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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, (θ − 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 20â—¦ 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 –
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, (θ − 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, (θ − 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 –
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
ï‚— Provide flexibility
ï‚— Store or absorb energy
ï‚—
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
ï‚—
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
ï‚— D
ï‚— J
ï‚— Lo
ï‚— p
ï‚—
Wire diameter
Mean coil diameter
Polar Moment of Inertia
Free length
Pitch
Helical Spring
d Wire diameter
ï‚— D Mean coil diameter
ï‚— Di Inside coil diameter
ï‚— Do Outside coil diameter
ï‚— J Polar Moment of Inertia
ï‚— Lo (or Lf) Free length
ï‚— Ls Solid length
ï‚— Nt Number of coils (turns)
ï‚—
Helical Spring
ï‚—
ï‚—
Helical coil spring with round wire
Equilibrium forces at cut section anywhere in the body of the
spring indicates direct shear and torsion
Fig. 10–1
Ends of Compression Springs
https://www.youtube.com/watch?v=wFhyZ
UvZSDA
Fig. 10–2
Formulas for Compression Springs With Different Ends
Table 10–1
Na is the number of active coils
Some Common Spring Steels
ï‚—
ï‚—
ï‚—
Hard-drawn wire (0.60-0.70C)
â—¦ Cheapest general-purpose
â—¦ Use only where life, accuracy, and deflection are not too important
Oil-tempered wire (0.60-0.70C)
â—¦ General-purpose
â—¦ Heat treated for greater strength and uniformity of properties
â—¦ Often used for larger diameter spring wire
Music wire (0.80-0.95C)
â—¦ Higher carbon for higher strength
â—¦ Best, toughest, and most widely used for small springs
â—¦ Good for fatigue
 The unified numbering system (UNS) is an alloy designation system
widely accepted in North America.
 American Iron and Steel Institute (AISI)
 American Society for Testing and Materials (ASTM)
Some Common Spring Steels
ï‚—
Chrome-vanadium
â—¦ Popular alloy spring steel
â—¦ Higher strengths than plain carbon steels
â—¦ Good for fatigue, shock, and impact
ï‚—
Chrome-silicon
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
ï‚—
ï‚—
Helical coil spring with round wire
Equilibrium forces at cut section anywhere in the body of the
spring indicates direct shear and torsion
Fig. 10–1
Stresses in Helical Springs
ï‚—
Torsional shear and direct shear
Additive (maximum) on inside fiber of
cross-section
ï‚—
Substitute terms
ï‚—
Fig. 10–1b
Stresses in Helical Springs
Factor out the torsional stress
d  8FD 

