The document discusses design considerations for machine elements subjected to fluctuating loads. It covers topics such as stress concentration, fatigue failure, endurance limit, factors affecting fatigue strength, and methods to reduce stress concentration and improve fatigue life. Stress concentration occurs due to discontinuities and can be reduced by avoiding abrupt changes in cross-section and providing fillets. Fatigue failure is caused by fluctuating stresses and depends on factors like the number of cycles and mean stress. The endurance limit is the maximum stress amplitude a material can withstand without failure under completely reversed loading. Surface finish, size, and mean stress affect the endurance limit.

1. DESIGN OF MACHINE ELEMENTS
UNIT 4
Design for Fluctuating Load
Dr. Somnath G Kolgiri (ME, PhD, Mechanical Engg.)
PGMCOE, Pune

2. Content
•Stress concentration - causes & remedies, fluctuating
stresses, fatigue failures, S-N curve, endurance limit, notch
sensitivity, endurance strength modifying factors, design for
finite and infinite life,
•Cumulative damage in fatigue failure, Soderberg, Gerber,
Goodman, Modified Goodman diagrams,
•Fatigue design of components under combined stresses:-
Theoretical treatment only.

3. STRESS CONCENTRATION
•In design of machine elements, the following three
fundamental equations are used,
•The above equations are called elementary equations.
These equations are based on a number of assumptions.
•One of the assumptions is that there are no discontinuities
in the cross-section of the component.
•However, in practice, discontinuities and abrupt changes in
cross-section are unavoidable due to certain features of the
component such as oil holes and grooves, keyways and
splines, screw threads and shoulders.
• Therefore, it cannot be assumed that the cross-section of
the machine component is uniform. Under these
circumstances, the ‘elementary’ equations do not give
correct results.

4. STRESS CONCENTRATION
 Whenever a machine component changes the shape of its cross-section, the
simple stress distribution no longer holds good. This irregularity in the stress
distribution caused by abrupt changes of form is called stress concentration.
 A stress concentration (stress raisers or stress risers) is a location in an object
where stress is concentrated. An object is strongest when force is evenly
distributed over its area, so a reduction in area, e.g., caused by a crack, results
in a localized increase in stress.
 A material can fail, via a propagating crack, when a concentrated stress exceeds
the material's theoretical cohesive strength. The real fracture strength of a
material is always lower than the theoretical value because most materials
contain small cracks or contaminants that concentrate stress.
 It occurs for all kinds of stresses in the presence of fillets, notches, holes,
keyways, splines, surface roughness or scratches etc.

5. •A plate with a small circular hole, subjected to tensile stress is
shown in Fig. The distribution of stresses near the hole can be
observed by using the Photo-elasticity technique.
•In this method, an identical model of the plate is made of epoxy
resin. The model is placed in a circular polariscope and loaded at the
edges.
•It is observed that there is a sudden rise in the magnitude of stresses
in thevicinity of the hole.
Definition: Stress concentration is
defined as the localization of high
stresses due to the irregularities present
in the component and abrupt changes of
the cross-section. Stress concentration
factor is used. It is denoted by Kt a

6. THEORETICAL OR FORM STRESS
CONCENTRATION FACTOR
 The theoretical or form stress concentration factor is defined as the
ratio of the maximum stress in a member (at a notch or a fillet) to the
nominal stress at the same section based upon net area.
 Mathematically, theoretical or form stress concentration factor,
 The value of Kt depends upon the material and geometry of the part.

7. Mechanical & Aerospace Engr., SJSU
CONCEPT OF STRESS CONCENTRATION
Theoretical stress
concentration factor, Kt
Maximum stress at the discontinuity
Nominal stress, max stress
with no discontinuity
Kt is used for normal
stresses and Kts for
shear stresses.

