The fixed-end moment is a column stress

The first moment-area theorem says that the area under a beam’s $M/EI$ diagram between two points is the change of slope between them. The second says that the first moment of that area about one point is how far that point deviates from the tangent at the other. On a cantilever the two numbers are a tip rotation and a tip deflection. On a member fixed at both ends they are both zero, because a fixed end forbids exactly those two things: the far end may not turn relative to the near one, and it may not move off the near one’s tangent.

Two theorems with the answer nothing are two equations, and a member fixed at both ends has two unknown end moments. Hardy Cross noticed in 1930 that those two equations are ones every engineer already solved daily, in a different setting, without thinking of them as equations at all.

Two theorems whose answers are nothing #

Release both fixed ends and the member becomes a simple span, with the familiar moment diagram $M_s$ under its load. Put the ends back and they each apply a moment, and since those two moments act on a span with no load between them, they add a moment that varies in a straight line from one end to the other. So the member’s real moment diagram is

$$M(x) = M_s(x) - m(x), \qquad m(x) \text{ linear}$$

and the whole problem is to find a straight line.

A straight line has two numbers in it, and the fixed ends supply two conditions. The first theorem, with its answer set to zero, says the area under $M/EI$ over the whole span is nothing:

$$\int_0^L \frac{M_s - m}{EI}\,dx = 0$$

The second, with its answer set to zero, says the first moment of that area is nothing too:

$$\int_0^L \frac{M_s - m}{EI}\,x\,dx = 0$$

That is a pair of simultaneous equations in the two numbers that define $m$, and it can be solved as a pair. Cross’s observation was that it does not need to be.

The same two equations, in a column #

Give a fictitious column a cross-section as long as the member, and make its width at each point $1/EI$. For a prismatic member that is a rectangle; for a haunched one it narrows where the member deepens. Call the column’s area $A = \int dx/EI$, find its centroid $\bar{x}$, and call its second moment about that centroid $I$.

Now load the column with the simple-span diagram, $M_s/EI$, as a distributed pressure. Its total is $P = \int M_s\,dx/EI$ and its moment about the centroid is $M_c = \int M_s(x-\bar{x})\,dx/EI$. Write the straight line with its origin at the column’s centroid, $m = \alpha + \beta(x - \bar{x})$, and the two fixed-end conditions become

$$P - \alpha A = 0, \qquad M_c - \beta I = 0$$

because the first moment of the column’s area about its own centroid is zero, and that cross term drops out of both. So $\alpha = P/A$ and $\beta = M_c/I$, and

$$m(x) = \frac{P}{A} + \frac{M_c\,(x-\bar{x})}{I}$$

That is the stress in a short column under an eccentric load, with $P/A$ for the direct part and $My/I$ for the bending part. The line the fixed ends add to the simple-span diagram is that stress, and the end moments are the column’s two edge stresses. The centroid of the analogous column is called the elastic centre, and it is the whole trick. A section’s neutral axis passes through its centroid for the same reason: measured from there, axial force and bending stop interfering with each other, and two simultaneous equations become two separate divisions.

A prismatic member, worked as a column #

The figure above is small enough to check by hand.

The column is a rectangle 6 long and 1 wide, so $A = 6$, its centroid is at 3, and $I = 6^3/12 = 18$. The load on it is the simple-span diagram for a load of 10 at 2: a triangle with a peak of $Pab/L = 13.33$ and an area of 40. The triangle’s centroid is at $(a + L)/3 = 2.667$, which is 0.333 to the left of the column’s centroid, so the load is eccentric and its moment about the centre is $40 \times 0.333 = 13.33$.

The column’s average stress is $P/A = 40/6 = 6.667$. Its bending stress at either edge is $M_c y/I = 13.33 \times 3/18 = 2.222$, adding on the side the load leans toward and subtracting on the other. So the edge stresses are $8.889$ at the left and $4.444$ at the right. The closed forms for a fixed-ended member are $Pab^2/L^2 = 10 \times 2 \times 16/36 = 8.889$ and $Pa^2b/L^2 = 4.444$. As fractions of $PL$ they are the 0.148 and 0.074 the figure prints.

Nothing in that calculation looked like compatibility. It was an area, a centroid, a second moment and an eccentricity, which are the four quantities a column check needs and the four a designer in 1930 computed without effort. The end nearer the load carries twice the moment of the far one because the load on the column leans toward that end, and the lean is the distance between two centroids.

The average height of a diagram #

The direct term has a reading of its own, and it turns two textbook formulae into one observation.