  1 

3 
2
D

d



Define Spring Index
Define Shear Stress Correction Factor
1
2C  1
Ks  1 

2C
2C
Maximum shear stress for helical spring
Curvature Effect
ï‚—
Can account for effect by replacing Ks with Wahl factor or
Bergsträsser factor which account for both direct shear and
curvature effect
Deflection of Helical Springs
Use Castigliano’s method to relate force and deflection
Fig. 10–1a
Critical Deflection for Stability
ï‚—
Buckling type of instability can occur in compression springs
when the deflection exceeds the critical deflection ycr
ï‚—
λeff 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.
ï‚—
a is the end-condition constant, defined on the next slide
ï‚—
C’1 and C’2 are elastic constants
End-Condition Constant
ï‚—
ï‚—
ï‚—
The a term in Eq. (10–11) is the end-condition constant.
It accounts for the way in which the ends of the spring are
supported.
Values are given in Table 10–2.
Table 10–2
Absolute Stability
ï‚—
Absolute stability occurs when, in Eq. (10–10),
2
C2 / eff
1
ï‚—
This results in the condition for absolute stability
ï‚—
For steels, this turns out to be
Strength of Spring Materials
ï‚—
ï‚—
ï‚—
ï‚—
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
10–4.
Constants for Estimating Tensile Strength
Table 10–4
Estimating Torsional Yield Strength
Since helical springs experience shear stress, shear yield strength
is needed.
ï‚— If actual data is not available, estimate from tensile strength
ï‚— Assume tensile yield strength is between 60-90% of tensile
strength
0.6Sut ï‚£ Ssy ï‚£ 0.9Sut
ï‚—
ï‚—
ï‚—
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 10–5)
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.
ï‚— The spring is made longer than needed, then compressed to solid
height, intentionally exceeding the torsional yield strength.
ï‚— This operation sets the spring to the required final free length.
ï‚— Yielding induces residual stresses opposite in direction to those
induced in service.
ï‚— 10 to 30 percent of the initial free length should be removed.
ï‚— 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.
ï‚—
Example 10–1
11 mm.
Example 10–1
Shigley’s Mechanical Engineering Design
Example 10–1
Shigley’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
ï‚— Preferred range for spring index
ï‚—
ï‚—
Preferred range for number of active coils
Shigley’s Mechanical Engineering Design
Helical Compression Spring Design for Static Service
ï‚—
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 ≤ 7/8 Fs
Define fractional overrun to closure as ξ where
ï‚—
This leads to
ï‚—
Solving the outer equality for ξ,
ï‚—
Thus, it is recommended that
ï‚—
ï‚—
ξ = 1/7 = 0.143 0.15
Shigley’s Mechanical Engineering Design
Summary of Recommended Design Conditions
ï‚—
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).
Shigley’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.
ï‚— The fom would be proportional to the relative material cost,
weight density, and volume
ï‚—
Shigley’s Mechanical Engineering Design
Design Flowchart for Static Loading
Choose d
Shigley’s Mechanical Engineering Design
Design Flowchart for Static Loading
Shigley’s Mechanical Engineering Design
Example 10–2
an 89 N
50.8 mm. Because of
25.4 mm and the free
101.6 mm. Design the spring.
Shigley’s Mechanical Engineering Design
Example 10–2
Shigley’s Mechanical Engineering Design
Shigley’s Mechanical Engineering Design
Practice Problems
Example (Problem -10-5)
Exercise (Problem- 27, 28, 30)
Shigley’s Mechanical Engineering Design
Lecture Slides
Rolling-Contact Bearings
Lect. 2.1
The McGraw-Hill Companies © 2012
Bearings
Why bearings?
https://www.youtube.com/watch?v=lgGd44
VRneY
Shigley’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.
Shigley’s Mechanical Engineering Design
Bearings
What affect the frictional characteristics of a bearing?
– Load,
– speed,
– and the operating viscosity of the lubricant
Shigley’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.
Shigley’s Mechanical Engineering Design
Bearings
The nomenclature of a ball bearing is
illustrated in Fig. 11–1, 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. 11–1
Shigley’s Mechanical Engineering Design
Types of Ball Bearings
Some of the various types of standardized bearings that are manufactured
are shown in Fig. 11–2.
The single-row deep-groove bearing (Fig. 11–2a) will take radial load as
well as some thrust load..
Shigley’s Mechanical Engineering Design
Types of Ball Bearings
The use of a filling notch (Fig. 11–2b) 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.
Shigley’s Mechanical Engineering Design
Types of Ball Bearings
The angular-contact bearing (Fig. 11–2c) provides a greater thrust
capacity.
Shigley’s Mechanical Engineering Design
Types of Ball Bearings
All these bearings may be obtained with shields (Fig. 11–2d) on one or
both sides. The shields are not a complete closure but do offer a measure
of protection against dirt.
Shigley’s Mechanical Engineering Design
Types of Ball Bearings
It has seals (Fig. 11–2e) that prevent lubricant from going out of the
bearing and dirt or any liquid from entering inside the bearing.
Shigley’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. 11–2f and
h) may be used.
Shigley’s Mechanical Engineering Design
Types of Ball Bearings
Double-row bearings (Fig. 11–2g) 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.
Shigley’s Mechanical Engineering Design
Types of Ball Bearings
The one-way ball thrust bearings (Fig. 11–2i) are made in many types
and sizes.
Shigley’s Mechanical Engineering Design
Types of Roller Bearings
Some of the large variety of standard roller bearings available are
illustrated in Fig. 11–3.
Straight roller bearings (Fig. 11–3a) 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.
.
Shigley’s Mechanical Engineering Design
Types of Roller Bearings
The straight-roller bearing (Fig. 11–3a) takes high radial load, but does
not take thrust load.
(Fig. 11–3)
Shigley’s Mechanical Engineering Design
Types of Roller Bearings
The spherical-roller thrust bearing (Fig. 11–3b) 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. 11–3)
Shigley’s Mechanical Engineering Design
Types of Roller Bearings
Needle bearings (Fig. 11–3d) are very useful where radial space is
limited.
They have a high load capacity when separators are used, but may be
obtained without separators.
Shigley’s Mechanical Engineering Design
Types of Roller Bearings
Tapered roller bearings (Fig. 11–3e, 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.
Shigley’s Mechanical Engineering Design
Lecture Slides
Lubrication and
Journal Bearings
Lect. 2.4
The McGraw-Hill Companies © 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.
Shigley’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
Shigley’s Mechanical Engineering Design
Types of Lubrication
Hydrodynamic
ï‚— Hydrostatic
ï‚— Elastohydrodynamic
ï‚— Boundary
ï‚— Solid film
ï‚—
Shigley’s Mechanical Engineering Design
Types of Lubrication
ï‚—
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.
Shigley’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.
ï‚—
Shigley’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 viscosity— any one of these
—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.
Shigley’s Mechanical Engineering Design
Viscosity
ï‚—
Shear stress in a fluid is proportional to the rate of change of
velocity with respect to y
ï‚—
m is absolute viscosity, also called dynamic viscosity
Fig. 12–1
Shigley’s Mechanical Engineering Design
Viscosity
ï‚—
For most lubricating fluids, the rate of shear is constant, thus
ï‚—
Fluids exhibiting this characteristic are called Newtonian fluids
Fig. 12–1
Shigley’s Mechanical Engineering Design
Units of Viscosity
ï‚—
ï‚—
ï‚—
ï‚—
Units of absolute viscosity
◦ ips units: reyn = lbf·s/in2
◦ SI units: Pa·s = N·s/m2
◦ cgs units: Poise =dyn·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:
Shigley’s Mechanical Engineering Design
Measurement of Viscosity
Saybolt Universal Viscosimeter used to measure viscosity
ï‚— 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
ï‚— Result is kinematic viscosity
ï‚— Unit is stoke = cm2/s
ï‚— Using Hagen-Poiseuille law kinematic viscosity based on seconds
Saybolt, also called Saybolt Universal viscosity (SUV) in seconds
is
ï‚—
where Zk is in centistokes (cSt) and t is the number of seconds
Saybolt
Shigley’s Mechanical Engineering Design
Measurement of Viscosity
ï‚—
In SI, kinematic viscosity n has units of m2/s
Conversion is
Eq. (12–3) in SI units,
ï‚—
To convert to dynamic viscosity, multiply v by density in SI units
ï‚—
ï‚—
where r is in kg/m3 and m is in pascal-seconds
Shigley’s Mechanical Engineering Design
Comparison of Absolute Viscosities of Various Fluids
Fig. 12–2
Shigley’s Mechanical Engineering Design
Petroff’s Lightly Loaded Journal Bearing
ï‚—
Fig. 12–3
Shigley’s Mechanical Engineering Design
Petroff’s Equation
Shigley’s Mechanical Engineering Design
Important Dimensionless Parameters
ï‚—
Some important dimensionless parameters used in lubrication
â—¦ r/c radial clearance ratio
â—¦ mN/P
â—¦ Sommerfeld number or bearing characteristic number
ï‚—
Interesting relation
Shigley’s Mechanical Engineering Design
Stable Lubrication
To the right of AB, changes in
conditions are self-correcting
and results in stable lubrication
ï‚— To the left of AB, changes in
conditions tend to get worse
and results in unstable
lubrication
ï‚— Point C represents the
approximate transition between
metal-to-metal contact and
thick film separation of the
parts
ï‚— Common design constraint for
point B,
ï‚—
Fig. 12–4
Shigley’s Mechanical Engineering Design
Thick Film Lubrication
ï‚—
Formation of a film
Fig. 12–5
Shigley’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
ï‚— Provide flexibility
ï‚— Store or absorb energy
ï‚—
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
ï‚—
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
ï‚— D
ï‚— J
ï‚— Lo
ï‚— p
ï‚—
Wire diameter
Mean coil diameter
Polar Moment of Inertia
Free length
Pitch
Helical Spring
d Wire diameter
ï‚— D Mean coil diameter
ï‚— Di Inside coil diameter
ï‚— Do Outside coil diameter
ï‚— J Polar Moment of Inertia
ï‚— Lo (or Lf) Free length
ï‚— Ls Solid length
ï‚— Nt Number of coils (turns)
ï‚—
Helical Spring
ï‚—
ï‚—
Helical coil spring with round wire
Equilibrium forces at cut section anywhere in the body of the
spring indicates direct shear and torsion
Fig. 10–1
Ends of Compression Springs
https://www.youtube.com/watch?v=wFhyZ
UvZSDA
Fig. 10–2
Formulas for Compression Springs With Different Ends
Table 10–1
Na is the number of active coils
Some Common Spring Steels
ï‚—
ï‚—
ï‚—
Hard-drawn wire (0.60-0.70C)
â—¦ Cheapest general-purpose
â—¦ Use only where life, accuracy, and deflection are not too important
Oil-tempered wire (0.60-0.70C)
â—¦ General-purpose
â—¦ Heat treated for greater strength and uniformity of properties
â—¦ Often used for larger diameter spring wire
Music wire (0.80-0.95C)
â—¦ Higher carbon for higher strength
â—¦ Best, toughest, and most widely used for small springs
â—¦ Good for fatigue
 The unified numbering system (UNS) is an alloy designation system
widely accepted in North America.
 American Iron and Steel Institute (AISI)
 American Society for Testing and Materials (ASTM)
Some Common Spring Steels
ï‚—
Chrome-vanadium
â—¦ Popular alloy spring steel
â—¦ Higher strengths than plain carbon steels
â—¦ Good for fatigue, shock, and impact
ï‚—
Chrome-silicon
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
ï‚—
ï‚—
Helical coil spring with round wire
Equilibrium forces at cut section anywhere in the body of the
spring indicates direct shear and torsion
Fig. 10–1
Stresses in Helical Springs
ï‚—
Torsional shear and direct shear
Additive (maximum) on inside fiber of
cross-section
ï‚—
Substitute terms
ï‚—
Fig. 10–1b
Stresses in Helical Springs
Factor out the torsional stress
d  8FD 

