Trusses

How Bridges Work

Chapter 4 · Trusses

How Bridges Work 2 showed that a deep beam is efficient because its material sits far from the neutral axis, and that an I-beam throws away the useless middle. A truss takes that idea to its limit. It replaces the solid web with a skeleton of straight bars, and arranges the bars in triangles so that every bar carries a simple pull or a simple push and nothing else. The result is a deep, light beam that can be made from timber, iron or steel, and that was widely used for railway bridges in the nineteenth century. This chapter explains why triangles work, names the classic layouts, and shows how to work out the force in every bar with a short program.

What was run

Every Python snippet in this chapter and in the three exercise solutions was executed, and the output shown in the comments is what Python printed. The truss solver uses only the standard library.

Why Triangles

Take four bars and join them at their ends with pins into a rectangle. Push on one corner and the rectangle collapses sideways into a parallelogram. Take three bars and pin them into a triangle, and nothing you do to the corners will change its shape without bending or breaking a bar. A triangle is the only polygon that is rigid on its own. Add a diagonal to the rectangle and it becomes two triangles, and it is rigid too. Every truss is a row of triangles.

In the model engineers use, the bars are joined at joints by frictionless pins, and the loads act only at the joints. Then each bar is pulled or pushed along its own length and does not bend: it is in tension (stretched) or compression (squeezed), the two simple forces of How Bridges Work 1. Real trusses have riveted, bolted or welded joints, but the pin model gives the right forces to a good approximation. For a plane truss with j joints and m bars, the structure is just rigid, with no spare bars, when m + 3 = 2j (the 3 stands for the support reactions). Fewer bars and it can collapse; more and it needs extra analysis.

The Parts of a Truss

Part	Where	Usually carries	Does the job of
Top chord	the upper horizontal members	compression (in a span with the deck loaded from above)	the top flange of an I-beam
Bottom chord	the lower horizontal members	tension	the bottom flange
Diagonals and verticals	between the chords	tension or compression, depending on layout	the web: they carry the shear
Joints	where members meet	no bending in the model	pins

This table matches How Bridges Work 2. Under a downward load the whole truss sags like a beam: the top is squeezed, the bottom is stretched, and the diagonals and verticals carry the shear between them.

Three Classic Layouts

Nineteenth-century engineers patented several arrangements. Here are the three you will meet most often, with the facts that the sources I checked agree on:

Layout	Patent	Diagonals (under gravity load)	Verticals
Howe	1840, William Howe	slope towards the centre; in compression	in tension
Pratt	1844, Thomas and Caleb Pratt	the reverse of the Howe; in tension	in compression
Warren	1848, England (credited to James Warren and Willoughby Monzani)	alternate between tension and compression	none required

(The sources differ on some details: whether the Pratt patent names Thomas alone or both Thomas and Caleb, how the second Warren name is spelt, and the patent drawing is said to differ from the form later called a Warren truss.) The Pratt was a clever match of materials and forces. Its long diagonals are in tension and its short verticals in compression. A compression bar can buckle (see below), so you want compression bars short; a tension bar cannot buckle, so it can be long and thin. That is why the earliest Pratt trusses paired wooden compression members with iron tension rods. The Warren needs fewer bars and fewer joints, and its simplicity has kept it in use ever since.

a small Warren truss (supports at A and C, load at B)

   D─────E
  ╱ ╲   ╱ ╲        top chord: D to E
 ╱   ╲ ╱   ╲
A─────B─────C      bottom chord: A to B to C
      ↓
     load
    

The Method of Joints

To find the force in every bar, treat each joint as a small object in balance. At every joint the forces must add up to zero in two directions, horizontal and vertical. Each bar pulls on the joint along its own line, so a bar's force can be split into horizontal and vertical parts using its direction. For a truss with j joints there are 2j equations, exactly enough for the m bar forces and 3 reactions. Doing them by hand takes pencil and patience; a computer can set them all up and solve them at once.

