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Geometry beats material — page 10

Essays 217 to 223 of 223 on this thread, in the same order.

Four ways to put six bolts in one plate. Six bolts inside a 150 × 150 mm field of bolt centres, no two closer than 60 mm, loaded through a point (200, 0) mm from the field's centre. The two-column layout carries 193.5 kN complete and 138.8 kN with its worst bolt missing. The ring carries 167.2 kN complete and 117.8 kN with its worst bolt missing. The strongest found carries 234.3 kN complete and 149.7 kN with its worst bolt missing. The most robust found carries 231.6 kN complete and 174.7 kN with its worst bolt missing. The layout found by maximising the complete capacity and the layout found by maximising the worst omission are different layouts, 1 per cent apart when complete and 17 per cent apart with a bolt missing, and both beat the ring — the most evenly spread of the four — on both counts. The dashed line runs from each group's centroid to the load: 200 mm, 200 mm, 180 mm, 188 mm. The ringed bolt is the one each group can least afford to lose.

The strongest layout leans on one bolt

Search a plate for the six bolt positions that carry most and the answer carries 234 kN — and loses 36 per cent of it if one particular bolt is missing. Move that one bolt fifty millimetres, into the corner the optimum had just left, and the group carries 232 kN and loses 25 per cent whichever bolt goes. Robustness here costs one per cent of strength, and a search for strength alone will never find it.

A tubular K joint with a 12.6 mm gap. A 168.3 × 8.0 mm chord with two 114.3 × 6.3 mm braces at 45°, their toes 12.6 mm apart on the chord's crown. The braces' working lines, carried down through the chord, meet 3.0 mm below the chord's axis — an eccentricity of 0.02 chord diameters, inside the band of −93 to +42 mm in which the joint rules let its moment be neglected. Nobody chose that number. It follows from the gap, the brace diameter and the angle, and the gap was chosen so both toes could be welded. The working lines would meet on the chord's axis at a gap of 6.7 mm, which is a joint no rule permits.

Connections

The joint whose lines may not meet

In a welded tubular truss the noding eccentricity is not a fabrication error. It is fixed by the brace diameter, the brace angle and the gap between the toes, and the gap is chosen for the welder. For a third of ordinary proportions the one gap at which the working lines would meet lies in a band no fabrication rule permits — so the concurrent joint the truss was analysed with is the one joint that cannot be built.

The same restraint, spread and gathered. The buckled shape of a 24.0 m compression chord with the same smeared restraint, 0.35 N/mm per mm, delivered by U-frames at two spacings. With frames every 2.0 m the chord buckles in half-waves of about 5.1 m that ignore the frames — the smeared shape — at 1892 kN, matching the smeared answer. With frames every 4.0 m it buckles between them, with a node at every frame, at 1548 kN: 18 per cent below the smeared 1897 and at the Euler load of one bay. The dots are the frames.

Stability

A row of frames is not a foundation

The top chord of a half-through girder is held sideways by U-frames, and the standard calculation smears them into a continuous elastic foundation. That is right while the frames are close and quietly wrong once they are not. The crossover sits at seven-tenths of the buckle's own half-wavelength, whatever the frames' stiffness — and beyond it the chord buckles between frames at a load no stiffening of the frames can raise.

One volume of steel, divided three ways. A Pratt truss of eight panels at a depth of 1, carrying 10 kN at each top joint, drawn three times with every member as wide as its area, the total volume the same in each. With equal areas the mid-span deflection is 43503; fully stressed, with area in proportion to force, 32702; with area in proportion to the square root of the product of each member's real and virtual forces — the division that makes mid-span as stiff as this steel can make it — 31318. The fully stressed truss is 4 per cent short of the stiffest; the equal-area truss is 39 per cent short. No member is allowed less than 10 per cent of the equal area; the deflections are in units of load × length / (E × volume).

Deflection

The truss that is stiff by accident

Give a truss a fixed volume of steel and ask how to divide it among the members. Sized for strength — every member at the same stress — its mid-span deflection comes within four per cent of the stiffest that steel can make, although stiffness was never asked about. The reason is an inequality, and the same inequality says where the accident stops: at the quarter point the strength design is sixty-nine per cent short of the best.

Equilibrium

Rigid by every test, and still folding

Counting the unknowns says whether a frame can be solved and the rank of its equations says whether it can stand, and both answers are yes or no. A frame a few degrees from a critical form passes both and is still nearly a mechanism — and of the ways that shows, the last to arrive is the one the rank test measures. Its own sag under load eats a tenth of its geometry at six degrees; sixteen-figure arithmetic would not notice anything until well below half a degree.

Four numbers read from a moment diagram. The moment diagram of the fixed-ended beam under a central point load, scaled to a largest value of one, with the four values the quarter-point formula reads: the peak, and the magnitudes at a quarter, a half and three quarters of the span — 1.00, 0.00, 1.00 and 0.00. The formula turns them into a gradient factor of 1.923. Solving the buckling problem for the whole diagram, on a beam 8 m between lateral restraints, fork-supported at its ends, gives 1.723: the formula is 12 per cent too high, on the unsafe side.

Stability

The formula that reads four numbers

The factor that credits a beam for the shape of its moment diagram is usually taken from a formula that reads the diagram at four places — its peak and its quarter points — and nowhere else. On the straight-line diagrams it was built around it is safe. On a fixed-ended beam under a central load, whose moment is zero exactly where the formula looks, it is twelve per cent unsafe, and it would give the same answer for a diagram that deserves twenty-two per cent less.

The toe that runs out of ground. A block 6.0 m square on ground that bears 300 kPa at most — 10800 kN over the whole base — pushed 8.0 m up, weighing 3000 kN, drawn at three stages of the push with the lean exaggerated eight times and the contact pressure under the base; the dashed line is the ground's bearing limit. At a push of 629 kN it touches the ground over 3.9 m of its 6.0 and has not yet brought the ground to its limit anywhere. At a push of 770 kN it touches the ground over 2.5 m of its 6.0 and has brought the ground to its limit under 0.8 m. At a push of 776 kN it touches the ground over 2.3 m of its 6.0 and has brought the ground to its limit under 1.1 m. The last is the tipping push: the heel has lifted, the toe is pressing the ground at its limit over a widening block, and the lever arm of the weight about that block is shrinking as the block grows.

Equilibrium

The ballast that helps it over

On rigid ground a heavier block is harder to tip, in exact proportion to its weight — that is the whole of the overturning check. On ground with a bearing capacity it is not. The toe presses the ground to its limit, the weight's lever arm shrinks as the yielded block under the toe grows, and the tipping push peaks when the block weighs half of what the ground under it can bear. Past that, every tonne of ballast added to steady it brings it closer to going over.

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