  1 

3 
2
D

d



Define Spring Index
Define Shear Stress Correction Factor
1
2C  1
Ks  1 

2C
2C
Maximum shear stress for helical spring
Curvature Effect
ï‚—
Can account for effect by replacing Ks with Wahl factor or
Bergsträsser factor which account for both direct shear and
curvature effect
Deflection of Helical Springs
Use Castigliano’s method to relate force and deflection
Fig. 10–1a
Critical Deflection for Stability
ï‚—
Buckling type of instability can occur in compression springs
when the deflection exceeds the critical deflection ycr
ï‚—
λeff 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.
ï‚—
a is the end-condition constant, defined on the next slide
ï‚—
C’1 and C’2 are elastic constants
End-Condition Constant
ï‚—
ï‚—
ï‚—
The a term in Eq. (10–11) is the end-condition constant.
It accounts for the way in which the ends of the spring are
supported.
Values are given in Table 10–2.
Table 10–2
Absolute Stability
ï‚—
Absolute stability occurs when, in Eq. (10–10),
2
C2 / eff
1
ï‚—
This results in the condition for absolute stability
ï‚—
For steels, this turns out to be
Strength of Spring Materials
ï‚—
ï‚—
ï‚—
ï‚—
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
10–4.
Constants for Estimating Tensile Strength
Table 10–4
Estimating Torsional Yield Strength
Since helical springs experience shear stress, shear yield strength
is needed.
ï‚— If actual data is not available, estimate from tensile strength
ï‚— Assume tensile yield strength is between 60-90% of tensile
strength
0.6Sut ï‚£ Ssy ï‚£ 0.9Sut
ï‚—
ï‚—
ï‚—
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 10–5)
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.
ï‚— The spring is made longer than needed, then compressed to solid
height, intentionally exceeding the torsional yield strength.
ï‚— This operation sets the spring to the required final free length.
ï‚— Yielding induces residual stresses opposite in direction to those
induced in service.
ï‚— 10 to 30 percent of the initial free length should be removed.
ï‚— 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.
ï‚—
Example 10–1
11 mm.
Example 10–1
Shigley’s Mechanical Engineering Design
Example 10–1
Shigley’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
ï‚— Preferred range for spring index
ï‚—
ï‚—
Preferred range for number of active coils
Shigley’s Mechanical Engineering Design
Helical Compression Spring Design for Static Service
ï‚—
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 ≤ 7/8 Fs
Define fractional overrun to closure as ξ where
ï‚—
This leads to
ï‚—
Solving the outer equality for ξ,
ï‚—
Thus, it is recommended that
ï‚—
ï‚—
ξ = 1/7 = 0.143 0.15
Shigley’s Mechanical Engineering Design
Summary of Recommended Design Conditions
ï‚—
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).
Shigley’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.
ï‚— The fom would be proportional to the relative material cost,
weight density, and volume
ï‚—
Shigley’s Mechanical Engineering Design
Design Flowchart for Static Loading
Choose d
Shigley’s Mechanical Engineering Design
Design Flowchart for Static Loading
Shigley’s Mechanical Engineering Design
Example 10–2
an 89 N
50.8 mm. Because of
25.4 mm and the free
101.6 mm. Design the spring.
Shigley’s Mechanical Engineering Design
Example 10–2
Shigley’s Mechanical Engineering Design
Shigley’s Mechanical Engineering Design
Practice Problems
Example (Problem -10-5)
Exercise (Problem- 27, 28, 30)
Shigley’s Mechanical Engineering Design
Lecture Slides
Rolling-Contact Bearings
Lect. 2.1
The McGraw-Hill Companies © 2012
Bearings
Why bearings?
https://www.youtube.com/watch?v=lgGd44
VRneY
Shigley’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.
Shigley’s Mechanical Engineering Design
Bearings
What affect the frictional characteristics of a bearing?
– Load,
– speed,
– and the operating viscosity of the lubricant
Shigley’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.
Shigley’s Mechanical Engineering Design
Bearings
The nomenclature of a ball bearing is
illustrated in Fig. 11–1, 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. 11–1
Shigley’s Mechanical Engineering Design
Types of Ball Bearings
Some of the various types of standardized bearings that are manufactured
are shown in Fig. 11–2.
The single-row deep-groove bearing (Fig. 11–2a) will take radial load as
well as some thrust load..
Shigley’s Mechanical Engineering Design
Types of Ball Bearings
The use of a filling notch (Fig. 11–2b) 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.
Shigley’s Mechanical Engineering Design
Types of Ball Bearings
The angular-contact bearing (Fig. 11–2c) provides a greater thrust
capacity.
Shigley’s Mechanical Engineering Design
Types of Ball Bearings
All these bearings may be obtained with shields (Fig. 11–2d) on one or
both sides. The shields are not a complete closure but do offer a measure
of protection against dirt.
Shigley’s Mechanical Engineering Design
Types of Ball Bearings
It has seals (Fig. 11–2e) that prevent lubricant from going out of the
bearing and dirt or any liquid from entering inside the bearing.
Shigley’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. 11–2f and
h) may be used.
Shigley’s Mechanical Engineering Design
Types of Ball Bearings
Double-row bearings (Fig. 11–2g) 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.
Shigley’s Mechanical Engineering Design
Types of Ball Bearings
The one-way ball thrust bearings (Fig. 11–2i) are made in many types
and sizes.
Shigley’s Mechanical Engineering Design
Types of Roller Bearings
Some of the large variety of standard roller bearings available are
illustrated in Fig. 11–3.
Straight roller bearings (Fig. 11–3a) 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.
.
Shigley’s Mechanical Engineering Design
Types of Roller Bearings
The straight-roller bearing (Fig. 11–3a) takes high radial load, but does
not take thrust load.
(Fig. 11–3)
Shigley’s Mechanical Engineering Design
Types of Roller Bearings
The spherical-roller thrust bearing (Fig. 11–3b) 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. 11–3)
Shigley’s Mechanical Engineering Design
Types of Roller Bearings
Needle bearings (Fig. 11–3d) are very useful where radial space is
limited.
They have a high load capacity when separators are used, but may be
obtained without separators.
Shigley’s Mechanical Engineering Design
Types of Roller Bearings
Tapered roller bearings (Fig. 11–3e, 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.
Shigley’s Mechanical Engineering Design
Lecture Slides
Rolling-Contact Bearings
Lect 2.2
The McGraw-Hill Companies © 2012
Bearing Life Definitions
ï‚—
ï‚—
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
â—¦ clean and properly lubricated,
â—¦ mounted and sealed against the entrance of dust and dirt,
â—¦ is maintained in this condition, and is operated at reasonable
temperatures,
then metal fatigue will be the only cause of failure.
Shigley’s Mechanical Engineering Design
Bearing Life Definitions
Common life measures are
• Number of revolutions until the first tangible evidence
of fatigue
• Number of hours of use at a standard angular speed until
the first tangible evidence of fatigue
Shigley’s Mechanical Engineering Design
Bearing Life Definitions
ï‚—
Bearing Failure: Spalling or pitting of an area of 0.01 in2
ï‚—
Life: Number of revolutions (or hours @ given speed) required for
failure.
â—¦ For one bearing
ï‚—
Rating Life: Life required for 10% of sample to fail.
â—¦ For a group of bearings
â—¦ Also called Minimum Life or L10 Life
ï‚—
Median Life: Average life required for 50% of sample to fail.
â—¦ For many groups of bearings
â—¦ Also called Average Life or Average Median Life
â—¦ Median Life is typically 4 or 5 times the L10 Life
Shigley’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 manufacturer’s rating life.
â—¦ Depends on type, geometry, accuracy of fabrication, and
material of bearing
â—¦ Also called Basic Dynamic Load Rating, and Basic Dynamic
Capacity
ï‚— Basic Load Rating, C:
A catalog load rating based on a rating life of 106 revolutions of the
inner ring.
â—¦ The radial load that would be necessary to cause failure at such a
low life is unrealistically high.
ï‚—
Shigley’s Mechanical Engineering Design
Load-Life Relationship
ï‚—
ï‚—
ï‚—
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,
â—¦ a = 3 for ball bearings
â—¦ a = 10/3 for roller bearings
(cylindrical and tapered
roller)
Fig. 11–4
Shigley’s Mechanical Engineering Design
Load-Life Relationship
ï‚—
Applying Eq. (11–1) to two load-life conditions,
ï‚—
Denoting condition 1 with R for catalog rating conditions, and
condition 2 with D for the desired design conditions,

