Structural Engineering & Statics
where the load goes
Equilibrium

Counting the unknowns, and finding out whether statics can answer

Two equations per joint, one unknown per member, one per restraint. Subtract, and the sign of the answer says whether the structure is a mechanism, solvable, or beyond what equilibrium alone can settle.
10 min read 7 figures One support too manyThe statics of things that do not move

Before a structure is analysed it is worth asking whether it can be. The answer comes from counting, it takes ten seconds, and it decides which body of theory the problem belongs to.

Count the unknowns: one axial force per member, one component per restrained direction at each support. Count the equations: two per joint in a plane. Subtract. The sign of what is left says everything.

Counting unknowns against equationsThree frames differing by one member. Two equilibrium equations per joint, one unknown per member and one per restraint: fewer unknowns than equations is a mechanism, equal is solvable by statics, more needs stiffness.m 4 + r 3 − 2j 8 = -1a mechanismm 5 + r 3 − 2j 8 = 0statically determinatem 6 + r 3 − 2j 8 = +1one member too manystatics can answer only the middle case
Fig. 1 Three frames differing by one member. Two equilibrium equations per joint, one unknown per member and one per restraint — and the difference decides whether statics can finish the job.

The three cases

Fewer unknowns than equations. The equations cannot all be satisfied, and the structure moves. It is a mechanism: a four-bar square with pinned corners folds into a rhombus without stretching anything, and no amount of member strength prevents it. This is the failure mode that kills people during construction, when a frame is complete but the bracing has not gone in.

Exactly as many. The system has a unique solution and statics finds it. The structure is statically determinate, and everything on this site that is solved by equilibrium alone lives here.

More unknowns than equations. The system has infinitely many solutions satisfying equilibrium, and equilibrium alone cannot choose between them. The structure is statically indeterminate, or redundant, and finishing the job needs to know how stiff each member is.

The counting rule for a plane pin-jointed truss is

m+r−2j,m + r - 2j,

with mm members, rr restraint components and jj joints. Negative is a mechanism, zero is determinate, positive is the degree of redundancy.

Counting unknowns against equationsThree frames differing by one member. Two equilibrium equations per joint, one unknown per member and one per restraint: fewer unknowns than equations is a mechanism, equal is solvable by statics, more needs stiffness.m 5 + r 3 − 2j 8 = 0statically determinatem 6 + r 3 − 2j 8 = +1one member too manystatics can answer only the middle case
Fig. 2 The determinate frame and the same frame with one more diagonal. The extra member adds an unknown without adding an equation, so the count goes from zero to one and the structure passes out of reach of statics.

Why an extra member is not free

The intuition that more members means a stronger structure is correct and incomplete. The extra diagonal in the second figure does make the frame stronger. It also makes it a different kind of problem.

With two diagonals rather than one, the load can travel by two routes, and equilibrium does not say how it splits. Any split that adds up is in equilibrium — all the load down one diagonal, all down the other, or any mixture. Choosing among them requires a further principle, and the principle is compatibility: the two diagonals share their end joints, so whatever the load does it must leave both members fitting the same deformed shape.

That converts the problem from a force problem into a force-and-displacement problem. The stiffer route takes more load, in proportion to stiffness, and stiffness depends on the material, the area and the length — none of which appeared anywhere in statics.

The practical consequence is that a redundant structure is sensitive to things a determinate one is not. A support that settles a few millimetres redistributes load in a redundant frame and does nothing at all in a determinate one. Temperature change does the same. A member fabricated slightly short is stressed before any load arrives.

What redundancy buys

Given the cost, redundancy is nonetheless normal, and for one reason: it survives losing something.

A determinate structure has exactly enough members, and every one of them can be found by statics. Remove one and the count goes negative — it becomes a mechanism, and it comes down. There is no alternative route because the count says there is none.

A redundant structure has a spare. Remove a member and it is still standing, with the load redistributed to the remaining routes, which is the load path being re-chosen under duress. That property is called robustness, and after the progressive collapse at Ronan Point in 1968 it became a design requirement rather than a bonus.

So the trade is explicit. Determinacy buys analysability, insensitivity to settlement and temperature, and clean force paths. Redundancy buys survival. Most real structures choose redundancy and pay the analysis cost, which since about 1960 has been paid by a computer.

One support too manyThe same uniformly loaded beam with three sets of restraints, and the bending moment in each. Adding restraint moves moment from mid-span to the supports and lowers the peak — but only the first case can be solved by statics.simply supportedstatics alonesag 32.0propped at one endneeds stiffnesssag 18.0hog 32.0built in at both endsneeds stiffnesssag 10.7hog 21.3the load never changes; only what is holding the endsthe built-in case peaks at two-thirds of the simple span's moment
Fig. 3 The same uniformly loaded beam with three sets of restraints. Only the first can be solved by statics; the other two need the stiffness, and both have lower peak moments as a result.

That figure shows the other thing redundancy buys, which is efficiency. Building in both ends drops the peak moment to two-thirds of the simply supported value, because the ends now take a share. The material saving is real, and it is unavailable to a determinate structure.

Counting a truss

The rule generalises with care. For a pin-jointed truss the formula above is right. For a rigid-jointed frame each member carries three unknowns rather than one, and the count changes accordingly.

