Failure Theories Predict capability of materials to withstand the infinite combination of non-standard loads Failure: Behavior of a member that renders it unsuitable for its intended function. Static loading (no impact, fatigue nor surface wear) How much distortion is too much ? World War II tanker broken in two by a brittle fracture, despite the normal ductility of the used steel. Impact loading, Sharp notches, Low temperatures, Can promote brittle fracture of usually ductile material Max. Normal-Stress-Theory (W.J.M. Rankine 1802-1872, engl. scientist & educator) simplest Failure occurs, when greatest tensile stress exceeds uniaxial tensile strength. +σ2 Sut +τ S +σ uc Sut +σ1 Suc Sut No Failure occurs within these bounds Suc No Failure occurs within this area Principle Mohr circles σ1 –σ2 plot Correlates well for brittle fracture Max. Shear-Stress-Theory (C.A. Coulomb 1736-1806, French scientist) oldest Also called: Tresca Theory or Guest’s Law Failure occurs, when maximum shear stress exceeds shear strength in uniaxial tension test. +σ 2 Sy +τ No Failure Sy Syt within + these bounds +σ σ1 σ1 − σ2 Sy = τmax = No Failure 2 2 within this area Principle Mohr circles σ1 –σ2 plot Correlates well for ductile yielding Max. Distortion-Energy-Theory (Maxwell 1856 Engl., Hueber 1904 Pol., best Mises 1913 & Hencky 1925 Ger./US) Also called: Max.-Octahedral-Shear Theory Failure occurs, when distortion-energy in unit volume (arbitrary load condition) equals distortion-energy in same volume for uniaxial yielding. P Energy stored in a spring U = ⋅ y 2 P P2 with spring rate k = U = y 2 ⋅ k 2 A ⋅E P ⋅L axially loaded bar: k = U = L 2A ⋅E U P2 per unit volume: u = u = ()V = A ⋅L 2A 2 ⋅E P σ2 stress: σ = u = A 2E δ ε ⋅ σ stress-strain rel.: E = u = ε 2 ε ⋅ σ ε ⋅ σ ε ⋅ σ 3-D: u = 1 1 + 2 2 + 3 3 2 2 2 Max. Distortion-Energy-Theory ε1 ⋅ σ1 ε2 ⋅ σ2 ε3 ⋅ σ3 δ νδ νδ u = + + ε = 1 − 2 − 3 σ 2 2 2 1 E E E δ νδ νδ with 3-D stress-strain rel. ε = 2 − 1 − 3 2 E E E 1 2 2 2 δ 3 νδ 1 νδ 2 uσ = []σ1 + σ2 + σ3 − 2ν()σ1σ2 + σ2σ3 + σ3σ1 ε 3 = − − 2E E E E actual average difference σ1 + σ2 + σ3 with 3-D averaged stress: σav = 2 3 3σ u = av []1− 2ν av 2E Distortion energy ud = uσ − uav ⎡ 2 2 2 ⎤ 1+ ν ()σ1 − σ2 + ()σ2 − σ3 + (σ3 − σ1) ud = ⎢ ⎥ 3E ⎣ 2 ⎦ Simple tension test: σ1 = Sy 1+ ν u = S 2 d 3E y ()σ − σ 2 + ()()σ − σ 2 + σ − σ 2 S > σ = 1 2 2 3 3 1 y e 2 Max. Distortion-Energy-Theory 3D: ()σ − σ 2 + ()()σ − σ 2 + σ − σ 2 S > σ = 1 2 2 3 3 1 y e 2 2D: (Juvinall p254) Homework Read Chapters 6.5 - 6.8 (6.0-6.1) Problem P6.28 (b) only Due: Monday 03/28/05 Exam Monday, 11 April Failure Theories Example: σ1 = 35 ksi Steel Sy=100 ksi σ2 = -25 ksi σ1= Sy= σ2= Load Point Shear Diagonal σmax<Sy 1.7 1.9 Load Line Max. Normal Sress Th. h. σ −σ <S T 1 2 y gy er 2.9 Max. Shear Stress nTh. En io rt to is . D ax M 2 2 σ1 +σ2 −σ1σ2 <Sy Sy=.