Lecture 4

Design For Static Strength:

➢ Failure Theories
➢ Factor of Safety
➢ Failure of Ductile Materials
➢ Stress Concentration Effects
➢ Stress Concentration Factors
 Design For Static Strength
By definition a Static Load is a stationary force
or moment acting on a member.
To be stationary the force or moment must have:
➢ Unchanging magnitude
➢ Unchanging point of application
➢ Unchanging direction
➢ Unchanging sense
The static load can be an axial tension or
compression, a shear load, a bending or
torsional moment, or any combination of these.
But the load should not change in any manner if
it is to be considered static.
 Design For Static Strength
Sometimes a load is assumed to be static even
when it is known that some variation is to be
expected.
This assumption is usually made to simplify the
design calculations and get the approximate
dimensions of the component when the variation
is considered minor.
The purpose of this lecture is to develop the
relations between strength and loads and to
achieve optimum component dimensions, with
the requirement that the component will not fail in
service.
 Design For Static Strength
It should be noted that nearly all of the numerical
data that we use in design have some
uncertainties.
These include:
➢ uncertainties in strength, and
➢ uncertainties in the magnitudes of the loads.
To account for all the uncertainties in design we
employ the concept of factor of safety (FS).
This assures that the loads acting on a
mechanical part, or the stresses resulting from
these loads are related to the strength of the part
to achieve a safe and trouble-free design.
 Design For Static Strength
The failures shown in Figs. 6-2 to 6-9 exemplify the need
for the designer to be well familiar with the knowledge
needed to achieve satisfactory performance of the parts
he/she designs and be very good in failure prevention.

Figure 6-2
Typical failure of a
stamped alternator
Failure due to bracket, which is due
residual stress to residual stresses
caused by cold-
forming operation
 Design For Static Strength
Fig. 6-3 shows a chain test fixture that failed in one cycle
due to brittle fracture initiated by a stress concentration.
a) Shows the two fractured pieces.
b) Enlarged view of one portion to show cracks that are
induced by stress concentration at the support-pin holes.
Fractured Stress
pieces concentration
 Design For Static Strength
Figure 6-4 shows;
a) Failure of a track drive-shaft due to corrosion fatigue.
b) Direct end view of the failure section.
Note that it was necessary to use a clear tape to hold the
pieces of the shaft in place when taking these pictures.

Corrosion
fatigue
 Design For Static Strength
Figure 6-5 shows an automotive rocker-arm articulation-
joint failed by fatigue.

Fatigue
failure
 Design For Static Strength
Figure 6-6 shows a valve-spring failure caused by spring
surge in an over-revved engine. The fractures exhibit the
classic 45o shear failure.

45o
shear
failure
 Design For Static Strength
Figure 6-7 demonstrates a brittle fracture of a lock-
washer produced in one cycle of load application. The
washer failed when it was installed (tightened by the nut).
 Design For Static Strength

Figure 6-8 Shows a failure of an overhead pulley

retaining bolt on a weight-lifting machine. A
manufacturing error produced a gap that forced the bolt to
take the entire moment load, and hence failure.
 Design For Static Strength
Figure 6-9 Showing a Bevel Gear failure of a 5.6 kW
American made outboard motor. The reason for the failure
was that the owner has replaced the safety shear pin with a
substitute pin, having greater strength than the original.

Bevel gear Bevel pinion

Out of 21 Out of 14 teeth
teeth 8 are all are broken.
broken.

Pinion
splines
 6.1 Static Strength
When designing any machine element the designer
should have at his/her disposal the results of many
strength tests of the material chosen.
These tests should have been conducted on specimens
having the same:
➢ Heat treatment
➢ Surface finish and
➢ Size as the proposed element.
This means that:
➢ If the part is to experience a bending load, it should be
tested with bending load,
➢ If it is subjected to combined bending and torsion, it
should be tested under combined bending and torsion.
 Static Strength
We can appreciate the following design situations:
➢ Failure of the part will endanger human life, and
the part is made in extremely large quantities;
Hence elaborate testing program is justified,
➢The part is made in large quantities so that a
moderate tests is feasible,
➢ The part is made in small quantities that test is
not justified at all, or the design must be completed
so rapidly that there is not enough time for testing,
➢ The part has already been designed, tested and
found to be unsatisfactory. Its is required to find
out why the part is unsatisfactory & how to improve it.
 Static Strength
In practice the designer will usually have only
published data of yield strength, ultimate
strength, and % elongation, such as those listed
in design tables, & little more from design books.

