The diagram is an integral, and that is why it can be drawn by eye

Three curves are drawn one above another for every beam ever analysed, and they are usually presented as three separate calculations. They are not. They are one function and its two integrals, and the relationship is exact.

$$\frac{dV}{dx} = -w, \qquad \frac{dM}{dx} = V.$$

The shear is the integral of the applied load and the moment is the integral of the shear. Everything about the shapes follows from those two statements, including the rules that let the diagrams be sketched without arithmetic.

Why the relations hold #

Take a slice of beam of length $\mathrm{d}x$ and write its equilibrium.

Vertically, the shear on one face, the shear on the other, and the load applied over the slice must balance. The difference between the two shears is the load carried over that length, which in the limit is $\mathrm{d}V = -w\,\mathrm{d}x$.

Rotationally, taking moments about one face, the moment on the far face differs from the moment on the near one by the shear times the distance, which in the limit is $\mathrm{d}M = V\,\mathrm{d}x$. The load’s own moment about the face is second-order in $\mathrm{d}x$ and drops out.

Two lines of algebra on an infinitesimal slice, and every rule below follows.

The rules that fall out #

Where there is no load, the shear is constant. The integral of zero is a constant, so the shear diagram is horizontal between point loads.

Under a uniform load, the shear slopes. The integral of a constant is a straight line, and its gradient is the load intensity.

At a point load, the shear jumps. A finite force applied over zero length is a step, and its size is the load. This is the one place the diagram is discontinuous.

The moment is one degree higher. Constant shear integrates to a linear moment; sloping shear integrates to a parabola. A beam under point loads has a moment diagram made of straight lines; the same beam under a uniform load has a parabola.

The moment peaks where the shear crosses zero. The moment is stationary where its derivative vanishes, and its derivative is the shear. This is the single most useful rule in the subject: to find where a beam is worst loaded, find where the shear changes sign.

The middle region of that second figure is worth a moment’s attention. Between two symmetric loads the shear is zero, so the moment is constant. That is pure bending: a length of beam with a moment and no shear, which is exactly what a four-point bending test creates deliberately in a laboratory.

The area rule #

Integration between limits is the area under a curve, so the change in moment between any two stations is the area under the shear diagram between them.

$$M_2 - M_1 = \int_{x_1}^{x_2} V \, dx.$$

That converts the moment diagram into an exercise in adding up rectangles and triangles, which is how it was done before it was done by machine, and how a check is still done by anyone who does not trust the output.

For a simply supported beam with a central point load $P$ and span $L$: the shear is $P/2$ over the first half, so the area under it is $P L/4$, and the moment at mid-span is $PL/4$. No calculus, one rectangle.

For a uniform load $w$: the shear falls linearly from $wL/2$ to zero at mid-span, so the area is a triangle of $wL^2/8$. Two familiar results in two lines each.

The same rule works backwards for checking. If a moment diagram is drawn with a peak in the wrong place, the shear diagram’s zero crossing will not agree with it, and one of the two is wrong.

A load that is not uniform #

The rules above are stated for point loads and uniform ones, which between them cover most of a floor. The general statement is a degree count, and it is worth having because the exceptions are the loads whose peaks are not where anyone expects.

A load varying as $x^n$ gives a shear of degree $n+1$ and a moment of degree $n+2$. Uniform is $n=0$, hence the linear shear and parabolic moment. A hydrostatic pressure on a retaining wall, a snow drift against a parapet, the self-weight of a tapering member: all are $n=1$, so the shear is a parabola and the moment a cubic.

Take a triangular load rising from nothing at the left to $w_0$ at the right over a span $L$. The total is $w_0L/2$, the reactions are a third and two thirds of it, and the shear is

$$V(x) = \frac{w_0 L}{6} - \frac{w_0 x^2}{2L}$$

which vanishes at $x = L/\sqrt{3} = 0.577L$. The moment there is $w_0L^2/(9\sqrt{3}) = 0.0642\,w_0L^2$.

