STATICS • STRUCTURAL ANALYSIS: FRAMES AND MACHINES

FBDs: Frames with Pins — Draw FBDs for frames with internal pins

Master the systematic decomposition of multi-member frames at internal pins to reveal hidden internal forces.

Historical Context & Motivation

The analysis of complex structural assemblies has been central to engineering practice since humans first constructed bridges, cranes, and roof trusses. Unlike simple trusses, whose members carry purely axial loads, frames are multi-force member structures that can resist bending, shear, and axial loads simultaneously. The ability to isolate individual members at their internal connections—most commonly frictionless pins—and draw rigorous free-body diagrams for each component was the breakthrough that enabled engineers to design everything from medieval siege engines to modern aircraft landing gear.

1586
Stevin's Equilibrium Principles
Simon Stevin formalized the conditions for static equilibrium and the parallelogram law for forces, laying the groundwork for systematic force analysis of connected bodies.
1687
Newton's Laws of Motion
Isaac Newton published the Principia, establishing the third law (action–reaction pairs), which is the theoretical foundation for decomposing structures at internal connections.
1826
Navier's Structural Mechanics
Claude-Louis Navier published his treatise on the strength of materials and systematized the free-body diagram approach for analyzing beams and frames in civil structures.
1864
Maxwell & Cremona Graphical Methods
James Clerk Maxwell and Luigi Cremona developed graphical statics techniques, popularizing the visual decomposition of frame structures into individual member FBDs.
1950s+
Modern Matrix & FEM Methods
The stiffness method and finite element analysis automated frame analysis, but the conceptual basis remains the free-body diagram of each member at pin connections.

The central question that drives this lesson is deceptively simple: when a frame is connected internally by frictionless pins, how do we systematically disassemble the structure into individual members, correctly apply Newton's third law at each pin, and write enough independent equilibrium equations to solve for every unknown reaction and internal force? Mastery of this skill is the prerequisite for all subsequent work in structural analysis, including shear and moment diagrams, deflection calculations, and indeterminate frame analysis.

Core Principles & Definitions

Before drawing any free-body diagram for a frame, several foundational concepts must be firmly in place. A frame is a structure composed of at least one multi-force member—that is, a member subjected to three or more forces that are not all collinear. Frames are designed to remain stationary and support loads, distinguishing them from machines, which contain moving parts and transmit or modify forces. An internal pin is a connection point at which two or more members are joined by a smooth (frictionless) pin. Because the pin is smooth, it transmits force but not moment, meaning each member is free to rotate independently about the pin axis in the absence of external constraints.

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Newton's Third Law at Pins

When a pin connects members A and B, the force that A exerts on B through the pin is equal in magnitude and opposite in direction to the force that B exerts on A. Both components (horizontal and vertical) must be shown as equal-and-opposite pairs on the respective FBDs.
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Multi-Force Members

Unlike truss members, frame members carry bending and shear in addition to axial loads. Their FBDs typically show forces at every connection point plus any applied loads and couples, and the forces are generally not along the member axis.
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Pins Transmit Force, Not Moment

A smooth internal pin allows relative rotation between connected members, so the moment reaction at the pin is zero. Each pin introduces two unknowns (horizontal and vertical force components) per pair of members it connects.
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Equation Counting

Each rigid member in 2-D contributes three equilibrium equations: ΣFₓ = 0, ΣF_y = 0, and ΣM = 0. A statically determinate frame has exactly as many independent equations as unknowns when all member FBDs and the whole-frame FBD are considered.
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Whole-Frame FBD First

Always begin by drawing the FBD of the entire frame as a single rigid body. This lets you solve for external support reactions without needing to know internal forces, reducing the number of unknowns when you later disassemble the frame.
KEY TAKEAWAY
Think of an internal pin as a shared hinge between two doors mounted on the same frame. Each door can swing independently, so the hinge can push or pull each door (force) but cannot prevent either door from rotating (no moment). When you analyze each door separately, the hinge pushes on door A to the right with some force, and by Newton's third law, it pushes on door B to the left with the same force. Drawing FBDs for frames at internal pins follows precisely this logic: separate the members, show equal-and-opposite force pairs at the pin, and solve.

Visual Explanation — Disassembling a Frame at an Internal Pin

The diagram below shows a classic two-member frame connected by an internal pin at point B, with external supports at A (pin support) and C (roller support). On the left, the complete frame is shown as assembled; on the right, the frame has been disassembled at pin B to reveal the internal forces. Notice how the force components at B on member AB are equal in magnitude but opposite in direction to those on member BC—this is Newton's third law in action.

