Interactive tools

What this workspace deliberately will not do

• It will not score the options. A weighted score hides the decision inside the weights, and the weights were chosen by whoever typed them.
• It will not tell you which constraints are real. Ask who owns each one; the ones that belong to nobody can usually be discarded.
• It will not tell you when to stop diverging. Convergent thinking that starts too early kills the option that would have won.

What to notice

• Three numbers are set and four are computed. The computed ones cannot be edited, because editing them would let you express a frame that does not exist.
• Set the pitch to zero and the ridge collapses onto the eaves. The rule still holds; the geometry has simply become a flat portal.
• The member length is not a design quantity. It is a comparison quantity, and its whole value is that it updates the instant a parameter moves.

What the proxy is and is not

• The lateral stiffness proxy is ½·n·r² for a uniform ring: the second moment of area of the perimeter about the plan centroid, per unit column area. It is a comparison number, not a stiffness.
• It says nothing about the bracing, which is what actually carries the shear. Two towers with the same proxy and different bracing behave very differently.
• Halving the top radius quarters the proxy. That is why a sharp taper is a structural decision as well as an aesthetic one.

The question this calculator cannot answer

• Is the process you are about to automate correct? Automating a check that omits a load case gives you a fast, consistent, thoroughly documented omission — applied to every project, with the authority of having come from a tool.
• Who will own it when its author leaves? An unowned tool becomes an unmaintained tool, and an unmaintained tool becomes a liability the year after it was written.

The distinction that matters

• Some faults change the answer now. Others — a hard-coded span, an unexplained constant — do not yet, and are waiting for the day somebody changes an input.
• A workbook can be entirely correct today and structurally unsafe as a document. Both classes have to be fixed, and only the first announces itself.
• A check row — a total computed a second way that must agree — would have caught three of these the moment they happened.

What the receiving model says on its own

• error: Nothing holds the model horizontally. It does not matter that there is no horizontal load: the matrix has zero stiffness in that direction and the solve divides by zero.

Still missing, in the order they matter

• The question this model exists to answer
• What would make this model wrong
• How the result will be checked
• What is deliberately excluded
• What it will be used to decide

What hand calculation discards, permanently

• Continuity
• Load sharing
• Everything three-dimensional

How the freedom counts are arrived at

• No model. One span, one formula.
• One frame taken out of the building: 5 columns × 7 levels, three freedoms each.
• Every column line at every level: 5 × 4 × 7 nodes, six freedoms each.
• The frame plus a 0.5 m mesh over 6 floors of 768 m²: 3072 nodes per floor.
• The same mesh carried through 0.25 m of thickness in 2 layers, floors only.

Beam (frame) carries

• Axial
• Bending about both axes
• Shear
• Torsion

Beam (frame) ignores

• Stress distribution within the section
• Local behaviour at holes, copes and connections
• Warping, unless the formulation includes it

Freedoms the model gave no stiffness

• simple: L rz, R rz
• propped: R rz
• fixed: none

What each model is blind to

• One-way strip: Two-way action, corner torsion, and the reaction that actually goes onto the long-side supports.
• Orthogonal grillage: Twisting moment, which is why it over-predicts the span moments.
• Plate theory (exact): Cracking, reinforcement, creep, and the supports actually being beams that deflect.
• Shell finite elements: Nothing plate theory does not — but it will happily report a peak moment at a point support that is a singularity, not a result.

What each model is

• Stick model, bending only: One vertical line element with the core's gross second moment of area. The cheapest possible model of a core.
• Stick model, with shear deformation: The same element, with the shear area included. For a squat core this is not a refinement.
• Coupled walls: The openings are admitted: the walls act together only as far as the coupling beams make them.

What each step is doing

• Steps 1–2: no physics is added by the transformation. The global matrix is the axial matrix seen from a rotated viewpoint.
• Step 3: assembly is addition, and nothing else. A node connected to nothing would leave a zero row here.
• Step 4: a restraint deletes a row and a column, because that displacement was already known.
• Step 5: displacements first, then everything else. The residual is the check that decides whether any of it is usable.

