Internal Forces via Section Cuts — Compute internal normal force, shear force, and bending moment using section cuts - Statics and Dynamics

SECTION 1

Historical Context & Motivation

The analysis of internal forces within structural members is one of the central problems of engineering mechanics. While external reactions at supports can be found by applying global equilibrium to an entire structure, engineers since the Renaissance have recognized that understanding how a beam, column, or frame behaves requires knowledge of the forces and moments that develop inside the member at every cross-section. The method of section cuts — conceptually slicing a body along an imaginary plane and enforcing equilibrium on the exposed free body — evolved over several centuries as mathematicians and engineers formalized the relationship between external loads and internal stress resultants.

1638 • Galileo's Cantilever Problem

In Dialogues Concerning Two New Sciences, Galileo analyzed a cantilever beam embedded in a wall, becoming the first to attempt a quantitative treatment of bending failure. Although his stress distribution was incorrect, his work framed the problem of internal bending moments.

1687 • Newton's Laws of Motion

Newton's Principia established the equilibrium conditions ΣF = 0 and ΣM = 0 that underpin the section-cut method. Every free-body diagram drawn from a section cut relies on Newton's third law: the internal forces on the two exposed faces are equal in magnitude and opposite in direction.

1826 • Navier's Beam Theory

Claude-Louis Navier published a comprehensive treatment of beam flexure, correctly relating bending moment to the linear stress distribution across a cross-section (σ = My/I). His work established the modern framework connecting internal moment to deformation.

1864 • Maxwell and Structural Analysis

James Clerk Maxwell introduced the method of sections for truss analysis and developed reciprocal theorems, providing powerful tools for determining internal forces in complex structures. These techniques remain standard in structural engineering coursework.

1930s • Systematic Shear-Moment Diagrams

By the early twentieth century, the plotting of shear force and bending moment diagrams using successive section cuts became a cornerstone of undergraduate engineering education, enabling rapid visualization of how internal forces vary along a structural member.

The fundamental question that motivates the section-cut method is deceptively simple: given a body in static equilibrium under known external forces and moments, what are the internal normal force, shear force, and bending moment at an arbitrary cross-section? Answering this question is the first step toward predicting stress, deformation, and ultimately structural safety — making it an indispensable skill for every practicing engineer.

SECTION 2

Core Principles & Definitions

Before performing any section-cut analysis, one must internalize several foundational ideas. The method rests entirely on Newton's third law and the conditions of static equilibrium. When an imaginary cut is made through a member, the material that has been removed is replaced by the internal force resultants that it was exerting on the remaining portion. These resultants — the normal force, shear force, and bending moment — are vector quantities resolved along and perpendicular to the member's longitudinal axis.

Internal Normal Force (N)

The component of the internal resultant force acting along the longitudinal axis of the member. A positive N indicates tension; a negative N indicates compression. It is found by summing forces parallel to the member's axis on the free body created by the cut.

Internal Shear Force (V)

The component of the internal resultant force acting perpendicular to the longitudinal axis in the plane of the cross-section. A positive V (by the beam sign convention) acts downward on the left face and upward on the right face of the cut.

Internal Bending Moment (M)

The internal couple that tends to rotate the cross-section. A positive M causes the beam to bend concave-up (sagging), compressing the top fibers and stretching the bottom fibers. It is obtained by summing moments about the centroid of the cut section.

Free-Body Diagram (FBD)

The sketch of the isolated portion of the member showing all external loads, reactions, and the unknown internal resultants at the cut face. Equilibrium equations are written for this FBD to solve for N, V, and M at the location of interest.

Sign Convention

A consistent sign convention is essential. The beam sign convention defines positive N as tension, positive V as causing clockwise rotation of the element, and positive M as producing sagging (concave-up bending). This convention must be applied before writing equilibrium equations.

✦

KEY TAKEAWAY

Think of a section cut like slicing a rope that two teams are pulling in a tug-of-war. The moment you cut the rope, you must replace the missing team's pull with an equivalent force to keep your side stationary. In a beam, that 'pull' has three components — a push/pull along the axis (N), a sideways push (V), and a twist (M) — all acting at the cut face to maintain equilibrium.