When the load on the column has no eccentricity — a symmetric load on a symmetric member — the bending term vanishes and both end moments are $P/A$. For a prismatic member $A$ is $L/EI$ and $P$ is the area of the simple-span diagram over $EI$, so $P/A$ is that area divided by the span. The fixed-end moment is the average height of the simple-span moment diagram. A central point load draws a triangle of height $PL/4$, whose average height is half that, $PL/8$. A uniform load draws a parabola of height $wL^2/8$, whose average height is two-thirds of that, $wL^2/12$. Neither formula needs to be remembered once the shapes of a triangle and a parabola are.

For a member whose stiffness varies, the average is weighted by $1/EI$, so the stiff parts of the member count for less. A haunch makes the ends stiff, and the ends are where the simple-span diagram is low; discounting the low parts of a diagram raises its average. That is the haunch result below, arrived at before any haunch is drawn.

The same reading says what happens when the load is not symmetric. The average height still sets the mean of the two end moments, and the eccentricity only divides that mean unequally between them. For the load at a third of the span, the average height of the triangle is $Pab/2L = 0.111\,PL$, and the two end moments, 0.148 and 0.074, average exactly that.

Why a column, and not the algebra #

The equations were always solvable, so it is fair to ask what the analogy adds.

It adds a choice of origin. Written about an arbitrary point, the two fixed-end conditions are coupled: each unknown appears in both. Written about the elastic centre they are not, and each unknown is found by one division. That is the same simplification that makes a section’s second moment about its centroid the useful one. Cross’s analogy imports it into a place nobody had thought to look for a centroid.

It also makes a haunched member no harder than a prismatic one. A member whose depth changes along its length has an $EI$ that varies, and the algebraic route turns into integrals of awkward functions coupled together. The column route needs the same four properties of a column whose width happens to vary, and computing the area, centroid and second moment of an irregular strip was ordinary work on a drawing board.

A haunch narrows the column #

Reinforced concrete frames are routinely deepened at their supports, where the hogging moment is largest, and that is exactly the member a prismatic formula cannot handle.

A straight haunch of rectangular section doubles the depth, which multiplies the second moment by eight, which divides the column’s width by eight at the very end and by less along the taper. The column has lost most of its area near its ends.

A symmetric member under a symmetric load puts no eccentricity on its column, so the bending term vanishes, the line is flat, and both end moments are simply $P/A$. The haunch removed width near the ends, which is where the simple-span parabola is smallest, so it cut the column’s area much more than it cut the load on it. $P/A$ rises. Taken off the same parabola, the result is a member that carries more at its ends and less in its span.

Moment goes to where the member is stiff #

The effect is monotonic and it is not small.

A fixed-ended member’s end moments go to where it is stiff, and a haunch makes the ends stiff. That is the general rule of a redundant structure, whose load divides between its routes in proportion to their stiffness, met inside a single member rather than between several.

It has a design consequence that runs against a common reading of the drawing. A haunch is often described as relieving the support section, and it does not: it increases the moment there. What it does is increase the section’s capacity faster than it increases the moment. At twice the depth a rectangular section has four times the section modulus, and the end moment has grown by 19 per cent, so the bending stress at the face of the support has fallen to about 30 per cent of the prismatic member’s. The mid-span section, which was not deepened, is the one genuinely relieved, by 38 per cent.

The two curves add to $wL^2/8$ at every depth, because statics fixes the total of the end moment and the mid-span moment for a symmetric member under a uniform load. Only the division between them is the member’s to decide, and stiffness is how it decides.

The curves also flatten. A column that has already lost most of its width near its ends has little left to lose, so each further increment of haunch depth changes its properties less. The first half-depth of haunch moves the end moment from 0.083 to 0.094. The step from twice to three times the depth moves it from 0.099 to 0.104.

The elastic centre moves the wrong way #

A symmetric haunch keeps the elastic centre at mid-span. A haunch at one end does not, and the direction it moves is the least intuitive result of the method.

The deep end is the stiff end, and the elastic centre moves away from it, because the column’s width is the reciprocal of the stiffness. The column is thin where the member is strong. A central load on the member is therefore an eccentric load on the column, leaning toward the left, and the left edge stress rises accordingly: 0.223 $PL$ at the haunched end against 0.083 at the plain one. The haunch has drawn 78 per cent more moment to its own end and taken a third off the other.