8. THE CAUSES OF STRESS CONCENTRATION
ARE AS FOLLOWS:
1. Variation in Properties of Materials In design of machine components, it is assumed
that the material is homogeneous throughout the component. In practice, there is
variation in material properties from one end to another due to the following
factors:
(a) internal cracks and flaws like blow holes;
(b) cavities in welds;
(c) air holes in steel components; and
(d) nonmetallic or foreign inclusions.
These variations act as discontinuities in the component and cause stress concentration.
2. Load Application Machine components are subjected to forces. These forces act
either at a point or over a small area on the component. Since the area is small,
the pressure at these points is excessive. This results in stress concentration. The
examples of these load applications are as follows:
(a) Contact between the meshing teeth of the driving and the driven gear
(b) Contact between the cam and the follower
(c) Contact between the balls and the races of ball bearing
(d) Contact between the rail and the wheel
(e) Contact between the crane hook and the chain

9. 3. Abrupt Changes in Section In order to mount gears, sprockets, pulleys and ball
bearings on a transmission shaft, steps are cut on the shaft and shoulders are
provided from assembly considerations. Although these features are essential,
they create change of the cross-section of the shaft. This results in stress
concentration at these cross-sections.
4. Discontinuities in the Component Certain features of machine components such
as oil holes or oil grooves, keyways and splines, and screw threads result in
discontinuities in the cross-section of the component. There is stress
concentration in the vicinity of these discontinuities.
5. Machining Scratches Machining scratches, stamp marks or inspection marks are
surface irregularities, which cause stress concentration

10. METHODS TO REDUCE STRESS
CONCENTRATION
• The presence of stress concentration can not be totally eliminated but it
may be reduced to some extent.
• A device or concept that is useful in assisting a design engineer to visualize
the presence of stress concentration and how it may be mitigated is that of
stress flow lines.
• The mitigation of stress concentration means that the stress flow lines shall
maintain their spacing as far as possible.
• Some of the changes adopted in the design in order to reduce the stress
concentration are as follows:
1. Avoid abrupt changes in cross section
2. Place additional smaller discontinuities adjacent to discontinuity
3. Improve surface finish

11.  In Fig. (a), we see that stress lines tend to bunch up and cut very close to
the sharp re-entrant corner. In order to improve the situation, fillets may
be provided, as shown in Fig. (b) and (c) to give more equally spaced flow
lines.
 It may be noted that it is not practicable to use large radius fillets as in case
of ball and roller bearing mountings. In such cases, notches may be cut as
shown in Fig. (d).

12. • Following figures show the several ways of reducing the stress concentration in
shafts and other cylindrical members with shoulders, holes and threads :
• The stress concentration effects of a press fit may be reduced by making more
gradual transition from the rigid to the more flexible shaft.

13. The stress concentration factors are determined by two methods, viz., the
mathematical method based on the theory of elasticity and experimental
methods like photo-elasticity. For simple geometric shapes, the stress
concentration factors are determined by photo-elasticity. The charts for
stress concentration factors for different geometric shapes and conditions
of loading were originally developed by RE Peterson. At present, FEA
packages are used to find out the stress concentration factor for any
geometric shape.
The chart for the stress concentration factor for a rectangular plate with a
transverse hole loaded in tension or compression is shown in Fig. 5.2.
The nominal stress so in this case is given by, where t is
the plate
thickness.
The values of stress concentration factor for a flat plate with a shoulder
fillet subjected to tensile or compressive force are determined from Fig.
5.3. The nominal stress so for this case is given by,

14. Flat plate with a hole
Flat Plate with Shoulder Fillet in Tension or
Compression

15. The charts for stress concentration factor for a round shaft with shoulder
fillet subjected to tensile force, bending moment, and torsional moment
are shown in Fig. 5.4, 5.5 and 5.6 respectively. The nominal stresses in
these three cases are as follows:
(i) Tensile Force
(ii) Bending Moment
(iii) Torsional Moment

18. Q1.A flat plate subjected to a tensile force of 5 KN is shown in Fig. The
plate material is grey cast iron FG 200 and the factor of safety is 2.5.
Determine the thickness of the plate.
Solution
Given P = 5 kN Sut = 200 N/mm2 (fs) = 2.5

