Load, shear, and moment relationships

Every beam that carries a transverse load — a floor joist, a bridge girder, a diving board — must resist that load internally by developing a shear force and a bending moment at each cross-section. Finding those internal quantities by repeatedly cutting the beam and writing equilibrium for every section is slow and error-prone once the loading gets complicated. There is a shortcut: the load intensity, the shear force, and the bending moment are tied together by two simple differential relations, so once you know how the load varies along the span you can sketch the entire shear-force and bending-moment picture by pure calculus bookkeeping. This article builds on the method of sections and the free-body idea already used to find $V$ and $M$ at a single cut. It develops the differential relations $dV/dx = -w(x)$ and $dM/dx = V(x)$ , shows how a concentrated force or couple produces a jump in the diagrams, and uses the condition $V=0$ to locate the section carrying the maximum bending moment — the section a designer actually cares about.

The core insight

The load intensity, shear force, and bending moment along a beam are linked by $dV/dx=-w(x)$ and $dM/dx=V(x)$ , so shear and moment diagrams can be built by tracking slopes and areas instead of cutting every section, with concentrated loads producing jumps in $V$ , concentrated couples producing jumps in $M$ , and the maximum moment always sitting where $V=0$ .

The core idea

Why a differential relation exists at all. Isolate a thin slice of beam of length $dx$ somewhere along the span, carrying a distributed load of intensity $w(x)$ (force per unit length, taken positive downward). Force equilibrium on that slice, and separately moment equilibrium about one face, connect the small change in shear $dV$ and the small change in moment $dM$ across the slice directly to $w(x)$ and $V(x)$ . Because the slice is infinitesimally short, those equilibrium statements collapse into derivatives rather than finite differences.

\frac{dV}{dx} = -w(x), \qquad \frac{dM}{dx} = V(x)

where: $V(x)$ is the internal shear force at position $x$ , $M(x)$ is the internal bending moment at $x$ , and $w(x)$ is the distributed load intensity (positive downward). The first equation says the shear diagram's slope at any point is minus the load intensity there; the second says the moment diagram's slope equals the shear there. Chain the two together and $d^2M/dx^2 = -w(x)$ — the moment diagram's curvature is set directly by the load.

Integrated (area) form. Because slope-equals-rate is just calculus, integrating between two sections $x_1$ and $x_2$ turns the differential relations into statements about areas, which is usually the faster way to work by hand:

V(x_2)-V(x_1) = -\int_{x_1}^{x_2} w(x)\,dx, \qquad M(x_2)-M(x_1) = \int_{x_1}^{x_2} V(x)\,dx

where: the first integral is the area under the load diagram between the two sections (with a sign flip), and the second is the area under the shear diagram between them. In words: the change in shear equals minus the area under the load curve, and the change in moment equals the area under the shear curve.

Jump conditions at concentrated actions. A concentrated force or couple acts over essentially zero length, so it cannot be represented by a smooth $w(x)$ — it shows up instead as a discontinuity. At a point where a downward concentrated force $P$ acts, the shear diagram drops by $P$ (a jump of $-P$ ), while the moment diagram stays continuous there, only changing slope. At a point where an applied couple $M_0$ acts, the moment diagram jumps by $M_0$ (sign per convention), while the shear diagram is completely unaffected, since a couple carries no net force.

Diagram shapes follow from the load shape. A zero load gives constant shear and a straight-line moment. A uniform load ( $w$ constant) gives a linearly varying shear and a parabolic moment. A linearly varying (triangular) load gives a parabolic shear and a cubic moment — each integration raises the polynomial degree by one. Recognising the shape in advance is a fast check on a hand-drawn diagram.

Figure

Shear and moment diagrams for a simply supported beam under a uniform load showing a straight line shear diagram crossing zero at midspan and a parabolic moment diagram peaking at midspan

Derivation ALocating and finding the maximum bending moment from the load

The goal is to show, starting only from the differential relation $dM/dx = V(x)$ together with the equilibrium reactions of a loaded beam, that the section of maximum bending moment always occurs exactly where the shear force is zero, and that its value can be obtained as the accumulated area under the shear diagram up to that section — without ever cutting the beam a second time.

Step 1. Start from $dM/dx = V(x)$ , the second differential relation. A function's derivative vanishes at its local maxima and minima, so any point where $M(x)$ is a stationary value (a candidate for the largest bending moment) must satisfy $dM/dx = 0$ , which by the relation above means $V(x) = 0$ at that same point. This links a purely geometric idea (a peak on the moment diagram) to a purely mechanical one (zero internal shear).

Step 2. Confirm it is a maximum, not a minimum or inflection, by checking the sign change of $V$ . Since $w(x)$ is typically positive (downward) over an ordinary span, $dV/dx = -w(x) < 0$ , so $V(x)$ is decreasing through the point where it crosses zero — it goes from positive to negative. Because $V = dM/dx$ , this means the slope of $M(x)$ goes from positive to negative there too, which is exactly the signature of a local maximum on the moment diagram.

