Trusses: Method of Sections — Analyze trusses using method of sections - Statics and Dynamics

SECTION 1

Historical Context & Motivation

The analysis of trusses stands as one of the foundational problems in structural engineering, driven by centuries of demand for lightweight yet rigid frameworks capable of spanning rivers, supporting roofs, and carrying heavy loads. Early bridge builders relied on empirical intuition and craft traditions, but as iron and steel replaced timber in the 18th and 19th centuries, engineers needed rigorous analytical methods to predict the internal forces in each member of a truss. The method of joints — analyzing equilibrium at each pin connection — was the first systematic technique, but it becomes tedious for large trusses when only a few specific member forces are needed. The method of sections emerged as an elegant shortcut: by making an imaginary cut through the entire truss, engineers could isolate a portion and apply equilibrium to determine member forces directly, bypassing dozens of intermediate calculations.

1826 • Navier's Structural Theory

Claude-Louis Navier published his treatise on the mechanics of structures, formalizing the relationship between external loads and internal member forces in frameworks — laying the groundwork for systematic truss analysis.

1862 • Ritter's Method of Sections

August Ritter introduced the method of sections (Rittersche Schnittmethode), demonstrating that a judicious cut through a truss allows direct calculation of member forces via moment and force equilibrium on the resulting free body.

1864 • Maxwell's Reciprocal Diagrams

James Clerk Maxwell developed graphical methods for truss analysis based on reciprocal force diagrams, providing a visual complement to the analytical section method and enabling rapid checks of member forces.

1873 • Cremona's Force Polygon Method

Luigi Cremona refined Maxwell's graphical approach into a standardized procedure using force polygons, which became a staple of engineering education alongside the method of sections throughout the late 19th century.

1950s–present • Finite Element Methods

Computational methods superseded hand calculations for complex structures, yet the method of sections remains essential in engineering education, preliminary design, and verification of numerical results.

The central question that the method of sections answers is deceptively simple: given a truss under known loads, how can we determine the force in a specific member without analyzing every joint in the structure? This efficiency makes the method indispensable in both academic problem-solving and professional practice, where rapid estimation of critical member forces guides material selection and cross-sectional sizing during the early stages of design.

SECTION 2

Core Principles & Definitions

Before applying the method of sections, it is essential to revisit the idealizations that define a simple truss. Every member is assumed to be a straight, two-force member connected at frictionless pin joints, loaded only at its endpoints. This means each member carries either pure tension (T) or pure compression (C), with the force directed along the member's longitudinal axis. External loads and support reactions act only at the joints. These assumptions are remarkably accurate for well-designed trusses and form the basis of both the method of joints and the method of sections.

Imaginary Cut

A straight (or curved) line is drawn through the truss, severing no more than three members whose forces are unknown. The cut divides the truss into two separate free bodies; either portion can be analyzed.

Exposed Internal Forces

At each cut member, the internal axial force is exposed and represented as an external force on the free-body diagram. By convention, unknown member forces are initially assumed in tension (pulling away from the joint).

Three Equations of Equilibrium

For a 2-D rigid body, three independent equilibrium equations are available: ΣFₓ = 0, ΣFᵧ = 0, and ΣM = 0. These equations can solve for up to three unknowns, which is why the cut should pass through at most three members with unknown forces.

Strategic Moment Points

Choosing the moment center at the intersection of two unknown forces eliminates them from the moment equation, yielding the third unknown directly. This strategic selection is the hallmark of an efficient section analysis.

Sign Convention

If the computed force is positive, the initial assumption of tension is correct; if negative, the member is in compression. This convention applies consistently throughout the method and eliminates ambiguity.

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KEY TAKEAWAY

Think of the method of sections like opening a book to a specific page rather than reading from the beginning. The method of joints reads every page (joint) sequentially, whereas the method of sections lets you cut directly to the chapter (member) you need. By slicing the truss and examining one portion as a free body, you access internal forces that were previously hidden — just as opening the book reveals the text on any page without having to read everything that came before.

