← Engineering Mechanics, SOM & Structural Analysis · AAI Manager (Civil)

Chapter 8 of 13

Structural Analysis

In the AAI Manager (Civil) syllabus under Engineering Mechanics, SOM & Structural Analysis · 4 parts

Determinacy, Indeterminacy & Stability · Statically Determinate Beams & Frames · Slope Deflection Method · Moment Distribution Method

CIVILGYAN TEST SERIES · civilgyantests.com — free study notes for civil engineering exams

📑 Contents (32 sections)

Part 1 of 4

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 4

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.)

Part 3 of 4

Slope Deflection Method

Last reviewed 16 Sept 2026 · 5 min read

The displacement approach

In the slope deflection method (G. A. Maney, 1915) the unknowns are joint displacements — rotations of rigid joints and, in sway frames, lateral translations. Member end moments are written in terms of these displacements, and joint equilibrium equations are solved for them. It is the basis of moment distribution and of the stiffness matrix method.

It is convenient when the kinematic indeterminacy $D_{k}$ is small, even if the static indeterminacy is large (a fixed beam with many spans, a multi-bay frame).

Assumptions: members are prismatic and linearly elastic; axial and shear deformations are neglected; joints are rigid (all members at a joint rotate by the same angle).

Sign convention

Moments at member ends: clockwise positive (acting on the member).
Rotations $θ$ of joints: clockwise positive.
Chord rotation $ψ = Δ/ L$ : clockwise positive (the far end moving so that the chord rotates clockwise).

Slope-deflection equations

For member AB of length $L$ and flexural rigidity $E I$ :

FormulaSlope-deflection equations

M_{A B} = M_{F A B} + \frac{2 E I}{L} (2 θ_{A} + θ_{B} - 3 ψ)

M_{B A} = M_{F B A} + \frac{2 E I}{L} (2 θ_{B} + θ_{A} - 3 ψ)

$M_{F A B}$ , $M_{F B A}$ = fixed-end moments due to loads on the span (clockwise positive: e.g. UDL gives $- w L^{2} /12$ at A and $+ w L^{2} /12$ at B); $ψ = Δ/ L$ with $Δ$ the relative transverse displacement of B with respect to A.

Where the equations come from

Superpose on a fixed-ended member: (1) the loads with ends fixed → fixed-end moments; (2) rotation $θ_{A}$ with B fixed → $4 E I θ_{A} / L$ at A and $2 E I θ_{A} / L$ at B (carry-over factor ½); (3) rotation $θ_{B}$ similarly; (4) relative translation $Δ$ with no rotation → $- 6 E I Δ/ L^{2}$ at both ends.

Modified equation for a pinned far end

If end B is a pin or roller (so $M_{B A} = 0$ ), eliminate $θ_{B}$ :

M_{A B} = (M_{F A B} - \frac{M _{F B A}}{2}) + \frac{3 E I}{L} (θ_{A} - ψ)

This saves one unknown per pinned end.

Procedure

Identify unknown joint rotations (and sway $Δ$ if the frame can sway). Fixed supports have $θ = 0$ .
Compute fixed-end moments for every loaded member.
Write slope-deflection equations for every member end.
Joint equilibrium: at each rigid joint with unknown rotation, $\sum M_{m e mb er e n d s} =$ external moment at the joint (often 0).
Shear (sway) equation for each sway degree: horizontal equilibrium of the storey, using column shears from their end moments.
Solve the simultaneous equations; substitute back for end moments.
Find reactions, shear and moment diagrams by statics.

Frames without and with sway

A frame does not sway when it is symmetric in geometry and loading, or when lateral movement is prevented (a brace or a support).

For a portal with columns AB and DC (feet A, D) and beam BC, with horizontal load $P$ at beam level, the shear equation is:

\frac{M _{A B} + M _{B A}}{h _{1}} + \frac{M _{D C} + M _{C D}}{h _{2}} + P = 0

(sign depends on the chosen sway direction; column shear $= - (M_{t o p} + M_{b o tt o m}) / h$ for clockwise-positive moments, plus any load on the column).

