77/3848–12FE Reference Handbook 10.6 · Statics · p. 98 Two-force body (equal, opposite, collinear)
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

Two-Force Member

A member with forces at only two points, and no moments applied, must have equal-and-opposite collinear forces along the line joining the points. Links, truss bars, and hydraulic cylinders are typical.

(N)
End forces

Step-by-step solved example

A 2 m link from (0,0) to (1.2, 1.6) m carries 5 kN compression. Give the support components at (0,0).

AB5 kN C3 kN4 kNu = 0.6 i + 0.8 jSupport on member: 3 i + 4 j kNReplay
Forces at the two pins are collinear with the member axis.

  1. 1. Direction

    Unit vector along the link = (0.6, 0.8).

  2. 2. Compression

    The support pushes the member from A toward B. That 5 kN force is 5(0.6 i + 0.8 j) = 3 i + 4 j kN on the member.

Answer: 3 kN (+x) and 4 kN (+y) on the member

10 practice questions

0/10 correct

1.Three-force members can carry

2.If a link has a force in the middle it is

3.Hydraulic cylinder (pins at both ends) is modeled as

4.A cable can take

5.Collinear equal-opposite forces produce

6.To prove a member is two-force, check

7.Direction of force in a two-force member is

8.A rigid body with forces at two points not collinear with the points cannot be in

9.Slider of a piston (ideal) transmits

10.Pin-ended column in statics is treated as