Failure Theories, Static Loads
Failure theories are generally divided into two categories, those that relate to ductile
materials, and those that relate to brittle materials. Shigley proposes the following as reasonable
criteria to determine whether a material is ductile or brittle.
Ductile materials experience a strain at failure,  F , in a uniaxial tensile test, greater than
0.05. Ductile materials possess an identifiable yield strength, S Y , that is frequently the same in
tension and compression. When the yield strengths in tension and compression are not the same,
they are denoted by S T and S C . By convention S C is positive.
Brittle materials experience a strain at failure,  F , in a uniaxial tensile test, less than
0.05. Brittle materials do not possess an identifiable yield strength, S Y , and frequently possess
different ultimate strengths in compression, S UC , and in tension, S UT . By convention, S UC is
positive. It is necessary to keep this in mind as we proceed through the theories of failure.
Concrete is a classic example of a brittle material that exhibits different ultimate strengths
in compression and tension. For typical concrete mixtures and curing times, S UC is on the order
of twenty-to-thirty times S UT . It is not a terribly important topic for mechanical engineers, but I
want you to understand the basic concept of pre-stressed concrete. Consider a free-standing
concrete wall. Vertical, reinforcing steel rods, fixed to the foundation at the base of the wall, are
placed in tension, e.g., by a big hydraulic jack. The concrete is then poured and allowed to cure.
At which point, the tension mechanism is removed. The steel rods try to return to their original
length, but cannot do so because they are bonded to the concrete. The result is steel rods in
vertical tension and concrete in vertical compression. This is the pre-stress. Suppose there is a
horizontal wind, perpendicular to the wall. The upwind side at the base of the wall would
experience tension, but because of the compressive pre-stress it does not fail.
MAE3501 - 18 - 1
Ductile Materials
There are three generally accepted failure theories for ductile materials.
Maximum-Shear-Stress Theory
The maximum-shear-stress (MSS) theory is also referred to as the Treska or Guest
theory.
This theory resulted from an examination of the results obtained in a uniaxial tensile test.
As the ductile material begins to yield, slip lines are observed oriented at approximately
45 relative to the axis of the test. When the ductile material finally fractures, the fracture lines
are also observed to be at approximately 45 relative to the axis of the test. The maximum shear
stress in the material occurs on planes oriented at 45 relative to the axis of the test. Therefore, it
seemed reasonable to postulate that a ductile material yields when the maximum shear stress in
the material,  MAX , is equal to the maximum shear stress in a uniaxial test when the longitudinal
stress is equal to the yield strength, S Y , of the material.
The maximum shear stress in a uniaxial test is equal to one half the longitudinal stress.
The maximum-shear-stress theory expressed mathematically.
 MAX 
1   3 S Y

