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Understanding Failure Theories: Brittle vs

. Ductile Materials
School
University of British Columbia**We aren't endorsed by this school
Course
MECH 260
Subject
Mechanical Engineering
Date
Dec 10, 2024
Pages
33
Uploaded by DeaconRose9451
Failure theoriesMECH 260
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Failure Theories►Used in design to specify limits for material strength, deformation and fracture.►A theory of failure is used to predict when an engineered component would fail under a given arrangement of external loading
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Examples of failures and their angles
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Brittle Materials►Materials that exhibit little or no yielding before failure are defined as brittle.►Yield stress and ultimate stress typically within 10% ►We need to avoid yield stress in devices and structures, but practically that means ultimate (failure) stress, because they are very close►Typically strong in compression –weak in tension►Examples: Ceramics, most of composite materials►Carbon steels are very sensitive to temperature and can be very brittle at low temperatures σfail=σultimateσallowed=σultimate/F.S.
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Ductile Materials ►Any material that is subjected to large strains before fracture is defined as ductile.►Yield stress and failure stress vary significantly►For ductile materials we need to avoid yield stress, since excessive deformations occur afterwards►Plastic strains lead to short fatigue lifeσfail=σyieldσallowed=σyield/F.S.
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Theories of Failure -Brittle MaterialsMaximum Normal Stress Criterion►Assumes that failure occurs when the maximum normal stressin the component reaches the value at which failure occurs in a uniaxial tension test of the same material. ►Failure is defined by ultimate normal stress of the material (typically very similar to yield stress for brittle mat.)►Applicable to brittle materials onlyTo avoid failure:σ1< σultandσ2< σult(we assume here F.S. = 1)
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Ductile Materials ►The most common form of yielding for ductile materials is slipping (Lüderlines)►Ductile materials fail in shear. (max shear stress at 45°)
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Theories of Failure –Ductile MaterialsMaximum Shear Stress Criterion (Tresca)►Assumes that failure occurs when the maximum shear stressin the component reaches the value of the shear stress at failure in a uniaxial tension or compression test of the same material. ►Ductile materials fail in shear (45°slipping)►Slightly conservative for ductile steels𝜏𝜏𝑚𝑚𝑚𝑚𝑚𝑚=𝜎𝜎𝑌𝑌2(See figure on the right)
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3D Stress State►In general, we deal with 3D stress state►We can orient the cube so that no shear occurs on any plane and we have 3 principal stresses, σ1, σ2, and σ3.►This loading is called triaxial loading. ►In the same manner as plane stress, we can orient the cube so that we have maximum shear stress acting on the cube.
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3D Principal –Triaxial Stress maxintminσσσ≥≥
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Is there a Mohr’s Circle Solution?►There is no Mohr’s circle solution for problems of triaxial stress state.►Solution for maximum principal stresses and maximum shear stress is analytical►Either closed form solution or numerical solution (or computer program) are used to solve the problem.►However, once the principal stresses, σ1σ2and σ3are found, the Mohr Circle can be used to find maximum shear stresses.►Online solvers can be found to assist in the solutions. Example:http://www.engapplets.vt.edu/Mohr/java/nsfapplets/MohrCircles2-3D/Applets/applet.htm
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Absolute Maximum Shear Stresses23max,12σστ−= ±13max,22σστ−= ±12max,32σστ−= ±Absolute max shear stress is numerically the largest between:NormalStress, σσ1σ3τy’z’, τabs maxσ2τx’y’τy’z’You don’t need to draw this in exams
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3D Stress StateA Case Study –The two principal stresses are of the same sign−σσ1σ3σ2ττabs max21maxστ=abs
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3D Stress StateA Case Study –The two principal stresses are of opposite signσσ1σ2ττabs max221maxσστ−=abs
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Example: triaxial stress state, not plane stress►Determine the maximum principal stresses and the maximum shear stress for the following triaxial stress state. σ=MPa=
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Solution of the problemσ==MPa1222222232xyzxyxzyzxyxzyzxyzxyxzyzxyzyxzzxyIIIσσσσ σσ σσ στττσ σ στ τ τσ τσ τσ τ=++=++−−−=+−−−= 20 + 30 –10 = 40 MPa= -3025 MPa= 89500 MPaSolve321230IIIσσσ−+−=I1, I2, andI3are known as stress invariantsas they do not change in value when the axes are rotated to new positions.These equations not needed on exam!
