CELE Structural Theory & Analysis Reviewer 2026

Answer: B. Joint displacements and rotations

Step 1 — Distinguish force vs. displacement methods: Force methods (e.g., three-moment equation, virtual work) take redundant forces/moments as unknowns. Displacement methods take joint displacements (translations, Δ) and rotations (θ) as unknowns. Step 2 — Once θ and Δ are found, member-end moments are calculated by back-substitution into the slope-deflection equation. Step 3 — Why the other options are wrong: 'Member forces and reactions' are the final results, not the primary unknowns. 'Fixed-end moments' are inputs (loading terms), not unknowns. 'Redundant reactions' are the unknowns in force methods, not displacement methods.

Answer: A. 30 kN·m

Step 1 — Recall the FEM formula for a UDL on a fully fixed beam: |FEM| = wL²/12. Step 2 — Substitute values: |FEM| = 10 × (6)² / 12 = 10 × 36 / 12 = 360 / 12 = 30 kN·m. Step 3 — Sign convention: FEM_AB = −30 kN·m (counterclockwise at A) and FEM_BA = +30 kN·m (clockwise at B), using the clockwise-positive convention. Step 4 — Why the other options are wrong: 45 kN·m = wL²/8 (midspan moment for a simply supported beam, not FEM). 60 kN·m = wL²/6 (incorrect formula). 15 kN·m = wL²/24 (not a standard FEM formula). Always use wL²/12 for UDL on a fixed-fixed beam.

Answer: B. PL/8

Step 1 — Identify the loading case: central point load P on a fixed-fixed beam. Step 2 — Apply the formula: |FEM| = PL/8. This is derived from the general off-center formula FEM_AB = Pab²/L² with a = b = L/2, giving P(L/2)(L/2)²/L² = PL/8. Step 3 — Why the other options are wrong: PL/4 is the midspan moment of a simply supported beam under P at center (not a FEM). PL/12 has no standard meaning for a central load FEM. PL/6 is also not a standard central-load FEM formula. Remember: FEM for central P = PL/8, while simply supported midspan M = PL/4.

Answer: B. The chord rotation due to relative settlement or sidesway (Δ/L)

Step 1 — Define chord rotation: ψ = Δ/L, where Δ is the relative transverse displacement (settlement or sidesway) of the two ends of the member, and L is the member length. Step 2 — When there is no support settlement and no sidesway (braced frame with symmetric loading), ψ = 0 and the equation simplifies to M_AB = (2EI/L)(2θ_A + θ_B) + FEM_AB. Step 3 — Physical meaning: If support B settles by Δ relative to A, the member chord tilts, adding a moment even if no loads exist. Step 4 — Why the other options are wrong: θ_A is the joint rotation at A. FEM coefficients are tabulated constants. Distribution factors belong to the moment distribution method, not the slope-deflection equation.

Answer: B. k = 3EI/L

Step 1 — Standard stiffness for a member with far end fixed: k = 4EI/L. This is the moment required at the near end to rotate it by 1 radian while the far end is fixed. Step 2 — Modified stiffness for a member with far end pinned: k = 3EI/L. This accounts for the fact that the far pin provides no moment restraint, reducing the effective rotational stiffness. Step 3 — Practical benefit: Using k = 3EI/L for a pin-ended member eliminates carry-over to that end (carry-over factor = 0), saving one iteration step. Step 4 — Why the other options are wrong: 2EI/L and EI/L are not standard stiffness values for beam members in moment distribution. Always check the boundary condition at the far end before choosing k.

Answer: C. 1/2

Step 1 — The carry-over factor (COF) for a prismatic beam with a far fixed end is always 1/2. This means that when a moment M is distributed at the near end, a moment of M/2 is automatically induced at the far fixed end. Step 2 — Physical basis: From the slope-deflection equation, fixing the far end and rotating the near end by θ produces M_near = 4EIθ/L and M_far = 2EIθ/L. The ratio M_far/M_near = 2/4 = 1/2. Step 3 — For a member with a far pinned end: COF = 0 (no carry-over, since a pin cannot carry moment). Step 4 — Why the other options are wrong: COF = 1 would mean full moment transfer (incorrect for a standard prismatic member). COF = 1/3 and 1/4 have no physical basis for standard prismatic beams.

