5.4 Structural Analysis

Determinacy & Reactions

A structure is analyzed by equilibrium: ΣFx = 0, ΣFy = 0, ΣM = 0 (three equations in a plane). A structure is statically determinate when the number of unknown reactions equals the available equilibrium equations, so reactions can be found from statics alone. When unknowns exceed equations, the structure is statically indeterminate and requires compatibility (deflection) equations.

For a planar truss the determinacy check is m + r = 2j (m = members, r = reactions, j = joints): equal → determinate, greater → indeterminate, less → unstable mechanism. For beams/frames, degree of indeterminacy = (reactions + internal forces) − equations.

Support types set the reaction count: a roller = 1 reaction, a pin = 2, a fixed support = 3. Always start a problem by drawing the free-body diagram and solving the global reactions; the FE Reference Handbook tabulates reactions and deflections for standard cases.

Shear & Moment Diagrams

Shear (V) and bending moment (M) diagrams trace internal forces along a member. They are linked by calculus:

• dV/dx = −w (slope of shear = negative of distributed load).
• dM/dx = V (slope of moment = shear).
• Area under the load diagram = change in shear; area under the shear diagram = change in moment.

Consequences tested constantly:

• Maximum moment occurs where shear = 0 (or crosses zero).
• A point load causes a step in shear and a kink (slope change) in moment.
• A point moment causes a step in the moment diagram.

Sign convention: sagging moment positive, hogging negative — keep it consistent.

Deflection, Influence Lines & Trusses

Deflection of beams comes from EI·d²y/dx² = M(x). Two graphical/energy tools are tested: the moment-area method (first theorem: change in slope = area of M/EI diagram; second: tangential deviation = moment of that area) and the conjugate-beam concept, where loading a fictitious beam with the M/EI diagram makes its shear = real slope and its moment = real deflection. For common cases use Handbook formulas, e.g., simple-beam midspan δ = 5wL⁴/(384EI).

Influence lines show how a single reaction or internal force varies as a unit load moves across the span — essential for placing moving (vehicle) loads to maximize an effect, central to bridge design.

Trusses carry only axial force in members (two-force members). Solve by the method of joints (ΣFx = ΣFy = 0 at each pin) or the method of sections (cut and apply ΣM to isolate one member force). A zero-force member carries no load and is identified by joint geometry.

Superposition & a Worked Shear/Moment Example

The principle of superposition states that for a linear-elastic structure, the response to several loads equals the sum of the responses to each load applied alone. It underlies indeterminate-beam methods (force/flexibility method) and lets you combine Handbook standard cases instead of re-deriving.

Worked example: A simply supported beam spans L = 20 ft and carries a uniform load w = 2 kip/ft.

• Reactions: R_A = R_B = wL/2 = 2 × 20 / 2 = 20 kip each.
• Shear is +20 kip at A, decreasing linearly to −20 kip at B, crossing zero at midspan (x = 10 ft).
• Maximum moment at midspan: M_max = wL²/8 = 2 × 20² / 8 = 100 kip·ft.

Check by area: moment at midspan = area of the shear diagram from A to midspan = ½ × 20 kip × 10 ft = 100 kip·ft. The two methods agreeing confirms the result — a fast self-check on the exam.

Frames, Indeterminate Methods & Section Properties

Beyond simple beams, the FE Civil exam tests frames (rigid beam–column joints transfer moment) and indeterminate structures. Two solution families are referenced in the Handbook:

• Force (flexibility) method — choose redundant reactions, remove them to get a determinate primary structure, then enforce compatibility (zero deflection) at each redundant using superposition of Handbook deflection cases.
• Displacement (stiffness/slope-deflection, moment-distribution) methods — solve unknown joint rotations/translations; these underlie all matrix structural-analysis software.

Member stiffness depends on EI and section geometry. Key section properties: moment of inertia I, section modulus S = I/c (elastic), and plastic modulus Z for steel design. For a rectangle, I = bh³/12 and S = bh²/6.

A cantilever carrying load P at its free end has reactions of a vertical force P and a fixed-end moment PL, with tip deflection δ = PL³/(3EI) — a Handbook standard worth memorizing because cantilever and propped-cantilever cases recur in both analysis and deflection questions. Recognize the support and load pattern, then read the closed-form result rather than re-deriving it under time pressure.

, three-hinged arches), each releasing one moment and adding one equation of condition (ΣM = 0 about the hinge), which can render an otherwise indeterminate structure determinate. When counting determinacy, add these condition equations to the three global equilibrium equations. Always verify stability before solving: a structure with enough reactions can still be a mechanism if the supports are collinear or concurrent (geometric instability), giving meaningless results.

The robust workflow for any FE Civil analysis problem is: (1) draw the free-body diagram, (2) check stability and determinacy, (3) solve global reactions, (4) section the member to find internal V and M, and (5) sketch the diagrams, locating M_max where V = 0.