The moment that goes round the corner

A simply supported beam’s moment diagram begins at zero and ends at zero, and everything between is bounded by those two facts. Weld the same beam to a column at each end and the diagram has nowhere left to end. It arrives at the last section of the beam carrying a real number, and that number has to go somewhere — not into a support, because there is no support at the top of a portal, but into the column.

So it turns the corner. The ribbon drawn on the tension face of the beam runs off the end of it and continues, unbroken and at the same magnitude, down the outside of the column. Nothing is spent in the corner and nothing is lost there: what happens at the knee is a change of direction, in a quantity that was never directional to begin with.

The moment is continuous, and the shear turns into axial force #

Two things happen at the knee, and only the first of them is usually said out loud.

The moment is continuous. Equilibrium of the block between the two members makes what leaves equal what arrives — that is $\Sigma M = 0$ on the joint and nothing more. What makes the quantity large rather than zero is the kinematic half: a rigid joint preserves the angle between the members, so the beam end and the column top rotate together and neither can shed its moment by turning. At the frame above, 65.2 kNm arrives and 65.2 kNm leaves.

The shear is not continuous — it changes its name. The beam’s end shear of 80.0 kN is perpendicular to the beam, so at this corner it is vertical, and the column is vertical: the same 80.0 kN, handed across the joint unaltered, is the column’s axial compression. The column’s shear of 24.5 kN arrives horizontally, as axial force in the beam — a portal beam under pure gravity load carries 24.5 kN of compression nobody applied, which is the horizontal thrust of the frame and the same quantity an arch pushes out with.

Neither transformation is a physical process. Shear and axial force are components of one internal force vector resolved on the member’s own axes, and a cut is what brings them into existence at all. The joint is where the axes turn, and turning them swaps the two names.

Which free body produced the number #

Every quantity above comes from one free body: the knee itself, cut through the beam just clear of the corner and through the column just below it, with everything outside the cuts removed and replaced by what it was applying.

Three equations, written on that block:

$$\Sigma F_y = 0 \;\Longrightarrow\; N_{\text{column}} = V_{\text{beam}} = 80.0\ \text{kN}$$

$$\Sigma F_x = 0 \;\Longrightarrow\; N_{\text{beam}} = V_{\text{column}} = 24.5\ \text{kN}$$

$$\Sigma M = 0 \;\Longrightarrow\; M_{\text{column}} = M_{\text{beam}} = 65.2\ \text{kNm}$$

The residuals are the check that the picture is a solution rather than an illustration, and they are not zero by construction: the end forces come out of a global stiffness solve of the whole frame, resolved into global axes and summed on this one block. Getting −2.8e−14 kNm out of quantities of order 65 means the assembly and the free body are the same structure read twice. Continuity of moment at a rigid joint is not a modelling convenience. It is an equation, and it closes to the last digit the arithmetic has.

Choosing this particular body is the whole trick, and choosing the free body well is most of the skill. Cut through the beam a metre away instead and the same three equations still close, but they now carry a metre of distributed load and a metre of moment gradient, and the equality of the two couples is buried.

Adding a corner did not create moment; it moved it #

The beam’s own statics is untouched by any of this: 94.8 kNm at midspan plus 65.2 kNm at the corner is 160.0 kNm, which is $wL^2/8$ for 20 kN/m over 8 m, to a residual of exactly zero.

That identity is neither a coincidence nor an approximation. Cut a free body of the whole beam between its two ends: the reactions and the applied load give a free moment diagram, the parabola of a simply supported span peaking at $wL^2/8$, and the end couples the columns apply give a fixing diagram, here a straight line at constant 65.2 kNm because both ends are equal. The real moment is their sum, so at midspan

$$M_{\text{mid}} = \frac{wL^2}{8} - M_{\text{corner}} = 160.0 - 65.2 = 94.8\ \text{kNm}.$$

Whatever the corner takes, midspan gives up, one for one. No amount of joint stiffness changes the total: the sum of the sagging peak and the hogging end is $wL^2/8$ for any end restraint whatever, including none. This is the bargain a continuous beam strikes over its supports reached from a different direction — there the end moment appears because the neighbouring span pulls the end down, here because a column will not let it rotate.

