The elastic centre is not on the frame

A member fixed at both ends is a short column under an eccentric load: a strip as long as the member and as wide as $1/EI$, carrying the simple-span moment diagram as a pressure, with the end moments as its two edge stresses. The strip is straight because the member is.

Nothing in that argument required it to be. The two conditions it rests on — that the area under $M/EI$ is a change of slope and that its first moment is a deviation — are properties of a curve rather than of a line, and a frame is a curve that happens to have corners in it.

The section is the frame’s own outline #

Draw the frame’s centreline and give it a width. At every point the width is $1/EI$ — wide where the frame is flexible, narrow where it is stiff — and the result is a thin section shaped like a three-sided box with its bottom missing.

That section has the four properties any section has. Its area is $A = \int ds/EI$, which for this frame is $2h/EI_c + L/EI_b$, or 16.0 for equal stiffnesses at these dimensions. Its centroid is the elastic centre. Its second moment about a horizontal axis through that centroid is $I_x = \int (y - \bar{y})^2\,ds/EI$. And it has a product of inertia, which is zero here because the frame is symmetric and which is what makes a non-symmetric frame harder rather than different.

The elastic centre comes out 3.44 m above the feet — 1.56 m below the beam, in the air inside the frame. It is not a point on the structure, nothing is there, and it is the origin every subsequent number is measured from.

Why an origin removes a simultaneous equation #

Cut the frame at the crown and it becomes two bent cantilevers, each fixed at its foot. The cut releases three actions; symmetry disposes of the shear; two remain, and they are an unknown moment and an unknown horizontal thrust.

Written about an arbitrary point those two are coupled: the moment produced by the thrust appears in the rotation condition, and the moment’s own effect appears in the displacement condition. Two unknowns, two equations, and the cross terms are what make it a system.

Applied at the elastic centre they are not coupled at all. The rotation condition is $\int M\,ds/EI = 0$, and a thrust applied at the centroid contributes $H\int(y-\bar{y})\,ds/EI$, which is zero by the definition of a centroid. The displacement condition is $\int M(y-\bar{y})\,ds/EI = 0$, and a pure couple contributes $M_0\int(y-\bar{y})\,ds/EI$, which is the same zero.

So each unknown is one division:

$$M_0 = -\frac{\int M_s\,ds/EI}{A}, \qquad H = -\frac{\int M_s\,(y-\bar{y})\,ds/EI}{I_x}$$

That is $P/A$ and $M y/I$ again, and it is the same simplification that makes a section’s neutral axis pass through its centroid. Cross’s contribution was to notice that a frame has a centroid too, if one is willing to call $1/EI$ a width.

The numbers fall out: $M_0 = 33.8$ kN·m and $H = 6.4$ kN for the frame above. The moment anywhere is then $M_s + M_0 + H(y - \bar{y})$ — the released diagram plus a constant plus a linear term.

Two routes that were never told about each other #

The hero figure’s caption ends by giving the same three moments from a stiffness solution: the frame assembled as three members with six degrees of freedom, solved as a linear system, and read out.

They agree to every figure printed. That agreement is the check the analogy needs, because the analogy’s whole claim is that a compatibility problem can be answered by computing four properties of a section, and an argument of that shape is exactly the kind that can be almost right. Neither route contains any part of the other: one integrates $1/EI$ around an outline, the other inverts a stiffness matrix.

The contraflexure that does not move #

Two exact relationships hide in those numbers and both are worth checking by eye.

The crown and the knee always add to the free moment. 17.8 and 27.2; 31.7 and 13.3; always 45.0. That is statics rather than stiffness — the beam is a simply supported span carrying $wL^2/8$ at its centre, and whatever hogging the knees supply is subtracted from it. Nothing about the stiffness ratio can change the sum, only the division.

And the base moment is exactly half the knee moment, opposite in sign. 10.6 against −21.2; 6.7 against −13.3; 53.3 against −106.7 on a squat frame of quite different proportions. The ratio is −0.500000 at every stiffness ratio, every height and every span tried.

That is not an accident of these numbers and it is not something the analogy was asked for. A column here carries no transverse load, so its moment is a straight line; its base cannot rotate; and the moment carried over to the fixed end of a prismatic member is half the moment applied at the other end, which is the carry-over factor moment distribution is built on. The point of contraflexure therefore sits at a third of the height from the base, and it sits there whatever the frame is made of.

