CEE 330 · Structural Analysis and Design-I · Week 1
Determinacy & Stability
Before you can analyse a structure you have to ask two questions about it, in order: will it stand up at all? and can equilibrium alone tell me the forces? Get these wrong and every calculation that follows is answering the wrong problem.
Two ideas get confused constantly, and they are not the same question. Stability asks whether the structure holds together at all. Determinacy asks whether the equations of statics are enough to find the forces in it.
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Stools. A two-legged stool falls over — no arrangement of forces holds it up. That is instability. A three-legged stool never wobbles, and you can work out the load in each leg from equilibrium alone: that is statically determinate. Now add a fourth leg. It is stronger and safer — but on an uneven floor, how much load does each leg carry? It depends on how stiff the legs are and exactly how uneven the floor is. Equilibrium can no longer answer it. That is statically indeterminate.
Unstable
Fewer restraints than motions
The structure can move as a rigid body or fold into a mechanism. No amount of clever analysis helps — it must be fixed, not solved.
Determinate
Exactly enough
Unknowns = equations. Solvable with ΣFx, ΣFy, ΣM alone. Forces don't depend on material or member size.
Indeterminate
More restraints than equations
Solvable, but only by also using compatibility — how much things stretch and bend. Forces now depend on E, I and A.
Core Idea
Check stability first, always. A determinacy count on an unstable structure is meaningless — you would be computing the forces in something that is already falling down.
02
What Equilibrium Can and Can't Do
A rigid body in a plane can do exactly three things: slide sideways, slide up, and spin. So statics gives you exactly three equations to stop it doing them.
Three equations per rigid body in 2-D. That is all statics will ever give you — the rest has to come from geometry or from the material.
The bookkeeping
Unknowns vs. equations
Count the unknowns (reaction components, member forces). Count the equations. If unknowns exceed equations by n, the structure is indeterminate to the nth degree, and n is the number of redundants.
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Why this decides your whole method. Determinate → you can go straight to shear and moment diagrams with statics. Indeterminate → you need virtual work, moment distribution, stiffness methods, or the approximate methods in Weeks 6–8. The count you do in this chapter tells you which door to walk through for the rest of the course.
03
Counting: Beams & Frames
Degree of static indeterminacy
\[ i = 3m + r - 3j - c \]
m = members, r = reaction components, j = joints (including supports), c = equations of condition (internal releases).
The logic is simpler than it looks. Every member carries three internal force components, so 3m counts the internal unknowns and r the external ones. Every joint gives you three equilibrium equations, so 3j counts what you can write down, and c adds the extra equations that internal hinges hand you for free.
Pin / hinge supportr = 2 (H, V)
Roller supportr = 1 (perpendicular only)
Fixed supportr = 3 (H, V, M)
Link / cabler = 1 (along its axis)
i < 0unstable — a mechanism
i = 0determinate (if geometrically stable)
i > 0indeterminate to degree i
Same formulaworks for beams — a beam is a one-member frame
Core Idea
A beam is just a frame that happens to be straight. The single formula \( i = 3m + r - 3j - c \) handles cantilevers, continuous beams, portal frames and arches alike — you never need a separate rule.
04
Counting: Trusses
Trusses get their own formula because their joints are different. An ideal truss joint is a frictionless pin and every member carries only axial force — no moment to speak of.
Plane truss
\[ i = m + r - 2j \]
One unknown per member (its axial force), and only two equations per joint (ΣFx, ΣFy) — a pin joint cannot carry moment, so ΣM gives nothing new.
Why 2j and not 3j
At a truss joint all member forces pass through one point, so they are automatically concurrent and their moment about that point is identically zero. ΣM = 0 is satisfied trivially and carries no information. That single difference is the whole reason trusses need their own count.
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Watch for the trap: counting alone can call a truss determinate while it is actually a mechanism. If m + r = 2j but the members are badly arranged — a rectangular panel with no diagonal, say — part of the truss folds even though the arithmetic looks perfect. Section 7 covers this.
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Equations of Condition
An internal hinge is a gift: it tells you something equilibrium of the whole structure cannot. At a hinge, the bending moment is known to be zero, and that is one extra equation.
Internal hinge, 2 members
c = 1
One moment release: \( \sum M = 0 \) at the hinge, taking either side alone.
Internal hinge, n members
c = n − 1
A hinge joining three members releases two moments, not one.
Internal roller / shear release
c = 2
Releases both moment and axial force at that section.
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A hinge is a free equation, not a free unknown. Students often try to count a hinge as an extra unknown and get the sign backwards. Think of it the other way round: the hinge tells you something — "no moment here" — so it goes on the same side of the ledger as ΣF = 0. More hinges, more information, less indeterminacy.