Now compare that with the uniform load carrying the same total weight, which gives $WL/8 = 0.0625\,w_0L^2$ at mid-span. The magnitudes differ by 2.7% and the locations differ by 8% of the span — and the second difference is the one that costs something. A designer who has replaced a triangular load by its uniform equivalent has a peak moment that is very nearly right and a peak position that is wrong by almost a tenth of the span, which is where the reinforcement stops, where the haunch ends, and where the splice was going to go.

The degree count also explains why sketching works as well as it does. Each integration smooths, so the moment diagram is two degrees gentler than the load that produced it: a load with a kink gives a shear with a kink and a moment with none visible. A moment diagram sketched from a roughly-known load is therefore more trustworthy than the shear diagram between them — which is the reverse of the usual anxiety, since the shear is the one carrying every discontinuity the loading has.

Two checks the diagram has to pass #

Because the diagrams are integrals, they have to close. That gives two tests that a drawn diagram either survives or fails, and a diagram that fails one of them is wrong in a way no amount of looking at it will reveal.

The shear must return to zero at the right-hand end. The shear at any station is the sum of everything vertical to the left of it. At the far end of the beam, everything vertical has been passed — every load and every reaction — and equilibrium of the whole beam says that sum is zero. So a shear diagram that ends anywhere other than on its baseline has lost a force, and the size of the gap is exactly the force lost.

The moment must return to its known end value. At a free end or a simple support the moment is zero, so the total area under the shear diagram from one end of the beam to the other must be zero too: the positive area and the negative area have to be equal. A moment diagram that does not come back to zero at a simply supported end has either a reaction in the wrong place or an arithmetic error in the area, and again the residue names its own size.

These are not stylistic preferences. They are the statement that the beam is in equilibrium, arriving at the far end of a calculation that assumed it at the beginning, and a check that can fail is the only kind worth having. Every figure on this site that draws a shear and moment pair was produced by summing free bodies station by station, so both checks are satisfied by construction — which means their value here is not as a check on the generator but as a check on any diagram sketched by hand beside it.

There is a third test available for a symmetric beam under a symmetric load, and it is the cheapest of the lot: the shear diagram must be antisymmetric about mid-span and the moment diagram symmetric. An asymmetry in either, for a symmetric problem, is a mistake and not a subtlety.

The third discontinuity, which is a couple #

The rules above name one discontinuity — the step in the shear at a point load. There is a second, and it is the one that gets missed, because nothing about integrating the load hints that it exists.

An applied couple — a moment applied at a station, with no net force — produces a step in the moment diagram and no change at all in the shear. The slice argument shows why immediately: the vertical equilibrium of a slice containing a pure couple has nothing new in it, so $V$ is untouched, while the rotational equilibrium of that slice now has an extra term of finite size acting over zero length, so $M$ jumps by exactly the applied couple.

Applied couples are less exotic than they sound. A column framing eccentrically into a beam applies one. So does a bracket welded to the side of a girder, a cantilevered sign, and any connection where the load’s line of action misses the member’s centroid. The moment step in each of those cases is the load times the eccentricity, and it appears in the diagram as a vertical wall.

The design consequence is worth stating because it is the opposite of the usual instinct. The section immediately either side of an applied couple carries two different moments, differing by the whole of the applied value, and the more highly stressed of the two is not necessarily the one on the side the bracket is on. Both have to be checked, and the diagram is the only thing that says which.

Drawing it round a corner #

Everything above concerns a straight member lying horizontally, where “up” and “down” are unambiguous. A frame has members that are vertical, inclined and sometimes upside down relative to each other, and the sagging-positive convention quietly stops being usable at the first corner.

The profession’s answer is to abandon signs and adopt a rule that survives rotation: draw the moment diagram on the tension face. The ordinate is plotted on whichever side of the member the bending puts into tension, and no sign is written at all.

The rule has three virtues that the sign convention does not. It is invariant — turning the drawing upside down does not change which face is in tension. It is directly useful, because reinforcement goes on the tension face and the diagram then indicates where the steel goes without an intervening translation. And it makes the check at a joint immediate: two members meeting at a rigid corner must have moments that balance there, and drawn on their tension faces the two ordinates either line up on the inside of the corner or they do not.

