Internal Forces via Section Cuts — Compute internal forces using section cuts - Statics

SECTION 1

Historical Context & Motivation

Structural analysis has always rested on a fundamental question: what happens inside a loaded member? External reactions at supports tell only part of the story. Engineers need to know the internal forces — normal force, shear force, and bending moment — acting on every cross-section, because material failure originates at internal stress concentrations, not at the supports themselves. The method of section cuts (sometimes called the free-body-diagram method for internal forces) was developed precisely to answer that question by exposing the internal resultants on an imaginary cut plane and enforcing equilibrium on the resulting sub-body.

1638 • Galileo's Cantilever Problem

In Two New Sciences, Galileo analyzed a cantilever beam embedded in a wall, recognizing that fracture depends on the internal resistance at the fixed cross-section — an early, qualitative version of a section cut.

1687 • Newton's Laws of Motion

Newton's second and third laws provided the rigorous equilibrium framework needed to sum forces and moments on any free body, including the sub-bodies created by section cuts.

1826 • Navier's Bending Theory

Claude-Louis Navier published a comprehensive theory relating internal bending moment to stress via M/I = σ/y, cementing the section-cut approach as the gateway to stress analysis in beams.

1864 • Culmann's Graphical Statics

Karl Culmann formalized graphical methods for constructing shear and moment diagrams from section-cut calculations, making internal-force analysis standard practice for 19th-century bridge engineers.

Modern • Finite Element Method

Today's FEA software automates section-cut logic by evaluating equilibrium at thousands of element boundaries, but every engineer must still understand the underlying hand-calculation method to validate computational results.

The central question this lesson addresses is straightforward yet powerful: given a structural member in static equilibrium with known external loads and support reactions, how do we determine the internal normal force N, shear force V, and bending moment M at an arbitrary cross-section? The answer lies in the disciplined application of an imaginary cut, the construction of a free-body diagram for one side, and the subsequent enforcement of the three scalar equilibrium equations in two dimensions.

SECTION 2

Core Principles & Definitions

Before executing a section cut, several foundational ideas must be clearly understood. The section-cut method is not an independent theory — it is a direct consequence of Newton's laws applied to a sub-body of the structure. Every step hinges on the principle that if the whole body is in equilibrium, then every part of the body is also in equilibrium, including any portion isolated by a hypothetical cutting plane.

Static Equilibrium of Sub-Bodies

If a structure is in equilibrium, any portion obtained by an imaginary cut must also satisfy ΣF_x = 0, ΣF_y = 0, and ΣM = 0 independently.

Internal Resultants at the Cut Face

The distributed internal stresses on the exposed cross-section are replaced by three resultant quantities: a normal force N (along the member axis), a shear force V (perpendicular to the axis), and a bending moment M.

Sign Conventions

Positive N produces tension; positive V causes clockwise rotation of the element; positive M causes the beam to bend concave upward (sagging). Consistent sign conventions are essential to avoid errors when comparing results from left and right sub-bodies.

Newton's Third Law at the Cut

The internal forces on the left face of the cut are equal in magnitude but opposite in direction to those on the right face. Analyzing either sub-body yields the same magnitudes for N, V, and M.

External Reactions First

Before making any section cut, the global support reactions must be determined from the whole-body free-body diagram. These reactions are external loads on the sub-body and are indispensable for the internal-force calculation.

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KEY TAKEAWAY

Think of a section cut like slicing a rope in a tug-of-war and asking: what tension must exist at the cut to keep each team stationary? The rope doesn't actually break — you merely imagine removing material so the internal forces become external forces on the sub-body, allowing you to solve for them with standard equilibrium equations.

SECTION 3

Visual Explanation — The Section-Cut Procedure

The diagram below illustrates the complete section-cut procedure on a simply supported beam carrying a concentrated load. On the top, the original beam with its support reactions is shown. A vertical cutting plane at a chosen location separates the beam into left and right sub-bodies. The lower portion of the figure shows the free-body diagram of the left sub-body with the three internal resultants — N, V, and M — drawn at the exposed face using the positive sign convention.

Top: a simply supported beam with a concentrated load P and reactions A_y and B_y. The dashed yellow line marks the cut plane at distance x from A. Bottom-left: the left sub-body FBD with the three internal resultants N, V, and M drawn in positive directions. Bottom-right: the sign convention used throughout this lesson.

