© D.J.DUNN 1
1
COMPLEX STRESS
TUTORIAL 4
THEORIES OF FAILURE
This short tutorial covers no known elements of the E.C. or Edexcel Exams
but should be studied as part of complex stress, structures and materials.
You should judge your progress by completing the self assessment
exercises.
These may be sent for marking or you may request copies of the solutions
at a cost (see home page).
On completion of this tutorial you should be able to do the following.
Explain the greatest principal stress theory (Rankine)
Explain the greatest principal strain theory (St. Venant)
Explain the maximum shear stress theory (Guest And
Coulomb)
Solve problems involving the above theories.
It is assumed that students doing this tutorial are already familiar with
complex stress theory.
INTRODUCTION
Modern CADD systems allow the engineer to calculate stress levels in a component
using finite stress analysis linked to the model. The reasons why a given material fails
however, is not something a computer can predict without the results of research being
added to its data bank. In some cases it fails because the maximum tensile stress has
been reached and in others because the maximum shear stress has been reached. The
exact combination of loads that makes a component fail depends very much on the
properties of the material such as ductility, grain pattern and so on. This section is about
some of the theories used to predict whether a complex stress situation is safe or not.
There are many theories about this and we shall examine three. First we should consider
what we regard as failure. Failure could be regarded as when the material breaks or
when the material yields. If a simple tensile test is conducted on a ductile material, the
stress strain curve may look like this.
Figure 1
The maximum allowable stress in a material is σ
max
. This might be regarded as the
stress at fracture (ultimate tensile stress), the stress at the yield point or the stress at the
limit of proportionality (often the same as the yield point). The Modulus of elasticity is
defined as E = stress/strain = σ/ε and this is only true up to the limit of proportionality.
Note that some materials do not have a proportional relationship at all. The maximum
allowable stress may be determined with a simple tensile test.
There is only one direct stress in a tensile test (σ = F/A) so it follows that σ
max
= σ
1
and it will have a corresponding strain ε
max
= ε
1
. Complex stress theory tells us that
there will be a shear stress τ and strain γ that has a maximum value on a plane at 45
o
to
the principal plane. It is of interest to note that in a simple tensile test on a ductile
material, at the point of failure, a cup and cone is formed with the sides at 45
o
to the
axis. Brittle materials often fail with no narrowing (necking) but with a flat fail plane at
45
o
to the axis. This suggests that these materials fail due to the maximum shear stress
being reached.
Figure 2
© D.J.DUNN 2
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In a complex stress situation, there are 3
principal stresses σ
1
, σ
2
and σ
3
. σ
1
is the
greatest and σ
3
is the smallest. Remember that
a negative stress is smaller than zero. There
are corresponding principal strains ε
1
, ε
2
and
ε
3
and shear strains.
Figure 3
1. THE GREATEST PRINCIPAL STRESS THEORY (RANKINE)
This simply states that in a complex stress situation, the material fails when the greatest
principal stress equals the maximum allowable value.
σ
1
= σ
max
σ
max
could be the stress at yield or at fracture depending on the definition of failure.
If σ
1
is less than σ
max
then the material is safe.
Safety Factor = σ
max
/σ
1
WORKED EXAMPLE No.1
A certain material fractured in a simple tensile test at a stress level of 800 MPa. The
same material when used as part of a structure must have a safety factor of 3.
Calculate the greatest principal stress that should be allowed to occur in it based on
Rankine’s theory.
SOLUTION
S.F. = 3 = σ
max
/σ
1
= 800/σ
1
σ
1
= 800/3 = 266.7 MPa
© D.J.DUNN 3
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2. THE GREATEST PRINCIPAL STRAIN THEORY (St. VENANT)
This states that in a complex stress situation, the material fails when the greatest
principal strain reaches the maximum allowable strain determined in a simple tensile
test.
