Structural Engineering & Statics
where the load goes
Series

Truss — the series

6 essays on one idea, from the one that introduces it to the one that assumes the rest.
  1. A Pratt truss of 6 panels. A Pratt truss under equal panel-point loads. The joint equilibrium equations were assembled and solved; 10 members came out in tension, 9 in compression and 2 carrying nothing.

    The triangle that cannot fold, and everything built out of it

    A square of pinned bars is a mechanism. A triangle is not, and that single fact is the reason trusses exist and the reason they look the way they do.

    part 1 · structures
  2. The joints are not pins, and this is what that costs. A 4-panel Pratt truss solved twice on the same stiffness matrix: once with a moment release at every member end, which is the pin-jointed idealisation, and once with the joints continuous, which is what welding them produces. The axial forces are the same to within a per cent; the bending the second solution adds is worst in member 0, where the bending stress reaches 24.3% of the axial stress. Members are shaded by that ratio.

    The joint that is not a pin

    Every truss on this site is analysed as though its joints were frictionless pins. Almost none are. The bending that follows is called secondary, which is a claim about size — and the claim is checkable.

    part 2 · structures
  3. A closed force polygon. The forces on a joint, laid tip to tail. Equilibrium is the statement that the polygon closes, and the gap when it does not is the out-of-balance force, to scale.

    One drawing solves the whole truss

    The method of joints solves a truss one joint at a time, and each solution is thrown away as soon as the next begins. Drawn instead of computed, the joints share their edges — every member's force appears once in a single figure, and the figure's own closure is the check.

    part 3 · structures
  4. A Pratt truss of 8 panels. A Pratt truss under equal panel-point loads. The joint equilibrium equations were assembled and solved; 14 members came out in tension, 13 in compression and 2 carrying nothing.

    Halving the panel buys a shorter strut

    Subdividing a truss into more panels of the same span and depth barely changes the chord forces, because the couple that carries the moment has not moved. What it changes is the length of every compression member, and a buckling capacity goes as the inverse square of a length.

    part 4 · structures
  5. 3 continuous spans against 3 simple ones. The bending moment in a continuous beam, solved by the stiffness method, drawn over the moment in the same spans made simply supported. The peak sagging moment falls from 28.1 to 18.0, and a hogging moment of 22.5 appears over the supports where there was none.

    The chord is a continuous beam

    A truss analysis reports one number per member, and for a chord running through eight panels that number is an axial force with no bending in it. The chord is a single piece of steel spanning between its own nodes, and anything landing between them bends it.

    part 5 · structures
  6. Influence line for the shear force at x = 10.5. The shear force at one fixed station, plotted against the position of a unit load walking across the span. The beam was re-solved at 301 load positions. The worst position is x = 10.56, giving 0.560.

    Two diagonals, one of which is absent

    A truss diagonal is sized for the shear in its panel, and near mid-span that shear changes sign depending on where the load stands. A member that can only pull cannot carry the reversed case, so the panel gets a second diagonal — and at any instant one of the pair is not there.

    part 6 · structures

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