19. Q1.A rectangular plate, 15 mm thick, made of a brittle material is shown
in Fig. Calculate the stresses at each of three holes of 3, 5 and 10 mm
diameter. [161.82, 167.33 and 200 N/mm2]

20. Q2. A plate, 10 mm thick, subjected to a tensile load of 20 kN is shown
in Fig. The plate is made of cast iron (Sut = 350 N/mm2) and the factor
of safety is 2.5. Determine the fillet radius. [2.85 or 3 mm]

21. Q2. A non-rotating shaft supporting a load of 2.5 kN is shown in Fig.
The shaft is made of brittle material, with an ultimate tensile strength of
300 N/mm2. The factor of safety is 3. Determine the dimensions of the
shaft.
Solution
Given P = 2.5 kN Sut = 300 N/mm2 (fs) = 3

23. Q1. A round shaft made of a brittle material and subjected to a bending
moment of 15 N-m is shown in Fig. The stress concentration factor at the
fillet is 1.5 and the ultimate tensile strength of the shaft material is 200
N/mm2. Determine the diameter d, the magnitude of stress at the fillet
and the factor of safety. [11.76 mm, 140.91 N/mm2, and 1.42]
Q2. A shaft carrying a load of 5 kN midway between two bearings is
shown in Fig. Determine the maximum bending stress at the fillet
section. Assume the shaft material to be brittle. [20.39 N/mm2]

24. FLUCTUATING STRESSES
.
•In the previous chapters, the external forces acting on a machine
component were assumed to be static.
•In many applications, the components are subjected to forces, which are
not static, but vary in magnitude with respect to time.
•The stresses induced due to such forces are called fluctuating stresses.
•It is observed that about 80% of failures of mechanical components are
due to ‘fatigue failure’ resulting from fluctuating stresses.
•There are three types of mathematical models for cyclic stresses—
fluctuating or alternating stresses, repeated stresses and reversed stresses
•Stress–time relationships for these models are illustrated in Fig

25. •The fluctuating or alternating stress varies in a sinusoidal manner with
respect to time. It has some mean value as well as amplitude value. It
fluctuates between two limits—maximum and minimum stress. The
stress can be tensile or compressive or partly tensile and partly
compressive.
•The repeated stress varies in a sinusoidal manner with respect to time,
but the variation is from zero to some maximum value. The minimum
stress is zero in this case and therefore, amplitude stress and mean
stress are equal
•The reversed stress varies in a sinusoidal manner with respect to time,
but it has zero mean stress. In this case, half portion of the cycle consists
of tensile stress and the remaining half of compressive stress. There is a
complete reversal from tension to compression between these two halves
and therefore, the mean stress is zero.
are maximum and minimum stresses, while are
called mean stress and stress amplitude respectively. It can be proved that

26. FATIGUE FAILURE
It has been observed that materials fail under fluctuating stresses at a
stress magnitude which is lower than the ultimate tensile strength of the
material. Sometimes, the magnitude is even lower than the yield strength.
Further, it has been found that the magnitude of the stress causing fatigue
failure decreases as the number of stress cycles increase. This
phenomenon of decreased resistance of the materials to fluctuating
stresses is the main characteristic of fatigue failure.
•Fatigue failure is defined as time delayed fracture under cyclic loading.
Examples of parts in which fatigue failures are common are transmission
shafts, connecting rods, gears, vehicle suspension springs and ball
bearings.
•The fatigue failure, however, depends upon a number of factors, such as
the number of cycles, mean stress, stress amplitude, stress concentration,
residual stresses, corrosion and creep.