It stops being true the moment a column stops being prismatic. Haunch the column at its knee and its carry-over is no longer a half — the essay below measured 0.463 one way and 0.743 the other on a member haunched at one end — and the third-height rule goes with it. A rule that holds for every stiffness ratio and fails for a taper is a rule about the member rather than about the frame, which is worth knowing before applying it to a portal with tapered legs.

The centre moves away from whatever is stiffened #

The direction is the one the essay below found on a straight member and it is easier to disbelieve here, because the frame is a picture of a structure and the intuition about stiff things attracting attention is strong.

The analogous column’s width is the reciprocal of the stiffness. Stiffen the beam and the section gets thinner along its top; a thinner top contributes less area to the centroid calculation; the centroid moves down. The elastic centre runs away from the stiff member, and it does so for the same reason a haunched member’s does.

The thrust falls with it, and the reason is a lever arm. The load acts on the beam; the frame resists its moment about the elastic centre; and the further the centre is from the beam, the longer the arm and the smaller the force that has to act on it. A portal with a very stiff beam and slender columns has a low elastic centre and almost no thrust — which is the pinned-base limit approached from an unexpected direction.

Height, which moves the area rather than the shape #

Taller columns add area to the section low down, which pulls the centroid further from the beam and lengthens the lever again. The thrust falls almost in proportion to the height, and the frame’s moments drift toward the simply supported ones.

That is the ordinary behaviour of a portal and it is worth having a mechanism for: a tall slender portal is nearly a beam on two pinned columns, and a squat stiff one is nearly a fixed-ended beam. The elastic centre’s height is where on that continuum a particular frame sits, and it is a single number computed from a section property rather than a judgement.

The whole of it, once, by hand #

The frame in the hero figure is small enough to run through completely, and doing it once is the argument for the method, because every step is a thing a designer already knows how to do.

The section. Two columns 5 m long and a beam 6 m long, all at $EI = 1$, so the band is 1 wide throughout and $A = 2(5) + 6 = 16$. Measuring height from the feet, the centroid is $[2(5)(2.5) + 6(5)]/16 = 55/16 = 3.4375$ m, which is 1.5625 m below the beam.

The second moment about that axis, treating each column as a strip and the beam as a lump: $2[(1.5625^3 + 3.4375^3)/3] + 6(1.5625)^2 = 29.62 + 14.65 = 44.27$. That is the arithmetic of a section’s own second moment with nothing new in it.

The load on the section. The released frame’s moment is $-wa^2/2$ along the beam, where $a$ is the distance from the crown, and a constant $-wL^2/8 = -45$ kN·m down each column. Its total is $-90 - 450 = -540$, and its first moment about the elastic centre is $+281.25$.

The two stresses. $M_0 = 540/16 = 33.75$ kN·m; $H = -281.25/44.27 = -6.35$ kN.

And the answers. At the feet, $-45 + 33.75 + 6.35(3.4375) = 10.6$ kN·m. At the knees, $-45 + 33.75 - 6.35(1.5625) = -21.2$. At the crown, $0 + 33.75 - 9.93 = 23.8$.

Six numbers, four of which are section properties, and no simultaneous equation anywhere. The check that catches a lost factor is the statics one: the crown and the knee have to add to $wL^2/8 = 45.0$, and $23.8 + 21.2$ does.

What it replaced, and what replaced it #

The analogy was published in 1930 and it is worth being clear about what the alternative was, because that is what makes a saving of this size a method rather than a curiosity.

The alternative was the force method: release three redundants, compute nine flexibility coefficients by integrating pairs of moment diagrams around the frame, and invert a three-by-three matrix. Each coefficient is a product integral of the kind the unit-load method uses, and on a frame each of them runs over three members. For a single-bay portal that is a long afternoon and for a multi-bay frame it is not attemptable.

The analogy’s economy is entirely the choice of origin. It computes the same flexibility matrix; it computes it about an axis where six of the nine terms are zero, so the matrix is diagonal and the inversion is three divisions. A matrix that is diagonal because of where its origin was put is the same trick as the neutral axis passing through a centroid, and it is what a section engineer had already internalised by 1930 without thinking of it as linear algebra.

