L 0 Ã¢Ë†â€™ 3d
2.06 Ã¢Ë†â€™ 3(0.037)
=
= 0.186 in
Na
10.5
Helical Compression Spring
Design for Static Service
The preferred range of spring index is 4 Ã¢â€°Â¤ C Ã¢â€°Â¤ 12, with the lower indexes being more
difficult to form (because of the danger of surface cracking) and springs with higher
indexes tending to tangle often enough to require individual packing. This can be the first
item of the design assessment. The recommended range of active turns is 3 Ã¢â€°Â¤ Na Ã¢â€°Â¤ 15.
To maintain linearity when a spring is about to close, it is necessary to avoid the gradual
touching of coils (due to nonperfect pitch). A helical coil spring force-deflection characteristic is ideally linear. Practically, it is nearly so, but not at each end of the force-deflection
curve. The spring force is not reproducible for very small deflections, and near closure,
nonlinear behavior begins as the number of active turns diminishes as coils begin to touch.
The designer confines the springÃ¢â‚¬â„¢s operating point to the central 75 percent of the curve
between no load, F = 0, and closure, F = Fs . Thus, the maximum operating force should
be limited to Fmax Ã¢â€°Â¤ 78 Fs . Defining the fractional overrun to closure as ÃŽÂ¾ , where
Fs = (1 + ÃŽÂ¾ )Fmax
(10Ã¢â‚¬â€œ17)
it follows that
Fs = (1 + ÃŽÂ¾ )Fmax = (1 + ÃŽÂ¾ )
7
Fs
8
.
From the outer equality ÃŽÂ¾ = 1/7 = 0.143 = 0.15. Thus, it is recommended that ÃŽÂ¾ Ã¢â€°Â¥ 0.15.
In addition to the relationships and material properties for springs, we now have
some recommended design conditions to follow, namely:
4 Ã¢â€°Â¤ C Ã¢â€°Â¤ 12
(10Ã¢â‚¬â€œ18)
3 Ã¢â€°Â¤ Na Ã¢â€°Â¤ 15
(10Ã¢â‚¬â€œ19)
ÃŽÂ¾ Ã¢â€°Â¥ 0.15
(10Ã¢â‚¬â€œ20)
n s Ã¢â€°Â¥ 1.2
(10Ã¢â‚¬â€œ21)
where ns is the factor of safety at closure (solid height).
When considering designing a spring for high volume production, the figure of
merit can be the cost of the wire from which the spring is wound. The fom would be
proportional to the relative material cost, weight density, and volume:
fom = Ã¢Ë†â€™(relative material cost)
ÃŽÂ³ Ãâ‚¬ 2 d 2 Nt D
4
(10Ã¢â‚¬â€œ22)
For comparisons between steels, the specific weight ÃŽÂ³ can be omitted.
Spring design is an open-ended process. There are many decisions to be made,
and many possible solution paths as well as solutions. In the past, charts, nomographs,
and Ã¢â‚¬Å“spring design slide rulesÃ¢â‚¬Â were used by many to simplify the spring design problem. Today, the computer enables the designer to create programs in many different
formatsÃ¢â‚¬â€direct programming, spreadsheet, MATLAB, etc. Commercial programs are
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Mechanical Springs
Figure 10Ã¢â‚¬â€œ3
529
STATIC SPRING DESIGN
Choose d
Helical coil compression spring
design flowchart for static
loading.
Over-a-rod
Free
In-a-hole
As-wound or set
As-wound
Set removed
As-wound or set
D = d rod + d + allow
Ssy = const(A) Ã¢Ââ€žd m Ã¢â‚¬Â
Ssy = 0.65A Ã¢Ââ€žd m
D = d hole Ã¢Ë†â€™ d Ã¢Ë†â€™ allow
C=
2Ã¢ÂÂ£ Ã¢â‚¬â€œ Ã¢ÂÂ¤
+
4Ã¢ÂÂ¤
Ssy
Ã¢ÂÂ£= n
s
Ã¢Ë†Å¡( )
2Ã¢ÂÂ£ Ã¢â‚¬â€œ Ã¢ÂÂ¤
4Ã¢ÂÂ¤
Ã¢ÂÂ¤=
2
Ã¢â‚¬â€œ
3Ã¢ÂÂ£
4Ã¢ÂÂ¤
D=
SsyÃ¢ÂÂ²d 3
8ns(1 + Ã¢ÂÂ°)Fmax