Worked examples

Worked ExampleExample 1 — two-span continuous beam

A beam ABC is fixed at A and simply supported at B and C. AB = 6 m carries a UDL of 20 kN/m; BC = 4 m carries a central load of 40 kN. $E I$ constant. Find the support moments.

Solution. Unknowns: $θ_{B}$ , $θ_{C}$ ( $θ_{A} = 0$ ).

Fixed-end moments: $M_{F A B} = - \frac{20 \times 36}{12} = - 60$ , $M_{F B A} = + 60$ ; $M_{F B C} = - \frac{40 \times 4}{8} = - 20$ , $M_{F C B} = + 20$ (kN·m).

Use the modified equation for BC (C is a simple support, $M_{C B} = 0$ ):

$M_{B C} = (- 20 - \frac{20}{2}) + \frac{3 E I}{4} θ_{B} = - 30 + 0.75 E I θ_{B}$

$M_{B A} = 60 + \frac{2 E I}{6} (2 θ_{B}) = 60 + 0.6667 E I θ_{B}$

$M_{A B} = - 60 + \frac{2 E I}{6} θ_{B} = - 60 + 0.3333 E I θ_{B}$

Joint B: $M_{B A} + M_{B C} = 0$ → $30 + 1.4167 E I θ_{B} = 0$ → $E I θ_{B} = - 21.18$

$M_{A B} = - 60 - 7.06 = - 67.06 kN \cdot m$ $M_{B A} = 60 - 14.12 = + 45.88 kN \cdot m$ , $M_{B C} = - 30 - 15.88 = - 45.88 kN \cdot m$ ✔ (equal and opposite)

So hogging moments: 67.06 kN·m at A and 45.88 kN·m at B.

Worked ExampleExample 2 — non-sway portal frame

A symmetric portal frame ABCD has fixed feet A and D, columns 4 m high and beam BC 6 m long carrying a UDL of 24 kN/m. All members have the same $E I$ . Find the moments.

Solution. Symmetry → no sway, and $θ_{C} = - θ_{B}$ .

FEM for BC: $M_{F B C} = - \frac{24 \times 36}{12} = - 72$ , $M_{F C B} = + 72$ .

$M_{A B} = \frac{2 E I}{4} θ_{B} = 0.5 E I θ_{B}$ ; $M_{B A} = E I θ_{B}$

$M_{B C} = - 72 + \frac{2 E I}{6} (2 θ_{B} + θ_{C}) = - 72 + \frac{E I}{3} θ_{B}$ (using $θ_{C} = - θ_{B}$ )

Joint B: $E I θ_{B} - 72 + \frac{E I θ _{B}}{3} = 0$ → $E I θ_{B} = 54$

$M_{B A} = 54 kN \cdot m$ , $M_{B C} = - 54 kN \cdot m$ , $M_{A B} = 27 kN \cdot m$

Mid-span moment of beam $= \frac{w L ^{2}}{8} - 54 = 108 - 54 = 54 kN \cdot m$ (sagging).

Worked ExampleExample 3 — effect of support settlement

A two-span continuous beam ABC (AB = BC = $L$ , constant $E I$ , simply supported at A and C and at B) has no load, but support B settles by $δ$ . Find the moment at B.

Solution. Chord rotations: AB: B moves down relative to A → $ψ_{A B} = + δ / L$ (clockwise); BC: B lower than C → $ψ_{B C} = - δ / L$ .

With pinned ends at A and C (modified equations), by symmetry $θ_{B} = 0$ :

$M_{B A} = \frac{3 E I}{L} (0 - ψ_{A B}) = - \frac{3 E I δ}{L ^{2}}$ ; $M_{B C} = \frac{3 E I}{L} (0 - ψ_{B C}) = + \frac{3 E I δ}{L ^{2}}$

Joint B balances ✔. Support moment at B $= \frac{3 E I δ}{L ^{2}}$ (sagging at B — the settling support "hangs" from the beam).

