Theories Of Failure

Theories of Failure The material properties are usually determined by simple

tension or compression tests. The mechanical members are subjected to biaxial or

triaxial stresses. To determine whether a component will fail or not, some

failure theories are proposed which are related to the properties of materials obtained from uniaxial tension or compression tests. Initially we will consider failure of a mechanical member

subjected to biaxial stresses

The Theories of Failures which are applicable for this situation are: • Max principal or normal stress theory (Rankine’s theory) • Maximum shear stress theory (Guest’s or Tresca’s theory) • Max. Distortion energy theory (Von Mises & Hencky’s theory) • Max. strain energy theory • Max. principal strain theory

Ductile materials usually fail by yielding and hence the limiting strength is the yield strength of material as determined from simple tension test which is assumed the same in compression also. For brittle materials limiting strength of material is ultimate tensile strength in tension or compression.

Max. Principal or Normal stress theory (Rankine’s Theory): • It is assumed that the failure or yield occurs at a point in a member when the max. principal or normal stress in the biaxial stress system reaches the limiting strength of the material in a simple tension test. • In this case max. principal stress is calculated in a biaxial stress case and is equated to limiting strength of the material.

Maximum principal stress  σx + σ y   + σ1 =   2 

2

 σx − σ y    + τ 2xy  2 

Minimum principal stress  σx + σ y σ 2 =   2

  σx − σy  −    2

2

  + τ 2xy 

•For ductile materials σ 1 should not

 S yt  exceed FOS   

in tension, FOS=Factor of safety

•For brittle materials σ 1 should not

 S ut    exceed FOS 

in tension

This theory is basically applicable for brittle materials which are relatively stronger in shear and not applicable to ductile materials which are relatively weak in shear.

+σ2 Syt

-σ1 Syc

o

Syt

+σ1

Syc -σ2

Boundary for maximum – normal – stress theory under bi – axial stresses

2. Maximum Shear Stress theory (Guest’s or Tresca’s theory): •The failure or yielding is assumed to take place at a point in a member where the max shear stress in a biaxial stress system reaches a value equal to shear strength of the material obtained from simple tension test. •In a biaxial stress case max shear stress developed is given by σ x −σ y  2 2 τ max =  

τ max =

 

2

τ yt

FOS

where τ

=

+ τ xy

S yt

2 × FOS This theory is mostly used for ductile materials. max

τ

τ

τ max

τ max σ o

σ2= σ3 =0

σ1

-σ2

o σ3 =0

Mohr’s circle for uni – axial tension

τ max Mohr’s circle for bi– axial stress condition

+σ1

σ

According to the Maximum shear stress theory, τ max = And also

τ max

S yt 2

Max. direct stress − Min. direct stress = 2

Assuming that σ1> σ2> σ3 and σ3 =0 CASE – 2 (Second quadrant) σ1 is -ve and σ2 is +ve ,Then σ − ( − σ 1 ) σ 2 + σ 1 2σ 1 τ max = 2 = = 2 2 2 S yt Then τ max = σ1 = 2 -σ1 CASE – 3 (Third quadrant) σ1 is -ve and σ2 is more -ve ,Then σ − (− σ 2 ) 0 + σ 2 τ m ax = 3 = 2 2 σ S yc Then τ m ax = = 2 2 i.e σ = S yc

Syc

σ1=Syc

CASE – 1 (First quadrant ) σ1 and σ2 are +ve

+σ2 Syt

o

σ1=Syt

Syt

Syc -σ2

σ 1 − σ 3 σ 1 + 0 σ 1 S yt τ max = = = = 2 2 2 2 i.e. σ 1 = S yt +σ1

CASE – 4 (Fourth quadrant) σ1 is +ve and σ2 is -ve ,Then σ − (− σ 2 ) σ 1 + σ 2 2σ τ max = 1 = = 2 2 2 S yt Then τ max = σ = 2

3.Max. Distortion energy theory (Von Mises & Hencky’s theory): •It is assumed that failure or yielding occurs at a point the member where the distortion strain energy (also called shear strain energy) per unit volume in a biaxial stress system reaches the limiting distortion energy (distortion energy at yield point) per unit volume as determined from a simple tension test. •The maximum distortion energy is the difference between the total strain energy and the strain energy due to uniform stress.

3.Max. Distortion energy theory (Von Mises & Hencky’s theory):

3.Max. Distortion energy theory (Von Mises & Hencky’s theory): • The criteria of failure for the distortion – energy theory is expressed as S yt =

[

1 (σ 1 − σ 2 ) 2 + (σ 2 − σ 3 ) 2 + (σ 3 − σ 1 ) 2 2

• Considering the factor of safety S yt FOS

=

[

1 (σ 1 − σ 2 ) 2 + (σ 2 − σ 3 ) 2 + (σ 3 − σ 1 ) 2 2

• For bi – axial stresses (σ3=0), S yt

FOS

(

= σ1 + σ 2 − σ1 σ 2 2

2

)

]

]

3.Max. Distortion energy theory (Von Mises & Hencky’s theory): • A component subjected to pure shear stresses and the corresponding Mohr’s circle diagram is τ Y

τ

τ -σ2

τ

o

σ1

X

τ

Element subjected to pure shear stresses

τ Mohr’s circle for pure shear stresses

σ

From the figure, σ1 = -σ2 =

τ and σ3=0

Substituting the values in the equation

S yt

We get

FOS

(

= σ1 + σ 2 − σ1 σ 2 2

2

)

S yt = 3τ Replacing τ by Ssy , we get

S sy =

S yt

= 0.577 S yt

3 In the biaxial stress case, principal stress σ 1, σ 2 are calculated based on σ x ,σ y & τ xy which in turn are used to determine whether the left hand side is more than right hand side, which indicates failure of the component.

