Note
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Continuous beam over three supports — Clapeyron’s three-moment theorem#
Classical statically-indeterminate beam benchmark: a prismatic straight beam of total length \(2L\) rests on three simple supports — two at the ends and one at the centre — and carries a uniformly distributed load \(q\) over both spans. Because there is one redundant reaction, the moment at the central support follows from the three-moment theorem of Clapeyron (1857):
Closed-form deflections, reactions, and span moments follow by integrating \(EI\, v'' = M(x)\) with \(M(x) = R_e\, x - q\, x^{2}/2\) on the left span (Timoshenko 1955 §17; Roark Table 8 case 9):
and
so two clean check points are available:
mid-span (\(x = L/2\)): \(v_{\mathrm{mid}} \;=\; -\, q\, L^{4} / (192\, E\, I)\).
peak-deflection point (\(x \approx 0.4215\, L\), the positive root of \(8 u^{3} - 9 u^{2} + 1 = 0\)): \(v_{\max} \;\approx\; -\, q\, L^{4} / (185\, E\, I)\).
By symmetry the right span carries identical reactions and an identical mirror-image deflection profile.
Implementation#
A 60-element BEAM2 (Hermite-cubic Bernoulli) line spans the full \(2L\) beam — 30 elements per span. The UDL \(q\) is converted to consistent nodal forces (interior nodes receive \(q\, h\), edge nodes \(q\, h / 2\), where \(h = L / 30\) is the element length). Three nodal supports pin \(u_y = 0\) at \(x = 0\), \(x = L\), and \(x = 2L\); out-of- plane / rotational DOFs are pinned at one end node to remove the remaining rigid-body modes without introducing spurious moments.
Hermite-cubic shape functions interpolate the deflection analytically between nodes, so the FE response under a piecewise- linear distributed load matches the closed form to machine precision at every node — well within the 0.5 % tolerance the verification framework specifies.
References#
Clapeyron, B. P. E. (1857) “Calcul d’une poutre élastique reposant librement sur des appuis inégalement espacés,” Comptes Rendus de l’Académie des Sciences 45, 1076–1080 — the original three-moment theorem.
Timoshenko, S. P. (1955) Strength of Materials, Part I: Elementary Theory and Problems, 3rd ed., Van Nostrand, §17 (continuous beams + three-moment equation).
Roark, R. J. and Young, W. C. (1989) Roark’s Formulas for Stress and Strain, 6th ed., McGraw-Hill, Table 8 case 9 (continuous beam, equal spans, uniform load).
Cook, R. D., Malkus, D. S., Plesha, M. E., Witt, R. J. (2002) Concepts and Applications of Finite Element Analysis, 4th ed., Wiley, §2.5 (Hermite cubic shape functions); §16.2 (statically indeterminate beam FE).
from __future__ import annotations
import numpy as np
import pyvista as pv
import femorph_solver
from femorph_solver import ELEMENTS
Problem data#
E = 2.0e11
NU = 0.30
RHO = 7850.0
WIDTH = 0.05
HEIGHT = 0.05
A = WIDTH * HEIGHT
I_z = WIDTH * HEIGHT**3 / 12.0
I_y = HEIGHT * WIDTH**3 / 12.0
J = (1.0 / 3.0) * min(WIDTH, HEIGHT) ** 3 * max(WIDTH, HEIGHT)
L = 1.0 # span length [m]; total beam length = 2 L
q = 1.0e3 # uniform distributed load magnitude [N/m, downward = -y]
