The area of a diagram is a rotation

A beam’s deflected shape is the second integral of $M/EI$, and nobody has ever found the integration hard. The difficulty is entirely in the two constants: one for the slope, one for the position, and both of them fixed by conditions at the ends rather than by anything in the integrand. Set the problem up as algebra and the constants arrive as two simultaneous equations that carry no meaning. Set it up as geometry and they arrive as two statements about a picture that is already drawn.

Two numbers have come out of that picture, and neither of them was integrated for. 180 is $PL^2/2$, and 720 is $PL^3/3$. The first is the tip rotation and the second the tip deflection, and both were obtained by measuring an area and finding where its centre of gravity sits.

The two theorems, and why they are one theorem #

Write the elastic curve as $v'' = M/EI$ and integrate once between two stations $A$ and $B$:

$$\theta_B - \theta_A = \int_A^B \frac{M}{EI}\,dx$$

That is the first theorem and it is a statement about area. The change in slope of the elastic curve between two points is the area under the $M/EI$ diagram between them — not proportional to it, not approximately it, but exactly it, because the diagram is the derivative of the slope.

Integrate a second time, and the constant is chosen by measuring the deviation from a tangent rather than a datum:

$$t_{B/A} = \int_A^B \frac{M}{EI}\,(x_B - x)\,dx$$

which is the second theorem: the vertical distance from $B$ on the elastic curve to the tangent drawn at $A$ is the first moment of that same area about $B$. One area, taken twice — once plain, once weighted by distance — and the second is the first with a lever arm on it, exactly as a section’s second moment is its area weighted twice.

For a cantilever the second theorem is the whole answer, because the tangent at a built-in end is horizontal by definition and a deviation from a horizontal tangent is a deflection. That is why the hero figure needs no construction at all.

Which free body produced the number #

Cut the cantilever at a station $x$ and take the piece between the cut and the free tip. The only force on it is the tip load $P$ at a distance $(L-x)$, so the moment on the cut is $-P(L-x)$: linear, zero at the tip, $-PL$ at the root. That is the diagram in the figure, and it was drawn from the free body rather than recalled.

Its area is $\tfrac{1}{2} \cdot L \cdot PL = PL^2/2 = 180$, and the centroid of a triangle sits one third of the way from its wide end — so at $L/3 = 2.00$ m from the root. The first moment about the tip is therefore $180 \times (6 - 2) = 720$, and the closed form $PL^3/3 = 10 \times 216/3$ agrees exactly.

The figure prints both routes side by side: 720 from the geometry, 720.000 from integrating the curvature twice and applying the boundary conditions algebraically. The agreement is not a check that the method works; it is the observation that the two are the same operation.

The construction the method is famous for #

A simply supported span has no horizontal tangent anywhere, so the second theorem does not hand over a deflection. It hands over a deviation, and the deflection has to be assembled from two of them.

The construction runs like this. Draw the tangent at $A$. The far support $B$ sits on the beam, so its deflection is zero, and its deviation from that tangent is therefore the whole rise of the tangent over the span: $t_{B/A} = 1706.67$, and dividing by the span gives the slope at $A$, $\theta_A = 213.33$. The closed form is $wL^3/24 = 10 \times 512/24$, and it agrees.

Now go to mid-span. The tangent at $A$ has risen $213.33 \times 4 = 853.33$ by the time it gets there; the beam has deviated from that tangent by $t_{C/A} = 320.00$; and the deflection is the difference, $533.33$. The closed form is $5wL^4/384EI = 5 \times 10 \times 4096/384$, and it agrees to the twelfth figure.

Neither 853.33 nor 320.00 is a deflection. They are two geometric quantities whose difference happens to be one, and every reader who has found this method slippery has found it slippery here. The compensation is that the whole calculation is a subtraction of two areas’ first moments, done on a diagram that had to be drawn anyway.

The first theorem earns its keep on the same beam in a different way. The total area is 426.67; the two end slopes are $-213.33$ and $+213.33$; their difference is 426.67. So the area is not “the rotation” of anything — it is the change of rotation, and on a symmetric beam the two rotations are equal and opposite and the area is twice either of them.

The method against the method it replaced #

Virtual work asks a different question and gets the same answer.

The two methods divide the labour differently, and the division is worth naming because it survived a century of being taught side by side. Virtual work is best when one number is wanted at one place — it needs a second analysis for the unit load and then produces the answer with no construction. Moment-area is best when the shape is wanted, or when the beam is not prismatic, because a change of $EI$ is a change to the diagram being measured rather than a new case in an integral.