27. ENDURANCE LIMIT
•The fatigue or endurance limit of a material is defined as the
maximum amplitude of completely reversed stress that the standard
specimen can sustain for an unlimited number of cycles without fatigue
failure. Since the fatigue test cannot be conducted for unlimited or
infinite number of cycles, cycles is considered as a sufficient number
of cycles to define the endurance limit.
•There is another term called fatigue life, which is frequently used with
endurance limit. The fatigue life is defined as the number of stress cycles
that the standard specimen can complete during the test before the
appearance of the first fatigue crack.

28. ENDURANCE LIMIT AND FATIGUE FAILURE
It has been found experimentally that when a material is
subjected to repeated stresses, it fails at stresses below the
yield point stresses. Such type of failure of a material is
known as fatigue.
The failure is caused by means of a progressive crack
formation which are usually fine and of microscopic size. The
failure may occur even without any prior indication.
The fatigue of material is effected by the size of the
component, relative magnitude of static and fluctuating loads
and the number of load reversals.

29. FACTORS TO BE CONSIDERED WHILE
DESIGNING MACHINE PARTS TO AVOID
FATIGUE FAILURE
• The following factors should be considered while designing
machine parts to avoid fatigue failure:
• The variation in the size of the component should be as gradual
as possible.
• The holes, notches and other stress raisers should be avoided.
• The proper stress de-concentrators such as fillets and notches
should be provided wherever necessary.
• The parts should be protected from corrosive atmosphere.
• A smooth finish of outer surface of the component increases the
fatigue life.
• The material with high fatigue strength should be selected.
• The residual compressive stresses over the parts surface
increases its fatigue strength.

30.  A standard mirror polished specimen, as shown in figure is rotated in a fatigue
testing machine while the specimen is loaded in bending.
 As the specimen rotates, the bending stress at the upper fibers varies from
maximum compressive to maximum tensile while the bending stress at the
lower fibers varies from maximum tensile to maximum compressive.
 In other words, the specimen is subjected to a completely reversed stress cycle.
This is represented by a time-stress diagram as shown in Fig. (a).

31.  Endurance or Fatigue limit (σe) is defined as maximum value of the
completely reversed bending stress which a polished standard specimen can
withstand without failure, for infinite number of cycles.
 It may be noted that the term endurance limit is used for reversed bending
only while for other types of loading, the term endurance strength may be
used when referring the fatigue strength of the material.
 It may be defined as the safe maximum stress which can be applied to the
machine part working under actual conditions.
 We have seen that when a machine member is subjected to a completely
reversed stress, the maximum stress in tension is equal to the maximum
stress in compression as shown in Fig.(a). In actual practice, many machine
members undergo different range of stress than the completely reversed
stress.
 The stress verses time diagram for fluctuating stress having values σmin and
σmax is shown in Fig. (c). The variable stress, in general, may be considered
as a combination of steady (or mean or average) stress and a completely
reversed stress component σv.

32.  The following relations are derived from Fig. (c):
a =
max min
2
Alternating stress
Mean stress
m =
max min
2
+

33. FACTORS AFFECTING ENDURANCE LIMIT
1) SIZE EFFECT:
• The strength of large members is lower than that of small specimens.
• This may be due to two reasons.
• The larger member will have a larger distribution of weak points than the
smaller one and on an average, fails at a lower stress.
• Larger members have larger surface Ares. This is important because the
imperfections that cause fatigue failure are usually at the surface.
 Effect of size:
• Increasing the size (especially section thickness) results in larger surface
area and creation of stresses.
• This factor leads to increase in the probability of crack initiation.
• This factor must be kept in mind while designing large sized components.

34.  2) SURFACE ROUGHNESS:
• Almost all fatigue cracks nucleate at the surface of the members.
• The conditions of the surface roughness and surface oxidation or corrosion
are very important.
• Experiments have shown that different surface finishes of the same material
will show different fatigue strength.
• Methods which Improve the surface finish and those which introduce
compressive stresses on the surface will improve the fatigue strength.
• Smoothly polished specimens have higher fatigue strength.
• Surface treatments. Fatigue cracks initiate at free surface, treatments can be
significant
• Plating, thermal or mechanical means to induce residual stress.
 3) EFFECT OF TEMPERATURE:
• When the mechanical component operates above the room temperature, its
ultimate tensile strength, and hence endurance limit decrease with increase
in temperature.