Observe how the internal forces N, V, and M at the exposed face serve as the replacement for all interactions between the removed material and the retained sub-body. Before the cut, those interactions were distributed stresses; after the cut, they are lumped into three equivalent resultant quantities. The equilibrium equations written for the left sub-body at a general position x (to the left of the load) yield N = 0, V = A_y, and M = A_y × x. Notice how the expressions change once the cut location passes the point of load application — a critical concept when constructing complete shear and moment diagrams.

SECTION 4

Mathematical Framework

The section-cut method in two dimensions reduces to three scalar equilibrium equations. Because most beams carry transverse loads (perpendicular to the longitudinal axis), the normal force N is often zero, leaving two non-trivial equations for V and M. Below are the governing equations and variable definitions that form the backbone of every section-cut calculation.

AXIAL EQUILIBRIUM

ΣFₓ = 0 → N + Σ(horizontal external forces on sub-body) = 0

N = internal normal force at the cut face (positive = tension). For beams with only transverse loading, N = 0.

TRANSVERSE EQUILIBRIUM

ΣF_y = 0 → V + Σ(vertical external forces on sub-body) = 0

V = internal shear force at the cut face. Solve for V by summing all vertical forces (reactions, applied loads) on the chosen sub-body.

MOMENT EQUILIBRIUM

ΣM_cut = 0 → M + Σ(moments of external forces about the cut) = 0

M = internal bending moment at the cut face. Summing moments about the cut point eliminates the unknown V from this equation, enabling direct solution for M.

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Strategic Tip — Moment Summation Point

Always sum moments about the point of the cut. This eliminates both N and V from the moment equation (their lines of action pass through that point), letting you solve for M directly without needing V first. Similarly, using ΣF_y = 0 solves for V independently of M. This decoupling is computationally efficient and reduces the chance of propagating algebraic errors.

DIFFERENTIAL RELATIONS (PREVIEW)

dV/dx = −w(x), dM/dx = V(x)

These differential relations between distributed load w(x), shear V(x), and moment M(x) are derived by applying section cuts at x and x + dx and taking the limit. They form the basis of shear and moment diagram construction, covered in subsequent lessons.

SECTION 5

Step-by-Step Section-Cut Procedure

While the underlying theory is straightforward, a disciplined procedure prevents the sign and bookkeeping errors that plague internal-force calculations. The following flowchart and enumerated steps codify the process. The second SVG diagram below depicts a more complex scenario — a beam with both a concentrated load and a uniformly distributed load — illustrating how the procedure adapts when the cut falls in a distributed-load region.

Step 1 — Draw the whole-body FBD. Include all external loads and support reactions. Use the whole-body equilibrium equations (ΣF_x = 0, ΣF_y = 0, ΣM = 0) to determine all unknown reactions.
Step 2 — Choose the cut location. Identify the cross-section where internal forces are needed. If the loading changes character (e.g., a point load or the start/end of a distributed load), you may need separate expressions for each segment.
Step 3 — Pass an imaginary cutting plane. Separate the beam into two sub-bodies. Select the sub-body with fewer external forces to minimize arithmetic.
Step 4 — Draw the sub-body FBD. Include all external forces on that piece and draw N, V, and M in their positive directions at the cut face.
Step 5 — Apply equilibrium equations. Write ΣF_x = 0 for N, ΣF_y = 0 for V, and ΣM_cut = 0 for M. A negative result means the actual direction is opposite to the assumed positive direction.
Step 6 — Interpret the results. Report N, V, and M with their signs. Positive V and M follow the sign convention established at the outset; these values feed into stress formulas (σ = N/A, τ = VQ/Ib, σ = My/I) in subsequent analysis.

A simply supported beam (L = 8 m) with a uniformly distributed load w = 2 kN/m spanning from x = 2 m to x = 6 m. The section cut is made at x = 3.5 m, which falls inside the loaded region. The left sub-body FBD shows only 1.5 m of the UDL (resultant 3 kN), along with the three unknown internal resultants at the cut face.

The critical subtlety in this example is that the cut falls inside the region of the distributed load. Only the portion of the distributed load that lies on the chosen sub-body enters the equilibrium equations — not the entire load. The resultant of that partial UDL is w × (length on the sub-body) and it acts at the centroid of that partial rectangle. Forgetting to truncate the distributed load at the cut plane is one of the most common errors in section-cut problems.

SECTION 6

Worked Example — Overhanging Beam with Point Load

Consider an overhanging beam of total length 9 m. A pin support is located at point A (x = 0 m) and a roller support at point B (x = 6 m). A concentrated load P = 12 kN acts downward at the free end C (x = 9 m). Determine the internal forces N, V, and M at cross-section D located 4 m from A (x = 4 m).

