The cut that needs a joint first

The method of sections answers one question about a truss in one equation. Cut the truss so that exactly three members are severed, keep one side as a free body, and take moments about the point where two of the three severed members meet. Those two pass through the point and have no moment about it, so the third is the only unknown left. Ritter published it in 1862 as a way of avoiding a whole graphical analysis to get one number, and it has been the fastest route to a single member force ever since.

The essay on the method said, near its end, that some frames have no three-member cut in some regions, and that for them the method has to be applied twice with an intermediate result, or abandoned. That sentence is true and too brief. The frames it describes are common — every K-braced panel in a lattice tower, a crane mast, a long-span roof girder — and the way the method has to be applied to them says something about what a moment centre is that a three-member cut never has to say.

A tower face with no Ritter point #

Take one face of a lattice tower — a triangulated frame standing on end — 18 m high in six panels of 3 m, 6 m wide at its feet and 3 m at its top. The legs are straight lines converging upward. At each level above the base a horizontal member runs between the legs, and it has a joint at its midpoint. In each panel two diagonals rise from the leg joints at the bottom of the panel to that midpoint at the top: a K lying on its side, or an inverted V, repeated up the tower.

The face is pinned at both feet. Wind puts 4 kN on the windward leg at every level, a further 20 kN at the top, and the headframe adds 60 kN of weight shared by the two top joints.

Now try to find the force in one diagonal of the third panel by the method of sections.

A horizontal cut through the panel severs the windward leg, the leeward leg, and both diagonals. Four unknowns. The free body above the cut has three equations of equilibrium. The count alone is enough to say the section cannot finish the job, but it is more useful to see why no equation isolates a member.

Moments about the midpoint of the horizontal at the top of the panel. Both diagonals pass through it, so both drop out. Both legs are left, each with a lever arm, in one equation.

Moments about the point where the legs’ lines meet. For this taper that is 36 m above the base. Both legs pass through it and drop out. Both diagonals are left.

Resolving horizontally. Every member has a horizontal component — the legs because they lean, the diagonals because they slope — so all four are in it.

There is no point in the plane through which three of the four severed members pass, because two of them meet at one point, the other two at another, and no point lies on three of the four lines. A three-member cut always has such a point, as a matter of geometry: any two of three lines meet somewhere, and the third member’s force is found by taking moments there. A four-member cut has two such points, each of which removes two unknowns and leaves two. The method of sections is not short of equations here; it is short of a point.

The joint that supplies the fourth equation #

The free body that the section lacks is the midpoint of the horizontal, cut out on its own.

Four members meet at that joint, and on its own it is hopeless: four unknowns and two equations. But one of the two equations is unusually clean. The horizontal member’s two halves lie along one line with no vertical component, and no load is applied at the joint. So resolving vertically involves the two diagonals and nothing else:

$$D_L \sin\alpha_L + D_R \sin\alpha_R = 0.$$

On a symmetric face the two diagonals make the same angle, so $D_R = -D_L$. Whatever the load, one diagonal of each K is a tie and the other a strut of exactly the same size. When the wind reverses, they swap.

That is the fourth equation. It does not give any force, only a relationship between two of them, and it is exactly the relationship the section was missing.

The section, now with three unknowns #

Return to the free body above the cut, and replace $D_R$ by $-D_L$ everywhere. Three unknowns remain: the two legs and one diagonal force. And the moment centre that was useless before is now decisive.

About the point where the legs’ lines meet, 36 m up and on the tower’s centreline, both legs drop out. What remains is the loads above the cut and the two diagonals — which are now one unknown.

The loads above a cut through the third panel are the wind at the four upper levels, at heights 9, 12, 15 and 18 m, the extra 20 kN at 18 m, and the headframe weight. The headframe’s two 30 kN shares sit symmetrically about the centreline and have no moment about a point on it. The horizontal loads have lever arms of 27, 24, 21 and 18 m below the meeting point:

$$4(27 + 24 + 21 + 18) + 20 \times 18 = 360 + 360 = 720 \text{ kNm}.$$

Each diagonal starts at the midpoint of the horizontal, 9 m up, and runs down to a leg joint at 6 m, 2.5 m either side of the centreline: a length of $\sqrt{2.5^2 + 3^2} = 3.905$ m. The moment of a diagonal force about a point 27 m directly above its upper end is that distance times the force’s horizontal component, $27 \times (2.5/3.905)D = 17.29D$. The two diagonals, one pulling each way along their own slopes with $D_R = -D_L$, add rather than cancel. So

$$2 \times 17.29\,D = 720, \qquad D = 20.83 \text{ kN}.$$

One equation, one unknown. The legs then come from the other two equations, and both are worth writing out because each has a small surprise in it.