ï‚—
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,
Shigley’s Mechanical Engineering Design
Load-Life Relationship
Solving Eq. (a) for FR, which is just another notation for the catalog load
rating,
ï‚—
ï‚—
ï‚—
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.
Shigley’s Mechanical Engineering Design
Load-Life Relationship
ï‚—
It is often convenient to define a dimensionless multiple of rating
life
Shigley’s Mechanical Engineering Design
Example 11–1
2 kN
2
16.2 kN
Shigley’s Mechanical Engineering Design
Reliability vs. Life
At constant load, the life measure distribution is right skewed.
ï‚— The Weibull distribution is a good candidate.
ï‚— Defining the life measure in dimensionless form as x = L/L10, the
reliability is expressed with a Weibull distribution as
ï‚—
Shigley’s Mechanical Engineering Design
Reliability vs. Life
ï‚—
An explicit expression for the cumulative distribution function is
Shigley’s Mechanical Engineering Design
Relating Load, Life, and Reliability
ï‚—
ï‚—
ï‚—
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. 11–5
Shigley’s Mechanical Engineering Design
Relating Load, Life, and Reliability
ï‚—
Along a constant reliability contour (BD), Eq. (11–2) applies:
Fig. 11–5
Shigley’s Mechanical Engineering Design
Relating Load, Life, and Reliability
ï‚—
Along a constant load line (AB), Eq. (11–4) applies:
ï‚—
Solving for xB,
Fig. 11–5
Shigley’s Mechanical Engineering Design
Relating Load, Life, and Reliability
ï‚—
Substituting xB into Eq. (a),
ï‚—
Noting that FB = C10, and including an application factor af,
ï‚—
Note that when RD = 0.90, the denominator equals one and the
equation reduces to Eq. (11–3).
Shigley’s Mechanical Engineering Design
Weibull Parameters
The Weibull parameters x0, q, and b are usually provided by the
catalog.
ï‚— Typical values of Weibull parameters are given on p. 608 at the
beginning of the end-of-chapter problems, and shown below.
ï‚— Manufacturer 1 parameters are common for tapered roller
bearings
ï‚— Manufacturer 2 parameters are common for ball and straight
roller bearings
ï‚—
Shigley’s Mechanical Engineering Design
Relating Load, Life, and Reliability
ï‚—
ï‚—
Eq. (11–6) can be simplified slightly for calculator entry. Note
that
where pf is the probability for failure
Thus Eq. (11–6) can be approximated by
Shigley’s Mechanical Engineering Design
Example 11–3
1840 N
(1.2) (1.84)
29.7 kN
Shigley’s Mechanical Engineering Design
Lecture Slides
Rolling-Contact Bearings
Lect.2.3
The McGraw-Hill Companies © 2012
Dimension-Series Code
ï‚—
ï‚—
ï‚—
ï‚—
ï‚—
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. 11–7
Shigley’s Mechanical Engineering Design
Representative Catalog Data for Ball Bearings (Table 11–2)
Representative Catalog Data for Cylindrical Roller Bearings
(Table 11–3)
Combined Radial and Thrust Loading
ï‚—
Equivalent Radial Load, Fe:
Is the load that does the same damage as the combined radial
and thrust loads together.
ï‚—
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).
â—¦ d = diameter of roller
â—¦ Used to check for permanent deformation
â—¦ Used in combining radial and thrust loads into an equivalent
radial load
Shigley’s Mechanical Engineering Design
Combined Radial and Thrust Loading
ï‚—
ï‚—
ï‚—
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.
â—¦ V = 1 for inner ring rotation
â—¦ V = 1.2 for outer ring rotation
Fig. 11–6
Shigley’s Mechanical Engineering Design
Combined Radial and Thrust Loading
ï‚—
The data can be approximated by
two straight lines
X is the ordinate intercept and Y is
the slope
ï‚— Basically indicates that Fe equals Fr
for smaller ratios of Fa/Fr, then
begins to rise when Fa/Fr exceeds
some amount e
ï‚—
Fig. 11–6
Shigley’s Mechanical Engineering Design
Combined Radial and Thrust Loading
ï‚—
It is common to express the two
equations as a single equation
where
i = 1 when Fa/VFr ≤ e
i = 2 when Fa/VFr > e
ï‚—
X and Y factors depend on geometry
and construction of the specific
bearing.
Fig. 11–6
Shigley’s Mechanical Engineering Design
Equivalent Radial Load Factors for Ball Bearings
X and Y for specific bearing obtained from bearing catalog.
 Table 11–1 gives representative values in a manner common to
many catalogs.
Table 11–1
ï‚—
Shigley’s Mechanical Engineering Design
Equivalent Radial Load Factors for Ball Bearings
Table 11–1
ï‚—
ï‚—
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.
Shigley’s Mechanical Engineering Design
Bearing Life Recommendations (Table 11–4)
Shigley’s Mechanical Engineering Design
Recommended Load Application Factors (Table 11–5)
Shigley’s Mechanical Engineering Design
Example 11–4
Shigley’s Mechanical Engineering Design
Example 11–4
Shigley’s Mechanical Engineering Design
Bearing Lubrication
ï‚—
The purposes of bearing lubrication
â—¦ To provide a film of lubricant between the sliding and rolling
surfaces
â—¦ To help distribute and dissipate heat
â—¦ To prevent corrosion of the bearing surfaces
â—¦ To protect the parts from the entrance of foreign matter
Shigley’s Mechanical Engineering Design
Bearing Lubrication
ï‚—
Either oil or grease may be used, with each having advantages in
certain situations.
Shigley’s Mechanical Engineering Design
Realized Bearing Reliability
ï‚—
Eq. (11–6) was previously derived to determine a suitable catalog
rated load for a given design situation and reliability goal.
ï‚—
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. (11–6) for the reliability,
ï‚—
ï‚—
Shigley’s Mechanical Engineering Design
Realized Bearing Reliability
ï‚—
Similarly for the alternate approximate equation, Eq. (11–7),
Shigley’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 11–2. Choose a bearing and
estimate the existing reliability in service.
Shigley’s Mechanical Engineering Design
Group Example (11.20)
An 02-series single-row deep-groove ball bearing with a 65-mm bore
(see Tables 11–1 and 11–2 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.
Shigley’s Mechanical Engineering Design
Machine Design II
Gears – 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.
Shigley’s Mechanical Engineering Design
Types of Gears
Spur
Bevel
Helical
Figs. 13–1 to 13–4
Worm
Shigley’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
Shigley’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
Shigley’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
Shigley’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
Shigley’s Mechanical Engineering Design
Nomenclature of Spur-Gear Teeth
Shigley’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.
Shigley’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
Shigley’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.
Shigley’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.
Shigley’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.
Shigley’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.
Shigley’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.
Shigley’s Mechanical Engineering Design
Tooth Size
Shigley’s Mechanical Engineering Design
Machine Design II
Gears – General
Lect. 3.2
Dr. Mir Md Maruf Morshed
Conjugate Action
ï‚—
When surfaces roll/slide against
each other and produce constant
angular velocity ratio, they are
said to have conjugate action.
Fig. 13–6
Shigley’s Mechanical Engineering Design
Conjugate Action
ï‚—
Forces are transmitted on
line of action which is
normal to the contacting
surfaces.
• 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 “line of
action” or the “pressure line”.
Fig. 13–6
Shigley’s Mechanical Engineering Design
Conjugate Action
ï‚—
Angular velocity ratio is
inversely proportional to the
radii to point P, the pitch
point.
ï‚—
Circles drawn through P
from each fixed pivot are
pitch circles, each with a
pitch radius.
Fig. 13–6
Shigley’s Mechanical Engineering Design
Involute Profile
ï‚—
The most common conjugate profile is the involute profile.
Fig. 13–8
Shigley’s Mechanical Engineering Design
Involute Profile
The most common conjugate profile is the involute profile.
ï‚— Can be generated by unwrapping a string from a cylinder, keeping
the string taut and tangent to the cylinder.
https://www.tec-science.com/wpï‚— Circle is called base circle.
content/uploads/2018/10/en-involute-gearï‚—
involute-circle-construction.mp4


