Example comapring failure criteria Stress analysis of a spacecraft structural member gives the state of stress as shown below. If the part is made from an alloy with Y = 500 MPa, check yielding according to Rankine, Tresca and von Mises criteria. What is its safety factor for each criterion? Maximum principal stress Criterion (Rankine) 0 ⎤ ⎡ 200 0 σ = ⎢⎢ 0 100 −30 ⎥⎥ MPa ⎢⎣ 0 −30 −50 ⎥⎦ • The principal stresses are: σ1 = 200; σ 2 = 105.77; σ3 = −55.77; • Maximum principal stress is 200MPa • Factor of safety FS=500/200=2.5 Maximum shear stress Criterion (Tresca) • Yield function f = σe − Y 2 • Maximum shear stress σe = σ1 − σ 3 2 = 200 + 55.78 = 127.89 MPa 2 • Shear stress for uniaxial tension Y = 250 MPa 2 f = 127.89 − 250 < 0 • Factor of safety FS=250/127.89=1.95 Distortional Energy Density Criterion (von Mises) • Yield function f = 1 2 ( σ1 − σ2 ) 2 + ( σ2 − σ3 ) + ( σ3 − σ1 ) − Y = 224 − 500 2 2 • Factor of safety FS=500/224=2.2 • Why are all the results so close? σ1 = 200; σ 2 = 105.77; σ3 = −55.77; Other yield criteria • For isotropic materials there is usually substantial difference between yield in tension and compression. Why? • For orthotropic materials there is also differences between the behavior along the three principal directions and between shear and normal stresses • Fortunately, for most applications we have at least transverse isotropy 4.5.1 Mohr-Coulomb yield criterion (Charles Augustin de Coulomb, 1736-1806, Christian Otto Mohr, 1835-1918) • Yield of materials like rock and concrete does change with hydrostatic pressure • Yield function written in terms of cohesion c and internal friction angle φ as f = σ 1 − σ 3 + (σ 1 + σ 3 ) sin φ − 2c cos φ σ1 ≥ σ 2 ≥ σ 3 • What would it look like for plane stress? From test results • For tension test σ 1 = YT , σ 2 = σ 3 = 0, f =0 ⇒ YT = 2 cos φ 1 + sin φ • For compression test σ 3 = −YC , σ 2 = σ 1 = 0, f = 0 ⇒ • Get c= YT 2 YT YC φ= ⎛ Y ⎞ − tan −1 ⎜⎜ T ⎟⎟ 2 ⎝ YC ⎠ π 2 cos φ YC = 1 − sin φ Simpler treatment for plane stress • If both stresses are of the same sign, then we compare the corresponding Y. If they are of different signs • Check if the two formulations are the same! f = σ1 σ 3 YT − YC −1 Drucker-Prager yield criterion (Dan Drucker, 1918-2001, William Prager 1903-1980) • Generalization of Von Mises to account for hydrostatic pressure f = α I1 + J2 − K • Gets outer bound of Coulomb Moher with 2sin φ 6c cos φ α= , K= 3 ( 3 − sin φ ) 3 ( 3 − sin φ ) • Inner bound with 2sin φ 6c cos φ α= , K= 3 ( 3 + sin φ ) 3 ( 3 + sin φ ) Pictorially Reading assignment Sections 4.6: Question: For what kind of beam cross-section there will not be much difference between the moment causing initial yield and the fully plastic moment? Source: www.library.veryhelpful.co.uk/ Page11.htm