A rigid corner in a portal frame is where this earns its keep. The beam’s hogging moment at the corner and the column’s moment at the top of the column are the same moment, transmitted through the joint, and the diagram runs continuously round the outside of the corner because the outside is the tension face on both members. If a drawn diagram shows the ordinate hopping from the outside of one member to the inside of the other, the joint is not in equilibrium and the analysis is wrong.

There is a cost to the convention, which is that it cannot be used inside a calculation — an integration needs signed values, and “tension face” is not a number. So the working is done in signs and the presentation is done on faces, and a good deal of confusion in the subject is the residue of moving between the two.

The cantilever, and the sign that stays negative #

A cantilever inverts most of the reading, and comparing it with a simple span is the fastest way to internalise the sign convention.

The free end has no reaction and nothing beyond it, so both shear and moment are zero there. Both grow toward the support, where the moment reaches $wL^2/2$ — four times the simply supported value for the same span and load.

The factor of four is why cantilevers are expensive and why they are used anyway: a cantilever needs no support at its far end, and eliminating a support is sometimes worth four times the moment. Moving the supports inboard is the compromise between the two, and it is a better deal than either extreme.

The moment being hogging throughout means the tension is in the top fibres for the entire length. A reinforced concrete cantilever has its steel in the top, and a balcony slab with the bars in the bottom is a well-known way to build something that falls off a building.

Reading the diagrams as instructions #

The diagrams are usually treated as results. They are more useful as instructions, because each one says what the beam needs at every station.

The moment diagram says how much section modulus is required. Where it peaks, the beam must be deepest or the reinforcement heaviest; where it is small, material can be removed. A haunched beam, thickened over its supports, is a moment diagram made physical.

The shear diagram says where the web works hardest and where shear reinforcement goes — which is near the supports, where the moment is small. The two demands peak in different places, which is convenient and not accidental.

The zero of the moment diagram — the point of contraflexure — says where the tension swaps faces. Reinforcement has to be lapped past it, and a splice placed exactly there is placed at the one station where the bar is doing nothing.

One more integration #

The chain does not stop at the moment. Integrating twice more gives the shape the beam takes.

So load, shear, moment, slope and deflection are one function and its four integrals, and the whole of elementary beam theory is that single chain read in either direction. Reading downward gives the analysis; reading upward — from a required deflected shape back to the load that produces it — is what precamber and form-finding do.

The free body is still doing all the work; the diagram is only a way of drawing several hundred of them at once.

Where the model stops #

Small deflections. The relations are written on the undeformed beam. Where deflection is large enough to change the geometry, an axial force starts contributing moment and the load amplifies itself.

No axial force. A beam-column has both, and the axial force starts contributing moment of its own.

Statically determinate. Every diagram on this page was obtained by statics. A continuous beam has diagrams that depend on the relative stiffness of its spans, and no amount of integration will produce them from the loads alone.

Concentrated loads are not concentrated. A point load is a fiction — a real load spreads over a bearing width, so the shear step is a steep ramp and the moment kink is a small curve. The fiction is harmless for the overall diagram and misleading right at the load, where the local stresses are the design case.

The figures share a specific distortion: the shear and moment diagrams are drawn at whatever vertical scale fits the canvas, and the two scales are unrelated. Comparing the height of a shear ordinate with that of a moment ordinate in these pictures means nothing at all, since the two quantities do not even have the same units. Only the shapes, the signs, and the positions of the peaks are comparable.

The ladder from here #

Later rungs: the differential relations derived from a slice. The area rule applied systematically. Diagrams for overhangs and internal hinges. Influence lines, which ask where a moving load produces the worst effect. Continuous beams and moment distribution. The relation between the moment diagram and the deflected shape. Plastic hinges and collapse mechanisms. And frames, where the same three quantities exist in every member and the diagrams are drawn around corners.

The relationships were known to Clapeyron by the 1850s and were being taught as graphical shortcuts long before anyone described them as calculus. The word “integral” appears in almost no nineteenth-century structural text, and the technique is entirely there.