How much goes round is a stiffness ratio, and nothing else #

If the columns were rigid walls, the beam would be built in at both ends and the corner would take $wL^2/12$ — 106.7 kNm here. It takes 65.2. The missing 41.5 kNm went back into the span, because the columns are not walls but springs, and a spring lets the beam end rotate.

How much rotation is decided by one number: the beam-to-column stiffness ratio

$$N = \frac{I_{\text{beam}}/L}{I_{\text{col}}/H},$$

which for this frame is 1.27. Slope-deflection then gives a closed form for a symmetric portal on fixed bases,

$$M_{\text{corner}} = \frac{wL^2}{12}\cdot\frac{2}{2+N},$$

and $106.7 \times 2/3.27 = 65.2$ kNm.

Two things about that curve deserve more than a glance, and the first is the agreement. The assembled solution comes from a global stiffness matrix that knows nothing about slope-deflection, and it reproduces a hand formula from the 1910s to within 4.1e−6 kNm — four millionths of a kilonewton-metre — across a factor of four hundred in $N$. An analysis that agrees with an independent derivation over its whole range has been tested; one that agrees at a single point has been calibrated.

The second is what is not in the formula. Not the span, not the height, not the load, not the modulus. Two portals of entirely different size share their corner moment in the same proportion if their $N$ matches — a 30 m shed and a 6 m garage put the same fraction of $wL^2/12$ into the knee.

That is a specific case of the rule the stiffest path takes the load: where members share a rotation rather than a force, relative stiffness decides the split. Stiffen the columns of a portal and the knee moment goes up, the reverse of the intuition that a stronger member is a safer one, and the same reversal a redundant beam produces when a support settles.

The ratio also assumes the joint between the two members is infinitely stiff, and that is a third spring nobody drew.

A knee detailed as rigid and built as something between pinned and rigid delivers a corner moment the frame analysis never computed — and the difference reappears at midspan, where nobody looked for it.

The panel is where the number lives #

So far the corner has been a point where two lines meet. At the scale of the steel it is a rectangle — a block of web 0.457 m deep, 0.254 m wide and 8.6 mm thick — and the moment does not turn through it by magic.

The beam hands its 65.2 kNm over as a couple in its flanges: 142.7 kN pulling on the top flange and 142.7 kN pushing on the bottom, 0.457 m apart. The column removes 24.5 kN of that as its own shear, and the difference has to cross the panel.

118.2 kN is on no member diagram. The beam’s shear diagram gives 80.0 kN at that section and the column’s gives 24.5. Every analysis package draws both, and neither contains the largest shear force anywhere near the joint. It exists only for the free body that is the panel alone, cut on all four sides — a body no member-level model has any reason to draw, because a connection modelled as a point has no interior to put it in.

It is the same phenomenon as the shear nobody draws inside a section — a flow demanded by the change of bending moment along a body. The difference is that here the change happens over 254 mm rather than over a span, so the flow is not a modest complement to the vertical shear but several times it.

Stiffeners in a moment connection are there for this number and for nothing else. Two horizontal plates at the flange levels stop the column flanges folding under 142.7 kN of point load; a diagonal stiffener or a doubler plate carries the 118.2 kN the panel cannot. A moment connection is not a stronger version of a shear connection. It is a shear connection with an extra force in it that the members never saw, which is the argument a moment crossing a gap makes from the bolts’ side.

Sway is the worse case, and the ratio nearly doubles #

Under gravity the beam’s shear is large, so the panel demand is 1.48 times a member shear that is already substantial. Take the gravity load off, push the frame sideways instead, and the arithmetic changes character.

Under sway the beam carries almost no shear: with no load on the span its shear is the sum of the two end moments divided by the length, 8.8 kN, against a column shear of 20.0 kN. The flange couple is still a couple, though — 35.4 kNm over 0.457 m is 77.4 kN — and 57.4 kN of it crosses the panel.

2.87 against 1.48 is the point. Sizing the members, gravity governs; sizing the panel, sway governs, in a load case where the beam’s own shear is almost nothing. The panel is not a scaled-up version of the member forces around it: it has its own governing case, visible only to whoever draws its own free body.

A haunch is not extra strength; it is a place to stop #

The corner moment is the largest in the frame and it lands where the connection has to be made, which is the worst possible arrangement. The traditional answer is a haunch: a wedge of deeper section running from the knee some way along the beam.

