Trusses
A truss is a structure bought cheaply. Three assumptions that are all slightly false turn a lattice of thirty members into thirty numbers you can find with a pocket calculator — and the whole method stands or falls on knowing what those assumptions cost.
A roof truss over a sports hall might have forty members and twenty-two joints. Analysed as a general rigid body it is a nightmare: every member could carry an axial force, a shear and a bending moment, and nothing in statics alone will separate them. Analysed as a truss it takes an afternoon, because a truss is defined by the assumptions that make it easy rather than by what it looks like. This page is about those assumptions, the two methods they licence, and the members you can solve by looking rather than by calculating.
The idealisations
Three assumptions define a truss, and all three are approximations. The joints are frictionless pins, so no joint can transmit a couple. Every load and every reaction is applied at a joint, never partway along a member. And the members are weightless, or light enough that their weight is either negligible or split half to each end joint.
None of this is literally true. Real joints are welded or bolted through gusset plates and they certainly resist rotation; real members weigh something. What rescues the idealisation is that the members of a truss are slender and their centrelines meet at a point. If the centrelines are concurrent at each joint and the loading is applied at the joints, then the bending the joint fixity does introduce is small — secondary stresses, in the trade, and typically under a tenth of the axial stress. Get the centrelines wrong, so that they miss each other at a joint, and the eccentricity produces bending that is not secondary at all. That is a real detailing failure and not a modelling nicety.
What the assumptions buy is one thing, and it is enormous. A member of a truss is loaded at its two end pins and nowhere else, so it is a two-force member. Force balance makes the two end forces equal and opposite; moment balance about one pin forces the other to have no moment about it, so it must act along the line joining the pins.
The direction of every member force is therefore known from the drawing, before any equation is written. All that is left of each member is a single scalar. A structure that would have carried three unknowns per member now carries one, and the entire analysis reduces to finding one number per member.
A last caution about the geometry. Only a simple truss — one built up from a base triangle by adding two new members and one new joint at a time — is guaranteed rigid and determinate. A four-bar quadrilateral with no diagonal is a mechanism: it will fold under load however strong its members are, and no amount of equilibrium algebra will tell you so if you do not look at the shape.
A real truss has gusset plates, bolt groups and heavy members.
Tension and compression
Because a member force is a scalar, it needs a sign, and the universal convention is that tension is positive. A member in tension is being stretched: the forces its neighbours apply at its two ends point away from each other, out along the member. A member in compression is being squashed, and the end forces point towards each other.
The confusion arrives when you draw the free-body diagram of a joint rather than of a member, which is what you actually do when solving. Newton’s third law reverses everything. A member in tension pulls its two joints inwards, towards the member, so on the joint the arrow points away from the joint and along the member. A member in compression pushes its joints apart, so on the joint the arrow points towards the joint.
Learn one of these and derive the other rather than memorising both. The one worth learning is the joint version, because it is the one you draw: on a joint, an arrow pointing away is tension.
The practical procedure is simpler still. Assume every unknown member is in tension, always — draw every arrow pointing away from the joint, without thinking about whether it is plausible. Then a positive answer means tension and a negative one means compression, and you have removed the single largest source of sign errors from the problem. Guessing sensibly member by member feels clever and is not: half the time you will guess right, and the bookkeeping needed to remember which half will cost more than the guessing saved.
Report the answers with the letter, not the sign: \(F_{AF} = 8.49\ \text{kN (C)}\) rather than \(-8.49\ \text{kN}\). The distinction is not cosmetic. A tie in tension can be a slender rod, while the same force in compression makes the member a strut, and a strut can buckle at a load far below the one that would crush it. Which of the two a member is decides how it is sized.
One member, pinned at each end, touched nowhere in between.
The method of joints
Every joint of a truss is acted on by concurrent forces — the member forces all pass through the pin, and so does any load applied there. Concurrent forces have no moment about the point they share, so the moment equation is identically satisfied and gives nothing. Each joint therefore yields exactly two useful equations:
Two equations means you can only start at a joint with two unknown member forces or fewer. That single constraint drives the whole procedure. Solve the support reactions from the whole truss first, so that an end joint has a known force at it, then work inwards, and after each joint is finished look for the next one that has been brought down to two unknowns.
Take the four-panel truss in the figure: 12 m in four 3 m panels, 3 m deep, with 12 kN applied at the middle of the bottom chord. Symmetry gives \(A_y = E_y = 6\ \text{kN}\). Joint \(A\) has the horizontal chord \(AB\) and the diagonal \(AF\), and only the diagonal has a vertical component, so the vertical sum solves on its own:
and then the horizontal sum hands over the chord, \(F_{AB} = -F_{AF}\cos 45^\circ = +6\ \text{kN}\), six kilonewtons in tension. Notice that the equations were never simultaneous. Choosing the axis that only one unknown touches is worth as much here as choosing the moment centre was for a rigid body.
