STATICS • INTERNAL FORCES AND SHEAR–MOMENT DIAGRAMS

Maximum Moment & Shear — Identify maximum moment and shear locations

Pinpoint where beams are most vulnerable to failure by locating critical shear and bending moment values.

Historical Context & Motivation

The ability to locate maximum internal shear and maximum bending moment within a structural member is one of the most consequential skills an engineer can develop. The entire history of structural mechanics has, in a sense, been driven by the need to understand where beams, bridges, and frames are most susceptible to failure. Ancient builders relied on empirical rules and generous proportions, but as the demand for longer spans, lighter sections, and more economical use of material grew throughout the Industrial Revolution, engineers required rigorous analytical tools to predict exactly where internal forces would reach their critical values.

1638
Galileo's Cantilever Analysis
In Two New Sciences, Galileo studied the breaking of a cantilever beam, correctly identifying that the critical section lies at the wall support — the point of maximum bending moment for a tip-loaded cantilever.
1826
Navier's Beam Theory
Claude-Louis Navier published his treatise on flexure, formally connecting bending moment M to stress through σ = My/I. This relationship made locating the maximum moment synonymous with locating maximum bending stress.
1867
Mohr's Graphical Methods
Otto Mohr developed systematic graphical techniques for constructing shear and moment diagrams, giving engineers a visual language for rapidly identifying critical locations without solving every cross-section algebraically.
1930s
Hardy Cross & Moment Distribution
Hardy Cross introduced the moment-distribution method for indeterminate structures, extending the concept of locating peak moments to complex frames and continuous beams used in modern building construction.

The central question this lesson addresses is deceptively simple: given a loaded beam with known support reactions, at which cross-section does the internal shear force reach its maximum, and at which cross-section does the bending moment reach its maximum? Answering this question is the essential prerequisite for sizing structural members, selecting cross-sections, and verifying that a design can safely carry its intended loads.

Core Principles & Definitions

Before locating maximum values, we need to establish the foundational relationships that govern how shear and moment vary along a beam. These principles arise directly from the equilibrium of an infinitesimal beam element and provide the mathematical backbone for constructing and interpreting shear–moment diagrams. A firm grasp of these relationships transforms the task of finding maximum internal forces from a tedious section-by-section analysis into a systematic, almost mechanical procedure.

1

Shear–Load Relationship

The slope of the shear diagram at any point equals the negative of the distributed load intensity at that point: dV/dx = −w(x). Where no load acts, the shear is constant; under a uniform load, the shear varies linearly.
2

Moment–Shear Relationship

The slope of the moment diagram at any point equals the shear at that point: dM/dx = V(x). This fundamental link means the moment has a local extremum wherever the shear diagram crosses zero.
3

Concentrated Force Discontinuities

At a point where a concentrated force P acts, the shear diagram experiences a sudden jump of magnitude P. The moment diagram, while continuous, exhibits a slope discontinuity (a kink) at the same location.
4

Concentrated Couple Discontinuities

An applied concentrated couple M₀ causes a sudden jump of magnitude M₀ in the moment diagram but has no effect on the shear diagram. The shear remains continuous through the couple.
5

Zero-Shear ↔ Extreme Moment

The maximum or minimum bending moment in a span occurs either where V(x) = 0, where a concentrated couple is applied, or at a support/boundary. This is the single most important rule for locating Mmax.
KEY TAKEAWAY
Think of the shear diagram as the speedometer of the moment diagram. When the speedometer reads zero — i.e., V(x) = 0 — the moment has momentarily stopped changing, meaning it has reached a peak or a valley. Just as you find the summit of a hill by noting where the road levels off (slope = 0), you find the extreme moment by scanning the shear diagram for its zero crossing.

Visual Explanation — Simply Supported Beam

The following diagram illustrates a simply supported beam carrying a single concentrated load P at a general position along its span. Below the beam schematic, the corresponding shear (V) and bending moment (M) diagrams are drawn to scale, with the locations of maximum shear and maximum moment clearly annotated. This canonical case forms the template from which more complex loading patterns are analyzed by superposition.

A simply supported beam of span L = a + b with a single concentrated load P. The shear diagram (cyan) shows constant regions with a jump at the load, and the absolute maximum shear equals Pb/L (assuming a < b). The moment diagram (violet) peaks at the load point where V crosses zero, giving Mmax = Pab/L.