Internal Forces at Section D (x = 4 m)

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Step 1 — Draw the Whole-Body FBD and Find Support Reactions

The beam has pin support at A (reactions A_x and A_y) and roller at B (reaction B_y). Since there are no horizontal external loads, A_x = 0. Sum moments about A: ΣM_A = 0 → B_y(6) − 12(9) = 0, so B_y = 18 kN (upward). Sum vertical forces: ΣF_y = 0 → A_y + 18 − 12 = 0, so A_y = −6 kN (i.e., 6 kN downward).

A_x = 0, A_y = −6 kN, B_y = 18 kN
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Step 2 — Select the Cut and Choose a Sub-Body

Pass an imaginary vertical cut at D (x = 4 m). The left sub-body (A to D) has only one external force: the reaction A_y = −6 kN (downward). This is the simpler side — only one force rather than two — so we analyze the left sub-body.
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Step 3 — Draw the Sub-Body FBD

On the left sub-body, show: A_y = 6 kN acting downward at x = 0 m. At the cut face (x = 4 m), draw N pointing to the right along the beam axis (positive tension), V pointing downward on the left face (positive shear), and M as a counterclockwise couple (positive sagging on a left face).
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Step 4 — Apply ΣFₓ = 0 (Normal Force)

Since there are no horizontal forces on the sub-body, the equation yields N = 0 directly.

N = 0
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Step 5 — Apply Σ $F_y$ = 0 (Shear Force)

Taking upward as positive: −6 − V = 0. (A_y acts downward, and V in its positive direction on a left-face cut also acts downward.) Therefore V = −6 kN. The negative sign indicates V actually acts upward on the left face, meaning the beam at this section experiences a shear force of magnitude 6 kN.

V = −6 kN
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Step 6 — Apply ΣM_cut = 0 (Bending Moment)

Sum moments about point D (the cut). A_y = 6 kN downward acts at a distance 4 m to the left of D. Its moment about D is 6 × 4 = 24 kN·m clockwise. Taking counterclockwise as positive for consistency with the positive-M convention: M − 6(4) = 0 → M + (−24) = 0 → M = 24 kN·m? Let us be precise: A_y is 6 kN downward at x = 0, creating a clockwise moment about D of magnitude 24 kN·m. In the equation ΣM_D = 0 (CCW positive): M + (−6)(4) = 0 → M = +24? No — the downward force to the left of D creates a negative (clockwise) moment: M − 24 = 0 only if A_y were upward. Since A_y = 6 kN downward, the force creates a clockwise moment: ΣM_D = M + 6(4) = 0 → M = −24 kN·m. The negative moment indicates hogging (concave downward) at this section, which is physically sensible because the overhang load pulls the beam upward at B and down at C, creating a hogging region between A and B.

M = −24 kN·m (hogging)

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Verification Check

You can verify these results by analyzing the right sub-body (D to C) instead. That sub-body carries B_y = 18 kN up at x = 6 m and P = 12 kN down at x = 9 m. Applying equilibrium to the right sub-body (with reversed sign conventions on the right face of the cut) yields the same magnitudes: V = −6 kN and M = −24 kN·m. This consistency confirms the calculation and demonstrates Newton's third law at the cut.

SECTION 7

Strengths, Limitations & Common Pitfalls

Aspect	Strength	Limitation / Pitfall
Conceptual clarity	The method is a direct application of Newton's laws — no new theory needed. Any student comfortable with free-body diagrams can execute section cuts.	For complex loadings with many segments, the number of distinct expressions for V(x) and M(x) can become large and error-prone.
Generality	Applicable to beams, frames, trusses, cables, and 3-D members. Works for any loading type: point loads, distributed loads, couples, and combinations.	For statically indeterminate structures, section cuts alone cannot determine internal forces — compatibility equations or energy methods are also required.
Sign convention	A consistent convention (tension-positive N, sagging-positive M) enables direct comparison of results from either sub-body and seamless integration with stress formulas.	Students frequently mix up the positive direction of V on left vs. right faces, leading to sign errors. The direction of M on a right face is clockwise, opposite to the left face.
Distributed loads	Distributed loads are handled by including only the portion on the sub-body and replacing it by its resultant force at its centroid.	Common error: including the entire distributed load instead of truncating it at the cut. Another pitfall is incorrect centroid location for triangular or trapezoidal load distributions.
Choosing the sub-body	Either side of the cut works. Choosing the side with fewer external forces simplifies arithmetic significantly.	Students sometimes analyze the wrong side and include forces that do not act on it, or omit the support reaction that has already been determined.

