Normal Force, Shear & Bending Moment

SECTION 1

Historical Context & Motivation

Structural failures throughout history have driven engineers to develop systematic methods for analyzing forces hidden inside loaded members. Ancient Roman engineers relied on empirical rules and geometric proportions to size their arches and aqueducts, but they lacked a formal framework to predict whether a beam would crack, buckle, or shear apart at a particular cross-section. The intellectual leap from treating structures as rigid wholes to exposing the internal force resultants at an imaginary cut transformed structural engineering from craft into science. Understanding this progression reveals why we define three distinct quantities — normal force, shear force, and bending moment — and why each captures a fundamentally different mode of internal resistance.

1638

Galileo's Beam Problem

In Dialogues Concerning Two New Sciences, Galileo Galilei analyzed a cantilever beam and attempted to predict its breaking load. Although his stress distribution was incorrect, he pioneered the concept of examining internal resistance at a cross-section.

1826

Navier's Beam Theory

Claude-Louis Navier published his lectures on the mechanics of structures, correctly deriving the linear stress distribution in bending and formalizing the relationship between bending moment and curvature that remains foundational in strength of materials.

1857

Jourawski's Shear Formula

D. I. Jourawski developed the transverse shear stress formula for beams, completing the picture of how internal shear force distributes across a cross-section and enabling engineers to check for shear failures independently of bending.

1864

Maxwell & the Method of Sections

James Clerk Maxwell formalized the method of sections for trusses, reinforcing the principle that equilibrium of a free body created by an imaginary cut yields the internal forces at that cut — a procedure equally applicable to beams and frames.

1930s

Shear–Moment Diagrams Standardized

Textbooks by Timoshenko, Den Hartog, and others standardized the graphical construction of shear and bending moment diagrams, making the internal-force concept a routine tool in every structural analysis course.

The central question that this lesson addresses is deceptively simple: if a structural member is in static equilibrium under external loads, what exactly happens inside the material at any given cross-section? We answer this by making an imaginary cut, isolating one side as a free body, and enforcing equilibrium. The resultants at the cut face are precisely the internal normal force, shear force, and bending moment — three quantities that together fully describe the internal loading state for planar (2-D) problems.

SECTION 2

Core Principles & Definitions

Before computing anything, one must internalize the conceptual framework. When a structural member carries external loads and support reactions, the material at every interior cross-section develops a distributed stress field that can be resolved into three resultant quantities. These resultants are not additional forces acting on the body; they are the net effect of all internal stresses across the exposed face, and they must satisfy Newton's laws applied to whichever portion of the member you have isolated. The following principles underpin every calculation in this lesson.

Method of Sections

Pass an imaginary cutting plane through the member at the location of interest. Separate the member into two free bodies. The internal resultants appear on the exposed face of each free body, equal in magnitude and opposite in direction (Newton's third law).

Internal Normal Force (N)

The component of the internal force resultant that acts perpendicular to the cross-section (i.e., along the longitudinal axis of the member). Positive N indicates tension; negative N indicates compression. It resists any tendency of the member to stretch or shorten axially.

Internal Shear Force (V)

The component of the internal force resultant that acts tangent to (in the plane of) the cross-section. It resists the tendency of one part of the member to slide laterally relative to the adjacent part. The sign convention typically assigns positive V when the force on the left face acts downward.

Internal Bending Moment (M)

The internal couple resultant about the centroid of the cross-section. It resists the tendency of the member to rotate or bend at the cut. Positive M is conventionally taken as the moment that causes the beam to sag (concave upward).

Equilibrium Enforcement

After exposing the internal resultants on the free-body diagram, apply ΣFₓ = 0, ΣFᵧ = 0, and ΣM = 0 to the isolated segment. These three equations yield N, V, and M at the cut location uniquely (for statically determinate members).

✦ KEY TAKEAWAY

Think of the imaginary cut like slicing a rope that two teams are pulling in a tug-of-war. The moment you slice it, the tension that was hidden inside the rope must appear as an exposed force on each cut end to keep each team's piece in equilibrium. In a beam, the internal "rope" carries three kinds of action simultaneously: an axial pull or push (N), a lateral sliding resistance (V), and a rotational resistance (M). Exposing them is nothing more than applying equilibrium to a cleverly chosen free body.

