Factor of safety in a truss bridge: how the design margin is found

Factor of safety in a truss: capacity over demand

A cable rated for 10 kN that hangs a 2 kN weight has a factor of safety of 5. Capacity is what the cable can carry before it fails, demand is what it is asked to carry, and the ratio of the two is the margin.

The four-panel Warren truss in the truss bridge designer gives the same ratio for a whole structure: it carries 20 kN at its centre joint and fails at 48.0 kN. Divide the second figure by the first and the result is 2.40, the margin the designer reports.

For a truss the question has more than one answer, because the truss has many members and each has its own ratio. The member with the highest ratio of force to capacity, called its utilisation, is the one that fails first, and it decides the whole structure. A chain is as strong as its weakest link, and the fifteen members of a Warren truss are no exception. How the forces themselves are found is the subject of truss analysis.

The clean part is the arithmetic. In a pin-jointed truss every force is proportional to the load, so doubling the load doubles every force. The largest utilisation therefore scales with the load too, and the factor of safety is simply one divided by it. Fail load divided by applied load gives the same number.

A worked factor of safety on the Warren truss

The forces below come from the solver on the Warren preset with 20 kN at joint C. Each member is the 60 x 3 mm steel tube, with E = 200 GPa and yield 250 MPa. Capacity in tension is the yield load, 250 MPa times the steel area of 537 mm², which is 134.3 kN. Capacity in compression is the smaller of that and the Euler buckling load, which depends on length: 48.0 kN for the 3.000 m chords and 38.4 kN for the 3.354 m diagonals.

Member	Force (kN)	Capacity (kN)	Utilisation
A-B, D-E (bottom, end)	5.0 T	134.3	3.7 %
B-C, C-D (bottom, centre)	15.0 T	134.3	11.2 %
F-B, G-C, C-H, D-I (diagonal)	11.18 T	134.3	8.3 %
F-G, H-I (top, outer)	10.0 C	48.0	20.8 %
A-F, B-G, H-D, I-E (diagonal)	11.18 C	38.4	29.1 %
G-H (top, centre)	20.0 C	48.0	41.7 %

Take G-H: 20.0 / 48.0 = 0.417, so the factor of safety is 1 / 0.417 = 2.40. The centre top chord governs. Next is 29.1 percent, on the four compression diagonals, and the most heavily loaded tension member sits at 11.2 percent, with plenty of room.

A useful experiment follows. Swap only G-H for a 90 x 4 mm tube. Its Euler capacity at 3 m is 219.6 kN, and the utilisation of G-H drops to 9.1 percent. Then the governing members become the compression diagonals at 29.1 percent, and the factor of safety rises to 3.43, from 2.40, for an extra 13 kg of steel in a frame of about 200 kg. Resizing the member that governs is the cheapest way to raise a margin, and resizing any other member does nothing for it.

How these figures were computed

Forces come from the solver behind the designer, run on the four-panel Warren preset: 12 m span, 3 m height, 20 kN at the centre bottom joint, pin and roller supports. Capacity follows the designer’s rules: yield for tension, the lesser of yield and Euler buckling for compression, pinned ends, exact second moment of area of a hollow tube, π/64 (D⁴ − d⁴). Each utilisation was checked by hand as force over capacity, and the resized case was re-run in the solver. The figures are for the idealised truss only. Open the Warren preset in the designer to repeat them.

Why compression members govern the margin

Look at the table again. The three highest utilisations, 41.7, 29.1 and 20.8 percent, all belong to compression members, and the busiest tension member sits at 11.2. This is typical, and it follows from tension and compression: a member in tension can use all its steel up to yield, while a member in compression loses much of it to buckling. G-H could take 134.3 kN of yield load, but its Euler capacity of 48.0 kN at 3 m arrives first.

Cutting the truss with the method of sections is one way to pin down that G-H force by hand, and the number it gives, 20 kN, is the demand in the ratio.

What a real design margin has to cover

Real codes set required margins by application and material, and they split the problem into several checks rather than one overall ratio. Because the requirements differ from one use to the next, no single required value is given here, and the 2.40 above is not a statement that a bridge is adequate. It is the margin of a model.

The model leaves things out. Self-weight is not included, and a deck and truss do weigh something, so the true demand on every member is higher than 20 kN of traffic would suggest. Joint strength is not checked: a glued or bolted connection can let go below the member’s own capacity. Real bars are not perfectly straight, and a small initial bend lowers a buckling load below the Euler figure.