Bending Moment Diagrams — Construct bending moment diagrams for beams - Statics

SECTION 1

Historical Context & Motivation

The need to understand how beams bend under load has driven structural engineering since antiquity. Early builders relied on empirical rules—trial, error, and catastrophic failure—to size timber and stone elements. It was not until the Renaissance that scholars began formulating mathematical descriptions of how internal forces distribute through a loaded member. The bending moment diagram emerged from this centuries-long effort as the definitive graphical tool for visualizing the internal moment at every cross-section of a beam, enabling engineers to locate critical sections and design members against flexural failure.

1638 • Galileo's Beam Analysis

In Two New Sciences, Galileo Galilei analyzed the fracture of a cantilever beam, establishing the first quantitative treatment of bending and identifying that failure depends on the moment arm of the applied load.

1826 • Navier's Flexure Formula

Claude-Louis Navier published his theory of beam bending, deriving the linear stress distribution σ = My/I, which directly connects the internal bending moment M to stress. This made quantifying M at every section essential.

1864 • Mohr's Graphical Methods

Otto Mohr introduced systematic graphical methods for structural analysis, popularizing shear and moment diagrams as standard engineering tools. His approach allowed engineers to quickly identify maximum moments without repeatedly solving equilibrium equations.

1930s • Hardy Cross and Moment Distribution

Hardy Cross developed the moment distribution method for indeterminate structures, extending the relevance of bending moment diagrams to continuous beams and frames in multi-story buildings.

Modern • Computational Verification

Today, finite element software generates moment diagrams automatically, yet hand-drawn diagrams remain indispensable for validating computer output, developing engineering intuition, and passing professional licensing examinations.

The central question that bending moment diagrams answer is deceptively simple: At every cross-section along a beam, what is the magnitude and sign of the internal bending moment? Knowing this distribution allows an engineer to determine maximum stresses, predict deflection shapes, and ensure that a structural member has adequate capacity throughout its span.

SECTION 2

Core Principles & Definitions

Constructing a bending moment diagram rests on a handful of foundational ideas drawn from equilibrium and the method of sections. Before diving into calculations, it is essential to internalize these principles, since every step of the construction procedure follows from them. The internal bending moment at a section is the resultant couple that the material on one side of an imaginary cut exerts on the material on the other side, tending to bend the beam about a transverse axis. Its sign convention, relationship to the shear force, and dependence on boundary conditions form the backbone of the analysis.

Method of Sections

An imaginary cut is made at position x along the beam. The exposed internal forces (normal force N, shear V, and moment M) are found by enforcing equilibrium on the free-body diagram of one side of the cut.

Sign Convention

The standard beam sign convention defines a positive bending moment as one that causes the beam to sag (concave up), producing compression on top and tension on the bottom.

Shear–Moment Relationship

The differential relationship dM/dx = V(x) means the slope of the moment diagram at any point equals the shear force at that point. This integral connection links the two diagrams.

Boundary Conditions

At a pin or roller support, the bending moment is zero (no rotational restraint). At a free end, M = 0 as well. At a fixed support, a reaction moment develops and must be computed from equilibrium.

Area Under the Shear Diagram

The change in bending moment between two points equals the area under the shear diagram between those points: ΔM = ∫V dx. This graphical integration is the fastest way to sketch M(x) once V(x) is known.

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KEY TAKEAWAY

Think of the bending moment diagram as a topographic map of internal stress along the beam. Just as a hiker reads a contour map to find the steepest climb (maximum gradient), an engineer reads the moment diagram to find the section of maximum internal moment—the critical location where flexural stress is greatest and where the beam is most likely to yield or fracture. The shear diagram acts like the slope field: its value at any point tells you how steeply the moment curve is rising or falling.

SECTION 3

Visual Explanation — Simply Supported Beam with Point Load

The most instructive starting point is a simply supported beam carrying a single concentrated load P at its midspan. This classic loading produces a triangular moment diagram that peaks directly beneath the load. The diagram below shows the beam, its support reactions, the resulting shear force diagram (SFD), and the bending moment diagram (BMD) stacked vertically for direct visual comparison.