SECTION 3

Visual Explanation — The Imaginary Cut

Part (a) shows the original beam loaded by a concentrated force P with supports at A (pin) and B (roller). A dashed line marks the imaginary cut at section C. Part (b) isolates the left segment AC and exposes the three internal resultants on the cut face: normal force N (axial, shown in violet), shear force V (transverse, shown in pink), and bending moment M (couple, shown in amber). Equilibrium of this free body yields N, V, and M at C.

The diagram above illustrates the fundamental procedure. Observe that the three internal resultants — N, V, and M — are drawn on the cut face of the left segment with assumed positive directions following the standard beam sign convention. If, after applying the three equilibrium equations, a quantity comes out negative, the actual sense is simply opposite to what was assumed. This convention makes results from different analysts directly comparable and allows us to construct consistent shear and moment diagrams later. Note that you could equally isolate the right segment BC and apply equilibrium; Newton's third law guarantees you will obtain the same magnitudes for N, V, and M but with opposite senses on the right face.

SECTION 4

Mathematical Framework

With the free-body diagram established, the mathematical extraction of internal resultants reduces to straightforward statics: sum forces in two orthogonal directions and sum moments about a convenient point. The equations below formalize the procedure for a planar (2-D) problem. For a member lying along the x-axis with transverse loads in the y-direction, cutting at position x and isolating the left segment yields the following relations.

INTERNAL NORMAL FORCE

N(x) = −ΣFₓ,left

Sum all external horizontal force components on the left segment. N is the axial force required to keep ΣFₓ = 0. Positive N acts away from the cut face (tension).

INTERNAL SHEAR FORCE

V(x) = ΣFᵧ,left

Sum all external vertical force components (including support reactions) on the left segment. V is defined positive when the net transverse force on the left segment acts upward, which corresponds to a clockwise couple tendency at the cut.

INTERNAL BENDING MOMENT

M(x) = ΣM_C,left

Sum the moments of all external forces on the left segment about the centroid of the cut cross-section (point C). M is positive when it produces sagging (concave upward). This convention is standard in most U.S. statics and mechanics of materials textbooks.

DIFFERENTIAL RELATIONSHIPS (PREVIEW)

dV/dx = −w(x) dM/dx = V(x)

These differential relationships, derived from equilibrium of an infinitesimal element, connect the distributed load w(x) to the shear force and the shear force to the bending moment. They are the basis for constructing shear and moment diagrams by integration and will be explored in depth in subsequent lessons.

⚠️ Sign Convention Note

The sign convention adopted here (positive N = tension, positive V = upward on the left face, positive M = sagging) is the most common in U.S. engineering education. Some European texts use the opposite convention for V. Always confirm the convention before comparing results or using published formulas.

SECTION 5

Detailed Sign Conventions & Positive-Face Diagrams

A consistent sign convention is essential because N, V, and M can each be positive or negative at any cross-section, and the sign tells us the physical action. The widely used deformation sign convention assigns signs based on the deformation each resultant would produce rather than on an arbitrary coordinate direction. The diagram below shows positive internal resultants acting on both the left face and the right face of an isolated infinitesimal beam element dx. Notice that on a positive face (outward normal in the +x direction) the positive resultants act in specific directions, while on the negative face (outward normal in the −x direction) they act in the opposite directions, consistent with Newton's third law.

The upper portion shows positive N, V, and M on both faces of an infinitesimal element. Positive N (violet) pulls outward on both faces → tension. Positive V (pink) creates a clockwise shear couple. Positive M (amber) bends the element concave upward (sagging). The lower row illustrates the corresponding deformations.

Standard beam sign convention (deformation-based)

Resultant	Positive on + Face	Positive on − Face	Physical Meaning
N	Acts in +x direction (outward)	Acts in −x direction (outward)	Tension — member elongates
V	Acts in −y direction (downward)	Acts in +y direction (upward)	Clockwise shear couple on element
M	Counter-clockwise on + face	Clockwise on − face	Sagging (concave upward)

SECTION 6

Worked Example — Simply Supported Beam with a Point Load

Consider a simply supported beam AB of length L = 6 m. A pin support at A provides reactions Aₓ and Aᵧ, and a roller support at B provides reaction Bᵧ. A concentrated downward load P = 12 kN acts at point D, which is 2 m from A. We wish to determine the internal normal force, shear force, and bending moment at section C located 4 m from A (i.e., 2 m to the right of D).