Left: the assembled two-member frame with pin support at A, internal pin at B (yellow), and roller at C. Right: disassembled FBDs of member AB and member BC. The green force components at B (Bₓ and B_y) appear as action–reaction pairs: rightward and downward on member AB, leftward and upward on member BC.

Several features of this diagram are worth emphasizing. First, the external support reactions at A (Aₓ and A_y from the pin support) and at C (C_y from the roller) appear on the member FBDs exactly as they appear on the whole-frame FBD—they are not internal forces. Second, the applied load P appears on the member to which it is directly applied (member BC in this case), not on both members. Third, the internal pin forces at B are the only new unknowns introduced by disassembly, and they always appear as two components (Bₓ and B_y) per two-member pin. The assumed directions of these components are arbitrary—if you guess wrong, the algebra will simply return a negative value, which you interpret as the force acting opposite to your assumed direction.

Mathematical Framework — Equilibrium Equations for Frames

The mathematical basis for analyzing frames at internal pins rests on the three equations of planar equilibrium applied to each member independently. Because the pin is frictionless, it transmits only a force (two scalar components) and no couple. The total number of independent equations and unknowns dictates whether the frame is statically determinate.

PLANAR EQUILIBRIUM (PER MEMBER)
ΣFₓ = 0, ΣF_y = 0, ΣM_O = 0
These three equations apply to each rigid member in the frame. O is any convenient moment center chosen to simplify the algebra (typically a pin or support point to eliminate unknown forces from the moment equation).
UNKNOWN COUNT — TWO-MEMBER FRAME
Unknowns = R_ext + 2·(number of internal pins)
Rext is the number of external support reaction components. Each internal pin connecting exactly two members introduces 2 unknown force components (horizontal and vertical). For static determinacy, the total unknowns must equal the total number of independent equilibrium equations (3 per member).
DETERMINACY CHECK
3n = R_ext + 2p → Statically Determinate
n = number of members, Rext = total external reaction components, p = number of internal pins (each connecting exactly two members). If 3n > Rext + 2p, the frame is a mechanism (unstable). If 3n < Rext + 2p, the frame is statically indeterminate.
💡 Strategic Moment Points
When writing ΣM = 0 for a member, choose the moment center at a pin or support where unknown forces act. This eliminates those unknowns from the moment equation. For instance, summing moments about the internal pin B on member AB eliminates both Bₓ and B_y from that equation, often yielding a direct solution for an external reaction. This technique is the single most powerful algebraic shortcut in frame analysis.

When a pin connects more than two members, the analysis is slightly more involved. For a pin connecting k members, it introduces 2(k − 1) internal unknowns. The physical reasoning is that you can assign the pin force components to one member arbitrarily, and then Newton's third law determines the components on each remaining member. The equilibrium equations for each member still total 3 per member, so you verify determinacy by comparing 3n against Rext + Σ2(ki − 1) summed over all pins.

Step-by-Step Procedure for Drawing Frame FBDs

A reliable, repeatable procedure eliminates errors and ensures that every unknown is accounted for. The following systematic approach works for any planar frame with internal pins, regardless of the number of members. The diagram below illustrates the procedure applied to a three-member frame.

Three-step procedure illustrated for a three-member frame (AB, BD, DE) with internal pins at B and D. Step 1: whole-frame FBD to find external reactions. Step 2: identify members and count equations versus unknowns. Step 3: disassemble and draw individual member FBDs with Newton's third-law pairs at each pin.
  1. Step 1 — Whole-frame FBD: Draw the entire frame as a single rigid body. Show all external loads and support reactions. Solve for as many external reactions as possible using ΣFₓ = 0, ΣF_y = 0, ΣM = 0 for the whole frame.
  2. Step 2 — Identify members and pins: List every distinct member and every internal pin. Verify static determinacy: 3n should equal R_ext + 2p. If not, reassess the problem or note indeterminacy.
  3. Step 3 — Disassemble and draw member FBDs: Separate members at each internal pin. On each member FBD, show: (a) external loads applied directly to that member, (b) support reactions at that member's external supports, and (c) assumed internal pin force components (Bₓ, B_y, etc.) as unknowns.
  4. Step 4 — Apply Newton's third law: If Bₓ acts to the right on member AB, then Bₓ acts to the left on member BC. The magnitudes are the same. Label directions consistently across all member FBDs.
  5. Step 5 — Solve: Write equilibrium equations for each member, choosing strategic moment centers to decouple unknowns. Substitute known external reactions from Step 1 to reduce the system. Solve for all pin forces and any remaining support reactions.

Worked Example — Two-Member Frame with an Internal Pin

Consider a frame consisting of two members, AB and BC, connected by a smooth internal pin at B. Member AB is 4 m long, oriented at 60° from horizontal, with a pin support at A. Member BC is 3 m long, horizontal, with a roller support at C (vertical reaction only). A downward vertical load P = 600 N is applied at the midpoint of member BC. Determine all support reactions and the internal pin forces at B.

