The worst direction is an eigenvector

The bracket pushed from the wrong side took a group of bolts whose load could come from any direction in their plane and found that its capacity is a closed curve round the directions, with a worst direction that no drawing shows. It also found a lower bound in closed form: the bolt furthest from the centroid can never carry more than the direct share plus the whole torque’s, so the group carries at least $R_b/(1/n + d\,r_{max}/J)$, whatever the direction. The bracket with no worst bolt then pulled the same kind of bracket off its column as it twisted, gave every bolt a tension as well as a shear, and found that the bolt worst in shear and the bolt worst in tension come apart as the load tilts — so that a hand check pairing the worst of each, as if they met in one bolt, under-reads the group.

That essay ended on whether the search over direction and pull-off distance can still be reduced to something in closed form, as the in-plane case was, or whether the interaction between shear and tension makes every bracket a search. The answer turns out to depend on which interaction, and in a direction few would guess.

A capacity that is a surface #

The bracket is the earlier essay’s: six M20 grade 8.8 bolts in three rows of two, 75 mm apart each way, on a plate 200 mm wide, loaded at a point 150 mm to the side of the group’s centroid and some distance in front of the column face. Each bolt’s shear comes from the elastic method — the direct share plus the torque’s share in proportion to its distance from the centroid — and its tension from the edge model, the plate pivoting on whichever edges the load pushes it against, each bolt pulled in proportion to its distance from those edges. A bolt is at its limit when its shear and tension reach the interaction: the ellipse $(V/F_v)^2 + (T/F_t)^2 = 1$, or the straight line $V/F_v + T/1.4F_t = 1$ with $T \le F_t$ that Eurocode 3 prints.

The capacity is lowest just off vertical — 16° from straight down, towards the offset side — and highest for a load pushing roughly along the line from the load point to the centroid, where the torque vanishes. Under the ellipse the least is 172.5 kN, under the straight line 147.1, the Eurocode’s rule sitting a seventh lower everywhere because it adds shear and tension where the ellipse combines them in quadrature. Both minima are at directions nobody would draw as a design load, which is the whole difficulty: a bracket that may be loaded from several directions has to be checked at its worst one, and the worst one moves with the distance the load stands off the column.

A quadratic form in the direction #

Write the load’s direction as a unit vector $u = (F_x, F_y)$. Every bolt’s shear is then linear in u: the direct share is $u/n$, and the torque about the centroid, with the load point offset $e_x$ to the side, is $e_x F_y$, which adds $e_x F_y(-y, x)/J$ to a bolt at $(x, y)$. So a bolt’s shear vector is $Au$ with

$$A = \begin{bmatrix} 1/n & -e_x y/J \\ 0 & 1/n + e_x x/J \end{bmatrix}.$$

Its tension is linear in u as well, but only within a quarter-turn of directions: which edges the plate pivots on depends on the signs of $F_x$ and $F_y$, and while those signs are fixed the tension is $c \cdot u$, with c set by the bolt’s distances from those edges and the pull-off distance $e_z$. Under the ellipse the demand on the bolt — how much of its interaction one kilonewton uses — is then

$$\sqrt{\left(\frac{|Au|}{F_v}\right)^2 + \left(\frac{c\cdot u}{F_t}\right)^2} = \sqrt{u^{\mathsf T} M u}, \qquad M = \frac{A^{\mathsf T}A}{F_v^2} + \frac{c\,c^{\mathsf T}}{F_t^2}.$$

A quadratic form over unit vectors is largest along its matrix’s eigenvector, and its largest value is the larger eigenvalue — the oldest result there is about quadratic forms, and the one behind every principal stress. So in each quadrant the worst direction for each bolt is either M’s eigenvector, if it points into that quadrant, or one of the quadrant’s two edges, if it does not. The group’s least capacity is one over the largest of these over its bolts and quadrants: six bolts, four quadrants, twenty-four 2 × 2 eigenvalues, none of which needs a direction to be tried.

The figure draws the demand of the bolt that governs, the top one on the offset side, round every direction. Under the ellipse it is four arcs, each a piece of an ellipse whose axes are M’s eigenvectors for that quadrant; the longest radius is the major axis of the lower right arc, at 16.0°. A sweep of every direction at a quarter of a degree finds the same 172.5 kN at the same angle. The sweep is not needed.

A bolt’s principal directions #

The construction is the same one that finds principal stresses. A stress state at a point gives a traction on every plane through it, and the largest normal stress is found not by trying planes but by diagonalising the stress tensor; the circle nobody draws is that diagonalisation drawn. Here the “tensor” is M, built from the bolt’s place in the group and the plate’s pivot edges, and the “planes” are directions of the load. The bolt has principal directions of loading — the direction in which a kilonewton costs it most, and at right angles the one in which it costs least — and M’s eigenvalues are the costs.