Fig. 13–8
Shigley’s Mechanical Engineering Design
Circles of a Gear Layout
Fig. 13–9
Shigley’s Mechanical Engineering Design
Sequence of Gear Layout (Fundamentals)
• Pitch circles in contact
• Pressure line at desired
pressure angle
• Base circles tangent to
pressure line
• Involute profile from
base circle
• Cap teeth at addendum
circle at 1/P from pitch
circle
• Root of teeth at
dedendum
circle at 1.25/P from
pitch circle
• Tooth spacing from
circular pitch, p = p / P
Fig. 13–9
Shigley’s Mechanical Engineering Design
Relation of Base Circle to Pressure Angle
Fig. 13–10
Shigley’s Mechanical Engineering Design
Tooth Action
First point of
contact at a
where flank of
pinion touches
tip of gear
ï‚— Last point of
contact at b
where tip of
pinion touches
flank of gear
ï‚— Line ab is line of
action
ï‚— Angle of action
is sum of angle
of approach and
angle of recess
ï‚—
Fig. 13–12
Shigley’s Mechanical Engineering Design
Tooth Action
Pitch circles
Fig. 13–12
Shigley’s Mechanical Engineering Design
Shigley’s Mechanical Engineering Design
Shigley’s Mechanical Engineering Design
Shigley’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
ï‚— The contact ratio mc is the ratio of the arc of action and the circular
pitch.
ï‚—
ï‚—
The contact ratio is the average number of pairs of teeth in contact.
Shigley’s Mechanical Engineering Design
Contact Ratio
ï‚—
Contact ratio can also be found from the length of the line of action
ï‚—
The contact ratio should be at least 1.2
Watch this following Video for Gear Nomenclatures:

Fig. 13–15
Shigley’s Mechanical Engineering Design
Forming of Gear Teeth
ï‚—
Common ways of forming gear teeth
â—¦ Sand casting
â—¦ Shell molding
â—¦ Investment casting
â—¦ Permanent-mold casting
â—¦ Die casting
â—¦ Centrifugal casting
â—¦ Powder-metallurgy
â—¦ Extrusion
â—¦ Injection molding (for thermoplastics)
â—¦ Cold forming
Shigley’s Mechanical Engineering Design
Cutting of Gear Teeth
ï‚—
Common ways of cutting gear teeth
â—¦ Milling
â—¦ Shaping
â—¦ Hobbing
â—¦ Milling a tooth gear Video:
https://www.youtube.com/watch?v=27qaZz
i3ZCU
Shigley’s Mechanical Engineering Design
Shaping with Pinion Cutter

