Statically determinate and indeterminate trusses

Why it matters

Trusses are everywhere in aerospace hardware: engine mounts, landing-gear bracing, launch-vehicle interstages, satellite frames and the Warren-truss fuselages of light aircraft. Knowing whether a truss is determinate tells you whether statics alone gives the member forces or whether stiffnesses, temperature and fit-up errors will also matter. The member forces then size every tube and every joint.

Key ideas

Ideal truss assumptions. Members are straight and joined by frictionless pins; loads and reactions act only at the joints; member self-weight is neglected or lumped at joints. Each member is then a two-force member carrying only axial tension or compression. Real welded or riveted joints carry small secondary bending moments that this model ignores.

Determinacy and stability of a plane truss. Each joint gives two equilibrium equations (ΣFx = 0, ΣFy = 0), so a truss with j joints gives 2j equations. The unknowns are the m member forces and the r support reactions. The degree of static indeterminacy is D = m + r − 2j.

• D < 0: too few unknowns; the truss is a mechanism (unstable).
• D = 0: statically determinate, provided the arrangement is stable.
• D > 0: statically indeterminate to degree D.

The count is necessary but not sufficient. A truss with D ≥ 0 can still be unstable if members are badly arranged (for example a panel without a diagonal) or if all reactions are parallel or meet at one point. External indeterminacy is r − 3; the remainder is internal (redundant members). For space trusses use D = m + r − 3j.

A simple truss is built from a basic triangle by adding two members and one joint at a time, so m = 2j − 3. With three suitable reactions it is determinate and stable.

Method of joints. Find the reactions, then go joint by joint, always choosing one with at most two unknown member forces. Assume all unknown forces are tensile (pulling away from the joint); a negative result means compression.

Method of sections. Cut through no more than three members whose lines of action do not all meet at one point, and use the three equations for the free body on one side. Take moments about the point where two of the cut members meet to get the third directly. It is the fastest way to find the force in one chosen member.

Zero-force members (by inspection, for unloaded joints):

• Two non-collinear members meet at a joint with no load: both are zero-force.
• Three members meet at an unloaded joint and two are collinear: the third is zero-force. Zero-force members are not useless; they brace compression members against buckling and carry load under other load cases.

Indeterminate trusses. Redundant member forces depend on member stiffnesses AE/L. Solve with equilibrium plus compatibility of joint displacements, or with energy methods (unit load or Castigliano). A determinate truss has no stresses from temperature change or from a member that is slightly too long; an indeterminate one does.

Formulas

D = m + r − 2·j Degree of static indeterminacy of a plane truss; m = members, r = reaction components, j = joints (all counts, dimensionless).

D = m + r − 3·j Same, for a space (3-D) truss.

m = 2·j − 3 Number of members of a simple (just-rigid) plane truss.

ΣFx = 0, ΣFy = 0 At each pin joint (method of joints); forces in N.

ΣFx = 0, ΣFy = 0, ΣM = 0 On a cut portion (method of sections); moments in N·m.

δ = F·L / (A·E) Elongation of a member with axial force F (N), length L (m), area A (m²), modulus E (Pa); used for joint displacements and compatibility.

Worked examples

Example 1 (standard). Truss ABC: pin at A (0, 0), roller at B (6 m, 0), apex C at (2 m, 3 m). A vertical load of 12 kN acts downwards at C. Find the member forces.

1. Count: m = 3, r = 3, j = 3, so D = 3 + 3 − 6 = 0 and the triangle is stable: determinate.
2. Moments about A: R_B × 6 = 12 × 2, so R_B = 4 kN (up). Vertical balance: R_A = 12 − 4 = 8 kN (up). No horizontal load, so A_x = 0.
3. Joint A. Length AC = √(2² + 3²) = √13 = 3.606 m. ΣFy: 8 + F_AC × (3/√13) = 0, so F_AC = −9.615 kN.
4. ΣFx at A: F_AB + F_AC × (2/√13) = 0, so F_AB = +5.333 kN.
5. Joint B. Length BC = √(4² + 3²) = 5 m. ΣFy: 4 + F_BC × (3/5) = 0, so F_BC = −6.667 kN.
6. Check ΣFx at B: −F_AB + F_BC × (−4/5) = −5.333 + 5.333 = 0. Correct.

Answer: AB = 5.33 kN tension, AC = 9.62 kN compression, BC = 6.67 kN compression.

Example 2 (GATE level). Three bars of equal A and E hang from a rigid ceiling and meet at one joint D. The middle bar is vertical with length L; the two outer bars are inclined at θ = 45° to the vertical, symmetric. A vertical load P = 100 kN acts at D. Find the bar forces.

1. Count: m = 3, r = 6 (two at each ceiling pin), j = 4, so D = 3 + 6 − 8 = 1. One redundant.
2. Equilibrium at D (by symmetry both outer bars carry F₂): F₁ + 2·F₂·cos θ = P.
3. Compatibility: D moves down by Δ. The vertical bar stretches Δ; each inclined bar stretches Δ·cos θ.
4. Force–deformation: F₁·L/(AE) = Δ and F₂·(L/cos θ)/(AE) = Δ·cos θ, so F₂ = F₁·cos² θ.
5. Substitute: F₁(1 + 2·cos³ θ) = P, so F₁ = P / (1 + 2·cos³ θ).
6. cos 45° = 0.7071, cos³ 45° = 0.35355; 1 + 2 × 0.35355 = 1.7071.
7. F₁ = 100 / 1.7071 = 58.58 kN; F₂ = 58.58 × 0.5 = 29.29 kN.
8. Check: 58.58 + 2 × 29.29 × 0.7071 = 100.0 kN.

Answer: middle bar 58.6 kN, each inclined bar 29.3 kN, all tension. The stiffer (shorter) middle bar takes the larger share, which a determinate analysis cannot show.

Common mistakes

• Declaring a truss determinate from D = 0 without checking that it is stable.
• Counting a pin support as one reaction (it is two) or a roller as two (it is one).
• Mixing sign conventions between joints. Assume tension everywhere and let the sign tell you.
• In the method of sections, cutting four or more unknown members, or cutting through a joint.
• Assuming zero-force members can be removed from the real structure.

For GATE AE

Expect questions on counting determinacy and spotting unstable configurations, finding the force in one member by sections, identifying zero-force members by inspection, and small indeterminate systems (three-bar or two-bar sets) solved by compatibility. Practise reading the geometry quickly, using moments about clever points, and checking every answer with a spare equilibrium equation.

Quick check

1. A plane truss has m = 9, j = 6, r = 3. What is D?
2. How many reaction components does a pin support give in a plane truss?
3. At an unloaded joint, members AB and AC are collinear and AD is not. What is the force in AD?
4. Does a temperature rise cause member forces in a determinate truss?

Answers: 1. 0 (determinate if stable). 2. Two. 3. Zero. 4. No.