The mid-span moment follows from statics. The simple span’s $PL/4$ less the average of the two end moments leaves 0.097 $PL$, about a fifth less than the prismatic member’s 0.125. Against the prismatic member, the haunched end has gained 0.098 $PL$, the plain end has lost 0.042 and the span 0.028. A one-sided haunch moves more moment away from the other end than away from the span, which a designer who deepens one support for its own sake should expect to see at the other.

The numbers a moment distribution needs #

Moment distribution needs two numbers for every member, at each end: the moment that turns that end through a unit rotation with the far end fixed, and the fraction of it that arrives at the far end. For a prismatic member they are $4EI/L$ and one half at both ends, and the method’s familiar arithmetic depends on it. A haunched member has neither.

The column analogy supplies them with nothing added to it. A unit rotation imposed at one end is a concentrated unit load on the column at that edge. The stress it produces at that edge is the stiffness, $1/A + \bar{x}^2/I$, and the stress at the other edge is the moment carried over.

The haunched end is stiffer, which nobody would doubt. The surprise is in the carry-over factors, and it is the elastic centre again. A moment applied at the shallow end carries 0.743 of itself over to the deep end, because the deep end is stiff and takes what it is offered. A moment applied at the deep end carries only 0.463 over to the shallow one. Neither is one half, and the direction matters.

What is the same in both directions is the product of stiffness and carry-over: $7.394 \times 0.463$ and $4.603 \times 0.743$ are both 3.42. That product is the moment arriving at the far end per unit rotation of the near one, and its symmetry is Maxwell’s reciprocal theorem, which says the far end’s response to turning the near end is the near end’s response to turning the far one. In the column it is simpler still. The stress at one edge from a unit load at the other is the same whichever edge carries the load, because $\bar{x}(L-\bar{x})$ does not care which factor comes first.

Engineers who designed haunched concrete frames by moment distribution used published tables of exactly these factors for standard haunch shapes. The tables were column properties, tabulated.

What the column extends to #

The analogy is not confined to a straight member. A closed frame, a portal with fixed feet or an arch fixed at its springings has three redundants rather than two, and its analogous column is the frame’s own outline drawn as a thin section with width $1/EI$ along it. The fixed-end conditions become a stress under axial load and bending about two axes, $P/A + M_x y/I_x + M_y x/I_y$ with a product of inertia if the section is not symmetric, and the three redundants are that column’s stresses at the supports. A support that settles or rotates, which produces moments with no load at all, enters the analogy as a load on the column too. An imposed movement gives the theorems a non-zero answer: a rotation $\theta$ of one end is a concentrated load $\theta$ on the column at that edge, and a settlement $\Delta$ of one end relative to the other is a pure couple of magnitude $\Delta$, because it changes the second theorem’s answer and not the first’s. A couple on a column produces bending stress and no direct stress. For a prismatic member the edge stress is $My/I = \Delta \cdot (L/2) \big/ (L^3/12EI) = 6EI\Delta/L^2$, equal and opposite at the two ends, which is the settlement moment of a fixed-ended beam, found in one line from the section modulus of a rectangle.

Where the model stops #

Bending flexibility only. The column’s width is $1/EI$ and nothing else. Shear deformation moves the moments of a redundant beam too, and a deep haunch is exactly where the shear term stops being negligible. A column with shear flexibility has extra terms that the simple stress formula does not carry.

A straight haunch of rectangular section. The second moment is taken as the cube of the depth, which is right for a solid rectangle. An I-section’s second moment grows more nearly as the square of its depth, because its flanges carry most of it, so a steel haunch of the same proportions attracts less moment than these figures show.

Elastic and uncracked. A reinforced concrete member’s stiffness depends on how cracked it is, and its supports usually crack first. A cracked support region is a wider column than the uncracked member drawn here, and the moment moves back toward the span. An average stiffness is not a safe one for the same reason.

Fixed means fixed. A real support that rotates a little is between the fixed end these figures assume and a pinned one. The analogy handles it as a load on the column, but only if the support’s own stiffness is known.

Still open: the portal frame drawn as a column section #

The member here is straight, and the analogy’s full reach is the closed frame. A single-bay portal with fixed feet, drawn as a three-sided column section, has an elastic centre inside the frame, nearer the beam or nearer the feet according to how stiff the columns are, and its horizontal thrust under a gravity load comes out of that section’s bending about the horizontal axis through the centre. After that come the fixed arch, whose elastic centre sits above its springings and whose thrust line is a column stress. The settlement of one foot of a portal, as a single concentrated load on its column section. And the conjugate beam and the column analogy set side by side, since the conjugate beam turns the two theorems into a beam and the analogy turns them into a column, and the question is which of the two a frame with several bays still leaves solvable by drawing.