2
2
(18.1)
1   3  S Y
(18.2)
For use in design, equation (18.2) can be modified to include a factor of safety, n.
1   3 
SY
n
(18.3)
Consider plane stress, but constantly keep in mind that plane stress is a special case of
three-dimensional stress, one in which all the non-zero stresses act on planes perpendicular to a
given plane. For plane stress, the third principal stress is the normal stress acting on the given
plane and is equal to zero.
We wish to denote the three principal stresses as  1 ,  2 , and  3 , such that the following
relationship is satisfied.
1   2   3
(18.4)
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This leads to three distinct cases, depending on the values of the two principal stresses
determined from a Mohr’s circle analysis of the plane stress. Denote these two principal
stresses as  A and  B , such that the following relationship is satisfied.
 A  B
(18.5)
 A  B  0
(18.6)
1   A
(18.7)
2  B
(18.8)
3  0
(18.9)
1   3   A  S Y
(18.10)
 A  0  B
(18.11)
1   A
(18.12)
2  0
(18.13)
3  B
(18.14)
1   3   A   B  S Y
(18.15)
0   A  B
(18.16)
1  0
(18.17)
2  A
(18.18)
3  B
(18.19)
1   3    B  S Y
(18.20)
Case I
Case II
Case III
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Shigley tells us that the results of these three cases can be expressed in the following
diagram.
Shigley points out that the three unlabeled line segments represent cases for which
 B   A and are not normally used.
This diagram is used in the following manner to determine a factor of safety. Suppose,
for some state of plane stress, point a is located at the  A   B coordinates for that state of
stress. If the material is linear, then, as the loads causing the plane stress are increased, the
 A   B coordinates will increase proportionally, moving along the load line until failure
occurs at point b. The factor of safety at point a can then be determined using the following
relationship.
n
Ob
Oa
(18.21)
MAE3501 - 18 - 4
Distortion Energy Theory
The distortion-energy (DE) theory yields precisely the same results as the von MisesHencky, the shear-energy, and the octahedral-shear-stress theories. Despite what Shigley tells
us, this theory is not “called” by any of these other names.
This theory resulted from the observation that ductile materials subjected to a hydrostatic
state of stress, do not fail regardless of the magnitude of the hydrostatic pressure. However, note
that this result is also predicted by the maximum-shear-stress theory.
The derivation of the distortion-energy theorem proceeds in the following manner.
For a general three-dimensional state of stress, the three principal stresses are determined.
Note that no shear stresses act on the planes on which the three principal stresses act.
The average of the three principal stresses is determined.
 AV 
1   2   3
3
(18.22)
The principal stress distribution is then expressed as the sum of two stress distributions,
one of which is a hydrostatic stress distribution that causes no shear stresses within the material.
The other stress distribution does cause shear stresses within the material, and distorts the
material. The internal strain energy within the material resulting from the distortional stress
distribution is then determined and compared to the distortional strain energy in a uniaxial test.
The result.
( 1   2 ) 2  ( 2   3 ) 2  ( 1   3 ) 2
 SY
2
(18.23)
The quantity on the left side of equation (18.23) is referred to as the von Mises stress,
and is denoted by   .
MAE3501 - 18 - 5
Consider plane stress.
(  A   B ) 2  (  A  0) 2  (  B  0) 2
 SY
2
(18.24)
 A2  B2  2 A B  A2  B2
 SY
2
(18.25)
 A2  B2   AB  SY
(18.26)
Equation (18.26) is represented graphically in the following figure.
The maximum-shear-stress theory is conservative, by up to about 15 % , depending on
the two-dimensional state of stress, when compared to the distortion-energy or other equivalent
theories of failure.
MAE3501 - 18 - 6
It is not critical, but Figure 5-10 on page 224 is very difficult to understand, and I want
you to understand what an octahedral plane is.
2
3
1
3
1
2
After the three principal stresses, and the three mutually perpendicular directions in
which they act, have been determined, create a local coordinate system within the material as
shown. Select three points, one on each coordinate axis, equidistant from the origin. The
equilateral triangle containing these three points is an octahedral plane, so named because there
are eight such planes.
MAE3501 - 18 - 7
Coulomb-Mohr Theory
There are ductile materials that do not have equal yield strengths in tension and
compression, although this phenomenon is much more common in brittle materials. The
justification for studying this failure theory is that, when suitably modified, it can also be applied
to brittle materials.
The Coulomb-Mohr theory, also referred to as the internal-friction theory, is a variation
of the Mohr theory
The Mohr theory of failure can be explained by examining the following figure.
The material is tested to failure in a uniaxial tensile test, a uniaxial compressive test, and
a pure shear test. Note that the only practical method of conducting a pure shear test is to
consider torsion of a circular section.
Three Mohr’s circles are constructed, and a failure envelope is constructed tangent to the
three circles.
This theory is complicated by the need to conduct three tests to failure, and the judgment
required to construct the failure envelope.
MAE3501 - 18 - 8
The Coulomb-Mohr theory assumes that the failure envelope described above is a straight
line. This assumption eliminates the need for a shear test, and requires only S T and S C .
The two Mohr’s circles representing uniaxial tensile and compressive tests are shown
dashed in the above figure.
Consider a three-dimensional state of stress in which the three principal stresses are
ordered such that  1   2   3 . If the largest Mohr’s circle, shown gray in the above figure, is
tangent to the failure line, Shigley tells us that the following relationship results.
1  3

1
ST SC
(18.27)
Note that the second term in equation (18.27) is positive.
I looked at the above figure, and at first didn’t have the faintest idea how it leads to
equation (18.27). The following rather-complicated analysis is what I finally came up with.
First, the location of point O depends only on the sizes of the two dashed circles, the
Mohr’s circles that represent the results of tensile and compressive tests to yield.
The tensile dashed circle extends from the origin to S T . C1 is the center of that circle.
C1 
ST
2
(18.28)
MAE3501 - 18 - 9
B1C1 is the radius of that circle.
B 1 C1 
ST
2
(18.29)
The compressive dashed circle extends from the origin to  S C . S C  0 . C 3 is the center
of that circle.
S
(18.30)
C3   C
2
B 3 C 3 is the radius of that circle.
B 3C 3 
SC
2
(18.31)
Triangle O  B 3  C 3 is similar to triangle O  B1  C1 .
O  C1