It should be noted that the simple tension test

provide an enormous amount of information
regarding the probable behavior of a mechanical
element when placed in services.

Therefore, the fundamental problem of the

designer is to use the simple tension test data
and relate them to the strength of the part,
regardless of the stress state or the loading
conditions.
 6.2 Static Loads and Factor of Safety
We learned that the FS is defined by either of the
equations Fu S
n= , or n= (6 - 1)
F 
Where: Fu represents the maximum load that will
still enable the part to perform its proper function.
Therefore Fu is the limiting value of F.
In the second term of the equation, S is the
strength and σ is the actual normal stress. Also in
this case S is the limiting value of σ.
If in Eq. (6-1) S is a shear strength - Ss, then σ
must be a shear stress - τ, that is the two must be
consistent.
 Factor of Safety
Sometimes it is convenient to define two FS:
 One of them ns is used to account for the
uncertainties in the strength, and
 The other nl accounts for the uncertainties
with regard to the load.
 Thus the total factor of safety is defined as
n = ns nl (6 - 2)
When ns is applied to the strength then we have
S
p = (6 - 3)
ns
where σp is called the permissible stress.
 Factor of Safety
When nl is applied to the limiting load Fu, it gives
Fu
Fp = (6 - 4)
nl
where Fp is called the permissible load.
Special case from Eq. (6-1) is the case of pure
tension when the second terms can be used in
the design, bypassing Eqs. (6-3) and (6-4).
In other cases the critical stress may result from
several loads: say F1, F2, and F3, and therefore the
uncertainties in these loads are accounted for by
the factors n1, n2, and n3, respectively.
 Factor of Safety
If n2, say, appears to be the largest of the three
factors, then it would be very inefficient design to
use only n2 in designing the element.

Under these conditions the use of Eq. (1-8) is

recommended.
 p = c. f .F .( x1 , x2 ,....xi ).(n1 F1 , n2 F2 ,....n j F j ) (1 - 8)
Where:
c = a constant, xi = dimensions of the part
f = a function of the geometry, Fj = external loads applied
F = a function of the loads, nj = FS used to account for
the individual variation
See Prob. 6-3, Tutorial 5. in loads
 6.2.1 Reliability
A mathematical concept closely related to the
factor of safety (FS) is the reliability.
The usefulness of the reliability approach
depends on having adequate information on the
statistical distribution of:
➢ The loading applied to the part in service, from
which the stress can be calculated,
➢ The strength of production runs of parts being
manufactured.
Figure 6-16 shows hypothetical distribution curves
for stress and for corresponding strength.
 Reliability
Fig. 6-16 shows that the mean value of strength is 70 and
that of stress is 40 MPa.
This means that if an “average” part is put in service under
“average” loading conditions, there will be a margin of
safety of 30 MPa.
However, the shaded area between curves shows that a
part with S=50 will fail if it is installed under stress 60 MPa.
 Reliability - Margin of Safety
Fig. 6-17 shows a corresponding plot of the distribution of
margin of safety. In most instances, interest would be
focused on the size of the shaded area at the left,
indicating failure.
Effects of Tensile Yield Strength Distribution on FS
Figure 5-4 a,b shows the
tensile strength & yield
strength distribution.

