Trusses

A single rigid body held by two supports is not much of a structure, and the equilibrium conditions of the previous lesson gave only the reactions from the ground. Real structures are assemblies, and the oldest efficient assembly is the truss: a framework of straight bars joined at their ends into a pattern of triangles.

Why triangles

Take four bars pinned into a quadrilateral. Nothing prevents it collapsing into a parallelogram and then into a flat line, because the bars keep their lengths while the angles change freely. Take three bars pinned into a triangle and the shape is fixed, since the three side lengths determine the three angles completely. That is the whole reason a truss is drawn as triangles.

The engineering payoff is not stiffness for its own sake. In a triangulated frame with the loads applied at the joints, every bar carries force along its own length and nothing else, which is the most efficient way a piece of material can be used: the stress is the same at every point of the cross section, so no part of the bar is idle. A beam of the same span and depth uses its material far less evenly, which is why a railway bridge of 60 metres is a truss and a floor joist of 6 metres is not. The truss wins when the span is long enough that the beam's wasted material outweighs the cost of making dozens of connections.

Trussed roofs appear in Roman basilicas, but the analysis is nineteenth century, driven by railway building. Squire Whipple published the first correct treatment in 1847, and the American railway trusses of the following decades are still known by the names of the people who patented their patterns: Pratt, Howe, Warren, Baltimore.

The idealisation, stated honestly

Three assumptions turn a real framework into a solvable one.

The bars are joined by frictionless pins, so a joint can transmit force but no moment. The members are straight and weightless, or at least their weight is small enough to be split between their end joints. And all loads are applied at the joints, never partway along a member.

Together these make each member a two-force member: the only forces on it act at its two ends, and for that bar to be in equilibrium those two forces must be equal, opposite, and along the line joining the ends. If they had any component across the bar, they would form a couple with nothing to balance it. So one number describes each member, positive for tension and negative for compression, and that sign convention is worth holding to rigidly because compression members behave very differently, as a later lesson shows.

None of the three assumptions is exactly true. Real joints are welded or bolted through gusset plates and do resist rotation, so members pick up small bending stresses, known as secondary stresses, typically under ten per cent of the axial stress in a well proportioned truss. Members do weigh something. Purlins do sometimes land between joints. The idealisation survives because the errors are small and, more importantly, because the axial forces it predicts are correct even when the extra bending is not captured: the frame still has to carry the load somehow, and the triangulated path is much the stiffest one available.

The method of joints

Every joint is a particle acted on by concurrent forces, so it obeys $\sum F_{x} = 0$ and $\sum F_{y} = 0$ : two equations per joint, no moment equation, because all the forces pass through the same point. Work joint by joint, always choosing one with at most two unknown member forces, and the whole truss unravels.

The bookkeeping that avoids sign errors is to assume every member is in tension. Then the force a member exerts on a joint points away from that joint, along the member, and a negative answer simply means compression. No redrawing is needed and no arrow has to be guessed.

Example. A truss consists of three bars: a horizontal bottom chord from $A$ at $(0, 0)$ to $B$ at $(6, 0)$ in metres, and two rafters meeting at an apex $C$ at $(3, 4)$ . There is a pin at $A$ and a roller at $B$ , and a load of $40$ kN hangs at $C$ . Find the force in every member.

Each rafter is $5$ m long, from the 3-4-5 triangle, so its direction cosines are $0.6$ horizontally and $0.8$ vertically. The loading is symmetric, so each reaction is $20$ kN upwards.

At joint $A$ the unknowns are $F_{A B}$ along $(1, 0)$ and $F_{A C}$ along $(0.6, 0.8)$ , and the reaction is $(0, 20)$ . Vertically, $0.8 F_{A C} + 20 = 0$ , so $F_{A C} = - 25$ kN, meaning $25$ kN of compression. Horizontally, $F_{A B} + 0.6 (- 25) = 0$ , so $F_{A B} = + 15$ kN, tension. By symmetry $F_{B C} = - 25$ kN.