SECTION 3

Visual Explanation — The Section-Cut Procedure

The diagram below illustrates the complete section-cut procedure applied to a simply supported beam carrying a concentrated load P at midspan. The beam is first shown intact with its external reactions, then an imaginary cut is made at a distance x from the left support. The left-hand portion is isolated as a free body, and the three unknown internal resultants — N, V, and M — are drawn on the exposed cut face in their assumed positive directions per the standard beam sign convention.

Figure 1 — (a) A simply supported beam of length L carries a concentrated load P at midspan. Reactions A_y and B_y are each P/2 by symmetry. (b) An imaginary cut at distance x from the left support exposes the internal resultants N (cyan), V (violet), and M (amber) on the cut face. Equilibrium of the left free body yields the internal forces for 0 < x < L/2.

Notice that the three equilibrium equations — ΣFₓ = 0, Σ $F_y$ = 0, and ΣM = 0 — are applied to the isolated left-hand free body. Because no horizontal loads exist on this beam, N = 0 everywhere. The shear force equals P/2 throughout the region to the left of the applied load, and the bending moment increases linearly with x, reaching a maximum of PL/4 directly under the load. You could equally well analyze the right-hand free body; the results must be identical at the cut location, providing a useful self-check.

SECTION 4

Mathematical Framework

The section-cut method reduces to applying the three scalar equilibrium equations of planar statics to a free-body diagram created by an imaginary transverse cut. For a two-dimensional beam or frame member lying in the x–y plane with the longitudinal axis along x, the three unknowns at the cut are the internal normal force N, shear force V, and bending moment M. The governing equations are straightforward, but rigorous application of the sign convention is critical.

EQUILIBRIUM — AXIAL DIRECTION

ΣFₓ = 0 → N + Σ(external axial forces on FBD) = 0

N is the internal normal force at the cut (positive in tension). Sum all horizontal external forces and reactions acting on the isolated free body. A positive result means the member is in tension at that section.

EQUILIBRIUM — TRANSVERSE DIRECTION

ΣF_y = 0 → V + Σ(external transverse forces on FBD) = 0

V is the internal shear force at the cut (positive per beam sign convention: downward on the left face, upward on the right face). All transverse loads, reactions, and distributed load resultants on the free body contribute to this sum.

EQUILIBRIUM — MOMENT ABOUT THE CUT

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

M is the internal bending moment at the cut (positive when it causes sagging). Taking moments about the centroid of the cut section eliminates N and V from the equation, allowing M to be solved directly. The moment arms are measured from each force's line of action to the cut location.

💡

Why sum moments at the cut?

You may sum moments about any point, but choosing the centroid of the cut section is strategically optimal because the unknown forces N and V pass through that point, so their moment arms are zero. This decouples the moment equation from the force equations, letting you find M independently.

DIFFERENTIAL RELATIONSHIPS (FOR REFERENCE)

dV/dx = −w(x), dM/dx = V(x)

These differential relationships connect the distributed load intensity w(x) to shear V and moment M. While not required for a single section cut, they are the basis for constructing full V and M diagrams and provide a consistency check: the slope of the moment diagram at any point equals the shear at that point.

SECTION 5

Detailed Breakdown — Common Loading Scenarios

Different external loading conditions produce distinctly different internal-force distributions. The table below summarizes the behavior of V(x) and M(x) for the most common beam load cases. Understanding these patterns allows you to anticipate the shape of shear and moment diagrams before performing detailed calculations, providing a powerful error-checking tool.

