Elastic-energy and compliance identity
For Hermit’s linear static solve, :func:hm.compliance is the applied-load work:
outputs.py includes distributed terms in its compliance form and adds the
direct point-load vector product separately. Stored elastic energy must therefore
equal one half of that work. This example checks the identity independently for
a pressure load and for total tip point loads split equally over the tip edge.
There is no external reference value: the zero-reference scalar is the largest relative identity residual, while the printed values make the convention visible.
"""Elastic-energy and compliance identity
For Hermit's linear static solve, :func:`hm.compliance` is the applied-load work:
``outputs.py`` includes distributed terms in its compliance form and adds the
direct point-load vector product separately. Stored elastic energy must therefore
equal one half of that work. This example checks the identity independently for
a pressure load and for total tip point loads split equally over the tip edge.
There is no external reference value: the zero-reference scalar is the largest
relative identity residual, while the printed values make the convention visible.
"""
import pathlib
import sys
import csdl_alpha as csdl
import numpy as np
import hermit as hm
sys.path.insert(0, str(pathlib.Path(__file__).parent))
from _geometry import rect_plate # noqa: E402
from _harness import Case, main # noqa: E402
L, W, H, E, NU = 10.0, 2.0, 0.2, 4.32e8, 0.0
def _identity(state):
compliance = float(hm.compliance(state).value[0])
energy = float(hm.elastic_energy(state).value[0])
return compliance, energy, abs(energy / (0.5 * compliance) - 1.0)
def solve_at(n):
"""Largest pressure/point-load energy-identity residual on an ``n x n/2`` mesh."""
mesh = rect_plate(L, W, nx=n, ny=n // 2, cell="quad")
rec = csdl.Recorder(inline=True)
rec.start()
domain = hm.ShellDomain(mesh, element="CG2CG1")
material = hm.isotropic(domain, E=E, nu=NU, thickness=H, density=1.0)
bcs = hm.clamp(domain, where=lambda x: np.isclose(x[0], 0.0))
pressure_state = hm.solve(domain, material, hm.pressure(domain, 2.0), bcs)
xyz = np.asarray(domain.node_coords)
edge = np.flatnonzero(np.isclose(xyz[:, 0], L))
point_loads = None
for k in edge:
term = hm.point_load(domain, at=xyz[k], force=[0.0, 0.0, -1.0 / len(edge)])
point_loads = term if point_loads is None else point_loads + term
point_state = hm.solve(domain, material, point_loads, bcs)
cp, up, ep = _identity(pressure_state)
cf, uf, ef = _identity(point_state)
rec.stop()
print(f" pressure: compliance={cp:.12e}, energy={up:.12e}, residual={ep:.3e}")
print(f" points : compliance={cf:.12e}, energy={uf:.12e}, residual={ef:.3e}")
return max(ep, ef)
CASE = Case(
name="Elastic energy equals half compliance",
quantity="largest relative energy-identity residual for pressure and point loads",
reference=0.0,
# Not a fitted tolerance: the identity is exact in exact arithmetic, so the
# only floor is round-off in the direct solve of a thin-shell operator.
# Measured at ~3e-9, and -- the point -- *identical* for a penalty clamp and a
# strong clamp (2.0e-10 / 3.0e-9 / 2.9e-9 vs 4.0e-10 / 3.1e-9 / 3.0e-9 at
# n = 4 / 8 / 12), which rules out the penalty parameter as the cause.
tolerance=1e-7,
citation="Hermit outputs.py: elastic_energy and compliance conventions",
levels=(4, 8, 12),
quick_level=4,
solve=solve_at,
monotone=False,
notes="residual is round-off in the direct solve, not a discretisation error: "
"it does not shrink under refinement and is the same for penalty and "
"strong clamps",
)
if __name__ == "__main__":
main(CASE)