Warren truss bridge: layout, member forces and what fails first

The centre top chord of this Warren truss bridge carries 20.0 kN in compression, exactly the load placed on the bridge, and that single number runs the whole design. Put 20 kN on the middle joint of a 12 m span and each support takes 10 kN. Everything else follows from those two figures and a few triangles.

Fifteen members, nine joints and no verticals

The Warren layout is two chords joined by diagonals that zigzag between them. Bottom joints sit at 0, 3, 6, 9 and 12 m; top joints sit at 1.5, 4.5, 7.5 and 10.5 m, midway between. Each diagonal joins a top joint to a bottom joint, so the bridge is built from triangles that point up and down alternately. Counting the four bottom chord segments, three top chord segments and eight diagonals gives 15 members and 9 joints.

That count satisfies m + r = 2j, with three reaction components: 15 + 3 = 18 = 2 × 9. The truss is statically determinate. Remove one member and it turns into a mechanism; add one and it becomes indeterminate. James Warren’s English patent of 1848 describes the form, and the basic version has no verticals at all.

The preset is 3 m high, so each triangle has a 3 m base and 3 m height, which makes the diagonals 3.354 m long. A strictly equilateral triangle on a 3 m base would be about 2.6 m high; the preset is a little taller than that.

What each member of the zigzag carries

Three groups of members stand out in the schedule. The chords carry the bending. At mid-span the top chord pushes with 20.0 kN and the bottom chord pulls with 15.0 kN, falling to 10.0 kN at the next top segment and 5.0 kN at the end of the bottom chord. The numbers step down in equal increments because the bending moment of a centre-loaded span rises in a straight line from each support.

The diagonals are the surprise: all eight carry 11.18 kN. Each side of the bridge carries a constant shear of 10 kN, and a diagonal 3.354 m long over a 3 m rise multiplies it by 3.354 ÷ 3 = 1.118. Four diagonals push, four pull, and they alternate along the span.

The top chord is the governing member. It is only 3 m long, but it carries the largest force in the bridge, and a pushed steel tube buckles. The Euler load of a 60 mm by 3 mm tube at 3 m is 48.0 kN. The applied load of 20 kN uses 41.7 percent of it, so the bridge fails when the load reaches 20 × 2.399 = 48.0 kN. The diagonals are longer, which weakens them (38.4 kN each at 3.354 m), but they carry only 11.18 kN, so they are used 29.1 percent. The tension members sit below 12 percent, so the bottom chord stays far from yielding.

Give the centre top chord member the next tube size, 90 mm by 4 mm, and the governing member moves to one of the compression diagonals, with the failure load rising to 68.7 kN and the frame mass from 202 kg to 215 kg. The next member to go is a pushed diagonal, 3.354 m long.

A Warren truss bridge as a first model

Choose a Warren truss bridge for a first model: the member count is low, every diagonal has the same length, and the forces can be found by hand with the method of joints in a few minutes. Compared on one load with the other layouts in the full comparison of truss bridge designs, a six-panel Warren fails at 81.9 kN with 253 kg of tube, ahead of the Pratt and Howe at 55.3 kN and 239 kg.

Compare it with the Pratt truss bridge, which adds verticals and makes the diagonals pull. At four panels the Warren has 9 joints against 12 for the six-panel Pratt, which matters in a model bridge, where each joint is a glued spot that can fail before the stick does. The preset in the truss bridge designer shows the centre top chord turning first, at 48.0 kN.