Failure TheoriesWhy 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 typeUniaxialBiaxial

Pure Shear

Material PropertyDuctile Brittle

Application of StressStatic

Dynamic

Static Loading Maximum Normal Stress

Modified MohrYield strength

Maximum shear stressDistortion energy

Dynamic LoadingGoodman

GerberSoderberg

Static Failure TheoriesThe 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 Brittle MaterialCharacteristic Failure Stress

Yield Stress Ultimate Stress

Important Theories

1. Maximum Shear Stress2. Maximum Octahedral Shear Stress

1. Maximum Normal Stress

2. Modified Mohr.

Ductile MaterialsMaximum Shear Stress Theory

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,

2maxyS

=τ

For a general state of stresses

2231

maxyS

=−

=σστ

This leads to an hexagonal failure envelop. A stress system in the interior of the envelop is considered SAFE

for design purposes, the failure relation can be modified to include a factor of safety (n):

31 σσ −= yS

n

The Maximum Shear Stress Theory for Ductile Materials is also known as the Tresca Theory.

Several cases can be analyzed in plane stress problems:

Case 1: 021 ≥≥σσIn this case σ3=0

y

y

S

S

≥

==−

=

1

131max 222σ

σσστ

Yielding condition

Case 2: 31 0 σσ ≥≥

y

y

S

S

≥−

=−

=

31

31max 22

σσ

σστ

Distortion Energy TheoryBased 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.

ResilienceResilience is the capacity of a material to absorb

energy when it is deformed elastically and then, upon unloading, to have this energy recovered.

∫=y dUr

εεσ

0Modulus of resilience Ur

If it is in a linear elastic region,

EEU yy

yyyr 221

21 2σσ

σεσ =⎟⎟⎠

⎞⎜⎜⎝

⎛==

For general 3-D stresses: ( )

( )( )13322123

22

21

332211

221

21

σσσσσσνσσσ

εσεσεσ

++−++=

++=

Eu

u

Applying Hooke’s Law

There are two components in this energy a mean component and deviatoric component. 33

321 zyxM

σσσσσσσ++

=++

=

MDMDMD σσσσσσσσσ −=−=−= 3,32,21,1

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

( )( )

( )[ ] ( )13322123

22

21

2

222

2226

2121321

221

σσσσσσσσσννσ

σσσσσσνσσσ

+++++−

=−=

++−++=

EEu

Eu

MMean

MMMMMMMMMMean

( )13322123

22

213

1 σσσσσσσσσν−−−++

+=−=

Euuu MeanD

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

( )13322123

22

21

2

31

31 σσσσσσσσσνν

−−−+++

==+

=E

uSE

u DyTensile

1323

21

13322123

22

21

σσσσ

σσσσσσσσσ

−+=

−−−++=

y

y

S

S

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.

( ) ( ) ( ) ( )

222

222222

3

26

xyyxyxVM

zxyzxyxzzyyxVM

τσσσσσ

τττσσσσσσσ

+−+=

+++−+−+−=

2D

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

VM

y

yVM

Sn

S

σ

σ

=

≥

Case of Pure Shear

yy

Max

yxyVM

SS

S

577.03

3

==

≥=

τ

τσ

Several theories have been developed to describe the failure of brittle materials, such as:

�Maximum Normal Stress Theory

�Coulomb-Mohr Theory

�Modified-Mohr Theory

Brittle Materials

Maximum Normal Stress TheoryFailure occurs when one of the three principal stresses reaches a permissible strength (TS). 21 σσ >Failure is predicted to occur when σ1=St and σ2<-Sc

Where St and Sc are the tensile and compressive strengthFor a biaxial state of stresses

σ1

σ2

St

St-Sc

-Sc

Coulomb-Mohr Theory or Internal Friction Theory (IFT)This theory is a modification of the maximum normal stress theory in the which the failure envelope is constructed by connecting the opposite corners of quadrants I and III.

The result is an hexagonal failure envelop.Similar to the maximum shear stress theory but also accounts for the uneven material properties of brittle material

Mohr’s TheoryThe theory predicts that a material will fail if a stress

state is on the envelope that is tangent to the three Mohr’s circles corresponding to:

a. uni-axial ultimate stress in tension, b. uni-axial ultimate stress in compression, and c. pure shear.

Modified Mohr’s Theory

112 ≤−TC σ

σσσ

This theory is a modification of the Coulomb-Mohr theory and is the preferred theory for brittle materials.

