LECTURE DATE:07/12/2009 PREPARED BY:EBRU SEMA KOΕAROΔLU & FARAJ KHALIKOV STATIC FAILURE THEORIES DUCTILE TRESCA CRITERIAN MSST:MAXIMUM SHEAR STRESS THEORY BRITTLE VON-MISES THEORY MDET:MAXIMUM DISTORTION ENERGY THEORY MOHRCOLOUMB THEORY MAXIMUM NORMAL STRESS THEORY(MNST) UNIAXIAL TEST BRITTLE ππ ππ DUCTILE MAXIMUM NORMAL STRESS THEORY For maximum normal stress theory, the failure occurs when one of the principal stresses (π1 , π2 πππ π3 ) equals to the yield strength. 1 π1 > π2 > π3 Failure occurs when either π1 =ππ‘ or π3 = −ππ ,where ππ‘ is strength in tension and ππ is strenght in compression. MOHR-COULOMB THEORY The Coulomb-Mohr theory or internal friction theory assumes that the critical shearing stress is related to internal friction. MAXIMUM DISTORTION ENERGY THEORY(VON-MISES THEORY) The maximum distortion energy theory ,also known as the Von Mises theory, was proposed by M.T.Huber in 1904 and further developed by R.von Mises(1913).In this theory failure by yielding occurs when at any point in the body ,the distortion energy per unit volume in a state of combined stress becomes equal to that associated with yielding in a simple tension test. STRAIN ENERGY Generally strain energy U is obtained by this equation. U= π1 = 1 (π − ππ2 − ππ3 ) πΈ 1 π2 = 1 (−ππ1 + π2 − ππ3 ) πΈ π3 = 1 (−ππ1 − ππ2 + π3 ) πΈ 1 2 σij εij Then, substituting these three equations in to general strain energy equation: 1 1 1 1 2 πΈ 2 πΈ U= π1 (π1 − ππ2 − ππ3 )+ π2 1 1 2 πΈ −ππ1 + π2 − ππ3 + π3 (−ππ1 − ππ2 + π3 ) HYDROSTATIC STRESS The hydrostatic stress(ππ )causes a change in the volume. 2 ππ = π1 + π2 + π3 3 Strain energy associated with the hydrostatic stress: ππ = 1 3 (1 − 2π) 2 [ππ 2 + ππ 2 + ππ 2 − 2π ππ ππ + ππ ππ + ππ ππ = ππ 2πΈ 2 πΈ Then distortional energy ππ = π − ππ From previous equations: ππ = 1+π 3πΈ π1 2 + π2 2 + π3 2 − π1 π2 − π2 π3 − π3 π1 Then yielding will occur at this condition: ππ = 1+π π 3πΈ π ππππ =ππ = ππππ = π1 2 + π2 2 + π3 2 − π1 π2 − π2 π3 − π3 π1 (π1 − π2 )2 + (π2 −π3 )2 + (π3 − π1 )2 2 For plane stress condition π3 = 0 ππππ = π1 2 + π2 2 + π1 π2 If the state stress is in this area then the material will not yield. For pure shear condition 3 π3 = −π1 ππππ = π1 2 + π3 2 − π1 π3 = 3π1 2 = 3ππππ₯ 2 =ππ ππππ₯ = 0.577ππ MAXIMUM SHEAR STRESS THEORY The maximum shearing stress theory is an outgrowth of the experimental observation that a ductile material yields as a result of slip or shear along crystalline planes. Yielding begins whenever the maximum shear stress in a part equals to the maximum shear stress in a tension test specimen that beings to yield. 