Check at the apex: the two rafters each push down on nothing and pull the joint towards $A$ and $B$ respectively with $- 25$ kN, giving vertical components $(- 25) (- 0.8) = 20$ kN each, which is $40$ kN upwards against the $40$ kN load.

Now you. The same three-bar truss has its apex at $(4, 3)$ with the supports at $(0, 0)$ and $(8, 0)$ , and carries $24$ kN at the apex. Find all three member forces.

Answer

The rafters are again $5$ m long, now with direction cosines $0.8$ horizontal and $0.6$ vertical, and the reactions are $12$ kN each. At the left support, $0.6 F = - 12$ gives $F = - 20$ kN, so both rafters carry $20$ kN of compression. Horizontally the bottom chord takes $- 0.8 (- 20) = + 16$ kN, tension.

Zero-force members

Some members carry nothing at all under a given load case, and spotting them by inspection saves a great deal of arithmetic. Two rules cover almost every occurrence.

If exactly two members meet at an unloaded joint and they are not collinear, both carry zero: resolve perpendicular to one member and the other's force must vanish, then resolve again. If three members meet at an unloaded joint and two of them are collinear, the third carries zero, for the same reason applied to the direction perpendicular to the pair.

A zero-force member is not useless. It is zero for this load case, and a different pattern of loading, wind on one side rather than snow on both, will generally give it work to do. It also braces the members it connects against buckling out of the plane, which is a service the force calculation never sees. Removing one because the analysis says zero is a standard way to build something that falls down under a load nobody analysed.

The method of sections

The method of joints is complete but sequential: to reach a member in the middle of a long truss you may have to solve twenty joints first. When only one or two member forces are wanted, cut straight through the truss instead.

Slice the truss along a line crossing at most three members whose forces are unknown, discard one side, and treat the remainder as a rigid body. It has three equilibrium equations, so three unknowns can be found, and the moment equation can be taken about the point where two of the cut members intersect, which leaves one unknown alone in its own equation.

Take a six-panel Pratt through-truss spanning $24$ m. The bottom chord runs along $y = 0$ with joints $L_{0}$ to $L_{6}$ every $4$ m, and the top chord runs at $y = 3$ m with joints $U_{1}$ to $U_{5}$ directly above $L_{1}$ to $L_{5}$ . Verticals join each $L_{i}$ to the $U_{i}$ above it, end diagonals run from $L_{0}$ to $U_{1}$ and from $L_{6}$ to $U_{5}$ , and the four interior diagonals slope down towards midspan, from $U_{1}$ to $L_{2}$ , $U_{2}$ to $L_{3}$ , $U_{5}$ to $L_{4}$ and $U_{4}$ to $L_{3}$ . A deck load puts $24$ kN at each of $L_{1}$ through $L_{5}$ . The total is $120$ kN and the truss is symmetric, so each reaction is $60$ kN.

Example. Find the force in the top chord member $U_{2} U_{3}$ .

Cut vertically between $x = 8$ m and $x = 12$ m. The cut crosses $U_{2} U_{3}$ , the diagonal $U_{2} L_{3}$ , and the bottom chord $L_{2} L_{3}$ . Keep the left portion, which carries the $60$ kN reaction at $L_{0}$ and the loads at $L_{1}$ and $L_{2}$ . Two of the cut members, the diagonal and the bottom chord, meet at $L_{3}$ at $(12, 0)$ , so take moments there.

Counting anticlockwise as positive, the reaction at $L_{0}$ is $12$ m to the left of $L_{3}$ and contributes $- 60 \times 12 = - 720$ kN m. The loads at $L_{1}$ and $L_{2}$ are $8$ m and $4$ m to the left and contribute $+ 24 \times 8 = 192$ and $+ 24 \times 4 = 96$ kN m. The top chord force $F$ acts horizontally at $3$ m above the moment point, contributing $- 3 F$ if it is tensile. Everything else in the cut passes through $L_{3}$ and drops out. So $- 720 + 192 + 96 - 3 F = 0$ , giving $F = - 144$ kN: the top chord carries $144$ kN of compression, which is what a top chord should do.