Case 2 (Second quadrant) σ1 is -ve and σ2 is +ve and equal to σ, then

S yt

FOS S yt FOS S yt

(σ (σ

=

2

1

=

1

2

2

Case 1 (First quadrant)

) +σ σ )

+ σ 2 − σ1 σ 2 + σ2

2

1

+σ2

o

Case 3 (Third quadrant) σ1 is -ve and σ2 is +ve and equal to σ, then

S yt

(

2

2

= σ1 + σ 2 − σ1 σ 2 FOS S yt σ= FOS

)

Syt

Syc -σ2

2

2

= σ1 + σ 2 − σ1 σ 2 FOS S yt σ= FOS

Syt

= 3σ 2 = 3σ FOS S yt σ = 0.577 FOS Syc

(

S yt

2

-σ1

σ1 and σ2 are +ve and equal to σ, then

)

+σ1 Case 4 (Fourth quadrant) σ1 is +ve and σ2 is -ve and equal toSσ, then yt

FOS S yt FOS S yt

=

=

(σ (σ

2

1

1

2

2

) +σ σ )

+ σ 2 − σ1 σ 2

+ σ2

2

= 3σ 2 = 3σ FOS S yt σ = 0.577 FOS

Boundary for distortion – energy theory under bi – axial stresses

1

2

MPa

MPa

MPa

4.

Max. Strain energy theory (Heigh’s Thoery):

•Failure is assumed to take place at a point in a member where strain energy per unit volume in a biaxial stress system reaches the limiting strain energy that is strain energy at yield point per unit volume as determined from a simple tension test. • Strain energy per unit volume in a biaxial system is

2σ 1σ 2  1  2 2 U1 = σ1 +σ 2 −  2E  m  • The limiting strain energy per unit volume for yielding as determined from simple tension test is

1  S yt  U2 =   2 E  FOS 

2

Equating the above two equations then we get

σ1 +σ 2 2

2

2σ 1 σ 2 − m

 S yt   =   FOS 

2

In a biaxial case σ 1, σ 2 are calculated based as σ x, σ y & τItxy will be checked whether the Left Hand Side of Equation is less than Right Hand Side of Equation or not. This theory is used for ductile materials.

5.Max. Principal Strain theory (Saint Venant’s Theory): •It is assumed that the failure or yielding occurs at a point in a member where the maximum principal (normal) strain in a biaxial stress exceeds limiting value of strain (strain at yield port) as obtained from simple tension test. • In a biaxial stress case

E max

Syt σ1 σ 2 = − = E mE FOS× E

•One can calculate σ 1 & σ 2 given σ x , σ y & τ xy and check whether the material fails or not, this theory is not used in general as reliable results could not be detained in variety of materials.

Example :1 •

The load on a bolt consists of an axial pull of 10kN together with a transverse shear force of 5kN. Find the diameter of bolt required according to 1. Maximum principal stress theory 2. Maximum shear stress theory 3. Maximum principal strain theory 4. Maximum strain energy theory 5. Maximum distortion energy theory Permissible tensile stress at elastic limit =100MPa and Poisson’s ratio =0.3

Solution 1 • Cross – sectional area of the bolt, A=

• Axial stress,

π 2 d = 0.7854d 2 4

P 10 12.73 2 σ1 = = = kN / mm A 0.7854d 2 d2

• And transverse shear stress, Ps 5 2 τ= = = 6 . 365 kN / mm A 0.7854d 2

According to maximum principal stress theory • Maximum principal stress, 2

σ  σ  σ1 =  x  +  x  + τ 2xy  2   2 

 σx + σ y   + σ1 =   2 

 12.73  σ1 =  + 2   2d  σ1 =

2

 σx − σ y    + τ 2xy  2 

 12.73  2  6.365  2  + 2    2   2d   d  

15365 2 N / mm d2

• According to maximum principal stress theory, Syt = σ1 15365 100 = ⇒ d = 12.4mm 2 d

According to maximum shear stress theory • Maximum shear stress, 2

 σx − σ y   + τ 2xy τ max =   2  2

 σx  τ max =   + τ 2xy  2 

 12.73  2  6.365  2  9 9000 2 =  2  +  2   = 2 kN / mm = 2 N / mm 2 d  d   d   d

• According to maximum shear stress,

S yt

9000 100 τ max = ⇒ 2 = 2 d 2 d = 13.42mm

According to maximum principal strain theory • The maximum principal stress,  σx + σ y   + σ1 =   2 

2

 σx − σ y    + τ 2xy  2  2

15365 σ  σ  σ1 =  x  +  x  + τ 2xy = d2  2   2 

• And minimum principal stress,  σx + σ y   − σ 2 =   2  2

2

 σx − σ y    + τ 2xy  2  2

12.73  σx   σx   12.73   6.365  2 σ 2 =   −   + τ xy = 2 −  + 2  2  d  2d   d   2   2  − 2635 2 σ2 = N / mm d2

2

• And according to maximum principal strain theory,

σ1 σ 2 Syt − = E mE E σ2 15365 2635 × 0.3 σ1 − = Syt ⇒ + = 100 2 2 m d d d = 12.7mm

• According to maximum strain energy theory 2σ1 σ 2 2 σ1 + σ 2 − = S yt m 2 2 15365 − 2635 15365   − 2635  2  d 2  +  d 2  − 2 × d 2 × d 2 × 0.3 = 100 • According todmaximum distortion theory = 12.78mm 2

2

S yt =

(σ

1

2

2

+ σ 2 − σ1 σ 2

)

 15365  2  − 2635  2 15365 − 2635   100 =   2  +  − × 2 2 2   d   d   d d   d = 13.4mm