{
 "title": "Bauschinger effect of linear kinematic hardening: tension, then back to zero strain",
 "kind": "theory",
 "source": "J2 plasticity with Prager's linear kinematic hardening in uniaxial stress (W. Prager, Ing.-Arch. 1955; Simo and Hughes, Computational Inelasticity, Springer 1998, §2.3): from σ_a = σ_y + E_t (ε_a − σ_y/E) the material unloads elastically by 2σ_y and then hardens with E_t, so at zero strain σ = σ_a − 2σ_y − E_t (ε_a − 2σ_y/E)",
 "quantities": [
  {
   "id": "residual_stress",
   "quantity": "axial stress back at zero strain after a pull to ε_a = 0.01",
   "measure": {
    "command": "solve.probe",
    "params": {
     "step": "Pull",
     "elements": [
      1
     ]
    },
    "pointer": "/elements/0/stress/0/0"
   },
   "value": -247.52475247524754,
   "tolerance": {
    "relative": 1e-06
   },
   "reason": "a homogeneous small-strain state the element represents exactly, with increments that end at the load path's turn; isotropic hardening would give −281.835 (it yields again only at −σ_a)"
  },
  {
   "id": "plastic_strain",
   "quantity": "accumulated equivalent plastic strain",
   "measure": {
    "command": "solve.probe",
    "params": {
     "step": "Pull",
     "elements": [
      1
     ]
    },
    "pointer": "/elements/0/plastic_strain/0"
   },
   "value": 0.016089108910891083,
   "tolerance": {
    "relative": 1e-06
   },
   "reason": "(σ_a − σ_y)/H forward plus E/(E + H) (ε_a − 2σ_y/E) in reverse"
  }
 ]
}