Fig. 13–17
Shigley’s Mechanical Engineering Design
Shaping with a Rack

Fig. 13–18
Shigley’s Mechanical Engineering Design
Hobbing a Worm Gear

Fig. 13–19
Shigley’s Mechanical Engineering Design
Standard and Commonly Used Tooth Systems for Spur Gears
Table 13–1
Shigley’s Mechanical Engineering Design
Tooth Sizes in General Use
Table 13–2
Shigley’s Mechanical Engineering Design
Machine Design II
Gears – General
Lect. 3.3
Dr. Mir Md Maruf Morshed
Interference
ï‚—
ï‚—
ï‚—
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.
Shigley’s Mechanical Engineering Design
Interference
ï‚—
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.
ï‚—
Shigley’s Mechanical Engineering Design
Interference of Spur Gears
ï‚—
On spur and gear with one-to-one gear ratio, smallest number of
teeth which will not have interference is
ï‚—
k =1 for full depth teeth. k = 0.8 for stub teeth
ï‚—
On spur meshed with larger gear with gear ratio mG = NG/NP = m,
the smallest number of teeth which will not have interference is
ï‚—
Largest gear with a specified pinion that is interference-free is
Shigley’s Mechanical Engineering Design
Interference of Spur Gears
ï‚—
k =1 for full depth teeth. k = 0.8 for stub teeth
ï‚—
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.
Shigley’s Mechanical Engineering Design
Interference
ï‚—
For 20º pressure angle, the most useful values from Eqs. (13–11)
and (13–12) 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
Shigley’s Mechanical Engineering Design
Interference
ï‚—
Increasing the pressure angle to 25º 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
Shigley’s Mechanical Engineering Design
Interference
ï‚—
ï‚—
ï‚—
ï‚—
ï‚—
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.
Shigley’s Mechanical Engineering Design
Standard and Commonly Used Tooth Systems for Spur Gears
Shigley’s Mechanical Engineering Design
Tooth Sizes in General Use
Shigley’s Mechanical Engineering Design
Machine Design II
Gears – General
Lect. 3.4
Dr. Mir Md Maruf Morshed
Gear Trains
ï‚—
The combination of two or more gears transmitting power from
one shaft to another is called gear train.
ï‚—
Types of Gear Trains
â—¦
â—¦
â—¦
â—¦
Simple Gear Train
Compound Gear Train
Reverted Gear Train
Epicyclic Gear Train
Shigley’s Mechanical Engineering Design
Simple Gear Train
â—¦ When there is one gear on each shaft, the arrangement is called simple gear
train
(a)
ï‚—
(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
Shigley’s Mechanical Engineering Design
Train Value
L: last F: First
Shigley’s Mechanical Engineering Design
Compound Gear Train
ï‚—
ï‚—
ï‚—
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.
Shigley’s Mechanical Engineering Design
Compound Gear Train
The design of gear trains to accomplish a specific train value is
straightforward.
ï‚— The numbers of teeth on gears must be integers (determine them
first then obtain pitch diameters).
ï‚— Determine the number of stages necessary to obtain the overall
ratio, then divide the overall ratio into portions to be accomplished
in each stage.
ï‚— 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.
Shigley’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.
ï‚— 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. 13–7).
ï‚— 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.
ï‚—
Shigley’s Mechanical Engineering Design
Example 13–3
Substituting m = 5.4772 and Ф = 20
o
15.848
Shigley’s Mechanical Engineering Design
Example 13–3
Shigley’s Mechanical Engineering Design
Example 13–4
Shigley’s Mechanical Engineering Design
Example 13–4
Substituting m = 6 and Ф = 20
15.96
N3 = 16
Substituting m = 5 and Ф = 20
15.76
o
o
N5 = 16
Shigley’s Mechanical Engineering Design
Compound Reverted Gear Train
ï‚—
ï‚—
A compound gear train with input and output shafts in-line
Geometry condition must be satisfied
Fig. 13–29
Shigley’s Mechanical Engineering Design
Example 13–5
Shigley’s Mechanical Engineering Design
Example 13–5
Shigley’s Mechanical Engineering Design
Example 13–5
Shigley’s Mechanical Engineering Design
Example 13–5
Shigley’s Mechanical Engineering Design
Planetary Gear Train
ï‚—
ï‚—
ï‚—
ï‚—
ï‚—
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. 13–30
Shigley’s Mechanical Engineering Design
Planetary Gear Trains
ï‚—
Train value is relative to arm
Fig. 13–31
Fig. 13–30
Shigley’s Mechanical Engineering Design
Example 13–6
Fig. 13–30
Shigley’s Mechanical Engineering Design
Example 13–6
Shigley’s Mechanical Engineering Design
Example 13–6
Shigley’s Mechanical Engineering Design
Machine Design II
Gears – General
Lect. 3.5
Dr. Mir Md Maruf Morshed
Force Analysis – Spur Gearing
ï‚—
In the Figure shown below
â—¦ Gear 2 (the pinion) is the driver
â—¦ Gear 3 is the driven
â—¦ The shafts are a and b for the pinion
and the gear, respectively.
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. 13–32
Shigley’s Mechanical Engineering Design
Force Analysis – Spur Gearing
Shigley’s Mechanical Engineering Design
Force Analysis – Spur Gearing
ï‚—
ï‚—
ï‚—
ï‚—
ï‚—
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.
Shigley’s Mechanical Engineering Design
Force Analysis – Spur Gearing
ï‚—
This figure shows the
FBD of the pinion
ï‚—
Transmitted load Wt is
the tangential load
ï‚—
It is the useful component
of force, transmitting the
torque
Fig. 13–33
Shigley’s Mechanical Engineering Design
Power in Spur Gearing
ï‚—
Transmitted power H
ï‚—
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
Shigley’s Mechanical Engineering Design
Power in Spur Gearing
ï‚—
Useful power relation in customary units,
ï‚—
In SI units,
Shigley’s Mechanical Engineering Design
Example 13–7
Fig. 13–34
Shigley’s Mechanical Engineering Design
Example 13–7
Shigley’s Mechanical Engineering Design
Example 13–7
Shigley’s Mechanical Engineering Design
ISBN: 0073529281
Author: Budynas / Nisbett
Title: Shigley’s Mechanical
Engineering Design, 8e
Front endsheets
Color: 2C (Black & PMS 540 U)
Pages: 2,3
ISBN: 0073529281
Author: Budynas / Nisbett
Title: Shigley’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
10–1
Stresses in Helical Springs
10–2
The Curvature Effect
10–3
Deflection of Helical Springs
10–4
Compression Springs
10–5
Stability
10–6
Spring Materials
10–7
Helical Compression Spring Design for Static Service
10–8
Critical Frequency of Helical Springs
10–9
Fatigue Loading of Helical Compression Springs
518
519
520
520
522
523
528
534
536
10–10
Helical Compression Spring Design for Fatigue Loading
10–11
Extension Springs
10–12
Helical Coil Torsion Springs
10–13
Belleville Springs
10–14
Miscellaneous Springs
10–15
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 machinery’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.
10–1
Stresses in Helical Springs
