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Failure Theories Why Do Mechanical Components Fail? Mechanical Components Fail Because the Applied Stresses Exceeds the Material’S Strength (Too Simple)

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Failure Theories Why Do Mechanical Components Fail? Mechanical Components Fail Because the Applied Stresses Exceeds the Material’S Strength (Too Simple) 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 Material Property Application of Stress Uniaxial Ductile Static Biaxial Brittle Dynamic Pure Shear Static Loading Maximum Normal Stress Dynamic Loading Modified Mohr Goodman Yield strength Gerber Maximum shear stress Soderberg Distortion energy 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 Brittle Material Characteristic Yield Stress Ultimate Stress Failure Stress Important 1. Maximum Shear Stress 1. Maximum Normal Theories 2. Maximum Octahedral Stress Shear Stress 2. Modified Mohr. Maximum Ductile Materials Shear Stress Failure occurs when the maximum shear stress in Theory the part exceeds the shear stress in a tensile test specimen (of the same material) at yield. Hence in a tensile test, S τ = y max 2 For a general state of stresses σ− σ S τ = 1 3 = y 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 S y modified to include a factor of safety (n): n = σ1− σ 3 Several cases can be analyzed in plane stress problems: Case 1: σ1≥σ 2 ≥0 In this case σ3=0 σ− σ σ S τ = 1= 3 1 = y max 2 2 2 σ1 ≥ S y Case 2: σ1≥0 ≥σ 3 Yielding condition σ− σ S τ = 1 3 = y max 2 2 σ1− σ 3 ≥S y Distortion Energy Theory distortion of stressed elements.Based on the consideration of angular 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. ε y Modulus of resilience Ur Ur = σ d ε ∫0 If it is in a linear elastic region, 2 1 1 σ⎛ ⎞ σ ⎜ y ⎟ y U r=σy ε y = y σ⎜ ⎟ = 2 2 ⎝EE⎠ 2 1 For general 3-D stresses: =uσ ε() + σ ε + σ ε 21 1 2 2 3 3 Applying Hooke’s Law 1 2 2 2 σ=u σ +() σ + ν −2 σ() σ + σ σ + σ σ 2E 1 2 3 1 2 2 3 3 1 There are two components in σ1+ σ 2 + σ 3 σx+σ y+σ z this energy a mean component σ M = = and deviatoric component. 3 3 σ1 ,D = σ 1−σ M σ2 ,D = σ 2−σ M σ3 ,D = σ 3−σ M The energy due to the mean stress (it gives a volumetric change but not a distortion: 1 σu= σ()2 + σ 2 + ν 2 2 σ −() σ+ σ + σ σ σ Mean 2E MMMMMMMMM 1 1− ν 2 u = 3[]σ2 () 1−σ ν2 = σ+()2 σ +2 22 σ+ σ 2 + σ σ 2 + σ σ Mean 2E M 6E 1 2 3 1 2 2 3 3 1 1+ν u= u −σ u = σ+( 2 σ +2 σ2 − σ − σ σ −) σ σ D Mean 3E 1 2 3 1 2 2 3 3 1 tensile test with the distortion Compare the distortion energy of a energy of the material. 