Determinate vs Indeterminate Structures & Moment Distribution
Degrees of freedom, moment distribution and slope-deflection — how frame analysis questions are actually asked in campus rounds.
Key formulas
Ds = r − 3 (2D rigid frames, external); add releases (hinges) to reduce
Static indeterminacy (beam/frame): r = support reactions count; internal indeterminacy adds for closed boxes/frames.
K = 4EI/L (far end fixed), K = 3EI/L (far end hinged)
Member stiffness: Distribution factors are K_i / ΣK at each joint — the core of moment distribution setup.
M_AB = (2EI/L)(2θ_A + θ_B − 3ψ) + FEM_AB, ψ = Δ/L
Slope-deflection equation: ψ is the chord rotation from support settlement or sway.
Cheat sheet
- ▸Static indeterminacy = number of unknowns − number of equilibrium equations; a 2D pin-jointed truss needs m + r = 2j. Indeterminacy adds redundancy but makes structures less sensitive to local failure.
- ▸Moment distribution (Hardy Cross): fix all joints, compute distribution factors from member stiffnesses K = 4EI/L (far end fixed) or 3EI/L (far end hinged), balance unbalanced moments, carry over half to the far end, repeat until moments are negligible.
- ▸Slope-deflection writes each member end moment as a function of end rotations and chord rotation: M = (2EI/L)(2θA + θB − 3Δ/L) + FEM — a displacement method.
- ▸Stiffness and carry-over: carry-over factor is 0.5 to a fixed far end, 0 to a hinged end. Distribution factor for a member = its stiffness / sum of stiffnesses at the joint.
- ▸Fixed end moments to memorise: UDL on SS span FEM = wL²/12; central point load FEM = WL/8.
- ▸Choose the method: few unknown displacements → slope-deflection; many joint rotations → moment distribution; complex geometry/thermal/settlement → matrix (stiffness) methods by software.
Determinacy quick checks
| Structure | Equation | Stable & determinate when |
|---|---|---|
| Plane truss (pin-jointed) | m + r = 2j | m + r = 2j exactly |
| 2D rigid frame | Ds = 3m + r − 3j − c | Ds = 0 |
| Beam (2D) | Ds = r − 3 | r = 3 |
| Portal frame with fixed bases | r = 6 → Ds = 3 | Needs member equations (indeterminate) |
Visual
Moment distribution cycle (joint B fixed-C fixed, load on AB): Step 1: FEMs M_AB = −wL²/12, M_BA = +wL²/12 Step 2: balance B: distribute −M_BA by DFs (BA: .5, BC: .5) Step 3: carry over ½ of each to the far ends Step 4: repeat until residual ≈ 0, sum columns for final moments
Top 4 campus interview questions
MCQ mastery quiz
1. A plane pin-jointed truss is statically determinate when…
2. The carry-over factor to a fixed far end is…
3. Fixed end moment for a UDL over a span with both ends fixed is…
4. Support settlement in a statically determinate beam causes…