35.  4) Effect of metallurgical variables;
• Fatigue strength generally increases with increase in UTS
• Fatigue strength of quenched & tempered steels (tempered martensitic
structure) have better fatigue strength
• Finer grain size show better fatigue strength than coarser grain size.
• Non-metallic inclusions either at surface or sub-surface reduces' the
fatigue strength.

36. S-N DIAGRAM
 Fatigue strength of material is determined by R.R. Moore rotating beam
machine. The surface is polished in the axial direction. A constant bending
load is applied.

37. The S–N curve is the graphical representation of stress amplitude (Sf )
versus the number of stress cycles (N) before the fatigue failure on a log-
log graph paper. The S–N curve for steels is illustrated in Fig. The S–N
diagram is also called Wöhler diagram, after August Wöhler, a German
engineer who published his fatigue research in 1870. The S–N diagram is
a standard method of presenting fatigue data.

38.  A record is kept of the number of cycles required to produce failure at a given
stress, and the results are plotted in stress-cycle curve as shown in figure.
 A little consideration will show that if the stress is kept below a certain value the
material will not fail whatever may be the number of cycles.
 This stress, as represented by dotted line, is known as endurance or fatigue
limit (σe).
 It is defined as maximum value of the completely reversed bending stress which
a polished standard specimen can withstand without failure, for infinite number
of cycles (usually 107 cycles).

39. FATIGUE STRESS CONCENTRATION
FACTOR
• When a machine member is subjected to cyclic or fatigue
loading, the value of fatigue stress concentration factor
shall be applied instead of theoretical stress
concentration factor.
• Mathematically, fatigue stress concentration factor,

40. NOTCH SENSITIVITY
Notch sensitivity is defined as the susceptibility of a material to
succumb to the damaging effects of stress raising notches in
fatigue loading.
Notch Sensitivity: It may be defined as the degree to which the
theoretical effect of stress concentration is actually reached.
Notch Sensitivity Factor “q”: Notch sensitivity factor is defined
as the ratio of increase in the actual stress to the increase in the
nominal stress near the discontinuity in the specimen.
Where, Kf and Kt are the fatigue stress concentration factor and
theoretical stress concentration factor.
The stress gradient depends mainly on the radius of the notch,
hole or fillet and on the grain size of the material.

43. RELATIONSHIP BETWEEN ENDURANCE
LIMIT AND ULTIMATE STRENGTH

44. There is an approximate relationship between the endurance limit and the
ultimate tensile strength (Sut) of the material. These relationships are
based on 50% reliability.

47. REVERSED STRESSES—DESIGN FOR
FINITE AND INFINITE LIFE
•There are two types of problems in fatigue design—(i) components
subjected to completely reversed stresses, and (ii) components subjected
to fluctuating stresses, the mean stress is zero in case of completely
reversed stresses.
•The design problems for completely reversed stresses are further
divided into two groups—(i) design for infinite life, and (ii) design for
finite life.
Case I: When the component is to be designed for infinite life, the
endurance limit becomes the criterion of failure. The amplitude stress
induced in such components should be lower than the endurance limit in
order to withstand the infinite number of cycles. Such components are
designed with the help of the following equations:

48. Case II: When the component is to be designed for finite life, the S–N
curve as shown in Fig. 5.27 can be used. The curve is valid for steels. It
consists of astraight line AB drawn from cycles to
cycles on a log-log paper. The design procedure for such
problems is as follows:
The fatigue strength corresponding to N cycles. The value of the fatigue
strength (Sf) obtained by the above procedure is used for the design
calculations.

