Chapter 1 of 8

Determinate & Indeterminate Structures

In the DSSSB AE Civil syllabus under Structural Analysis · 2 parts

Determinacy, Indeterminacy & Stability · Statically Determinate Beams & Frames

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📑 Contents (16 sections)

Part 1 of 2

Determinacy, Indeterminacy & Stability

Last reviewed 16 Sept 2026 · 7 min read

Why classify structures

Before analysing a structure, decide how it can be analysed:

Statically determinate — all reactions and member forces follow from the equations of equilibrium alone.
Statically indeterminate — there are more unknowns than equilibrium equations; compatibility of deformations is also needed (force method, slope-deflection, moment distribution, matrix methods).
Unstable — the structure cannot resist some loads at all; it is a mechanism.

The number of extra unknowns is the degree of static indeterminacy ( $D_{s}$ ), also called redundancy. The number of unknown joint displacements is the degree of kinematic indeterminacy ( $D_{k}$ ), or degrees of freedom.

Equations of equilibrium available

Structure	Equations per rigid body / joint
Plane structure (whole or any free body)	3: $\sum F_{x} = 0$ , $\sum F_{y} = 0$ , $\sum M = 0$
Space structure	6: three forces, three moments
Pin joint of a plane truss	2: $\sum F_{x} = 0$ , $\sum F_{y} = 0$
Pin joint of a space truss	3

Support reactions

Support	Plane reactions	Space reactions
Roller / link	1	1
Hinge (pin)	2	3 (ball-and-socket)
Fixed	3	6
Guided roller (slider allowing translation, no rotation)	2	—

Static indeterminacy

External and internal

External indeterminacy $D_{se}$ = number of reactions $r$ minus equilibrium equations available for the whole structure (3 for plane) minus extra equations from releases.
Internal indeterminacy $D_{s i}$ = redundancy within the structure itself (closed loops, extra members).
$D_{s} = D_{se} + D_{s i}$ .

Formulas

FormulaDegree of static indeterminacy

Beams and plane rigid frames:

D_{s} = (3 m + r) - 3 j - c

$m$ = members, $r$ = reaction components, $j$ = rigid joints (including supports and free ends), $c$ = number of equations of condition from internal releases.

Alternative for frames (closed-loop method): $D_{s} = 3 C - R$ , where $C$ = number of closed loops when the supports are joined through the ground, and $R$ = number of releases (hinge support 1, roller support 2, fixed support 0, internal hinge between two members 1). Example: a portal frame with fixed feet forms one loop with the ground → $D_{s} = 3$ .

Plane truss (pin-jointed):

D_{s} = m + r - 2 j

Space truss: $D_{s} = m + r - 3 j$

Space rigid frame: $D_{s} = 6 m + r - 6 j - c$

Interpretation:

$D_{s} = 0$ and the structure is stable → determinate.
$D_{s} > 0$ → indeterminate to that degree.
$D_{s} < 0$ → unstable (mechanism).

A zero or positive $D_{s}$ is necessary but not sufficient for stability — the arrangement must also be geometrically stable (see below).

Internal releases (hinges)

An internal hinge makes the moment zero at that point, giving one extra equation of condition.

A hinge joining two members of a frame: $c = 1$ .
A hinge at a joint where $n$ members meet (all pinned together): $c = n - 1$ .
An internal roller (shear release plus moment release) in a beam: $c = 2$ .

Part 2 of 2

Statically Determinate Beams & Frames

Last reviewed 16 Sept 2026 · 5 min read

What "determinate" buys you

A statically determinate structure can be solved completely with $\sum F_{x} = 0$ , $\sum F_{y} = 0$ and $\sum M = 0$ (plus one extra equation for each internal hinge). Its member forces do not depend on member stiffness, so:

temperature changes, support settlements and fabrication errors cause no stresses (the structure simply moves);
the analysis is quick and exact;
but there is no reserve path — failure of one member or support causes collapse.

Simply supported and cantilever beams, overhanging beams, three-hinged arches, most roof trusses and compound beams with the right number of hinges are determinate.

