1. Yielding And Fracture Under Combine Stresses
PDT 351 – Material Failure Analysis
By : Wan Mohd Faizal Bin Wan Abd Rahim

2. Review On Previous Class
Type of failure
Method to determine failure
Failure Report
Concept of Strain and Stress
Testing Method
Principle direction strain and stress

3. Principle Normal Direction

4. Principle Normal Direction

5. Principle Normal Direction

6. Principle Normal Direction

7. Calculation Principle Direction

8. Calculation Principle Direction

9. Principle Normal Direction

10. Principle Normal Stress Direction
Previous Class Only see on Cubic Particle

11. How to avoid this condition?

12. Content For This Topic General form of failure criteria
Safety factor again failure, 𝑋= 𝜎 0 𝜎
Effective stress (Resultant Stress) , 𝜎
Yield Strength (material Characteristic) , 𝜎 0
Principle Shear Stress and Maximum Shear Stress
Maximum Normal Stress Fracture Criterion
Maximum Shear Stress Yield Criterion (Tresca)
Octahedral Shear Stress Yield Criterion (Von Mises)

13. Failure Criterion
(Von Mises)
(Tresca)

14. General form of failure Criteria
Engineering component often having complex loading
Tension
Compression
Bending
Torsion
Pressure
Or combination all of these
Occur more than 1 direction
Combination all these stress can cause the material to yield and fracture

15. General form of failure Criteria
In the book, to many criteria been discuss
Predict failure by yield (yield criteria)
Predict failure by fracture (fracture criteria)
Now we start with understand failure criteria by calculate effective value of stress , 𝝈 that characterizes the combined stress.
Then compared with the yield or fracture strength of material, 𝝈 𝟎
So we can judge, our study condition for given material may fail by either yielding or fracture

16. Fundamental Stress Criteria
Uniaxial tension
Tension with Transverse compression
Biaxial tension
Hydrostatic tension

17. General form of failure Criteria
Safety Failure, X
Failure criteria for isotropic material :-
𝒇 𝝁 𝑰 , 𝝁 𝑰𝑰 , 𝝁 𝑰𝑰𝑰 = 𝝈 𝒄 or 𝒇 𝝈 𝟏 , 𝝈 𝟐 , 𝝈 𝟑 = 𝝈 𝒄 (FAILURE)
Failure will occur if the principle normal stress is equal to the failure strength of the material
𝜎 𝑐 can be
Yield Strength , 𝜎 𝑜
Ultimate Strength, 𝜎 𝑢𝑡 𝑜𝑟 𝜎 𝑢𝑐
Depend on interest

18. General form of failure Criteria
Safety Failure, X
Considering engineering component having principle normal stress function, 𝑓( 𝜎 1 , 𝜎 2 , 𝜎 3 ) and where the material property 𝜎 𝑐 is known.
We will define, effective stress, 𝝈 , which is single numerical value that characterizes the state of applied stress.
𝜎 =𝑓( 𝜎 1 , 𝜎 2 , 𝜎 3 )
Now our problem is? How to find effective stress 𝝈

19. General form of failure Criteria
Safety Failure, X
Thus the failure will occur when,
𝝈 = 𝝈 𝒄 (at failure)
Failure is not expected if 𝜎 is less than 𝜎 𝑐
𝝈 < 𝝈 𝒄 (no failure)
Also we find the safety factor against failure is
𝑿= 𝝈 𝒄 𝝈

20. Content For This Topic General form of failure criteria
Safety factor again failure, 𝑋= 𝜎 0 𝜎
Effective stress (Resultant Stress) , 𝜎
Yield Strength (material Characteristic) , 𝜎 0
Principle Shear Stress and Maximum Shear Stress
Maximum Normal Stress Fracture Criterion
Maximum Shear Stress Yield Criterion (Tresca)
Octahedral Shear Stress Yield Criterion (Von Mises)

