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) ⎡⎤200 0 0 ⎢⎥ σ=⎢⎥0100−30MPa ⎣⎦⎢⎥03−−050 • The principal stresses are: σ=12200; σ=105.77; σ3=−55.77; • Maximum principal stress is 200MPa • Factor of safety FS=500/200=2.5 Maximum shear stress Criterion (Tresca) Y • Yield function f =−σe 2 σσ− 200 + 55.78 σ ==13 =127.89 MPa • Maximum shear stress e 22 • Shear stress for uniaxial tension Y = 250 MPa f =−127.89 250 <0 2 • Factor of safety FS=250/127.89=1.95 Distortional Energy Density Criterion (von Mises) • Yield function f = 1 σ −σ 22+ σ −σ + σ −σ 2−Y =224 −500 2 ()12 ()23 (31) • Factor of safety FS=500/224=2.2 • Why are all the results so close? σ=12200; σ=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 fc=−σ13σσ+()1+σ3sinφ−2 cosφ σ1≥σ2≥σ3 • What would it look like for plane stress? From test results • For tension test 2cosφ σσ==Yf,0σ=,=0⇒Y= 12TT3 1s+ inφ • For compression test 2cosφ σσ=−Yf,0=σ= ,=0⇒ Y= 32CC1 1s− inφ •Get ⎛⎞ YYTT π −1 YT c ==φ −tan ⎜⎟ 22YYCC⎝⎠ 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! σ σ f = 1 −−3 1 YYTC Drucker-Prager yield criterion (Dan Drucker, 1918-2001, William Prager 1903-1980) • Generalization of Von Mises to account for hydrostatic pressure fI= α 12+−JK • Gets outer bound of Coulomb Moher with 2sinφ 6c cosφ α ==, K 33()−−sinφ 3()3 sinφ • Inner bound with 2sinφ 6c cosφ α ==, K 33()++sinφ 3()3 sinφ 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.