B 1C1
S
ST
O C
2  O  C3 
2
ST
SC
B 3C 3
2
2
O
(18.32)
After a few lines of elementary algebra, I ended up with the following relationship.
O
ST SC
SC  ST
(18.33)
I “checked” my work by examining the above figure. In that figure, S C  2 S T , leading
to O  2 S T . This appears reasonable.
The gray circle extends from  1 on the right to  3 on the left. C 2 is the center of that
circle.
C2 
1   3
2
(18.34)
B1C1 is the radius of that circle.
B 1 C1 
1   3
2
(18.35)
MAE3501 - 18 - 10
Triangle O  B1  C1 is similar to triangle O  B 2  C 2 .
S TS C
S TS C
  3
S
 T
 1
2
2
O  C1 S C  S T
O  C2 SC  ST



ST
1   3
B 1 C1
B 2C 2
2
2
(18.32)
After about a half page of algebra, I ended up with equation (18.27).
Either yield strengths or ultimate strengths can be used in equation (18.27), depending on
how failure is defined.
I always feel better after I consider special cases in which I think I know the answer.
Consider the special case in which S C  S T . Both dashed Mohr’s circles are the same diameter,
and the failure line is horizontal. The diameter of the gray circle is then equal to the diameter of
either of the dashed circles.
1   3  S T  SC
(18.28)
For this special case, the Coulomb-Mohr theory degenerates into the maximum-shearstress theory.
MAE3501 - 18 - 11
This again leads to three distinct cases, depending on the values of the two principal
stresses determined from a Mohr’s circle analysis of the plane stress. Denote these two
principal stresses as  A and  B , such that the following relationship is satisfied.
 A  B
(18.29)
 A  B  0
(18.30)
1   A
(18.31)
2  B
(18.32)
3  0
(18.33)
A
1
ST
(18.34)
 A  0  B
(18.35)
1   A
(18.36)
2  0
(18.37)
3  B
(18.38)
A B

1
ST SC
(18.39)
0   A  B
(18.40)
1  0
(18.41)
2  A
(18.42)
3  B
(18.43)
Case I
Case II
Case III

B
1
SC
(18.44)
MAE3501 - 18 - 12
The Coulomb-Mohr theory is represented graphically in the following figure.
MAE3501 - 18 - 13
Brittle Materials
Brittle materials do not generally exhibit an onset of plasticity (yielding). Therefore,
failure theories of brittle materials generally involve ultimate strengths, S UC and S UT . These
ultimate strengths are frequently not equal.
Maximum-Normal-Stress Theory
The maximum-normal-stress (MNS) theory states that failure occurs when one of the
principal stresses reaches the material strength. Mathematically, failure occurs when either of
the following two equations is satisfied.
 1  S UT
(18.44)
 3   S UC
(18.45)
Consider plane stress. Failure occurs when either of the following two equations is
satisfied.
 A  S UT
(18.46)
 B   S UC
(18.47)
For plane stress, the maximum-normal-stress theory is represented graphically in the
following figure.
Shigley does not recommend the maximum-normal-stress theory, because it does a poor
job in the fourth quadrant, where both plane-stress principal stresses are negative.
MAE3501 - 18 - 14
Shigley discusses two modifications to the Mohr theory for brittle materials. We will
discuss only one.
Brittle-Coulomb-Mohr Theory
The brittle-Coulomb-Mohr (BCM) theory replaces the strengths in the Coulomb-Mohr
relationships with ultimate strengths.
Case I
 A  B  0
(18.48)
A
1
S UT
(18.49)
 A  0  B
(18.50)
A

 B 1
S UT S UC
(18.51)
0   A  B
(18.52)
B
1
S UC
(18.53)
Case II
Case III

Homework
Skim Sections 6 - 1 through 6 - 4, 6 - 7, 6 - 8 , and 6 - 15.
Problems 5-1 (a) (b), 5-19 (a) BCM only
MAE3501 - 18 - 15