Step 3. Compute the value of $M$ at that section using the area form of the relation rather than re-cutting the beam. Starting from a known value $M(x_1)=0$ at a free end or simple support, integrate the shear diagram up to the section $x^\ast$ where $V=0$ : $M(x^\ast) = M(x_1) + \int_{x_1}^{x^\ast} V(x)\,dx$ , which is just the area enclosed between the shear diagram and the axis over that stretch.

Step 4. Apply it to the uniformly loaded simply supported beam of span $L$ carrying $w$ : the shear starts at the reaction $V(0)=wL/2$ and falls linearly at rate $-w$ , so $V(x)=wL/2-wx$ , which is zero at $x^\ast=L/2$ — the midspan, matching intuition for a symmetric case. The enclosed triangular area up to that point gives $M_{max}=\tfrac{1}{2}\cdot\tfrac{wL}{2}\cdot\tfrac{L}{2}=\dfrac{wL^2}{8}$ , the standard result used to size beams under uniform loading. This general procedure — find where $V=0$ , then take the area under the shear curve up to that point — replaces cutting the beam entirely.

Worked example

A simply supported beam of span $L = 6\ \text{m}$ carries a uniformly distributed load $w = 4\ \text{kN/m}$ over its entire length. Find the reactions, sketch the shear force, and confirm the maximum bending moment.

Step 1. By symmetry each support carries half the total load: $R_A = R_B = \dfrac{wL}{2} = \dfrac{(4)(6)}{2} = 12\ \text{kN}$ .

Step 2. Taking $x$ from the left support, the shear starts at $V(0^+) = R_A = 12\ \text{kN}$ and falls at the constant rate $dV/dx = -w = -4\ \text{kN/m}$ dictated by the differential relation, giving $V(x) = 12 - 4x$ .

Step 3. Setting $V(x)=0$ gives $x = 3\ \text{m}$ , the midspan — exactly where Step 1 of the derivation predicts the moment is stationary. The maximum moment is the triangular area under the shear diagram from $0$ to $3$ : $M_{max}=\int_0^3(12-4x)\,dx=\left[12x-2x^2\right]_0^3=36-18=18\ \text{kN}\,\cdot\,\text{m}$ , matching the formula $wL^2/8=(4)(36)/8=18\ \text{kN}\,\cdot\,\text{m}$ from the derivation.

Step 4. As a companion case with no distributed load, a cantilever of length $L=3\ \text{m}$ fixed at the left end and carrying a concentrated load $P=10\ \text{kN}$ at its free right end has constant shear $V(x)=-10\ \text{kN}$ (no $w$ , so no slope) and a linearly varying moment $M(x)=-P(L-x)$ , reaching $M(0)=-PL=-(10)(3)=-30\ \text{kN}\,\cdot\,\text{m}$ at the fixed support, where the section is largest because that is where all the shear area has accumulated.

The beam sees a maximum shear of $12\ \text{kN}$ at the supports and a maximum bending moment of $18\ \text{kN}\,\cdot\,\text{m}$ at midspan, exactly the section where the shear diagram crosses zero.

Common traps

Forgetting that a concentrated load causes a jump (discontinuity) in the shear diagram, while the moment diagram stays continuous there and only changes slope.
Forgetting that a concentrated couple causes a jump in the moment diagram but leaves the shear diagram completely unaffected.
Assuming the maximum bending moment always sits at midspan or at a support — it occurs wherever $V=0$ , which depends entirely on the specific loading pattern.
Mixing sign conventions: which direction of load, shear, and moment counts as positive changes the signs in $dV/dx=-w$ and $dM/dx=V$ , so state one convention and hold it throughout a problem.

Ask the tutor: If the jump conditions at point loads and couples, or building a full shear and moment diagram for a mixed-loading beam, still feel shaky, ask the AI tutor to walk through one step by step with you.

Why does the shear diagram jump at a point load but the moment diagram does not?

A concentrated force changes the internal shear instantly, because equilibrium of a free body cut just past the load must include that extra force — hence the jump of magnitude equal to the load. The moment, however, is the accumulated area under the shear diagram up to that point; a single point contributes zero area no matter how large the shear jump is there, so the moment stays continuous, though its slope changes abruptly since $dM/dx=V$ changed.

Do I always need to integrate to build a shear or moment diagram, or can I just use the area rule?

The area rule is usually faster for hand sketching: start from a known reaction, and add up areas under the load diagram to track $V$ , then areas under the shear diagram to track $M$ , applying jumps at point loads and couples as you go. Formal integration is really the same operation, just written with calculus notation — use whichever is quicker for the load shape in front of you, but for triangular or curved loads, writing $w(x)$ and integrating directly is often cleaner than estimating areas by eye.

What happens to these relations if the load acts upward instead of downward?

Nothing about the relations themselves changes, only the sign of $w(x)$ that you substitute in. If you have defined downward loading as positive $w$ , an upward load is simply a negative $w(x)$ in $dV/dx=-w(x)$ , which flips the sign of the resulting shear slope and, through $dM/dx=V$ , the resulting moment behaviour too. The one thing that matters is picking a sign convention at the start of a problem and applying it consistently to every load, reaction, and diagram.

Load, shear, and moment relationships

The core idea

Worked example

Common traps

Before this

Where this leads