Loading Type	V(x) Shape	M(x) Shape	Key Feature
Concentrated force P	Constant between loads; jumps by P at the point of application	Linear (straight segments) between loads	V diagram has a discontinuity (step) at each point load; M has a 'kink' (slope change)
Concentrated couple M₀	No change in V	Discontinuity (jump) of magnitude M₀	Shear is unaffected; moment diagram steps up or down by M₀
Uniform distributed load w₀	Linear (straight line)	Quadratic (parabolic)	Max M occurs where V = 0; V changes at rate −w₀
Linearly varying load (triangular)	Quadratic (parabolic)	Cubic	Each integration raises the polynomial order by one

Internal force distributions for common beam loading types

Figure 2 — A cantilever beam fixed at the left end carrying a uniform distributed load w₀. Part (a) shows the loaded beam, (b) shows the left free body after a section cut at distance x. The shear diagram is linear (violet) and the moment diagram is parabolic (amber), consistent with dV/dx = −w₀ and dM/dx = V.

The cantilever example in Figure 2 demonstrates several key ideas. First, the fixed support supplies both a vertical reaction R_A = w₀L and a moment reaction M_A = w₀L²/2. Second, the shear decreases linearly because the distributed load adds progressively from left to right; V = 0 at the free end, confirming that no transverse force acts there. Third, the moment diagram is parabolic, consistent with the relation dM/dx = V. The maximum moment magnitude occurs at the wall (x = 0) and equals w₀L²/2, which is the critical design value for sizing the beam's cross-section.

SECTION 6

Worked Example — Simply Supported Beam with Two Loads

Consider a simply supported beam of length 6 m. A downward concentrated force of 12 kN acts at x = 2 m from the left support A, and a downward concentrated force of 6 kN acts at x = 4 m. Determine the internal normal force, shear force, and bending moment at x = 3 m (a section between the two loads).

Internal Forces at x = 3 m

1
Step 1 — Determine External Reactions

The beam is simply supported with a pin at A (x = 0) and a roller at B (x = 6 m). First, sum moments about B to find A_y: ΣM_B = 0 → A_y(6) − 12(4) − 6(2) = 0. This gives 6A_y = 48 + 12 = 60.

A_y = 10 kN (↑), B_y = 12 + 6 − 10 = 8 kN (↑)
2
Step 2 — Make the Section Cut at x = 3 m

Cut the beam at x = 3 m and isolate the left-hand free body. This portion (from x = 0 to x = 3 m) includes: the support reaction A_y = 10 kN (↑) at x = 0, and the 12 kN downward load at x = 2 m. On the cut face (right side of the left FBD), draw the three unknowns in their assumed positive directions: N pointing to the right (tension), V pointing downward, and M as a counterclockwise couple.

FBD of left portion has two external forces and three unknowns (N, V, M) at the cut.
3
Step 3 — Apply ΣFₓ = 0 to Find N

There are no horizontal external forces on the beam. The only horizontal force is the internal normal force N at the cut face. Therefore: ΣFₓ = 0 → N = 0. This is typical for beams loaded only by transverse forces.

N = 0 kN
4
Step 4 — Apply Σ $F_y$ = 0 to Find V

Taking upward as positive on the left FBD: ΣF_y = 0 → +10 − 12 − V = 0. Solving: V = 10 − 12 = −2 kN. The negative sign means the actual shear acts opposite to the assumed positive direction, which corresponds to the shear acting upward on the left face — physically consistent since the net downward load on the left FBD exceeds the upward reaction.

V = −2 kN (shear acts upward on the left face at x = 3 m)
5
Step 5 — Apply ΣM_cut = 0 to Find M

Sum moments about the cut location (x = 3 m), taking counterclockwise as positive: ΣM_cut = 0 → M − A_y(3) + 12(1) = 0. Substituting: M − 10(3) + 12(1) = 0 → M − 30 + 12 = 0.

M = 18 kN·m (positive → sagging at x = 3 m)
6
Step 6 — Verify with the Right-Hand Free Body

As a check, isolate the right portion (x = 3 m to x = 6 m). This FBD includes the 6 kN load at x = 4 m and B_y = 8 kN at x = 6 m. By Newton's third law, the internal forces on the right face are −N, −V, and −M (i.e., the negatives of the left-face values). ΣF_y = 0 → +V' − 6 + 8 = 0 → V' = −2 kN. ΣM_cut = 0 → −M' + 6(1) − 8(3) = 0 → −M' = −18 → M' = 18 kN·m. Both match, confirming our solution.

