Graphs
Force in the beam as the truck crosses (positive = pulled). Dashed red = limit.
Largest sag of any joint; dashed = L/360 service limit.
How hard every beam works (worst case). Tap a bar to pick it.
The science
Why triangles?
A frame of pinned sticks with 4 sides can lean over; a diagonal makes two triangles and it cannot change shape without stretching or squashing a stick. For j joints, m beams and r support reactions to be rigid: m + r ≥ 2j.
K u = f K = sum (E A / L) d d^T
Tension, compression, buckling
A pulled beam fails at ft A. A pushed beam fails at the smaller of fc A and the Euler load. Safety factor SF = capacity / |N|; the colours show u = 1/SF.
P_cr = pi^2 E I / L^2 N_cap = min(f_c A, P_cr)
Method of joints
Draw one pin and the beams that meet it. A pulled beam drags the pin toward its far end, a pushed one shoves it away; the load and the support reaction act too. The pin is at rest so the forces cancel.
sum Fx = 0 sum Fy = 0 force on pin = N (cos a, sin a)
Hand check, A-frame (4 m deck, 1.5 kN/m, apex 1.5 m): hanger 3 kN, slanted beams 2.5 kN each (2 C x 0.6 = 3), chord 2 kN.
Arches
A parabolic arch loaded at its joints is a funicular: every beam only pushes. In a tied arch the bottom chord holds the ends together.
H = w L^2 / (8 h) 12 m, rise 3 m, 1.5 kN/m => 9 kN
Cables and suspension
A cable can only pull. If a load would push it, it goes slack: the solver removes it and solves again. A suspension bridge hangs the deck from a main cable over two towers (self-anchored here: back-stays pull the deck ends inward).
H = w l^2 / (8 f) T = H sqrt(1 + slope^2)
Cables are pre-tensioned to carry the dead load. Wire rope: E = 120 GPa, 700 MPa on 60 % metal area.
What is simplified
Straight pin-jointed beams, small movements (sag is drawn bigger, labelled), no wind, shaking or fatigue, perfect joints. Typical material values, not for real design. Independent educational simulation; not affiliated with any brand.
Run checks (independent numbers)