Elastic-plastic thick-walled cylinder (inner pressure)

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This example evaluates the residual stresses in the circumferential direction of a thick-walled elastic-plastic cylinder after applying an inner pressure.

import felupe as fem
import numpy as np
import matplotlib.pyplot as plt

mesh = fem.Rectangle(a=(0, 20), b=(100, 60), n=(81, 41))
dy = mesh.y.max() - mesh.y.min()
mesh.points[:, 1] = ((mesh.y - mesh.y.min()) / dy) ** 2 * dy + mesh.y.min()

region = fem.RegionQuad(mesh)
mask = np.logical_and(mesh.y == mesh.y.min(), mesh.x <= mesh.x.max() * 1.0)
hole = fem.RegionQuadBoundary(mesh, mask=mask, ensure_3d=True)
field = fem.FieldContainer([fem.FieldAxisymmetric(region, dim=2)])
edge = fem.FieldContainer([fem.FieldAxisymmetric(hole, dim=2)])

boundaries = fem.dof.symmetry(field[0], axes=(True, False))
umat = fem.LinearElasticPlasticIsotropicHardening(E=2.1e5, nu=0.3, sy=355, K=0)
ax = umat.plot(ux=fem.math.linsteps([1, 1.01, 1.002, 1.005], num=16), ps=None, bx=None)

solid = fem.SolidBody(umat=umat, field=field)
pressure = fem.SolidBodyPressure(edge)
table = fem.math.linsteps([0, 355, 0], num=2)

ramp = {pressure: table}
step = fem.Step(items=[solid, pressure], ramp=ramp, boundaries=boundaries)
job = fem.Job(steps=[step]).evaluate()

fig, ax = plt.subplots(figsize=(12, 5), ncols=2, width_ratios=[2, 1])

solid.imshow("Stress", component=2, ax=ax[0])
stress = fem.topoints(solid.evaluate.stress(), region)

middle = mesh.x == 0
ax[1].plot(stress[middle, 2, 2], mesh.y[middle], "-", lw=2, label=r"$\sigma_\theta$")

ax[1].set_ylabel(r"Radial Position $x_r$ in mm")
ax[1].set_xlabel(r"Stress $\sigma(z=0)$ in MPa")
ax[1].legend()

[adtzlr]

A linear elastic - (ideal) plastic material formulation is used here.