Answer: D. 0.50

Step 1 — Recall the distribution factor formula: DF = k / Σk, where k is the stiffness of the member in question and Σk is the sum of stiffnesses of all members meeting at the joint. Step 2 — Since k_BA = k_BC (equal stiffness), Σk = k_BA + k_BC = 2k. Step 3 — Calculate: DF_BA = k_BA / (k_BA + k_BC) = k / 2k = 0.50. Step 4 — Verify: DF_BA + DF_BC = 0.50 + 0.50 = 1.0 ✓ (distribution factors at any joint must always sum to 1.0). Step 5 — Why the other options are wrong: DF = 1.0 applies only to a fixed end (no distribution) or a single-member joint. 0.25 and 0.75 arise from unequal stiffness ratios.

Answer: B. −32.0 kN·m

Step 1 — Identify the formula: FEM_AB = −Pab²/L². Step 2 — Substitute: P = 36 kN, a = 2 m, b = 4 m, L = 6 m. Step 3 — Calculate: FEM_AB = −36 × 2 × (4)² / (6)² = −36 × 2 × 16 / 36 = −1152 / 36 = −32.0 kN·m. Step 4 — The negative sign indicates a counterclockwise (hogging) moment at end A under the clockwise-positive convention. Step 5 — Why the other options are wrong: −16.0 kN·m results from using incorrect exponent placement. −24.0 kN·m and −48.0 kN·m arise from arithmetic errors in the multiplication or denominator.

Answer: B. The member-end moment M = 0

Step 1 — A simple support (pin or roller) provides vertical reaction but offers no moment restraint. Therefore, the moment at that end equals zero: M = 0. Step 2 — This boundary condition is substituted into the slope-deflection equation to solve for the unknown rotation θ at that end. Step 3 — Example: For a propped cantilever with pin at B, set M_AB or M_BA = 0 (depending on which end is the pin) and solve for θ_B. Step 4 — Why the other options are wrong: θ = 0 applies at a fully fixed support (no rotation). Δ = 0 applies at any support that prevents vertical movement (not unique to simple supports). V = 0 applies at a free end (cantilever tip), not a simple support.

Answer: B. Compute the fixed-end moments for all loaded spans

Step 1 — The correct sequence in moment distribution is: (1) Compute FEMs for all loaded spans, (2) Calculate distribution factors (DF = k/Σk) at all joints, (3) Balance each joint by distributing the unbalanced moment, (4) Carry over half the distributed moment to the far fixed end, (5) Repeat steps 3–4 until convergence, and (6) Sum all moments for the final member-end moments. Step 2 — FEMs must be computed first because they are the starting unbalanced moments that drive the entire distribution process. Step 3 — Why the other options are wrong: Distribution factors are calculated in step 2, not step 1. Carry-over comes in step 4. Summing moments is the final step.

Answer: B. A unit load moves across the entire span of the structure.

Step 1 — Recall the definition: an influence line (IL) is constructed by keeping the section (and the response function) fixed, then rolling a unit load (1 kN or 1 unit) across the structure from one end to the other. Step 2 — At each position of the unit load, the value of the chosen response is computed and plotted against the load position. Step 3 — This produces a diagram whose ordinate at any point gives the response value when the unit load is at that point. Step 4 — Option A describes a shear or moment diagram (fixed load, varying section) — the opposite concept. Options C and D do not correspond to IL construction at all. The key distinction: IL = fixed section, moving load; BM/SF diagram = fixed load, varying section.

Answer: C. The deflected shape of the structure after the restraint corresponding to that response is released and a unit displacement is imposed.

Step 1 — The Müller-Breslau principle is a powerful tool for determining IL shapes qualitatively (and quantitatively for determinate structures). Step 2 — The procedure: (a) Remove the restraint associated with the desired response (e.g., remove the pin at A to get the IL for reaction R_A); (b) Apply a unit displacement (or unit rotation) in the direction of that response. Step 3 — The resulting deflected shape of the structure is geometrically similar to the influence line for that response. Step 4 — Options A and B describe loading cases for shear/moment diagrams, not the IL. Option D (self-weight deflection) is unrelated to the principle.