1+ν 1+ν u = S2=σ u = σ+( 2 σ +2 σ2 − σ − σ σ −) σ σ Tensile 3E y D 3E 1 2 3 1 2 2 3 3 1 2 2 2 σ=S y σ +1 σ2 +3 σ1 − 2 σ 2 − σ 3 σ 3 − 1 σ σ 2 2 S=yσ1 + σ3 3 − σ 1 σ Plane Stress Von Mises effective stress : Defined as the uniaxial ensile stress t 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 σVM≥ S y S n = y σVM 2 2 2 σ− σ() + σ −( σ) +() σ − σ6(2 + 2 τ + 2) τ + τ σ = x y y z z x xy yz zx VM 2 2 2 2 2D σ=VM σx + σ y − x σ y σ3 + xy τ Case of Pure Shear σVM=3 xy τ ≥S y S τ =0y . = 577S Max 3 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 Failure occurs when one of the three principal stresses reaches a permissible strength (TS). σ1> σ 2 Failure is predicted to occur when σ1=St and σ2<-Sc σ2 Where St and Sc are the tensile and compressive strength St For a biaxial state of stresses -Sc St σ1 -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 Theory The 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 σ σ 2− 1 ≤1 σCTσ This theory is a modification of the Coulomb-Mohr theory and is the preferred theory for brittle materials. Maximum Normal-Strain Theory Also known as the Saint-Venant’s Theory. Applies only in the elastic range. Failure is predicted to occur if σ1−νσ = 2 ±S y or σ 2 −νσ 1= ±S y Where S is the yield strength. y σ For a biaxial state of stress 2 Sy -Sy Sy σ1 -Sy Maximum Strain-Energy Theory Yielding is predicted to occur when the total strain energy in a given volume is greater than or exceeds the strain energy in the same volume corresponding to the yield strength in tension or compression. 2 The strain energy stored per unit volume (u ) S y s u = during uniaxial loading is s 2⋅ E ε σ⋅ ε⋅ σ In a biaxial state of stress u =1 1+ 2 2 σ 2 2 1 =u σ() +2 σ2 −2 ν ⋅ σ ⋅ ⋅ σ σ 2E 1 2 1 2 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 30SY = MPa σ10x = MPa σ y = − MPa10τ xy = MPa 0 σ1 =10MPa σ 3 = −10MPa τ Max =10MPa Is Al ductile or brittle? Ductile Use either the Maximum Shear Stress Theory (MSST) or the Distortion Theory (DT) -10 10 MSST Theory S 30 30 n = y = = =1.5 σ1 −σ 3 10 − (−10) 20 2 2 2 DT Theory σVM = σ x +σ y −σ xσ y + 3τ xy = 300 =17.32MPa S 30MPa n = y = =1.73 σVM 17.32MPa b. 0.2%C Carbon Steel 65SY = Ksi σ x5= − Ksi σ y = − Ksi35τ xy = Ksi 10 In the plane XY the principal stresses are - 1.973Ksi and -38.03Ksi with a maximum shear stress in the XY plane of 18.03Ksi In any orientation σ1 = 0Ksi σ 2 = −1.973Ksi σ 3 = −38.03Ksi τ =19.01Ksi Max Ductile S 65 MSST Theory n = y = =1.71 σ1 −σ 3 0 − (−38.03) σ = σ 2 +σ 2 −σ σ = 38.03Ksi DT Theory VM 1 3 1 3 S 65Ksi n = y = =1.71 σVM 38.03MPa C. Gray Cast Iron 30S ut = Ksi120 Suc = σ Ksix = − 35Ksi σ y = Ksi τ xy =10Ksi 0 σ1 =10Ksi σ 2 = 0Ksi σ 3 = −35Ksi τ Max = 22.5Ksi -35 10 Brittle Use Maximum Normal Stress Theory (MNST), Internal Friction Theory (IFT), Modified Mohr Theory (MMT) S 30 MNST Theory (tensile) n = ut = = 3.0 σ1 10 S 120 MNST Theory (compression) n = uc = = 3.4 σ 3 35 σ σ IFT 2− 1 ≤1 th σCTσ 0σ 1≥ σ3 0 ≤ quadrant4 _ Suc line _ equation σ=3 −Suc + σ1 Sut 1 σ σ 10 − 35 =1 −3 = −0= . 625 nut S ucS 30 120 n 1= . 6 MMT th 0σ 1≥ σ3 0 ≤ quadrant4 _ Sut 1 SSuc ut σ1 − σ 3 = Suc− S ut SSuc n − ut 1 0= . 54 30 1 ( 120 )( 30 ) 10 − (− 35 = ) n 120− 30(n 120− 30 ) n1= . 84 Example 1 The cantilever tube shown is to be made of 2014 aluminum alloy treated to obtain a specified minimum yield strength of 276MPa.
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