Structural analysis: determinate and indeterminate structures

In short

Structural analysis finds the reactions, internal forces (axial force, shear force, bending moment, twisting moment) and deflections in a structure under given loads. A structure is statically determinate if equilibrium equations alone give all forces, and statically indeterminate (redundant) if extra equations from compatibility of displacements are needed. Almost every real Indian structure — a continuous RCC beam, a building frame, a fixed slab, a continuous box-girder bridge — is indeterminate, so the classical indeterminate methods (consistent deformation, slope-deflection, moment distribution) remain the backbone of design office checks and of every structural software result. It is one of the reliably scoring themes in GATE Civil — Structural Analysis usually carries roughly 6–10 marks — and 20–30 marks in university structural analysis papers, and it is a staple of technical rounds at L&T, Tata Projects, AFCONS, Shapoorji Pallonji, NHAI/RVNL consultancies and design firms like AECOM, STUP and TCE.

Explained

Equilibrium, determinacy and stability

• A planar structure has three independent equilibrium equations: ΣFx = 0, ΣFy = 0, ΣM = 0. A space structure has six.
• Statically determinate: number of unknown reactions/forces = number of equilibrium equations available. Forces are independent of material, cross-section and settlement.
• Statically indeterminate: unknowns exceed equations; the excess count is the degree of static indeterminacy (Ds). Extra equations come from compatibility (deflection/rotation conditions), which bring in E, I and A.
• Stability: a structure must be restrained against translation and rotation. Three parallel reactions, or three concurrent reactions (all meeting at a point), give geometric instability even if counts match.
• Unstable / deficient / redundant frames: if m + r < 2j the pin-jointed plane frame is deficient (a mechanism); if equal, determinate; if greater, redundant.
• Equation of condition: every internal hinge in a beam adds one equation (ΣM = 0 at the hinge); an internal roller or link adds equations too. Hence for a plane beam Ds = r − (3 + c) with r reactions and c conditions; for a frame this count gives only the external part.
• Common supports and their reaction counts: roller 1, hinge/pin 2, fixed 3, guided roller 2, link 1 (plane case).

Degrees of indeterminacy

• Static indeterminacy, plane truss: Ds = (m + r) − 2j, where m = members, r = reactions, j = joints. Space truss: Ds = (m + r) − 3j.
• Static indeterminacy, plane rigid frame: Ds = 3m + r − 3j, or by the cut method Ds = 3C − rr, where C = number of closed boxes and rr = released reactions/conditions. For a plane frame a simple rule is Ds = 3 × (number of closed loops) − (number of conditions).
• External indeterminacy Dse = r − 3 (plane); internal indeterminacy Dsi = Ds − Dse.
• Kinematic indeterminacy (degree of freedom) Dk: Dk = 3j − r for a plane frame (2j − r for a plane truss, 6j − r for a space frame), where j includes all joints. If members are treated as axially rigid (inextensible), subtract the number of such members: Dk = 3j − r − m.
• Determinate means Ds = 0, not Dk = 0. A simply supported beam has Ds = 0 but Dk = 2 (two end rotations, axial deformation ignored; 3 if the roller's horizontal movement is counted) — it is kinematically indeterminate.
• Choice of method: small Ds → force (flexibility) methods; small Dk → displacement (stiffness) methods. Software uses the stiffness route.

Determinate structures and their analysis

• Beams: simply supported, overhanging, cantilever, and compound beams with internal hinges. Draw SFD and BMD from reactions; point of contraflexure is where BM changes sign.
• Plane trusses — method of joints: start at a joint with not more than two unknowns; apply ΣFx = 0, ΣFy = 0. Identify zero-force members first (two non-collinear members at an unloaded joint carry zero force; with three members, two collinear, the odd one is zero).
• Method of sections: cut through not more than three unknown members, take moments about a point that eliminates two unknowns — ideal for a single member force in a long truss.
• Tension coefficient method for space trusses; graphical (Maxwell diagram) for quick checks.
• Three-hinged arch: determinate because the crown hinge gives a fourth equation. Horizontal thrust H = (bending moment at the crown section of a simply supported beam of the same span) / rise, for symmetric level supports; BM at any section = beam moment − H·y.
• Cables and suspension systems: H = wL²/(8h) for a parabolic cable with UDL w, span L, central dip h; maximum tension at the support.
• Determinate structures are unaffected by support settlement, temperature change and fabrication error — they only move, they do not develop stresses. This is their main advantage, and why a determinate scheme is sometimes chosen where support conditions are uncertain.

