Note
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Clamped-clamped beam under a central point load#
Statically-indeterminate first-order beam with both ends fully clamped, carrying a single transverse point load \(P\) at midspan. Symmetry plus compatibility at the clamps gives the elementary closed form (Roark & Young 1989, Table 8 case 5; Timoshenko 1955 §40):
with the mid-span deflection (the indeterminate clamping moments stiffen the beam by a factor of 4 vs. the simply-supported case)
and the mid-span bending moment \(M_\mathrm{mid} = +PL/8\) (numerically equal to the end-moment magnitude because the moment diagram is anti-symmetric about midspan).
Implementation#
A 20-element BEAM2 (Hermite-cubic Bernoulli) line spans the beam. Both end nodes are clamped (all 6 DOFs); a single \(-P\,\hat y\) force is applied at the midspan node.
The Hermite cubic basis recovers Euler-Bernoulli kinematics exactly under nodal point loads, so every node hits the analytical value to machine precision and the clamping reactions / moments come out cleanly.
References#
Timoshenko, S. P. (1955) Strength of Materials, Part I, 3rd ed., Van Nostrand, §40 (statically indeterminate beams).
Roark, R. J. and Young, W. C. (1989) Roark’s Formulas for Stress and Strain, 6th ed., McGraw-Hill, Table 8 case 5 (clamped-clamped beam, central point 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.
from __future__ import annotations
import numpy as np
import pyvista as pv
from vtkmodules.util.vtkConstants import VTK_LINE
import femorph_solver
from femorph_solver import ELEMENTS
Problem data#
E = 2.0e11 # Pa (steel)
NU = 0.30
RHO = 7850.0 # kg/m^3
b = 0.050 # square cross-section side [m]
A = b * b
I = b**4 / 12.0 # noqa: E741
J = 2.0 * I
L = 1.0 # span [m]
P = 5.0e3 # central point load magnitude [N], applied downward (-y)
N_ELEM = 20 # even so a node lands exactly at midspan
# Closed-form quantities -----------------------------------------------
R_published = P / 2.0
M_end_published = P * L / 8.0
delta_mid_published = P * L**3 / (192.0 * E * I)
print(f"P = {P:.1f} N, L = {L:.2f} m, EI = {E * I:.3e} N m^2")
print(f"R_L = R_R = P/2 = {R_published:.4f} N")
print(f"|M_end| = P L / 8 = {M_end_published:.4f} N m (clamping moment)")
print(f"δ_mid = P L^3/(192 EI) = {delta_mid_published:.4e} m")
print("stiffness ratio C-C / S-S = 4x (192 / 48 = 4)")
P = 5000.0 N, L = 1.00 m, EI = 1.042e+05 N m^2
R_L = R_R = P/2 = 2500.0000 N
|M_end| = P L / 8 = 625.0000 N m (clamping moment)
δ_mid = P L^3/(192 EI) = 2.5000e-04 m
stiffness ratio C-C / S-S = 4x (192 / 48 = 4)
Build a 20-element BEAM2 line#
points = np.array(
[[i * L / N_ELEM, 0.0, 0.0] for i in range(N_ELEM + 1)],
dtype=np.float64,
)
cells_list: list[int] = []
for i in range(N_ELEM):
cells_list.extend([2, i, i + 1])
cells = np.asarray(cells_list, dtype=np.int64)
cell_types = np.full(N_ELEM, VTK_LINE, dtype=np.uint8)
grid = pv.UnstructuredGrid(cells, cell_types, points)
m = femorph_solver.Model.from_grid(grid)
m.assign(
ELEMENTS.BEAM2,
material={"EX": E, "PRXY": NU, "DENS": RHO},
real=(A, I, I, J),
)
Boundary conditions: both ends clamped#
Apply the central point load#