The same count settles whether the truss can be solved at all. With \(j\) joints there are \(2j\) equations; the unknowns are \(m\) member forces and \(r\) reaction components. So
is the condition for a statically determinate truss. Fewer unknowns than equations and it is a mechanism that will collapse; more, and it is indeterminate and needs the members’ stiffnesses, not just their geometry. The truss in the figure has \(m = 13\), \(r = 3\) and \(j = 8\), and \(13 + 3 = 16 = 2(8)\). As with any counting rule this is necessary and not sufficient — a truss can satisfy it and still be unstable if the extra bar is in the wrong panel, leaving one region over-braced and another a quadrilateral.
Working a truss by hand is the only way to learn where the equations come from, but it is also where an afternoon disappears into arithmetic. Once the method is in your hands, the truss calculator will build the same joint equations for a truss you enter and report the force in every member — including telling you which of the three failures above you have, rather than returning a number for a structure that cannot stand.
Reactions first. Symmetry splits the 12 kN load in half.
Zero-force members
Some members carry nothing at all, and two rules find them by inspection. They are worth applying before any arithmetic, because every zero they reveal is an unknown deleted from a later joint, and a joint that had three unknowns becomes one you can start at.
The first rule: if only two members meet at a joint, the joint is unloaded, and the two members are not collinear, then both carry zero. The proof is to set up axes along one member. The other member is the only thing with a component across that axis, so its force must vanish; then the first is the only thing left along the axis, so it vanishes too.
The second rule: if three members meet at an unloaded joint and two of them are collinear, the third carries zero. Take the axis perpendicular to the collinear pair. Neither of them has a component on it, so the third member is alone, and alone it must be zero. Notice the corollary — the two collinear members then carry equal forces, so the force passes straight through the joint unchanged.
In the figure, joints \(B\), \(D\) and \(G\) are all bare, and each has a collinear pair plus one odd member, so the verticals \(BF\), \(CG\) and \(DH\) all carry nothing. Three of the thirteen members are settled before a single equation is written.
Two warnings. A joint carrying a load or a reaction is not unloaded, and the rules do not apply there, however innocent the geometry looks. And zero-force does not mean useless. A member is zero for this loading; move the 12 kN to another panel point, or load the truss from the other side, and it will carry force. Those members also brace the compression chord against buckling out of plane, which is a stability job that a statics calculation cannot see at all. Nobody deletes them from the drawing.
Only joint C carries a load. Every other joint is bare.
The method of sections
The method of joints is exhaustive and, for one member deep inside a long truss, wasteful: to reach the top chord at mid-span you would work through four joints you did not care about. The method of sections goes straight there. It is the free-body boundary argument used at full strength — draw the curve so it crosses the members you want to find, and nothing else.
Cut an imaginary line clean through the truss, crossing the member you want. Discard one side. The piece you keep is a rigid body, not a particle, so it gives three equations, and the forces in the cut members are now external to it, still acting along their own members because those members are still two-force members.
Three equations means the cut should generally pass through no more than three members. And the real economy is in choosing where to take moments: pick the point where two of the three cut members intersect, and both drop out, leaving one equation in one unknown.
Cutting the figure’s truss through \(FG\), \(FC\) and \(BC\), and keeping the left-hand piece, the diagonal and the bottom chord meet at \(C\). Moments about \(C\), taking clockwise as positive, leave only the reaction and the top chord:
One equation, and the member is done. Take moments about \(F\) instead, where the top chord and the diagonal cross, and the bottom chord falls out just as directly: \(F_{BC} = +6\ \text{kN}\). The vertical force sum, which the two chords cannot touch, gives the diagonal: \(F_{FC} = +8.49\ \text{kN}\). Three members, three independent equations, no simultaneous solving.
A cut through more than three members is not automatically hopeless — if all but one of them are concurrent, moments about that common point still isolate the odd one. And the two methods are meant to be mixed: a section to reach the middle of a long truss, then joints to walk outwards from there.
All of this rests on every member being a two-force member, which rested in turn on loads arriving only at the joints. Put a load halfway along a member — hang a hoist from the middle of a bottom chord — and that member is no longer a two-force member, its direction is no longer known, and neither method applies to it. The structure has become a frame, and it has to be taken apart instead.