Structural Engineering & Statics
where the load goes
Series

Force couple — the series

3 essays on one idea, from the one that introduces it to the one that assumes the rest.
  1. A general force system is a screw, not a force. Two forces that are neither parallel, nor concurrent, nor coplanar. Their resultant is (30, 0, -60) and no point in space reduces them to it alone: the moment left on the central axis is (14.40, 0.00, -28.80), which is parallel to the resultant and has a pitch of 0.4800. The simplest object equivalent to the system is therefore a force and a couple about the same line — a wrench — and a planar drawing can never produce one, because a coplanar system has its moment perpendicular to its resultant by construction.

    Moving a force, and what it costs

    Every free body begins by putting a force somewhere convenient. That move is free along the force's own line, costs a couple across it, and in three dimensions leaves behind something no choice of point can remove.

    part 1 · equilibrium
  2. Two load sets with the same resultant and different work. The two ways of putting a uniform load of 10 kN/m onto a beam element 6.00 m long. Both put 30.0 kN at each node, so both have the same resultant and the same moment about any point — they are equivalent for a rigid body. The consistent set adds a couple of 30.0 kN·m at each end, in opposite senses, which is what makes it do the same virtual work over the element's shape functions as the real load does. The couples cancel in the resultant, which is exactly why the resultant cannot see them, and they are the whole difference between an exact answer and one that is a third out.

    Equivalent in work, not in resultant

    Two force systems with the same resultant and the same moment about every point are interchangeable — for a rigid body. A finite element is not a rigid body, and substituting one for the other on a beam element leaves the tip of a cantilever a third too low with no warning of any kind.

    part 2 · equilibrium
  3. A downward pressure that pulls the corners up. The nodal forces that do the same work as a uniform pressure on one eight-node serendipity quadrilateral, as fractions of the whole load, each node's share being the integral of its shape function over the element. The shares are −1/12, −1/12, −1/12, −1/12, 1/3, 1/3, 1/3, 1/3 at its eight nodes, adding up to one. The four corner forces are negative: the pressure pushes down and the work-equivalent forces at the corners pull up.

    The corners the pressure pulls up

    Replace a uniform pressure on a beam element by its work-equivalent nodal loads and the ends acquire couples the resultant cannot see. Do the same on an eight-node plate element and the corners acquire forces pointing the wrong way: a pressure pushing down is represented by the corners being pulled up, a twelfth of the load each. It is correct, it is the vector that makes the solution the best one available, and it is the reason a printout of nodal forces is not a picture of where a load goes.

    part 3 · equilibrium

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