Theories of failure

Introduction

Theories of failure are those theories which help us to determine the safe dimensions of

a machine component when it is subjected to combined stresses due to various loads

acting on it during its functionality.

Some examples of such components are as follows:

1. I.C. engine crankshaft

2. Shaft used in power transmission

3. Spindle of a screw jaw

4. Bolted and welded joints used under eccentric loading

5. Ceiling fan rod

Theories of failure are employed in the design of a machine component due to the

unavailability of failure stresses under combined loading conditions.

Theories of failure play a key role in establishing the relationship between stresses

induced under combined loading conditions and properties obtained from tension test

like ultimate tensile strength (Sut) and yield strength (Syt).

Examples:

1.

d

Syt = 200 MPa

Sut = 300 MPa

Directly we can get (d) without using any failure

theory because only uniaxial load (P)

𝜎1 ≤ Syt

4P πd2 ≤ Syt

P

2.

Various Theories of Failure

1. Maximum Principal Stress theory also known as RANKINE’S THEORY

2. Maximum Shear Stress theory or GUEST AND TRESCA’S THEORY

3. Maximum Principal Strain theory also known as St. VENANT’S THEORY

4. Total Strain Energy theory or HAIGH’S THEORY

5. Maximum Distortion Energy theory or VONMISES AND HENCKY’S THEORY

1. Maximum Principal Stress theory (M.P.S.T)

According to M.P.S. T

P

So, different scientists give relationships between

Stresses induced under combined loading conditions and (Syt and Sut) obtained using

tension test which are called theories of failure.

T

d

Member is subjected to both Twisting moment and

uniaxial load, hence combined loading conditions.

We cannot determine (d) directly in this case

because failure stresses under combined loading

conditions are unknown.

Condition for failure is,

Maximum principal stress ( 1) failure stresses (Syt or Sut )

and Factor of safety (F.O.S) = 1

If 1 is +ve then Syt or Sut

1 is –ve then Syc or Suc

Condition for safe design,

Factor of safety (F.O.S) > 1

Maximum principal stress ( 1) ≤ Permissible stress ( per)

where permissible stress =

Failure stress =

Syt or

Sut

Factor of safety N N

1 ≤ Syt

or Sut

N N Eqn (1)

Note:

1.This theory is suitable for the safe design of machine components made of brittle

materials under all loading conditions (tri-axial, biaxial etc.) because brittle materials are

weak in tension.

2.This theory is not suitable for the safe design of machine components made of ductile

materials because ductile materials are weak in shear.

3.This theory can be suitable for the safe design of machine components made of ductile

materials under following state of stress conditions.

2

1 (i) Uniaxial state of stress (Absolute max = )

2

1 (ii) Biaxial state of stress when principal stresses are like in nature (Absolute max = )

(iii) Under hydrostatic stress condition (shear stress in all the planes is zero).

2. Maximum Shear Stress theory (M.S.S.T)

Condition for failure,

Maximum shear stress induced at a critical

point under triaxial combined stress

Yield strength in shear under tensile

test

Absolute max (Sys)T.T or Syt

2

unknown therefore use Syt

Condition for safe design,

≤ Permissible shear stress (τper) Maximum shear stress induced at a critical

tensile point under triaxial combined stress

where,

Permissible shear stress = Yield strength in shear under tension test

Factor of safety = (Sys)T.T

N = Syt

2N

Absolute max ≤ (Sys)T.T

N or Syt

2N

For tri-axial state of stress,

larger of [| σ1 - σ2

2 | , | σ2 - σ3

2 | , |

σ3 - σ1

2 |] ≤ Syt

2N

larger of [ | σ1 – σ2|, | σ2 – σ3|, | σ3 – σ1| ] ≤ Syt

N

For Biaxial state of stress, σ3 = 0

1

2 | | or | σ σ - σ 1 2

2 | ≤ Syt

2N

| σ1| ≤ Syt

N when σ1, σ2 are like in nature Eqn (2)

|σ1 – σ2| ≤ Syt

N when σ1, σ2 are unlike in nature Eqn (3)

Note:

1.M.S.S.T and M.P.S.T will give same results for ductile materials under uniaxial state of

stress and biaxial state of stress when principal stresses are like in nature.

2. M.S.S.T is not suitable under hydrostatic stress condition.

3.This theory is suitable for ductile materials and gives oversafe design i.e. safe and

uneconomic design.

3. Maximum Principal Strain theory (M.P.St.T)

Condition for failure,

Maximum Principal strain (ε1) Yielding strain under tensile test (ε Y.P.)T.T

ε1 (ε Y.P.)T.T or Syt

E

where E is Young’s Modulus of Elasticity

Condition for safe design,

Maximum Principal strain ≤ Permissible strain

where Permissible strain = = Yielding strain under tensile test (ε Y.P.)T.T Syt

Factor of safety N = EN

ε1 ≤ Syt

EN

E 1 [σ1 - µ(σ2 + σ3)] ≤

Syt

EN

σ1 - µ(σ2 + σ3) Syt

≤ N

for biaxial state of stress, σ3 = 0

σ1 - µ(σ2) ≤ Syt

N Eqn (4)

4. Total Strain Energy theory (T.St.E.T)

Condition for failure,

Total Strain Energy per unit volume

(T.S.E. /vol)

Strain energy per unit volume at yield

point under tension test (S.E /vol) Y.P.] T.T

Condition for safe design,

Total Strain Energy per unit volume ≤ Strain energy per unit volume at yield point

under tension test. Eqn (5)

σE.L

εE.L

Total Strain Energy per unit volume = 1

σ1 ε1 + 1

σ2 ε2 + 1

σ3 ε3 2 2 2

Eqn (6)

(triaxial)

Strain energy per unit volume up

to Elastic limit (E.L) = 2 1

σE.L εE.L

Condition for failure,

Maximum Distortion Energy/volume Distortion energy/volume at yield point

(M.D.E/vol) under tension test (D.E/vol) Y.P.] T.T

Condition for safe design,

Maximum Distortion Energy/volume ≤ Distortion energy/volume at yield point

under tension test (11)

T.S.E/vol = Volumetric S.E/vol + D.E/vol

D.E/vol = T.S.E/vol - Volumetric S.E /vol (12)

D.E/vol = 0 Under hydrostatic stress condition,

and

Under pure shear stress condition, Volumetric S.E/vol = 0

From equation (8)

2E T.S.E/vol = 1 [σ12 + σ22 + σ32 - 2µ (σ1 σ2 + σ2 σ3 +σ3 σ1)]

1 Volumetric S.E/vol = 2 (Average stress) (Volumetric strain)

1 = 2 (

σ1 + σ2 + σ3

3 ) [(

1-2µ E ) (σ1 + σ2 + σ3) ]

Vol S.E/vol = 1-2µ

(σ1 + σ2 + σ3)2 (13) 6E

From equation (12) and (13)

D.E/vol = 1+µ

[(σ1 - σ2)2 + (σ2 - σ3)2 + (σ3 - σ1)2] (14) 6E

To get [(D.E/vol) Y.P.] T.T ,

N Substitute σ1 = σ = Syt

, σ2 = σ3 = 0 in equation (14)