Plane trusses: method of joints and method of sections

Why it matters

Trusses carry large loads with little material: crane booms, gantry and pick-and-place frames, antenna and sensor masts, conveyor support bridges and the lattice of many robot cells. Because each ideal truss member carries only axial tension or compression, you can find every member force from statics alone and then size members for tension, or check them for buckling in compression. The two hand methods below are also what you use to sanity-check a finite-element model.

Key ideas

Ideal truss assumptions.

• Members are straight and joined at their ends by frictionless pins.
• Loads and reactions act only at the joints.
• Member weights are neglected or split equally between the two end joints. Under these assumptions every member is a two-force member: the force in it acts along its axis and is either tension (pulling on the joints) or compression (pushing on the joints). Real welded or bolted joints add small secondary bending, but the pin model gives the primary forces well.

Stability and determinacy. A plane truss built from triangles is rigid. 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), needs deformation compatibility;
• m + r < 2j: a mechanism, collapses. The count is necessary but not sufficient: a truss with the right count can still be unstable if members are arranged badly (for example a square panel with no diagonal elsewhere compensating).

Method of joints. Isolate one joint at a time. The forces are concurrent, so only two equations are available: ΣF_x = 0 and ΣF_y = 0. Start at a joint with at most two unknown member forces (usually a support, after finding the reactions) and work across the truss. Assume every unknown member force is tensile (arrow pulling away from the joint); a negative result means compression. The method finds all member forces but errors carry forward from joint to joint.

Method of sections. Cut the truss through not more than three members whose forces are unknown (and which are not all concurrent or all parallel), and treat either part as a rigid body with three equations. The trick is the moment centre: take moments about the point where two of the cut members meet, so that the third member's force comes out directly from one equation. A vertical-force equation gives the diagonal in a truss with parallel chords. Use it when only a few member forces are wanted.

Zero-force members. Spot these before calculating:

• At an unloaded, unsupported joint with only two non-collinear members, both members carry zero force.
• At an unloaded, unsupported joint with three members, two of them collinear, the third member carries zero force. Zero-force members are not useless: they brace compression members against buckling, carry load under other load cases, and keep the truss stable.

Behaviour to remember. In a simply supported truss with gravity loads, the top chord is in compression and the bottom chord in tension, like the compression and tension sides of a beam. Chord forces are largest near mid-span (where the bending moment is largest), and diagonal forces are largest near the supports (where the shear is largest).

Formulas

m + r = 2j

• m: number of members; r: number of independent reaction components; j: number of joints. Condition for a statically determinate plane truss.

ΣF_x = 0, ΣF_y = 0 at each joint

• Method of joints; two equations per joint, so at most two unknowns per joint.

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

• Method of sections; choose O at the intersection of two cut members.

F_chord = M / h, F_diag·sin θ = V

• For a truss with parallel chords: M is the bending moment of the equivalent beam at the moment centre (N·m), h the truss depth (m), V the shear force in the cut panel (N), θ the diagonal's angle with the horizontal. Quick checks for chord and diagonal forces.

Worked examples

Example 1 (standard): triangular frame by the method of joints. Given: a symmetric triangular truss ABC with span AB = 6 m and apex C 3 m above mid-span, pin at A, roller at B, 10 kN downward at C. Find all member forces.

1. Determinacy: m = 3, r = 3, j = 3, so m + r = 6 = 2j. Determinate.
2. Reactions by symmetry: R_A = R_B = 5 kN upward; no horizontal reaction because there is no horizontal load.
3. Geometry: AC and BC rise 3 m over 3 m, so θ = 45°.
4. Joint A: ΣF_y = 0: 5 + F_AC·sin 45° = 0, so F_AC = −7.07 kN (compression).
5. Joint A: ΣF_x = 0: F_AB + F_AC·cos 45° = 0, so F_AB = +5.00 kN (tension).
6. By symmetry F_BC = −7.07 kN. Check at C: 2 × 7.07 × sin 45° = 10.0 kN, matching the load. Answer: AC and BC 7.07 kN compression, tie AB 5.00 kN tension.

Example 2 (GATE level): Pratt truss by the method of sections. Given: a Pratt truss with four 3 m panels (span 12 m) and depth 4 m. Bottom joints L0 to L4, top joints U1 to U3 above L1 to L3. Pin at L0, roller at L4. Diagonals U1L2 and U3L2 slope down towards mid-span. Loads 20 kN at L1, L2 and L3. Find the forces in U1U2, L1L2 and U1L2.

1. Count: m = 13, r = 3, j = 8: m + r = 16 = 2j. Determinate.
2. Reactions by symmetry: R_L0 = R_L4 = 30 kN.
3. Cut through U1U2, U1L2 and L1L2 and keep the left part (joints L0, L1, U1). Assume all three cut forces are tensile.
4. Moments about L2 (where U1L2 and L1L2 meet): 30 × 6 − 20 × 3 + F_U1U2 × 4 = 0, so F_U1U2 = −30 kN.
5. Moments about U1 (where U1U2 and U1L2 meet): F_L1L2 × 4 − 30 × 3 = 0, so F_L1L2 = +22.5 kN.
6. Diagonal: length √(3² + 4²) = 5 m, vertical component 4/5 of its force. ΣF_y = 0: 30 − 20 − 0.8·F_U1L2 = 0, so F_U1L2 = +12.5 kN.
7. Check ΣF_x: −30 + 22.5 + 0.6 × 12.5 = 0. Correct.
8. Quick check with beam analogy: moment at L2 is 30 × 6 − 20 × 3 = 120 kN·m, and 120/4 = 30 kN in the chord. Shear in panel 2 is 30 − 20 = 10 kN, and 10/0.8 = 12.5 kN in the diagonal. Answer: U1U2 = 30 kN compression, L1L2 = 22.5 kN tension, U1L2 = 12.5 kN tension. Bonus: at joint U2, members U1U2 and U2U3 are collinear and there is no load, so the vertical U2L2 is a zero-force member. At L1, the 20 kN load must be carried by U1L1 alone, so U1L1 = 20 kN tension.

Common mistakes

• Cutting four or more members with unknown forces and then trying to solve with three equations.
• Taking moments about a point that does not remove two unknowns, creating needless simultaneous equations.
• Mixing up signs: if you assume tension everywhere, a negative answer means compression; do not also flip the arrow.
• Forgetting the support reactions before starting joints or sections.
• Declaring a member zero-force when the joint carries an external load or a support reaction.
• Using m + r = 2j as proof of stability; check the geometry too.

For GATE ME

Expect member forces in small trusses (triangles, simple Warren or Pratt panels, cantilever trusses from a wall), zero-force member identification, determinacy counting, and quick method-of-sections questions asking for one chord or diagonal force. Practise spotting zero-force members first, then picking the moment centre that gives the answer in a single equation.

Quick check

1. A plane truss has 9 members, 6 joints and is supported by a pin and a roller. Is it determinate?
2. Two non-collinear members meet at an unloaded, unsupported joint. What are their forces?
3. In a simply supported truss under gravity loads, which chord is in compression?
4. In the method of sections, at most how many unknown member forces should be cut?
5. A parallel-chord truss 2 m deep has a beam bending moment of 80 kN·m at a panel point. What is the chord force there?

Answers: 1. Yes: m + r = 9 + 3 = 12 = 2 × 6. 2. Both are zero. 3. The top chord. 4. Three. 5. 80/2 = 40 kN.