4 τmax = SYS = yield strength in shear = τmax = SY 2 σ 1 −σ 3 2 Then , ππ = π1 − π3 Elasticity -brittle Materials π1 (ππ₯ ) = ππππ₯ = ππ π where N is safety factor (MNST) -ductile (MSST) π1 −π3 2 (MDET) = π1− π3 = ππππ = πππ (yield strength in shear ) ππ 2 ππ π ππ π Strain Energy: ππ ππ 5 1 π π =π 2 ππ ππ ππ −π πππ π Design Margin=M= ππ EXAMPLE:A circular shaft of tensile strength ππ =350 MPa is subjected to a combined state of loading defined by bending moment M=8 kN.m and torque T=24kN.m.Calculate the required shaft diameter d in order to achieve a factor of safety N=2.Use a) the maximum shearing stress theory(MSST-Tresca) b)the maximum distortion energy theory(MDET –Von Mises) ππ₯π¦ ππ₯ ππ₯π¦ ππ₯ c= π 2 ππ₯ = Then,ππ₯ = ππ πΌ = 32π ππ 3 ππ 2πΌ and ππ₯π¦ = and ππ₯π¦ = ππ π½ = ππ 2π½ 16π ππ 3 We need to find principle stresses: 6 a) For maximum shearing stress theory 16 π1,2 = (π β π2 + π 2 ) ππ 3 π 32 π1 − π2 = π = ππ₯ 2 + 4ππ₯π¦ 2 = 3 π2 + π 2 π ππ Then, substituting the numerical values of M,T, ππ and N: d=113.8mm b) For maximum distortion energy theory ππππ = π (π1 − π2) 2 + π1 2 + π2 2 = ππ₯ 2 + 3ππ₯π¦ 2 = π = 16 π ππ 3 4π2 + 3π 2 Then, substituting the numerical values of M,T, ππ and N: d=109mm FRACTURE MODES OPENING SLIDING TEARING Fracture is defined as the separation of a part into two or more pieces.The mechanisms of brittle fracture are the concern of fracture mechanics ,which is based on a stress analysis in the vicinity of a crack or defect of unknown small radius in a part. 7 MULTIPLE FRACTURES P P Stress Intensity Factor: In the fracture mechanics approach a stress intensity factor,πΎπΌ , is evaluated. This can be thought of as a measure of the effective local stress at the crack root. πΎπΌ = π½π ππ 8 where π=normalstress, π½ = ππππππ‘ππ¦ ππππ‘ππ π€ππππ πππππππ ππ π π€ , π = πππππ πππππ‘π (or half crack length) ,w=member width(or half width of member) Fracture Toughness: In a toughness test of a given material, the stress – intensity factor at which a crack will propagate is measured. This is the critical stress intensity factor , referred to as the fracture toughness and denoted by the symbol πΎπΌπΆ . πΎπΌπΆ N= πΎπΌ (N=factor of safety) EXAMPLE: For a shape with width 12m , crack length 65mm , thickness 30mm and applied loading 50MPa.Find the factor of safety.( πΎπΌπΆ =28.3MPa π, ππ =240MPa) 50MPa 65mm 30mm 240 π1 = 50 12m = 4.8 From the ratio of a/w: π 65 × 10−3 = = 0.0054 π€ 12 9 From table β≅ 1. π½ = πΎπΌ = π ππ=50 π 65 2 πΎπΌ π ππ 10−3 =15.97MPa π πΎπΌπΆ For safety factor: π2 = πΎπΌ = 28.3 15.97 = 1.77 π2 is better than π1 in order to obtain a safe structure; therefore, to have a safe structure, loading should not be used to calculate safety factor. 10 Reference List: 1. Lecture Notes AEE 361, Demirkan ÇÖKER), 2009, “Static Failure Theories” 2. Static Failure theories, Ansel C. UGURAL & Saul K. Fenster, 2007, “Advanced Strength and Applied Elasticity”, fourth edition. 3. Lecture 5 & 6, The University of Jenessee at Martin School of Engineering, 2009, “Machine Design Notes”. 4. Static Failure theories, J. Keith Nisbett, 2009, “Shigley’s Mechanical Engineering Design”, eighth edition. 11