Howe truss bridge: why diagonals take compression and verticals pull

The end post of the Howe preset pushes with 12.0 kN and buckles at 33.2 kN, so a bridge loaded with 20 kN fails at 55.3 kN. The Howe truss bridge is easy to recognise: the diagonals lean up toward the centre of the span, and every vertical is a thin rod. William Howe patented the arrangement in 1840, with timber for the members that push and wrought-iron rods, threaded and tightened with nuts, for the verticals. Early railroads built a great many bridges on that plan.

The Pratt frame with its diagonals turned around

The Howe preset has the same frame as the Pratt: a 12 m span, 3 m high, six panels of 2 m, parallel chords, five verticals and inclined end posts. Four diagonals cross the interior panels, but here they slope up toward the centre. The count is 21 members and 12 joints, statically determinate: 21 + 3 = 24 = 2 × 12. The load is 20 kN in five 4 kN loads on the interior bottom joints, and each support takes 10 kN.

The geometry matches the Pratt member for member, which is why the two make such a good pair. Only the diagonals change direction. Everything the Pratt does with tension, the Howe does with compression.

Why the diagonals of a Howe truss bridge push and the rods pull

The shear in the left half is 6 kN in the panel beside the first loaded joint and 2 kN in the next. A diagonal leaning up toward the centre can only resist that shear by pushing: it is the strut that props the loaded joint upward. The vertical at that joint then has to hang the load and the push of the diagonal on it, so it pulls.

Joint B shows it well. A 4 kN load acts down, the diagonal B to I pushes down and to the left with a vertical component of 6.0 kN, and the rod B to H pulls up with 4 + 6 = 10.0 kN. That is the rod holding the loaded joint, and it is why the end verticals carry the most.

Rods, struts and the end post that still governs

The rods pull with 10.0 kN at the ends, 6.0 kN next, and 4.0 kN at the middle. The most heavily used is the first, B to H, at 7.4 percent of the yield capacity of the 60 mm by 3 mm tube, which is 134.3 kN. The chords reverse their roles from the Pratt in a small way: the bottom chord, pulling, peaks at 12.0 kN in the middle, and the top chord pushes with 10.7 kN.

The diagonals push. B to I and F to K carry 7.2 kN each and C to J and E to J carry 2.4 kN. Each is 3.606 m long, and the Euler load of that tube is 33.2 kN, so the heaviest diagonal is used 21.7 percent. In the Pratt the same diagonals pull, and they use 5.4 percent of their yield capacity.

The failure still comes at the end post. A to H uses 36.2 percent of its 33.2 kN, and the bridge fails at 20 × 2.764 = 55.3 kN, exactly the Pratt number, because the end post is the same member in both bridges. Once it is strengthened the Howe’s long pushed diagonal B to I takes over, at 92.1 kN; the Pratt truss bridge page gives the other half of that experiment.

Timber, steel and a card model

A Howe truss bridge makes sense in timber. Wood handles compression well and loses most of its strength in tension joints, so putting the push in the diagonals and the pull in rods is a sound use of two materials. Steel tubes behave differently: they buckle in compression, and long pushed members are what costs steel (see tension and compression). The Pratt layout is the reverse, keeping the long members in tension.

In a model, card strips and sticks are strong in tension and buckle when pushed, much like steel tubes, so a Howe truss bridge in card repeats the steel problem. The overall comparison of truss bridge designs puts the Howe next to the other three on the same load. The rod at joint B, 10.0 kN in tension, is the busiest member of the whole Howe frame that pulls.