The program below does exactly that. It takes the joint positions, the list of bars, the loads and the supports, and writes the 2j equations into a table (a matrix). It then solves them by Gaussian elimination, the textbook way of solving simultaneous equations. A positive answer means a bar in tension; a negative one means compression.

import math

def solve_truss(joints, members, loads, supports):
    """Method of joints as one linear system. Tension is positive."""
    names = list(joints)
    n = len(members) + len(supports)
    rows = {}
    for j in names:
        rows[(j, "x")] = [0.0] * n
        rows[(j, "y")] = [0.0] * n
    for k, (a, b) in enumerate(members):
        (xa, ya), (xb, yb) = joints[a], joints[b]
        length = math.hypot(xb - xa, yb - ya)
        cx, cy = (xb - xa) / length, (yb - ya) / length
        rows[(a, "x")][k] += cx
        rows[(a, "y")][k] += cy
        rows[(b, "x")][k] -= cx
        rows[(b, "y")][k] -= cy
    for k, (j, d) in enumerate(supports):
        rows[(j, d)][len(members) + k] = 1.0
    keys = list(rows)
    a = [rows[key] + [-(loads.get(key[0], (0, 0))[0 if key[1] == "x" else 1])] for key in keys]
    for c in range(n):                                   # Gaussian elimination
        p = max(range(c, n), key=lambda r: abs(a[r][c]))
        a[c], a[p] = a[p], a[c]
        for r in range(n):
            if r != c:
                f = a[r][c] / a[c][c]
                a[r] = [x - f * y for x, y in zip(a[r], a[c])]
    return [a[i][n] / a[i][i] for i in range(n)]

joints = {"A": (0, 0), "B": (4, 0), "C": (8, 0), "D": (2, 3), "E": (6, 3)}
members = [("A", "B"), ("B", "C"), ("D", "E"), ("A", "D"), ("D", "B"), ("B", "E"), ("E", "C")]
loads = {"B": (0, -40)}
supports = [("A", "x"), ("A", "y"), ("C", "y")]
res = solve_truss(joints, members, loads, supports)
for (a, b), f in zip(members, res):
    print(a + b, round(f, 2), "T" if f > 1e-9 else ("C" if f < -1e-9 else "-"))
print([round(r, 2) for r in res[len(members):]])
# AB 13.33 T
# BC 13.33 T
# DE -26.67 C
# AD -24.04 C
# DB 24.04 T
# BE 24.04 T
# EC -24.04 C
# [0.0, 20.0, 20.0]
    

The last line gives the three reactions: no horizontal reaction at A, and 20 kN up at each support, which is half the 40 kN load each (the load is in the middle). Read the table: the bottom chord (AB and BC) is in tension at 13.33 kN, the top chord (DE) in compression at 26.67 kN, and the outer diagonals AD and EC in compression, with the two diagonals that meet at the loaded joint, DB and BE, in tension. You can check one by hand. At joint A the 20 kN reaction must be balanced by the vertical part of AD: AD points up and to the right at a slope of 3 in 2, so 20 + F×(3/3.606) = 0, giving F = −24.04 kN (compression). The program agrees. It is worth doing that check on every program you write: a single hand-verified value catches a sign error that would otherwise corrupt every answer.

Deeper Is Lighter

The top and bottom chords carry the beam's bending moment, just as the flanges of an I-beam do (How Bridges Work 2): the force in a chord is roughly the beam moment divided by the depth of the truss, F = M / depth. The code below applies that to a 40 m truss under a uniform load of 20 kN/m:

w, span = 20, 40                      # kN/m uniform load, m
m = w * span**2 / 8                   # beam moment, kN m
print(m)                              # 4000.0
for depth in (4, 5, 8, 10):
    print(depth, m / depth)           # chord force = M / depth
# 4 1000.0
# 5 800.0
# 8 500.0
# 10 400.0
    

The beam moment is 4,000 kN·m. A truss 4 m deep needs chords carrying 1,000 kN; at 10 m deep each chord carries only 400 kN. A deeper truss has lighter chords, but longer diagonals and verticals, and more of them, and a deeper structure is harder to build and to keep stiff against the wind. Truss depth is therefore a compromise.