ε
1
= ε
max
ε
max
is the value determined in a simple tensile test. If the maximum allowable stress is
taken as the value at the limit of proportionality, we may further develop the theory
using the modulus of elasticity. ε
max
= σ
max
/E
From 3 dimensional relationships (covered in other tutorials) we have:
( ){ }
( ){ }
( ){ }
321max
321
max
1
max
3211
σσνσ than less is σ
whenfails material theand
σσνσ
σ
ε
ε
factor Safety
σσνσ
E
1
ε
+−
+−
==
+−
⎟
⎠
⎞
⎜
⎝
⎛
=
WORKED EXAMPLE No.2
A certain material fractured in a simple tensile test at a stress level of 600 MPa. The
same material when used as part of a loaded structure must has principal stresses of
600, 400, and 200 MPa. Determine the safety factor at this load based on the
greatest principal strain theory. Take Poisson’s ratio as 0.28.
SOLUTION
( ){ }
( ){ }
103.1
2004000.28600
600
σσνσ
σ
factor Safety
321
max
=
−
=
+−
=
The component is just safe as the safety factor is larger than 1.
© D.J.DUNN 4
4
3. THE MAXIMUM SHEAR STRESS THEORY (GUEST and COULOMB)
This states that in a complex stress situation, the material fails when the greatest shear
strain in the material equals the value determined in a simple tensile test. Applying
complex stress theory to a tensile test gives this as τ
max
=
½
σ
max
In a simple tensile test, σ
max
could be what ever stress is regarded as the maximum
allowable.
In a 3 dimensional complex stress situation the maximum shear strain is τ
=
½
(σ
1
 σ
3
)
If this is less than τ
max
then the material is safe.
31
maxmax
σσ
σ
τ
τ
factor Safety
−
==
On the limit when the safety factor is 1 it follows that σ
max
= (σ
1
 σ
3
)
Put into words, failure occurs when the maximum allowable stress is equal to the
difference between the greatest and the smallest principal stresses. Note negative
stresses are smaller than zero.
WORKED EXAMPLE No.3
Show that based on the greatest shear stress theory, the structure in W.E. No.2
should have failed.
SOLUTION
75.0
)200(600
600
σσ
σ
τ
τ
factor Safety
31
maxmax
=
−−
=
−
==
The component should have failed as the safety factor is less than 1.
© D.J.DUNN 5
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© D.J.DUNN 6
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SELF ASSESSMENT EXERCISE No.1
1. A certain steel fails in a simple tensile test when the stress is 30 MPa. The same
steel is used in a complex stress situation and the principal stresses are 11 MPa, 3
MPa and 0 MPa. Determine the factor of safety based on the three theories. ν = 0.3
(Answers 2.73, 2.97 and 2.73)
2. A certain steel fails in a simple tensile test when the stress is 30 MPa. The same
steel is used in a complex stress situation and the principal stresses are 11 MPa, 0
MPa and 3 MPa. Determine the factor of safety based on the three theories. ν = 0.3
(Answers 2.73, 2.52 and 2.14)
3. A certain steel failed in a simple tensile test when the stress is 460 MPa. The same
steel is used in a complex stress situation and the principal stresses are 200 MPa,
150 MPa and 100 MPa. Determine the factor of safety based on the three theories.
ν = 0.3
(Answers 2.3, 1.53 and 2.49)
4. The results from a 60
o
strain gauge rosette are
ε
A
= 600 µε
ε
B
= 200 µε
ε
C
= 400 µε
ν = 0.3 E = 205 GPa
Determine the principal strains and stresses.
The same material failed at a stress level of 300 MPa in a tensile test. Calculate the
safety factor of the complex situation based on the three theories.
(Answers ε
1
= 747 µε, ε
2
= 214 µε, σ
1
= 157 MPa, σ
2
= 0.61 MPa, 1.91, 1.912
and 1.91).
4. STRAIN ENERGY
Another theory for failure of materials is based on strain energy. This simply states that
a component will fail when the total strain energy reaches a critical level. This is not
covered here but strain energy is covered in the tutorial of that name.
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