Force Systems

Force systems classify how forces act together: collinear, concurrent, parallel and general. Learn resultants and their location, distributed loads, equivalent force–couple systems, and truss analysis by joints and sections.

Structures rarely carry a single force. Wheel loads on a bridge, wind pressure on a wall, soil pressure on a retaining wall and member forces meeting at a truss joint are all systems of forces that must be combined or balanced.

Replacing a complicated system by its resultant — or solving a network of forces in a truss — is a core skill for structural engineers.

An analogy

A group of people pushing a car from different points: the car responds as if one big push acted at one particular point — the resultant.

Types of force systems

Collinear (same line — add algebraically), concurrent (meet at a point — resultant through that point, e.g. truss joints), parallel (e.g. gravity loads on a beam — resultant plus location), and general non-concurrent (resultant force plus couple).

Distributed loads

Uniform loads (slab, self-weight) have resultant w·L at mid-length. Triangular loads (water or soil pressure) have resultant ½w_maxL at the centroid, one third from the high end. Trapezoidal loads split into a rectangle and a triangle.

Trusses

Trusses are triangulated frameworks whose members carry only axial force (ideal pin joints, loads at joints). Method of joints: ΣF_x = ΣF_y = 0 at each joint, starting where only two unknowns exist. Method of sections: cut through ≤ 3 members and take moments about the intersection of two of them to find the third directly.

Equivalent systems and zero-force members

Any force system reduces to a single force and couple at a chosen point (wrench in 3D). Zero-force members are identified by inspection (e.g. two non-collinear members at an unloaded joint); they still brace against buckling. Space trusses and cable nets use the same principles in 3D with matrix methods.

Formulas

• Parallel forces: R = ΣF_i, x̄ = ΣF_i x_i / ΣF_i
• Distributed load: R = ∫w dx (area), acting at the centroid of the load diagram
• Triangular load: R = ½ w_max L, at 2L/3 from the zero end
• Truss determinacy: m + r = 2j (members + reactions = 2 × joints)

Worked example: Resultant of parallel wheel loads

Given: 10 kN at 1 m, 20 kN at 3 m, 15 kN at 5 m

Find: Resultant and its location.

• R = 45 kN
• x̄ = (10 × 1 + 20 × 3 + 15 × 5)/45 = 3.22 m

Answer: 45 kN at 3.22 m

Worked example: Truss chord force by sections

Given: 3-panel truss, 4 m panels, depth 3 m; 30 kN at each interior bottom joint (x = 4, 8 m)

Find: Top-chord force at mid-span.

• Reactions 30 kN each
• Cut the middle panel; moments about the bottom joint at x = 4 m: M = 30 × 4 = 120 kN·m
• F_top = M/h = 120/3 = 40 kN (compression); the bottom chord, by moments about the top joint at x = 6 m (30 × 6 − 30 × 2 = 120), carries 40 kN tension

Answer: 40 kN compression

Common mistakes

• Wrong: Placing a triangular load's resultant at mid-length. — Right: It acts at the centroid — L/3 from the high end.
• Wrong: Assuming member forces' sense wrongly and ignoring the sign. — Right: Assume tension; a negative answer means compression.

Applications

• Roof & bridge trusses: Pratt, Howe, Warren and K-trusses analysed by joints and sections.
• Moving loads: Axle groups on bridges reduced to resultants to find worst positions.
• Retaining structures: Triangular earth/water pressure replaced by resultants for stability checks.

Key takeaways

• Classify: collinear, concurrent, parallel, general.
• Resultant = area of load diagram, at its centroid.
• Trusses: joints for all members, sections for a few.
• Assume tension; negative = compression.

Going further

• Matrix truss analysis: Direct stiffness method assembles member stiffnesses to solve large or indeterminate trusses.
• Influence lines for trusses: Show how a member force varies as a unit load crosses the bridge deck.
• Cable structures: Cables under UDL take a parabolic shape; thrust H = wL²/(8 sag).