Moment distribution belongs to the same family of methods and answers a third question — the moments in a redundant frame — by iterating rather than integrating. All three date from the same fifty years, and all three exist because the alternative was solving simultaneous equations by hand.

The beam the method was made for #

A prismatic beam has closed forms and does not need a method. A stepped one does, and here the geometric formulation costs almost nothing: the diagram acquires a discontinuity, the area is computed in two pieces and added, and the centroid is the weighted mean of two centroids. The arithmetic that replaces it in an algebraic treatment is two integrations with two more constants and two matching conditions at the step.

The area lost is exactly half the tip triangle — $\tfrac{1}{2}\cdot 3 \cdot 30 = 45$, halved by the doubled stiffness, so $180 - 22.5 = 157.5$. And the centroid moves toward the root, because the stiffened half now contributes less. A haunched beam, a plated beam, a beam with a cover plate that stops short: all of them are this figure, and all of them are why the method outlived the slide rule that made it necessary.

Stiffening the outer half further does less than the first doubling did, and the diagram says why before any arithmetic is done.

The check that reaches for something else #

An exact method is worth checking against a second exact method, because what such a check tests is the arithmetic rather than the theorem.

The sweep also contains something nobody put in it. The deflection at mid-span under a load at $a$ traces a curve that is symmetric about mid-span in value while the underlying beam is not symmetric in any other respect — a load at 1 m and a load at 7 m give the same mid-span deflection to the last figure.

That is Maxwell’s reciprocal theorem arriving without being asked for. The deflection at $C$ from a load at $A$ equals the deflection at $A$ from a load at $C$, so the mid-span deflection under a load at $a$ equals the deflection at $a$ under a load at mid-span — and that function is symmetric because the beam and the load case are. The reciprocity was not built into either route; it fell out of both.

The moment-area construction is one of the few methods in which the reason for that is visible rather than merely provable. Both deflections are the first moment of the same product of two diagrams, and a product does not know which of its factors came first.

The agreement figure is worth one further reading. Its two curves are not two approximations to a third thing; they are two evaluations of the same integral by routes with different rounding behaviour. Where they disagree, at the thirteenth figure, the disagreement is the trapezium rule applied to a diagram with a kink in it under the point load — and it shrinks as the fourth power of the sampling interval, which is a statement about the arithmetic and about nothing structural at all.

Finding where the maximum is, which no closed form offers #

There is a question the standard formulae mostly decline to answer, and the geometric formulation answers it directly: where along the beam is the deflection largest?

The maximum is where the slope is zero, and the first theorem says the slope changes by the area under $M/EI$. So the station of maximum deflection is the one at which the accumulated area from the left support exactly equals the slope at that support — a “find where the running total reaches a given value” question, which on a drawn diagram is a matter of stepping along it and adding.

For a symmetric load it is mid-span and nobody needs the construction. For an asymmetric one it is not, and it is not where the moment peaks either. Under a single point load at distance $b$ from the far support, the maximum deflection sits at

$$x = \sqrt{\frac{L^2 - b^2}{3}}$$

from the near support — a station that depends on where the load is and coincides with it only when the load is central.

The numbers are the reason the question is usually ignored, and they are worth knowing rather than assuming. As the load moves from mid-span toward a support, the point of maximum deflection moves the other way — toward mid-span — and it never gets further than about 7.7 per cent of the span from the centre, however extreme the load position. And the deflection at mid-span is never less than 97.4 per cent of the true maximum, at any load position at all.

So the profession’s habit of quoting mid-span deflection for every case is not an approximation anybody should worry about; it is exact to within 2.6 per cent by a theorem. What the moment-area construction supplies is the reason that is safe — and the machinery to answer the same question on a beam where it is not, which is any beam whose $EI$ steps, whose supports are unequal, or which carries several loads whose diagram has no symmetry left in it.

What the picture is for #

There is a temptation to file this as a hand method superseded by arithmetic, and to keep it for the same reason graphic statics is kept: as history. That undersells it in one specific respect.

Which matters most where compatibility is the missing equation rather than a convenience.

The second theorem’s output is a geometric compatibility statement, and compatibility is the missing equation in every indeterminate structure. Setting the deviation of the prop from the tangent at the fixed end to zero gives $R = 3wL/8$ in one line of geometry, with no integration, no unit load and no simultaneous equations — and the same construction with a non-zero deviation gives the effect of a support that has settled directly, since a settlement is a prescribed deviation and nothing else.

