Analysis of statically indeterminate structures

Why it matters

Aircraft structures are deliberately redundant: multi-spar wings, multi-cell boxes, frames with many stringers and fail-safe fittings all have more load paths than statics alone can resolve. Redundancy is what lets a damaged structure keep flying, but it means internal loads depend on the relative stiffness of the paths. You must be able to find them by adding compatibility to equilibrium.

Key ideas

Determinate and indeterminate. A structure is statically determinate if the equations of equilibrium alone give all reactions and member forces. If there are more unknowns than independent equilibrium equations, it is statically indeterminate, and the excess is the degree of indeterminacy (the number of redundants). Each redundant needs one extra equation, which comes from compatibility of displacements.

Degree of indeterminacy.

• Plane pin-jointed truss: r = m + R − 2j, with m members, R reaction components and j joints. r = 0 determinate (if stable), r > 0 indeterminate, r < 0 a mechanism. A count of r ≥ 0 is necessary but not sufficient: the members must also be arranged stably.
• Plane rigid-jointed frame: r = 3m + R − 3j.
• Beams without internal hinges: number of reaction components minus 3 (minus 2 when only transverse loads act and horizontal reactions are ignored). A propped cantilever is once redundant; a fixed-fixed beam under transverse load is twice redundant.
• A closed thin-walled cell is once redundant in shear: the constant shear flow q₀ cannot be found by statics and needs the twist (compatibility) condition.

Force (flexibility) method.

1. Choose redundants Xᵢ and remove them to leave a stable determinate released structure.
2. Find the displacement Δᵢ₀ of the released structure at each release, in the direction of Xᵢ, under the real loads (unit load method).
3. Find flexibility coefficients fᵢⱼ: displacement at i due to a unit value of Xⱼ. By Maxwell, fᵢⱼ = fⱼᵢ.
4. Compatibility: the real structure has no gap at the releases, so Δᵢ₀ + Σ fᵢⱼ·Xⱼ = 0.
5. Solve for X, then superpose: final force = force from real load + Σ Xⱼ × force from unit Xⱼ.

Displacement (stiffness) method. Unknowns are joint displacements; equilibrium at the joints is written in terms of them (the basis of the finite element method). Both methods give the same answer for linear elastic structures.

Energy view. For a linear structure the redundants make the total complementary energy stationary: ∂U/∂Xᵢ = 0 (Castigliano's least work, provided the supports of the redundants do not yield).

Physical consequences.

• Load goes preferentially to the stiffer path. Changing a member's area changes the force in every member.
• Temperature change, lack of fit and support settlement produce stresses in an indeterminate structure even with no external load; a determinate structure simply moves.
• Redundancy gives fail-safety: if one path fails, the load redistributes.

Formulas

r = m + R − 2·j

• Plane truss: m members, R reaction components, j joints.

Δ₁₀ + f₁₁·X₁ = 0

• One redundant. Δ₁₀: displacement at the release under the real load (m), f₁₁: displacement at the release due to X₁ = 1 (m/N), X₁: redundant (N).

Δᵢ₀ = Σ P₀·pᵢ·L/(A·E), fᵢⱼ = Σ pᵢ·pⱼ·L/(A·E)

• Truss form. P₀: member forces in the released structure from real loads (N), pᵢ: member forces due to unit Xᵢ.

Δᵢ₀ = ∫ M₀·mᵢ/(E·I) dx, fᵢⱼ = ∫ mᵢ·mⱼ/(E·I) dx

• Beam and frame form, bending only.

∂U / ∂X = 0

• Least work, linear elastic, rigid supports at the redundants.

σ = −E·α·ΔT

• Bar fixed between rigid walls heated by ΔT (K); α: coefficient of thermal expansion (1/K), compression negative.

Standard results: propped cantilever with UDL w over span L: prop reaction 3·w·L/8, fixed-end moment w·L²/8. Beam fixed at both ends with UDL: end moments w·L²/12, mid-span moment w·L²/24, reactions w·L/2.