50. INFINITE-LIFE PROBLEMS (REVERSED LOAD)
Example 1. A plate made of steel 20C8 (Sut = 440 N/mm2) in hot rolled
and normalised condition is shown in Fig. It is subjected to a completely
reversed axial load of 30 kN. The notch sensitivity factor q can be taken
as 0.8 and the expected reliability is 90%. The size factor is 0.85. The
factor of safety is 2. Determine the plate thickness for infinite life.

52. Q2. A rod of a linkage mechanism made of steel 40Cr1 (Sut = 550
N/mm2) is subjected to a completely reversed axial load of 100 kN. The
rod is machined on a lathe and the expected reliability is 95%. There is
no stress concentration. Determine the diameter of the rod using a factor
of safety of 2 for an infinite life condition.

53. Q3.A component machined from a plate made of steel 45C8 (Sut = 630
N/mm2) is shown in Fig. It is subjected to a completely reversed axial
force of 50 kN. The expected reliability is 90% and the factor of safety is
2. The size factor is 0.85. Determine the plate thickness t for infinite life,
if the notch sensitivity factor is 0.8.

54. Q4. A 25 mm diameter shaft is made of forged steel 30C8 (Sut = 600
N/mm2). There is a step in the shaft and the theoretical stress
concentration factor at the step is 2.1. The notch sensitivity factor is 0.84.
Determine the endurance limit of the shaft if it is subjected to a reversed
bending moment. [59.67 N/mm2]
Q5. A 40 mm diameter shaft is made of steel 50C4 (Sut = 660 N/mm2)
and has a machined surface. The expected reliability is 99%. The
theoretical stress concentration factor for the shape of the shaft is 1.6 and
the notch sensitivity factor is 0.9. Determine the endurance limit of the
shaft. [112.62 N/mm2]

55. FINITE-LIFE PROBLEMS (REVERSED LOAD)
Q1. A rotating bar made of steel 45C8 (Sut = 630 N/mm2) is subjected to
a completely reversed bending stress. The corrected endurance limit of
the bar is 315 N/mm2. Calculate the fatigue strength of the bar for a life
of 90,000 cycles.

56. Q2. A forged steel bar, 50 mm in diameter, is subjected to a reversed
bending stress of 250 N/mm2. The bar is made of steel 40C8 (Sut = 600
N/mm2). Calculate the life of the bar for a reliability of 90%.
Solution:-
Given:- Sf = Sb = 250 N/mm2 , Sut = 600 N/mm2, R = 90%

57. Q3. A rotating shaft, subjected to a non rotating force of 5 kN and simply
supported between two bearings A and E is shown in Fig. 5.32(a). The
shaft is machined from plain carbon steel 30C8 (Sut = 500 N/mm2) and
the expected reliability is 90%. The equivalent notch radius at the fillet
section can be taken as 3 mm. What is the life of the shaft?
Solution :- Given P = 5 kN Sut = 500
N/mm2, R = 90%, r = 3 mm
Step I Selection of failure-section
Taking the moment of the forces about
bearings A and E, the reactions at A
and E are 2143 and 2857 N
respectively. The bending moment
diagram is shown in Fig. 5.32(b). The
values of the bending moment shown
in the figure are in N-m. The
possibility of a failure will be at the
three sections B, C and D. The failure
will probably occur at the section B rather than at C or D. At the section

58. there is no stress concentration. At the section D, the diameter is more
and the bending moment is less compared with that of section B.
Therefore, it is concluded that failure will occur at the section B.

59. Q4. The section of a steel shaft is shown in Fig. 5.34. The shaft is
machined by a turning process. The section at XX is subjected to a
constant bending moment of 500 kN-m. The shaft material has ultimate
tensile strength of 500 MN/m2, yield point of 350 MN/m2 and
endurance limit in bending for a 7.5 mm diameter specimen of 210
MN/m2. The notch sensitivity factor can be taken as 0.8. The theoretical
stress concentration factor may be interpolated from following tabulated
values: where rf is the fillet radius and d is the
shaft diameter. The reliability is 90%.
Determine the life of the shaft.

