Test of ORNL plasticity theory under biaxial loading

This example verifies the ORNL plasticity theory (Oak Ridge, 1981) model in Abaqus under conditions of plane stress with biaxial stressing.

An exact solution is developed to verify the Abaqus results. The problem involves a state of uniform plane stress, so the geometric model is a single element, constrained to respond uniformly (ORNL – Oak Ridge National Laboratory Constitutive Model).

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Problem description

The virgin material properties are given by:

Young's modulus 207 GPa (30 × 106 lb/in2)
Virgin yield stress 207 MPa (30000 lb/in2)
Virgin work hardening slope 10.3 GPa (1.5 × 106 lb/in2)
10th cycle yield stress 234 MPa (34000 lb/in2)
10th cycle work hardening slope 10.3 GPa (1.5 × 106 lb/in2)

Biaxial loading

The case is set up with the same geometric and virgin material model as in Case 2 in Uniformly loaded, elastic-plastic plate. The plate is first loaded elastically to the virgin yield surface in the x-direction and then loaded into the plastic range in uniaxial tension in the x-direction to a stress, σ0, of 276 MPa (40000 lb/in2). Biaxial loading then follows, with σx and σy prescribed, as shown in Figure 1, so that σy=σ0-σx. This loading is defined by an amplitude curve (Amplitude Curves). Abaqus reads in two files (ORNL2.AMP and ORNL3.AMP) of values, which are calculated in the small program AMP (see ornlbiaxialload_ampdata.f).

“Exact” solution

An “exact” solution is developed by first defining the total strain rates, dεx and dεy, as functions of the stress rates dσx and dσy. The resulting rate equations are then integrated numerically with high accuracy to give a reference solution.

Ziegler's kinematic hardening gives, under isothermal conditions,

dσ=[Del-1dDel:nn:Del]:dε,

where

n=32σ0(σ-13Itrace(σ)-α+13Itrace(α)),

where σ0 is the yield stress and

d=nT:D:n+C,

where C is the slope of the stress versus plastic strain curve under uniaxial loading conditions.

Under plane stress conditions (σz=0) and with σxy=0,

n={nxny}={(σx-σy/2-αx+αy)/2σ0(σy-σx/2-αy+αx)/2σ0}

and

D=E1-ν2[1νν1]=[ABBA],

where A=E/(1-ν2) and B=Eν/(1-ν2).

Hence,

d=nT:D:n+C=Anx2+2Bnxny+Any2+C

and

D:nnT:D={PQ}(PQ)=[P2PQPQQ2],

where P=Anx+Bny, Q=Bnx+Any.

The stress rate-strain rate relation is, therefore,

{dσxdσy}=[A~B~B~C~]{dεxdεy},

where A~=A-P2/d, B~=B-PQ/d, and C~=A-Q2/d.

Inverting this relationship gives the total strain rates as

{dεxdεy}=1B~2-A~C~[-C~B~B~-A~]{dσxdσy}.

The center of the yield surface translates according to Ziegler's kinematic hardening rule, so that

dα=Cdε¯plσ0(σ-α),

where

dε¯pl=nT:D:dεd.

Hence,

dε¯P=1d(PQ){dεxdεy}=(Pdεx+Qdεy)/d,

and the translation rate at the center of the yield surface is given in components by

{dαxdαy}={C(Pdεx+Qdεy)(σx-αx)/σ0dC(Pdεx+Qdεy)(σy-αy)/σ0d}.

Given the values of the variables σx, σy, εx, εy, αx and αy, at the beginning of the increment, together with the prescribed stress increments dσx and dσy, the total strain rate equation and the translation rate equation for the center of the yield surface provide the values of dεx, dεy, dαx, and dαy.

A small program for calculating the required variables is given in ornlbiaxialload_exact.f. The main program provides prescribed stress increments Δ σ x and Δ σ y equal to those used in the finite element analysis. Each of these increments is then split into 1000 subincrements, and the total strain rate equation and the translation rate equation for the center of the yield surface are integrated over each subincrement to provide virtually exact values of Δ ε x , Δ ε y , Δ α x , and Δ α y , corresponding to the prescribed values of Δ σ x and Δ σ y used in the analysis. In each of the subincrements, a test is made to determine if α : d ε p l < 0. When this test is satisfied, the yield surface is expanded from the virgin properties to the 10th cycle properties so that σ 0 is increased from its virgin value of 207 MPa (30000 lb/in2) to its 10th cycle value of 234 MPa (34000 lb/in2). This value of σ 0 is used in each subincrement following the initial satisfaction of the test α : d ε p l < 0 , according to the ORNL plasticity algorithm.