Try these

• Release every restraint. The model is free to translate and rotate, and the factorisation fails on the first free degree of freedom.
• Hold only Ax and Ay. Two restraints prevent translation and not rotation about A — the structure spins.
• Hold Ax, Ay and By. Three restraints, determinate, and the member forces are what statics gives.
• Add Bx as well. It still solves, the member forces change, and the change came from a restraint the real structure may not have.

The point of separating the two amplifiers

• Set the imperfection to zero and the total is just the amplified lateral deflection. Most engineers stop there.
• Put the imperfection back and watch its contribution grow much faster than the lateral one as α falls. At α = 4 the load amplifier is a modest 1.33 and the imperfection is growing by a third of itself.
• Set F to zero and the structure still deflects — the imperfection alone is amplified. A perfectly-loaded imperfect column deflects; a perfectly-straight loaded one does not.

What to notice

• Change k and P_crit moves with it. Change L and P_crit moves with it. Change nothing else and the material never enters.
• The geometric contribution is proportional to P/L. Doubling the section area would double k and leave it untouched.
• α_cr = 11 is where ignoring second-order effects costs exactly 10 %. The convention quoted in practice is 10, where the cost is 11.1 %.

Why it converges from above

• A finite element model of a continuum is a CONSTRAINED version of it: the displacement field can only take the shapes the elements allow.
• A constrained structure is stiffer, so the computed buckling load is always too high — and always in the unsafe direction.
• The same is not true of a static analysis with cubic elements, which is exact for a prismatic member under end loads. Same model, different question, different adequacy.

Why this breaks bracketing

• Sweep the imposed load from one end of its range to the other and the response rises, peaks, and falls again. It is not monotonic.
• So the three-run bracket of Module 14 — worst case at a corner — does not apply. Two runs at the ends would report a comfortable answer and miss the peak entirely.
• The worst case for a vibration problem is often the LIGHTEST one, because there is least mass to resist the excitation. That is the opposite of every static intuition.

Things worth trying

• Hardening at 500 kN: the first correction is the LINEAR answer, and it overshoots by nearly three times. That is what nonlinearity means in practice.
• Switch to modified Newton and every row shows the same tangent. It takes many more iterations and each is far cheaper.
• Yielding at 200 kN: above the plateau of 150 kN there is no displacement that balances the load, and no number of iterations finds one.
• Snap-through at 25 kN: above the limit point of 24.1 kN, load control either fails or lands on the far branch — a completely different state of the structure.

Reading the plot

• Red dots mark the negative-tangent branch: real equilibrium states that load control cannot sit on.
• Orange markers are limit points, where the tangent is horizontal and the structure carries the most it can in that configuration.
• Move the attempted load above the first limit point and watch what load control does. Then note that displacement control traced the whole curve without difficulty.

What the picture is not

• The mass is fictitious. It is chosen for convergence speed, not from the structure, so nothing about the transient is physical.
• Reduce the damping and the path oscillates for far longer before settling — on exactly the same answer.
• The peak overshoot is larger than the final displacement. Reading a value off an intermediate step would be reading a number the method invented.

The three things this controls

• The step is set by the smallest element in the model, not the average. Four sliver elements out of a million set it for all of them.
• A softer material has a slower wave and therefore a longer step. Timber tolerates a coarser time discretisation than steel at the same element size.
• Halving the element size doubles the step count AND doubles the element count, so the cost grows faster than element count alone would suggest.

The technique

• Compare a quantity that depends on stiffness with one that does not. In a determinate structure the moment does not.
• Moment right, deflection wrong: stiffness. Both wrong: geometry or load. Both right, frequency wrong: mass.
• The powers of length are what set the factors — 10³ for a dimension, 10⁶ for an area, 10¹² for a second moment of area.

Try this in order

• Start with pin and roller and note the bottom chord tension. That is the force the member has to be designed for.
• Switch to pinned both ends. A horizontal reaction appears from nowhere and the bottom chord tension collapses. Nothing warns you.
• Increase the depth and watch the effect grow — the further the support sits from the neutral axis, the stronger the arching.
• Turn on the coincident node with the supports back to pin and roller, and watch the residual stop balancing.