A simply supported beam of span L carries a concentrated load P at midspan. The reactions are each P/2. The shear force diagram (SFD) is a step function that jumps from +P/2 to −P/2 at the load point. The bending moment diagram (BMD) is triangular with a peak value of PL/4 at midspan. Notice that the linear segments of the BMD correspond to the constant shear in each half, consistent with dM/dx = V.

Several features of this canonical diagram deserve explicit attention. First, the moment is zero at both supports because pin and roller connections transmit forces but not couples. Second, the BMD is piecewise linear because the shear is piecewise constant (no distributed load). Third, the maximum moment occurs where the shear crosses zero, which is directly beneath the applied load. This observation generalizes: wherever V(x) = 0 (and dM/dx therefore equals zero), the moment diagram reaches a local extremum. Understanding this one diagram deeply provides the template for constructing BMDs for more complex loadings.

SECTION 4

Mathematical Framework

The mathematical foundation of bending moment diagrams rests on the equilibrium equations applied to an infinitesimal beam element of length dx subjected to a distributed load w(x). Summing forces vertically and moments about one end of the element yields two coupled ordinary differential equations that relate the applied load, the internal shear, and the internal moment. These relations, along with appropriate boundary conditions, allow the engineer to construct V(x) and M(x) either analytically or by graphical integration.

LOAD–SHEAR RELATION

dV/dx = −w(x)

V = internal shear force, w(x) = distributed load intensity (positive upward), x = position along the beam. Integrating: V(x₂) − V(x₁) = −∫ w(x) dx over [x₁, x₂].

SHEAR–MOMENT RELATION

dM/dx = V(x)

M = internal bending moment. The slope of the moment diagram at any point equals the shear at that point. Integrating: M(x₂) − M(x₁) = ∫ V(x) dx over [x₁, x₂], i.e., the change in moment equals the area under the shear diagram.

COMBINED LOAD–MOMENT RELATION

d²M/dx² = −w(x)

Combining the two first-order relations yields a single second-order ODE. For a uniformly distributed load w = const, M(x) is a second-degree polynomial (parabola). For no load (w = 0), M(x) is linear.

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Jump Discontinuities

At a concentrated load P, the shear diagram has a jump discontinuity of magnitude P, while the moment diagram has a slope discontinuity (a kink). At an applied couple M₀, the moment diagram has a jump discontinuity of magnitude M₀, while the shear diagram is unaffected. These rules are essential when sketching diagrams for beams with multiple concentrated loads and couples.

Practically, the construction procedure follows a systematic workflow. First, compute all support reactions using global equilibrium (ΣF = 0, ΣM = 0). Next, draw the shear diagram from left to right using the load–shear relation: the SFD starts at the left reaction, decreases under downward distributed loads, and jumps at concentrated forces. Finally, integrate the SFD to produce the BMD using the shear–moment relation. Alternatively, one can write the moment function M(x) directly by taking moments about the cut section for each loading segment—an approach that is especially transparent for beams with only point loads.

SECTION 5

Shape Rules for Common Loadings

An experienced engineer can sketch accurate moment diagrams almost by inspection by memorizing the shape rules that link loading type to diagram shape. Since dV/dx = −w and dM/dx = V, the degree of M(x) is always one higher than that of V(x), which is itself one higher than the degree of w(x). The table below catalogs these relationships, and the SVG diagram illustrates the three canonical cases side by side.

Loading Type	w(x)	V(x) Shape	M(x) Shape
No load (unloaded segment)	0	Constant (horizontal line)	Linear (straight sloped line)
Uniform distributed load (UDL)	w₀ = constant	Linear (sloped line)	Parabolic (2nd degree)
Triangular distributed load	Linear in x	Parabolic (2nd degree)	Cubic (3rd degree)
Concentrated point load P	Dirac delta	Jump of magnitude P	Slope change (kink)
Applied couple M₀	—	No change	Jump of magnitude M₀

Degree-ladder: load → shear → moment

Three canonical beam cases illustrating the degree-ladder rule: no load → constant V → linear M; constant load → linear V → parabolic M; linear load → parabolic V → cubic M. The curvature of the parabolic moment diagram under a UDL is concave toward the load.

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Curvature Rule

Under a downward distributed load, the moment diagram is concave downward (opening toward the load) when using the positive-sagging convention with M plotted below the baseline. This follows directly from d²M/dx² = −w: for w > 0 (downward), the second derivative is negative, meaning the curve is concave down. Use this as a quick check when sketching parabolic segments.

