CELE Structural Theory & Analysis Reviewer

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.