Frames and machines

A truss keeps things simple: every member is a two-force member, pinned at its ends with nothing else pushing on it in between, so each one is either in pure tension or pure compression. A frame or machine breaks that simplicity on purpose. At least one member — an L-shaped bracket, a lever arm, a plier jaw — carries a load somewhere other than at its end pins, or is pinned at three or more points. That single member can no longer be treated as a straight force line; it needs a full free-body diagram with independent horizontal and vertical pin forces, just like any rigid body in equilibrium. This article extends plane-truss analysis to structures containing such multi-force members. You will learn to tell a two-force member from a multi-force member on sight, disassemble a frame or machine into individual member FBDs, and solve for pin reactions and internal forces using whole-structure and then member-by-member equilibrium — the same three equations you already know, applied more carefully and more often.

The core insight

A frame or machine contains at least one multi-force member (loaded or pinned at three-plus points, so its pin forces have independent $x$ and $y$ components rather than lying along the member), and you solve it by applying equilibrium first to the whole structure for external reactions, then separately to each member's own free-body diagram, keeping Newton's-third-law pin forces consistent between adjoining members.

The core idea

Frame versus machine. A frame is a rigid structure built to support a load — a bracket, a crane arm — and stays in one fixed configuration while loaded. A machine contains moving parts designed to transmit or modify a force, such as pliers, a toggle clamp, or a hoist; the structure as a whole is not rigid, but at any single instant you freeze its configuration and analyze it exactly like a frame. Both are built from pin-connected members, and both mix two types of member.

Two-force members, recapped. A two-force member has load applied at exactly two points and nothing in between. Equilibrium forces the resultant at each pin to lie along the straight line joining the two points — pure axial tension or compression, one unknown per member, the entire logic of truss analysis.

Multi-force members, the new case. A multi-force member carries force at three or more points, or has a distributed load or applied couple along its length. Nothing forces its pin reactions onto a single line, so each pin contributes two independent unknowns, $x$ - and $y$ -components. You cannot shortcut this with an axial-force guess; you must draw the member's own FBD and apply full equilibrium to it.

The solution procedure. First, treat the entire frame or machine as one rigid body: apply $\sum F_x = 0$ , $\sum F_y = 0$ , $\sum M_O = 0$ to the whole-structure FBD to find whatever external reactions you can — internal pin forces cancel here because every internal pin appears twice, once on each adjoining member, as an action–reaction pair. Second, disassemble the structure into its individual members; a two-force member's unknown is a single axial force. Third, for every multi-force member draw a complete FBD showing pin-force components at each connection plus any directly applied load. Fourth, apply the three equilibrium equations to each multi-force member's FBD, keeping the force one member exerts on a shared pin equal and opposite to the force the neighbour exerts back. Fifth, solve, often one member at a time, choosing moment centres that eliminate as many unknowns as possible.

\sum F_x = 0, \qquad \sum F_y = 0, \qquad \sum M_O = 0 \qquad \text{(applied to the whole structure, then to each member)}

where: the sums run over every external force or applied couple acting on the free body being analyzed (whole structure first, then one member's own FBD); $O$ is any convenient moment centre, chosen to make as many unknown pin forces as possible pass through it and so drop out of the moment equation.

Figure

A frame with an L-shaped bracket ABC pinned to a wall at A and a straight two-force link BD pinned at B and D, with a downward load at C

Derivation ASolving a frame by whole-structure and member equilibrium

The goal is to find every pin reaction and internal member force in a frame consisting of a multi-force bracket and a two-force link, using nothing beyond the three planar equilibrium equations applied twice — once to a convenient sub-body and once to the full member FBD — and to show precisely why an unlucky choice of moment centre wastes algebra while a good one solves the problem in a single equation.

Step 1. Classify every member before writing a single equation. Link $BD$ carries force only at its two end pins with nothing along its length, so it is a two-force member: its internal force $F_{BD}$ must act along the straight line from $B$ to $D$ , one unknown. Bracket $ABC$ is pinned at $A$ , pinned to $BD$ at $B$ , and carries the external load at $C$ — three force locations, so it is a multi-force member whose pin at $A$ needs two independent unknowns $A_x, A_y$ .

Step 2. Because $BD$ is a two-force member, its line of action is fixed entirely by the coordinates of $B$ and $D$ — never guess this angle, compute it. The horizontal and vertical run $\Delta x, \Delta y$ between them fix the direction cosines $\cos\alpha = \Delta x/L$ and $\sin\alpha = \Delta y/L$ , where $L = \sqrt{\Delta x^2+\Delta y^2}$ . If $BD$ is assumed in tension, it pulls on bracket $ABC$ at $B$ with components $F_{BD}\cos\alpha$ horizontally and $F_{BD}\sin\alpha$ vertically, directed from $B$ toward $D$ .

Step 3. Isolate bracket $ABC$ as its own free body: it carries the unknown pin reaction $A_x, A_y$ at $A$ , the link's pull at $B$ , and the known applied load at $C$ . Take moments about $A$ rather than summing forces first, because a moment sum about pin $A$ automatically eliminates both unknowns $A_x$ and $A_y$ , leaving one equation in the single unknown $F_{BD}$ .