Maximum Normal-Strain TheoryAlso known as the Saint-Venant’s Theory.Applies only in the elastic range.Failure is predicted to occur if yy SS ±=−±=− 1221 or νσσνσσ

Where Sy is the yield strength. For a biaxial state of stress

σ1Sy-Sy

σ2

Sy

-Sy

Maximum Strain-Energy TheoryYielding is predicted to occur when the total strain energy in a given volume is greater than or exceeds the strain energy in the samevolume corresponding to the yield strength in tension or compression. The strain energy stored per unit volume (us) during uniaxial loading is E

Su y

s ⋅=

2

2

In a biaxial state of stress

( )2122

21

2211

221

22

σσνσσ

σεσε

σ

σ

⋅⋅⋅−+=

⋅+

⋅=

Eu

u

This theory is no longer used

Example:Given the material SY , σx , σv and τxy find the safety factors for all the applicable criteria. a. Pure aluminum

MPaMPaMPaMPaS xyyxY 0 10 10 30 =−=== τσσ

10-10

MPaMPaMPa Max 10 10 10 31 =−== τσσIs Al ductile or brittle? Ductile

Use either the Maximum Shear Stress Theory (MSST) or the Distortion Theory (DT)

5.12030

)10(1030

31

==−−

=−

=σσ

ySn

MSST Theory

DT Theory

73.132.17

30

32.173003 222

===

==+−+=

MPaMPaS

n

MPa

VM

y

xyyxyxVM

σ

τσσσσσ

b. 0.2%C Carbon SteelKsiKsiKsiKsiS xyyxY 10 35 5 65 =−=−== τσσ

In the plane XY the principal stresses are -1.973Ksi and -38.03Ksi with a maximum shear stress in the XY plane of 18.03KsiIn any orientation

KsiKsiKsiKsi

Max 01.1903.38 973.1 0 321

=−=−==

τσσσ

71.1)03.38(0

65

31

=−−

=−

=σσ

ySnMSST Theory

DT Theory71.1

03.3865

03.38312

32

1

===

=−+=

MPaKsiS

n

Ksi

VM

y

VM

σ

σσσσσ

Ductile

C. Gray Cast Iron

KsiKsiKsiKsiSKsiS xyyxucut 0 10 35 120 30 ==−=== τσσ

Brittle10-35

KsiKsiKsiKsi

Max 5.2235 0 10 321

=−===

τσσσ

Use Maximum Normal Stress Theory (MNST), Internal Friction Theory (IFT), Modified Mohr Theory (MMT)

0.31030

1

===σ

utSnMNST Theory (tensile)

4.335

120

3

===σ

ucSnMNST Theory (compression)

MMT

84.1

54.01

=

=

nn

)30120()30)(120(1)35(

301203010

1_4 0 0

31

31

−=−

−−

−=

−−

≤≥

n

SSSS

nSSS

quadrant

utuc

utuc

utuc

ut

th

σσ

σσ

IFT

6.1

625.0120

3530101 31

=

=−

−=−=

nSSn ucut

σσ

13

31

_

_4 0 0

σσ

σσ

ut

ucuc

th

SSSequationline

quadrant

+−=

≤≥112 ≤−

TC σσ

σσ

Example 1The cantilever tube shown is to be made of 2014 aluminum alloy treated to obtain a specified minimum yield strength of 276MPa. We wish to select a stock size tube (according to the table below). Using a design factor of n=4. The bending load is F=1.75kN, the axial tension is P=9.0kN and the torsion is T=72N.m. What is the realized factor of safety?

Consider the critical area ( top surface).

IMc

AP

x +=σ

Maximum bending moment = 120F

I

dkNxmm

AkN

x

⎟⎠⎞

⎜⎝⎛×

+= 275.1120

9σ

Jd

J

d

JTr

zx362

72=

⎟⎠⎞

⎜⎝⎛×

==τ

( ) 2122 3 zxxVM τσσ +=

GPaGPanS y

VM 0690.04276.0

==≤σ

For the dimensions of that tube57.4

06043.0276.0

===VM

ySn

σ

Example 2:A certain force F is applied at D near the end of the 15-in lever, which is similar to a socket wrench. The bar OABC is made of AISI 1035steel, forged and heat treated so that it has a minimum (ASTM) yield strength of 81kpsi. Find the force (F) required to initiate yielding. Assume that the lever DC will not yield and that there is no stress concentration at A.Solution:

1) Find the critical section

The critical sections will be either point A or Point O. As the moment of inertia varies with r4

then point A in the 1in diameter is the weakest section.

Fd

inFd

dM

IMy

x 6.1421432

64

234 =××

=⎟⎠⎞

⎜⎝⎛

==ππ

σ

2) Determine the stresses at the critical section

Fin

inFd

dT

JTr

zx 4.76)1(1516

32

234 =

××=

⎟⎠⎞

⎜⎝⎛

==ππ

τ

3) Chose the failure criteria.

The AISI 1035 is a ductile material. Hence, we need to employ the distortion-energy theory.

lbfS

F

F

VM

y

zxxxyyxyxVM

4165.194

81000

5.19433 22222

===

=+=+−+=

σ

τστσσσσσ

Apply the MSS theory. For a point undergoing plane stress with only one non-zero normal stress and one shear stress, the two non-zero principal stresses (σA and σB) will have opposite signs (Case 2).

( ) ( )( )lbfF

FF

S

S

zxxzxx

yBA

zxxyBA

3884.7646.14281000

42

2

222

2122

2222

22

max

=×+=

+=+⎟⎠⎞

⎜⎝⎛=≥−

+⎟⎠⎞

⎜⎝⎛±==

−=

τστσσσ

τσσστ

Example 3:A round cantilever bar is subjected to torsion plus a transverse load at the free end. The bar is made of a ductile material having a yield strength of 50000psi. The transverse force (P) is 500lb and the torque is 1000lb-in applied to the free end. The bar is 5in long (L) and a safety factor of 2 is assumed. Transverse shear can be neglected. Determine the minimum diameter to avoid yielding using both MSS and DET criteria.Solution

1) Determine the critical section

The critical section occurs at the wall.

