A king-post truss spans 8-15 m by turning the load into compression in two diagonal struts and tension in the bottom chord and king post, and the pegged joint at the strut foot, not the timber, is usually what fails first. With a rise of half the half-span (tan 0.5), a king-post load P produces about 1.12 P in each strut and 1.0 P in the tie; cut the rise to a quarter of the half-span and they rise to about 2 P in both. Build the tension joints with iron bolts or straps where possible, and keep the rise generous.
Part of Bridge Types & Construction
A triangulated timber frame, king-post or queen-post, that carries a deck over gaps too wide for a beam.
A Truss Works Because Each Member Takes Only Push Or Pull
A beam bends, and bending stresses the whole section. A triangle cannot change shape without a side changing length, so each member of a truss is loaded only along its length, in compression or tension. The parent describes the king-post form: a horizontal bottom chord spans the gap, a vertical king post rises from its centre, and two struts run from the top of the post down to the chord ends. The struts are in compression, the bottom chord and the king post are in tension. The king-post form is the simplest truss for the 8-15 m gaps that a multi-beam deck cannot reach. A queen-post truss uses two verticals and a flat panel between them, and the parent gives 10-15 m for it.
Three things can fail: a strut can buckle, a tension joint can pull apart, and the chord can bend between its supports. For a deck hung from the chord, the chord also bends under the deck load. The arithmetic below covers the first two.
What You Need
- Timber of the same species and moisture in the chord, struts and post.
- A scaled layout — draw the truss full-size on flat ground, then cut to the drawing.
- Hardwood pegs, iron bolts or iron straps for the joints. The parent lists mortise-and-tenon joints pinned with hardwood pegs, iron bolts, or multiple wraps of heavy wire.
- Abutments that hold the chord ends level and do not spread.
Method: Lay Out And Check A King-Post Truss
Step 1: Fix the rise Choose the height of the king post. A rise of 2.5 m over a 10 m span (half-span 5 m) gives a strut angle of 26.6 degrees from horizontal.
Step 2: Find the forces For a load P hung at the foot of the king post: each strut carries P / (2 sin θ), the chord carries P / (2 tan θ), and the post carries P.
Step 3: Check the strut for buckling A strut under compression bows sideways before the wood crushes. Use Euler’s formula P(cr) = π² E I / L².
Step 4: Check the joints Work out the force each peg or bolt must carry. This usually governs.
Worked Example: A 10 m Span, 2.5 m Rise, 15 kN On The King Post
Inputs, mine: half-span 5 m, rise 2.5 m, so tan θ = 0.5 and θ = 26.6 degrees; sin θ = 0.447; strut length √(5² + 2.5²) = 5.59 m. King-post load P = 15 kN (about 1,530 kgf; you would get P from Dead Load vs Live Load). Struts of 15 × 15 cm square, modulus of elasticity E = 9 GPa, a round figure recalled from the US Forest Products Laboratory Wood Handbook and not checked here.
- Strut force C = 15 / (2 × 0.447) = 16.8 kN.
- Chord tension T = 15 / (2 × 0.5) = 15.0 kN.
- Second moment of area I = 0.15⁴ / 12 = 4.22 × 10⁻⁵ m⁴.
- P(cr) = π² × 9 × 10⁹ × 4.22 × 10⁻⁵ / 5.59² = 3.75 × 10⁶ / 31.2 = 120 kN, for a pinned-end strut.
- Ratio P(cr) / C = 120 / 16.8 = 7.2. Buckling is not the problem for this strut, provided it is straight and braced.
Now the tension joint. The chord pulls 15.0 kN across the joint at the strut foot. If it is held by 2 hardwood pegs of 25 mm diameter in single shear, the area per peg is π × 0.025² / 4 = 491 mm², and the shear stress is 15,000 / (2 × 491) = 15 MPa. Oak shear strength along the grain is roughly 10 MPa (recalled from the Wood Handbook range and not checked here), so 2 pegs are at or past the limit. With 4 pegs the stress is 7.6 MPa, still with a margin of only about 1.3. This is the weak link.
How the rise changes the forces. Strut force is P × √(1 + r²) / (2r) and tie force is P / (2r), where r = rise / half-span.
| Rise / half-span r | Strut force | Tie force |
|---|---|---|
| 0.25 (a quarter) | 2.06 P | 2.0 P |
| 0.33 | 1.60 P | 1.5 P |
| 0.5 | 1.12 P | 1.0 P |
| 0.75 | 0.83 P | 0.67 P |
A flat truss is cheaper in height and far more expensive in joint load.
Common Mistakes
- Relying on pegs alone for the tie — use iron bolts or straps, or lengthen the joint and add more pegs.
- A flat rise to save timber — the forces double. Keep r near 0.5.
- Crooked struts — a bow lowers the buckling load. Use straight timber and brace the strut midway.
- Forgetting chord bending — the deck load between supports bends the chord. Keep the deck lightweight and the chord stiff.
Safety
A truss joint can fail suddenly
A pegged tension joint fails by shear with little warning. Do not let people stand under the span during first loading. Load-test with weights from the bank and check every joint for slip before opening.
What’s Next
- Learn next: Truss Structures — the general theory
- Combine with: Multi-Beam Bridge — the shorter-span alternative