Figure 10–1a 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. 10–1b), 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. (3–23), p. 89, with V = F and the torsional shear stress given
by Eq. (3–37), p. 101. The result is
Ï„max =
Tr
F
+
J
A
F
F
Figure 10–1
(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 = FDå…¾2
F
(b)
F
D
(a)
(a)
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Mechanical Springs
519
at the inside fiber of the spring. Substitution of Ï„max = Ï„ , T = F D/2, r = d/2, J =
πd 4/32, and A = πd 2 /4 gives
Ï„=
8F D
4F
+
3
Ï€d
Ï€d 2
(b)
Now we define the spring index
C=
D
d
(10–1)
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
Ï„ = Ks
8F D
Ï€d 3
(10–2)
where K s is a shear stress-correction factor and is defined by the equation
Ks =
2C + 1
2C
(10–3)
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.
10–2
The Curvature Effect
Equation (10–2) 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. (10–2) 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 − 1 0.615
+
4C − 4
C
(10–4)
KB =
4C + 2
4C − 3
(10–5)
The first of these is called the Wahl factor, and the second, the Bergsträsser factor.2
Since the results of these two equations differ by the order of 1 percent, Eq. (10–6)
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 Samónov, “Some Aspects of Design of Helical Compression Springs,” Int. Symp. Design and
Synthesis, Tokyo, 1984.
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Mechanical Engineering Design
effect of the direct shear. Thus, using Eq. (10–5) with Eq. (10–3), the curvature correction factor is found to be
KB
2C(4C + 2)
Kc =
=
(10–6)
Ks
(4C − 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
Ï„ = KB
to predict the largest shear stress.
10–3
8F D
Ï€d 3
(10–7)
Deflection of Helical Springs
The deflection-force relations are quite easily obtained by using Castigliano’s theorem.
The total strain energy for a helical spring is composed of a torsional component and
a shear component. From Eqs. (4–18) and (4–20), p. 162, the strain energy is
U=
T 2l
F 2l
+
2G J
2AG
(a)
Substituting T = F D/2, l = π D N , J = πd 4/32, and A = πd 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 Castigliano’s theorem, Eq. (4–26),
p. 165, to find total deflection y gives
y=
∂U
8F D 3 N
4F D N
=
+ 2
∂F
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)
(10–8)
The spring rate, also called the scale of the spring, is k = F/y, and so
. d4G
k=
8D 3 N
10–4
(10–9)
Compression Springs
The four types of ends generally used for compression springs are illustrated in Fig. 10–2.
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 10–1 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 10–1 are often
3
For a thorough discussion and development of these relations, see Cyril Samónov, “Computer-Aided
Design of Helical Compression Springs,” ASME paper No. 80-DET-69, 1980.
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Mechanical Springs
521
Figure 10–2
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 10–1
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 − a)d
where a varies, with an average of 0.75, so the entry d Nt in Table 10–1 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, “Accurate Spring Heights,” Machine Design, vol. 56, no. 2, January 26, 1984.
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10–5
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
1− 1− 2
λeff
(10–10)
where ycr is the deflection corresponding to the onset of instability. Samónov5 states that
this equation is cited by Wahl6 and verified experimentally by Haringx.7 The quantity
λeff in Eq. (10–10) is the effective slenderness ratio and is given by the equation
λeff =
αL 0
D
(10–11)
C1 and C2 are elastic constants defined by the equations
C1 =
E
2(E − G)
C2 =
2π 2 (E − G)
2G + E
Equation (10–11) contains the end-condition constant α. This depends upon how the
ends of the spring are supported. Table 10–2 gives values of α for usual end conditions.
Note how closely these resemble the end conditions for columns.
Absolute stability occurs when, in Eq. (10–10), the term C2 /λ2eff is greater than
unity. This means that the condition for absolute stability is that
L0 < Table 10–2 π D 2(E − G) 1/2 α 2G + E End Condition End-Condition Constants α for Helical Compression Springs* (10–12) 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 ∗ Ends supported by flat surfaces must be squared and ground. 5 Cyril Samónov “Computer-Aided Design,” op. cit. 6 A. M. Wahl, Mechanical Springs, 2d ed., McGraw-Hill, New York, 1963. 7 J. A. Haringx, “On Highly Compressible Helical Springs and Rubber Rods and Their Application for Vibration-Free Mountings,” I and II, Philips Res. Rep., vol. 3, December 1948, pp. 401– 449, and vol. 4, February 1949, pp. 49–80. bud29281_ch10_517-568.qxd 12/16/2009 7:14 pm Page 523 pinnacle 203:MHDQ196:bud29281:0073529281:bud29281_pagefiles: Mechanical Springs 523 For steels, this turns out to be L 0 < 2.63 D α (10–13) For squared and ground ends α = 0.5 and L 0 < 5.26D. 10–6 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 10–3. 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
(10–14)
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 10–4.
In Eq. (10–14) when d is measured in millimeters, then A is in MPa · mmm and when
d is measured in inches, then A is in kpsi · 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 strength—
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 ≤ Ssy ≤ 0.52Sut
(10–15)
for steels.
For wires listed in Table 10–5, 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 ≥ 0.50Sut .
Many nonferrous materials (not shown) as a group have Ssy ≥ 0.35Sut . In view of this,
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Mechanical Engineering Design
Table 10–3
High-Carbon and Alloy
Spring Steels
Source: From Harold C. R.
Carlson, “Selection and
Application of Spring
Materials,” Mechanical
Engineering, vol. 78, 1956,
pp. 331–334.
Name of
Material
Similar
Specifications
Music wire,
0.80–0.95C
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 120°C
(250°F) or at subzero temperatures.
Oil-tempered wire,
0.60–0.70C
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
180°C (350°F) or at subzero
temperatures.
Hard-drawn wire,
0.60–0.70C
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
120°C (250°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 220°C (425°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 250°C
(475°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 10–4
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
0.004–0.256
201
†
OQ&T wire
A229
0.187
0.020–0.500
147
Hard-drawn wire‡
A227
0.190
0.028–0.500
140
Chrome-vanadium wire
A232
0.168
0.032–0.437
