Frames and machines

Most pinned structures are not trusses. A crane jib with a cylinder halfway along it, a pair of pliers, a folding chair, the linkage on a digger — all of them are assemblies of pinned members, and none of them obeys the rule that made a truss easy. The analysis that replaces it is not harder in principle, only longer and far less forgiving of sign errors, and it rests entirely on drawing the right free bodies.

Not a truss

The definition is blunt: a frame or a machine is a pin-connected structure containing at least one multi-force member — a member acted on at three or more points, or at two points by forces that are not collinear. One such member anywhere in the assembly is enough, and it does not matter how truss-like the rest of it looks.

The distinction between a frame and a machine is about purpose rather than method. A frame is stationary and fully constrained, and its job is to support loads: a bracket, a roof frame, a shelving stanchion. A machine has moving parts and its job is to transmit a force and usually to change it: pliers, tongs, a hydraulic press, a bicycle brake. Both are dismembered and analysed the same way. The only real difference is that a machine is analysed in one position at a time, because the geometry — and therefore the answer — changes as it moves.

What is lost is the thing the truss idealisation bought. In a truss every member is a two-force member, so its force acts along its own axis and one scalar describes it. A multi-force member has no such line. The force it receives at a pin can point anywhere, so it must be carried as two unknown components, and the member also carries shear and bending along its length, which is why frame members are I-sections and beams while truss members are angles and rods.

Practically, this rules out both truss methods at a stroke. The method of joints assumed the forces at a joint were all directed along known member axes; here they are not. The method of sections assumed the forces exposed by the cut lay along their members; here they do not, and cutting through the middle of a member exposes an axial force, a shear and a bending moment. Neither method survives.

The compensation, for a machine, is mechanical advantage. A pair of pliers with 100 mm handles and 25 mm jaws is two levers on a common pin; taking moments about that pin turns 40 N of grip into 160 N at the jaw. The output is the input scaled by a ratio of moment arms, and that ratio is the whole reason the thing exists.

Dismembering

The free body of the complete frame is still worth drawing first, because internal pin forces do not appear on it and the supports often come out immediately. But it is rarely enough. Three equations cannot resolve four support components, and in any case it says nothing about what the members carry.

So take the frame apart. Draw a separate free-body diagram of every member, and of every pin that connects three or more members, and of any pulley you have to break out. Each part is then an ordinary rigid body with its own three equations, and the internal forces that were invisible on the whole structure are external forces on the pieces — the same shift of the boundary that makes the method of sections work.

Two rules govern the drawing. Every force applied at a pin must be assigned to exactly one part: if a 2 kN load hangs from the pin at \(B\), put it on the diagram of one member and not on both, or you have doubled the load. And the weight of a part, if it is not being neglected, goes at its centre of gravity on that part’s diagram alone.

Then solve in the order that avoids simultaneous equations. Look for a part with three or fewer unknowns and start there, and within it take moments about a pin where two unknowns act so that both vanish. In the three-hinged frame in the figure, moments about \(A\) for the whole structure give the reaction at \(C\) on its own, and everything else follows from force sums.

One habit is worth insisting on because it is the difference between a tidy solution and an hour of confusion: label the pin forces once, at the start, and never re-label them. Every part that touches pin \(B\) refers to the same pair \(B_x\), \(B_y\), with the sense reversed on the second part and nowhere else.

Forces at a pin

The pin is where dismembering goes wrong, and it goes wrong in the same way for everybody. A pin transmits force between two members; by Newton’s third law, whatever member \(AB\) exerts on member \(BC\) through the pin, member \(BC\) exerts back on \(AB\) with the same magnitude and the opposite sense.

So the procedure is mechanical. On one part, assume components \(B_x\) and \(B_y\) in whatever direction you like. On the other part, draw the same two symbols pointing the opposite way. Not new symbols; not a fresh guess. The pair is a single unknown pair appearing twice, and that is what makes the six equations of the dismembered frame solvable against six unknowns rather than eight.

Assume the same sense on both parts, which is the classic slip, and the algebra quietly invents a force. The pin is then pushing the structure in some direction all by itself, the equilibrium equations become inconsistent, and the numbers that come out are wrong without being obviously wrong. It is worth checking the pairing explicitly on every pin before writing a single equation.

The initial guess itself does not matter in the least. If \(B_x\) comes out negative, the true force points the other way — and both arrows reverse together, so the third-law pairing is untouched. What matters is only that you never change the assumed sense partway through, and that when you substitute a solved value into a later equation you carry its sign with it.

A pin joining three or more members deserves a free-body diagram of its own, treated as a small particle with the forces from every member on it. Trying to lump such a pin onto one of the members is where the bookkeeping usually collapses. The same goes for a pulley: draw it separately, with the cable tension the same on both sides if it turns freely, and the pin force at its axle as a proper pair.

Two-force members inside

A frame is defined by having at least one multi-force member, not by having only those. Most frames contain links, struts, ties and hydraulic rams that are pinned at exactly two points and touched nowhere in between, and each of those is still a two-force member with all the advantages that carries.

The test takes a moment and pays for itself. Go through the members one at a time and ask what touches each: if it is pinned at exactly two places, carries no load between them, and its weight is neglected, then its force acts along the line joining the two pins. Two unknown components collapse to one unknown magnitude on a known line.

In the figure’s three-hinged frame, member \(AB\) carries 800 N in its middle and is multi-force, but \(BC\) is touched only at its two pins. So the reaction at \(C\) must act along \(CB\), which is enough to solve the whole frame without ever dismembering it:

Three unknowns, three equations, no dismembering at all. Miss the two-force member and the same frame takes six equations across two parts to reach exactly the same answer.

Two traps. A member that looks like a link but has a pin somewhere along it, or a cable attached partway down, is not a two-force member however slender it is. And a curved or bent two-force member still has its force along the straight line joining the two pins, not along its own curve — the moment argument cares about the two points, not about the shape between them.

Counting the parts

The determinacy check for a frame is done after dismembering, not before, and it is nothing more than adding up. With \(n\) separate parts in a plane there are \(3n\) equations:

Against that, count two unknown components for every pin connecting two members, one unknown for every two-force member, and the support reactions. Equal totals and the frame is statically determinate; more unknowns and it is indeterminate, and the extra load path has to be found from how much each member stretches rather than from statics.

The three-hinged frame is the standard illustration. Whole, it has four unknown reactions against three equations and looks indeterminate. Dismembered into two members it has six equations against \(A_x\), \(A_y\), \(B_x\), \(B_y\), \(C_x\), \(C_y\) — six unknowns, and it closes. The apparent indeterminacy was an artefact of refusing to take it apart.

The count relies on the pin pair being counted once, which is exactly what the third-law bookkeeping guarantees. Count \(B_x\) and \(B_y\) twice, once for each member, and you will convince yourself that a perfectly ordinary frame is indeterminate.

As with any determinacy count this one is necessary and not sufficient. A frame can balance its equations and still be a mechanism: three pins in a line, or a panel with no diagonal, will move under load however the totals add up. And a machine is deliberately a mechanism — it is held in place by the applied forces rather than by its supports, which is why a machine problem must always state the position it is analysed in, and why the answer changes as the jaws close.