Core Idea
This is exactly why the three-hinged arch is determinate while the two-hinged arch is not. Both have four reaction components against three equations — but the crown hinge supplies the fourth equation, and the count closes: \( 3(2)+4-3(3)-1 = 0 \).
06
Live Determinacy Calculator
Pick a standard structure to load its counts, or enter your own. The calculator shows the arithmetic term by term, so you can see where a number came from rather than just the verdict.
07
When Counting Lies
The formulas count how many restraints there are. They cannot see where those restraints point. A structure can pass the count and still collapse.
Failure 1
All reactions parallel
Three vertical rollers give r = 3 and the count says determinate. But nothing resists horizontal load — ΣFx = 0 can never be satisfied. The structure slides away.
Failure 2
All reactions concurrent
If the lines of action of every reaction meet at one point, the structure can rotate about that point. Taking moments there gives 0 = 0 — no restraint at all against spin.
Failure 3
Internal mechanism
The supports are fine but part of the structure folds — a truss panel with no diagonal, or too many hinges in a row in a beam. Locally unstable, globally counted as stable.
Three Structures That Pass the Count and Still Fall Down
Every one of these satisfies its counting formula. Arithmetic is necessary but never sufficient — after you count, look at the structure and ask whether anything can move.
Core Idea
The count is a necessary condition, not a sufficient one. Passing it means "not obviously a mechanism"; it does not mean stable. Always follow the arithmetic with the physical question: can any part of this move without stretching a member? If yes, it is unstable no matter what the formula said.
08
Why Indeterminacy Is Usually Good
Determinate structures are easier to analyse, so beginners often assume they are better. In practice most real structures are deliberately indeterminate.
In favour of indeterminate
Redundancy and stiffness
Extra load paths mean losing one member need not mean collapse. Moments redistribute, deflections are smaller, and the structure is generally lighter for the same load. A continuous beam beats a series of simply supported spans on every count except analysis effort.
In favour of determinate
Indifference to movement
A determinate structure develops no stress from support settlement, temperature change, or fabrication error — it simply moves. An indeterminate one fights those movements and picks up real internal forces. That is why long bridges get expansion joints and why three-hinged arches are chosen on poor ground.
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The engineering judgement. Good foundations and a controlled environment → go indeterminate and enjoy the efficiency. Settlement-prone ground or big thermal swings → determinacy buys you immunity from movements you cannot predict. This trade-off is exactly why the three-hinged arch in the next chapter exists at all.
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Worked Examples
Example 1
Portal Frame with Fixed Bases
Problem: A single-bay portal frame has two columns and one beam, rigidly connected, with both column bases fixed. Classify it.
Count reactions: each fixed base gives H, V and M → r = 3 + 3 = 6.
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Conditions: all joints are rigid, no internal hinges → c = 0.
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\( i = 3(3) + 6 - 3(4) - 0 = 9 + 6 - 12 = 3 \)
Answer: Statically indeterminate to the 3rd degree. Pin the bases instead (r = 4) and it drops to 1st degree — the classic result, and the reason pinned bases are chosen when foundations may rotate.
Example 2
Three-Hinged Arch
Problem: An arch is pinned at both abutments and has a hinge at the crown. Show that it is determinate, and explain why the two-hinged version is not.
1
Model it: the crown hinge splits the arch into two members → m = 2, with joints at each abutment and one at the crown → j = 3.
2
Reactions: two pins → r = 2 + 2 = 4 — one more than the three equations of statics.
3
Condition: the crown hinge joins 2 members → c = 2 − 1 = 1. It supplies the missing equation: \( \sum M = 0 \) at the crown, taking one half alone.
Remove the crown hinge: now m = 1, j = 2, c = 0, giving \( i = 3 + 4 - 6 = 1 \) → indeterminate to 1st degree.
Answer: Three-hinged → i = 0; two-hinged → i = 1. One hinge is the entire difference, and it is why three-hinged arches are used where abutments may settle or spread: with i = 0, support movement causes no stress at all.
Example 3
A Truss That the Count Gets Wrong
Problem: A truss has m = 13, j = 8, r = 3, and the count says determinate. On inspection one rectangular panel has no diagonal. What is the true classification?
1
The count: \( i = m + r - 2j = 13 + 3 - 16 = 0 \) → apparently determinate.
2
Look at the structure. A four-bar rectangular panel with pinned corners and no diagonal is a parallelogram linkage — it racks sideways with no member changing length.
3
What that means. Deformation with zero member strain requires zero force, so there is no internal resistance to it. That panel is a mechanism.
4
Reconcile the arithmetic. The total is right but the distribution is wrong: one region has a surplus member (locally indeterminate) while the open panel has a deficit (locally unstable). The two errors cancel in the sum.
Answer: The truss is unstable, despite i = 0. Add a diagonal to the open panel and remove the surplus member elsewhere. This is the standard warning that counting is necessary but never sufficient.