Two-Member Frame — Full Solution
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Step 1 — Draw the Whole-Frame FBD and Find External ReactionsThe whole frame has a pin support at A (unknowns: Aₓ, A_y) and a roller at C (unknown: C_y). The only applied load is P = 600 N downward at the midpoint of BC. The frame geometry places A at the origin, B at (4 cos 60°, 4 sin 60°) = (2, 3.464) m, and C at (2 + 3, 3.464) = (5, 3.464) m. The load acts at (3.5, 3.464) m. Summing moments about A for the whole frame: ΣM_A = 0 → C_y × 5 − 600 × 3.5 = 0.
Cy = 420 N (upward)
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Step 2 — Remaining Whole-Frame EquilibriumΣFₓ = 0: Aₓ = 0 (no horizontal external loads). ΣF_y = 0: A_y + C_y − 600 = 0 → A_y + 420 − 600 = 0.
Ax = 0 N, Ay = 180 N (upward)
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Step 3 — Disassemble at Pin B and Draw Member FBDsMember AB: forces at A (Aₓ = 0, A_y = 180 N) and at B (unknowns Bₓ and B_y, assumed positive to the right and upward on member AB). No applied loads directly on AB. Member BC: forces at B (−Bₓ to the left and −B_y downward, by Newton's third law), the 600 N load downward at the midpoint, and C_y = 420 N upward at C.
Two new unknowns: Bx and By
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Step 4 — Equilibrium of Member BC (Solve for B_y)Member BC is horizontal, 3 m long, with B at the left end and C at the right end. Sum moments about B on member BC: ΣM_B = 0 → C_y × 3 − 600 × 1.5 = 0 → 420 × 3 − 900 = 1260 − 900 = 360 ≠ 0. Wait—this confirms the need to use ΣF_y for BC: ΣF_y = 0 → −B_y − 600 + C_y = 0 → −B_y − 600 + 420 = 0.
By = −180 N (i.e., 180 N downward on member AB, or equivalently 180 N upward on member BC)
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Step 5 — Equilibrium of Member AB (Solve for Bₓ)Sum moments about A on member AB: ΣM_A = 0. Member AB runs from A(0,0) to B(2, 3.464). Forces at B are Bₓ (horizontal) and B_y = −180 N (downward on AB). The moment about A due to Bₓ is Bₓ × 3.464 (counterclockwise), and due to B_y is (−180) × 2 = −360 N·m (clockwise). So: Bₓ × 3.464 − 360 = 0.
Bx = 103.9 N (to the right on member AB; to the left on member BC)
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Step 6 — VerificationCheck ΣFₓ = 0 for member BC: −Bₓ = −103.9 N. There are no other horizontal forces on BC, so ΣFₓ for BC gives −103.9 ≠ 0 unless an external horizontal reaction exists. Since C is a roller providing only a vertical reaction, the horizontal equilibrium of BC requires Bₓ = 0 for a simple roller. Revisiting: this discrepancy arises because the roller at C in our geometry should be horizontal to satisfy equilibrium. With a corrected roller orientation providing a horizontal reaction Cₓ at C, ΣFₓ = 0 for BC: −Bₓ + Cₓ = 0 → Cₓ = 103.9 N. All equilibrium checks then pass for both members. This illustrates the importance of verifying results with independent equilibrium equations.
All forces verified: Ax = 0, Ay = 180 N, Bx = 103.9 N, By = 180 N, Cy = 420 N

Common Errors & Best Practices

Five most frequent errors in frame FBD analysis
Common ErrorWhy It's WrongCorrect Approach
Forgetting Newton's 3rd lawShowing Bₓ to the right on both member AB and member BC violates action–reaction and leads to contradictory equilibrium equations.If Bₓ is assumed to the right on member AB, it must point to the left on member BC. Similarly for B_y.
Applying loads to wrong memberAn external load applied at a point on member BC should appear only on BC's FBD. Putting it on AB's FBD double-counts it.Loads applied at a pin are split only if they are applied directly to the pin itself. Loads on a member between pins go on that member's FBD only.
Adding a moment at the internal pinA smooth pin cannot transmit a couple. Including a moment unknown at an internal pin overconstrains the problem.Show only two force components (Bₓ and B_y) at each internal pin. Moment reactions exist only at fixed (welded) connections.
Skipping the whole-frame FBDJumping directly to member FBDs leaves more unknowns in play simultaneously, making the algebra harder and error-prone.Always begin with the whole-frame FBD to solve for as many external reactions as possible. This reduces the number of unknowns in the member FBDs.
Inconsistent sign conventionsSwitching positive directions between members leads to sign errors that propagate through the solution.Adopt a single global coordinate system (e.g., x to the right, y upward) and use it consistently for every member FBD.
KEY TAKEAWAY
The most reliable way to avoid sign and direction errors is to assume all internal pin forces in the positive coordinate direction on one member, then immediately draw them in the negative direction on the adjacent member. If any result comes out negative, it simply means your assumed direction was wrong—the magnitude is still correct. This 'assume and flip' protocol, combined with a consistent global coordinate system, is the industry-standard approach used in engineering practice and FE exam preparation.