21. Principle Shear Stress and Maximum Shear Stress
Before this we already discuss,
Principle Strain Direction
Principle Normal Stress Direction
In this principle, we introduce New Coordinate System – Principle Axes, or 𝐼,𝐼𝐼,𝐼𝐼𝐼
No Shear Stress are present

22. Principle Shear Stress and Maximum Shear Stress
𝜎 1 , 𝜎 2 , 𝜎 3 = Principle Normal Stress
𝜏 1 , 𝜏 2 , 𝜏 3 = Principle Shear Stress
𝜎 𝜏1 , 𝜎 𝜏2 , 𝜎 𝜏3 = Principle Normal
Stress on two shear plane
Principle Shear Stress is the equivalent state of stress is
found 450 rotation about any of the 1,2 or 3 axes.

23. Principle Shear Stress and Maximum Shear Stress
𝜏 1 = 𝜎 2 − 𝜎 , 𝜏 2 = 𝜎 1 − 𝜎 , 𝜏 3 = 𝜎 1 − 𝜎 2 2
Principle Normal Stress On Stress Plane
𝜎 𝜏1 = 𝜎 2 + 𝜎 3 2 , 𝜎 𝜏2 = 𝜎 1 + 𝜎 3 2 , 𝜎 𝜏3 = 𝜎 1 + 𝜎 2 2
Maximum Normal Stress
𝜏 𝑀𝐴𝑋 =𝑀𝐴𝑋( 𝜏 1 , 𝜏 2 , 𝜏 3 )

24. Principle Shear Stress and Maximum Shear Stress
Revisit Previous Problem, 𝜎 1 =704.3𝑀𝑃𝑎 𝜎 2 =195.7𝑀𝑃𝑎 𝜎 3 =0 𝑜𝑢𝑡𝑠𝑖𝑑𝑒 𝑎𝑛𝑑 −20 𝑀𝑃𝑎 𝑖𝑛𝑠𝑖𝑑𝑒 By using this formula, 𝜏 𝑀𝐴𝑋 =𝑀𝐴𝑋 𝜎 2 − 𝜎 3 2 , 𝜎 1 − 𝜎 3 2 , 𝜎 1 − 𝜎 2 2 ,
𝜏 𝑀𝐴𝑋 =𝑀𝐴𝑋( 𝜏 1 , 𝜏 2 , 𝜏 3 )
𝜏 𝑚𝑎𝑥 =352.1 𝑀𝑃𝑎 (𝑜𝑢𝑡𝑠𝑖𝑑𝑒)
𝜏 𝑚𝑎𝑥 =362.1 𝑀𝑃𝑎 (𝑖𝑛𝑠𝑖𝑑𝑒)
(𝑨𝑵𝑺 )𝝉 𝒎𝒂𝒙 =𝟑𝟔𝟐.𝟏 𝑴𝑷𝒂 (𝒊𝒏𝒔𝒊𝒅𝒆)
𝝈 𝟏 𝒂𝒏𝒅 𝝈 𝟑 control the choice

25. Content For This Topic General form of failure criteria
Safety factor again failure, 𝑋= 𝜎 0 𝜎
Effective stress (Resultant Stress) , 𝜎
Yield Strength (material Characteristic) , 𝜎 0
Principle Shear Stress and Maximum Shear Stress
Maximum Normal Stress Fracture Criterion
Maximum Shear Stress Yield Criterion (Tresca)
Octahedral Shear Stress Yield Criterion (Von Mises)

26. This 3 Criterion to determine effective Stress, 𝝈

27. Maximum Normal Stress Fracture Criterion
Simplest failure criterion
Failure is expected when the largest principle normal stress reach the uniaxial strength of the material.
Application → predicting fracture for brittle material under
tension-dominated loading
For such material, a maximum normal stress fracture criterion can be express by :
𝜎 𝑢 =𝑀𝐴𝑋 ( 𝜎 1 , 𝜎 2 , 𝜎 3 ) at fracture
Effective Stress, 𝝈 𝑵
𝜎 𝑁 =𝑀𝐴𝑋 ( 𝜎 1 , 𝜎 2 , 𝜎 3 )
Safety Factor Against Failure,
𝑋= 𝜎 𝑐 𝜎 𝑁