SECTION 6

Worked Example — Simply Supported Beam with UDL and Point Load

Consider a simply supported beam of length L = 6 m. It carries a uniformly distributed load w = 2 kN/m over its entire span and a concentrated load P = 6 kN applied at x = 4 m from the left support A. Support A is a pin and support B (at x = 6 m) is a roller. We wish to determine the support reactions, draw the shear force and bending moment diagrams, and identify the location and magnitude of the maximum bending moment.

BMD Construction — Combined Loading

1
Step 1 — Compute Support Reactions

Draw the global free-body diagram with reaction forces A_y (upward at A) and B_y (upward at B). Sum moments about A: ΣM_A = 0 → B_y × 6 − (2 × 6) × 3 − 6 × 4 = 0 → 6B_y = 36 + 24 = 60 → B_y = 10 kN. Then ΣF_y = 0 → A_y + 10 − 12 − 6 = 0 → A_y = 8 kN.

A_y = 8 kN ↑, B_y = 10 kN ↑
2
Step 2 — Construct the Shear Force Diagram

Starting from the left: at x = 0⁺, V jumps to +8 kN (reaction A_y). The UDL of 2 kN/m causes V to decrease linearly. At x = 4⁻ m, V = 8 − 2(4) = 0 kN. At x = 4 m, the concentrated load P = 6 kN acts downward, so V drops by 6 kN to V = 0 − 6 = −6 kN. From x = 4 to x = 6, V continues to decrease linearly: V(6⁻) = −6 − 2(2) = −10 kN. At B, the reaction B_y = +10 kN closes V to zero. ✓

V crosses zero at x = 4 m (from above, just before the point load).
3
Step 3 — Construct the Bending Moment Diagram (0 ≤ x ≤ 4 m)

M(0) = 0 (pin support). The moment is obtained by integrating the shear: M(x) = 8x − 2x²/2 = 8x − x² for 0 ≤ x ≤ 4. At x = 4: M(4) = 8(4) − (4)² = 32 − 16 = 16 kN·m. Since V is positive and decreasing linearly in this interval, M is a concave-down parabola increasing from 0 to 16 kN·m. The shear reaches zero at x = 4, confirming this is a local maximum of M in this segment.

M(4) = 16 kN·m (local maximum)
4
Step 4 — Continue the BMD (4 m < x ≤ 6 m)

For x > 4, rewrite using a coordinate shift or continue integrating. At x = 4⁺ the moment is still 16 kN·m (moment is continuous even at a point load). For 4 ≤ x ≤ 6, M(x) = 16 + ∫₄ˣ V(t) dt where V(t) = −6 − 2(t − 4). So M(x) = 16 − 6(x − 4) − (x − 4)². At x = 6: M(6) = 16 − 6(2) − (2)² = 16 − 12 − 4 = 0 kN·m. ✓ (This matches the roller boundary condition at B.) In this interval, V is negative throughout, so M decreases. The moment curve is again parabolic (concave down).

M(6) = 0 kN·m ✓ — boundary condition satisfied
5
Step 5 — Identify the Maximum Bending Moment

The shear diagram shows that V = 0 only at x = 4 m (ignoring the endpoints). Therefore, the maximum bending moment occurs at x = 4 m with a magnitude of M_max = 16 kN·m. This is the critical section for flexural design. The parabolic shape in each segment can be verified by checking the curvature: d²M/dx² = −w = −2 kN/m < 0, confirming a concave-down parabola throughout.

M_max = 16 kN·m at x = 4 m

SECTION 7

Strengths, Limitations, & Common Pitfalls

Strengths	Limitations	Common Pitfalls
Provides a complete visual of M(x) along the entire span, making it easy to locate the critical section.	Assumes linear elastic behavior and small deformations; does not capture plasticity or large-deflection nonlinearities.	Forgetting to include the reaction moment at a fixed support, leading to incorrect starting values.
Graphical integration from the SFD is fast and reveals qualitative features (concavity, extrema) at a glance.	Applicable only to statically determinate beams by statics alone; indeterminate beams require compatibility equations.	Mixing up sign conventions (especially plotting positive M above vs. below the baseline).
Directly usable with the flexure formula σ = My/I to find maximum normal stresses.	Does not show shear stress distribution or axial effects; a separate SFD and normal force diagram are needed.	Treating a couple (moment) load as a force—couples cause a jump in M, not in V.
Helps identify points of inflection where the moment changes sign, important for reinforcement placement in concrete design.	For complex geometries or 3-D frames, hand-drawn diagrams become unwieldy; FEA is preferred.	Incorrect parabola concavity under distributed loads—always verify with d²M/dx² = −w.

