SimuVeda

Engineering · Simulator 095

Truss Forces

A truss is a frame of straight bars joined at their ends, used in roofs and bridges. Pick a truss, set its span and height, and hang loads on it. Every bar is either pulled (tension) or squeezed (compression); see which is which, how big the forces are, and why shallow trusses work harder.

How the model works

In an ideal truss the bars are joined by frictionless pins and loads act only at the joints, so each bar carries a pure axial force F: positive for tension, negative for compression. At every joint the forces must balance:

Σ F_x = 0, Σ F_y = 0 at each joint

That gives 2J equations for J joints. With M bars and 3 support reactions (a pin at the left, a roller at the right), the truss is statically determinate when M + 3 = 2J, and the simulator solves the equations all at once (the method of joints).

A useful check: at mid-span, the chords act like the flanges of a beam, so the chord force is roughly the bending moment divided by the truss height, F ≈ M / h. Halving the height doubles the chord forces. In a Pratt truss the diagonals slope down toward the middle and are in tension; in a Howe truss they slope the other way and are in compression. Some bars carry nothing for a given load case (zero-force members) but stop the truss from buckling or collapsing under other loads.

Ideal pin-jointed trusses with loads only at joints, no self-weight, and no buckling check: real compression members must also be checked for buckling, which this model does not do.

Guided experiments

  1. A-frame. 4 m span, 2 m high, 10 kN at the apex. Prediction: each rafter is in compression with 7.07 kN (10 / (2 sin 45°)) and the tie is in tension with 5 kN.
  2. Flatter roof. The same frame only 0.5 m high. Prediction: the rafters now carry about 20.6 kN and the tie 20 kN, four times as much, because the bars meet at a shallow angle.
  3. Pratt bridge. 12 m span in 6 panels, 2 m high, 10 kN at each inner bottom joint. Prediction: the top chord at mid-span carries 45 kN of compression, M/h = 90/2, and the bottom chord up to 40 kN of tension. All the diagonals are in tension, and the end verticals hang 10 kN each.
  4. Howe versus Pratt. The same bridge with the diagonals reversed. Prediction: the chord peaks swap (45 kN tension at the bottom, 40 kN compression at the top), the diagonals switch to compression and every vertical is in tension. Steel bridges usually prefer the Pratt layout, which keeps the long diagonals in tension.

Challenge

A Pratt truss with 6 panels is loaded by a single 30 kN load at the middle bottom joint. Which members carry no force at all? Predict first, then look for bars drawn in grey (load case "Single joint", joint 3, 30 kN).

Reveal the explanation

Three verticals: the two next to the supports (at 2 m and 10 m) and, perhaps surprisingly, the middle one. At the joints next to the supports, the vertical is the only bar with a vertical component and there is no load, so it must carry zero. At the loaded middle joint the two diagonals meet the bottom chord, and they carry the 30 kN up between them (21.2 kN each), so the middle vertical has nothing to do.

References

  • Hibbeler, R. C., Engineering Mechanics: Statics, 14th ed. (2016), ch. 6.
  • Beer, F. P., Johnston, E. R. et al., Vector Mechanics for Engineers: Statics, 12th ed. (2019), ch. 6.
  • Whipple, S. (1847), A Work on Bridge Building (early analysis of truss forces).