Energy principles and theorems

• Strain energy: U = ∫M² dx /(2EI) in bending, ∫P² L/(2AE) for axial members, ∫T² dx/(2GJ) for torsion, ∫ (kV²/2GA) dx for shear.
• Castigliano's first theorem: ∂U/∂δ = P. Second theorem: δ = ∂U/∂P (deflection in direction of load P); θ = ∂U/∂M. Apply a dummy load where no load exists, differentiate, then set the dummy to zero.
• Unit load method (virtual work): δ = Σ (P·k·L)/(AE) for trusses, δ = ∫ (M·m dx)/EI for beams/frames, where k and m are forces/moments from a unit load at the point and direction wanted.
• Theorem of least work (Castigliano's theorem of minimum strain energy): redundant forces take values that minimise U — ∂U/∂R = 0, the basis of the force method.
• Maxwell's reciprocal theorem: deflection at B due to unit load at A = deflection at A due to unit load at B (δAB = δBA). Betti's law is the general form: the work of system 1 loads through system 2 displacements equals the reverse.
• Müller-Breslau principle: the influence line for a redundant/reaction is the deflected shape when the corresponding restraint is released and a unit displacement imposed (used for influence lines — detailed separately).
• Moment-area theorems: θAB = area of M/EI diagram between A and B; deflection = moment of that area about the point. Conjugate beam method re-loads a fictitious beam with M/EI; its shear = slope, moment = deflection.

Force (flexibility) methods for indeterminate structures

• Method of consistent deformation (compatibility method): choose the redundants, release them to get a determinate "primary" structure, compute displacements due to loads (ΔL) and due to unit redundant (δ11), and solve ΔL + R·δ11 = 0 (or = known settlement/slip).
• Matrix form: [δ]{R} = −{ΔL}; the flexibility matrix [δ] is symmetric by Maxwell's theorem.
• Standard results: propped cantilever with UDL → prop reaction 3wL/8, fixed-end moment wL²/8 at the fixed end; fixed beam with UDL → end moments wL²/12, mid-span moment wL²/24; fixed beam with central point load → end moments WL/8, mid-span WL/8.
• Theorem of three moments (Clapeyron): for two adjacent spans of a continuous beam with constant EI, MA L1 + 2MB(L1 + L2) + MC L2 + 6(A1x̄1/L1 + A2x̄2/L2) = 0 — solved span by span.
• Column analogy and elastic centre methods are used for fixed arches and closed rings.
• Two-hinged arch: Ds = 1; H = ∫(M·y dx/EI) / ∫(y² dx/EI); for a parabolic arch with UDL over the whole span H = wL²/(8h), the same as the three-hinged value. Rib shortening, temperature rise and yielding of abutments change H in a two-hinged arch but not in a three-hinged one.
• Temperature and settlement: in an indeterminate structure these produce real internal forces — a key exam and interview point. H due to temperature in a two-hinged arch = αTL / ∫(y²dx/EI).

Displacement (stiffness) methods

• Slope-deflection method: unknowns are joint rotations and sway. For member AB of length L and flexural rigidity EI, MAB = (2EI/L)(2θA + θB − 3Δ/L) + MFAB. Write equations for all members, enforce ΣM = 0 at each joint and a shear (sway) equation for each independent sway, solve for θ and Δ, back-substitute.
• Number of simultaneous equations = Dk, so it suits frames with few joints.
• Moment distribution method (Hardy Cross): iterative, no simultaneous equations.
  1. Compute fixed-end moments (FEM) for all members.
  2. Compute relative stiffness k = I/L (use 3I/4L or 0.75 k for a far end that is simply supported/hinged).
  3. Compute distribution factors DF = k / Σk at each joint; at a fixed end DF = 0, at a free/hinged end DF = 1.
  4. Balance each joint (apply the negative of the unbalanced moment × DF), carry over half of the distributed moment to the far end (COF = 1/2 for a prismatic member fixed at the far end, 0 to a hinged far end).
  5. Iterate until the carry-overs are negligible; sum each column for final moments. Add a sway correction pass for unsymmetrical frames or loading.
• Kani's method (rotation contribution) is a related iterative scheme popular for multi-storey frames; substitute frame method (IS 456 permits sub-frame analysis) is the practical office simplification.
• Stiffness/matrix formulation is the computer method — the detail belongs to matrix methods, but note the duality: flexibility deals with forces and compatibility; stiffness deals with displacements and equilibrium.
• Approximate methods for lateral load: portal method and cantilever method for building frames — used for quick checks before running software.