Results and discussion

The loading path in stress space is shown in Figure 1. When the stress contacts point A, the yield surface starts to translate so that at point B the yield surface occupies the position shown by the dashed curve. At point B the stress path changes direction, and elastic loading along path BC occurs. At point C the stress point pierces the yield surface, and since α:dεpl<0, the ORNL algorithm prescribes an expansion of the yield surface from the virgin properties to the 10th cycle properties. The expanded yield surface is indicated in Figure 1 by the dashed and dotted curve. Continuing loading along path CD produces an elastic response since point C lies inside the 10th cycle yield surface. At point D the stress point contacts the expanded yield surface, and active plastic yielding occurs along path DE. A comparison between the “exact” results and the finite element results in Table 1 and Table 2 shows very close agreement.

References

  1. Nuclear Standard NE F9–5T, Guidelines and Procedures for Design of Class 1 Elevated Temperature Nuclear System Components, USDOE Technical Information Center, Oak Ridge, Tennessee, March 1981.

Tables

Table 1. Comparison of “exact” and numerical results for biaxial plate using ORNL plasticity theory—stresses and strains in the x-direction.
Numerical solution "Exact" solution Increment
σx, MPa (103 lb/in2) εx(%) σx, MPa (103 lb/in2) εx(%) type
206.84 (30.00) 0.1000 206.84 (30.00) 0.1000 Elastic
224.08 (32.50) 0.2656 224.08 (32.50) 0.2655 Plastic
241.32 (35.00) 0.4312 241.32 (35.00) 0.4311 Plastic
258.55 (37.50) 0.5968 258.55 (37.50) 0.5968 Plastic
275.79 (40.00) 0.7624 275.79 (40.00) 0.7624 Plastic
258.55 (37.50) 0.7516 258.55 (37.50) 0.7516 Elastic
68.95 (10.00)* 0.6324 68.95 (10.00) 0.6324 Elastic
51.71 (7.50) 0.6216 51.71 (7.50) 0.6214 Elastic
34.48 (5.00) 0.4702 34.47 (5.00) 0.4744 Plastic
17.24 (2.50) 0.2999 17.24 (2.50) 0.3076 Plastic
3.65 (0.53) 0.1191 0.00 (0.00) 0.1292 Plastic
−17.24 (−2.50) −0.0709 −17.24 (−2.50) −0.0591 Plastic
−34.48 (−5.00) −0.2687 −34.47 (−5.00) −0.2558 Plastic
−51.71 (−7.50) −0.4731 −51.71 (−7.50) −0.4598 Plastic
*The yield surface is pierced at σx= 68.95 MPa (10000 lb/in2), σy= 206.84 MPa (30000 lb/in2. The next increment is elastic due to the expansion of the yield surface.
Table 2. Comparison of “exact” and numerical results for biaxial plate using ORNL plasticity theory—stresses and strains in the y-direction.
Numerical solution "Exact" solution Increment
σy, MPa (103 lb/in2) εy(%) σy, MPa (103 lb/in2) εy(%) type
0.00 (0.00) −0.0300 0.00 (0.00) −0.0300 Elastic
0.04 (0.01) −0.1111 0.00 (0.00) −0.1108 Plastic
0.04 (0.01) −0.1923 0.00 (0.00) −0.1922 Plastic
0.04 (0.01) −0.2734 0.00 (0.00) −0.2734 Plastic
0.04 (0.01) −0.3545 0.00 (0.00) −0.3545 Plastic
17.24 (2.50) −0.3437 17.24 (2.50) −0.3437 Elastic
206.84 (30.00)* −0.2245 206.84 (30.00) −0.2245 Elastic
224.08 (32.50) −0.2137 224.08 (32.50) −0.2135 Elastic
241.32 (35.00) 0.0313 241.32 (35.00) 0.0319 Plastic
258.55 (37.50) 0.2914 258.55 (37.50) 0.2930 Plastic
275.79 (40.00) 0.5537 275.79 (40.00) 0.5564 Plastic
293.03 (42.50) 0.8172 293.03 (42.50) 0.8211 Plastic
310.26 (45.00) 1.0811 310.26 (45.00) 1.0860 Plastic
327.51 (47.50) 1.3450 327.50 (47.50) 1.3508 Plastic
*The yield surface is pierced at σx= 68.95 MPa (10000 lb/in2), σy= 206.84 MPa (30000 lb/in2. The next increment is elastic due to the expansion of the yield surface.

Figures

Figure 1. Biaxial stress path in σxσy plane for ORNL plasticity solution.