The discriminator

• Look at the RATIO column, not the value column. A converging sequence has ratios tending to 1; a singular one has a constant ratio above 1.
• One extra refinement settles which you are looking at, and it is the cheapest question you can ask of a mesh.
• For the singular case the honest fix is not a finer mesh. It is a different quantity: the punching shear at a flat-slab column is the difference in column axial force above and below, which has no mesh in it at all.

About these cases

• Every case is invented. None is a retelling of a real project, and none is taken from any source.
• Where a real failure is relevant it is named once, in Module 13, with a pointer to the published investigation — not reconstructed.
• 80 cases are written against a target of 80. The categories still short are listed in the coverage matrix.

Where each range comes from

• Span: Grid set by the architect; construction tolerance and a possible late grid shift of one column line.
• Section depth: The serial sizes available within the agreed floor zone.
• Young's modulus: Structural steel. One of the few genuinely tight inputs, and its narrow range is the point.
• Imposed load present in service: What is actually on the floor when it vibrates, not the ultimate design value. The widest range here, and usually the one that decides the answer.
• End restraint stiffness: A nominally pinned connection behaves as continuous at vibration strains and as a pin at ultimate. The range spans both.

Two things this study is doing at once

• Verifying: for deflection the elasticities must be +4 for span, −3 for depth and −1 for modulus, because those are the exponents in the formula. Anything else means the model is not doing what beam theory says.
• Exploring: the tornado ordering directs effort at the inputs with the largest swing, which are not the ones with the largest elasticity. Young's modulus has an elasticity of −1 and almost no swing, because it is known to ±2 %.
• And it cannot find interactions. Varying one input while holding the others at base values is silent about combinations, and no amount of finer sampling fixes that.

What the surface shows that two sweeps cannot

• If the contours were straight parallel bands, the two variables would act independently and two separate sweeps would tell you everything.
• Where the contours curve or fan out, the effect of one variable depends on the value of the other — an interaction, and one-at-a-time sampling is blind to it.
• Five variables at nine values each is 59 049 runs for a full factorial against 45 for one-at-a-time. That gap is why Module 16 exists.

What the load path does not include

• A compression member also has to be stable, so its real volume is larger than the load path suggests. Compare like with like, or design the members.
• Connections are not in this number. Where connections drive member size — timber especially — the ranking can reverse.
• It assumes member area is proportional to force, which is true for ties and only approximately true for struts.
• The lightest structure is not automatically the cheapest, the lowest-carbon, the simplest or the most robust.

Reading it

• Chord force is moment over depth, so the shallow end is dominated by chord material.
• Diagonals lengthen in proportion to depth, so the deep end is dominated by web material — and by web flexibility, which is why the deflection turns round too.
• An interior optimum means a genuine trade-off, and a design pushed to either extreme is worse than one in the middle.
• If the truss is deflection-controlled, the material optimum is not the answer — and no single-objective optimiser will mention that. 'Deeper is always stiffer' is only true if you re-size the members as you go.

Try this

• Determinate, stress only: converges in one or two passes with every member fully stressed. The forces do not depend on the sizes.
• Switch to indeterminate: sizing changes the forces, so the loop has to run several times. That is the snowball.
• Add a deflection limit: the governing constraint changes and the volume rises. Stress-only optimisation can violate serviceability without saying so.
• Switch to the catalogue: utilisations scatter, volume rises, and fabrication gets simpler.

Why the shape is not an aesthetic choice

• Fixing q = F/L for every cable makes the equilibrium equations linear in the coordinates, so the shape comes from a single linear solve.
• The result is funicular: it carries its load in pure axial tension, with no bending anywhere.
• Invert it and you have the funicular arch for that loading — which is why hanging models were used to design masonry vaults long before anyone could solve the equations.
• Set the load to zero and the cable becomes straight: with no transverse load there is nothing for it to hang under.

What this method is doing

• Follows the slope downhill from where it started. It cannot leave the basin it began in, and it does not know there is anything outside it.