Chrome-silicon wire
A401
0.108
0.063–0.375
0.146
0.013–0.10
169
0.263
0.10–0.20
128
Material
§
302 Stainless wire
#
A313
Diameter,
in
0.478
Phosphor-bronze wire**
∗ Surface
B159
A,
kpsi inm
Relative
Diameter,
A,
Cost
mm
MPa mmm of Wire
0.10–6.5
2211
2.6
0.5–12.7
1855
1.3
0.7–12.7
1783
1.0
169
0.8–11.1
2005
3.1
202
1.6–9.5
1974
4.0
0.3–2.5
1867
7.6–11
2.5–5
2065
0.20–0.40
90
5–10
2911
0
0.004–0.022
145
0.1–0.6
1000
0.028
0.022–0.075
121
0.6–2
0.064
0.075–0.30
110
2–7.5
8.0
913
932
is smooth, free of defects, and has a bright, lustrous finish.
†
Has a slight heat-treating scale which must be removed before plating.
‡
Surface is smooth and bright with no visible marks.
§
Aircraft-quality tempered wire, can also be obtained annealed.
Tempered
# Type
to Rockwell C49, but may be obtained untempered.
302 stainless steel.
∗∗ Temper
CA510.
Joerres8 uses the maximum allowable torsional stress for static application shown in
Table 10–6. For specific materials for which you have torsional yield information use
this table as a guide. Joerres provides set-removal information in Table 10–6, that
Ssy ≥ 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 . Samónov discusses the problem of
allowable stress and shows that
Ssy = Ï„all = 0.56Sut
(10–16)
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. (10–2) is used without stresscorrection factor.
8
Robert E. Joerres, “Springs,” 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 10–5
Mechanical Properties of Some Spring Wires
Material
Elastic Limit,
Percent of Sut
Tension Torsion
Music wire A228
65–75
HD spring A227
60–70
45–60
45–55
G
E
Diameter
d, in
0.125
28.0
193
11.6
80.0
0.125
Oil tempered A239
85–90
45–50
28.5
196.5
11.2
77.2
Valve spring A230
85–90
50–60
29.5
203.4
11.2
77.2
Chrome-vanadium A231
88–93
65–75
A232
88–93
Chrome-silicon A401
29.5
203.4
11.2
77.2
29.5
203.4
11.2
77.2
85–93
65–75
29.5
203.4
11.2
77.2
A313*
65–75
45–55
28
193
10
69.0
17-7PH
75–80
55–60
29.5
208.4
11
75.8
414
65–70
42–55
29
200
11.2
77.2
420
65–75
45–55
29
200
11.2
77.2
431
Stainless steel
72–76
50–55
30
206
11.5
79.3
Phosphor-bronze B159
75–80
45–50
15
103.4
6
41.4
Beryllium-copper B197
70
50
17
117.2
6.5
44.8
75
50–55
19
131
7.3
50.3
65–70
40–45
31
213.7
11.2
77.2
Inconel alloy X-750
*Also includes 302, 304, and 316.
Note: See Table 10–6 for allowable torsional stress design values.
Table 10–6
Maximum Allowable
Torsional Stresses for
Helical Compression
Springs in Static
Applications
Source: Robert E. Joerres,
“Springs,” 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
60–70
Hardened and tempered
carbon and low-alloy
steel
50
65–75
Austenitic stainless
steels
35
55–65
Nonferrous alloys
35
55–65
Material
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Mechanical Springs
527
EXAMPLE 10–1
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 A–28, the wire diameter is d = 0.037 in. From Table 10–4, we find
A = 201 kpsi · inm and m = 0.145. Therefore, from Eq. (10–14)
Sut =
A
201
=
= 324 kpsi
dm
0.0370.145
Then, from Table 10–6,
Ssy = 0.45Sut = 0.45(324) = 146 kpsi
Answer
7
− 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. (10–6),
KB =
4C + 2
4 (10.8) + 2
=
= 1.124
4C − 3
4 (10.8) − 3
Now rearrange Eq. (10–7) replacing τ with Ssy , and solve for F:
F=
Answer
Ï€d 3 Ssy
Ï€(0.0373 )146(103 )
=
= 6.46 lbf
8K B D
8(1.124) 0.400
(c) From Table 10–1, Na = 12.5 − 2 = 10.5 turns. In Table 10–5, G = 11.85 Mpsi,
and the scale of the spring is found to be, from Eq. (10–9),
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 10–1,
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. (10–13) and Table 10–2 give
L 0 < 2.63 D 0.400 = 2.63 = 2.10 in α 0.5 bud29281_ch10_517-568.qxd 528 12/16/2009 7:14 pm Page 528 pinnacle 203:MHDQ196:bud29281:0073529281:bud29281_pagefiles: 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 α 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 10–1, the pitch of the body coil is p= Answer 10–7 L 0 − 3d 2.06 − 3(0.037) = = 0.186 in Na 10.5 Helical Compression Spring Design for Static Service The preferred range of spring index is 4 ≤ C ≤ 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 ≤ Na ≤ 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 spring’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 ≤ 78 Fs . Defining the fractional overrun to closure as ξ , where Fs = (1 + ξ )Fmax (10–17) it follows that Fs = (1 + ξ )Fmax = (1 + ξ ) 7 Fs 8 . From the outer equality ξ = 1/7 = 0.143 = 0.15. Thus, it is recommended that ξ ≥ 0.15. In addition to the relationships and material properties for springs, we now have some recommended design conditions to follow, namely: 4 ≤ C ≤ 12 (10–18) 3 ≤ Na ≤ 15 (10–19) ξ ≥ 0.15 (10–20) n s ≥ 1.2 (10–21) 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 = −(relative material cost) γ π 2 d 2 Nt D 4 (10–22) For comparisons between steels, the specific weight γ 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 “spring design slide rules” were used by many to simplify the spring design problem. Today, the computer enables the designer to create programs in many different formats—direct programming, spreadsheet, MATLAB, etc. Commercial programs are bud29281_ch10_517-568.qxd 12/16/2009 7:14 pm Page 529 pinnacle 203:MHDQ196:bud29281:0073529281:bud29281_pagefiles: Mechanical Springs Figure 10–3 529 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) ⁄d m † Ssy = 0.65A ⁄d m D = d hole − d − allow C= 2␣ – ␤ + 4␤ Ssy ␣= n s √( ) 2␣ – ␤ 4␤ ␤= 2 – 3␣ 4␤ D= Ssy␲d 3 8ns(1 + ␰)Fmax 8(1 + ␰)Fmax ␲d2 D = Cd C = D ⁄d KB = (4C + 2) ⁄ (4C − 3) ␶s = K B8(1 + ␰)FmaxD ⁄ (␲d 3) ns = Ssy ⁄ ␶s OD = D + d ID = D − d Na = Gd 4 ymax/(8D3Fmax) Nt: Table 10 –1 Ls: Table 10 –1 L O: Table 10 –1 (LO)cr = 2.63D/␣ fom = −(rel. cost)␥␲ 2d 2Nt D ⁄ 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 † const is found from Table 10–6. 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. 10–3. 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. bud29281_ch10_517-568.qxd 530 12/16/2009 7:14 pm Page 530 pinnacle 203:MHDQ196:bud29281:0073529281:bud29281_pagefiles: 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. (10–3) with τ = Ssy /n s , C = D/d, K B from Eq. (10–6), and Eq. (10–17), Ssy 8Fs D 4C + 2 8(1 + ξ ) Fmax C = KB = (a) ns πd 3 4C − 3 πd 2 Let α= Ssy ns (b) β= 8 (1 + ξ ) Fmax πd 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 2α − β 2α − β 2 3α C= + − (10–23) 4β 4β 4β EXAMPLE 10–2 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 • Music wire, A228; from Table 10–4, A = 201 000 psi-inm; m = 0.145; from Table 10–5, E = 28.5 Mpsi, G = 11.75 Mpsi (expecting d > 0.064 in)
• Ends squared and ground
• Function: Fmax = 20 lbf…
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