Pratt Truss — Method of Joints, Force Flow Through 12 Pins
A six-panel Pratt truss with 12 pin joints and 21 bars. Click any joint to drop a load and watch every bar recolor as forces redistribute through the lattice.
Show the joint-by-joint free-body diagram
A real bridge truss in twelve pins
The lattice above is a six-panel Pratt truss — the most common bridge-truss configuration in use, named after Caleb and Thomas Pratt who patented it in 1844. It has 12 pin joints, 21 bars, and two supports (a pin on the left, a roller on the right). That's the smallest truss that's worth calling "a structure" rather than just a triangle, and it's the workhorse of pedestrian bridges, train bridges, and roof rafters.
What's special about a Pratt truss is the diagonal direction: every diagonal slants from the top chord downward toward the centre of the span. Under a downward load, that means every diagonal ends up in tension and every vertical ends up in compression. The Howe truss is the mirror image — diagonals slant outward from the centre — and you get the opposite force pattern. Choosing one over the other is mostly historical: cast iron handled compression well, so the Howe truss made sense in the 1800s; modern steel handles tension better, so the Pratt won out.
Source: Wikipedia — Truss bridge, Pratt truss.
Why the load travels through every bar
Try clicking different joints in the visualization above. Watch what happens when you put the load at L1 (the joint just inside the left support) versus L3 (the centre). The load is the same magnitude, applied at the same height — but the colour pattern across the entire truss changes. Bars on the far side of the structure suddenly get more or less stressed.
That redistribution is the heart of a truss. The load can't just teleport from where it's applied to the supports — it has to walk through the lattice, joint by joint, with every pin in the path obeying:
Source: LibreTexts — Mechanics Map §5.4 Method of Joints.
For 12 joints that's 24 equations. Match that against the 21 unknown bar forces plus the 3 unknown support reactions (R_A_x, R_A_y, R_B_y), and the problem is exactly determinate: one solution, no slack. That's what the visualization solves in real time when you move the load.
How to read the colours and widths
- Blue = tension. The bar is being stretched. It pulls inward on each pin it touches.
- Red = compression. The bar is being squeezed. It pushes outward on each pin it touches.
- Grey = essentially zero force right now. Some Pratt-truss bars are zero-force members under certain loadings — they don't help carry the load at all in that geometry.
- Bar width is proportional to force magnitude. The thickest bar is the one that needs the most material to survive; the thinnest are nearly idle.
In a Pratt truss with a centre load, the most-stressed members tend to be the end diagonals (compression) and the central bottom-chord bars (tension). Move the load off-centre and you'll see which bars wake up to redistribute it.
Two-force members are still the trick
Each of the 21 bars is a two-force member: it connects exactly two pins, and nothing else acts on it (no transverse load, no couple, weight neglected). From elementary statics, the two end forces on such a member must be equal, opposite, and directed along the bar. So each bar's force is one number — tension positive, compression negative.
That's what makes the matrix solvable. With 21 such numbers plus 3 reaction components, you have 24 scalar unknowns. Twelve joints, two equilibrium equations each, and the system closes.
Geometry levers
The depth of the truss is the most important design knob you have. Try sliding it:
- Deep truss (large ). Diagonals are nearly vertical. The end diagonals do most of the work efficiently — small forces in a short load path. Bottom and top chords get an easier ride. Material-efficient, but visually obtrusive.
- Shallow truss (small ). Same load, same span, but every member must work much harder. The end diagonals point almost flat, so they need a huge force to deliver any vertical resistance. The bottom chord has to pull harder to keep the geometry. This is why deep trusses dominate long-span bridges — depth is structural leverage.
The reaction split between R_A and R_B follows the simplest possible lever rule, summing moments about A:
where is the distance from the left support to the loaded joint. Move the load right and the right support takes more.
