Structural Engineering & Statics
where the load goes
Concept

Distributed load — where it appears

A load spread along a length or over an area, replaced for equilibrium by a resultant at its centroid and by nothing for bending. Replacing it by its resultant is legitimate for finding reactions and illegitimate for finding internal forces, because the two produce different bending along the member.

Named by 6 essays across 2 fields — each of them below, with the objects they name alongside it.

A triangular load and the force that replaces it. A triangular distributed load with its resultant computed by integration: an area of 24.0 acting at 5.33 from the left. The two moment diagrams below show what the substitution costs — the reactions are identical and the peak moment is not.

The load that is spread out, and the force that replaces it

A distributed load can be swapped for a single force at its centroid. The reactions come out identical and the bending moment does not, and knowing which side of the cut the swap is legitimate on is most of the skill.

equilibrium · Distributed load Where a beam's load comes from. A 8 × 6 m panel carrying 5 kN/m², divided at 45° from the corners. The long beams take a trapezoid of 15.0 m² each and the short beams a triangle of 9.0 m²; the four areas sum to 48.0 m², which is the panel, so no load has been invented or lost. The line loads quoted are the uniform equivalents; the real distributions peak at 15.00 kN/m at midspan.

The load a beam is given is a decision

Every beam calculation so far has started with a load per metre, handed over as though it were a property of the beam. It is not. It is the answer to a prior question nobody draws, and two defensible answers to it differ by sixty per cent on the same floor.

equilibrium · Tributary area The wind pushes on one face and pulls on three. A 30 × 20 m building in plan, with the measured pressure coefficient on each face and the arrows drawn in the direction the pressure acts. Only the windward face is pushed; the other three are sucked, and the side faces are sucked hardest of all at c_p = -0.7. The horizontal resultant is 842 kN at a velocity pressure of 0.9 kPa, and the arithmetic of it is the whole point: the leeward suction pulls the building downwind, so it ADDS, supplying 38% of the answer, while the two side faces cancel each other exactly and supply none of it. The coefficients are wind tunnel data; what is computed is the free body they are applied to.

Most of it is suction

A wind load is drawn as arrows pressing on the windward face, which is where about three fifths of it comes from. The rest is a pull on the back. The two side faces carry the largest suctions on the building and contribute nothing at all to the answer — and the inside of the building, which nobody draws, decides whether the roof stays on.

equilibrium · Wind pressure The snow that left the roof is standing against the wall. A roof in section with a 1.0 m obstruction at its downwind end, drawn with the vertical scale exaggerated 4 times because a metre of snow on twenty metres of roof is thinner than the line it would be drawn with. The balanced layer is 0.48 kN/m² everywhere; the wedge against the wall holds the snow that 30% of the 20 m upwind gave up, so its area is fixed by conservation of mass rather than chosen. That makes it 0.85 m deep and 3.4 m long, with a peak load of 2.60 kN/m² — 5.4 times the balanced value, and still only 27% of the snow on the roof. An intensity several times the design load, made of a quantity nobody would notice had moved.

The load that arrives where the wind stops

Snow is the one load a structure is given rather than subjected to. What falls is spread evenly over a whole region; what a member carries is whatever the wind left above it, and the wind piles it against whatever gets in the way. The heaviest patch on a roof is usually a quarter of the snow on it.

equilibrium · Snow drift The bearing is one length and the web is loaded over another. A load applied over a stiff bearing of 200 mm on the flange of a girder with a 1200 × 8 mm web. The flange bends under it and the yield lines that form spread the load along the web over 659 mm — 3.3 times the bearing, and 70% of the yield resistance is that spread rather than the bearing. The effective length is not a decision anybody made: it is what the flange's own bending stiffness against the web's own strength works out to.

The support that is not a point

A reaction is drawn as a single arrow because the equilibrium equations only need its total. Underneath the arrow is a bearing of some width, delivering a pressure over that width, and almost everything a designer would like to know about the region near a support is a consequence of the width the arrow does not have.

internal-forces · Support width 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.

equilibrium · Force couple

Named alongside it

The objects these essays reach for when they reach for this one.

Load pathFree bodyResultantSelf-weightBending momentContinuityTributary areaBase shearBearingCentroidCladdingCompatibility

All concepts