Internal Resultants at Section C

Step 1 — Draw the FBD and Find Support ReactionsIsolate the entire beam. Since there are no horizontal loads, Aₓ = 0. Sum moments about A: ΣM_A = 0 → Bᵧ × 6 − 12 × 2 = 0, giving Bᵧ = 4 kN (upward). Then ΣFᵧ = 0 → Aᵧ + Bᵧ − 12 = 0, giving Aᵧ = 8 kN (upward).

Aₓ = 0, Aᵧ = 8 kN ↑, Bᵧ = 4 kN ↑

Step 2 — Cut at C (x = 4 m) and Isolate the Left SegmentThe left segment (from A to C) includes the pin support at A (reactions Aₓ and Aᵧ) and the concentrated load P at D (2 m from A, which is 2 m to the left of the cut). On the cut face at C, expose the unknowns N, V, and M with assumed positive senses: N acting to the right, V acting downward on the positive face, and M acting counter-clockwise (sagging convention).

Step 3 — Apply ΣFₓ = 0 → Solve for NHorizontal equilibrium: Aₓ + N = 0 → 0 + N = 0.

N = 0 (no axial force — expected, since all loads are transverse)

Step 4 — Apply ΣFᵧ = 0 → Solve for VVertical equilibrium on the left segment: Aᵧ − P − V = 0 → 8 − 12 − V = 0.

V = −4 kN (negative means V actually acts upward on the positive face at C, opposite to the assumed direction)

Step 5 — Apply ΣM_C = 0 → Solve for MSum moments about the cut point C (taking counter-clockwise positive): Aᵧ × 4 − P × 2 − M = 0 → 8(4) − 12(2) − M = 0 → 32 − 24 − M = 0.

M = +8 kN·m (positive → the beam sags at this section, which is physically intuitive for a downward load)

Step 6 — Verify Using the Right SegmentAs a check, isolate the right segment CB (length 2 m) with only Bᵧ = 4 kN acting upward at B. The exposed resultants on the left face of this segment are −N, −V, −M by Newton's third law. ΣFᵧ = 0 on right: V + Bᵧ = 0 → V = −4 kN ✓. ΣM_C = 0 on right: −M + Bᵧ × 2 = 0 → M = 8 kN·m ✓. Both values agree.

Verified: N = 0, V = −4 kN, M = +8 kN·m at section C.

💡 Pro Tip

Always verify your answers by checking the other segment. If both segments give the same N, V, and M (with proper sign adjustments for Newton's third law), you can be confident your support reactions and equilibrium equations are correct. This self-check is especially valuable during exams.

SECTION 7

Comparing N, V, and M — Roles and Physical Effects

Although N, V, and M are computed from the same free-body diagram, each resultant drives a different stress distribution and a different failure mode. Understanding these distinctions is critical when you move from statics into mechanics of materials, where you will compute the actual stresses and predict structural failure. The table below summarizes the key differences.

Comparison of the three internal resultants

Property	Normal Force (N)	Shear Force (V)	Bending Moment (M)
Direction	Along the member axis (perpendicular to cut)	In the plane of the cut (transverse)	Couple about the centroidal axis of the cut
Stress produced	Uniform normal stress σ = N/A	Non-uniform shear stress τ = VQ/(It)	Linear normal stress σ = −My/I
Deformation	Axial elongation or shortening	Lateral sliding / angular distortion	Curvature (bending) of the member
Typical failure mode	Tensile fracture or buckling (compression)	Shear rupture or web buckling	Flexural cracking or yielding at extreme fibers
Common in	Truss members, columns, cables	Short beams, bolted/riveted connections	Beams, frames, any flexural member

✦ KEY TAKEAWAY

Imagine holding a long breadstick at both ends. If you pull the ends apart, the breadstick resists with internal tension (N). If you push one end up and the other down, it resists the sideways sliding with shear (V). If you bend it into a curve, the top compresses and the bottom stretches — that resistance is the bending moment (M). Real structural members experience all three simultaneously, and the engineer's job is to ensure that the stresses each one produces remain within safe limits.

