Yielding Under Multi-axial Stress and Elastic-Plastic Stress-Strain Relations 2.002 Mechanics and Materials II March 29, 2004Uniaxial tension/compression: initial linear elastic response, as axial stress, σ, is increased up to the uniaxial “yield condition”: Suppose that, at some location in a body made of the same material, the state of stress is multi-axial, with cartesian components σij; QUESTION: Will plastic deformation occur under this state of stress?Approach: we need to define a non-negative scalar, stress-valued function of [all] the stress components, such that it can consistently generalize the uniaxial yield criterion, |σ | < σy Observation # 1: pressure insensitivity of uniaxial yielding Suppose that a uniaxial test is performed under fixed superposed hydrostatic pressure, p, so the cartesian stress components are Plastic deformation is observed to commence when | σ | = σy, essentially independent of the value of p This suggests that yielding is ~ independent of the mean normal stress given by Σ= (σ11 + σ22 + σ33)/3Recall the stress deviator tensor, whose components are given by Clearly, the stress deviator tensor is independent of the mean normal stress The Mises equivalent tensile stress is defined, for any state of stress, σij, in terms of the components of the corresponding stress deviator tensor by The yield condition for general multiaxial states of stress can be expressed asIs our general criterion for multiaxial yielding consistent with our previously-established uniaxial yield criterion |σ | = σy? Mises stress measure: Uniaxial stress: Stress deviator: Mises yield specializes to the uniaxial yield Condition under uniaxial stressEquivalent Expressions for Mises Equivalent Tensile Stress deviator components: In terms of stress In terms of stress components: In terms of principal stress values:EXAMPLE: Combined tension and torsion of a thin-walled tube: Stress components and relation to loads and tube geometry: Stress deviator components: Evaluate Mises stress and compare to Uniaxial yield strength The Mises yield condition for this stress state can be represented as an ellipse in a 2D space whose axes are ``σ’’ and ``τ’’EXAMPLE (continued) A tube of wall thickness t = 3 mm and mean radius R = 30 mm is made of a material having tensile yield strength σy = 500 MPa and is preloaded to an axial force F = 200 kN What is the maximum torque that can be applied without causing yield in the tube? rearrange Mises yield: load/stress/geometry: algebra… numerical values & units ANSWER:EXAMPLE A tube of axial length L= 200 mm, wall thickness t = 3 mm and mean radius R = 30 mm is made of a material having tensile yield strength σy = 500 MPa and is preloaded to an axial force F = 200 kN The torque is increased to its initial yield value (previously-determined value: Mt =3.46 kNm) with F held constant. Then, with dF=0, the torque is further incremented by an amount dMt = 0.1 kNm. The Young’s modulus is E=208 GPa, ν=0.3, and the initial value of the plastic hardenig moduus is h = 3 GPa QUESTION: •Evaluate the stress increment, d σij •Evaluate the strain increment, d εij•Evaluate the increment in length of the tube, d L•Evaluate the increment in the end-to-end rotation of the•Two ends of the tube, d φStress increment: Matrix form:Strain increment: Equivalent plastic strain increment: Increment in Mises equivalent stress: Formal definition: alternate derivation:Strain increment: Axial strain increment: Shear strain increment: Radial and hoop strain increments: Note: dε(e) = 0; σ(dev)= - σ(dev)/2; etc.rr rr zzStrain/displacement relations: Axial tube elongation increment: Tube end-to-end rotation