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

SECTION 1

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.

SECTION 2

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.

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.

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.

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.

Equation Counting

Each rigid member in 2-D contributes three equilibrium equations: ΣFₓ = 0, Σ

F_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.

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.

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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.

SECTION 3

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.

SECTION 4

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)

R_ext 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, R_ext = total external reaction components, p = number of internal pins (each connecting exactly two members). If 3n > R_ext + 2p, the frame is a mechanism (unstable). If 3n < R_ext + 2p, the frame is statically indeterminate.

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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 R_ext + Σ2(k_i − 1) summed over all pins.

SECTION 5

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.

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, Σ $F_y$ = 0, ΣM = 0 for the whole frame.
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.
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.
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.
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.

SECTION 6

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

1
Step 1 — Draw the Whole-Frame FBD and Find External Reactions

The 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: Σ $M_A$ = 0 → $C_y$ × 5 − 600 × 3.5 = 0.

C_y = 420 N (upward)
2
Step 2 — Remaining Whole-Frame Equilibrium

ΣFₓ = 0: Aₓ = 0 (no horizontal external loads). Σ $F_y$ = 0: $A_y$ + $C_y$ − 600 = 0 → $A_y$ + 420 − 600 = 0.

A_x = 0 N, A_y = 180 N (upward)
3
Step 3 — Disassemble at Pin B and Draw Member FBDs

Member 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: B_x and B_y
4
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: Σ $M_B$ = 0 → $C_y$ × 3 − 600 × 1.5 = 0 → 420 × 3 − 900 = 1260 − 900 = 360 ≠ 0. Wait—this confirms the need to use Σ $F_y$ for BC: Σ $F_y$ = 0 → − $B_y$ − 600 + $C_y$ = 0 → − $B_y$ − 600 + 420 = 0.

B_y = −180 N (i.e., 180 N downward on member AB, or equivalently 180 N upward on member BC)
5
Step 5 — Equilibrium of Member AB (Solve for Bₓ)

Sum moments about A on member AB: Σ $M_A$ = 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.

B_x = 103.9 N (to the right on member AB; to the left on member BC)
6
Step 6 — Verification

Check 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: A_x = 0, A_y = 180 N, B_x = 103.9 N, B_y = 180 N, C_y = 420 N

SECTION 7

Common Errors & Best Practices

Common Error	Why It's Wrong	Correct Approach
Forgetting Newton's 3rd law	Showing 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 member	An 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 pin	A 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 FBD	Jumping 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 conventions	Switching 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.

Five most frequent errors in frame FBD analysis

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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.

SECTION 8

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.

Feature	Statics FBD (This Lesson)	Advanced Analysis
Unknowns at pins	Fₓ, $F_y$ (force components only)	Fₓ, $F_y$ , plus internal moment for rigid joints; FBDs also include deformation compatibility
Number of equations	3 per member (equilibrium only)	3 per member + compatibility conditions (deformation equations)
Solution method	Direct algebra from equilibrium	Stiffness matrix, moment distribution, or energy methods
Applicability	Statically determinate frames	Determinate and indeterminate frames, dynamic loading, nonlinear analysis

Statics FBDs vs. Advanced Structural 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.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

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?

Show answer

A fixed support provides three reaction components (Aₓ,

A_y

, and

M_A

). A smooth internal pin transmits two force components (Bₓ and

B_y

) but no moment. A roller provides one reaction component (perpendicular to the rolling surface). Total unknowns: 3 + 2 + 1 = 6. With two members, there are 3 × 2 = 6 independent equilibrium equations. Since unknowns equal equations, the frame is statically determinate.

Question 2: BASIC 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.

Show answer

Starting with the whole-frame FBD: pin at A provides Aₓ and

A_y

; roller at D provides Dₓ only. The 900 N load acts at C. Σ

M_A

= 0 for the whole frame: 900 × 6 − Dₓ × 0 = ... but Dₓ has zero moment arm about A if D is directly below B. We must use the member FBD. For member BD (vertical, 3 m), taking moments about B: Σ

M_B

= 0 → Dₓ × 3 = 0, so Dₓ = 0 N (no horizontal loads or offsets exist to generate a horizontal reaction). Then Σ

F_y

for BD gives

B_y

on BD = 0, so by Newton's 3rd law,

B_y

on ABC from BD is also 0. The beam ABC then behaves as a simply supported beam with a pin at A:

A_y

= 900 N. The internal pin at B transmits only vertical force in this configuration.

Question 3: INTERMEDIATE

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.

Show answer

By symmetry of geometry (but not loading), use the whole-frame FBD first. The base is 6 m, each member is 5 m, so the apex height is √(5² − 3²) = 4 m. Whole-frame: ΣFₓ = 0 → Aₓ + Bₓ + 1200 = 0. Σ

M_A

= 0 →

B_y

× 6 + 1200 × 4 = 0 →

B_y

= −800 N (downward). Σ

F_y

= 0 →

A_y

B_y

= 0 →

A_y

= 800 N. For member AC, take moments about A: Σ

M_A

= 0 →

C_y

× 3 + Cₓ × 4 + 1200 × 4 = 0 (where Cₓ and

C_y

are pin forces on member AC). For member BC, Σ

M_B

= 0 → −

C_y

× 3 + Cₓ × 4 = 0 (the 1200 N acts at C on the frame, assign it to one member). Solving these simultaneous equations: Cₓ = −600 N,

C_y

= −800 N. Then Aₓ = 600 N and Bₓ = −1800 N. Verification: ΣFₓ = 600 − 1800 + 1200 = 0 ✓.

Question 4: APPLIED

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.

Show answer

The diagonal brace AC is a two-force member (pinned at both ends with no intermediate loads), so it carries only an axial force along its line of action from A to C. Let the horizontal member AB be 2 m with B at the free end, and let C be directly below A at a distance of, say, 1.5 m (forming a right triangle with hypotenuse AC = √(2² + 1.5²) = 2.5 m). The FBD of member AB shows the 5 kN downward load at B and the force F_AC from the brace at an angle θ = arctan(1.5/2) = 36.87°. Taking moments about A on member AB: Σ

M_A

= 0 → −5 × 2 + F_AC sin(36.87°) × 2 = 0 → F_AC = 10/1.2 = 8.33 kN in tension (the brace pulls on pin A, resisting the downward load). The fixed support at A provides the remaining equilibrium forces and the fixing moment.

Question 5: CRITICAL 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?

Show answer

A two-force member is one that is loaded only at its two endpoints (no intermediate loads, supports, or couples). For such a member, equilibrium requires that the resultant forces at its two endpoints be equal in magnitude, opposite in direction, and collinear along the line connecting the two endpoints. This reduces the member's contribution from 4 unknowns (two components at each end) to just 1 unknown (the magnitude of the axial force along its axis, with direction known). The net reduction is 3 unknowns (from 4 to 1), which brings the total from 10 to 7 unknowns. Since the two-force member's three equilibrium equations are automatically satisfied by the collinearity condition, the effective system reduces to 6 equations and 7 unknowns—still one extra. However, if there is also a geometric constraint (e.g., the two-force member direction implies a known relationship among reactions), determinacy may be achieved. The key geometric condition is that the two-force member must connect two pins with no other loads between them. Recognizing two-force members before writing equations is a critical skill that dramatically simplifies frame analysis.

SUMMARY

Summary — FBDs for Frames with Internal Pins

Summary

Drawing free-body diagrams for frames with internal pins is a systematic process built on three pillars: the equations of planar equilibrium (ΣFₓ = 0, Σ $F_y$ = 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 = R_ext + 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.