Step 4. Solve that moment equation for $F_{BD}$ ; only components with a nonzero moment arm about $A$ contribute (a component whose line of action passes through $A$ drops out automatically). Substitute $F_{BD}$ back into $\sum F_x = 0$ and $\sum F_y=0$ for bracket $ABC$ , which now contain only $A_x$ and $A_y$ and so solve directly. A negative result simply means the true direction is opposite to the one assumed — not an error, just a sign to report honestly. Method check: whole-structure equilibrium alone gives only external reactions at the wall; it never reveals the internal force in $BD$ , because that pin force is internal to the assembled structure and cancels out of a whole-structure FBD. Only by isolating one member — here bracket $ABC$ — and applying moment equilibrium about a pin that kills the unwanted unknowns do the internal forces emerge.

Worked example

A frame consists of member $ABC$ , an L-shaped bracket pinned to a wall at $A(0,0)$ , with $B(0,2)$ two metres above $A$ and $C(1.5,2)$ 1.5 m horizontally from $B$ . Member $BD$ is a straight two-force link from $B$ to $D(1.0,3.5)$ . A downward load of 300 N acts at $C$ . Find the pin reaction at $A$ and the force in link $BD$ .

Step 1. Classify the members. Link $BD$ is pinned at both ends with no other load, so it is a two-force member carrying a single axial force $F_{BD}$ along line $BD$ . Bracket $ABC$ is pinned at $A$ , connected to $BD$ at $B$ , and loaded at $C$ , so it is a multi-force member needing a full FBD with components $A_x, A_y$ at $A$ .

Step 2. Find the geometry of $BD$ . From $B(0,2)$ to $D(1.0,3.5)$ : $\Delta x = 1.0$ m, $\Delta y = 1.5$ m, so length $L = \sqrt{1.0^2+1.5^2} = 1.803$ m, giving $\cos\alpha = 1.0/1.803 = 0.5547$ and $\sin\alpha = 1.5/1.803 = 0.8321$ . Assuming tension, $BD$ pulls on $ABC$ at $B$ with components $F_{BD}(0.5547)$ horizontal and $F_{BD}(0.8321)$ vertical, both directed from $B$ toward $D$ .

Step 3. Take moments about $A$ on bracket $ABC$ 's FBD, which eliminates $A_x$ and $A_y$ immediately. Since $B$ sits directly above $A$ , the vertical component of the link force has zero moment arm about $A$ ; only its horizontal component, at a 2 m arm, and the 300 N load at $C$ , at a 1.5 m arm, contribute:

\sum M_A = 0: \quad F_{BD}\cos\alpha\,(2) - 300(1.5) = 0 \;\Rightarrow\; F_{BD} = \frac{450}{2(0.5547)} = 405.6\ \text{N (tension)}

Step 4. Substitute $F_{BD} = 405.6$ N into the force-equilibrium equations for bracket $ABC$ :

\sum F_x = 0: \; A_x = -405.6(0.5547) = -225.0\ \text{N}, \qquad \sum F_y = 0: \; A_y = 300 - 405.6(0.8321) = -37.5\ \text{N}

Both negative signs mean the pin reaction at $A$ points opposite to the directions originally assumed: $A_x = 225.0$ N and $A_y = -37.5$ N (in the corrected sense), and the link $BD$ carries $405.6$ N in tension.

Common traps

Treating a multi-force member as if it were two-force. If a member is loaded or pinned at three or more points, its end pin forces are not generally along its axis — always draw the full FBD with independent $x$ - and $y$ -components; do not shortcut with an assumed axial direction.
Losing Newton's third law between adjoining FBDs. The force member 1 exerts on a shared pin must be exactly equal and opposite to the force member 2 exerts back on that same pin — this bookkeeping slip is the single most common error in frame and machine problems.
Stopping at whole-structure equilibrium. That step gives only external reactions; internal pin forces (like $F_{BD}$ above) only appear once you disassemble the structure and isolate a member.
Guessing a two-force member's line of action from a sketch instead of computing it from pin coordinates, as done explicitly in Step 2 above — an eyeballed angle produces silently wrong moment arms.

Ask the tutor: if disassembling a frame into separate member FBDs still feels confusing, ask the AI tutor to walk through matching pin-force directions step by step for your specific diagram.

How do I quickly tell if a member is two-force or multi-force just from a diagram?

Count the points where force is applied to that one member. If exactly two points carry force and nothing else touches the member along its length, it is two-force, and its internal force lies along the line joining those two points. The moment a third force, a distributed load, or an applied couple shows up anywhere on that member, it becomes multi-force and needs a full FBD with independent pin components — no shortcut. Pin location alone does not decide it; what matters is how many distinct places carry load.

Why did the worked example assume tension in BD, and what if the answer had come out negative?

Assuming tension is just a bookkeeping convention: you draw the link pulling outward on the member it connects to, then let the algebra confirm or correct that guess. A positive result confirms tension as drawn. A negative result, as happened with $A_x$ and $A_y$ above, simply means the true direction is opposite to what was assumed — not a mistake, just a sign to report honestly rather than a reason to re-solve.

Is a machine like pliers analyzed any differently from a rigid frame?

No — the method is identical. At the instant you analyze it, freeze the machine's moving parts in one configuration and treat each part (each plier arm, hinged at the pivot) like a frame member: draw its FBD, apply the three equilibrium equations, and match pin forces between parts by Newton's third law. What differs is only the physical picture — a machine amplifies a force between input and output, so lever-arm ratios show up directly in the moment equation — but the algebra is the same procedure throughout.

Frames and machines

The core idea

Worked example

Common traps

Before this

Where this leads