Verified: V = −2 kN and M = 18 kN·m from the right FBD.

SECTION 7

Strengths, Limitations, and Practical Considerations

Aspect	Strengths	Limitations
Generality	Works for any statically determinate structure — beams, frames, trusses, shafts — regardless of loading type or support configuration.	For statically indeterminate structures, equilibrium alone is insufficient; compatibility equations and material laws (e.g., Euler–Bernoulli beam theory) are also needed.
Complexity	Requires only basic statics — three equilibrium equations. No advanced mathematics is needed for individual section cuts.	Generating full V(x) and M(x) expressions for beams with many load discontinuities requires multiple cuts and piecewise functions, which can be tedious without systematic tools.
Physical Insight	The FBD of the cut section directly shows force flow through the structure, building excellent engineering intuition about load paths.	Provides resultant forces and moments at the centroid — does not directly yield the stress distribution across the section (that requires beam bending theory).
Error Detection	Results can be checked by analyzing the opposite portion of the cut or by verifying consistency between V and M diagrams using dM/dx = V.	Sign convention errors are common. Inconsistent application of the beam sign convention vs. equilibrium sign convention is the most frequent source of mistakes.

Strengths and limitations of the section-cut method

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PRACTICAL NOTE

In practice, engineers rarely compute N, V, and M at a single section in isolation — they generate shear force diagrams (SFD) and bending moment diagrams (BMD) by applying section cuts across all regions of the member. These diagrams reveal the critical sections (maximum V or M) that govern design. The section-cut method at a single point is the fundamental building block of that process.

SECTION 8

Connection to Advanced Theory

The section-cut technique for computing N, V, and M is the gateway to several advanced topics in solid mechanics. Understanding how this fundamental method scales and connects to more sophisticated analyses is essential for the engineering student progressing through courses in mechanics of materials, structural analysis, and finite element methods.

Section-Cut Analysis (This Course)	Advanced Extension
Internal moment M at a section → single scalar value	Mechanics of Materials: M relates to the bending stress distribution σ = My/I (Navier's formula), enabling computation of maximum stress in the beam.
Internal shear V at a section → single scalar value	Mechanics of Materials: V relates to the shear stress distribution τ = VQ/(Ib) (shear formula), critical for web design in I-beams and connection design.
2D analysis with three internal resultants (N, V, M)	3D Statics: A general 3D section cut yields six internal resultants — N, $V_y$ , $V_z$ (shear in two planes), $M_x$ (torque), $M_y$ , $M_z$ (bending about two axes).
Statically determinate structures — equilibrium suffices	Structural Analysis: Indeterminate structures require compatibility and force-displacement methods (slope-deflection, moment distribution, matrix stiffness) in addition to equilibrium.
Hand calculation at discrete sections	Finite Element Analysis: The FEA approach automates section cuts by discretizing the structure into elements, assembling stiffness matrices, and computing internal forces at every node.

How section-cut analysis connects to advanced topics

It is worth emphasizing that no matter how sophisticated the analysis tool — whether you are using the flexibility method, the stiffness method, or a commercial FEA package — the underlying concept is identical to the method of section cuts. The software effectively makes virtual cuts at every element boundary and enforces equilibrium and compatibility simultaneously. Mastering the hand-calculation version builds the physical intuition needed to interpret, verify, and troubleshoot computational results in professional practice.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

When you make an imaginary section cut through a beam and draw the free-body diagram of the left portion, three internal resultants appear on the cut face: N, V, and M. Explain why these three quantities are sufficient to represent the entire effect of the removed right portion on the remaining left portion. Why not four or two?

Show answer

In two-dimensional planar statics, the most general force system that one portion of a body can exert on another at a cross-section is equivalent to a single resultant force (resolved into components along and perpendicular to the axis, giving N and V) plus a resultant couple moment (M). These three quantities — two force components and one moment — correspond to the three independent equations of 2D equilibrium (ΣFₓ = 0, Σ

F_y

= 0, ΣM = 0). Two quantities would be insufficient to maintain general equilibrium, and four would be redundant in a planar problem. In 3D, six internal resultants are needed (three force components and three moment components), corresponding to six equilibrium equations.