28. Maximum Normal Stress Fracture Criterion
Failure locus for maximum normal stress fracture criterion for plane stress
𝜎 1 = ± 𝜎 𝑢
𝜎 2 = ± 𝜎 𝑢
𝜎 3 = ± 𝜎 𝑢
𝑋= 𝐷 𝑁 𝐷

29. Maximum Normal Stress Fracture Criterion
From Previous Example, Consider the pipe with closed end (previous example), with wall thickness 10mm and inner diameter 0.60mm, subjected to 20MPa internal pressure and torque of 1200kN∙m. What is the safety factor against yields at the inner pipe is made of 18 Ni maraging steel by using Maximum Normal Stress Criterion?
𝜎 𝑁 =𝑀𝐴𝑋 ( 𝜎 1 , 𝜎 2 , 𝜎 3 )
𝑋= 𝜎 𝑐 𝜎 𝑁
ANS = 2.54

30. Content For This Topic General form of failure criteria
Safety factor again failure, 𝑋= 𝜎 0 𝜎
Effective stress (Resultant Stress) , 𝜎
Yield Strength (material Characteristic) , 𝜎 0
Principle Shear Stress and Maximum Shear Stress
Maximum Normal Stress Fracture Criterion
Maximum Shear Stress Yield Criterion (Tresca)
Octahedral Shear Stress Yield Criterion (Von Mises)

31. Maximum Shear Stress Yield Criterion (Tresca criterion)
Yielding of ductile material → often predicted to occur when the maximum shear stress on any plane reach a critical value 𝜏 0 , which is a material property :
𝝉 𝟎 = 𝝉 𝒎𝒂𝒙 (at yielding)
This idea we call, maximum shear stress yield criterion (TRESCA CRITERION)
Recall from previous topic, Maximum Shear Stress Criterion
And yield criterion can be stated as follows:
𝜏 𝑜 =𝑀𝐴𝑋 𝜎 2 − 𝜎 , 𝜎 1 − 𝜎 , 𝜎 1 − 𝜎 (at yielding)
𝜏 1 = 𝜎 2 − 𝜎 , 𝜏 2 = 𝜎 1 − 𝜎 , 𝜏 3 = 𝜎 1 − 𝜎 2 2
𝜏 0 is only can be determine from the test and very
rare available in the table if compare to the 𝜎 0 is commonly
available in the table material property.

32. Maximum Shear Stress Yield Criterion (Tresca criterion)
Now, we need to calculate 𝜏 0 from 𝜎 0 .
In uniaxial test tension test, stress at yield strength can define
𝜎 1 = 𝜎 0 𝑎𝑛𝑑 𝜎 2 = 𝜎 3 =0
Substitution to the yield criterion
equation
𝜏 0 = 𝜎 2 2

33. Maximum Shear Stress Yield Criterion (Tresca criterion)
In the uniaxial test, note that the maximum shear stress occurs on planes oriented at 45o with respect to the applied stress axis.
So that, 𝜎 𝑜 we can written as :
Effective stress in Tresca Criterion, 𝜎 𝑆
Safety Factor Against Failure, X
𝜎 0 2 =𝑀𝐴𝑋 𝜎 2 − 𝜎 , 𝜎 1 − 𝜎 , 𝜎 1 − 𝜎 2 2
𝜎 𝑆 =𝑀𝐴𝑋 𝜎 2 − 𝜎 3 , 𝜎 1 − 𝜎 3 , 𝜎 1 − 𝜎 2
𝑋= 𝜎 𝑐 𝜎 𝑆

34. Maximum Shear Stress Yield Criterion (Tresca criterion)
Graphical Representation of the Maximum Shear Stress Criterion
For plane stress, 𝜎 3 =0 and Tresca Criterion can be represented on plot 𝜎 1 versus 𝜎 2
The region of no yielding , where 𝜎 𝑆 < 𝜎 0 is region bounded by the lines,
𝜎 1 − 𝜎 2 =± 𝜎 0, 𝜎 2 =± 𝜎 0, 𝜎 1 =± 𝜎 0
We can get,
𝜎 𝑆 =𝑀𝐴𝑋 𝜎 1 − 𝜎 2 , 𝜎 2 , 𝜎 1