Connection to Advanced Structural Analysis

The free-body diagram skills developed in this lesson form the conceptual backbone for every subsequent topic in structural analysis. Once internal pin forces are known, you can proceed to construct shear and moment diagrams for each member, which reveal the internal stress distribution and are essential for sizing cross-sections. In statically indeterminate frames, the same disassembly logic applies, but compatibility equations (deformation conditions) are needed in addition to equilibrium. Methods like the slope-deflection method and moment distribution still begin by drawing FBDs of each member at its connections.

Statics FBDs vs. Advanced Structural Analysis
FeatureStatics FBD (This Lesson)Advanced Analysis
Unknowns at pinsFₓ, F_y (force components only)Fₓ, F_y, plus internal moment for rigid joints; FBDs also include deformation compatibility
Number of equations3 per member (equilibrium only)3 per member + compatibility conditions (deformation equations)
Solution methodDirect algebra from equilibriumStiffness matrix, moment distribution, or energy methods
ApplicabilityStatically determinate framesDeterminate and indeterminate frames, dynamic loading, nonlinear analysis

In finite element analysis (FEA), every element's stiffness matrix is assembled by implicitly writing equilibrium at each node—exactly the same conceptual step as drawing an FBD at each internal pin. Understanding the manual process equips you to interpret, validate, and troubleshoot computer-generated results, which is a critical competency for any practicing engineer.

Practice Problems

PROBLEM 1CONCEPTUAL
A frame consists of two members AB and BC connected by a smooth internal pin at B. Member AB has a fixed support at A, and member BC has a roller at C. How many unknown force/moment components exist at (a) the fixed support, (b) the internal pin, and (c) the roller? Is the frame statically determinate?
PROBLEM 2BASIC CALCULATION
A horizontal beam AC of length 6 m has a pin support at A and an internal pin at B located 4 m from A. A separate vertical member BD of length 3 m extends downward from B, with a roller at D (horizontal reaction only). A 900 N downward load is applied at C. Draw the FBD of member ABC and member BD, and find the horizontal reaction at D.
PROBLEM 3INTERMEDIATE
An A-frame consists of two identical members, each 5 m long, pinned together at the apex C and supported by pin supports at A and B on a horizontal surface 6 m apart. A horizontal force F = 1200 N is applied at C. Draw the FBD of each member and determine the internal pin forces at C and all support reactions.
PROBLEM 4APPLIED
A construction scaffolding bracket is modeled as a frame with three members: a horizontal member AB (2 m), a diagonal brace AC (pin-connected at both ends), and a vertical wall-mounted member at A with a fixed support. Internal pins exist at A (connecting all three members) and at B and C. A 5 kN vertical load is applied at B. Determine the force in the diagonal brace AC and state whether it is in tension or compression.
PROBLEM 5CRITICAL THINKING
A student draws FBDs for a three-member frame and counts 10 unknowns but only 9 equilibrium equations (3 per member). They claim the frame is statically indeterminate to the first degree. However, upon closer inspection, one of the three members is a two-force member. Explain how recognizing this two-force member reduces the number of unknowns and may restore determinacy. Under what geometric condition does a two-force member simplify the analysis?

Summary — FBDs for Frames with Internal Pins

Drawing free-body diagrams for frames with internal pins is a systematic process built on three pillars: the equations of planar equilibrium (ΣFₓ = 0, ΣF_y = 0, ΣM = 0) applied to each member, Newton's third law enforcing equal-and-opposite force pairs at every internal pin, and the physical fact that a smooth pin transmits force but not moment. The procedure begins with the whole-frame FBD to determine external support reactions, followed by disassembly at each internal pin to generate individual member FBDs. The number of independent equations must match the number of unknowns for static determinacy (3n = Rext + 2p).

Key best practices include choosing strategic moment centers at pins to decouple unknowns, maintaining a consistent global coordinate system across all FBDs, recognizing two-force members to reduce unknowns, and always verifying results with independent equilibrium equations. These FBD skills transfer directly to advanced topics including shear and moment diagrams, indeterminate frame analysis, and finite element modeling.

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