A Pratt truss of 6 panelsA Pratt truss under equal panel-point loads. The joint equilibrium equations were assembled and solved; 10 members came out in tension, 9 in compression and 2 carrying nothing.tensioncompression2 carrying nothing
Fig. 4 A six-panel Pratt truss. Twenty-one members, three restraint components and twelve joints — twenty-four unknowns against twenty-four equations, which is why every member force in this figure could be solved.
A Warren truss of 6 panelsA Warren truss under equal panel-point loads. The joint equilibrium equations were assembled and solved; 10 members came out in tension, 11 in compression and 2 carrying nothing.tensioncompression2 carrying nothing
Fig. 5 A Warren truss of the same span. A different member arrangement with the same count, and the same conclusion — the difference between the two shows up in which members pull and which push, not in whether they can be found.

Both are determinate, and the solver behind these figures proves it by returning a unique answer. When a frame is redundant the same solver has to fall back on a least-squares solution, which is a warning sign rather than an answer.

There is a subtlety the count cannot see. A structure can have the right number of members and still be a mechanism if they are arranged badly — three collinear members restraining a joint, or a panel with no diagonal while another has two. The count is necessary and not sufficient, and the honest test is whether the equations are independent, which is what the solver actually checks.

What the count does not tell

Determinacy is a statement about solvability, not about strength, stiffness, or whether the structure is any good.

A determinate truss can be hopelessly under-designed. A redundant frame can be a mechanism in disguise. A structure with a count of zero and one member ten times too small collapses at a tenth of the intended load, and the count is entirely silent about it.

The count also says nothing about which arrangement is better. The Pratt and the Howe have identical counts and put their diagonals in tension and compression respectively — a difference that decides which one to build in steel and which in timber, and one the count cannot express.

The other thing the count decides

Redundancy is not only an analysis problem. It changes what the structure does under load.

One support too manyThe same uniformly loaded beam with three sets of restraints, and the bending moment in each. Adding restraint moves moment from mid-span to the supports and lowers the peak — but only the first case can be solved by statics.simply supportedstatics alonesag 32.0propped at one endneeds stiffnesssag 18.0hog 32.0built in at both endsneeds stiffnesssag 10.7hog 21.3the load never changes; only what is holding the endsthe built-in case peaks at two-thirds of the simple span's moment
Fig. 6 The same beam with three sets of restraints. Adding restraint lowers the peak moment and moves part of it to the supports, which is a real efficiency and not a bookkeeping artefact.

A built-in beam peaks at two-thirds of the simply supported moment and deflects a fifth as far, which for a member sized by how far it moves is the difference between two sections. That gain is unavailable to a determinate structure, and it is bought with the analysis cost the count predicts.

The deflected shape is the moment, integrated twiceA loaded beam with its deflected shape above and its bending-moment diagram below. The shape was obtained by integrating the moment twice and fitting the constants to the supports; the vertical scale is exaggerated enormously.the largest movement, at x = 4.00momentdrawn at roughly three hundred times the real deflection —a beam at its serviceability limit moves about a three-hundredth of its span
Fig. 7 A deflected shape, obtained by integrating the moment diagram twice. For a redundant structure the moment diagram itself depends on this curve, which is why the two have to be solved together.

The circularity in that last sentence is the whole difficulty. In a determinate structure the forces come first and the deflections follow; in a redundant one each depends on the other, and the solution has to satisfy both at once.

Where the model stops

Pin joints. The formula above assumes joints that transmit no moment, which is the truss idealisation. Real connections are bolted or welded and do transmit moment, which adds unknowns and usually adds redundancy. A truss analysed as pin-jointed and built with welded joints has secondary bending stresses that the analysis does not contain.

Plane structures. In three dimensions there are three equations per joint and the formula becomes m+r−3jm + r - 3j. Space frames are counted differently and fail differently.

Small deflections. The count is a statement about the equations of the undeformed structure. A cable net has a count that says mechanism and stands up anyway, because its geometry changes under load until it can carry it — the analysis is genuinely nonlinear and the linear count does not apply.

No internal releases. A hinge inserted in a member adds an equation, and a three-pinned arch is determinate for exactly that reason. Any count that ignores releases will call it redundant.

The figures have a limitation of their own: they show the count as an arithmetic caption beside a drawing, which makes it look like a property that can be seen. It cannot. Two frames that look nearly identical can differ in count by one member, and the difference in what can then be said about them is total.

The ladder from here

Later rungs: internal releases and how they change the count. The three-pinned arch, determinate by design. Rigid-jointed frames and their different formula. Space frames. Kinematic indeterminacy, which counts degrees of freedom rather than forces. The force method and the compatibility equations that resolve redundancy. The displacement method, which is what every structural program actually does. Robustness and progressive collapse. And prestress in a mechanism, which is how a cable net or a tensegrity stands up while counting as a mechanism throughout.

Maxwell wrote the counting rule down in 1864 in a paper about reciprocal figures, as a side remark. It carries his name and is the least of what that paper contains.

Where this goes next

1289.510.5ΣM about one support gives the other reaction; ΣF then gives the first Everything adds to nothing, and that is the whole of statics A structure that stays put obeys two statements — the forces on it sum to zero, and so do the moments. Every number in the subject comes out of those two sentences. simply supportedstatics alonesag 32.0propped at one endneeds stiffnesssag 18.0hog 32.0built in at both endsneeds stiffnesssag 10.7hog 21.3the load never changes; only what is holding the endsthe built-in case peaks at two-thirds of the simple span's moment One support too many, and what it costs to know Add a redundant restraint and the load has two routes to the ground. Equilibrium cannot say how it splits, and the answer turns out to depend on stiffness — which is a different kind of question. tensioncompression2 carrying nothing The triangle that cannot fold, and everything built out of it A square of pinned bars is a mechanism. A triangle is not, and that single fact is the reason trusses exist and the reason they look the way they do.