Plane trusses: method of joints and method of sections

Why it matters

Roof trusses, bridge girders, transmission towers, crane booms and machine frames are built from slender members joined at their ends. Treating them as trusses lets you find the axial force in every member with nothing more than equilibrium, and those forces then size the members against yielding (tension) and buckling (compression).

Key ideas

Ideal plane truss assumptions.

1. All members and loads lie in one plane.
2. Members are straight and joined at their ends by frictionless pins.
3. Loads and reactions act only at the joints.
4. Self-weight is neglected, or shared half to each end joint.

With these, every member is a two-force member: it carries only an axial force, tension (pulling on the joints) or compression (pushing on them), and no bending. Real trusses have bolted or welded joints, so they carry small secondary bending moments, but the ideal analysis gives the primary forces well.

Simple truss. Built by starting from a triangle and adding two members and one new joint at a time. Triangles are rigid; a rectangle of four pinned members is not.

Determinacy and stability. With m members, j joints and r independent reaction components:

• m + r = 2j is necessary for a statically determinate, stable truss;
• m + r > 2j: statically indeterminate (redundant members), needs compatibility of deformations;
• m + r < 2j: a mechanism (unstable). The count is necessary but not sufficient: members or supports badly arranged (for example all reactions parallel or concurrent) can still make the truss unstable.

Method of joints. Each joint is a concurrent force system, so it gives two equations (ΣFx = 0, ΣFy = 0). Start at a joint with at most two unknown member forces (usually a support, after finding reactions) and move from joint to joint. Assume every unknown member force is tension (arrow pointing away from the joint); a negative answer means compression. Best when you need all member forces.

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 take either part as a free body. It gives three equations. Take moments about the point where two of the cut members meet to get the third directly. Best when you need a few members in the middle of a large truss.

Zero-force members. Spot them before calculating:

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

Link to beams. In a parallel-chord truss under vertical loads, the chords carry the bending moment (chord force ≈ M/h, h = truss depth) and the web members carry the shear. This is the same idea as flanges and web in an I-beam.

Formulas

m + r = 2j (necessary condition for a determinate plane truss)

• m = number of members, r = number of independent reaction components, j = number of joints.

ΣFx = 0, ΣFy = 0 at each joint (method of joints)

ΣFx = 0, ΣFy = 0, ΣM = 0 on a cut portion (method of sections)

F_chord = M / h (parallel-chord truss, section through a panel)

• M = bending moment of the external forces at the moment centre (N·m), h = depth between chords (m), F in N.

F_diagonal · sin θ = V (parallel chords; vertical component of the diagonal balances the panel shear)

• V = shear force in the panel (N), θ = angle of the diagonal to the horizontal.

Worked examples

Example 1 (method of joints). A triangular truss has joints A (0, 0) with a pin, B (6 m, 0) with a roller and apex C (3 m, 4 m). A vertical load of 30 kN acts downward at C. Find all member forces.

1. Determinacy: m = 3, r = 3, j = 3, so m + r = 6 = 2j.
2. Reactions: by symmetry R_A = R_B = 15 kN upward; horizontal reaction at A is zero.
3. Geometry of AC: length √(3² + 4²) = 5 m, so sin θ = 0.8, cos θ = 0.6.
4. Joint A, ΣFy = 0 (tension assumed): F_AC × 0.8 + 15 = 0, so F_AC = −18.75 kN.
5. Joint A, ΣFx = 0: F_AB + F_AC × 0.6 = 0, so F_AB = 18.75 × 0.6 = 11.25 kN.
6. By symmetry F_BC = F_AC.

Answer: F_AC = F_BC = 18.75 kN (compression), F_AB = 11.25 kN (tension).

Example 2 (GATE level, method of sections). A Pratt truss has bottom-chord joints L0 to L4 at 4 m spacing (span 16 m) and top-chord joints U1, U2, U3 vertically above L1, L2, L3 at a height of 3 m. The end posts are L0–U1 and U3–L4, the diagonals U1–L2 and U3–L2, and there are verticals at L1, L2 and L3. L0 is pinned, L4 is on a roller, and 20 kN acts downward at each of L1, L2 and L3. Find the forces in U1U2, L1L2 and U1L2.

1. Reactions: total load 60 kN, symmetric, so R_L0 = R_L4 = 30 kN.
2. Cut through U1U2, U1L2 and L1L2; take the left part (joints L0, L1, U1).
3. U1U2 – moments about L2 (8, 0), where the other two cut members meet: moment of external forces M = 30 × 8 − 20 × 4 = 160 kN·m. F_U1U2 = M/h = 160/3 = 53.33 kN, compression (the top chord is pushed, as in a sagging beam).
4. L1L2 – moments about U1 (4, 3): M = 30 × 4 = 120 kN·m. F_L1L2 = 120/3 = 40 kN, tension.
5. U1L2 – vertical forces. Diagonal length √(4² + 3²) = 5 m, sin θ = 0.6. Panel shear V = 30 − 20 = 10 kN. F_U1L2 × 0.6 = 10, so F_U1L2 = 16.67 kN, tension.
6. Check ΣFx on the part: −53.33 + 40 + 16.67 × 0.8 = −53.33 + 40 + 13.33 = 0. ✓
7. Bonus: joint U2 has three members, U1U2 and U2U3 collinear plus the vertical U2L2, and no load, so F_U2L2 = 0 (a zero-force member under this loading).

Answer: U1U2 = 53.3 kN (C), L1L2 = 40 kN (T), U1L2 = 16.7 kN (T).

Common mistakes

• Cutting four or more unknown members with one section.
• Forgetting to find the reactions before starting at a support joint.
• Mixing up sign conventions: decide once that tension is positive with arrows away from the joint, and keep it.
• Loading members between joints and still treating them as two-force members.
• Using m + r = 2j alone to declare a truss stable.
• Ignoring zero-force members; spotting them first saves time and avoids errors.

For GATE PI

Expect short numericals asking for the force in one member of a simple truss (often by sections), identification of zero-force members, and determinacy counts. Practise choosing the moment centre at the intersection of two cut members, and confirming tension or compression by physical reasoning (top chord compressed and bottom chord stretched under downward loads on a simply supported truss).

Quick check

1. A plane truss has 7 joints and 3 reaction components. How many members does a determinate simple truss need?
2. Two non-collinear members meet at an unloaded, unsupported joint. What are their forces?
3. In the method of sections, how many unknown member forces can usually be found from one cut?
4. Under downward loads on a simply supported truss, is the bottom chord in tension or compression?

Answers: 1. 11 members (m = 2j − r = 14 − 3). 2. Both zero. 3. Three. 4. Tension.