Codes, practice and limits

• IS 456:2000 (RCC) permits elastic analysis of continuous frames, substitute frames and a limited redistribution of moments up to 30% at ultimate limit state for continuous beams, with conditions on equilibrium and neutral axis depth; moment coefficient tables are given for continuous beams with nearly equal spans.
• IS 800:2007 (steel) permits elastic analysis, second-order (P-Δ) analysis and plastic analysis for members of plastic sections; it governs stability and sway classification.
• IS 875 (Parts 1–5) gives dead, imposed, wind and other loads; IS 1893 (Part 1):2016 gives seismic loads and requires a defined lateral load path; IS 1343 for prestressed and IRC 6 / IRC 112 for bridge loads.
• Indeterminate structures have advantages: smaller moments and deflections for the same span, redundancy against local failure (alternate load path), stiffer behaviour, material economy.
• Limitations: stresses from settlement, temperature, shrinkage, creep and fabrication error; more complex analysis; forces depend on relative stiffness, so a change of section changes the answers.
• Validation discipline: any software output is checked against hand equilibrium — ΣV of reactions = total load, free-body moment checks at a cut, and comparison with standard coefficients.

Key points

• Static indeterminacy (Ds) - excess of unknown forces over available equilibrium plus condition equations.
• Kinematic indeterminacy (Dk) - number of independent joint displacements (unknown slopes and translations).
• Equation of condition - an internal hinge gives one extra equation (ΣM = 0 there); it reduces Ds by one.
• Plane truss test - m + r − 2j: negative = unstable/deficient, zero = determinate, positive = redundant.
• Plane frame - Ds = 3m + r − 3j; equivalently 3 × closed loops − conditions.
• Determinate structures - forces independent of E, I, A; immune to settlement, temperature and lack of fit.
• Indeterminate structures - develop internal forces from settlement, temperature, shrinkage and creep; forces depend on relative stiffness.
• Force method - release redundants, use compatibility: ΔL + δ11 R = 0; number of equations = Ds.
• Displacement method - unknowns are rotations/sway, use equilibrium at joints; number of equations = Dk.
• Slope-deflection - MAB = (2EI/L)(2θA + θB − 3Δ/L) + MFAB; one sway equation per independent sway.
• Moment distribution - FEM → distribution factors (k = I/L) → balance → carry over ½ → iterate.
• Three-hinged arch - determinate; BM = beam moment − H·y; H = beam moment at crown / rise.
• Two-hinged arch - one redundant; H = ∫(My dx/EI)/∫(y²dx/EI); affected by rib shortening and temperature.
• Castigliano's second theorem - δ = ∂U/∂P; use a dummy load if no load acts at the point.
• Maxwell's reciprocal theorem - δAB = δBA; makes flexibility and stiffness matrices symmetric.
• Standard fixed-beam results - UDL: wL²/12 ends, wL²/24 centre; central point load: WL/8 ends and centre.
• Propped cantilever with UDL - prop reaction 3wL/8, fixed-end moment wL²/8, maximum sagging 9wL²/128.
• IS 456 redistribution - up to 30% of elastic moment may be redistributed in continuous RCC beams, with code conditions.

Formulas and examples

Example 1 — Degrees of indeterminacy of a two-bay portal frame A single-storey, two-bay frame: 3 columns fixed at the base, 2 horizontal beams on top. Members m = 5, joints j = 6 (3 bases + 3 tops), reactions r = 3 × 3 = 9.

• Ds = 3m + r − 3j = 15 + 9 − 18 = 6. Check by loops: the frame has 2 closed boxes → 3 × 2 = 6. ✔
• Dse = r − 3 = 9 − 3 = 6; Dsi = 6 − 6 = 0 (no internal redundancy).
• Dk = 3j − r = 18 − 9 = 9. Treating all 5 members as axially rigid: Dk = 18 − 9 − 5 = 4 — i.e. 3 joint rotations + 1 sway. Force method would need 6 equations, displacement method 4 → slope-deflection/moment distribution is the cheaper route here.

Example 2 — Propped cantilever by consistent deformation Span L = 6 m, UDL w = 20 kN/m, fixed at A, roller prop at B, constant EI.