Moments about the midpoint of the horizontal, 9 m up. The wind above has lever arms of 0, 3, 6 and 9 m about it, so its moment is $4(0 + 3 + 6 + 9) + 20 \times 9 = 252$ kNm, and again the headframe’s symmetric weight contributes nothing. Both diagonals pass through the point. Each leg leans inward by 0.25 m in its 3 m panel, so its line passes 2.24 m from the centreline at that height and its moment arm about the point is 2.24 m. The windward leg pulls and the leeward leg pushes, and both resist the overturning, so $2.24(F_{\text{wind}} - F_{\text{lee}}) = 252$ and the legs differ by 112.4 kN.

Resolving vertically. The diagonals’ vertical components cancel — that was the joint’s result, and it holds on the section as well as at the joint — so the only vertical forces on the free body are the legs and the 60 kN headframe. The legs are nearly vertical, and their sum is $-60/0.9965 = -60.2$ kN.

Sum and difference give the windward leg 26.1 kN of tension and the leeward leg 86.3 kN of compression. The surprise in the second equation is that the K’s diagonals are invisible to the tower’s vertical equilibrium at every cut: however large the shear, the diagonals add no net vertical force, so the legs alone carry the weight and the overturning. That is not true of a single-diagonal panel, whose diagonal pulls down on one side of every cut. The whole-face solution, which assembled and solved forty equations and knew nothing about cuts, gives the same four numbers.

Why the joint has to come first #

Stated as linear algebra, the order does not matter. The joint’s two equations and the section’s three are five equations in the section’s four unknowns plus the two halves of the horizontal, and any method of solving them gives the same answer.

Stated as a hand calculation, it matters completely, and the reason is the whole technique of the method of sections. Ritter’s method is valuable because each equation it writes has one unknown in it. Taken in the order joint then section, that holds at every step: the joint’s vertical equation has the diagonals’ ratio as its only content; the section’s moment about the legs’ meeting point then has one diagonal force in it; the moment about the mid-joint has the legs’ difference; the vertical resolution has their sum.

Take the section first and no equation it offers has fewer than two unknowns. The calculation becomes simultaneous equations, which is exactly what the method existed to avoid. The joint is not a second use of the method; it is what turns a four-member cut back into a three-member one.

The same pattern turns up wherever a frame has four-member cuts. Culmann’s construction finds three member forces by pairing the resultant of the loads with one severed member and the other two with each other, and on a K panel it stalls for the same reason: with four severed members there is always one force too many for the pairing, until the joint has tied two of them together. The graphical method needs the joint first for the same reason the algebraic one does.

The legs are carrying some of the shear #

The number 20.83 kN is interesting in itself, and the way to see why is to compare it with the shear it is supposed to carry.

The horizontal loads above the cut add up to $4 \times 4 + 20 = 36$ kN. In a frame with parallel legs, the legs are vertical and have no horizontal component, so all 36 kN has to be carried by the diagonals. Each diagonal’s horizontal component is then 18 kN.

On the parallel-legged tower the diagonals carry 25.46 kN. The legs’ meeting point has gone to infinity, and “take moments about a point at infinity” means “resolve at right angles to the direction it lies in” — the same limit a mechanism turning about a centre at infinity reaches when its rotation becomes a translation. Resolving horizontally, both vertical legs drop out and the diagonals’ horizontal components must supply the whole shear: $2 \times 25.46 \times 3/4.24 = 36$.

On the tapered tower the diagonals carry 20.83 kN, and their horizontal components add to 26.7 kN. The other 9.3 kN of the shear is carried by the legs, because a leaning leg in compression or tension has a horizontal component, and the two legs of a tapered tower lean in opposite directions while carrying forces of opposite sign. Their horizontal components add.

That is not a small effect, and it is the reason the moment centre at the legs’ meeting point is the right way to think about the diagonals. In a tapered tower the diagonals do not carry the shear. They carry the moment of the loads about the point where the legs meet, divided by their own lever arm about it.

A diagonal lightest at the base #

The consequence is visible the moment the diagonal force is plotted up the tower.

With parallel legs the diagonals follow the shear, heaviest at the base where all the wind above has accumulated. With a top half as wide as the base the diagonals are nearly uniform. With a top a quarter as wide the lightest diagonals in the tower are at the bottom and the heaviest are at the top — the opposite of what the shear diagram suggests.

The reason is the lever arms about the meeting point. Near the base, the diagonals are far below the meeting point and have a long lever arm about it, and the loads above are mostly close to it. Near the top, the meeting point is close above the diagonals and their lever arm is short, while the loads above them have nearly the same arms as their own.