Based on yield strength

distribution such as the
one in Fig. 5-4b, the FS
→ ns should be about :

➢ 1.2 for reliability of 90%

➢ 1.4 for reliability of 99%

If the part should operate

under adverse conditions,
say, high temperature or
corrosive environment,
the above SF’s may have
to be adjusted to much
higher values.
 Selection of numerical values for the FS
The selection of numerical values of the FS is
based primarily on the following conditions:
➢ Degree of uncertainties about the loading
➢ Degree of uncertainties about the strength
➢ Uncertainties in relating the loads to material
strength
➢ Consequences of failure – human safety and
economics
➢ Cost of providing a large FS
A key point in selecting the FS is balance. All
parts of the machine should have consistent FS.
 6.3 Failure Theories
In designing parts we always assure that the
internal stress does not exceed the strength of
the material.
It should be remembered that the strengths of
ductile materials are about the same in tension
and compression.
Now consider the two-dimensional stress element
shown in Fig. 6-10.
The problem now is how to relate a stress state
such as that in Fig. 6-10 to a single strength, such
as the yield strength, in order to achieve safety.
 Failure Theories
Fig. 6-11a shows Mohr’s circle for simple tension,
while Fig. 6-11b, Mohr’s circle for pure torsion, as
well as the principal stress element induced.
 Maximum-Normal-Stress Theory MNST
This theory has only historical value. Its
predictions for ductile materials do not agree with
experiments, giving results on the unsafe side.
The MNST states that: failure occurs whenever
the largest principal stress equals the strength.
Suppose that we arrange the principal stresses for
any stress state in the form
1   2   3
If yielding is the criterion then failure would occur
whenever
 1 = S yt , or  3 = − S yc (6 - 5)
Maximum Normal Stress Theory (MNST)
where Syt and Syc are the tensile and compressive
yield strengths, respectively.
According to MNST factor of safety is defined as
S yt S yc
n= =−
1 3
For pure torsion (Fig. 6-11b) σ1= - σ3= τ and
σ2=0. MNST theory predicts that a part would fail
in torsion when τ = Syt
But experiments show that parts loaded in torsion
failed when τ = 0.6 Syt,
that is when max shear stress is about 60% of
the yield strength.
6-5 Maximum Shear Stress Theory (MSST)
This theory is used to predict yielding and hence it applies
only to ductile materials.
The MSST states that:
‘yielding begins whenever the max shear stress in a
mechanical element becomes equal to the max shear
stress in a tension-test specimen of same material
when that specimen begins to yield’
Max shear stress for simple tension (Fig. 6-11a) is found as
1 S
 max = = (a)
2 2

Max shear stress for pure torsion (Fig. 6-11b) is determined

1 −  3
 max = (b)
2
 Maximum Shear Stress Theory
The MSST predicts that failure occurs whenever
Sy 1 −  3 Sy
 max = , hence = therefore   1 −  3 = S y (6 - 7)
2 2 2

The MSST also states that the yield strength in

shear is given by the equation
Sy
S sy = 0.50S y = (6 - 8)
2
By using MSST the FS is defined as
S sy Sy Sy Sy
n= = = =
 max 2 max  1 −  3  1 −  3
2 
 2 
 6-6 Distortion Energy Theory (DET)
This theory is also called shear-energy theory,
and the Von Mises-Hencky theory, or DET.
This is the best theory to use for ductile materials.
Like MSST the DET is defined to predict yielding.
 Distortion Energy Theory
The criterion for yielding based on DET for 3D
stress state is
2S y2 = ( 1 −  2 ) + ( 2 −  3 ) + ( 3 −  1 )
2 2 2
(6 - 11)

For 2D stress state, if σA and σB are the nonzero

stresses, then the criterion for yielding is
S y2 =  A2 −  A B +  B2 (6 - 12)

For pure torsion σA=-σB and τ max= σA therefore

S sy = 0.577 S y (6 - 13)
For design purposes we define the Von Mises
stress as   =  A2 −  A B +  B2 (6 - 14)
 Distortion Energy Theory
In case of combined bending and torsion, then it
is possible to bypass the Mohr’s circle, when
determining the Von Mises stress by using the
equation
  =  x2 + 3 xy2 (6 - 16)

where σx and τxy are the stresses acting on the

stress element at the critical point.
The DET defines the factor of safety as follows
Sy
n=

 6-7 Failure of Ductile Materials
For biaxial stress state, we designate the nonzero
principal stresses as σ1=σA and σ1= σB, with σ3=0.
A plot of the MNST, MSST and DET on σA and σB
coordinate axes is shown in Fig. 6-13.
 Failure of Ductile Materials
The MSST is employed when:
➢ The dimensions need not be held too close,
➢ If a quick size estimate is needed, or
➢ If the FS is known to be generous.