Strengths and common pitfalls of the section-cut method

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KEY TAKEAWAY

The section-cut method is the universal first step in structural internal-force analysis, much like dissection is to anatomy: you cannot study the internal workings without first exposing them. Its power lies in its simplicity — only Newton's laws and careful bookkeeping are needed. Its danger also lies in that simplicity: sloppy sign conventions, misplaced resultants, or forgotten reactions lead to incorrect results that can propagate into unsafe designs.

SECTION 8

Connection to Advanced Theory — From Discrete Cuts to Continuous Diagrams

A single section cut provides internal forces at one specific location. In practice, engineers need to know how N, V, and M vary along the entire length of a member to identify critical cross-sections where forces are maximum. This motivates the construction of shear and moment diagrams — continuous graphical representations of V(x) and M(x) — which are the subject of the next lesson. The section-cut method provides the conceptual and computational foundation for these diagrams.

Feature	Section Cut (This Lesson)	Shear–Moment Diagrams (Next Lesson)
Output	N, V, M at a single cross-section	V(x) and M(x) as continuous functions over the entire beam
Procedure	One FBD, three equilibrium equations	Multiple section cuts at variable x, or integration of differential relations dV/dx = −w, dM/dx = V
Use case	Spot-checking a known critical location, homework/exam problems with a specified cut	Full structural design: finding absolute maximum V and M to size members
3-D extension	Six internal resultants: N, $V_y$ , $V_z$ , T (torque), $M_y$ , $M_z$	Separate diagrams for each resultant along the member axis

Beyond shear and moment diagrams, internal-force results feed directly into the stress-analysis equations of Mechanics of Materials: normal stress σ = N/A for axial members, flexural stress σ = My/I for beams, and shear stress τ = VQ/(Ib) for transverse shear. In three dimensions, a section cut exposes six resultants — three forces and three moments — which are the gateway to combined-loading analysis and Mohr's circle. Mastering the two-dimensional version in this lesson is therefore an investment that pays dividends throughout the engineering curriculum and into professional practice.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

A simply supported beam is in static equilibrium under several transverse loads. You make an imaginary cut at some interior point and analyze the left sub-body. Your classmate analyzes the right sub-body at the same cut. Will you both obtain the same values of V and M at the cut? Explain why or why not, referencing a fundamental law of mechanics.

Show answer

Yes, both sub-bodies yield identical magnitudes for V and M at the cut section. This is a direct consequence of Newton's third law: the internal forces on the left face of the cut are equal in magnitude and opposite in direction to those on the right face. When each sub-body is in equilibrium, the equilibrium equations automatically enforce this condition. The sign of V and M will be the same if both students use the same sign convention (e.g., positive V causes clockwise rotation of the element, positive M causes sagging). This mutual consistency also serves as a powerful self-check: if analyzing both sides yields different magnitudes, an error has been made in the reaction calculation or the equilibrium of one sub-body.

Question 2: BASIC CALCULATION

A simply supported beam AB has length L = 10 m with a pin at A and roller at B. A single concentrated load P = 20 kN acts downward at x = 4 m from A. Find the internal shear force V and bending moment M at a section 3 m from A.

Show answer

First, find reactions. ΣM_A = 0: B_y(10) − 20(4) = 0 → B_y = 8 kN. ΣF_y = 0: A_y = 12 kN. Cut at x = 3 m (left of the load). Left sub-body carries only A_y = 12 kN upward. ΣF_y = 0: 12 − V = 0 → V = 12 kN. ΣM_cut = 0: M − 12(3) = 0 → M = 36 kN·m (sagging). The shear is constant at 12 kN between A and the load, and the moment increases linearly from 0 at A to 48 kN·m at the load.

Question 3: INTERMEDIATE

A cantilever beam (fixed at A, free at B) has length L = 6 m. It carries a uniformly distributed load w = 3 kN/m over its entire length and a concentrated load P = 10 kN at the free end B. Determine the internal shear force V and bending moment M at a section 2 m from the free end B (i.e., x = 4 m from A).

Show answer

For this problem, choosing the right sub-body (from the cut to the free end B) avoids needing the support reactions. The right sub-body (2 m long) carries: P = 10 kN downward at B and a UDL of w = 3 kN/m over 2 m (resultant = 6 kN at 1 m from the cut). At the cut, draw N, V, and M in positive directions on the right face (N points left, V points up, M is clockwise). ΣF_x = 0: N = 0. ΣF_y = 0: V − 6 − 10 = 0 → V = 16 kN. ΣM_cut = 0 (CCW positive on the right sub-body): −M − 6(1) − 10(2) = 0 → M = −26 kN·m. The negative moment confirms hogging, as expected for a cantilever loaded downward. The shear of 16 kN acts to resist the total downward load between the cut and the free end.