Part 4 of 4

Moment Distribution Method

Last reviewed 16 Sept 2026 · 6 min read

The idea

The moment distribution method (Hardy Cross, 1930) solves indeterminate beams and frames by successive corrections instead of simultaneous equations.

Imagine every joint locked against rotation; loaded members develop fixed-end moments.
Release one joint at a time. The unbalanced moment at the joint is shared among the members meeting there in proportion to their stiffnesses (distribution).
Each distributed moment induces half of itself (for a fixed far end) at the far end of the member (carry-over).
Repeat until the carried-over moments are negligible.

Each step is an exact equilibrium statement, and the process converges quickly. It is a displacement method, equivalent to solving slope-deflection equations iteratively.

Definitions

DefinitionStiffness, distribution and carry-over

Stiffness $K$ — moment needed at the near end to rotate it by one radian.
- Far end fixed: $K = \frac{4 E I}{L}$
- Far end pinned/roller (simple end): $K = \frac{3 E I}{L}$
- Symmetric member in a symmetric frame (far end rotates equal and opposite): $K = \frac{2 E I}{L}$
- Antisymmetric case (far end rotates equally in the same sense): $K = \frac{6 E I}{L}$
Relative stiffness — use $I / L$ (fixed far end) and $\frac{3}{4} I / L$ (pinned far end) when $E$ is common.
Distribution factor at a joint: $D F_{i} = \frac{K _{i}}{\sum K}$ ; the factors at a joint add to 1. At a fixed support $D F = 0$ ; at a free pinned end $D F = 1$ .
Carry-over factor — ratio of the moment induced at the far end to the moment applied at the near end: $+ \frac{1}{2}$ when the far end is fixed; $0$ when the far end is pinned.

Sign convention and fixed-end moments

Use clockwise end moments positive (same as slope deflection). Fixed-end moments for a span loaded with a UDL $w$ : $- w L^{2} /12$ at the left end, $+ w L^{2} /12$ at the right end; central point load: $\mp W L /8$ ; eccentric load: $- W a b^{2} / L^{2}$ , $+ W a^{2} b / L^{2}$ .

Procedure

Compute stiffness and distribution factors at every joint.
Compute fixed-end moments.
Balance each joint: unbalanced moment $U$ = algebraic sum of end moments at the joint (plus any external moment); distribute $- U \times D F$ to each member end.
Carry over half of each distributed moment to the far end (if that end is fixed or continuous).
Repeat balancing and carrying over; stop when changes are small.
Sum each column to get final end moments; check that the moments at each internal joint balance.

Exam TipPinned outer ends

Two ways: (a) use stiffness $3 E I / L$ for the member and do not carry over to the pinned end after releasing it once; or (b) use $4 E I / L$ and balance the pinned end to zero in every cycle. Method (a) converges faster.

Overhangs

The moment at the support due to the overhang is statically known ( $=$ load × arm). Treat that support like a pinned end carrying a known external moment; the overhang itself has zero stiffness.

Frames with sway

If a frame can sway (unsymmetrical geometry or loading, lateral loads):

Non-sway analysis: prevent sway with an imaginary horizontal restraint; distribute as usual; find the restraint force $R$ from horizontal equilibrium of column shears.
Sway analysis: allow an arbitrary sway $Δ$ with joints locked. Fixed-end moments due to sway in each column $= - \frac{6 E I Δ}{L ^{2}}$ (with far end fixed) or $- \frac{3 E I Δ}{L ^{2}}$ (far end pinned). Choose convenient round values in the correct ratio; distribute; find the corresponding lateral force $S$ .
Combine: final moments $= M_{n o n - s w a y} + \frac{R}{S} M_{s w a y}$ (with signs so the restraint force is cancelled).

For columns of equal height and fixed feet, sway FEMs are in the ratio of $I / L^{2}$ .

Support settlement

A settlement $δ$ of one support produces fixed-end moments $- 6 E I δ / L^{2}$ at both ends of each affected span (sign by chord rotation). Distribute them like any other fixed-end moments.

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