The DET should be used when:

➢ The margin of safety is to be held to very close
limits, or
➢ The cause of an actual part’s failure is being
investigated.
 6-9 Stress Concentration
In developing the basic stress equations for tension,
compression, bending and torsion it was assumed that no
irregularities occurred in the parts.

➢ In practice most of the parts require: shoulders, key slots,

splines, oil grooves, threads, notches, etc.

“Any discontinuity in a machine part alters the

stress distribution in the neighborhood of the
discontinuity, so that the basic stress equations no
longer describe the stress state in the part.”

➢ Such discontinuities are called stress raisers, and the

regions in which they occur are called areas of stress
concentration.
 Stress Concentration
A theoretical, or geometric, stress-concentration factor Kt
or Kts is used to relate actual maximum stress at the
discontinuity to the nominal stress.
These factors are defined as followed
 max  max
Kt = , or K ts = (6 - 21)
0 0
where Kt is used for normal stress, and Kts for shear stress.
The nominal stresses σ0, and τ0, are calculated by using the
elementary stress equations and the net area, but
sometimes the gross section is used instead.
The subscript t in Kt means that the value of the stress-
concentration factor depends on the geometry of the part
only. That is why it is called a theoretical, or geometric
stress-concentration factor.
 6-10 Determination of the Stress-
Concentration Factors
The values of the stress-concentration factors are
determined by using the methods of the theory of
elasticity. Fig. 6-18 represents an infinite plate
stressed uniformly in tension by a stress σ0.
Determining Stress-Concentration Factors
A small elliptical hole drilled in the plate will have a
stress at the edge defined by the equation
 2b 
 max =  0 1 +  (6 - 22)
 a 
If a = b, then the ellipse becomes a circle and Eq.
(6-22) reduces to
 max = 3 0 (6 - 23)
Therefore the value of the Kt is
 2b 
K t = 1 +
 a


 Kt=3.

Here the plate is infinite and the nominal tensile

stress σ0 is at a point remote from the discontinuity.
 Other Methods for Determining Stresses
6-10.1 The method of Photoelasticity
The part being investigated is made of transparent
material having double-refraction properties.
➢ Then the model is placed in a loading frame, and
a beam of polarized light is directed through it and
projected onto a screen, or photographic plate.
➢ When the model is loaded, fringes of colored
light originate at the points of maximum stress.
A certain stress is associated with each fringe so
that one can determine the stresses at the edges
merely by counting the fringes as they originate.
 The Method of Photoelasticity
Figures 6-19 and (15-19) are pictures, taken by
photoelastic methods, of the fringes on a gear tooth loaded
by forces Wr and F, respectively.
It is seen that stress concentration exists at the point of
application of forces and at both fillets of the root of the
teeth.
 The Finite-Element Method
This method is a powerful approach for determining
stresses, made possible by the advances in
computers & computer-aided design methods.
The finite elements used may be:
➢ Lines,
➢ Triangles, or
➢ Any convenient geometric shapes.

The member to be analyzed is:

➢ Divided into a large number of finite elements,
which may be of different sizes.
➢ Starting from the known loading, boundary
configuration & constrains, a computer analysis is
made and iterated until all conditions are satisfied.
 Finite-Element Method
Fig. 6-20 illustrates the method applied for the analysis of
gear-tooth stresses. A force F acts at a junction of
elements 1 and 2 at the tip of the tooth. In general
triangular final elements are useful for 2D stress analysis.