76/3848–12FE Reference Handbook 10.6 · Statics · p. 98 Plane truss: method of joints (ΣFH = 0, ΣFV = 0)
1. Moment & Equilibrium2. Centroid of a Composite Area3. Coulomb Friction4. Force Resultant5. Rectangle Centroidal Moment of Inertia6. Parallel-Axis Theorem7. Distributed Load Resultant8. Method of Joints9. Two-Force Member10. Ideal Pulley11. Moment of a Force12. Simply Supported Beam Reactions13. Load–Shear–Moment Relations14. Method of Sections (Truss)15. Sagging Cable (Small Sag)16. Overhanging-Beam Reactions17. Radius of Gyration18. Triangle Area Moment of Inertia19. Triangular Distributed Load Resultant20. Three-Force Member21. Couple Moment22. Frames: Multi-Force Members23. Zero-Force Truss Members

Handbook formula

Method of Joints

Ideal truss members are two-force members. Cut a joint, assume tension (away from the joint), and write ΣFx = ΣFy = 0. Start at a joint with at most two unknowns. Positive result = tension.

(N)
Member axial force (tension +)
(N)
Components at the joint

Step-by-step solved example

Simple triangle: base 4 m, height 3 m. Top joint has 12 kN downward. Find the two equal rafter forces.

12 kN7.21 kN C7.21 kN C2 C (3/√13) = 12Each rafter C = 7.21 kNReplay
Ideal truss joint: two-force members, ΣFx = ΣFy = 0.

  1. 1. Geometry

    Half-base = 2 m and height = 3 m, so each rafter is √13 = 3.61 m. The rafters push on the loaded joint, so they are in compression.

  2. 2. ΣFy

    2 C (3/√13) = 12 → C = 2√13 = 7.21 kN compression.

Answer: Each rafter 7.21 kN compression

10 practice questions

0/10 correct

1.A two-force member carries

2.At an unloaded joint (no external load and no support reaction) with two collinear members plus one extra, the zero-force member is

3.Assume tension; a negative answer means

4.Method of joints cannot start if a joint has

5.Method of sections is better when you need

6.A simple plane truss with m members and j joints is determinate if

7.At a pinned support the joint includes

8.Counters (crossing diagonals) indicate

9.ΣFx = 0 at a joint does not involve

10.A vertical 10 kN load at a joint with a horizontal member and a 45° member: the 45° force is

45°10 kN
Joint with one horizontal member, one member at 45°, and a 10 kN vertical downward load.