3432

64

2dPL

d

dPL

IMc

x ππσ =

⎟⎠⎞

⎜⎝⎛

==34

16

32

2dT

d

dT

JTc

xy ππτ =

⎟⎠⎞

⎜⎝⎛

==

( ) ( )

( )

( ) ⎥⎦⎤

⎢⎣⎡ +×±×=

⎥⎦⎤

⎢⎣⎡ +±=⎟

⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛±=

+⎟⎠⎞

⎜⎝⎛±⎟

⎠⎞

⎜⎝⎛=+⎟⎟

⎠

⎞⎜⎜⎝

⎛ −±⎟⎟⎠

⎞⎜⎜⎝

⎛ +=

2232,1

223

2

3

2

332,1

22

22

2,1

10005500550016

16161616

2222

d

TPLPLdd

TdPL

dPL

xyxx

xyyxyx

πσ

ππππσ

τσστσσσσ

σ

indn

Sdd

yMAX

MAX

031.1

000,252

500002

4.137152

)8.980(264502

31

3331

≥

==≤=−

=−−

=−

=

τσσ

σστ

3126450

d=σ 32

8.980d

−=σ The stresses are in the wrong order.. Rearranged to

3126450

d=σ 33

8.980d

−=σ

MSS

indnS

d

dddd

yVM

VM

025.12

5000026950

8.980264508.98026450

3

33

2

3

2

3312

32

1

≥

=≤=

⎟⎠⎞

⎜⎝⎛−⎟⎠⎞

⎜⎝⎛−⎟

⎠⎞

⎜⎝⎛−+⎟

⎠⎞

⎜⎝⎛=−+=

σ

σσσσσDET

Example 4:

In the wheel suspension of a car, the spring motion is provided by a torsion bar fastened to the arm on which the wheel is mounted. The torque in the torsion bar is created by the 2500N force acting on the wheel from the ground through a 300mm long lever arm. Because ofspace limitations, the bearing holding the torsion bar is situated 100mm from the wheel shaft. The diameter of the torsion bar is 28mm. Find the stresses in the torsion bar at the bearing by using the DET theory.

Solution

The stresses acting on a torsion bar are:

1. Torsion

2. bending

( )( )( )

MPaPad

dlengtharmF

JTc 174

028.0014.03.0250032

32

)2

)(_(44 =

×=

×==

ππτ

( ) ( )( )( )

MPaPadI

116028.0

64

64

44 ====ππ

σ

dlengthbearingFMc 014.01.025002

_×⎟

⎠⎞

⎜⎝⎛×

( ) ( )

( )

MPaMPa

xyxx

xyyxyx

4.125 4.241

0.1742

0.1162

0.116

2222

21

22

2,1

22

22

2,1

−==

+⎟⎠⎞

⎜⎝⎛±=

+⎟⎠⎞

⎜⎝⎛±⎟

⎠⎞

⎜⎝⎛=+⎟⎟

⎠

⎞⎜⎜⎝

⎛ −±⎟⎟⎠

⎞⎜⎜⎝

⎛ +=

σσ

σ

τσστσσσσ

σ

The principal stresses are:

( ) ( ) ( )( )

nS

MPa yVM

VM

≤=

−−−+=−+=

6.322

4.1254.2414.1254.241 2231

23

21

σ

σσσσσ

DET

MSS

( )

nSMPa

MPa

YMax

Max

≤=×

=−−

=−

=

8.3662

4.1832

4.1254.2412

31

τ

σστ

Example 5:The factor of safety for a machine element depends on the particular point selected for the analysis. Based upon the DET theory, determine the safety factor for points A and B.

This bar is made of AISI 1006 cold-drawn steel (Sy=280MPa) and it is loaded by the forces F=0.55kN, P=8.0kN and T=30N.m

Solution:Point A 2324

432

464

2dP

dFl

dP

d

dFl

AreaP

IMc

x ππππσ +=+

⎟⎠⎞

⎜⎝⎛

=+=

( )( )( )( )

( )( )( )

MPax 49.9502.01084

02.01.01055.032

2

3

3

3

=+=ππ

σ

( )( )

MPadT

JTr

xy 10.19020.0301616

33 ====ππ

τ

( ) ( )[ ]77.2

1.101280

1.1011.19349.953 212222

===

=+=+=

VM

y

xyxVM

Sn

MPa

σ

τσσ

Point B ( )( )( )

( )( )

( )( )( )

( )[ ]22.6

02.45280

02.4543.21347.25

43.2102.0

43

1055.0402.03016

3416

47.2502.010844

2122

2

3

33

2

3

2

==

=+=

=⎟⎠⎞

⎜⎝⎛

+=+=

===

n

MPa

MPaAV

dT

MPadP

VM

xy

x

σ

πππτ

ππσ