Indeterminate Structures: The Criterion Behind the Code, With a Worked Check

Draw a simple beam, a pin at one end and a roller at the other, put a uniform load on it, and the two reactions come straight out of statics. Weld one end into a column, or drop a third support under it, and something quietly changes: there are now more unknown forces than the three equations of plane statics can solve. The structure is statically indeterminate, and equilibrium alone will not give you the answer.

Almost every real steel frame is indeterminate on purpose. The useful question for a working engineer is not whether a structure is indeterminate, it is by how much, because that single number, the degree of static indeterminacy, is the criterion that decides everything downstream: how much the structure can redistribute its load, how much lighter it can be than a determinate one, and, in the age of solvers, how you check an answer you did not compute by hand. This article is about that criterion and how to use it as a check. We take one beam, load it three ways along the indeterminacy scale, solve all three in the real CalcSteel FEM engine, and confirm every number against closed-form theory to three decimals.

Static indeterminacy is decided by arithmetic, not intuition. In a plane, a rigid body has exactly three independent equilibrium equations: the sum of horizontal forces, the sum of vertical forces, and the sum of moments, each equal to zero. Count your unknowns against those three and one of three things is true.

• Fewer unknowns than equations: the structure is a mechanism. It moves. It is not a structure.
• Exactly as many as equations: statically determinate. Statics solves it outright, and the internal forces depend only on geometry and load, never on the section.
• More unknowns than equations: statically indeterminate. The surplus is the degree of static indeterminacy (DSI), and each unit of it is one redundant, a force statics cannot find.

For a plane beam whose only unknowns are its reactions, the criterion is a subtraction, DSI = r − 3 − c, where r is the number of reaction components and c is the number of internal releases (an internal hinge sets the moment there to zero and adds one equation). A pin gives two reaction components, a roller one, a fixed support three. For a rigid plane frame you also count members and joints, DSI = (3m + r) − 3n − c. Run the criterion on the three beams in this article and you get 0, 1 and 3. That is the whole spectrum we will work through.

"The criterion behind the code" has two honest readings, and the same DSI number sits behind both.

Behind the software code. A modern solver never asks you which forces are redundant. It assembles a global stiffness matrix and solves K d = F for the joint displacements, then recovers every member force from those displacements. Indeterminacy is handled implicitly: the moment the stiffnesses are assembled, K already encodes how each member shares load with its neighbours, so a DSI-0 beam and a DSI-24 frame go through the exact same solve. The criterion never appears on screen, but it is doing the work, and it tells you how many independent conditions the solver quietly satisfied on your behalf. For how the hand methods make those conditions explicit, and where they stop being practical, see our companion piece on why the hand method stops and the matrix method starts.

Behind the design code. Steel design standards, AISC 360, ABNT NBR 8800, Eurocode EN 1993-1-1, are built on top of the assumption that most frames are indeterminate. That is why they contain rules for moment redistribution, why they reward redundancy with better robustness and progressive-collapse behaviour, and why the analysis method you are allowed to use, elastic, elastic with redistribution, or plastic, is tied to how ductile and how redundant the structure is. Redundancy is not an accident the code tolerates; it is a property the code is written around.

If indeterminacy makes the analysis harder, why build it in? Because the same continuity that adds redundants pays for three things you want, and each one shows up in the worked cases below.

Smaller peak moments. Fixity spreads bending toward the supports instead of piling it up at midspan. The same beam that peaks at wL²/8 when simply supported peaks at only wL²/12 when both ends are fixed, a third less moment for the identical span and load. Less moment means a smaller required section modulus, which means less steel.

Stiffness. Continuity fights deflection hard. Fix both ends of a uniformly loaded beam and it deflects only one fifth as much as the same beam simply supported. For the floor beams that are governed by deflection, and most are, that is often the difference between a section that works and one that does not.

Redundancy, which is to say safety. A redundant is literally an alternate load path. Overload one region of an indeterminate structure and the moment redistributes to the parts that still have capacity; it warns you before it falls. A determinate structure has no spare path: form one hinge and it becomes a mechanism. This is why codes lean on redundancy for robustness.

To see the criterion do real work, we hold everything constant except the supports. The beam spans L = 6 m and carries a uniform load of w = 20 kN/m, so wL = 120 kN and wL² = 720 kN·m² throughout. Only the ends change, and with them the degree of indeterminacy:

• Simply supported, a pin and a roller: r = 3, so DSI = 0. Determinate.
• Propped cantilever, fixed at one end, roller at the other: r = 4, so DSI = 1. One redundant.
• Fixed-fixed, built in at both ends: r = 6, so DSI = 3. Under a symmetric transverse load one of the three is a degenerate axial condition that carries no force, so two are active in bending.

All three were built and solved in the shipping CalcSteel FEM engine, each span subdivided into 48 elements, and every reaction and moment below matches the classical closed form to three decimals. Watch the governing bending moment fall as the criterion climbs.