Observe the direct correspondence between the two diagrams. In the shear diagram, the horizontal segments indicate that no distributed load acts between the supports and the concentrated force. The shear is positive (upward resultant on the left face) from A to the load, then negative from the load to B. The moment diagram is piecewise linear (since V is constant in each segment), increasing with a positive slope of +Pb/L from A to the load, then decreasing with a slope of −Pa/L from the load to B. The peak of the triangle coincides precisely with the point where the shear crosses zero — this is not a coincidence but a direct consequence of the relationship dM/dx = V(x). For design purposes, the maximum bending stress will occur at the cross-section under the load, and the maximum shear stress will occur at either support reaction.

Mathematical Framework

The differential relationships between load, shear, and moment form a trio of equations that are derived from the equilibrium of an infinitesimal beam element of length dx subjected to a distributed load w(x). These relationships may also be expressed in integral form, which is especially convenient for constructing diagrams graphically. Together they provide a complete analytical framework for locating critical sections.

DIFFERENTIAL RELATIONSHIPS
dV/dx = −w(x) and dM/dx = V(x)
V = internal shear force, M = internal bending moment, w(x) = distributed load intensity (positive downward), x = position along the beam axis.
INTEGRAL RELATIONSHIPS
V(x₂) − V(x₁) = −∫[x₁ to x₂] w(x) dx and M(x₂) − M(x₁) = ∫[x₁ to x₂] V(x) dx
The change in shear between two sections equals the negative of the area under the load diagram; the change in moment equals the area under the shear diagram between those sections.
CONDITION FOR EXTREME MOMENT
dM/dx = 0 ⟹ V(x) = 0
By the first-derivative test, an interior maximum or minimum of M(x) occurs wherever the shear passes through zero. If V(x) changes sign by passing through zero (rather than jumping), the moment at that point is a smooth local extremum. If V(x) is discontinuous (concentrated load), the moment has a kink, and one must check the moment value on either side.

A critical subtlety arises for beams with distributed loads: the location where V(x) = 0 may fall between supports or load application points, requiring you to solve the algebraic equation V(x) = 0 for x. For example, consider a uniformly distributed load w₀ over the full span L of a simply supported beam. The shear varies linearly as V(x) = w₀L/2 − w₀x, which equals zero at x = L/2. Substituting back, Mmax = w₀L²/8, occurring at midspan — a result that should be committed to memory.

SIMPLY SUPPORTED BEAM — UNIFORM LOAD
M_max = w₀L² / 8 (at x = L/2)
w₀ = uniform load intensity (force/length), L = span. This is one of the most frequently referenced results in structural engineering.
⚠️ Don't Forget the Boundaries
The absolute maximum moment on a beam does not always occur at an interior V = 0 crossing. For a cantilever beam, the maximum moment occurs at the fixed support, which is a boundary — not a zero-shear point. Always compare interior extrema with boundary values to determine the absolute maximum.

Locating Maxima for Common Loading Cases

Practicing engineers often work from a catalog of known loading cases, each with a well-established location for maximum shear and maximum moment. Knowing these standard results accelerates the design process and provides a reliable sanity check for computational analyses. The following table summarizes the most common configurations encountered in statics and introductory structural analysis courses.

Standard maximum shear and moment results for common beam–load configurations
Configuration|V|_max Location|V|_max Value|M|_max Location|M|_max Value
SS beam, central point load PAt either supportP/2Midspan (x = L/2)PL/4
SS beam, uniform load w₀At either supportw₀L/2Midspan (x = L/2)w₀L²/8
SS beam, off-center load P at a from ASupport nearer to PPb/L (a < b)Under the loadPab/L
Cantilever, tip load PAt fixed supportPAt fixed supportPL
Cantilever, uniform load w₀At fixed supportw₀LAt fixed supportw₀L²/2
SS beam, triangular load (zero at A, w₀ at B)At support Bw₀L/3x = L/√3 from Aw₀L²/(9√3)
Cantilever beam with uniformly distributed load w₀. The shear diagram (cyan) is linear, reaching its maximum magnitude w₀L at the fixed support. The moment diagram (violet) is parabolic, with the maximum magnitude w₀L²/2 also occurring at the fixed support. In this case, both |V|max and |M|max occur at the same location — the boundary.

Notice that for the cantilever the maximum shear and maximum moment both occur at the fixed support. This is fundamentally different from the simply supported beam case, where Vmax and Mmax generally occur at different locations. The parabolic shape of the moment diagram in the cantilever case arises because the shear varies linearly — integrating a linear function yields a quadratic. This is a direct application of the integral relationship M(x₂) − M(x₁) = ∫V dx.

