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

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

Principle Normal Direction

Principle Normal Direction

Principle Normal Direction

Principle Normal Direction

Calculation Principle Direction

Calculation Principle Direction

Principle Normal Direction

Principle Normal Stress Direction Previous Class Only see on Cubic Particle

How to avoid this condition?

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)

Failure Criterion (Von Mises) (Tresca)

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

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

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

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

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

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

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)

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, 1-2-3 or 𝐼,𝐼𝐼,𝐼𝐼𝐼 No Shear Stress are present

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.

Principle Shear Stress and Maximum Shear Stress 𝜏 1 = 𝜎 2 − 𝜎 3 2 , 𝜏 2 = 𝜎 1 − 𝜎 3 2 , 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 )

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

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)

This 3 Criterion to determine effective Stress, 𝝈

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

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

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

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)

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 3 2 , 𝜎 1 − 𝜎 3 2 , 𝜎 1 − 𝜎 2 2 (at yielding) 𝜏 1 = 𝜎 2 − 𝜎 3 2 , 𝜏 2 = 𝜎 1 − 𝜎 3 2 , 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.

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

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 3 2 , 𝜎 1 − 𝜎 3 2 , 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

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

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

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)

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 = 1 2 ( 𝜎 1 − 𝜎 2 ) 2 + ( 𝜎 2 − 𝜎 3 ) 2 + ( 𝜎 3 − 𝜎 1 ) 2 Effective Stress, 𝜎 𝐻 𝜎 𝐻 = 1 2 ( 𝜎 1 − 𝜎 2 ) 2 + ( 𝜎 2 − 𝜎 3 ) 2 + ( 𝜎 3 − 𝜎 1 ) 2 Safety Failure against failure 𝑋= 𝜎 𝑐 𝜎 𝐻

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

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 = 1 2 ( 𝜎 1 − 𝜎 2 ) 2 + ( 𝜎 2 − 𝜎 3 ) 2 + ( 𝜎 3 − 𝜎 1 ) 2 𝑋= 𝜎 𝑐 𝜎 𝐻 ANS = 2.78

Failure Criterion (Von Mises) (Tresca)

Failure Criterion - Polymers

Failure Criterion – Biaxial Fracture

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.