Things worth trying

• On the smooth landscape, compare gradient descent's evaluation count with everything else. It wins outright — right answer, a fifth of the budget.
• Switch to the rugged landscape without changing anything. The same method is now trapped.
• Move the start to (6, 3) on the rugged landscape and gradient descent finds the global optimum. Nothing about the method changed.
• Grid search always evaluates 441 times regardless. In five variables at the same resolution it would be 4 084 101.

Knee: Concrete band beam and slab

• The knee is where the trade-off curve bends most sharply. It is a good place to start a conversation and a bad place to end one: it depends on the axes you chose and their scaling, and it knows nothing about who is paying or what they value.

Try this

• On the four real schemes, plot carbon against depth: three are on the front and the composite steel frame is dominated by the concrete band beam on both axes.
• Now switch the vertical axis to mass. The front collapses to a single design — the CLT hybrid dominates everything. Nothing about the schemes changed; the question did.
• Switch to the concave demonstration. All three designs are non-dominated, and the balanced middle one wins at none of the 41 weightings.

What confidence is and is not

• It is the proportion of the k nearest training examples that voted for the winning class. Nothing more.
• It is NOT a probability that the answer is right. Move the point far outside every cluster and the confidence can still be 100 %, because the nearest neighbours still agree with each other.
• That is the same failure as the surrogate's: a fitted method is defined everywhere, and has no representation of 'outside my experience'.
• k-nearest-neighbour is used here because every prediction is traceable to specific examples — the circled points are the ones that decided it.

Why this is the normal case, not the pathological one

• The rarest class in an archive is usually the one worth detecting — anomalies, defects and unusual failures are rare by definition.
• So the class most likely to be starved in training is the class the model exists to find.
• And overall accuracy is least sensitive to exactly that class, because it is the smallest share of the test set.
• With more classes it gets worse: a class collapsing costs 16 accuracy points out of four classes and 3 points out of twenty.

What the engineer has to do

• Record the range of every input in the training set. That is bookkeeping, not cleverness, and it is the only defence.
• Check every query against it, and refuse or flag anything outside.
• Confirm the promising candidates with the deterministic analysis. The surrogate searches; the analysis answers.
• Try the cubic function: when the truth is in the surrogate's own family, extrapolation is nearly free. That is a fact about that function, not a general licence.

What to try

• Push the degree to 10 and watch the training error collapse while the test error rises. A model chosen on training error alone would pick this.
• Set the noise to zero. The two curves converge, because there is no noise for the extra parameters to learn.
• Increase the training points. The best degree moves back towards 1 — more data shifts the balance towards the true relationship.
• The polynomial fit itself is a least-squares solve, and at high degree the normal-equations matrix becomes ill-conditioned — the same arithmetic problem as Module 12's stiffness range.

Why not

• A deterministic check exists for this and has not been run. Confidence is not evidence where a calculation is available.
• No human review recorded.

Try this

• Set the confidence to 100 % with the check unrun. It still cannot be acted on — confidence is not a substitute for a calculation that could have been done.
• Set the check to 'none exists'. Now confidence and evidence and a review are what you have, which is the situation where a recommendation genuinely helps.
• Remove the evidence. Without it there is nothing for a reviewer to examine, so the review would be a formality.
• Reject at review. A recorded rejection is as useful as an acceptance — it says the question was considered.

Note the mass ranking

• At most spans the lightest scheme by mass is not the lowest-carbon one. Change the span and watch the two rankings separate — carbon is quantity times factor, and the factor varies by more than an order of magnitude between materials.

What is not in these figures

• Foundations and substructure — which would widen the lightest scheme's advantage.
• Cladding, which follows from the structural depth and is often more carbon-intensive per square metre than the frame.
• Cost, programme and buildability, none of which correlates reliably with carbon.
• Sequestration in the timber options, which is excluded — the conservative choice, and it is stated rather than assumed.

What none of these steps supplies

• An expectation of what the structure does. Every check depends on having one, and it comes from Structural Behaviour rather than from any procedure.
• The judgement that the brief has already given away the answer — which is what the module's worked project turned on.
• How much checking is enough. Module 13 said it should follow the consequence; judging the consequence is not on any list.