Start with the simply supported beam, the reference every other case is measured against. Statics alone gives the reactions, wL/2 = 60 kN at each end, and the bending moment is a pure sagging parabola peaking at midspan:

M_max = wL²/8 = 90 kN·m, at x = 3.0 m

The engine returns 60.000 kN at each support, a peak shear of 60.000 kN at the ends, and a midspan moment of 90.000 kN·m, exactly wL²/8. Because the beam is determinate, none of this depends on the section; change the profile and the diagram does not move. Hold on to the 90: it is the largest the governing moment will ever be for this beam and load, and it becomes the upper bound of the check we build at the end.

Fix the left end and leave a roller under the right, and the beam becomes a propped cantilever, DSI 1. Statics gives three equations but there are four reactions, so one force, call it the prop reaction X₁, cannot be found from equilibrium. The criterion tells us exactly how many extra conditions we need: one, because DSI = 1. That condition is compatibility, the physical fact that the prop holds its end at zero deflection.

Release the prop and you have a plain cantilever. Under the load it would sag at the free end by δ₁₀ = wL⁴/8EI; under a unit upward force there it would rise by δ₁₁ = L³/3EI. The real prop allows neither, so the two must cancel, δ₁₀ = X₁·δ₁₁, which gives the one redundant directly:

X₁ = (wL⁴/8EI) / (L³/3EI) = 3wL/8 = 45 kN

EI cancels, so for a single prismatic member the redundant is exact. Statics finishes the job: the fixed end carries 5wL/8 = 75 kN and a hogging fixing moment of wL²/8 = 90 kN·m; the bending moment crosses zero, the point of contraflexure, at x = L/4 = 1.5 m, and the span sags to a peak of 9wL²/128 = 50.625 kN·m at x = 5L/8 = 3.75 m.

Now the check. The CalcSteel engine returns a prop reaction of 45.000 kN, a fixed-end reaction of 75.000 kN, a hogging fixed-end moment of −90.000 kN·m, contraflexure at x = 1.500 m, and a span peak of +50.625 kN·m at x = 3.750 m. Every number matches. Notice what the one redundant did to the diagram: the peak moment is still 90 kN·m, but it has moved off midspan and onto the support as hogging, and the largest sagging moment in the span has collapsed to barely more than half the determinate value.

Build the same beam in at both ends and it becomes fixed-fixed, r = 6 and DSI = 3. Under the symmetric load the axial redundant carries nothing, so two redundants are active, and symmetry does half the work: both ends must take the same moment and both reactions must be wL/2 = 60 kN. The two compatibility conditions are that neither built-in end is allowed to rotate. Enforcing them gives the classic result:

M_end = −wL²/12 = −60 kN·m (hogging), M_mid = +wL²/24 = +30 kN·m (sagging)

The engine agrees to the decimal: reactions 60.000 kN, end moments −60.000 kN·m, midspan moment +30.000 kN·m, with the bending moment crossing zero at x = 1.27 m and x = 4.73 m. Compared with the simply supported beam, the governing moment has dropped from 90 to 60 kN·m, a full third, purely because two more redundants let the ends take their share. This is the redistribution the design code is built to exploit, quantified.

Line the three peak moments up and a pattern falls out that is more useful than any single number. For this beam and load the governing bending moment went 90, 90, 60 kN·m as the supports stiffened, and it will never leave that range. Both extremes are closed form and easy to remember:

• Upper bound, both ends pinned: wL²/8 = 90 kN·m. No end can shed moment to a support, so midspan takes all of it.
• Lower bound, both ends fully fixed: wL²/12 = 60 kN·m. The ends take the most they can, so the field moment is smallest.

Any real single-span beam under a uniform load sits between these two, because real end conditions are somewhere between a perfect pin and a perfect fixity. That gives you a check you can run in your head on any solver output: for a uniformly loaded prismatic span, the peak bending moment must land between wL²/12 and wL²/8. If your model reports 130 kN·m for this beam, or 40, the model is wrong before you read another number, a released support, a doubled load, a units slip. The criterion does not just count redundants; it brackets the answer.

The same bracketing logic is why an indeterminate member can be lighter. The required elastic section modulus is proportional to the design moment, so dropping the governing moment from 90 to 60 kN·m cuts the required modulus by a third, and with it the steel. Redundancy pays for itself in kilograms.

You will almost never solve a real indeterminate frame by hand again. The skill that matters now is auditing the solver, and the criterion tells you exactly what to audit. Four checks, in order, catch the overwhelming majority of modelling errors.

1. Count the degree first. Before you trust a single force, run DSI = (3m + r) − 3n − c on the model. It confirms the structure is stable (DSI ≥ 0 and properly arranged) and tells you how many redundants the solver had to resolve. A model you expected to be DSI 3 that counts out at DSI 0 has a missing support or a stray release.