SECTION 3

Visual Explanation — Cutting a Pratt Truss

A Pratt truss with vertical loads P at joints G and H. The red dashed line represents section cut a–a, which passes through three members: top chord GH, diagonal BG, and bottom chord BC. The internal forces F_GH, F_BG, and F_BC are exposed as external forces on the left free body. Supports at A (pin) and E (roller) are also shown.

In the diagram above, the section cut a–a has been drawn so that it intersects exactly three members: the top chord GH, the diagonal BG, and the bottom chord BC. Because only three unknowns are exposed and three equilibrium equations are available for a planar rigid body, the system is solvable. The choice of which side to analyze (left or right of the cut) is a matter of convenience — select the side with fewer external forces and reactions to simplify arithmetic. Notice how the exposed forces are drawn along the axes of their respective members, reflecting the two-force member assumption that is fundamental to all truss idealizations.

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WHY ONLY THREE MEMBERS?

In two dimensions, any free body has three independent equilibrium equations. If a section cut passes through four or more members with unknown forces, the system becomes indeterminate with respect to those equations alone. In such cases, you would need to combine the method of sections with supplementary equations (e.g., from the method of joints at an adjacent pin) or choose a different cut.

SECTION 4

Mathematical Framework

The mathematical machinery behind the method of sections is the same set of equilibrium equations used throughout statics, applied to a well-chosen free-body diagram. The power of the method lies not in new equations but in strategic selection of moment centers and force-summation directions that isolate individual unknowns. The procedure begins by determining the external support reactions for the entire truss (treated as a rigid body) and then applying equilibrium to the portion of the truss on one side of the section cut.

EQUILIBRIUM — FORCE SUMMATION (HORIZONTAL)

ΣFₓ = 0

The sum of all horizontal force components acting on the free body equals zero. This includes horizontal components of support reactions, applied loads, and the resolved components of exposed member forces.

EQUILIBRIUM — FORCE SUMMATION (VERTICAL)

ΣFᵧ = 0

The sum of all vertical force components acting on the free body equals zero. For trusses with predominantly vertical loads, this equation is often the most useful for determining diagonal member forces.

EQUILIBRIUM — MOMENT SUMMATION

ΣM_O = 0

The sum of moments about any point O equals zero. By choosing O at the intersection of the lines of action of two of the three unknown member forces, only the third unknown appears in this equation, yielding a direct solution. The moment of a force F about O is M = F × d, where d is the perpendicular distance from O to the line of action.

The strategic selection of the moment center is the defining skill of the method of sections. Consider a cut that exposes three unknowns: F₁, F₂, and F₃. If the lines of action of F₁ and F₂ intersect at point O, then taking moments about O eliminates both F₁ and F₂ from the equation, leaving F₃ as the sole unknown. Similarly, a different moment point can isolate each of the other two forces. In many problems, one or two forces can also be found using force summation equations (ΣFₓ or ΣFᵧ) when the member orientations are favorable — for instance, when one unknown acts horizontally and another acts at an angle.

FORCE RESOLUTION FOR INCLINED MEMBERS

Fₓ = F cos θ, Fᵧ = F sin θ

For any member inclined at angle θ to the horizontal, the force F in the member resolves into horizontal and vertical components. The angle θ is determined from the geometry of the truss (i.e., from the coordinates of the joints at each end of the member).

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PROCEDURAL CHECKLIST

1. Draw the free-body diagram of the entire truss and solve for support reactions. 2. Identify the member(s) whose force you need. 3. Pass a section cut through the truss that intersects those members plus no more than two additional unknown-force members (three unknowns total). 4. Draw the FBD of one portion, showing all external forces/reactions and the exposed member forces (assumed in tension). 5. Choose moment centers and summation directions that isolate each unknown. 6. Solve and interpret sign: positive = tension, negative = compression.

SECTION 5

Geometric Considerations & Cut Strategy

The effectiveness of the method of sections depends critically on the geometry of the cut and the spatial relationships between exposed member forces. A well-chosen cut reduces a complex truss to a straightforward system of equations, while a poorly chosen cut can lead to coupled equations or indeterminate situations. This section examines the geometric factors that govern cut selection and moment-point strategy for several common truss configurations.

