Structural Engineering & Statics
where the load goes
Series

Method of sections — the series

3 essays on one idea, from the one that introduces it to the one that assumes the rest.
  1. A truss cut through panel 3. The truss severed through one panel, with everything to the right removed and the three cut member forces drawn on the exposed faces. Taking moments about the marked joint removes two of the three unknowns, so one equation gives the third: -52.94.

    Answering one question without solving the rest

    A truss of fifty members can be interrogated about one of them. Cut through three, take moments about the point where two of them meet, and the third falls out in a single line.

    part 1 · equilibrium
  2. The section, once the joint has supplied the fourth equation. Everything above a cut through panel 3. Four severed members; three equations. Moments about the mid-joint of the horizontal remove both diagonals and leave both legs. Moments about the point where the legs' lines meet, 36.0 m up, remove both legs and leave both diagonals. With the joint's result that the diagonals are equal and opposite, that second equation has one unknown: 20.83 kN in each diagonal. The legs follow: 26.1 T and 86.3 C.

    The cut that needs a joint first

    The method of sections works because a cut through three members leaves three unknowns and a point about which two of them have no moment. A K-braced tower has no such cut anywhere: every section severs two legs and two diagonals. One joint in the middle of a horizontal supplies the missing equation, and only in that order does each step have one unknown — after which the diagonals turn out to be carrying not the shear but the moment about the point where the legs would meet.

    part 2 · equilibrium
  3. The forces the girder can hold with nothing on it. The one self-stress state of a K-braced girder of eight 3 m panels, 3 m deep: member forces in equilibrium at every joint with no load and no reaction. It lives entirely in the two middle panels — twelve members: the four diagonals meeting at the middle joint of the central vertical in tension, and the four central chord members and the half-verticals at the outer side of each middle panel in compression, at 0.89 and 0.45 of the diagonal force. It is symmetric about mid-span, so a symmetric load can call on it as freely as any other: symmetry does not supply the missing equation. Every other member of the girder is idle in it, drawn faint.

    The redundancy only the sun can find

    A K-braced girder that is symmetric about mid-span has one member more than statics can resolve, and it is at the centre, where two K's meet at one joint. Symmetry does not remove it, because the forces it allows are symmetric themselves. Loads barely touch it — a section calculation that ignores it is exact for a load at almost any joint. What finds it is a millimetre of misfit or a sunlit top chord, which put forces into the middle of the girder that no load calculation contains.

    part 3 · equilibrium

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