Buckling: Why Compression Bars Are Dangerous

A tension bar fails only when the material tears. A compression bar can fail much earlier, by suddenly bowing sideways, called buckling. Try pushing the two ends of a plastic ruler together: it bows out long before it crushes. The load at which a slender, pin-ended strut buckles was worked out by Euler: P = π²EI / L², where E is the stiffness of the material, I is the same second moment of area as in How Bridges Work 2 (for the weakest direction of the cross-section), and L is the length.

import math

E = 200e3                             # MPa

def euler_kn(i_mm4, length_m):
    """Buckling load (kN) of a pin-ended strut."""
    return math.pi**2 * E * i_mm4 / (length_m * 1000) ** 2 / 1000

i = 8.0e6                             # mm4, a modest tube or angle
for length in (2, 4, 8):
    print(length, round(euler_kn(i, length)))
# 2 3948
# 4 987
# 8 247
    

Doubling the length divides the buckling load by four: a 2 m strut carries 3,948 kN, a 4 m strut 987 kN, an 8 m strut only 247 kN. This is the reason compression members are kept short, braced against the sideways movement, or made as hollow tubes (which have a large I for little material). The great cantilever arms of the Forth Bridge are a famous example: its big compression members are steel tubes, and its tension members are lattice (this is general knowledge, and How Bridges Work 6 looks at it properly).

Common mistakes

Using the pin model on a truss with a long beam-like member that carries load between joints (it bends, and the model is wrong); taking a negative answer as an error instead of reading it as compression; and checking a compression bar only for crushing. Always check buckling, which usually fails first.

Where This Leads

Trusses give a light, deep beam for spans up to a few hundred metres, but all that steel still has to be made, delivered and kept free from rust. The next chapter steps back from shapes to the stuff bridges are made of: timber, stone, cast and wrought iron, steel and concrete, and why the Forth Bridge is painted.

Exercise 1

Using the solver and the truss of this chapter (joints A, B, C at x = 0, 4, 8 m on the bottom and D, E at (2, 3) and (6, 3) on top), solve it for (a) 60 kN downward at the top joint E and (b) 40 kN at B and 60 kN at E together. Print the force in each bar, whether it is in tension or compression, and the reactions, and check that the vertical reactions add up to the load. Which bar changes from compression to tension between (a) and (b)?

📄 View solution

Exercise 2

Build a Pratt truss with joints A to E on the bottom chord at x = 0, 4, 8, 12, 16 m and F, G, H on top at (4, 3), (8, 3), (12, 3). Members: bottom AB, BC, CD, DE; top FG, GH; end posts AF and HE; verticals BF, CG, DH; diagonals FC and HC. A pin at A and a roller at E carry 50 kN down at each of F, G and H. Check that m + 3 = 2j, solve it, and say which bars are in tension, which in compression, and which carry no force.

📄 View solution

Exercise 3

A 40 m truss carries 20 kN/m. Its top chord is a pin-ended strut with I = 8.0×10⁶ mm⁴ and a panel length of 4 m (E = 200,000 MPa). For depths of 4, 5, 8 and 10 m work out the chord force (moment divided by depth) and say whether the strut passes with a factor of safety of 2 against buckling. Then find the smallest depth, in 0.5 m steps, that does.

📄 View solution

Quick Reference

Triangle: the only rigid polygon; every truss is built from triangles
Pin model: loads at joints only, so each bar is purely tension or compression
Rigidity count: m + 3 = 2j for a plane truss with no spare bars
Chords: the top is usually in compression, the bottom in tension; chord force ≈ M / depth
Pratt: diagonals in tension, verticals in compression; Howe: the reverse; Warren: alternating diagonals
Method of joints: forces balance at every joint, giving 2j equations; positive is tension
Buckling: P = π²EI/L²; doubling the length cuts it to a quarter

Coming next

How Bridges Work 5 is about materials: timber, stone, cast and wrought iron, steel and concrete, how fatigue and corrosion wear a bridge down, and why the Forth Bridge is painted.