Metal Plasticity - Initial Yield Stress

[IMOTEC]

I am trying to do ABAQUS example problems. But many problems used an initial yield stress while entering material property. Passage from the example is below.

"The blank is made of a steel alloy that is assumed to satisfy the Ramberg-Osgood relation for true stress and logarithmic strain,
ε=(σ/K)^(1/n)
with a reference stress value (K) of 763 MPa and a work-hardening exponent of 0.245. Isotropic elasticity is assumed, with a Young's modulus of 211 GPa and a Poisson's ratio of 0.3. An initial yield stress of 200 MPa is obtained with these data. The stress-strain behavior is defined by piecewise linear segments matching the Ramberg-Osgood curve up to a total (logarithmic) strain level of 140%, with von Mises yield and isotropic hardening."

My question is how can i calculate the initial yield stress with these data?

 

[IMOTEC]

Nobody can help me!!!

 

[corus]

Perhaps the initial yield stress is defined as that leaving 1 or 2% plastic strain, as per normal material definitions. Check the formula to see if that's what it gives.

 

[IMOTEC]

But another examples:

"The blank is made of aluminum-killed steel, which is assumed to satisfy the Ramberg-Osgood relation between true stress and logarithmic strain,
ε=(σ/K)^(1/n)
with a reference stress value (K) of 513 MPa and a work-hardening exponent of 0.223. Isotropic elasticity is assumed, with a Young's modulus of 211 GPa and a Poisson's ratio of 0.3. An initial yield stress of 91.3 MPa is obtained from these data. The stress-strain behavior is defined by piecewise linear segments matching the Ramberg-Osgood curve up to a total (logarithmic) strain level of 107%, with Mises yield, isotropic hardening, and no rate dependence."

If this example we take plastic strain 0.01, then the initial yield stress is 183 MPa nearly.And there is conflict. Because last example, they use ε as plastic strain. But previous example they use ε as true strain and then calculate plastic strain.