- Length of each panel (span ÷ 6)
- Distance of the load from the left support
- Angle of an end diagonal from the bottom chord
- Vertical component of the end-diagonal direction
- Horizontal component of the end-diagonal direction
- Vertical reaction at the right (roller) support, from Σ M_A = 0
- Vertical reaction at the left (pin) support, from Σ F_y = 0
- End-diagonal force when the load sits at the centre joint L3 (closed-form check from joint L0 equilibrium)
- F_{L_0 U_1}\bigm|_{k \;=\; 3} = -\,(P/2) / \sin\theta \;=\; 3} = -\,(\textcolor{#dc4124}{\left(10000\right)}/2) / \sin\theta \;=\; -6403\;\mathrm{N}
- Bottom-chord end-tension when the load sits at L3
- F_{L_0 L_1}\bigm|_{k \;=\; 3} = (P/2) \cdot \cot\theta \;=\; 3} = (\textcolor{#dc4124}{\left(10000\right)}/2) \cdot \cot\theta \;=\; 4000\;\mathrm{N}
Load placed at the centre joint L3 of a 12 m span (P = 10,000 N). By symmetry each support takes half: R_A = R_B = 5,000 N. With panel length ℓ = 2 m and depth h = 2.5 m the end-diagonal angle is θ = arctan(2.5/2) ≈ 51.34°. Joint L0 equilibrium gives F_{L0U1} = −R_A/sinθ ≈ −6,403 N (compression) and F_{L0L1} = R_A·cotθ = 4,000 N (tension).
| Quantity | Expected | Computed | Δ |
|---|---|---|---|
| R_A | 5000 | 5000 | 0 |
| R_B | 5000 | 5000 | 0 |
| F_L0U1_centerload | -6403 | -6403 | 2.374 \times 10^{-4} |
| F_L0L1_centerload | 4000 | 4000 | 4.547 \times 10^{-13} |
| theta_deg | 51.34 | 51.34 | 8.254 \times 10^{-6} |
Same span and depth but the load shifts to L1 (one panel from the left). Σ M_A = 0 gives R_B = P·a/L = 10,000·2/12 ≈ 1,667 N, then Σ F_y gives R_A = 8,333 N. Most of the load goes straight to the near support, but the bars in between must rearrange themselves to deliver it — the engineering view in the viz shows the redistribution.
| Quantity | Expected | Computed | Δ |
|---|---|---|---|
| R_A | 8333 | 8333 | 3.333 \times 10^{-4} |
| R_B | 1667 | 1667 | 3.333 \times 10^{-4} |
- [1] J. Moore et al.. LibreTexts — Mechanics Map §5.4 Method of Joints, eq. At every joint, ΣF_x = 0 and ΣF_y = 0; positive member force = tension (pulls joint toward the bar); negative = compression (pushes joint away). (eng.libretexts.org)
- [2] D. Baker, W. Haynes. LibreTexts — Engineering Statics: Open & Interactive §6.4 Method of Joints, eq. A truss is a frame of two-force members joined at frictionless pins; method of joints applies ΣF = 0 at each pin to find member forces. (eng.libretexts.org)
- [3] Wikipedia. Truss bridge — Pratt truss configuration, eq. In a Pratt truss the diagonal members run downward from the top chord toward the centre of the span. Under a downward load the diagonals are in tension and the verticals are in compression — the opposite of a Howe truss. (en.wikipedia.org)
- All members are two-force members (pin-jointed at both ends, no transverse load on a member). Each member is therefore in pure tension or pure compression. (LibreTexts §6.4)
- All joints are frictionless pins — no moments are transmitted through joints.
- Self-weight of members is neglected; only the applied load P at the chosen joint matters.
- Left support is a pin (resists both horizontal and vertical), right support is a roller (vertical only). With a vertical-only applied load, the horizontal pin reaction is zero.
- Six-panel Pratt configuration is fixed in this concept (12 joints, 21 members). The geometry — span L and depth h — is adjustable.