Worked Example — Simply Supported Beam with Mixed Loading

Consider a simply supported beam of span L = 6 m. A uniformly distributed load of w₀ = 10 kN/m acts over the left half (0 ≤ x ≤ 3 m), and a concentrated load P = 20 kN is applied at x = 4.5 m. Determine the locations and magnitudes of the maximum internal shear force and maximum bending moment.

Mixed-Loading Beam Analysis
1
Step 1 — Compute Support ReactionsSum moments about A (taking counterclockwise as positive). The resultant of the UDL is W = 10 × 3 = 30 kN acting at x = 1.5 m. The concentrated load P = 20 kN acts at x = 4.5 m. Therefore: ΣMA = 0 → RB(6) − 30(1.5) − 20(4.5) = 0 → RB = (45 + 90)/6 = 22.5 kN. From ΣFy = 0: RA = 30 + 20 − 22.5 = 27.5 kN.
RA = 27.5 kN ↑, RB = 22.5 kN ↑
2
Step 2 — Write Shear Equations by RegionRegion 1 (0 ≤ x ≤ 3 m): V(x) = 27.5 − 10x. Region 2 (3 < x < 4.5 m): the UDL has ended, so V(x) = 27.5 − 30 = −2.5 kN (constant). Region 3 (4.5 < x ≤ 6 m): after the concentrated load, V(x) = −2.5 − 20 = −22.5 kN (constant).
V(x) = {27.5 − 10x for 0≤x≤3; −2.5 for 3<x<4.5; −22.5 for 4.5<x≤6}
3
Step 3 — Locate V = 0 (Interior Extremum of M)In Region 1: V(x) = 27.5 − 10x = 0 → x = 2.75 m. The shear does not cross zero in Regions 2 or 3 (V remains negative throughout). Therefore, the only interior zero-shear crossing occurs at x = 2.75 m.
V = 0 at x = 2.75 m
4
Step 4 — Compute Moment at Critical LocationsAt x = 2.75 m: M = 27.5(2.75) − 10(2.75)²/2 = 75.625 − 37.8125 = 37.8125 kN·m. At x = 3 m (end of UDL): M = 27.5(3) − 10(3)²/2 = 82.5 − 45 = 37.5 kN·m. At x = 4.5 m (just left of P): M = 37.5 − 2.5(1.5) = 33.75 kN·m. At x = 0 and x = 6 m (supports, simple supports with no couple): M = 0. Comparing all values: the maximum moment occurs at x = 2.75 m.
Mmax = 37.81 kN·m at x = 2.75 m
5
Step 5 — Identify Maximum ShearEvaluate the absolute value of shear at all discontinuities and boundaries. At A (x = 0⁺): |V| = 27.5 kN. Just left of x = 3: |V| = |27.5 − 30| = 2.5 kN. Just right of x = 4.5: |V| = 22.5 kN. At B (x = 6⁻): |V| = 22.5 kN. The largest value is at the left support.
|V|max = 27.5 kN at x = 0 (support A)
🔧 Design Implication
In this example, the critical section for bending design is at x = 2.75 m, while the critical section for shear design is at the left support. A structural engineer would check both the flexural capacity (using σ = Mc/I with M = 37.81 kN·m) and the shear capacity (using τ = VQ/Ib with V = 27.5 kN) at their respective critical locations.

Strengths and Limitations of the V = 0 Rule

The rule that maximum bending moment occurs where the shear crosses zero is extraordinarily powerful for statically determinate beams with well-defined loading, but it is important to understand its scope and the situations where it must be applied with additional care. The table below contrasts the strengths and limitations of this approach.

Strengths and limitations of using V = 0 to find M_max
StrengthsLimitations / Caveats
Directly derived from calculus (dM/dx = V); mathematically rigorous for continuous regions.Does not automatically identify the absolute maximum if the beam has multiple V = 0 crossings — each candidate must be evaluated and compared.
Works for any loading pattern (point, distributed, triangular, trapezoidal) on determinate beams.Concentrated couples cause moment jumps that may produce the absolute maximum even though V ≠ 0 at that point.
Enables rapid estimation: just scan the shear diagram for zero crossings.For cantilevers and overhanging beams, the maximum moment often occurs at a boundary (fixed support or intermediate support), not at an interior zero-shear point.
Easily combined with the area method (change in M = area under V) for quick numerical results.For indeterminate beams, the shear diagram itself requires advanced analysis (compatibility equations, moment distribution) before V = 0 can be located.
KEY TAKEAWAY
The V = 0 rule is like using the first-derivative test in calculus to find a function's maximum — it locates local extrema. But just as you must also check endpoint values on a closed interval to find the absolute maximum, you must compare interior V = 0 moment values with boundary moments (at supports, free ends, and points of applied couples) to determine the true governing value for design.