Worked examples

Example 1 (standard): propped cantilever by the force method. Given: cantilever fixed at A, simply supported (propped) at B, span L = 6 m, UDL w = 10 kN/m, uniform EI.

1. Redundant: the prop reaction R_B. Released structure: a cantilever fixed at A.
2. Tip deflection of the released cantilever under the UDL (downward): Δ₁₀ = w·L⁴/(8·EI).
3. Tip deflection due to unit upward load at B: f₁₁ = L³/(3·EI), upward.
4. Compatibility (no deflection at B): R_B·L³/(3·EI) = w·L⁴/(8·EI), so R_B = 3·w·L/8.
5. Numbers: R_B = 3 × 10 × 6 / 8 = 22.5 kN. Then R_A = w·L − R_B = 60 − 22.5 = 37.5 kN.
6. Fixed-end moment: M_A = w·L²/2 − R_B·L = 180 − 135 = 45 kN·m (hogging), which equals w·L²/8. Answer: R_B = 22.5 kN, R_A = 37.5 kN, M_A = 45 kN·m hogging.

Example 2 (GATE level): three-bar suspension. Given: a load P = 100 kN hangs from joint D. A vertical bar of length L = 2 m runs up from D to the ceiling, and two side bars, each at θ = 45° to the vertical, run up from D symmetrically, so each has length L/cos θ. All bars have the same AE, with A = 200 mm² and E = 200 GPa. Unknowns: F_c (central) and F_s (each side), two unknowns but one useful equilibrium equation, so once redundant.

1. Equilibrium (vertical): F_c + 2·F_s·cos θ = P.
2. Compatibility: D moves down by δ; the central bar stretches δ, each side bar stretches δ·cos θ.
3. Hooke's law: F_c·L/(AE) = δ and F_s·(L/cos θ)/(AE) = δ·cos θ, so F_s = F_c·cos²θ.
4. Substitute: F_c·(1 + 2·cos³θ) = P. With cos 45° = 0.7071, cos³θ = 0.3536, so F_c = 100 / 1.7071 = 58.58 kN.
5. Side bars: F_s = 58.58 × 0.5 = 29.29 kN. Check: 58.58 + 2 × 29.29 × 0.7071 = 100.0 kN.
6. Deflection of D: δ = F_c·L/(AE) = 58 580 × 2 / (200 × 10⁻⁶ × 200 × 10⁹) = 2.93 × 10⁻³ m. Answer: F_c = 58.6 kN, F_s = 29.3 kN (both tension), δ = 2.93 mm.

Common mistakes

• Choosing releases that leave a mechanism. The released structure must be stable and determinate.
• Getting the sign of Δ₁₀ and f₁₁ inconsistent: both must be measured in the direction of the assumed positive redundant.
• Using r = m + R − 2j for frames or 3-D trusses; the count changes with the type of structure.
• Assuming forces split equally between parallel paths. They split in proportion to stiffness.
• Forgetting thermal or lack-of-fit stresses: in a redundant structure they exist without external load.
• Using the deformation of the whole side bar instead of its component along the bar (δ·cos θ).

For GATE AE

Expect degree-of-indeterminacy counts for trusses and frames, propped cantilevers and fixed beams by compatibility, three-bar or parallel-bar problems, thermal stress in restrained bars, and the redundant shear flow of a closed cell. Practise one-redundant problems with the unit load method until the Δ₁₀ + f₁₁·X = 0 setup is automatic.

Quick check

1. A plane truss has 9 members, 6 joints and 4 reaction components. What is its degree of indeterminacy?
2. In the force method, what equation determines each redundant?
3. A steel bar (E = 200 GPa, α = 12 × 10⁻⁶ /K) is fixed between rigid walls and heated by 40 K. What is the stress?
4. Why does a stiffer member in a redundant structure attract more load? Answers: 1. r = 9 + 4 − 12 = 1; 2. compatibility at the release, Δᵢ₀ + Σfᵢⱼ·Xⱼ = 0; 3. 96 MPa compression; 4. compatibility forces equal (or related) deformations, so force is proportional to stiffness.