A haunch does not reduce the corner moment — 65.2 kNm still arrives. It moves the section that has to carry the plain beam’s capacity away from the corner, to where the moment has already collapsed.

It falls that fast because the beam’s shear at the corner is the largest it will ever be, 80.0 kN, and shear is the slope of the moment diagram. Ten per cent of the span buys an eighty-eight per cent reduction, and the return is steepest at the start.

The number that matters for detailing is the contraflexure point: 0.92 m from the corner along the beam, and 1.33 m above the base in the column. Those are the sections where the moment is zero, and a splice put there has almost nothing to carry. A haunch is not a strengthening measure. It is a statement about where the plain section is allowed to begin.

The column’s point earns a second look, because the corner moment can be recovered from it: the column’s shear times the height up to it, $24.5 \times (4 - 1.33) = 65.4$ kNm, which is 65.2 to the rounding of those two figures. The lever arm is not the column’s height, and nothing but a solve says where it is.

The method that was nothing but this equation, repeated #

The formula for $N$ above came from slope-deflection, which is a way of writing joint equilibrium in terms of joint rotations and solving the resulting equations. For a portal that is a couple of equations and can be done by hand. For a six-storey frame it is not.

Hardy Cross’s answer, published in 1930 in about ten pages, was to enforce joint continuity iteratively rather than simultaneously. Clamp every joint and compute the fixed-end moments. Release one joint, let it rotate until its moments balance, distribute the imbalance among the members in proportion to their stiffnesses — the same stiffnesses that appear in $N$ — carry half of each correction to the far end, clamp it again and move on.

What the method converges to is the state in which every joint’s moments sum to zero — the free body in the second figure on this page, insisted on at every joint at once. Passing the imbalance around is joint equilibrium applied one joint at a time until the passing stops. The distribution factors are relative stiffness, so anyone using the method was manipulating $N$ by hand, for every joint in the building, for some thirty years.

Where the model stops #

The joint is a point in the analysis and a rectangle in the steel. The frame solution puts the peak moment where the two centrelines cross, 127 mm inside the column face. The moment at the face — where the connection actually is — is lower by the shear times that offset: conservative for the connection, and misleading about where the beam’s plain section can start.

The panel is assumed rigid. A real panel shears, and its distortion adds directly to the frame’s sway. A frame whose panels have been sized to the last megapascal is a frame that sways more than the analysis says, which matters because sway makes its own load worse.

The panel is assumed not to buckle. 54 MPa over an 8.6 mm web is comfortable; the same panel in a deeper beam with a thinner column web is a plate in shear, and plates in shear ripple at a stress falling with the square of the slenderness.

Everything here is elastic. A frame designed plastically puts hinges at the knees on purpose, at which point the corner moment is $M_p$ and does not depend on $N$ at all: after the first yield the stiffness ratio stops deciding anything.

The columns are assumed identical and the frame symmetric. Two different columns rotate the joint under gravity load — which the closed form does not cover and the assembled solve does.

What the pictures on this page cannot show is the deformation that produces all of it. Every figure here is a force diagram, and the mechanism behind the numbers is a rotation: the beam end and the column top turning through one shared angle. The column’s slope-deflection equation gives it from the corner moment,

$$\theta = \frac{M_{\text{corner}}H}{4EI_{\text{col}}} = \frac{65.2 \times 4}{4 \times 23{,}940} = 2.72\ \text{milliradians},$$

which is the value the frame solve carries in its displacement vector, and which is invisible at any scale that keeps the frame on the page. Equilibrium says the two couples are equal; it does not say what either of them is. Only that rotation says that, and it is the one quantity in this essay that no figure on this page draws.

The ladder from here #

Later rungs on this anchor: the slope-deflection equations derived, and the corner moment obtained without a matrix. The unequal portal, where the two columns differ and the joint translates as well as rotates. The knee under combined gravity and sway, where panel and members are governed by different load combinations. Panel-zone flexibility as an explicit spring. The plastic knee, with the hinge placed in the beam beyond the haunch on purpose. Haunch proportioning against the moment curve rather than by rule of thumb. Column web stiffening, and when a doubler plate is cheaper than a heavier column. The interior joint, where two beams arrive and the panel demand roughly doubles for the same member forces. And the corner in reinforced concrete, where the reinforcement must be detailed for the diagonal tension a steel web takes in one piece.