A Warren truss with section cut s–s through three members. Moment point O₁ is located at the bottom chord joint where F_bot passes through, so taking moments about O₁ eliminates F_bot and allows F_top to be solved directly. Similarly, moment point O₂ at the upper joint isolates F_bot.

The Warren truss in the diagram above illustrates a common situation: the top and bottom chords are parallel, and the diagonals alternate in direction. When the cut s–s is made, three member forces are exposed: F_top (top chord), F_diag (diagonal), and F_bot (bottom chord). Since the two chord forces are horizontal and parallel, they do not intersect at a finite point. However, because F_bot passes through the bottom joint (O₁ in the diagram), taking moments about O₁ eliminates F_bot from the equation. The perpendicular distance from O₁ to the top chord is the truss height h, so the moment arm for F_top is simply h. The diagonal force F_diag can then be isolated using the vertical force equilibrium equation ΣFᵧ = 0, since it is the only exposed force with a vertical component (the chord forces being horizontal).

Parallel chord trusses (Pratt, Howe, Warren): The chord forces are horizontal, so ΣFᵧ = 0 directly yields the vertical component of the diagonal, and moments about a chord joint yield the opposite chord force.
Non-parallel chord trusses (e.g., scissors or cambered trusses): Chord forces have both horizontal and vertical components, making moment-point selection even more critical. The intersection of the lines of action of two inclined members can sometimes lie far from the truss, requiring careful geometry.
K-trusses and complex configurations: When a single section cut inevitably severs more than three unknown-force members, combine the method of sections with the method of joints at adjacent nodes, or use multiple section cuts.

SECTION 6

Worked Example — Pratt Truss Under Concentrated Loads

Consider a symmetric Pratt truss spanning 16 m with four equal panels of 4 m each, a height of 3 m, pin-supported at joint A (left) and roller-supported at joint E (right). Vertical loads of 20 kN act downward at each of the two interior upper-chord joints (G and H). We wish to determine the forces in members GH (top chord), BC (bottom chord), and BG (diagonal) using the method of sections.

Method of Sections — Pratt Truss

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Step 1 — Determine Support Reactions

Treat the entire truss as a rigid body. By symmetry (symmetric geometry and symmetric loading), each vertical reaction equals half the total load. The total downward load is 2 × 20 kN = 40 kN. Therefore A_y = E_y = 20 kN (upward). Joint A is a pin, but the horizontal loads are zero, so A_x = 0.

A_y = 20 kN ↑, E_y = 20 kN ↑, A_x = 0
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Step 2 — Make the Section Cut

Pass a vertical section cut between joints B and C (bottom chord) and between joints G and H (top chord). This cut severs three members: top chord GH, diagonal BG (running from joint B on the bottom chord up to joint G on the top chord), and bottom chord BC. Analyze the left portion of the truss, which includes joints A, B, F, and G, the reaction A_y = 20 kN, and the load P = 20 kN at G.

Three unknowns exposed: F_GH, F_BG, F_BC
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Step 3 — Find F_GH (Moment about B)

Take moments about joint B (at the bottom chord, x = 4 m, y = 0). Forces F_BC and F_BG both pass through B, so their moments about B are zero. The remaining forces are: A_y = 20 kN at x = 0 (moment arm = 4 m, CCW), the load 20 kN ↓ at G (x = 4 m, directly above B, moment arm = 0 for the horizontal distance, but G is at height 3 m above B — the 20 kN is vertical and passes through x = 4, same x as B, so moment arm = 0). F_GH acts horizontally at height 3 m, so its moment arm about B is 3 m. Setting ΣM_B = 0 (taking CCW as positive): +20(4) + F_GH(3) = 0. Note that F_GH is assumed in tension (pointing to the right, away from the cut on the left portion), and its moment about B is clockwise (negative if CCW is positive). Correcting: +20(4) − F_GH(3) = 0 → F_GH = 80/3 ≈ 26.67 kN.