What replaced it is a computer, which does not care whether a matrix is diagonal. That is a fair exchange and it costs the thing this essay is about: moment distribution and the column analogy both leave a reader with a picture of where a frame’s moment goes, and the answer to a stiffness matrix leaves a reader with six numbers. The elastic centre’s height is a design quantity — it says how near a frame is to pinned-base behaviour — and no modern output prints it.

Which free body produced the number #

The free body is half the frame, cut at the crown, and it is worth being careful about what is on its cut face.

Three actions cross a cut through a frame: an axial force, a shear and a moment. Symmetry kills the shear at the crown of a symmetric frame under a symmetric load, because the two halves must push on each other equally and there is no vertical action left over. The axial force at the crown is the thrust $H$, and the moment is $M_0$ — but $M_0$ as computed above is the moment at the elastic centre, not at the crown, and the two differ by $H$ times the distance between them.

That is the one bookkeeping trap in the method and it is worth stating plainly: the redundants are defined at a point that is not on the structure, so every value read off has to be transported back to wherever it is wanted. The transport is the $H(y - \bar{y})$ term, and it is the whole of the bending-stress half of the analogy.

The comparison figures load the released frame with the applied load alone. Three things are assumed by that and each is a real restriction:

Bending flexibility only. The section’s width is $1/EI$ and nothing else, so axial shortening of the columns and shear deformation of the beam are both absent. For a portal of ordinary proportions both are small; for a squat one the shear term is not.

Prismatic members. The width is constant along each member, which is what makes the integrals arithmetic rather than calculus. A haunch turns them back into integrals — soluble, and the reason the essay below spends its time on exactly that case.

And symmetric. An asymmetric frame has a product of inertia, and its two redundants are coupled again through a $I_{xy}$ term — the same $P/A + M_x y/I_x + M_y x/I_y$ with a cross term that a section with no axis of symmetry always carries.

What the picture cannot show #

Nothing here sways. Every load is symmetric, so the frame’s third redundant stays at zero and the whole analysis is two numbers. A frame that leans has a horizontal load or an asymmetric one, the shear at the crown is no longer zero, and the analogous column is being asked for a stress under a load with an eccentricity about both axes.

The feet are fixed, exactly. A portal on pad foundations is not, and a base that rotates a little sits between the fixed case here and a pinned one — which the analogy handles by adding a flexibility at the foot, as an extra area at the end of the section, and which nothing in these figures does. A base fixity of eighty per cent is common and it moves the base moment by more than any stiffness ratio drawn here.

And the section is a fiction with no thickness. The band drawn beside each frame is $1/EI$ wide and its width is in units of a reciprocal bending stiffness, which is not a length. Two frames drawn side by side at the same scale have bands whose widths mean nothing comparable unless their $EI$ are in the same units — the width is a weighting, and drawing it as a width is exactly the visual pun the method is built on.

The assumption that makes the pun work #

The analogy is not a metaphor; it is an observation that two problems have the same algebra. What makes them the same is that in both, a quantity is distributed along a line, two weighted integrals of it are set to zero, and the weights are $1$ and the distance from an origin.

In a column the quantity is a stress and the integrals are force and moment. In a frame the quantity is a curvature and the integrals are a rotation and a deviation. Neither problem knows about the other, and every property that carries across — the centroid removing the cross term, the second moment setting the gradient, the edge stress being the answer — carries across because of that shared shape and for no other reason.

Which is also the limit. A property of columns with no counterpart in the algebra does not transfer: a column has a kern, a limit beyond which one edge goes into tension, and an analogous column’s “tension” is a change of sign in a bending moment with no significance whatever.

Still open: the arch, whose elastic centre is above its springings #

The frame here has three straight members. Curve them and nothing in the method changes: the analogous column is still the centreline at a width of $1/EI$, and the integrals are still an area, a centroid and a second moment — computed round a curve instead of along three lines.

A fixed arch is that case, and its elastic centre sits above its springings, generally below the crown, on the axis of symmetry. The thrust line is then a stress on that section, which is where this method meets the graphical thrust line from the other side of the collection. After it come the settlement of one foot, which is a concentrated load on the analogous column rather than a load case at all; and the conjugate beam set beside the analogy, since one turns the two theorems into a beam and the other turns them into a column, and a frame of several bays leaves only one of them solvable by drawing.