Connection to Indeterminate Beams and Moving Loads

The principles developed for determinate beams extend naturally into more advanced structural analysis topics. In statically indeterminate beams (e.g., propped cantilevers, continuous beams over multiple supports), the fundamental relationship dM/dx = V(x) still holds, and maximum moment still occurs where V = 0 or at boundaries with moment fixity. The difference is that computing the reactions and constructing the shear diagram requires solving compatibility equations or using methods like moment distribution, slope-deflection, or matrix stiffness analysis.

Comparison between determinate beam analysis and advanced topics
FeatureDeterminate Beams (This Lesson)Indeterminate / Moving Loads (Advanced)
Reaction computationEquilibrium equations alone (ΣF = 0, ΣM = 0)Equilibrium + compatibility + constitutive relations
V = 0 rule applicabilityDirectly applicable after reactions are foundStill valid, but reactions require advanced methods first
Number of M_max candidatesTypically 1–3 locationsMay be numerous; negative moments at intermediate supports must also be checked
Moving loadsFixed loads → fixed critical sectionsInfluence lines used to find the load position that produces absolute M_max
Design toolV and M diagrams drawn by hand or simple softwareMoment and shear envelopes covering all possible load positions

Another important extension involves influence lines, which graphically depict how the moment or shear at a fixed section varies as a unit load traverses the span. The maximum ordinate of the influence line for moment at a particular section indicates how sensitive that section is to load placement. By combining influence lines with actual load patterns (such as a truck crossing a bridge), engineers construct moment envelopes and shear envelopes that capture the worst-case internal forces at every section — a critical step in bridge and infrastructure design. The conceptual foundation for all of this, however, remains exactly what you have learned in this lesson: the interplay between V and M as expressed by dM/dx = V.

Practice Problems

PROBLEM 1CONCEPTUAL
Explain, using the relationship dM/dx = V(x), why the bending moment reaches a local maximum or minimum at a point where the shear force equals zero. Under what circumstances could the absolute maximum moment occur at a location where V ≠ 0?
PROBLEM 2BASIC CALCULATION
A simply supported beam of span L = 8 m carries a single concentrated load P = 24 kN at midspan. Determine the maximum bending moment and the maximum shear force, and state where each occurs.
PROBLEM 3INTERMEDIATE
A simply supported beam of length L = 10 m carries a uniformly distributed load w₀ = 6 kN/m over its entire span plus a concentrated load P = 30 kN at x = 7 m from the left support. Find the location and magnitude of the maximum bending moment.
PROBLEM 4APPLIED
A cantilever beam of length 4 m is fixed at the left end. It supports a triangular distributed load that varies linearly from w = 0 at the fixed end (x = 0) to w = 12 kN/m at the free end (x = 4 m). The load acts downward. Determine the maximum shear force and maximum bending moment, and explain their locations.
PROBLEM 5CRITICAL THINKING
A simply supported beam of span L carries two concentrated loads: P₁ at x = a and P₂ at x = L − a (placed symmetrically but with P₁ ≠ P₂). Prove that the location of maximum moment lies between the two loads and derive an expression for this location in terms of P₁, P₂, a, and L. Discuss what happens when P₁ = P₂.

Lesson Summary

Identifying the locations of maximum shear and maximum bending moment is the essential bridge between constructing shear–moment diagrams and performing structural design. The core mathematical relationships — dV/dx = −w(x) and dM/dx = V(x) — establish that the bending moment reaches a local extremum wherever the shear force crosses zero. The absolute maximum shear typically occurs at supports or immediately adjacent to concentrated loads, and is found by evaluating |V| at every discontinuity.

For design, always compare interior V = 0 moment values with boundary moments (at fixed supports, ends, or points of applied couples) to determine the true governing value. Cantilever beams frequently have their maxima at the fixed support rather than at interior points. Mastering this procedure for determinate beams provides the foundation for advanced topics including indeterminate analysis, influence lines, and moment envelopes for moving loads.

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