F_GH = +26.67 kN → positive → Tension
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Step 4 — Find F_BC (Moment about G)

Take moments about joint G (at x = 4 m, y = 3 m). Forces F_GH passes through G (moment arm = 0), and F_BG also passes through G (moment arm = 0). So only A_y, the 20 kN load at G, and F_BC contribute. The 20 kN load at G acts through G, so its moment about G is zero. A_y = 20 kN ↑ at x = 0, moment arm = 4 m (horizontal distance from G), produces a CW moment. F_BC is horizontal at y = 0, moment arm = 3 m (vertical distance from G). ΣM_G = 0 (CCW positive): −20(4) + F_BC(3) = 0 → F_BC = 80/3 ≈ 26.67 kN.

F_BC = +26.67 kN → positive → Tension
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Step 5 — Find F_BG (Vertical Force Equilibrium)

Apply ΣFᵧ = 0 to the left free body. Vertical forces: A_y = +20 kN (up), load at G = −20 kN (down), and the vertical component of F_BG. The chord forces F_GH and F_BC are horizontal, contributing zero to ΣFᵧ. Member BG connects B (4, 0) to G (4, 3), making it a vertical member. Therefore F_BG is entirely vertical. Assuming tension (pointing from the cut upward toward G on the left FBD is away from joint B, i.e., upward): +20 − 20 + F_BG = 0 → F_BG = 0 kN.

F_BG = 0 kN → Zero-force member
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Step 6 — Verify and Interpret

Check ΣFₓ = 0 on the left FBD: F_GH acts to the right (assumed tension, pointing right away from the left portion) and F_BC also acts to the right. Wait — if both are tension, they both point to the right on the left FBD, giving ΣFₓ = F_GH + F_BC ≠ 0. Re-examining: on the left free body, F_GH (top chord, assumed tension) points to the right (away from the left portion toward the right), but re-checking Step 3: A_y creates a CCW moment about B, and F_GH (pointing right at height 3 m above B) creates a CW moment. This is consistent. For ΣFₓ: F_GH → right (+), F_BC → right (+). Since no horizontal external forces exist on the left FBD, ΣFₓ = 0 requires F_GH + F_BC = 0. This means one must be negative. Reconsidering: F_GH in compression means it pushes on the left portion (pointing left), so F_GH = −26.67 kN (compression) and F_BC = +26.67 kN (tension). ΣFₓ: −26.67 + 26.67 = 0 ✓. The top chord is in compression and the bottom chord is in tension — as expected for a simply supported truss under downward loads.

F_GH = 26.67 kN (C), F_BC = 26.67 kN (T), F_BG = 0 kN ✓

SECTION 7

Method of Sections vs. Method of Joints

The method of sections and the method of joints are complementary tools for truss analysis. Neither is universally superior — the choice depends on the nature of the problem and what information is sought. Understanding the strengths and limitations of each method ensures that you select the most efficient approach, or combine both when needed.

Criterion	Method of Joints	Method of Sections
Number of unknowns per step	2 (ΣFₓ = 0, ΣFᵧ = 0 at each pin)	Up to 3 (ΣFₓ, ΣFᵧ, ΣM = 0 on a rigid section)
Best suited for	Finding all member forces in a truss, or starting from a joint with only two unknowns	Finding forces in specific members without solving the entire truss
Efficiency for large trusses	Becomes tedious — must work sequentially through many joints	Highly efficient — jumps directly to the members of interest
Moment equation use	Not used (concurrent forces at a point)	Essential — strategic moment centers isolate unknowns
Limitation	Must begin at a joint with ≤ 2 unknown member forces	Section cut must not pass through more than 3 unknown-force members
Error propagation	Errors at early joints propagate through subsequent calculations	Each cut is independent — errors do not propagate

Comparison of the two primary analytical methods for simple trusses

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WHEN TO USE WHICH METHOD

In engineering practice, the method of sections is often used first to determine the forces in the most critical members — typically the ones with the longest span or the highest expected loads. The method of joints can then be used to fill in the remaining member forces if a complete analysis is needed. Many exam problems explicitly state 'use the method of sections,' but in design work, engineers routinely blend both methods within a single analysis. A useful analogy: the method of joints is like computing a spreadsheet row by row, whereas the method of sections is a targeted cell lookup — both give valid answers, but the right tool depends on whether you need the whole table or just one value.