Static solve + reaction extraction#
res = m.solve()
dof = m.dof_map()
def _react(node: int, dof_idx: int) -> float:
rows = np.where((dof[:, 0] == node) & (dof[:, 1] == dof_idx))[0]
return float(res.reaction[rows[0]]) if len(rows) else 0.0
R_left = _react(1, 1)
R_right = _react(right, 1)
M_left = _react(1, 5) # ROTZ reaction = clamp moment
M_right = _react(right, 5)
print()
print("reactions (femorph-solver) → (analytical)")
print(f" R_left_UY = {R_left:+.4f} → {+R_published:+.4f} N")
print(f" R_right_UY = {R_right:+.4f} → {+R_published:+.4f} N")
print(f" M_left_RZ = {M_left:+.4f} → {+M_end_published:+.4f} N m")
print(f" M_right_RZ = {M_right:+.4f} → {-M_end_published:+.4f} N m")
# Vertical equilibrium and exact reaction match
assert np.isclose(R_left + R_right, P, rtol=1e-12), "vertical equilibrium failed"
assert np.isclose(R_left, R_published, rtol=1e-12), "left reaction off"
assert np.isclose(R_right, R_published, rtol=1e-12), "right reaction off"
# Clamp moment-reactions are equal in magnitude, opposite in sign — both
# act to resist the downward load at midspan. M_left = +PL/8 (CCW about
# +z) lifts the right end against the load; M_right = -PL/8 (CW)
# mirrors it on the other side.
assert np.isclose(M_left, +M_end_published, rtol=1e-12), "left clamp moment off"
assert np.isclose(M_right, -M_end_published, rtol=1e-12), "right clamp moment off"
assert np.isclose(M_left + M_right, 0.0, atol=1e-9), "moment anti-symmetry violated"
reactions (femorph-solver) → (analytical)
R_left_UY = +2500.0000 → +2500.0000 N
R_right_UY = +2500.0000 → +2500.0000 N
M_left_RZ = +625.0000 → +625.0000 N m
M_right_RZ = -625.0000 → -625.0000 N m
Mid-span deflection#
mid_uy = float(res.displacement[np.where((dof[:, 0] == mid_node) & (dof[:, 1] == 1))[0][0]])
err_mid = (mid_uy - (-delta_mid_published)) / (-delta_mid_published)
print()
print(f"δ_mid computed = {mid_uy:+.4e} m")
print(f"δ_mid published = {-delta_mid_published:+.4e} m")
print(f"relative error = {err_mid * 100:+.6f} %")
assert abs(err_mid) < 1e-8, "mid-span deflection error too large for Bernoulli BEAM2"
δ_mid computed = -2.5000e-04 m
δ_mid published = -2.5000e-04 m
relative error = -0.000000 %
Render the deflected line#
pts = np.asarray(grid.points)
disp_y = np.zeros(N_ELEM + 1)
for i in range(N_ELEM + 1):
rows = np.where((dof[:, 0] == i + 1) & (dof[:, 1] == 1))[0]
if len(rows):
disp_y[i] = float(res.displacement[rows[0]])
warped = grid.copy()
warped.points = pts + np.column_stack([np.zeros(N_ELEM + 1), 200.0 * disp_y, np.zeros(N_ELEM + 1)])
warped["uy"] = disp_y
plotter = pv.Plotter(off_screen=True, window_size=(720, 280))
plotter.add_mesh(grid, color="grey", line_width=2, opacity=0.5)
plotter.add_mesh(warped, scalars="uy", line_width=5, cmap="viridis")
plotter.add_points(
np.array([[0.0, 0.0, 0.0], [L, 0.0, 0.0]]),
render_points_as_spheres=True,
point_size=18,
color="black",
label="clamps",
)
plotter.add_points(
np.array([[0.5 * L, 0.0, 0.0]]),
render_points_as_spheres=True,
point_size=14,
color="#d62728",
label=f"P = {P:.0f} N",
)
plotter.view_xy()
plotter.camera.zoom(1.05)
plotter.show()
Total running time of the script: (0 minutes 0.183 seconds)