Now you. Using the same truss and the same cut, find the force in the bottom chord $L_{2} L_{3}$ by taking moments about $U_{2}$ at $(8, 3)$ .

Answer

The diagonal $U_{2} L_{3}$ and the top chord both pass through $U_{2}$ , so only the bottom chord force survives from the cut. The reaction contributes $- 60 \times 8 = - 480$ kN m, the load at $L_{1}$ contributes $+ 24 \times 4 = 96$ kN m, and the load at $L_{2}$ sits directly below $U_{2}$ and contributes nothing. A tensile bottom chord force $F$ acting $3$ m below the moment point contributes $+ 3 F$ . So $- 480 + 96 + 3 F = 0$ and $F = + 128$ kN, tension.

Solving the whole truss gives a pattern worth reading. The bottom chord runs $80$ , $80$ , $128$ , $128$ , $80$ , $80$ kN in tension from one end to the other, and the top chord runs $128$ , $144$ , $144$ , $128$ kN in compression. Both track the bending moment of an equivalent solid beam, divided by the depth of the truss: the chords are doing the beam's bending and the web members are doing its shear. The end diagonals carry $100$ kN of compression and the diagonals nearer midspan only $60$ and $20$ kN, because shear is largest at the supports and smallest in the middle. The vertical $L_{3} U_{3}$ at midspan carries exactly zero, since the two diagonals meeting at $L_{3}$ each contribute $20 \times 0.6 = 12$ kN of upward pull, together taking the whole $24$ kN load at that joint.

Counting members and joints

Before analysing anything, count. Each joint gives two equations in a plane, so a truss with $j$ joints supplies $2 j$ equations. The unknowns are the $m$ member forces plus the $r$ reaction components. Then

m + r = 2 j

is the condition for a statically determinate truss, $m + r < 2 j$ marks a mechanism with too few bars, and $m + r > 2 j$ marks an indeterminate truss with more bars than statics can sort out.

Example. Check the six-panel truss above.

Its joints are $L_{0}$ to $L_{6}$ and $U_{1}$ to $U_{5}$ , so $j = 12$ . Its members are six bottom chords, four top chords, five verticals, two end diagonals and four interior diagonals, so $m = 21$ . A pin and a roller give $r = 3$ . Then $m + r = 24$ and $2 j = 24$ : determinate, as the section calculations assumed.

Now you. A planar truss has $9$ joints, $16$ members, a pin at one support and a roller at the other. Classify it.

Answer

$m + r = 16 + 3 = 19$ while $2 j = 18$ . There is one member more than statics can handle, so the truss is statically indeterminate to the first degree.

As in the previous lesson, the count is necessary and not sufficient. A truss can satisfy $m + r = 2 j$ and still be a mechanism if the bars are arranged badly, with a redundant bar in one panel paying for a missing one somewhere else. The count catches the common mistakes and the geometry has to be looked at anyway.

What the truss analysis still does not tell you

At the end of all this every member has a number attached: $128$ kN of tension in one bottom chord, $100$ kN of compression in an end diagonal. Nothing yet says whether any of them survives.

Two gaps remain. The first is that a force is not a verdict: $100$ kN would snap a bicycle spoke and would barely register in a bridge chord, so the force has to be compared against the size of the member. That comparison is stress, two lessons away. The second is that tension and compression are not mirror images. A slender bar in compression can fail by bowing sideways at a load far below anything the material would object to, which is the subject of the buckling lesson and the reason truss compression members are stocky while ties can be thin cables.

Before either, there is the more common structural element that a truss was invented to replace over short spans. A beam takes its load anywhere along its length rather than only at joints, so the force it carries varies from point to point, and describing that variation is the next step.