Whipple’s method of joints (1847)

American railway bridges of the 1840s were built from rules of thumb, and some fell down. Squire Whipple, in A Work on Bridge Building (1847), showed that the force in every member of a pin-jointed truss follows from statics alone. Each joint is in equilibrium, so its two equations, \(\sum F_x = 0\) and \(\sum F_y = 0\), give the forces in two members. Start at a support and work across the truss, two unknowns at a time (method_of_joints()). The example solves a six-panel Pratt truss under a train of panel loads and checks the chord forces against the method of sections: at any vertical cut, the chord force is the bending moment divided by the truss depth.

import matplotlib.pyplot as plt
import numpy as np

from engineeringkit.structural_engineering import influence_line, method_of_joints
from engineeringkit.structural_engineering.visualizers import plot_truss

A Pratt truss with six panels

Joints 0-6 run along the bottom chord, joints 7-11 along the top. The left end is pinned and the right end is on a roller.

panels, a, depth, P = 6, 4.0, 4.0, 50e3
bottom = [(a * k, 0.0) for k in range(panels + 1)]
top = [(a * k, depth) for k in range(1, panels)]
nodes = np.array(bottom + top)
T = {k: panels + k for k in range(1, panels)}  # top joint above bottom joint k
members = [(k, k + 1) for k in range(panels)]  # bottom chord
members += [(T[k], T[k + 1]) for k in range(1, panels - 1)]  # top chord
members += [(0, T[1]), (panels, T[panels - 1])]  # end posts
members += [(k, T[k]) for k in range(1, panels)]  # verticals
members += [(T[k], k + 1) for k in range(1, panels // 2)] + [(T[k], k - 1) for k in range(panels // 2 + 1, panels)]  # diagonals

loads = np.zeros(2 * len(nodes))
loads[2 * np.arange(1, panels) + 1] = -P  # a load at every interior bottom joint
fixed = [0, 1, 2 * panels + 1]
result = method_of_joints(nodes, members, loads, fixed)
left, right = result.reactions[1], result.reactions[2 * panels + 1]
print(f"{len(members)} members, {len(nodes)} joints; reactions {left / 1e3:.0f} and {right / 1e3:.0f} kN")
print("joints used in the order", result.order)
summary = result.summary()
print(f"{summary['tension_members']} members in tension, {summary['compression_members']} in compression")

21 members, 12 joints; reactions 125 and 125 kN
joints used in the order (0, 1, 6, 5, 7, 2, 8, 9, 3, 4, 10, 11)
12 members in tension, 8 in compression

Check: the method of sections

Cut the truss through the panel next to top joint \(k\), on the side its diagonal runs to. The top chord and the diagonal both pass through that joint, so moments about it give the bottom-chord tension as the bending moment there divided by the depth. At midspan, moments about the bottom joint give the top-chord compression the same way. The moment is the sum of the loads times the moment influence line at the section.

L = panels * a
for k in range(1, panels):
    moment = sum(P * influence_line(L, a * j, "moment", at=a * k) for j in range(1, panels))
    if k == panels // 2:
        name, chord = "top chord", -result.member_forces[panels + k - 2]
    else:
        name, chord = "bottom chord", result.member_forces[k if k < panels // 2 else k - 1]
    print(f"x = {a * k:4.0f} m: M/h = {moment / depth / 1e3:6.1f} kN, {name} {abs(chord) / 1e3:6.1f} kN")
    assert np.isclose(chord, moment / depth)
assert np.isclose(left + right, (panels - 1) * P)

fig, ax = plt.subplots(figsize=(8, 3))
plot_truss(nodes, members, forces=result.member_forces, ax=ax)
for (i, j), force in zip(members, result.member_forces):
    mx, my = nodes[[i, j]].mean(axis=0)
    ax.annotate(f"{force / 1e3:.0f}", (mx, my), fontsize=7, ha="center", va="center", backgroundcolor="w")
ax.set_title("Pratt truss: tension red, compression blue [kN]")
fig.tight_layout()

plt.show()

x =    4 m: M/h =  125.0 kN, bottom chord  125.0 kN
x =    8 m: M/h =  200.0 kN, bottom chord  200.0 kN
x =   12 m: M/h =  225.0 kN, top chord  225.0 kN
x =   16 m: M/h =  200.0 kN, bottom chord  200.0 kN
x =   20 m: M/h =  125.0 kN, bottom chord  125.0 kN