The shape all of these numbers describe is one nobody has ever seen. A real 8 m beam at its serviceability limit has moved by about a three-hundredth of its span, which is thinner than the line any of these figures is drawn with, and every deflected shape on this page is exaggerated by two or three orders of magnitude before it becomes a picture at all.

Where the two constants went #

It is worth being explicit about the accounting, because the whole claim of the method is that it does not lose anything.

The double integral of $M/EI$ produces two constants. Algebraically they are fixed by two boundary conditions, written as equations and solved. Geometrically they are fixed by two choices: which tangent to measure from, and which point’s deviation to measure. Those are the same two pieces of information wearing different clothes, and the reason the geometric version feels like it has fewer steps is that the tangent is chosen before any arithmetic starts, at a place where the answer is already known.

For a cantilever the tangent is at the root and is horizontal, so one of the two choices is free and the method has one step. For a simply supported span neither end has a known slope, so both choices have to be made and the answer is a difference. The number of steps tracks the number of unknown end slopes exactly, which is a property no algebraic treatment makes visible.

The scaling is legible in the construction rather than in the formula. The diagram’s height goes as $wL^2$; its width as $L$; the lever arm in the first moment as $L$ again. Four factors of length, arrived at by counting rather than by expanding an integral, and the fourth power is why spans are short.

Where the model stops #

The diagram must be of $M/EI$, not of $M$. Every statement above is about the curvature diagram, and using the moment diagram instead is correct only for a prismatic beam of one material. The step in the figure above is the whole reminder: nothing about the moment changed there.

Curvature is $M/EI$ only while the material is elastic and the section is uncracked. Past first yield the diagram is not proportional to the moment at all, and the curvature has to come from the section rather than from a division. A cracked reinforced concrete member has an $EI$ that varies with the moment it is carrying, so the diagram to be measured depends on the answer.

Small deflections throughout. $v'' = M/EI$ is the linearised curvature; the exact expression divides by $(1+v'^2)^{3/2}$, and everything on this page is the first term of that. For the deflections a beam is allowed the approximation is superb, and for a member that is buckling it is not.

Shear is absent. The double integral of a moment does not contain the racking of the section, and for a deep member that missing term is most of the answer. Moment-area inherits the omission exactly, because it is the same integral.

Sign conventions bite harder here than anywhere. An area is positive or negative according to the sagging convention, a first moment takes its sign from both the area and the lever arm, and a construction that subtracts two of them has four opportunities to be wrong. The generator behind these figures integrates the sampled diagram directly and prints the plain double integral beside the geometric result, so a sign slip has somewhere to show.

What the pictures cannot show #

The shaded areas are drawn to a scale chosen so that the diagram fills its panel, and the number printed beside them is in the units of $M/EI$ — which for $EI = 1$ are the moment’s units and for any real beam are not. No area on this page has an area’s dimensions. The first theorem’s output is a rotation, in radians; the second theorem’s is a length. A reader who measures the shaded region with a ruler learns nothing.

The centroid mark is the second distortion. It is drawn as a point on the diagram, which suggests it is a station along the beam where something happens. Nothing happens there: it is a property of a shaded region, and moving the load moves it without any physical event occurring at the old position or the new one.

And the tangent construction, which is the whole method for a simply supported span, appears in none of these drawings — because a tangent to a curve exaggerated ten thousand times is a line whose slope is exaggerated ten thousand times, and drawing it would put it off the page.

The ladder from here #

Later rungs on this anchor: the conjugate beam, which turns both theorems into a single fictitious beam whose shear is the real beam’s slope and whose moment is its deflection, and the strange support conditions that fiction requires. Moment-area applied to frames, where the tangent has to be transported round a corner. The method on a beam of continuously varying depth, where the area is an integral again and the geometry buys only the constants. Elastic weights and the column analogy, which is the same idea pushed until it solves indeterminate frames. And the reason Otto Mohr found this at all: he was working on the graphical solution of trusses, where a displacement diagram is a drawing, and the beam theorems are that habit applied to a continuum.

Mohr published in 1868 and Charles Greene arrived at the same construction independently in the 1870s; both were teaching, and both wanted a method a class could carry out on a drawing board. That origin explains the shape of it. It is not an approximation made for speed — it is exact arithmetic reorganised so that the difficult parts are ones a draughtsman is already good at.