SECTION 8

Connection to Advanced Structural Analysis

The method of sections, while complete for statically determinate trusses, represents the starting point of a broader continuum of structural analysis methods. As structures become more complex — featuring redundant members, rigid connections, or distributed loads — the assumptions of the simple truss model break down, and more sophisticated techniques are required. Understanding how the method of sections connects to these advanced methods provides valuable context for courses in structural analysis, matrix methods, and finite element analysis.

Feature	Method of Sections (Simple Trusses)	Advanced Methods
Determinacy	Statically determinate (m + r = 2j)	Handles indeterminate structures (m + r > 2j) via compatibility and force/displacement methods
Connections	Frictionless pins only — two-force members	Rigid (moment-resisting) joints, semi-rigid connections → members carry bending and shear
Loading	Concentrated loads at joints only	Distributed loads, thermal effects, settlement, dynamic loads
Solution approach	Hand calculation via equilibrium equations	Stiffness matrix assembly [K]{d} = {F}, solved computationally
Output	Axial forces only (T or C)	Axial force, shear, bending moment, deflections at every point

Simple truss analysis vs. advanced structural methods

Despite the power of computational tools, the method of sections retains its relevance for several important reasons. First, it builds the physical intuition needed to interpret and validate computer output — if a finite element model predicts tension in a top chord of a simply supported truss under gravity load, an engineer trained in the method of sections immediately recognizes this as an error. Second, preliminary design often involves sizing members for worst-case load combinations, and the method of sections provides rapid estimates without the overhead of building a full computational model. Third, many professional engineering exams (the FE and PE exams in the United States, for instance) prominently feature method-of-sections problems, making mastery of this technique essential for licensure.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

Explain why a section cut through a planar truss should pass through no more than three members with unknown forces. What happens if the cut severs four unknown-force members, and how might you still proceed with the analysis?

Show answer

A two-dimensional rigid body in static equilibrium provides exactly three independent equations: ΣFₓ = 0, ΣFᵧ = 0, and ΣM = 0. If the cut exposes three unknowns, these three equations form a solvable 3 × 3 system. With four unknowns and only three equations, the system is underdetermined from the section alone. However, you can often resolve this by first using the method of joints at an adjacent joint to reduce the number of unknowns to three, or by making a second section cut that shares one or more of the unknowns, creating a system of equations across both cuts.

Question 2: BASIC CALCULATION

A simply supported Pratt truss has a span of 12 m (three panels of 4 m each) and a height of 4 m. A single vertical load of 30 kN is applied at the center upper-chord joint. Using the method of sections, determine the force in the top chord member immediately adjacent to the loaded joint.

Show answer

By symmetry, each vertical support reaction is 15 kN. Make a section cut through the top chord, a diagonal, and the bottom chord at one panel from the center. Taking moments about the bottom-chord joint directly below the cut (eliminating the bottom chord force and the diagonal force from the moment equation), the top chord force F_top satisfies: 15(4) − F_top(4) = 0, giving F_top = 15 kN. Since the moment analysis shows this force acts to push on the cut section, the top chord is in compression: F_top = 15 kN (C).

Question 3: INTERMEDIATE

A Warren truss (without verticals) spans 20 m with five panels of 4 m each and a height of 3 m. It is pin-supported at the left end and roller-supported at the right end. Vertical loads of 10 kN act downward at each of the four interior top-chord joints. Using the method of sections, find the force in the diagonal member in the second panel from the left.

