Sizing a
footbridge truss
A fabricator has one shot at quoting this job. Get the geometry wrong and the bridge fails inspection; get the arithmetic wrong and the firm loses money. You'll use Pythagoras, trigonometry and percentages to find a design that passes both tests.
The job
Read the spec like an engineerAshfield Steel Fabrication has been asked to build a pedestrian footbridge over a canal. It has two identical side frames, each a Pratt truss made from square hollow section (SHS) steel.
- Clear span 12.0 m, divided into equal panels
- Diagonal members must sit between 35° and 55° to the horizontal — BS EN 1993 buckling rule
- SHS 80×80×5 weighs 11.4 kg per metre
- Steel costs £1,450 per tonne, plus 8% added for offcuts and waste
- Client's budget for steel: £1,800
Your question: how deep should the truss be, and how many panels, so it passes the angle rule and comes in under budget?
Explore the drawing
Drag the chord, step the panelsDrag the grip on the top chord and step the panel count. Only a narrow band of designs satisfies both rules at once — find it by hand, then prove it with algebra below.
Worked solution
Open one step at a timeTake the 6-panel, 1.50 m deep design and check it properly.
The 12 m span is split into 6 equal panels.
Each panel is a right-angled triangle: base 2.00 m, height 1.50 m, hypotenuse = the diagonal member.
A 1.5 : 2 : 2.5 triangle — a scaled 3-4-5. Spotting that saves you time in an exam.
The mark for the method is for 2² + 1.5² written down. Write it even if you can do it on the calculator — you keep 1 mark if you slip on the arithmetic.
You know the opposite (1.50) and the adjacent (2.00), so use tan.
36.9° sits inside the 35°–55° window. Passes.
Rounding to 37° then using it later is how marks leak away. Keep the full value in your calculator and round only at the end.
One frame contains: two chords, a vertical at each of the 7 nodes, and one diagonal per panel.
£1767.39 against a £1,800 budget — the job is on, with £32.61 to spare.
Use the multiplier × 1.08 in one step. Working out 8% separately and adding it costs you time and invites a slip.
Push the depth up and the angle improves but the steel bill climbs; pull it down and you save money until the diagonals flatten past 35° and fail inspection.
1.50 m sits near the middle of that window. That is what engineering maths is for: not one answer, but a range you can defend.
Your turn
Answers checked instantlyA different bridge uses 4 panels across the same 12 m span, with a truss depth of 1.50 m. Find the length of one diagonal member, to 2 decimal places.
For that same 4-panel bridge, find the angle of the diagonal to the horizontal, to 1 decimal place. Then decide whether it meets the 35°–55° rule.
An 8-panel design must have diagonals at exactly 55° to the horizontal. Panel width is 12 ÷ 8 = 1.50 m. Find the truss depth needed, to 2 decimal places.
Exam-style question
Full mark schemeA second footbridge spans 10 m in 5 equal panels with a truss depth of 1.60 m. It has two identical frames, a vertical at every node and one diagonal per panel. The steel weighs 11.4 kg/m and costs £1,450 per tonne, with 8% added for waste.
Work out the total cost of the steel, to the nearest pound.
Mark scheme — 5 marks
- Panel width 10 ÷ 5 = 2.00 m and diagonal √(2.00² + 1.60²) = 2.5613 m M1 A1
- One frame: 2×10 + 6×1.60 + 5×2.5613 = 20 + 9.6 + 12.806 = 42.406 m M1
- Two frames: 84.813 m → mass 84.813 × 11.4 = 966.87 kg = 0.96687 t M1
- 0.96687 × 1450 × 1.08 = £1514.15… A1
Accepted answer: £1514. Watch the node count — 5 panels means 6 verticals, not 5. That single slip is the most common lost mark on this question type.
Produced for EAL engineering learners.
Maps to GCSE Mathematics (9–1) — geometry, trigonometry and number.