61. Q5. A cantilever beam made of cold drawn steel 20C8 (Sut = 540 /mm2)
is subjected to a completely reversed load of 1000 N as shown in Fig.
The notch sensitivity factor q at the fillet can be taken as 0.85 and the
expected reliability is 90%. Determine the diameter d of the beam for a
life of 10000 cycles.
Step I Selection of failure section The failure will occur either at the
section A or at the section B. At section A, although the bending moment
is maximum, there is no stress concentration and the diameter is also
more compared with that of the section B. It is, therefore, assumed that
the failure will occur at the section B.

62. Step III Diameter of beam From above Fig.

63. CUMULATIVE DAMAGE IN FATIGUE
In certain applications, the mechanical component is subjected to
different stress levels for different parts of the work cycle. The life of
such a component is determined by Miner’s equation. Suppose that a
component is subjected to completely reversed stresses
cycles, and so on. Let N1 be the number of stress cycles before fatigue
failure, if only the alternating stress is acting. One stress cycle will
consume of the fatigue life and since there are n1 such cycles at this
stress level, the proportionate damage of fatigue life will be
Similarly, the proportionate damage at stress level will be
Adding these quantities, we get

64. Q1. The work cycle of a mechanical component subjected to completely
reversed bending stresses consists of the following three elements:
(i) ± 350 N/mm2 for 85% of time (ii) ± 400 N/mm2 for 12% of time
(iii) ± 500 N/mm2 for 3% of time The material for the component is
50C4 (Sut = 660 N/mm2) and the corrected endurance limit of the
component is 280 N/mm2. Determine the life of the component.
Solution :- Given Sut = 660 N/mm2 Se = 280 N/mm2
Step II Calculation of N1, N2 and N3 From above Fig.

65. Q2.A solid circular shaft made of steel Fe 620 (Sut = 620 N/mm2 and Syt
= 380 N/mm2) is subjected to an alternating torsional moment, which
varies from –200 N-m to + 400 N-m. The shaft is ground and the
expected reliability is 90%. Neglecting stress concentration, calculate the
shaft diameter for infinite life. The factor of safety is 2. Use the
distortion energy theory of failure. [29.31 mm]

66. SODERBERG AND GOODMAN LINES
When a component is subjected to fluctuating stresses as shown in Fig.
(a), there is mean stress (sm) as well as stress amplitude (sa). It has been
observed that the mean stress component has an effect on fatigue failure
when it is present in combination with an alternating component.

67. The fatigue diagram for this general case is shown in Fig. 5.39. In this diagram,
the mean stress is plotted on the abscissa. The stress amplitude is plotted on the
ordinate. The magnitudes of (sm) and (sa) depend upon the magnitudes of
maximum and minimum force acting on the component. When stress
amplitude (sa) is zero, the load is purely static and the criterion of failure is Sut
or Syt. These limits are plotted on the abscissa. When the mean stress (sm) is
zero, the stress is completely reversing and the criterion of failure is the
endurance limit Se that is plotted on the ordinate. When the component is
subjected to both components of stress, viz., (sm) and (sa), the actual failure
occurs at different scattered points shown in the figure. There exists a border,
which divides safe region from unsafe region for various combinations of (sm)
and (sa). Different criterions are proposed to construct the borderline dividing
safe zone and failure zone. They include Gerber line, Soderberg line and
Goodman line.
Gerber Line A parabolic curve joining Se on the ordinate to Sut on the
abscissa is called the Gerber line.
Soderberg Line A straight line joining Se on the ordinate to Syt on the
abscissa is called the Soderberg line.
Goodman Line A straight line joining Se on the ordinate to Sut on the

68. • straight line connecting the endurance limit (σe) and the yield
strength (σy), as shown by the line AB in following figure, follows
the suggestion of Soderberg line.
• This line is used when the design is based on yield strength. the line
AB connecting σe and σy, as shown in following figure, is called
Soderberg's failure stress line.
SODERBERG METHOD FOR COMBINATION
OF STRESSES
If a suitable factor of safety (F.S.) is
applied to the endurance limit and yield
strength, a safe stress line CD may be
drawn parallel to the line AB.