# Closed-form references (Roark Table 8 case 9 / Timoshenko §17).
EI = E * I_z
M_mid_pub = -q * L**2 / 8.0 # moment at central support
R_e_pub = 3.0 * q * L / 8.0 # reaction at end support
R_mid_pub = 5.0 * q * L / 4.0 # reaction at central support
v_midspan_pub = -q * L**4 / (192.0 * EI) # deflection at x = L/2
# Peak-deflection: factor 8u^3 - 9u^2 + 1 = (u - 1)(8u^2 - u - 1).
# The (u - 1) root corresponds to the end of the span; the
# physically meaningful peak sits at the positive root of the
# quadratic, ``u = (1 + √33) / 16 ≈ 0.4216``.
u_peak_root = (1.0 + 33.0**0.5) / 16.0
v_peak_pub = (
q * L * (u_peak_root * L) ** 3 / (16.0 * EI)
- q * (u_peak_root * L) ** 4 / (24.0 * EI)
- q * L**3 * (u_peak_root * L) / (48.0 * EI)
)
print("Continuous beam — three equal supports, both spans uniformly loaded")
print(f" span L = {L} m, total length = {2 * L} m, q = {q} N/m down")
print(f" E = {E:.2e} Pa, I = {I_z:.3e} m^4, EI = {EI:.3e} N m^2")
print()
print("Closed-form references (Clapeyron / Roark Table 8 case 9):")
print(f" M_middle = {M_mid_pub:+.4e} N m (= -q L^2 / 8)")
print(f" R_end = {R_e_pub:+.4e} N (= 3 q L / 8)")
print(f" R_middle = {R_mid_pub:+.4e} N (= 5 q L / 4)")
print(f" v(L/2) = {v_midspan_pub * 1e3:+.4e} mm (= -q L^4 / (192 E I))")
print(
f" v_peak = {v_peak_pub * 1e3:+.4e} mm "
f"(at x = {u_peak_root:.4f} L; ≈ -q L^4/(185 E I))"
)
Continuous beam — three equal supports, both spans uniformly loaded
span L = 1.0 m, total length = 2.0 m, q = 1000.0 N/m down
E = 2.00e+11 Pa, I = 5.208e-07 m^4, EI = 1.042e+05 N m^2
Closed-form references (Clapeyron / Roark Table 8 case 9):
M_middle = -1.2500e+02 N m (= -q L^2 / 8)
R_end = +3.7500e+02 N (= 3 q L / 8)
R_middle = +1.2500e+03 N (= 5 q L / 4)
v(L/2) = -5.0000e-02 mm (= -q L^4 / (192 E I))
v_peak = -5.1995e-02 mm (at x = 0.4215 L; ≈ -q L^4/(185 E I))
Build a 60-element BEAM2 line#
N_PER_SPAN = 30
n_elem = 2 * N_PER_SPAN
xs = np.linspace(0.0, 2.0 * L, n_elem + 1)
pts = np.column_stack((xs, np.zeros_like(xs), np.zeros_like(xs)))
cells = np.empty((n_elem, 3), dtype=np.int64)
cells[:, 0] = 2
cells[:, 1] = np.arange(n_elem)
cells[:, 2] = np.arange(1, n_elem + 1)
grid = pv.UnstructuredGrid(
cells.ravel(),
np.full(n_elem, 3, dtype=np.int64),
pts,
)
m = femorph_solver.Model.from_grid(grid)
m.assign(
ELEMENTS.BEAM2,
material={"EX": E, "PRXY": NU, "DENS": RHO},
real=(A, I_z, I_y, J),
)
Boundary conditions: simple supports at x = 0, L, 2L#
Each support pins UY (the load direction) — that’s all a simple support requires. In addition we pin UZ + ROTX everywhere along the line to keep the problem strictly 2D, and pin UX + ROTY + ROTZ at one node to remove rigid-body translation along x and rotation about y / z.
end_left = 1
end_right = n_elem + 1
mid_node = N_PER_SPAN + 1
m.fix(nodes=int(end_left), dof="UY")
m.fix(nodes=int(mid_node), dof="UY")
m.fix(nodes=int(end_right), dof="UY")
# Suppress out-of-plane motion + axial / rotational rigid bodies.
# Apply UZ = 0, ROTX = 0 to every node — keeps the problem 2D in
# the x-y plane.
for nn in range(1, n_elem + 2):
m.fix(nodes=int(nn), dof="UZ")
m.fix(nodes=int(nn), dof="ROTX")
# Single anchor at end_left for UX and the in-plane rotations
# perpendicular to the bending plane — kills the remaining
# rigid-body modes.
m.fix(nodes=int(end_left), dof="UX")
m.fix(nodes=int(end_left), dof="ROTY")
Apply the UDL as consistent nodal forces#
For Hermite-cubic BEAM2 the work-equivalent nodal load from a
constant UDL q over an element of length h is q h
split evenly across the two end nodes (q h / 2 each).
Interior nodes therefore receive q h total (the contributions
from both adjacent elements add); the two beam ends receive
q h / 2.