35. Maximum Shear Stress Yield Criterion (Tresca criterion)
From Previous Example, Consider the pipe with closed end (previous example), with wall thickness 10mm and inner diameter 0.60mm, subjected to 20MPa internal pressure and torque of 1200kN∙m. What is the safety factor against yields at the inner pipe is made of 18 Ni maraging steel by using Tresca Criterion?
𝜎 𝑆 =𝑀𝐴𝑋 𝜎 2 − 𝜎 3 , 𝜎 1 − 𝜎 3 , 𝜎 1 − 𝜎 2
𝑋= 𝜎 𝑐 𝜎 𝑆
ANS = 2.47

36. Content For This Topic General form of failure criteria
Safety factor again failure, 𝑋= 𝜎 0 𝜎
Effective stress (Resultant Stress) , 𝜎
Yield Strength (material Characteristic) , 𝜎 0
Principle Shear Stress and Maximum Shear Stress
Maximum Normal Stress Fracture Criterion
Maximum Shear Stress Yield Criterion (Tresca)
Octahedral Shear Stress Yield Criterion (Von Mises)

37. Octahedral Shear Stress Yield Criterion (von Mises criterion)
Yielding of ductile material → often predicted to occur when shear stress on the octahedral planes reach the critical value,
𝝉 𝒉 = 𝝉 𝒉𝟎 (at yielding)
This idea we call, octahedral shear stress yield criterion
(von Mises CRITERION or Distortion Energy CRITERION)
Yielding Criterion
𝜎 0 = ( 𝜎 1 − 𝜎 2 ) 2 + ( 𝜎 2 − 𝜎 3 ) 2 + ( 𝜎 3 − 𝜎 1 ) 2
Effective Stress, 𝜎 𝐻
𝜎 𝐻 = ( 𝜎 1 − 𝜎 2 ) 2 + ( 𝜎 2 − 𝜎 3 ) 2 + ( 𝜎 3 − 𝜎 1 ) 2
Safety Failure against failure
𝑋= 𝜎 𝑐 𝜎 𝐻

38. Octahedral Shear Stress Yield Criterion (von Mises criterion)
Graphical Representation of the Octahedral Shear Stress Criterion
Same with the Tresca, plane stress by assume that 𝜎 3 =0 and plot 𝜎 1 versus 𝜎 2 .
Inserting the boundary condition above to the main equation, we can get elliptical shape as shown below

39. Maximum Shear Stress Yield Criterion (von Mises criterion)
From Previous Example, Consider the pipe with closed end (previous example), with wall thickness 10mm and inner diameter 0.60mm, subjected to 20MPa internal pressure and torque of 1200kN∙m. What is the safety factor against yields at the inner pipe is made of 18 Ni maraging steel by using von Mises Criterion?
𝜎 𝐻 = ( 𝜎 1 − 𝜎 2 ) 2 + ( 𝜎 2 − 𝜎 3 ) 2 + ( 𝜎 3 − 𝜎 1 ) 2
𝑋= 𝜎 𝑐 𝜎 𝐻
ANS = 2.78

40. Failure Criterion
(Von Mises)
(Tresca)

41. Failure Criterion - Polymers

42. Failure Criterion – Biaxial Fracture

43. Assignment
A Solid shaft of diameter d is made of AISI 1020 steel (as rolled) and is subjected to a tensile axial force of 200kN and torque of 1.50kN∙m. Refer to figure below for details explanation.
What is the safety factor against yielding if the diameter is 50mm
For the situation of (a), what adjusted value of diameter is required to obtain a safety factor against yielding 2.0?
Using all 3 criterion to calculate problem above and explain the result diffrent.
Explain the significant finding and advantage/disadvantage for all the 3 criterion.