Structural Engineering

A. Analysis of statically determinate beams, columns, trusses, and frames

Analysis of Statically Determinate Structures

Find reactions from equilibrium, then internal axial force, shear, and moment in determinate beams, frames, and trusses by sections.

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B. Deflection of statically determinate beams, trusses, and frames

Beam Deflection and Column Analysis

Standard deflection formulas, superposition, moment-area and virtual work for beams and trusses, plus Euler buckling with effective-length factors.

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D. Structural determinacy and stability analysis of beams, trusses, and frames

Determinacy, Stability, and Indeterminate Structures

Count reactions, members, and joints to classify beams, trusses, and frames, then solve a single-degree indeterminate structure by the force method.

Before you can analyze a structure you must know whether equilibrium alone can solve it. A statically determinate structure has exactly as many unknown reactions and member forces as equilibrium equations; an indeterminate one has more, and the extra unknowns—the redundants—require compatibility of deformation to find. Get the count wrong and every later number is meaningless. The NCEES FE Reference Handbook gives the determinacy inequalities for trusses and frames in its Civil Engineering chapter and an intro to the force method; the exam reliably asks you to classify a structure and to solve a one-degree indeterminate beam. Points vanish when candidates miscount reactions or forget that an internal hinge adds a condition equation.

Stability comes first

A structure must be stable before determinacy even matters. External instability arises when reactions are too few or badly arranged—three parallel reactions, or three concurrent reactions, cannot resist a general load even though the count looks adequate. Internal instability (a mechanism) arises when members are arranged so part of the structure can move. Always confirm a proper arrangement of supports and members; the determinacy formulas assume stability.

\text{Stable} \;\Rightarrow\; \text{reactions non-parallel, non-concurrent, sufficient in number}

Counting unknowns: the building blocks

Four quantities drive every classification:

m

= number of members,

r

= number of independent reaction components,

j

= number of joints, and

c

= number of condition equations from internal releases (an internal hinge that transmits no moment gives

c=1

; an internal hinge connecting several members gives one condition per extra member). A pin support contributes

r=2

, a roller

r=1

, a fixed support

r=3.

m,\ r,\ j,\ c \;\rightarrow\; \text{inputs to the determinacy test}

Trusses

A pin-jointed plane truss has two equilibrium equations per joint, so the test compares member-plus-reaction unknowns against

2j

. Equality means determinate; an excess means the truss is internally and/or externally indeterminate, with the degree equal to the surplus. A deficit means it is a mechanism—unstable—regardless of how the bars are arranged.

m + r = 2j \ (\text{determinate}), \qquad m + r > 2j \ (\text{indeterminate})

Beams and frames

A rigid-jointed plane frame has three equilibrium equations per joint and each member carries three internal actions, giving the test below. Treat a single beam as a one-member frame: the formula then reduces to comparing the reactions against

3+c

. The degree of static indeterminacy (DSI) is the left side minus the right side. Internal hinges raise

c

and lower the degree, which is why a Gerber (hinged) beam can be made determinate on purpose.

3m + r = 3j + c \ (\text{determinate}), \qquad \text{DSI} = (3m + r) - (3j + c)

The force (flexibility) method

For a singly indeterminate structure, choose one reaction as the redundant and remove it to get a stable, determinate primary structure. Compute the deflection at the released point under the real load,

\Delta_0

, and the deflection there per unit of the redundant,

\delta_{11}

(the flexibility coefficient). Compatibility demands the true deflection there be zero (or the known support movement), which solves for the redundant in one equation.

\Delta_0 + R\,\delta_{11} = 0 \quad\Rightarrow\quad R = -\frac{\Delta_0}{\delta_{11}}

Moment distribution for indeterminate frames

Moment distribution is an iterative alternative that needs no simultaneous equations. Start with fixed-end moments (FEMs) from the handbook tables, then release each joint in turn: the unbalanced moment is distributed to members in proportion to their distribution factors

DF = (EI/L)/\sum(EI/L)

, and half of each distributed moment carries over to the far fixed end. Iterating to convergence yields the member end moments. It is the classic hand method for continuous beams and braced frames.

DF_i = \frac{(EI/L)_i}{\sum (EI/L)}, \qquad \text{COF} = \tfrac{1}{2}

Exam strategy

Always test stability before determinacy, and watch the support arrangement for the parallel/concurrent reaction traps. For the count, write

m,r,j,c

explicitly—miscounting reaction components is the single biggest error. Remember each internal hinge adds to

c

and drops the degree by one. For a one-degree indeterminate beam, the fastest path is superposition: pick the redundant, use two handbook deflection cases, and enforce zero net deflection. Memorize the propped-cantilever result

R=3wL/8

and the fixed-end moments—they appear again and again.

Key equations

Truss classification

m + r \;\lessgtr\; 2j

$<2j$ unstable; $=2j$ determinate; $>2j$ indeterminate. $m$ = bars, $r$ = reactions, $j$ = joints.

Frame classification

3m + r \;\lessgtr\; 3j + c

$<$ unstable; $=$ determinate; $>$ indeterminate. $c$ = condition (release) equations.

Degree of static indeterminacy (frame)

\text{DSI} = (3m + r) - (3j + c)

Number of redundants; the count of extra equations (compatibility) needed beyond statics.

Degree of indeterminacy (truss)

\text{DSI} = (m + r) - 2j

Surplus of unknowns over the $2j$ joint-equilibrium equations.