SECTION 8

Connection to Advanced Theory — From Resultants to Diagrams and Stress

The definitions established in this lesson form the foundation for two major extensions. First, by computing V(x) and M(x) at every position along the member (not just one cross-section), you construct the shear and bending moment diagrams. These graphical tools reveal where V and M reach their maximum values — information essential for design. Second, once the internal resultants are known, the stress formulas from mechanics of materials convert them into actual stress values that can be checked against material strength. The table below maps each concept in this lesson to its advanced counterpart.

Roadmap from internal resultants to stress and deformation analysis

This Lesson (Statics)	Advanced Extension	Where You'll See It
N, V, M at a single cross-section	N(x), V(x), M(x) as functions → shear & moment diagrams	Later in this statics course and in Mechanics of Materials
Sign convention for V and M	Differential relations: dV/dx = −w(x), dM/dx = V(x)	Graphical integration for shear–moment diagrams
Internal normal force N	Axial stress σ = N/A; axial deformation δ = NL/(AE)	Mechanics of Materials — axial loading chapter
Internal shear force V	Shear stress τ = VQ/(Ib); shear flow q = VQ/I	Mechanics of Materials — transverse shear chapter
Internal bending moment M	Flexure formula σ = −My/I; beam deflection via EIy″ = M	Mechanics of Materials — bending and deflection chapters

For three-dimensional problems, the internal resultants at a cross-section expand from three to six: a normal force N, two shear force components V_y and V_z, two bending moment components M_y and M_z, and a twisting (torsional) moment T. The planar definitions you learned here are the 2-D subset of this general framework. Mastery of the 2-D case, including careful free-body-diagram construction and sign-convention discipline, transfers directly to the 3-D context that you will encounter in advanced structural analysis and machine design courses.

SECTION 9

Practice Problems

PROBLEM 1 — CONCEPTUAL

A horizontal beam is loaded only by vertical forces and supported by a pin and a roller. Explain why the internal normal force N is zero at every cross-section, even though N is one of the three internal resultants we define at every cut.

PROBLEM 2 — BASIC CALCULATION

A simply supported beam of length L = 10 m carries a single concentrated downward load P = 20 kN at a distance of 3 m from the left support A. Determine the internal shear force V and bending moment M at a cross-section located 5 m from A.

PROBLEM 3 — INTERMEDIATE

A cantilever beam (fixed at the left end A, free at the right end B) has length L = 4 m and carries a uniformly distributed load w = 3 kN/m over its entire length. Determine N, V, and M at a section 1 m from the free end B (i.e., at x = 3 m from A).

PROBLEM 4 — APPLIED

A horizontal beam AC is 8 m long and simply supported at A and C. A hoist (modeled as a concentrated load W = 50 kN) travels along the beam. Health and safety regulations require that the maximum bending moment at the midpoint never exceed 120 kN·m. Determine whether the beam is safe when the hoist is located at x = 3 m from A, and find the hoist position that produces the absolute maximum moment at the midpoint.

PROBLEM 5 — CRITICAL THINKING

Using the differential relationships dV/dx = −w(x) and dM/dx = V(x), prove that the bending moment M(x) reaches a local maximum or minimum at a cross-section where V(x) = 0. Then discuss a scenario where V = 0 does not correspond to a global maximum of |M| along the beam.

SUMMARY

Lesson Summary

At any cross-section of a loaded structural member, the method of sections reveals three internal resultants that maintain equilibrium. The internal normal force N acts along the member axis and resists axial stretching or compression. The internal shear force V acts tangent to the cross-section and resists lateral sliding between adjacent parts. The internal bending moment M is a couple that resists the tendency of the member to bend or curve at the cut location.

To compute these quantities, isolate one segment of the member via an imaginary cut, draw all external forces and reactions on that segment, and then enforce the three planar equilibrium equations (ΣFₓ = 0, ΣFᵧ = 0, ΣM = 0). A clear sign convention — positive N for tension, positive V for a clockwise shear couple, positive M for sagging — ensures consistency and enables the later construction of shear and bending moment diagrams. These internal resultants bridge statics and mechanics of materials: once N, V, and M are known, the corresponding stress and deformation formulas complete the structural analysis.