Question 4: APPLIED

A horizontal boom of a construction crane is modeled as a simply supported beam with a pin at A (x = 0) and roller at B (x = 8 m). A triangular distributed load acts from A to B with intensity zero at A and w₀ = 6 kN/m at B. Determine the internal shear force V and bending moment M at the midpoint (x = 4 m). The resultant of a triangular load of base L and peak w₀ is ½w₀L acting at ⅔L from the zero-intensity end.

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Total resultant of the full triangular load: R = ½(6)(8) = 24 kN, acting at x = ⅔(8) = 16/3 m from A. Reactions: ΣM_A = 0 → B_y(8) = 24(16/3) → B_y = 16 kN. A_y = 24 − 16 = 8 kN. At x = 4 m, the load intensity is w(4) = 6(4/8) = 3 kN/m (linearly interpolated). The portion of the triangular load from x = 0 to x = 4 is itself a triangle with base 4 m and peak 3 kN/m. Its resultant: ½(3)(4) = 6 kN at ⅔(4) = 8/3 m from A. Left sub-body: ΣF_y = 0 → 8 − 6 − V = 0 → V = 2 kN. ΣM_cut = 0 → M − 8(4) + 6(4 − 8/3) = 0 → M = 32 − 6(4/3) = 32 − 8 = 24 kN·m. Thus V = 2 kN and M = 24 kN·m (sagging) at midspan.

Question 5: CRITICAL THINKING

A beam AB (L = 10 m, pin at A, roller at B) carries a UDL w = 4 kN/m over the first 5 m (from A to midpoint C) and a concentrated moment (couple) M₀ = 30 kN·m clockwise applied at point D (x = 7 m). (a) Find the reactions. (b) Determine V and M just to the left and just to the right of point D. (c) Explain physically why the bending moment exhibits a discontinuity at D while the shear does not.

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(a) UDL resultant: 4 × 5 = 20 kN at x = 2.5 m. ΣM_A = 0: B_y(10) − 20(2.5) − 30 = 0 → B_y = 8 kN. A_y = 20 − 8 = 12 kN. (b) Cut just left of D (x = 7⁻ m). Left sub-body: external forces are A_y = 12 kN and UDL resultant 20 kN. ΣF_y = 0: 12 − 20 − V = 0 → V = −8 kN. ΣM_cut = 0: M − 12(7) + 20(4.5) = 0 → M = 84 − 90 = −6 kN·m. Cut just right of D (x = 7⁺ m): the couple M₀ = 30 kN·m CW now lies on the left sub-body. ΣF_y is unchanged → V = −8 kN (no new vertical force). ΣM_cut = 0: M − 12(7) + 20(4.5) − 30 = 0 → M = −6 − 30 = −36 kN·m? Let's re-examine: the 30 kN·m clockwise couple is negative (CW = negative if CCW is positive): M − 84 + 90 − 30 = 0 → M = 24 kN·m. So M jumps from −6 kN·m to 24 kN·m (a change of 30 kN·m, equal to the applied couple). (c) An applied concentrated moment contributes no net force to the ΣF_y equation, so V is continuous across it. However, the couple directly enters the ΣM equation, causing an instantaneous jump in M equal to the magnitude of the applied couple. This is analogous to how a concentrated point load causes a jump in the shear diagram but not in the moment diagram.

SUMMARY

Lesson Summary

Summary

The method of section cuts exposes internal forces by passing an imaginary cutting plane through a structural member and analyzing the equilibrium of the resulting sub-body free-body diagram. At the cut face, three internal resultants are revealed: the normal force N (along the axis), the shear force V (transverse to the axis), and the bending moment M. These are computed using the three scalar equilibrium equations ΣF_x = 0, ΣF_y = 0, and ΣM_cut = 0 applied to the chosen sub-body.

Successful execution of the method requires: (1) determining support reactions from the whole-body FBD first, (2) using a consistent sign convention (tension-positive N, sagging-positive M), and (3) including only the external forces that act on the selected sub-body — particularly truncating distributed loads at the cut plane. Section-cut results at discrete points serve as the foundation for constructing continuous shear and moment diagrams and ultimately for computing stresses in Mechanics of Materials.