Plane trusses: method of joints and method of sections

Why it matters

Trusses carry load efficiently because every member works only in tension or compression. You see them in bridges, roof structures, crane booms, transmission towers, vehicle chassis sub-frames and space-frame race-car chassis. Finding member forces quickly, and spotting the members that carry nothing, is a standard GATE and interview skill and the basis for sizing members against yielding and buckling.

Key ideas

Idealised plane truss. A plane truss is a set of straight members in one plane, joined at their ends. The analysis rests on four assumptions:

• members are connected by frictionless pins;
• loads and reactions act only at the joints;
• member self-weight is neglected or lumped at the joints;
• deformations are small, so the original geometry is used.

With these, every member is a two-force member and carries only an axial force, tension (pulling on the joints) or compression (pushing on them). Real trusses have bolted or welded gusset joints, so members also carry small "secondary" bending moments; the pin-jointed model gives the primary forces used in design.

Stability and determinacy. A simple truss is built from a basic triangle by adding two members and one joint at a time. For a plane truss with m members, j joints and r independent support reactions:

• m + r = 2j: statically determinate (if also geometrically stable);
• m + r > 2j: statically indeterminate (redundant members);
• m + r < 2j: a mechanism (unstable). The count is necessary but not sufficient: a truss can satisfy it and still be unstable if members are badly arranged or all reactions are parallel or concurrent.

Sign convention. Assume every unknown member force is tensile, drawn as an arrow pulling away from the joint (or away from the cut). A positive answer is tension, a negative answer is compression. Keep this convention throughout; do not switch arrow directions halfway.

Method of joints. Each joint is a concurrent force system, so it gives two equations: ΣF_x = 0 and ΣF_y = 0. Start at a joint with no more than two unknown member forces (often a support, after finding reactions), solve, and move on joint by joint. Good for finding all member forces. Errors carry forward, so check with the last joint, which should balance automatically.

Method of sections. Cut the truss through the members you want (generally no more than three with unknown forces, not all meeting at one point and not all parallel). Either part is a rigid body in equilibrium, so three equations apply. Take moments about the point where two of the cut members intersect to get the third directly. For a parallel-chord truss:

• top or bottom chord force = bending moment at the opposite joint ÷ truss depth;
• diagonal force from ΣF_y = shear in that panel ÷ sin θ. This is the fastest way to find a few member forces in a large truss.

Zero-force members. Spot them before calculating:

• At an unloaded joint with only two non-collinear members, both are zero.
• At an unloaded joint with three members, two of them collinear, the third is zero. They are not useless: they brace compression members against buckling (reducing effective length), keep the shape stable during construction and carry load under other load cases.

Formulas

m + r = 2j

• m: number of members; r: number of independent reactions; j: number of joints. Determinacy check for a plane truss.

ΣF_x = 0, ΣF_y = 0 at every joint

• Two equations per joint (method of joints).

ΣF_x = 0, ΣF_y = 0, ΣM = 0 for a cut portion

• Three equations per section (method of sections).

F_chord = M / h

• M: bending moment of the external forces about the moment centre (N·m); h: perpendicular distance from the moment centre to the chord (m). Parallel-chord trusses.

F_diag = V / sin θ

• V: panel shear force, i.e. the sum of external vertical forces on one side of the cut (N); θ: angle of the diagonal with the horizontal. Parallel-chord trusses with vertical loads.

Member stress: σ = F / A

• A: member cross-section area (m²). Check compression members for buckling as well (see the column topic).

Worked examples

Example 1 (standard): method of joints on a triangular truss. Given: truss ABC with A pinned and B on rollers, AB = 6 m horizontal, apex C 4 m above the mid-point of AB, a 30 kN vertical load at C. Find all member forces.

1. Reactions by symmetry: R_A = R_B = 15 kN upward; horizontal reaction at A is zero.
2. Geometry: AC = √(3² + 4²) = 5 m, so sin θ = 4/5 and cos θ = 3/5 for AC with the horizontal.
3. Joint A, all members assumed in tension: ΣF_y = 0: 15 + F_AC·(4/5) = 0, so F_AC = −18.75 kN.
4. ΣF_x = 0: F_AB + F_AC·(3/5) = 0, so F_AB = −(−18.75)(0.6) = +11.25 kN.
5. By symmetry F_BC = F_AC.
6. Check at C: vertical components 2 × 18.75 × 0.8 = 30 kN, balancing the load. Answer: AC and BC: 18.75 kN compression; AB: 11.25 kN tension.

Example 2 (GATE level): method of sections on a Pratt truss. Given: bottom-chord joints L0 to L4 at 3 m spacing (span 12 m), top joints U1, U2, U3 directly above L1, L2, L3 at height 4 m. End posts L0–U1 and U3–L4; verticals U1L1, U2L2, U3L3; diagonals U1L2 and U3L2. Supports: pin at L0, roller at L4. Loads 20 kN at each of L1, L2, L3. Find the forces in U1U2, U1L2 and L1L2.

1. Determinacy: j = 8, m = 13, r = 3; m + r = 16 = 2j.
2. Reactions by symmetry: R_L0 = R_L4 = 30 kN.
3. Cut through U1U2, U1L2 and L1L2 and keep the left part (L0, L1, U1). Assume tension in all three.
4. U1U2, moments about L2 (where the other two cut members meet): F·h = M_L2 with M_L2 = 30 × 6 − 20 × 3 = 120 kN·m. The top chord is pushed, so F_U1U2 = 120 / 4 = 30 kN compression.
5. L1L2, moments about U1: M_U1 = 30 × 3 = 90 kN·m (the 20 kN at L1 passes through U1's vertical), so F_L1L2 = 90 / 4 = 22.5 kN tension.
6. U1L2, ΣF_y on the left part: 30 − 20 − F·sin θ = 0, with sin θ = 4/5 for the 3 m by 4 m diagonal, so F_U1L2 = 10 / 0.8 = 12.5 kN tension.
7. Side note: joint U2 is unloaded with U1U2 and U2U3 collinear, so U2L2 is a zero-force member. Answer: U1U2 = 30 kN (C), L1L2 = 22.5 kN (T), U1L2 = 12.5 kN (T).

Common mistakes

• Starting the method of joints at a joint with three unknowns. Find reactions first and start where there are two.
• Flipping the assumed arrow at the second joint. A member force acts in opposite directions on its two end joints; keep the "tension = away from the joint" rule everywhere.
• Using the sloping length instead of the perpendicular distance in a moment equation.
• Cutting through four unknown members, or three that meet at one point, and expecting a solution.
• Treating zero-force members as removable for all load cases.
• Forgetting that m + r = 2j is necessary, not sufficient, for stability.

For GATE ME

Questions ask for the force in one named member (method of sections is usually fastest), for zero-force members in a drawn truss, and for determinacy and stability checks. Practise spotting zero-force members first, reading geometry quickly (3-4-5 and 45° triangles) and using the chord-force = M/h shortcut.

Quick check

1. A truss has 9 joints and 3 support reactions. How many members make it just determinate?
2. At an unloaded joint, members P and Q are collinear and member R is not. What is the force in R?
3. Why do truss members carry only axial force in the ideal model?
4. In Example 2, what is the force in vertical U1L1?

Answers: 1. m = 2j − r = 15. 2. Zero. 3. Each member is a two-force member (pin ends, loads only at joints), so its force acts along its axis. 4. 20 kN tension (joint L1: the 20 kN load is carried by U1L1, since L0L1 and L1L2 are horizontal).