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Solution to Example-800000-600000-400000-2000000200000400000600000-100-80-60-40-20020406080100Stress (MPa)Sigma (MPa)-51.8 MPa65.3 MPa26.5 MPa
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Results𝜎𝜎1= 65.3𝑀𝑀𝑀𝑀𝑀𝑀𝜎𝜎2= 26.5𝑀𝑀𝑀𝑀𝑀𝑀𝜎𝜎3=−51.8𝑀𝑀𝑀𝑀𝑀𝑀𝜏𝜏𝑚𝑚𝑚𝑚𝑚𝑚= (65.3 + 51.8)/2= 58.5𝑀𝑀𝑀𝑀𝑀𝑀221maxσστ−=abs
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Mohr’s circles for this exampleNormalStress, σ (MPa)σ1=63.5σ3= -51.8τy’z’, τabs max=58.5σ2=26.5Shear (MPa)
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Theories of Failure –Ductile MaterialsMaximum Shear Stress Criterion (Tresca)►For plane stress, we deal with two situations where failure occurs: ►A) Principle stresses have the same sign►B) Principle stresses have the opposite sign221maxσστ−=abs21maxστ=abs=𝜎𝜎𝑌𝑌2=𝜎𝜎𝑌𝑌2( to have failure)( to have failure)Definition: Yielding of the material begins when the absolute maximum shear stress in the material reaches the shear stress that causes the same material to yield when it is subjected only to axial tension.
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Theories of Failure –Ductile MaterialsMaximum Shear Stress Criterion (Tresca)►Failure occurs when principle stresses have the values:►A graph of these equations is shown on the right►Safe zone is inside the envelope (pink region)𝜎𝜎1=𝜎𝜎𝑌𝑌𝜎𝜎2=𝜎𝜎𝑌𝑌If σ1and σ2have the same sign𝜎𝜎1− 𝜎𝜎2=𝜎𝜎𝑌𝑌If σ1and σ2have opposite signs
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ExampleThe solid shaft has a radius of 0.5 in and is made of steel having a yield stress of 36 ksi. Determine if the loadings cause the shaft to fail according to maximum-shear-stress theory (Tresca).222122,xyyxyxτσσσσσσ+−±+=
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Example cont.The solid shaft has a radius of 0.5 in and is made of steel having a yield stress of 36 ksi. Determine if the loadings cause the shaft to fail according to maximum-shear-stress theory (Tresca).
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Theories of Failure –Ductile MaterialsMaximum Distortion Energy Criterion (Von Mises)►Originated from observation that ductile materials stressed hydrostatically (equal principal stresses) exhibited yield strengths larger than expected.►Theorizes that if strain energy is divided into hydrostatic volume changing energy and angular distortion energy, the yielding is primarily affected by the distortion energy.►Theory: Yielding occurs when the distortion strain energyper unit volume reaches the distortion strain energy per unit volume for yieldin simple tension or compression of the same material.
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Theories of Failure –Ductile MaterialsMaximum Distortion Energy Criterion (Von Mises)►Utilize the fact that the material absorbs internal energy.►General stress and strain relationship.►Substituting these equations in the energy gives the strain energy density.StrainStressStrain energy “u”
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Theories of Failure –Ductile MaterialsMaximum Distortion Energy Criterion (Von Mises)►The strain-energy density can be considered the sum of two parts:Energy needed to cause a volume change with no change in shape. This is caused by the application of the average stress.The remaining portion of the stress will distort the element and causes the energy of distortion.(σ1–σave),(σ2–σave) and(σ3–σave)
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Theories of Failure –Ductile MaterialsMaximum Distortion Energy Criterion (Von Mises)►Substitute (σ1–σave),(σ2–σave) and(σ3–σave) into strain energy density equation to yield distortion energy density.►In case of plane stress σ3=0 giving►When we perform a simple tensile test to the yield point, (σ1= σy, σ2= σ3= 0) the distortion energy becomes ►Distortion Energy theory predicts failure when , thus for case of plane stress we have at the point of failure231YdEuσν⋅+=)(31222121σσσσν+−+=EudYdduu)(=2222121Yσσσσσ=+−
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Theories of Failure –Ductile MaterialsMaximum Distortion Energy Criterion (Von Mises)►The last equation is an ellipse seen on the right ►The left-hand side is called equivalent stressor von Mises stress►Von Mises stress has to be less than yield stress of the material to prevent yielding (failure)►In terms of xycomponents, for plane stress𝜎𝜎12− 𝜎𝜎1𝜎𝜎2+𝜎𝜎22=𝜎𝜎𝑌𝑌2𝜎𝜎𝑚𝑚2− 𝜎𝜎𝑚𝑚𝜎𝜎𝑦𝑦+𝜎𝜎𝑦𝑦2+ 3𝜏𝜏𝑚𝑚𝑦𝑦2=𝜎𝜎𝑌𝑌2
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Theories of Failure –Ductile MaterialsMaximum Distortion Energy Criterion (Von Mises)►Figure on the right shows comparison of von Mises stress with experimental data ►Von Mises failure criterion is commonly used in analysis of ductile materials (steel, aluminum)►Maximum Shear Stress theory useful for design situations where higher safety factoris desired
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Theories of Failure –Summary
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Revisit of previous exampleThe solid shaft has a radius of 0.5 in and is made of steel having a yield stress of 36 ksi. Determine if the loadings cause the shaft to fail according to maximum-distortion-energy theory (von Mises).
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Example 2The plate is made of copper which yields at 105 ksi. Using the maximum-shear-stress theory (Tresca), determine the tensile stress σxthat can be applied to the plate if a tensile stress σy= 0.5 σxis also applied. Furthermore, solve the problem using maximum-distortion-energy theory (Von Mises).
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Example 2
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