70. • A straight line connecting the
endurance limit (σe) and the
ultimate strength (σu), as shown by
line AB in figure given below
follows the suggestion of Goodman.
• A Goodman line is used when the
design is based on ultimate strength
and may be used for ductile or
brittle materials.
GOODMAN METHOD FOR COMBINATION
OF STRESSES:

71. • In the design of components subjected to fluctuating stresses, the
Goodman diagram is slightly modified to account for the yielding
failure of the components, especially, at higher values of the mean
stresses.
• The diagram known as modified Goodman diagram and is most
widely used in the design of the components subjected to
fluctuating stresses.
MODIFIED GOODMAN DIAGRAM:
For the purpose of design, the problems are classified into two groups:
(i) components subjected to fluctuating axial or bending stresses; and
(ii) components subjected to fluctuating torional shear stresses.
The modified Goodman diagram for fluctuating axial or bending stresses
is shown in Fig. 5.40. In this diagram, the Goodman line is ‘modified’ by
combining fatigue failure with failure by yielding. In this diagram, the
yield strength Syt is plotted on both the axes—abscissa and ordinate—
and a yield line CD is constructed to join these two points to define
failure by yielding. The line CD is inclined at 45° to the abscissa.

72. Similarly, a line AF is
constructed to join Se on the
ordinate with Sut on the
abscissa, which is the
Goodman line discussed in the
previous article. The point of
intersection of these two lines
is B.
The magnitudes of Pa and Pm can be
determined from maximum and minimum
forces acting on the component.
The area OABC represents the region of safety for components subjected
to fluctuating stresses. The region OABC is called modified Goodman
diagram. While solving a problem, a line OE with a slope of tan is
constructed in such a way that

73. •The modified Goodman diagram for fluctuating torsional shear stresses
is shown in Fig. 5.41. In this diagram, the torsional mean stress is plotted
on the abscissa while the torsional stress amplitude on the ordinate. The
torsional yield strength Ssy is plotted on the abscissa and the yield line is
constructed, which is inclined at 45° to
the abscissa.

74. Example .1 A cantilever beam made of cold drawn steel 4OC8 (Sut =
600 N/mm2 and Syt = 380 N/mm2) is shown in Fig. The force P acting
at the free end varies from –50 N to +150 N. The expected reliability is
90% and the factor of safety is 2. The notch sensitivity factor at the fillet
is 0.9. Determine the diameter ‘d’ of the beam at the fillet cross-section.

75. Step III Permissible stress amplitude Refer to above Fig. The
coordinates of the point X are determined by solving the following two
equations simultaneously.

76. Example.2 A transmission shaft of cold drawn steel 27Mn2 (Sut = 500
N/mm2 and Syt = 300 N/mm2) is subjected to a fluctuating torque which
varies from –100 N-m to + 400 N-m. The factor of safety is 2 and the
expected reliability is 90%. Neglecting the effect of stress concentration,
determine the diameter of the shaft. Assume the distortion energy theory
of failure.

78. COMBINED LOADING
All four components of stress exist,
xa alternating component of normal stress
xm mean component of normal stress
xya alternating component of shear stress
xym mean component of shear stress
Calculate the alternating and mean principal stresses,
1a, 2a = (xa /2) ± (xa /2)2
+ (xya)2
1m, 2m = (xm /2) ± (xm /2)2
+ (xym)2

79. COMBINED LOADING
Calculate the alternating and mean von Mises stresses,
a′ = (1a + 2a - 1a2a)1/2
2 2
m′ = (1m + 2m - 1m2m)1/2
2 2
Fatigue design equation
nf
Se
1
=
Sut
′a ′m
+ Infinite life

80. THANK YOU