Reaction components

r_{\text{roller}}=1,\ r_{\text{pin}}=2,\ r_{\text{fixed}}=3

Independent reaction components contributed by common 2-D supports.

Compatibility (force method)

\Delta_0 + R\,\delta_{11} = 0

$\Delta_0$ = deflection at the release under real load; $\delta_{11}$ = deflection per unit redundant $R$ .

Propped-cantilever redundant

R_B = \dfrac{3wL}{8}

Roller reaction of a fixed-pinned beam under full-span UDL $w$ ; classic single-degree result.

Distribution factor

DF_i = \dfrac{(EI/L)_i}{\sum (EI/L)}

Fraction of joint unbalance taken by member $i$ in moment distribution; sums to 1 at a joint.

Fixed-end moment (UDL)

\text{FEM} = \dfrac{wL^2}{12}

Magnitude of the end moment of a fixed-fixed beam under uniform load; starting value for moment distribution.

Worked examples

Classifying a propped cantilever and a truss

Problem. (a) A beam is fixed at $A$ and supported on a roller at $B$ (a propped cantilever), with no internal hinge. (b) A plane truss has $m=14$ members, $r=3$ reactions, and $j=8$ joints. Classify each.

Solution. (a) Treat the beam as one member:

m=1,\ j=2,\ c=0

. Reactions: fixed end

r=3

, roller

r=1

, so

r=4

.

\text{DSI} = (3m+r) - (3j+c) = (3+4) - (6+0) = 1

The beam is stable and indeterminate to the first degree (one redundant). (b) Truss test:

m+r = 14+3 = 17

versus

2j = 16

.

\text{DSI} = 17 - 16 = 1

Stable and statically indeterminate to the first degree. Sanity check: both have exactly one more unknown than equilibrium provides, so each needs one compatibility equation—consistent with the force method handling a single redundant.

\text{(a) DSI}=1,\qquad \text{(b) DSI}=1

Redundant reaction by the force method

Problem. A propped cantilever (fixed at $A$ , roller at $B$ ) spans $L=20\ \text{ft}$ and carries a uniform load $w=3\ \text{kip/ft}$ . Find the roller reaction $R_B$ , the reaction at $A$ , and the fixing moment $M_A$ .

Solution. Remove the roller to get a cantilever fixed at

A

(the primary structure). Under the UDL, the free-end (point

B

) deflection is downward:

\Delta_0 = \frac{wL^4}{8EI}.

The redundant

R_B

acting upward at the tip deflects it up by

\delta = R_B L^3/(3EI)

. Compatibility requires zero net deflection at

B

:

\frac{wL^4}{8EI} = \frac{R_B L^3}{3EI} \;\Rightarrow\; R_B = \frac{3wL}{8} = \frac{3(3)(20)}{8} = 22.5\ \text{kip}

Vertical equilibrium:

R_A = wL - R_B = 60 - 22.5 = 37.5\ \text{kip}

. The fixing moment is the standard propped-cantilever result

M_A = wL^2/8 = 3(20)^2/8 = 150\ \text{kip-ft}

. Sanity check:

R_A+R_B = 60\ \text{kip} = wL

, and

R_B=3wL/8

matches the tabulated redundant—both confirm the solution.

R_B = 22.5\ \text{kip},\ R_A = 37.5\ \text{kip},\ M_A = 150\ \text{kip-ft}

Common pitfalls

•Miscounting reaction components: a fixed support is $r=3$ , a pin $r=2$ , a roller $r=1$ . The most common classification error is treating a fixed support as a pin.
•Ignoring internal hinges. Each internal hinge adds a condition equation $c$ and lowers the degree of indeterminacy by one—forget it and a determinate Gerber beam looks indeterminate.
•Confusing determinacy with stability. Meeting the count ( $m+r=2j$ or $3m+r=3j+c$ ) does not guarantee stability—reactions can be parallel or concurrent, giving a count-OK but unstable structure.
•Choosing a redundant whose removal leaves an unstable primary structure. The released structure must remain stable and determinate.
•Sign errors in the compatibility equation: the deflection from the real load and from the redundant must be measured positive in the same direction so they correctly cancel.
•In moment distribution, forgetting the carry-over (half of each distributed moment goes to the far end) or letting distribution factors at a joint not sum to one.

References

NCEES FE Reference Handbook — Civil Engineering: Stability, Determinacy, and Classification of Structures — Truss ( $m+r$ vs $2j$ ) and frame ( $3m+r$ vs $3j+c$ ) classification criteria.
NCEES FE Reference Handbook — Civil Engineering: Elementary Statically Indeterminate Structures by Force Method — Compatibility statement and the propped-cantilever redundant example; member fixed-end moments table.
NCEES FE Reference Handbook — Mechanics of Materials — Cantilever and simple-beam deflection formulas used as superposition cases in the force method.
Hibbeler, Structural Analysis — Determinacy/stability counting and moment-distribution procedure.

F. Loads, load combinations, and load paths

Loads, Load Combinations, and Load Paths

Define gravity and lateral loads, combine them with ASCE 7 LRFD and ASD equations, reduce live load by tributary area, and trace the load path to the foundation.

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G. Design of steel components

Design of Steel Components

AISC tension, compression, and flexural member design—yield and rupture, block shear, the column curve, and the plastic moment with resistance factors.

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H. Design of reinforced concrete components

Design of Reinforced Concrete Components

ACI strength design of singly-reinforced beams—stress block, nominal moment, strain-based phi, tension-controlled limits, and one-way shear.

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