8(1 + Ã¢ÂÂ°)Fmax
Ã¢ÂÂ²d2
D = Cd
C = D Ã¢Ââ€žd
KB = (4C + 2) Ã¢Ââ€ž (4C Ã¢Ë†â€™ 3)
Ã¢ÂÂ¶s = K B8(1 + Ã¢ÂÂ°)FmaxD Ã¢Ââ€ž (Ã¢ÂÂ²d 3)
ns = Ssy Ã¢Ââ€ž Ã¢ÂÂ¶s
OD = D + d
ID = D Ã¢Ë†â€™ d
Na = Gd 4 ymax/(8D3Fmax)
Nt: Table 10 Ã¢â‚¬â€œ1
Ls: Table 10 Ã¢â‚¬â€œ1
L O: Table 10 Ã¢â‚¬â€œ1
(LO)cr = 2.63D/Ã¢ÂÂ£
fom = Ã¢Ë†â€™(rel. cost)Ã¢ÂÂ¥Ã¢ÂÂ² 2d 2Nt D Ã¢Ââ€ž 4
Print or display: d, D, C, OD, ID, Na , Nt , L s , LO, (LO)cr , ns , fom
Build a table, conduct design assessment by inspection
Eliminate infeasible designs by showing active constraints
Choose among satisfactory designs using the figure of merit
Ã¢â‚¬Â
const is found from Table 10Ã¢â‚¬â€œ6.
also available.9 There are almost as many ways to create a spring-design program as
there are programmers. Here, we will suggest one possible design approach.
Design Strategy
Make the a priori decisions, with hard-drawn steel wire the first choice (relative material cost is 1.0). Choose a wire size d. With all decisions made, generate a column of
parameters: d, D, C, OD or ID, Na , L s , L 0 , (L 0 )cr , n s , and fom. By incrementing
wire sizes available, we can scan the table of parameters and apply the design recommendations by inspection. After wire sizes are eliminated, choose the spring design
with the highest figure of merit. This will give the optimal design despite the presence
of a discrete design variable d and aggregation of equality and inequality constraints.
The column vector of information can be generated by using the flowchart displayed
in Fig. 10Ã¢â‚¬â€œ3. It is general enough to accommodate to the situations of as-wound and
9
For example, see Advanced Spring Design, a program developed jointly between the Spring Manufacturers
Institute (SMI), www.smihq.org, and Universal Technical Systems, Inc. (UTS), www.uts.com.
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Mechanical Engineering Design
set-removed springs, operating over a rod, or in a hole free of rod or hole. In as-wound
springs the controlling equation must be solved for the spring index as follows. From
Eq. (10Ã¢â‚¬â€œ3) with Ãâ€ž = Ssy /n s , C = D/d, K B from Eq. (10Ã¢â‚¬â€œ6), and Eq. (10Ã¢â‚¬â€œ17),
Ssy
8Fs D
4C + 2 8(1 + ÃŽÂ¾ ) Fmax C
= KB
=
(a)
ns
Ãâ‚¬d 3
4C Ã¢Ë†â€™ 3
Ãâ‚¬d 2
Let
ÃŽÂ±=
Ssy
ns
(b)
ÃŽÂ²=
8 (1 + ÃŽÂ¾ ) Fmax
Ãâ‚¬d 2
(c)
Substituting Eqs. (b) and (c) into (a) and simplifying yields a quadratic equation in C.
The larger of the two solutions will yield the spring index
2ÃŽÂ± Ã¢Ë†â€™ ÃŽÂ²
2ÃŽÂ± Ã¢Ë†â€™ ÃŽÂ² 2 3ÃŽÂ±
C=
+
Ã¢Ë†â€™
(10Ã¢â‚¬â€œ23)
4ÃŽÂ²
4ÃŽÂ²
4ÃŽÂ²
EXAMPLE 10Ã¢â‚¬â€œ2
Solution
A music wire helical compression spring is needed to support a 20-lbf load after being
compressed 2 in. Because of assembly considerations the solid height cannot exceed
1 in and the free length cannot be more than 4 in. Design the spring.
The a priori decisions are
Ã¢â‚¬Â¢ Music wire, A228; from Table 10Ã¢â‚¬â€œ4, A = 201 000 psi-inm; m = 0.145; from
Table 10Ã¢â‚¬â€œ5, E = 28.5 Mpsi, G = 11.75 Mpsi (expecting d > 0.064 in)

Ã¢â‚¬Â¢ Ends squared and ground

Ã¢â‚¬Â¢ Function: Fmax = 20 lbf…

Purchase answer to see full

attachment