Show answer

First, find the reactions: total load = 4 × 10 = 40 kN. By ΣM about the right support: R_left(20) = 10(16) + 10(12) + 10(8) + 10(4) = 400, so R_left = 20 kN. Cut through the second panel, exposing the top chord, diagonal, and bottom chord. The diagonal connects the bottom-chord joint at x = 4 m to the top-chord joint at x = 8 m and y = 3 m, giving θ = arctan(3/4) = 36.87°. Using ΣFᵧ = 0 on the left portion (which includes R_left = 20 kN up and one load of 10 kN down at the first interior top joint): 20 − 10 + F_diag sin 36.87° = 0. F_diag(0.6) = −10, so F_diag = −16.67 kN. The negative sign indicates compression: F_diag = 16.67 kN (C).

Question 4: APPLIED

A highway sign truss spans 24 m (six panels at 4 m each) with a height of 5 m. It carries a uniform wind load equivalent to concentrated horizontal forces of 8 kN at each upper-chord joint (six joints, 48 kN total). The truss is pinned at the left base and roller-supported (horizontal roller) at the right base. Using the method of sections, determine the force in the bottom chord member at the third panel (midspan).

Show answer

With horizontal loading and a horizontal roller at the right support, the right support provides only a vertical reaction, while the left pin provides both horizontal and vertical reactions. Taking ΣFₓ = 0 for the whole truss: A_x = 48 kN (to the right, opposing the wind). Take ΣM about A to find E_y: the horizontal forces at the upper chord (height 5 m) produce overturning moments. E_y(24) = 8(5)(6 joints, at various x-positions). Computing the full moment and solving: E_y = 8 × 5 × (0 + 4 + 8 + 12 + 16 + 20)/24 = 40 × 60/24 = 100 kN. Then A_y = −100 kN (downward by ΣFᵧ = 0). Cut at midspan, take moments about the top chord joint to isolate F_bot. The moment arm of F_bot about the top joint is 5 m. Including A_x, A_y, and three horizontal wind loads on the left portion, solve for F_bot. The result depends on the precise geometry but will typically yield a large tension force in the bottom chord, reflecting the truss's role in resisting the wind-induced bending moment.

Question 5: CRITICAL THINKING

A Howe truss is identical in geometry to a Pratt truss (same span, height, and panel count), but the diagonals slope in the opposite direction. For the same loading, explain qualitatively why the Pratt diagonal carries tension while the Howe diagonal carries compression, and discuss which configuration is preferable if the members are slender steel rods. Support your argument using the method of sections and the concept of buckling.

Show answer

Using the method of sections and ΣFᵧ = 0 on the left free body, the vertical component of the diagonal force must balance the net vertical shear (reaction minus any loads between the support and the cut). In a Pratt truss, the diagonals slope downward toward the midspan, so for the diagonal to provide an upward vertical component (opposing the net downward shear), it must be in tension. In a Howe truss, the diagonals slope upward toward the midspan; to provide the same upward vertical component, the force must act in compression (pushing the joint upward). For slender steel rods, tension is preferable because compression members are susceptible to Euler buckling (P_cr = π²EI/L²), which limits their load-carrying capacity. Consequently, Pratt trusses are preferred for steel construction because the long diagonal members carry tension, while the shorter vertical members (which are less prone to buckling) carry compression. Howe trusses were historically used in timber construction, where compression members could be made from robust wooden beams without excessive buckling risk.

SUMMARY

Lesson Summary

Summary

The method of sections is a powerful technique for determining internal member forces in a statically determinate truss without analyzing every joint. The method proceeds by making an imaginary section cut through the truss — severing no more than three unknown-force members — and applying the three equilibrium equations (ΣFₓ = 0, ΣFᵧ = 0, ΣM = 0) to the resulting free body. The hallmark skill is strategic moment-point selection: choosing a point through which two of the three unknown forces pass, thereby isolating the third unknown in a single equation.

Compared to the method of joints, the method of sections excels when only specific member forces are needed, offering a direct path to the answer without sequential joint-by-joint analysis. Both methods rest on the same two-force member assumption and the requirement that the truss be statically determinate (satisfying m + r = 2j). Mastery of the method of sections builds the physical intuition needed to interpret results from advanced computational tools and remains a cornerstone of every structural engineering curriculum.