Strengths, limitations, and pitfalls of bending moment diagrams

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ENGINEERING JUDGMENT

Bending moment diagrams are to structural engineers what electrocardiograms are to cardiologists: a single graph that reveals the health of the entire member. Just as an irregular heartbeat spike signals a problem, an unexpectedly large moment peak tells the engineer exactly where to add material or reinforcement. However, both tools require a trained eye—the diagram itself is only as reliable as the equilibrium analysis and loading assumptions that produced it. Always sanity-check the BMD by verifying boundary conditions (M = 0 at pins, rollers, and free ends) and confirming that the net area under the SFD accounts for the total moment change.

SECTION 8

Connection to Advanced Theory

The bending moment diagram developed in statics is the foundation for nearly every subsequent topic in structural and solid mechanics. In mechanics of materials, the flexure formula σ = My/I uses M(x) from the BMD to compute normal stresses at any fiber of the cross-section. In beam deflection theory, the Euler–Bernoulli equation EI·d²v/dx² = M(x) is integrated twice (with boundary conditions) to produce the deflection curve v(x). Thus, a correctly drawn BMD is the essential input to computing both stress and displacement.

Concept in Statics	Advanced Extension	Key Relationship
M(x) from equilibrium	Flexural stress analysis (Mechanics of Materials)	σ = My/I — bending stress is proportional to M
Shear–moment relation dM/dx = V	Euler–Bernoulli beam deflection	EI d²v/dx² = M(x), double integration for deflection
Shape rules for determinate beams	Moment distribution for indeterminate structures	Fixed-end moments + carry-over factors extend BMDs to continuous beams
Location of M_max	Plastic analysis and plastic hinge formation	Plastic hinges form where M reaches the plastic moment $M_p$
Principle of superposition for loadings	Influence lines and moving loads	Influence line ordinate × load = contribution to M at a section

From statics to advanced structural analysis

Looking forward, students will encounter moment envelopes in structural design courses, which show the range of bending moments a section can experience under all possible load combinations. Mastering the construction of BMDs for individual load cases is the prerequisite for generating these envelopes, whether by hand (superposition) or by computer (load combination analysis). The conceptual toolkit built here—sign conventions, shape rules, and the integral connection between V and M—transfers directly.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

A simply supported beam carries a uniformly distributed load over its entire span. Without performing any calculations, describe the shape of the bending moment diagram and explain why the maximum moment occurs at midspan. What mathematical property of M(x) confirms this location?

Show answer

Under a UDL on a simply supported beam, the shear diagram is linear, starting positive at the left reaction and ending negative at the right reaction. The moment diagram, being the integral of a linear function, is a second-degree parabola (concave down). The maximum moment occurs where the shear is zero. By symmetry of loading and supports, V = 0 at midspan (x = L/2). Mathematically, dM/dx = V = 0 at midspan, and d²M/dx² = −w < 0 confirms this is a maximum. The peak value is wL²/8.

Question 2: BASIC CALCULATION

A simply supported beam of length L = 8 m carries a single concentrated load P = 12 kN at a distance a = 3 m from the left support. Determine the support reactions and the maximum bending moment, and state where it occurs.

Show answer

Taking moments about A:

B_y

× 8 = 12 × 3, so

B_y

= 4.5 kN. From Σ

F_y

= 0:

A_y

= 12 − 4.5 = 7.5 kN. For a single point load on a simply supported beam, the maximum moment occurs directly under the load. Using the left segment: M(3) =

A_y

× 3 = 7.5 × 3 = 22.5 kN·m. Alternatively, M_max = Pab/L = 12 × 3 × 5 / 8 = 22.5 kN·m. The BMD is triangular with the peak at x = 3 m.

