Teacher and presenter guide · Statics you can watch fail, on a real direct-stiffness solver
Truss Bridge Builder
Demo 8 of 17 · Part four, Example projects · Engineering · 30 minutes for stages 1–3, plus the Sandbox
You need: index.html open in any modern browser — projector,
Chromebook or phone, mouse or touch. No network, no install, no accounts; nothing is fetched at run time
and nothing leaves the machine.
Learning goals
One per stage, in the words the learner reads on the Takeaway card. Each is what its stage demonstrates,
and each stage's check-yourself item tests exactly that. The game underneath all four is the same:
carry the load for less than par.
- Mechanisms. A frame with a four-sided panel folds without straining any member; a
triangle cannot change shape without changing a member's length — and the solver finds the fold as a
rank deficiency, not as a rule someone typed in.
- Compression buckles. Compression capacity falls as 1/L² — 180 kN at 2 m, 80 kN at 3 m,
45 kN at 4 m — while tension stays at 300 kN at any length, which is why cheap designs shorten their
struts or hang the deck so the long members are pulled.
- Load paths. Load splits by stiffness, not by counting: a stiffer path attracts more
force, a mid-span pier turns one 16 m span into two 8 m ones and cuts the chord forces far more than in
half, and a member drawn grey is steel you paid for that carries nothing.
- Sandbox. Every result on this span comes from the same three facts: triangles hold
shape, compression capacity falls with L², and load follows stiffness.
Where this sits on the path
The second of Part four's AI-built classroom tools and the second of its games, after Orbital Mechanics
Golf. Demo 7 hands forward one habit — commit a prediction before you run anything — and this demo makes
it structural: before every load test the learner calls which member goes first, and the engine's own
record answers. The solver is the real direct stiffness method, so every collapse is a computed result.
One question for the session
Ask it before anyone touches a control; return to it at the debrief.
"When a bridge fails, which piece goes first — and could you have called it before the truck
rolled on?"
Most of the room expects the thinnest-looking member to snap. The demo's answer is that the frame usually
folds before anything snaps, and when something does go it is a long pushed member, not a pulled
one. Both surprises are the statics.
Running it
Open ⚙ Settings → Presentation mode first: it scales the page for the back row and adds a
Notes button that reopens this guide. The Guide card opens on load; dismiss
it with Start with Mechanisms. Reset (in Settings) puts the demo back to Level 1 in place
— no reload — and leaves the class leaderboard alone; the two-step clears under
Class data are the only way to empty that.
- 0–4Stage 1, Level 1. Build only the flat roadway.
Take the room's call under the canvas — most say "it holds" — and press ▶ Load test. It
folds. Name what they saw: joint, member, tension, compression, support, mechanism. Add a post and two
struts; run it again, and read the cost against par.
- 4–10Pairs play Level 1 with X-ray on. Have them use
Check a joint: every force meeting there and their sum, always zero. That panel
is the method of joints.
- 10–18Level 2 — the lesson. It opens with a rectangular frame
already built. The Notice line under the canvas says the stiffness matrix is singular; the dashed
ghost rocking on the canvas is the fold the solver found. Call it, test it, watch it fold into a
parallelogram. Let them add diagonals and watch the cue flip to Rigid. Par forces a second pass:
which members do nothing?
- 18–25Stage 2, Level 3 — Heavy Haul. Four times the load.
Point them at Weakest link under the canvas and the debrief's buckles at
column: compression rating depends on length and tension rating does not.
- 25–30Stage 3. Project a student's debrief. Load Pratt, Howe
and Warren from the Gallery at equal depth and compare where the forces go. Post the
leaderboard. Levels 4–5 and the Sandbox make good homework — the gate opens for free play once a stage
has been tested once.
Stage 1 · Mechanisms (Levels 1–2)
Question. Why does a flat roadway fold under a car before any member is overloaded?
Predict prompt. Lay the flat roadway alone and call it: does it hold, snap, buckle, or fold?
Activity. ~2 minDraw the deck; call; test. ~4 minLevel 2's frame:
call; test; add diagonals until the cue reads Rigid; test again.
Expected observation. The flat deck and the rectangle both fold with every member at 0% of
its limit. The debrief's How it moved table names the joints that ran away. One diagonal per
four-sided panel removes exactly one fold; the cue counts them down.
Check-yourself key. The fold changed no member's length, and a member only resists a
change in its own length. "Too weak" and "loose joints" are both refuted: strength never entered it,
and the pins are ideal.
Teaching limit. Every joint is an ideal pin. Real connections carry some bending and can
fail; glued or nailed models usually fail at the joint, which this solver cannot show.
Stage 2 · Compression buckles (Level 3)
Question. The same 4 m member holds 300 kN pulled. Why does it fail at 45 kN pushed?
Predict prompt. A 125 kN truck over 10 m: call the first failure of your first design.
Activity. ~3 minBuild or load a Gallery truss; turn on Numbers;
call; test. ~4 minFlip a working design under the deck and test again.
Pcr = π²EI / L² ≈ 720 kN·m² / L², capped at 180 kN
Expected observation. The first member to go is a long blue one, and the weakest-link line
reports it buckling at a fraction of the 300 kN the same member would hold pulled. The regression suite
measures the propped design peaking at 37% of limit and the same steel hung at 32%.