2. Close global equilibrium. Equilibrium is necessary whether or not a structure is indeterminate, and it is the fastest sanity check. Sum the vertical reactions: for the propped cantilever, 75 + 45 = 120 kN, exactly wL. Sum the horizontals; sum the moments. If the reactions do not balance the applied load, nothing downstream is worth reading.

3. Close one compatibility condition per redundant. This is the check equilibrium cannot give you, and the criterion tells you how many you owe: DSI of them. For the propped cantilever that is one, the prop deflection is zero, and recomputing the released-structure flexibility reproduces the prop reaction 3wL/8 = 45 kN that the solver reports. For the fixed-fixed beam it is the two zero end rotations, visible as the symmetric −60 kN·m at each end.

4. Bracket the peaks. Check the governing moment against the closed-form bounds: wL²/12 to wL²/8 for a uniform span, PL/8 to PL/4 for a central point load, and so on. A result outside the bracket is a modelling error, not a surprise about the structure.

Run those four and you are not trusting the black box, you are holding it to the same physics you would have used by hand, at a fraction of the effort.

The calculator below runs the same direct stiffness method as the worked cases, in your browser. Build the simply supported beam and read 60 and 90; fix one end for the propped cantilever and watch 45, 75 and the hogging 90 appear; fix both ends and see the peak fall to 60. Then run the checks yourself: sum the reactions to wL = 120 kN, and confirm the peak moment stays inside the wL²/12 to wL²/8 bracket every time.

It is free, no login for the analysis, and it does not stop at DSI 1. Add supports and members until you have a structure no hand method could touch, and it still solves and still balances in the same instant.

The redundancy the criterion counts is exactly what design codes turn into allowable design savings. Because an indeterminate structure can shed moment from an overstressed region to a stiffer one before it fails, the standards let you account for some of that redistribution instead of designing every section for its peak elastic moment.

AISC 360 permits a limited redistribution of negative moments in continuous compact-section beams, lowering the support moments and raising the span moments to match, provided the section can rotate enough to deliver it. ABNT NBR 8800 and Eurocode EN 1993-1-1 take the same idea further, tying the amount of elastic redistribution, and whether plastic global analysis is allowed at all, to the cross-section class: the more compact and ductile the section, the more of its redundancy you are allowed to bank. In every case the ceiling on redistribution is ductility, the ability to form and hold a plastic hinge without local buckling, because redistribution is only real if the overstressed region can yield and hold while the rest catches up. That is the criterion closing the loop: you count the redundants to know they exist, and the code lets you spend them only to the extent the steel can actually deliver.

Reading the degree off the supports alone. Internal hinges and releases subtract from the count. A beam that looks over-supported can be determinate, or even a mechanism, once its releases are subtracted. Always include the c term.

Confusing indeterminate with unstable. Both are a mismatch between unknowns and equations, in opposite directions. Too few, or badly arranged, is a mechanism; too many is indeterminate. Three reactions that are all parallel or all pass through one point are unstable even though the count looks right, so check the arrangement, not just the number.

Assuming the section does not matter. For a determinate structure it does not. The instant a structure is indeterminate, the internal forces depend on relative stiffness, on E, I and length; change a member and the moments redistribute. Two solvers that disagree on an indeterminate frame usually disagree because they were given different sections or releases, not because one is wrong.

Is an indeterminate structure automatically stronger? Not automatically stronger at the ultimate limit, but stiffer and more robust: the alternate load paths and moment redistribution give a real margin before collapse, which is why codes value redundancy.

Do I still need to know this if the solver handles it? Yes, precisely because the solver handles it silently. You cannot audit a number you cannot bracket. Counting the degree, closing equilibrium, and checking the wL²/12 to wL²/8 range is what catches the wrong support or the missing release that turns a right method into a wrong answer.

Does redistribution need plastic design? No. Some elastic redistribution is allowed for compact sections under the standards; full plastic analysis is a separate, more demanding route. Both depend on the same thing, enough ductility to deliver the redistribution the redundancy promises.

• The degree of static indeterminacy is a count, DSI = r − 3 − c for a plane beam and (3m + r) − 3n − c for a plane frame. It is the criterion behind both the solver and the design code.
• The same beam (L = 6 m, w = 20 kN/m) peaks at 90 kN·m simply supported (DSI 0), still 90 kN·m but relocated to the support as a propped cantilever (DSI 1), and only 60 kN·m fixed-fixed (DSI 3). More redundancy, a smaller governing moment, and one fifth the deflection.
• Every number here was computed by the shipping CalcSteel FEM engine and matches closed-form theory to three decimals: reactions 60/60, 75/45 and 60/60 kN; peak moments 90, 90 and 60 kN·m; span moments 90, 50.625 and 30 kN·m.
• Use the criterion as a check: count the degree, close global equilibrium, close one compatibility condition per redundant, and confirm the peak moment sits inside the wL²/12 to wL²/8 bracket. A result outside the bracket is a modelling error.
• Redundancy is not overhead. It buys smaller moments, stiffer members and alternate load paths, and the design codes are written to let you spend it, up to the ductility the section can deliver.