Failure Theories Why do mechanical components fail? Mechanical components fail because the applied stresses exceeds the material’s strength (Too simple). What kind of stresses cause failure? Under any load combination, there is always a combination of normal and shearing stresses in the material.

What is the definition of Failure? Obviously fracture but in some components yielding can also be considered as failure, if yielding distorts the material in such a way that it no longer functions properly Which stress causes the material to fail? Usually ductile materials are limited by their shear strengths. While brittle materials (ductility < 5%) are limited by their tensile strengths. Stress at which point?

Stress at which point?

Failure Theories Load type Uniaxial Biaxial Pure Shear

Material Property Ductile Brittle

Static Loading Maximum Normal Stress Modified Mohr Yield strength Maximum shear stress Distortion energy

Application of Stress Static Dynamic

Dynamic Loading Goodman Gerber Soderberg

Static Failure Theories The idea behind the various classical failure theories is that whatever is responsible for failure in the standard tensile test will also be responsible for failure under all other conditions of static loading.

Ductile Material Characteristic Yield Stress Failure Stress Important Theories

Maximum Shear Stress Theory

Brittle Material Ultimate Stress

1. Maximum Shear Stress 1. Maximum Normal Stress 2. Maximum Octahedral Shear Stress 2. Modified Mohr.

Ductile Materials Failure occurs when the maximum shear stress in the part exceeds the shear stress in a tensile test specimen (of the same material) at yield. Hence in a tensile test,

τ max =

Sy 2

For a general state of stresses σ1 − σ 3 S y τ max = = 2 2 This leads to an hexagonal failure envelop. A stress system in the interior of the envelop is considered SAFE

The Maximum Shear Stress Theory for Ductile Materials is also known as the Tresca Theory. for design purposes, the failure relation can be modified to include a factor of safety (n):

n=

Sy

σ1 − σ 3

Several cases can be analyzed in plane stress problems: Case 1: σ 1 ≥ σ 2 ≥ 0 In this case σ3=0 σ1 − σ 3 σ1 S y τ max = = = σ1 ≥ S y

2

2

2

Case 2: σ 1 ≥ 0 ≥ σ 3 Yielding condition

τ max =

σ1 − σ 3

2 σ1 − σ 3 ≥ S y

=

Sy 2

Distortion Energy Theory Based on the consideration of angular distortion of stressed elements. The theory states that failure occurs when the distortion strain energy in the material exceeds the distortion strain energy in a tensile test specimen (of the same material) at yield.

Resilience Resilience is the capacity of a material to absorb energy when it is deformed elastically and then, upon unloading, to have this energy recovered. Modulus of resilience Ur If it is in a linear elastic region,

εy

U r = ∫ σ dε 0

1 1 ⎛σ y ⎞ σ y U r = σ yε y = σ y ⎜⎜ ⎟⎟ = 2 2 ⎝ E ⎠ 2E

2

1 For general 3-D stresses: u = (σ 1ε 1 + σ 2ε 2 + σ 3ε 3 ) 2 Applying Hooke’s Law u = 1 σ 2 + σ 2 + σ 2 − 2ν (σ σ + σ σ + σ σ ) 1 2 3 1 2 2 3 3 1 2E

(

There are two components in this energy a mean component and deviatoric component.

)

σM =

σ 1, D = σ 1 − σ M

σ1 + σ 2 + σ 3 3

=

σ 2, D = σ 2 − σ M

σ x +σ y +σ z 3 σ 3, D = σ 3 − σ M

The energy due to the mean stress (it gives a volumetric change but not a distortion: u Mean u Mean

(

)

1 = σ M2 + σ M2 + σ M2 − 2ν (σ M σ M + σ M σ M + σ M σ M ) 2E 1 1 − 2ν 2 2 = 3σ M (1 − 2ν ) = σ 1 + σ 22 + σ 32 + 2σ 1σ 2 + 2σ 2σ 3 + 2σ 3σ 1 2E 6E

[

]

(

)

u D = u − u Mean

1 +ν 2 = σ 1 + σ 22 + σ 32 − σ 1σ 2 − σ 2σ 3 − σ 3σ 1 3E

(

)

Compare the distortion energy of a tensile test with the distortion energy of the material.

uTensile

1 +ν 2 1 +ν 2 = S y = uD = σ 1 + σ 22 + σ 32 − σ 1σ 2 − σ 2σ 3 − σ 3σ 1 3E 3E

(

S y = σ 12 + σ 22 + σ 32 − σ 1σ 2 − σ 2σ 3 − σ 3σ 1 S y = σ + σ − σ 3σ 1 2 1

2 3

Plane Stress

Von Mises effective stress : Defined as the uniaxial tensile stress that creates the same distortion energy as any actual combination of applied stresses.

)

This simplifies the approach since we can use the following failure criterion

σ VM ≥ S y n=

σ VM =

(σ

(

Sy

σ VM

2 2 2 ) ( ) ( ) − σ + σ − σ + σ − σ + 6 τ + τ + τ x y y z z x xy yz zx 2

2

σ VM = σ x 2 + σ y 2 − σ xσ y + 3τ xy2

2

2 2D

)

Case of Pure Shear

σ VM = 3τ xy ≥ S y τ Max =

Sy 3

= 0.577 S y

Brittle Materials Several theories have been developed to describe the failure of brittle materials, such as: Maximum Normal Stress Theory Coulomb-Mohr Theory Modified-Mohr Theory

Maximum Normal Stress Theory

σ1 > σ 2

Failure occurs when one of the three principal stresses reaches a permissible strength (TS). Failure is predicted to occur when σ1=St and σ2