Check-yourself key. No — a pushed member fails by buckling sideways, at a load that
falls with the square of its length. Second item: the long members are now in tension, so the
300 kN limit applies.
Teaching limit. Euler with a 0.6 cap is a stand-in for real column curves; effective length
is taken as the whole unbraced run with K = 1; the section values are the authors' choices (Details labels
which are sourced).
Stage 3 · Load paths (Levels 4–5)
Question. How does a wheel's weight find its way to the ground when you cannot build below
the deck, or when a pier appears mid-span?
Predict prompt. Level 4 forbids building below the deck: call the first failure of your first
design there.
Activity. ~4 minGallery bowstring or Pratt at 2–3 m depth; Check a joint
on a deck joint; call; test. ~3 minLevel 5: build two short bridges that share a deck.
Expected observation. In a Pratt the verticals are blue and the diagonals red; in a Howe the
reverse; in a Warren the diagonals alternate. On Level 5 the pier drops par from $3,600-class spans to the
forces of two 8 m ones. Grey members are the debrief's under-5% rows.
Check-yourself key. Not necessarily — a stiffer added path attracts more force and can
overload its neighbour. Second item: the verticals carry each panel's shear to the top chord; erase
one and its panel becomes a rectangle.
Teaching limit. The moving load is quasi-static — no dynamics, no impact — and the deck
members carry wheels straight to their two joints with no bending.
Stage 4 · Sandbox (Level 6)
Question. With no par and any vehicle, which of your ideas about trusses survive a 20 t crane?
Predict prompt. Pick a Gallery truss and a vehicle; call it before the test.
Activity. Free build on 18 m; vary Depth and Panels; swap
vehicles. Nothing is saved to the leaderboard.
Expected observation. Deeper trusses lower the chord forces; more panels shorten the struts;
the crane finds the long compression members first.
Check-yourself key. Shortening the compression members, or hanging the deck so they pull
instead of push. "Shallower" is refuted: chord force is roughly bending over depth.
Teaching limit. The vehicles are invented, sized to make par interesting.
Misconceptions
"More members always makes it stronger." Load shares by stiffness, so an added
member can pull force onto a neighbour and overload it, and every member costs steel and a joint. The
debrief flags members under 5% of limit. Try it: add one to a design that just made par and watch
the peak move.
"The verticals do nothing — they're spacers." In a Pratt they carry each panel's shear
in compression; in a Howe, in tension. A member genuinely carrying nothing is drawn grey, and that is a
different thing. Try it: erase one vertical with X-ray on.
"Tension and compression are just plus and minus." The same 4 m member holds 300 kN
pulled and 45 kN pushed. The sign selects a different failure mode. Try it: rebuild a design upside
down under the deck.
"It failed because the steel was too weak." On Levels 1 and 2 the frame folds with
every member at 0% of its limit. Strength cannot fix a mechanism; a diagonal can.
Why connections are priced
Costing follows a fabricator's bill: steel by weight (9.4 kg/m at $4.80/kg, about $45 a
metre) plus $180 per joint for the gusset, bolts and labour. A flat price per metre makes
extra panels free, which hides a real design driver; here connections run roughly 45% of a typical bill, so
bracing a panel you already have costs steel only while adding a panel costs a joint too. The figures are
illustrative; the shape of the bill is the point.
Static determinacy, plainly
Count members m, reactions r, joints j; each joint gives two equations.
m + r < 2j: a mechanism — it folds and strength cannot help. m + r = 2j: determinate — every
member is critical. m + r > 2j: indeterminate — load now splits by stiffness, which is why the game
solves K·u = F instead of walking joint to joint. The count is necessary, not sufficient: members in the
wrong places leave a local mechanism, and the solver checks the matrix, not the count.
Discussion questions
- The bare rectangle folded without any member changing length. Where did the load go, and why could no
member resist it?
- Two designs, same cost and same joints: one props the deck from above, one hangs it from below. Predict
which is safer, then check. What in Pcr = π²EI/L² explains it?
- A Warren's diagonals alternate tension–compression along its length. Take one bottom joint and use
ΣFy = 0 to show why its two diagonals must have opposite signs.
- Why is the bottom chord's force greatest at mid-span and near zero at the supports?
- One member snapped and the whole bridge came down. What does that say about its spare load paths, and
what would a spare path cost?
- Turn on Dead load and retest a design that just made par. What happens, and why does it
matter more as spans grow?
Levels, par and logistics
Every par is beatable by a triangulated design and by no un-triangulated ladder at any depth
or panel count — verified by test-physics.js, 86 checks, deterministic.
Presenter controls
- ⚙ Settings: Open Presenter Notes (this guide), Presentation
mode, Reset (Level 1, in place, leaderboard kept), then Show par budgets,
Keyboard editing, pick-the-vehicle demos, and the two-step leaderboard clears.
- Details in each stage intro: what is live, how it works, assumptions, sources, boundary.
- Keys: 1 2 3 Draw / Erase / Check a joint · Ctrl+Z undo ·
Space test (once a call is committed) · X X-ray · F numbers.
- Boards and saved designs live in each machine's browser storage.