Howe truss: the Pratt inverted
Take a panelled bridge truss and slope the diagonals up toward the center instead of down, and every member in the web changes sign: the verticals become hangers in tension and the diagonals become struts in compression — the exact inverse of the Pratt. We solve every member of a 16 m Howe through-bridge by the method of joints, watch the signature appear at the very first vertical, then put the two trusses side by side to see why the geometry was a material decision all along.
- Span
- 16 m (4 panels × 4 m)
- Height
- 3 mthe 3-4-5 panel: cos θ = 0.8, sin θ = 0.6
- Loads
- 20 kN ↓ at each deck node (2, 3, 4)
- Supports
- Pin @ 1 · Roller @ 5
- Members
- 13bolted joints drawn at every node
Step 1 — Reactions, and reading the joints
Symmetry splits 60 kN evenly — and the drawing tells you what the joints are.
Three 20 kN deck loads hang from the bottom chord, so the two supports together must push up with 60 kN. Truss and loading are both symmetric about midspan, so nothing favours either end:
Before the numbers, read the drawing: every node is drawn with a bolted joint — the bolt pairs on each member end are how the real chord splices would be made — but the analysis still idealises every joint as a frictionless pin, exactly as the method of joints requires. And the count says statics will be enough: 13 members + 3 reactions = 16 = 2 × 8 joints — the truss is statically determinate, so the joint-by-joint walk will unwind it completely.
Step 2 — The end panel
Joint 1 gives the rafter and the chord; joint 6 hands back a vertical in tension.
Each panel is 4 m across and the truss stands 3 m tall, so every sloping member is a 3-4-5 triangle’s hypotenuse: length 5 m, , . Those two numbers carry the whole walk. Start at joint 1, where only the end rafter 1-6 and the bottom chord 1-2 meet the 30 kN reaction — two unknowns, two equations:
The rafter leans into the support as a strut and shoves the bottom chord outward into tension — so far, any truss. Now climb the rafter to joint 6, where it meets the top chord 6-7 and the vertical 2-6, with no load applied:
Stop at that second result. A vertical in tension — this is the Howe signature. The vertical is not a post propping the top chord up; it is a hanger, carrying the deck load at joint 2 up to the top chord, where the compression diagonals take over. Watch it repeat at every vertical in the truss.
Step 3 — The loaded joint
Joint 2 carries a deck load: the diagonal turns out to be a strut.
Joint 2 is where a 20 kN deck load hangs. Four members meet there, but two are already solved — the chord 1-2 (+40) and the vertical 2-6 (+30) — leaving the diagonal 2-7 and the chord 2-3. The vertical equation balances the known +30 kN hanger against the load and the diagonal’s vertical component:
Negative again: the diagonal is in compression, bracing up toward the crown, and its horizontal push is what steps the bottom-chord tension up from +40 to +53.333 kN toward midspan. Hanger in tension, diagonal in compression — the Howe pattern is now fully on display.
Step 4 — Symmetry closes the truss
One check at the crown, and the mirror writes the right half for free.
By symmetry the right-hand diagonal matches its twin, F₄₋₇ = −16.667 kN, and both arrive at the crown joint 7 from below. Their horizontal components are equal and opposite, so ΣF_x cancels itself and the top chord runs through unchanged, F₇₋₈ = F₆₋₇ = −40 kN. The vertical equation then hands over the center hanger:
Tension once more — the center vertical hangs the midspan deck load straight up to the crown. The rest is the mirror: F₄₋₅ = +40, F₃₋₄ = +53.333, F₈₋₅ = −50, F₄₋₈ = +30 kN. All 13 member forces are on the table, and the results figure below reads as a two-color rule.
Step 5 — Howe vs Pratt
Slope the diagonals the other way and every sign in the web flips.
Put this page next to the Pratt bridge example and the two trusses read as photographic negatives of each other. Same panelled through-bridge, same deck loads hanging from the bottom chord — but the Pratt slopes its diagonals down toward midspan, and in that example the long diagonals came out as tension ties (+53.03 and +17.68 kN) while the interior vertical worked as a compression post (−12.5 kN). Here the diagonals slope up toward the center, and every sign in the web flips: each interior diagonal is a −16.667 kN strut, and every vertical is a hanger in tension — +30, +20, +30 kN. The chords never notice; bottom tension and top compression are the same story in both.
The proof
Hand calculation vs the solver, all 13 members.
| Quantity | By hand | StructureCalcs | |
|---|---|---|---|
| Reactions R_1 / R_5 | 30 / 30 kN | 30 / 30 kN | |
| End rafters (1-6, 8-5) | −50 kN (C) — R/0.6 | −50 kN | |
| Top chords (6-7, 7-8) | −40 kN (C) | −40 kN | |
| Verticals (2-6 / 3-7 / 4-8) | +30 / +20 / +30 kN (T) | +30 / +20 / +30 kN | |
| Diagonals (2-7, 4-7) | −16.667 kN (C) | −16.667 kN | |
| Bottom chords (1-2, 4-5 / 2-3, 3-4) | +40 / +53.333 kN (T) | +40 / +53.333 kN |
Every value was worked by hand with the classical method, then checked against this site’s solver — the same engine the Try it button opens. This agreement is re-run automatically on every build.
Now make it yours
Open this exact model in the calculator — then change a load, drag a support, and watch every diagram update in real time. The best way to build intuition is to break it and see what happens.
Take it with you
Export this worked example as a PDF, or download it as a .screport and open it in the Report Builder — the model travels inside the file, so you can reconstruct it, re-solve, and build your own report from it.