SECTION 1
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
1826 • Navier's Beam Theory
1867 • Mohr's Graphical Methods
1930s • Hardy Cross & Moment Distribution
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.
SECTION 2
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.
Shear–Load Relationship
dV/dx = −w(x). Where no load acts, the shear is constant; under a uniform load, the shear varies linearly.Moment–Shear Relationship
dM/dx = V(x). This fundamental link means the moment has a local extremum wherever the shear diagram crosses zero.Concentrated Force Discontinuities
Concentrated Couple Discontinuities
Zero-Shear ↔ Extreme Moment
KEY TAKEAWAY
SECTION 3
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.
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.
SECTION 4
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
V = internal shear force, M = internal bending moment, w(x) = distributed load intensity (positive downward), x = position along the beam axis.
INTEGRAL RELATIONSHIPS
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
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
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
SECTION 5
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.
| Configuration | |V|_max Location | |V|_max Value | |M|_max Location | |M|_max Value |
|---|---|---|---|---|
| SS beam, central point load P | At either support | P/2 | Midspan (x = L/2) | PL/4 |
| SS beam, uniform load w₀ | At either support | w₀L/2 | Midspan (x = L/2) | w₀L²/8 |
| SS beam, off-center load P at a from A | Support nearer to P | Pb/L (a < b) | Under the load | Pab/L |
| Cantilever, tip load P | At fixed support | P | At fixed support | PL |
| Cantilever, uniform load w₀ | At fixed support | w₀L | At fixed support | w₀L²/2 |
| SS beam, triangular load (zero at A, w₀ at B) | At support B | w₀L/3 | x = L/√3 from A | w₀L²/(9√3) |
Standard maximum shear and moment results for common beam–load configurations
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.
SECTION 6
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 Reactions
Sum 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 Region
Region 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 Locations
At 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 Shear
Evaluate 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
SECTION 7
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 | Limitations / 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. |
Strengths and limitations of using V = 0 to find M_max
KEY TAKEAWAY
SECTION 8
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.
| Feature | Determinate Beams (This Lesson) | Indeterminate / Moving Loads (Advanced) |
|---|---|---|
| Reaction computation | Equilibrium equations alone (ΣF = 0, ΣM = 0) | Equilibrium + compatibility + constitutive relations |
| V = 0 rule applicability | Directly applicable after reactions are found | Still valid, but reactions require advanced methods first |
| Number of M_max candidates | Typically 1–3 locations | May be numerous; negative moments at intermediate supports must also be checked |
| Moving loads | Fixed loads → fixed critical sections | Influence lines used to find the load position that produces absolute M_max |
| Design tool | V and M diagrams drawn by hand or simple software | Moment and shear envelopes covering all possible load positions |
Comparison between determinate beam analysis and advanced topics
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.
SECTION 9
Practice Problems
Question 1: CONCEPTUAL
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Question 2: BASIC CALCULATION
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Question 3: INTERMEDIATE
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Question 4: APPLIED
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Question 5: CRITICAL THINKING
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SUMMARY
Lesson Summary
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.