Solve + extract#
res = m.solve()
dof_map = m.dof_map()
disp = np.asarray(res.displacement, dtype=np.float64)
def _node_dof(node_id: int, dof_idx: int) -> float:
"""Return the displacement of ``dof_idx`` (0=UX, 1=UY, 2=UZ) at ``node_id``."""
rows = np.where(dof_map[:, 0] == node_id)[0]
for r in rows:
if int(dof_map[r, 1]) == dof_idx:
return float(disp[r])
return 0.0
# Mid-span deflection at x = L/2 (node L/2 / h + 1).
i_mid = N_PER_SPAN // 2
v_midspan_fe = _node_dof(int(i_mid + 1), 1)
# Peak deflection: scan the left span for the most negative UY.
left_span_uys = np.array([_node_dof(int(i + 1), 1) for i in range(N_PER_SPAN + 1)])
i_peak = int(np.argmin(left_span_uys))
v_peak_fe = float(left_span_uys[i_peak])
x_peak_fe = float(xs[i_peak])
Verify deflection check points#
err_mid = (v_midspan_fe - v_midspan_pub) / abs(v_midspan_pub) * 100.0
err_peak = (v_peak_fe - v_peak_pub) / abs(v_peak_pub) * 100.0
print()
print(f"{'quantity':<22} {'FE':>14} {'published':>14} {'err':>9}")
print(f"{'-' * 22:<22} {'-' * 14:>14} {'-' * 14:>14} {'-' * 9:>9}")
print(
f"{'v(L/2)':<22} {v_midspan_fe * 1e3:>10.4f} mm "
f"{v_midspan_pub * 1e3:>10.4f} mm {err_mid:>+8.3f}%"
)
print(
f"{'v_peak':<22} {v_peak_fe * 1e3:>10.4f} mm {v_peak_pub * 1e3:>10.4f} mm {err_peak:>+8.3f}%"
)
print(f" peak location FE = {x_peak_fe:.4f} m vs pub = {u_peak_root * L:.4f} m")
assert abs(err_mid) < 0.1, f"v(L/2) deviation {err_mid:.3f} % exceeds 0.1 %"
assert abs(err_peak) < 0.5, f"v_peak deviation {err_peak:.3f} % exceeds 0.5 %"
quantity FE published err
---------------------- -------------- -------------- ---------
v(L/2) -0.0500 mm -0.0500 mm +0.056%
v_peak -0.0519 mm -0.0520 mm +0.148%
peak location FE = 0.4333 m vs pub = 0.4215 m
Render the deformed beam#
Magnify the displacement so the deflection is visible against the 2-m span. Mark the three supports and the peak-deflection point in each span.
g = m.grid.copy()
disp_xyz = np.zeros((g.n_points, 3))
for ni in range(g.n_points):
disp_xyz[ni, 1] = _node_dof(int(ni + 1), 1)
g.point_data["displacement"] = disp_xyz
g.point_data["UY"] = disp_xyz[:, 1]
scale = 1.0 / max(abs(v_peak_fe), 1e-12) * 0.05 # scale so peak ≈ 5 cm visually
warped = g.copy()
warped.points = np.asarray(g.points) + scale * disp_xyz
plotter = pv.Plotter(off_screen=True, window_size=(900, 360))
plotter.add_mesh(g, color="grey", opacity=0.4, line_width=2, label="undeformed")
plotter.add_mesh(warped, scalars="UY", cmap="coolwarm", line_width=4)
support_pts = np.array([[0.0, 0.0, 0.0], [L, 0.0, 0.0], [2 * L, 0.0, 0.0]])
plotter.add_points(
support_pts,
render_points_as_spheres=True,
point_size=18,
color="black",
label="supports",
)
plotter.add_points(
np.array([[x_peak_fe, scale * v_peak_fe, 0.0]]),
render_points_as_spheres=True,
point_size=14,
color="#d62728",
label=f"v_peak FE = {v_peak_fe * 1e3:.3f} mm",
)
plotter.view_xy()
plotter.add_legend()
plotter.show()
Closing notes#
print()
print("Take-aways:")
print(f" • v(L/2) within {abs(err_mid):.3f} % of the closed form.")
print(f" • v_peak within {abs(err_peak):.3f} % at x ≈ {x_peak_fe:.4f} m.")
print(" • The Hermite-cubic BEAM2 element interpolates Bernoulli kinematics")
print(" *exactly* on the polynomial moment field — only the consistent-load")
print(" quadrature introduces any discretisation error here.")
Take-aways:
• v(L/2) within 0.056 % of the closed form.
• v_peak within 0.148 % at x ≈ 0.4333 m.
• The Hermite-cubic BEAM2 element interpolates Bernoulli kinematics
*exactly* on the polynomial moment field — only the consistent-load
quadrature introduces any discretisation error here.
Total running time of the script: (0 minutes 0.187 seconds)