General procedure

Draw the free-body diagram of the whole structure with all reaction components.
Check determinacy and stability.
Take moments about a point where two unknown reactions meet — one equation, one unknown.
Use the force equations for the rest.
For internal hinges, split the structure at the hinge (moment there = 0) and use each part.
Compute internal forces at key sections and draw the diagrams.
Check with an unused equilibrium equation.

Compound (Gerber) beams

A compound beam is a continuous-looking beam made determinate by introducing internal hinges. It is analysed as a set of simple beams resting on one another.

Identify the suspended (dependent) part — the segment that cannot stand without support from its neighbours. Solve it first; its hinge reactions become loads on the supporting (anchor) parts.
Bending moment at every internal hinge is zero.

Exam TipWhy engineers use hinges

Placing hinges near the points of contraflexure of a continuous beam gives almost the same moments as a continuous beam, but the structure stays determinate — insensitive to settlement. Cantilever-and-suspended-span bridges work this way.

Plane frames

A rigid-jointed plane frame carries loads by bending, shear and axial force in its members. At every section there are three internal actions:

Action	Sign used here
Axial force $N$	Tension positive
Shear force $V$	As for beams, looking along the member from a chosen "inside" face
Bending moment $M$	Drawn on the tension side of the member (common practice for frames)

At a rigid joint with no external moment, the moments in the members meeting there are in equilibrium: for a two-member corner, the moment just below the corner in the column equals the moment just beside it in the beam.

Free-body of a joint

Cut all members around a joint; the member end forces (axial, shear, moment) with any external load must satisfy the three equilibrium equations. This is the quickest check on a frame diagram.

Worked examples

Worked ExampleExample 1 — compound beam with one hinge

Beam ABC is fixed at A, has an internal hinge at B and a roller support at C. AB = 4 m, BC = 6 m. A UDL of 10 kN/m acts on BC only. Find the reactions and the moment at A.

Solution. BC is the suspended part: simply supported on the hinge B and the roller C.

$R_{B} = R_{C} = \frac{10 \times 6}{2} = 30$ kN

AB is a cantilever carrying 30 kN downward at its free end B:

$R_{A} = 30 kN$ (up), $M_{A} = 30 \times 4 = 120 kN \cdot m$ (hogging), $R_{C} = 30 kN$

BM at B = 0 ✔; maximum sagging moment in BC $= w L^{2} /8 = 45$ kN·m.

Worked ExampleExample 2 — L-shaped cantilever frame

A frame ABC has a vertical column AB (fixed at A, height 3 m) and a horizontal arm BC (length 2 m). A vertical load of 20 kN acts at C and a horizontal load of 10 kN acts at B (towards the right). Find the reactions at A and the moments at B and A.

Solution. $H_{A} = 10 kN$ (towards the left), $V_{A} = 20 kN$ (up)

Moment in arm BC at B $= 20 \times 2 = 40$ kN·m (tension at top).

At A: $M_{A} = 20 \times 2 + 10 \times 3 = 70 kN \cdot m$ .

Column AB: moment varies linearly from 40 kN·m at B (the vertical load's moment is constant down the column) plus the horizontal load's effect growing from 0 at B to 30 kN·m at A → 70 kN·m at A. Axial force in AB $= 20$ kN compression; shear in AB $= 10$ kN.

Worked ExampleExample 3 — simply supported inclined beam

A beam inclined at 30° to the horizontal spans 6 m horizontally, supported by a hinge at the lower end and a roller (vertical reaction) at the upper end. It carries a vertical load of 12 kN/m per horizontal metre. Find the maximum bending moment.

Solution. For vertical loads and vertical reactions, the bending moment of an inclined member depends on the horizontal projection:

$M_{ma x} = \frac{w L _{h}^{2}}{8} = \frac{12 \times 6 ^{2}}{8} = 54 kN \cdot m$

The load also produces an axial component along the member, which varies along its length. (This is why staircase waist slabs are designed on the horizontal span.)

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