Question 2: BASIC CALCULATION

A simply supported beam of length 8 m carries a single downward concentrated load of 20 kN at a distance of 3 m from the left support. Using a section cut, determine the internal shear force V and bending moment M at a cross-section located 5 m from the left support.

Show answer

First, find reactions. Σ

M_B

= 0:

A_y

(8) − 20(5) = 0 →

A_y

= 12.5 kN. Σ

F_y

= 0:

B_y

= 20 − 12.5 = 7.5 kN. Cut at x = 5 m and analyze the left FBD, which includes

A_y

= 12.5 kN (up) at x = 0 and the 20 kN load (down) at x = 3 m. Σ

F_y

= 0: 12.5 − 20 − V = 0 → V = −7.5 kN. The negative sign means V acts upward on the left face. ΣM_cut = 0: M − 12.5(5) + 20(2) = 0 → M = 62.5 − 40 = 22.5 kN·m. Check with the right FBD: Σ

F_y

= V' + 7.5 = 0 → V' = −7.5 kN ✓. The section at x = 5 m has V = −7.5 kN and M = 22.5 kN·m (sagging).

Question 3: INTERMEDIATE

A cantilever beam of length 4 m is fixed at its left end. It carries a uniformly distributed load of w₀ = 5 kN/m over the entire span and an additional concentrated downward load of 10 kN at the free end (x = 4 m). Use a section cut to determine N, V, and M at x = 2 m from the fixed end.

Show answer

First, compute the fixed-end reactions. Σ

F_y

= 0:

R_A

= 5(4) + 10 = 30 kN (up). Σ

M_A

= 0:

M_A

= 5(4)(2) + 10(4) = 40 + 40 = 80 kN·m (counterclockwise). Cut at x = 2 m; analyze the left FBD. On the left portion, external loads are:

R_A

= 30 kN (up),

M_A

= 80 kN·m (CCW), and the distributed load over 0 to 2 m with resultant = 5(2) = 10 kN (down), acting at x = 1 m. ΣFₓ = 0: N = 0. Σ

F_y

= 0: 30 − 10 − V = 0 → V = 20 kN. ΣM_cut = 0: M + 80 − 30(2) + 10(1) = 0 → M + 80 − 60 + 10 = 0 → M = −30 kN·m. The negative moment indicates hogging at x = 2 m, which is expected for a cantilever. Results: N = 0, V = 20 kN, M = −30 kN·m.

Question 4: APPLIED

A horizontal beam AB is 10 m long, simply supported at A and B. It supports a linearly varying distributed load that increases from 0 at A (x = 0) to w_max = 12 kN/m at B (x = 10 m). The load intensity at any point x is w(x) = 1.2x kN/m. Determine the internal shear force and bending moment at x = 4 m using a section cut.

Show answer

Total load = ½(12)(10) = 60 kN, acting at 2L/3 = 20/3 m from A. Σ

M_B

= 0:

A_y

(10) − 60(10 − 20/3) = 0 →

A_y

(10) = 60(10/3) = 200 →

A_y

= 20 kN.

B_y

= 60 − 20 = 40 kN. Cut at x = 4 m. On the left FBD, the distributed load from 0 to 4 m has intensity w(x) = 1.2x. The resultant of this triangular load on the left portion is ½(1.2 × 4)(4) = ½(4.8)(4) = 9.6 kN, acting at 2/3 of 4 m = 8/3 m from A. Σ

F_y

= 0: 20 − 9.6 − V = 0 → V = 10.4 kN. ΣM_cut = 0: M − 20(4) + 9.6(4 − 8/3) = 0 → M − 80 + 9.6(4/3) = 0 → M − 80 + 12.8 = 0 → M = 67.2 kN·m. At x = 4 m: V = 10.4 kN, M = 67.2 kN·m (sagging).

