© D.J.DUNN 1

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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

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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

200-4000.28600

600

σσνσ

σ

factor Safety

321

max

=

−

=

+−

=

The component is just safe as the safety factor is larger than 1.

© D.J.DUNN 4

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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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