Question 3: INTERMEDIATE

A cantilever beam of length 5 m is fixed at its left end. It carries a UDL of w = 3 kN/m over the first 3 m (from the fixed end) and a concentrated load P = 4 kN at the free end (x = 5 m). Draw the shear and moment diagrams and find the maximum bending moment and its location.

Show answer

Starting from the free end (often easier for cantilevers): At x = 5 m, V = 0, M = 0. Moving leftward, V increases by P = 4 kN at x = 5, so V = −4 kN (using standard sign convention, or +4 kN if analyzing from the right). From x = 5 to x = 3 (length 2 m, no load), V is constant at −4 kN and M increases linearly: M(3) from the right = −4 × 2 = −8 kN·m. From x = 3 to x = 0, the UDL adds: V(0) = −4 − 3(3) = −13 kN. M(0) from the right = −8 − 4(3) − 3(3)(1.5) = −8 − 12 − 13.5 = −33.5 kN·m. The maximum bending moment magnitude is 33.5 kN·m at the fixed support (x = 0). The moment diagram is linear from x = 5 to x = 3 and parabolic from x = 3 to x = 0. The reaction at the fixed end provides V = 13 kN and M = 33.5 kN·m (hogging).

Question 4: APPLIED

A floor beam in a warehouse spans 10 m between pin supports. It carries a dead load (UDL) of 5 kN/m and a live load consisting of a concentrated force of 20 kN that can be placed anywhere on the span. Using superposition, determine the absolute maximum bending moment and its location. Which cross-section governs the flexural design?

Show answer

By superposition, M_total = M_UDL + M_point. For the UDL alone: M_max(UDL) = wL²/8 = 5(10)²/8 = 62.5 kN·m at midspan. For a single moving point load on a simply supported beam, the maximum moment occurs when the load is at midspan: M_max(P) = PL/4 = 20(10)/4 = 50 kN·m. By superposition, when P is at midspan, M_total(5 m) = 62.5 + 50 = 112.5 kN·m. To verify this is the absolute maximum, one can check the influence line for moment at midspan, which peaks precisely when the load is at midspan. Therefore, the critical section is at x = 5 m with M_max = 112.5 kN·m. This is the section that governs flexural design; the required section modulus is S ≥ M_max/σ_allow.

Question 5: CRITICAL THINKING

A simply supported beam carries two equal concentrated loads P, each placed symmetrically at a distance a from the nearer support (a < L/2). Show that the bending moment between the two loads is constant and equal to Pa. Explain the physical significance of this constant-moment region and why it is exploited in four-point bending tests of material specimens.

Show answer

By symmetry, each reaction is P. The shear diagram: V = +P from x = 0 to x = a, then V = 0 from x = a to x = L − a (between the loads), then V = −P from x = L − a to x = L. The moment diagram rises linearly from 0 to Pa over [0, a], remains constant at Pa over [a, L − a], and drops linearly back to 0 over [L − a, L]. Between the loads, V = 0, so dM/dx = 0 and M is constant at Pa. This constant-moment region is physically significant because it means there is pure bending with zero shear—the beam experiences only normal (flexural) stresses with no shear stresses in this zone. The four-point bending test exploits this: by testing a specimen in this configuration, the material's flexural strength is measured under a state of pure bending, eliminating the complicating effect of transverse shear and producing uniform stress across the test section for more reliable material characterization.

SUMMARY

Lesson Summary

Summary

A bending moment diagram is a graph of the internal bending moment M(x) plotted along the length of a beam, revealing exactly where and how severely the beam is being bent. Its construction relies on the method of sections and two fundamental differential relations: dV/dx = −w(x) and dM/dx = V(x). Together, these relations establish the degree-ladder rule: each integration step raises the polynomial degree by one—no load yields constant shear and linear moment; a constant (uniform) load yields linear shear and parabolic moment.

The systematic procedure is: (1) find support reactions from global equilibrium, (2) draw the shear force diagram using the load–shear relation, and (3) integrate the SFD to obtain the BMD, checking that boundary conditions are satisfied (M = 0 at pins, rollers, and free ends). The maximum bending moment occurs where the shear crosses zero or at a fixed support, and this critical section governs the flexural design of the member via σ = My/I. Mastering these diagrams builds the essential foundation for stress analysis, deflection computation, and advanced structural design.