• Primary structure: cantilever fixed at A, prop removed. Free-end deflection downward δ_L = wL⁴/(8EI) = 20 × 6⁴/(8EI) = 25,920/(8EI) = 3,240/EI (↓).
• Deflection at B from unit upward load: δ11 = L³/(3EI) = 216/(3EI) = 72/EI (↑).
• Compatibility (net deflection at B = 0): R × 72/EI = 3,240/EI → R_B = 45 kN.
• Check against the standard formula: 3wL/8 = 3 × 20 × 6/8 = 45 kN ✔
• Reaction at A = 20 × 6 − 45 = 75 kN; M_A = 20 × 6²/2 − 45 × 6 = 360 − 270 = 90 kN·m = wL²/8 = 20 × 36/8 = 90 ✔
• Point of zero shear from B: x = 45/20 = 2.25 m → max sagging moment = 45 × 2.25 − 20 × 2.25²/2 = 101.25 − 50.625 = 50.63 kN·m (= 9wL²/128 = 9 × 20 × 36/128 = 50.63 ✔).

Example 3 — Continuous beam by moment distribution Two equal spans AB = BC = 6 m, same I, A and C fixed, UDL 24 kN/m on span AB only.

• FEM: M_AB = −wL²/12 = −24 × 36/12 = −72 kN·m; M_BA = +72 kN·m; M_BC = M_CB = 0.
• Stiffness: k_BA = k_BC = I/6 → DF_BA = DF_BC = 0.5 each (joint B); DF at fixed A and C = 0.
• Unbalanced moment at B = +72 → balancing moment −72: distribute −36 to BA and −36 to BC.
• Carry over ½: −18 to AB, −18 to CB. No further unbalance at A or C (fixed).
• Final moments: M_AB = −72 − 18 = −90 kN·m; M_BA = 72 − 36 = +36 kN·m; M_BC = −36 kN·m; M_CB = −18 kN·m. Joint B balances: +36 − 36 = 0 ✔
• Span AB end shears (clockwise-positive end moments): V_A = wL/2 − (M_AB + M_BA)/L = 72 − (−90 + 36)/6 = 72 + 9 = 81 kN; V_B(AB) = 144 − 81 = 63 kN. Span BC carries only the end moments: shear = (36 + 18)/6 = 9 kN (constant), so a span with no load still carries force — the signature of indeterminacy.

Example 4 — Three-hinged parabolic arch Span L = 20 m, central rise h = 4 m, point load W = 100 kN at 5 m from left support A; hinges at A, B (right support) and crown C.

• Beam reactions: V_A = 100 × 15/20 = 75 kN, V_B = 25 kN.
• Beam moment at crown (x = 10 m) = 75 × 10 − 100 × 5 = 750 − 500 = 250 kN·m → H = 250/4 = 62.5 kN.
• Arch ordinate at x = 5 m: y = 4 × 4 × 5 × 15/20² = 1,200/400 = 3.0 m.
• Beam moment at x = 5 m = 75 × 5 = 375 kN·m → arch BM = 375 − 62.5 × 3.0 = 375 − 187.5 = +187.5 kN·m.
• Normal thrust and radial shear at x = 5 m: slope dy/dx = 4h(L − 2x)/L² = 16 × 10/400 = 0.4 → θ = 21.8°, cos θ = 0.928, sin θ = 0.371. Shear there (left of load) V = 75 kN. N = H cos θ + V sin θ = 62.5 × 0.928 + 75 × 0.371 = 58.0 + 27.8 = 85.8 kN; radial shear = V cos θ − H sin θ = 69.6 − 23.2 = 46.4 kN.

Example 5 — Truss joint deflection by unit load method A determinate plane truss; three members contribute to the vertical deflection at joint D. Member data (P from actual load, k from a 1 kN unit load at D):

• Σ(PkL/A) = 0.192 + 0.090 + 0.2083 = 0.4903 kN·m/mm².
• E = 2 × 10⁵ N/mm² = 200 kN/mm². δ = Σ(PkL)/(AE) = 0.4903/200 m = 0.002452 m = 2.45 mm downward.