Question 5: CRITICAL THINKING

Consider a simply supported beam with an overhang. The beam extends from A (pin support at x = 0) to B (roller support at x = 6 m) and then overhangs to C at x = 9 m. A concentrated downward load of 18 kN acts at the free end C. (a) Determine the reactions at A and B. (b) Using section cuts, find V and M at x = 3 m and at x = 7.5 m. (c) Identify where the bending moment changes sign and explain the physical significance of this location.

Show answer

(a) Σ

M_A

= 0:

B_y

(6) − 18(9) = 0 →

B_y

= 27 kN (up). Σ

F_y

= 0:

A_y

+ 27 − 18 = 0 →

A_y

= −9 kN, meaning

A_y

actually acts downward (the pin pulls the beam down to prevent the overhang from tipping). (b) At x = 3 m (between A and B): Left FBD has only

A_y

= −9 kN (down). Σ

F_y

= 0: −9 − V = 0 → V = −9 kN. ΣM_cut = 0: M − (−9)(3) = 0 → M + 27 = 0 → M = −27 kN·m (hogging). At x = 7.5 m (overhang region, between B and C): Use the right FBD with the 18 kN load at C. Σ

F_y

= 0: V − 18 = 0 → V = 18 kN (on the right portion, which must be negated for the left-face convention: V = −18 kN). Alternatively from the left: V = −9 + 27 − V' = 0 → V' = 18 kN. Wait — let us be careful. Analyzing the right FBD (from x = 7.5 to x = 9): the only load is 18 kN down at x = 9. Σ

F_y

= 0 on right portion: +V_right − 18 = 0 → V_right = 18 kN. By Newton's third law, V on the left face = −V_right = −18 kN. ΣM_cut on right: −M_right + 18(1.5) = 0 → M_right = 27, so M on the left face = −27 kN·m. Wait: let me recompute from the left side for x = 7.5: Σ

F_y

: −9 + 27 − V = 0 → V = 18 kN. ΣM_cut: M + 9(7.5) − 27(1.5) = 0 → M + 67.5 − 40.5 = 0 → M = −27 kN·m. So V = 18 kN, M = −27 kN·m at x = 7.5. (c) M = 0 at x = 0 (free moment at pin) and M = 0 at x = 9 (free end). In between, M is negative throughout the span (hogging everywhere). The physical significance is that the entire beam bends concave-down due to the overhang load, meaning the top fibers are in tension — opposite to what one might expect for a conventional simply supported beam. The location where M = 0 coincides with the supports at A and C. Note: at B, M = −9(6) + 0 = −54 kN·m... Let me recheck: M(x=6) from left: M + 9(6) = 0 → M = −54 kN·m. Actually, the beam moment varies linearly from 0 at A to −54 kN·m at B, then linearly from −54 kN·m at B to 0 at C. M changes sign nowhere between the supports — it is negative (hogging) throughout. The physical significance is that the overhanging load reverses the curvature relative to a standard simply supported beam, putting the top surface in tension everywhere — a critical design consideration.

SUMMARY

Lesson Summary

Summary

The method of section cuts is the fundamental technique for determining the internal normal force (N), internal shear force (V), and internal bending moment (M) at any cross-section of a structural member. The procedure involves four steps: (1) solve for external reactions using global equilibrium, (2) make an imaginary cut at the section of interest and isolate one portion, (3) draw the free-body diagram of the isolated portion showing all external forces and the unknown internal resultants in their assumed positive directions, and (4) apply the three planar equilibrium equations ΣFₓ = 0, Σ $F_y$ = 0, and ΣM = 0 to solve for N, V, and M.

A consistent sign convention is essential: positive N indicates tension, positive V causes clockwise rotation of a beam element, and positive M produces sagging (concave-up bending). The differential relationships dV/dx = −w(x) and dM/dx = V(x) connect the distributed load to the shear and moment distributions, enabling construction of full shear force and bending moment diagrams. Mastery of this method is the foundation for all subsequent topics in mechanics of materials — including bending stress, shear stress, and deflection calculations — and provides the physical intuition necessary to interpret the output of modern computational tools.