Trusses, Frames, and Machines

Truss Assumptions and Determinacy

A truss is a structure of straight members joined at their ends. The idealizing assumptions every FE truss problem relies on are:

1. Members connect through frictionless pins (no moment transfer at joints).
2. Loads are applied only at joints, never along a member.
3. Members are therefore two-force members — each carries force only along its own axis.
4. Each member is purely in tension (T) or compression (C).
5. Member self-weight is neglected, or lumped half to each end joint.

Before solving, check static determinacy using member count m, joint count j, and reaction count r:

The 2j term comes from writing two equilibrium equations (ΣFx, ΣFy) at each of the j joints. If the count says indeterminate, statics alone is not enough; if it says unstable, no equilibrium solution exists.

Method of Joints

The Method of Joints isolates one pin at a time and applies ΣFx = 0 and ΣFy = 0 (only two equations, because all forces pass through the joint so moments are automatically satisfied).

Procedure:

1. Find the support reactions from overall equilibrium.
2. Start at a joint with two or fewer unknown members.
3. Assume every unknown member is in tension (arrow pulling away from the joint).
4. Solve ΣFx = 0 and ΣFy = 0. A positive result confirms tension; a negative result means compression.
5. Carry known forces to adjacent joints and repeat.

Worked Example — Method of Joints

Consider a joint where member AB is horizontal and member AC rises at 30° above horizontal toward the joint, with a single downward 5 kN external load applied at the joint. Assume both unknowns in tension.

ΣFy = 0: F_AC sin 30° − 5 = 0 → F_AC = 5 / 0.5 = 10 kN (tension).

ΣFx = 0: F_AB + F_AC cos 30° = 0 → F_AB = −10(0.866) = −8.66 kN → 8.66 kN compression.

The negative sign on F_AB is the exam's built-in flag that the member is compressed — never discard the sign, report tension/compression based on it.

Method of Sections

The Method of Sections cuts an imaginary line through the truss and treats one side as a free body, using all three equilibrium equations (ΣFx, ΣFy, ΣM). Because you have three equations, cut through at most three members whose forces are unknown.

Steps: (1) find reactions, (2) slice through the target members, (3) draw the FBD of one portion, (4) take moments about a point where two unknowns intersect to isolate the third directly. This is far faster than working joint-by-joint across the whole truss when only one or two member forces are wanted.

Zero-Force Members

A zero-force member carries no load under the current loading. Identifying them first removes unknowns before any calculation:

• Rule 1: If exactly two non-collinear members meet at an unloaded joint, both are zero-force members.
• Rule 2: If three members meet at an unloaded joint and two are collinear, the third (non-collinear) member is zero-force.

Zero-force members are not pointless — they brace the truss against buckling and may carry load under different loading patterns. The exam often draws several so you can prune them quickly.

Frames and Machines

Unlike trusses, frames (stationary load-supporting structures) and machines (which transmit or modify forces, often with moving parts) contain multi-force members — members loaded at more than two points or by applied moments. These members carry axial force, shear force, AND bending moment, so they are not two-force members.

Analysis method:

1. Find external reactions on the whole assembly.
2. Disassemble the structure at its connecting pins.
3. Draw a separate FBD for each member.
4. Apply the three equilibrium equations to each member.
5. At each shared pin, enforce Newton's third law: the force one member exerts on another is equal and opposite.

Trap: Do not assume a diagonal member in a frame is a two-force member unless it is genuinely pin-connected at exactly two points with no intermediate load. If a load or pin attaches mid-span, the member carries shear and moment and the two-force shortcut fails.

Worked Example — Method of Sections

A Pratt truss spans 12 m with three 4 m panels and supports a 20 kN downward load at the first interior bottom joint. The left support reaction is found from overall equilibrium to be 13.3 kN upward. To find the force in the top chord member crossing the second panel, cut vertically through the top chord, a diagonal, and the bottom chord.

Taking moments about the bottom joint where the diagonal and bottom chord intersect eliminates two unknowns, leaving the top-chord force alone: the top chord carries the moment of the 13.3 kN reaction divided by the truss height. With a 4 m height, F_top ≈ (13.3 × 4)/4 = 13.3 kN in compression (top chords of simply supported trusses typically compress). This single moment equation delivers the answer that the Method of Joints would need three joints to reach.

Common Truss Mistakes

Working through these checks before computing keeps an FE truss question to its 3-minute budget and avoids the most penalized errors.