It has a limiting case that was built. If the legs are shaped so that at every height their tangents meet on the line of action of the resultant of the wind above that height, the moment of the loads about the meeting point is zero at every level, and the diagonals carry nothing at all. The legs take the whole wind by themselves. That is the rule the engineers of the tower on the Champ de Mars described for its profile in 1889, and it is why the tower’s lattice between its legs is so light for its size: they chose a shape in which the members this essay has been solving for are, for the design wind, idle.

The design reading follows directly. A tapered lattice trades diagonal force for leg force: at the cut above, tapering to 1.5 m cut the diagonals from 25.5 kN to 16.0 while raising the leeward leg from 72.0 to 98.0. Diagonals are long, slender, and governed by buckling; legs are short between panel points and governed by strength. Moving force from the first to the second is usually a good trade, which is why nearly every lattice tower tapers.

What reaches the feet #

The same two equations, written at the base instead of in the third panel, are the foundation design, and they give a number that the member forces above never show directly.

At a cut just above the feet the wind’s moment about the base is $4(3 + 6 + 9 + 12 + 15 + 18) + 20 \times 18 = 612$ kNm, carried by legs 6 m apart as a couple of $612/6 = 102$ kN. The headframe’s 60 kN adds 30 kN of compression to each. So the leeward foot pushes down with 132 kN and the windward foot pulls up with 72 kN: the weight of the headframe is nowhere near enough to hold that corner of the tower down, and the holding-down bolts at the windward foot are in tension under the design wind.

Every quantity in that paragraph came from a free body that does not know what the bracing is. A K, an X or a single diagonal gives the same feet reactions, because the feet reactions of a determinate tower depend only on the loads and the positions of the feet. What the bracing pattern decides is which members above carry the shear on its way down — and the joint-first section is what reads that off one panel at a time, without solving the rest.

It also closes the loop with the count that decides determinacy. The face has 20 joints, 40 equations, 36 members and 4 reaction components: determinate, which is why the joint-then-section route and the whole-face solution have to agree exactly rather than approximately, and why a K is not a redundant brace even though every cut through it severs four members. A K panel adds one joint and two members to a single-diagonal panel, and the count is unchanged.

What the K is for #

The joint’s result also explains why a K is drawn at all, rather than a single diagonal across each panel.

Compare the parallel-legged tower’s K with a single diagonal across the same 6 by 3 m panel, which is the trade halving a panel makes in a bridge truss, made across the panel’s width instead of along its length. The single diagonal is 6.71 m long and carries the whole 36 kN shear alone, which along its slope is 40.2 kN — as a strut when the wind blows one way. The K’s diagonals are 4.24 m each and carry 25.46 kN, one as a tie and one as a strut. So the strut carries under two-thirds of the force over a length 0.63 of the single diagonal’s, and its buckling capacity, which goes with the inverse square of its length, is two and a half times higher. When the wind turns round, the tie and the strut swap, and the same argument applies to the other.

The price is the joint in the middle of the horizontal, whose own equilibrium puts force into both halves of the horizontal member: 15.3 kN of compression on one side and 11.3 of tension on the other, in a member that a single-diagonal panel might have left almost unloaded. The horizontal has become a working member, and its midpoint connection has to be designed for the difference.

What the free bodies leave out #

Pin joints. The legs of a lattice tower are continuous through the panel points and the connections are bolted, so real members carry some bending. The axial forces from the pin-jointed analysis are the ones that govern design; the secondary moments are real and smaller, and the chord that runs through its joints behaves as a beam as well as a strut.

A plane face. A real tower has four faces, and wind at an angle loads two of them at once. Each face is solved as a plane truss under its share, and a leg belongs to two faces, so its force is the sum of two such analyses.

Symmetry of the K. The equal and opposite diagonal forces rely on the two diagonals meeting the horizontal at the same angle. An offset joint gives unequal diagonal forces in a fixed ratio, which is still enough to supply the fourth equation, and the method goes through unchanged.

Loads at joints. The wind is applied at the leg joints. Wind on a leg between joints bends it, which the section cannot report, as with any truss.

Small deflections. Every lever arm is measured on the tower as drawn. A tower that has swayed under the wind has moved its loads relative to its base, and for a tall slender mast that shift is a second-order load the equilibrium here does not contain.

Still open: the panel where two K’s share one joint #

The tower here is determinate because every K has a joint of its own. A K-braced bridge girder that is symmetric about midspan has, at its centre, two K’s opening toward one another and meeting at the same midpoint of the same vertical — and counting members against joints shows that it then has one more member than equilibrium can resolve. The joint that rescued the section here is overloaded there, with four diagonals instead of two, and its vertical equation relates four forces instead of fixing a ratio. Whether that central panel is solved by symmetry, by a stiffness, or by designing the joint out of existence is a question about what a section can do when the missing equation is not in any joint at all.