Where it is used

• Design consultancies (TCE, STUP, AECOM India, L&T Design Centre): continuous RCC/steel frames, transfer girders and sub-frame analysis per IS 456, with hand checks against moment-distribution results.
• Bridges: three-hinged and two-hinged arches, bowstring girders (determinate truss check), continuous box girders and integral bridges analysed under IRC 6 live loads; RVNL, NHAI and Konkan Railway works.
• Industrial and PEB structures: pin-based versus fixed-base portal frames — the choice directly follows from the determinacy/settlement trade-off; roof trusses analysed as determinate plane trusses.
• Site and construction engineering: staging, formwork and launching girders, where temporary structures are deliberately made determinate so that differential settlement of the ground does not induce stresses.
• Job roles: structural design engineer, proof-checking engineer for government projects, bridge rehabilitation engineer, STAAD.Pro/ETABS/MIDAS modeller, and "analysis validation" roles where software output must be reconciled with equilibrium.
• Exams and recruitment: GATE Civil, ESE (UPSC Engineering Services), state PSC AE/JE, NTPC/NHPC/BHEL/ISRO structural posts and GATE-based PSU hiring all test determinacy counts and classical methods.

Interview questions

• Q: How do you decide whether a structure is determinate? A: Count unknown reactions and internal forces against the available equilibrium and condition equations — for a plane frame, Ds = 3m + r − 3j; if Ds = 0 and the structure is stable, it is determinate.
• Q: Can a structure have Ds = 0 and still be unusable? A: Yes — it may be geometrically unstable, for example three parallel or three concurrent reactions, which cannot resist all load directions despite the correct count.
• Q: Why does support settlement stress an indeterminate structure but not a determinate one? A: A determinate structure has just enough restraints, so it simply moves to accommodate the settlement; an indeterminate one is over-restrained, so the imposed displacement must be resisted, generating internal forces proportional to EI.
• Q: Difference between static and kinematic indeterminacy? A: Static indeterminacy counts unknown forces beyond equilibrium; kinematic indeterminacy counts unknown joint displacements — a simply supported beam is Ds = 0 but Dk = 2 (axial deformation ignored).
• Q: When would you choose moment distribution over slope-deflection? A: When the structure has several joints but little or no sway — moment distribution avoids simultaneous equations; with sway or few joints, slope-deflection is quicker.
• Q: Why is the distribution factor at a hinged end 1 and at a fixed end 0? A: A hinge can carry no moment, so the whole unbalanced moment must be released there; a fixed end has infinite rotational restraint and absorbs moment without rotating.
• Q: What is the carry-over factor and why is it 0.5? A: It is the ratio of the moment induced at the far end to the moment applied at the near end; for a prismatic member fixed at the far end, the slope-deflection relation gives exactly one-half.
• Q: Why is a three-hinged arch preferred on poor or settling soil? A: It is statically determinate, so unequal settlement of the abutments, temperature change and rib shortening cause no internal stresses.
• Q: State and use Maxwell's reciprocal theorem. A: The deflection at B due to a unit load at A equals the deflection at A due to a unit load at B; it halves influence-coefficient computations and makes the flexibility and stiffness matrices symmetric.
• Q: What are zero-force members and why keep them? A: Members carrying no force under the given load case; they are retained to reduce slenderness, brace the frame and carry other load cases such as wind or reversed live load.
• Q: How much moment redistribution does IS 456 allow and why? A: Up to 30% of the elastic moment in continuous RCC beams, subject to equilibrium and neutral axis depth limits, because ductile sections can rotate and shed moment to less-stressed regions.
• Q: How do you sanity-check a STAAD/ETABS result? A: Verify ΣV of reactions equals the total applied load, check a free-body cut for ΣM = 0, and compare a typical span moment with standard coefficients such as wL²/12 or wL²/8.

Common mistakes

• Using Ds = 3m + r − 3j for a truss and (m + r) − 2j for a frame — the two formulas are not interchangeable.
• Forgetting the equation of condition at an internal hinge, overstating Ds by one for every hinge missed.
• Omitting the hinge/base joints when counting j, or counting only the top joints of a frame.
• Treating Ds = 0 as proof of a safe structure and ignoring geometric instability (parallel or concurrent reactions).
• Assuming temperature change and settlement cause no stress in a continuous beam or fixed arch — they do.
• Carrying over to a hinged far end (carry-over factor is 0, not 0.5) or using full k instead of 0.75 k for such a member.
• Sign errors in slope-deflection and moment distribution: fixed-end moments and final moments must use one consistent convention (clockwise positive) throughout, and each internal joint must finally sum to zero.

Related topics

• Influence lines and moving loads · Matrix methods of structural analysis · Strength of materials: stress and strain · Reinforced concrete design (IS 456) · Steel structures design (IS 800) · Structural dynamics basics