That reading makes two things obvious that the sweep only shows. The worst direction depends on the geometry and on the ratio $F_v/F_t$ and on nothing else: doubling the load changes the capacity and not the direction, and the matrix can be written once for a layout and reused. And the four quadrants are four different tensors, because the pivot edges change between them; a load that crosses a quadrant boundary passes from one tensor to another, which is why the polar curve has corners where the arcs meet, and why the worst direction can sit on a quadrant’s edge — a pure pull-off or a purely horizontal push — when the eigenvector of the neighbouring arc points out of its quadrant.

The straight line has nothing to diagonalise #

Under the straight line the same bolt’s demand is

$$\frac{|Au|}{F_v} + \frac{c\cdot u}{1.4F_t},$$

a norm plus a linear function of u, and the dashed curve shows what that looks like round the directions: a lopsided oval that is no piece of any ellipse. It has no eigenvector. Its largest value in a quadrant can be found by setting the derivative of that expression along the unit circle to zero, which leads to a quartic in the angle’s tangent, and in practice by trying directions — cheaply, because there are only four quadrants and the function inside each is smooth, but by trying them.

The earlier essay asked whether the interaction’s curvature makes every bracket a search. It is the other way round. The ellipse’s curvature is what keeps the problem closed; the straight line’s lack of it is what opens it. Squaring and adding the two utilisations is exactly the operation that turns a pair of linear functions of the direction into a quadratic form. Adding them as they stand does not.

What the hand floor costs #

The floor a designer would reach for without either result is the one the in-plane essay found, extended: the largest shear any bolt can have in any direction, $1/n + e_x r_{max}/J$, set against the largest tension any bolt can have in any direction, $e_z\sqrt{a^2 + b^2}$, a and b being the corner bolt’s distances from the pivot edges over their sums of squares. It is safe by construction, because it lets the worst of everything meet in one bolt in one direction.

In the plate’s plane the bracket carries 181 kN at its worst direction, and the floor says 164. Pulled 150 mm off the column it carries 172 under the ellipse and 147 under the straight line, against floors of 150 and 124. At 300 mm, 150 and 124 against 122 and 100.

The floor is 0.90 of the truth in the plane, because even there the direct share and the torque’s share at the furthest bolt are not quite aligned in any one direction — the in-plane lower bound is a bound and not an exact answer, for a rectangular group. It falls to 0.82 once the bracket is pulled 300 mm off the column, where the worst shear and the worst tension are both large and arrive at directions furthest apart, and recovers a little beyond, where tension dominates and the corner bolt is worst at both. A bracket checked by its floor is designed for a load up to a fifth larger than it can be shown to need. That is a fifth of the bolts, or a bolt size, on every bracket so checked.

A procedure that fits on one sheet #

The result turns a bracket check into arithmetic a designer can do by hand, using nothing beyond the elastic method’s own terms — the direct share and the torque over the polar moment that the bolt that carries more than its share set out. For each bolt that could govern — in practice the four corner bolts, which are both furthest from the centroid and furthest from the pivot edges — write A from its coordinates and the offset, and c for each quadrant from its distances to that quadrant’s pivot edges and the pull-off distance. Form M, take its larger eigenvalue, check that the eigenvector lies in the quadrant, and take the quadrant’s edges if it does not. The smallest capacity among those sixteen values is the bracket’s. A spreadsheet does it in sixteen rows, and so does a pencil.

It also changes how layouts are compared. The strongest layout leans on one bolt searched for bolt positions numerically, because the in-plane capacity over directions had to be swept for every candidate layout; with the eigenvalue, each candidate’s least capacity is a handful of 2 × 2 problems rather than a sweep, and a search over layouts becomes cheap enough to be exhaustive. And because the floor is a closed form too, the difference between the two — what checking by the floor costs a given layout — can be read off before choosing it.

The direction swings with the pull #

In the plate’s plane the worst direction is 14° off vertical, where the direct share and the torque’s share at the top bolt on the offset side come closest to adding. As the pull grows the tension’s share of the demand grows, and the tension is largest for a load pushing the plate against both a bottom edge and a side edge at once — a diagonal — so the worst direction swings towards 45°. Under the ellipse it has swung to 43.5° by 300 mm; under the straight line, which weights tension less heavily against shear, it is still at 20° there and reaches 44° only by 600 mm.

That swing is why a bracket cannot be checked at one conventional direction. A design that checks the vertical load and a horizontal one, the two a drawing would show, misses a direction 14° or 45° off them where the capacity is lowest — by a few per cent in the plane, and by more as the bracket stands further off its column. The eigenvector finds the direction for each pull-off distance without anyone having to guess it.

Other layouts, other floors #

The floor’s cost depends on the layout. A compact square of four loses only 8 per cent, because its four corner bolts are each the furthest from the centroid and the most pulled. A tall column of five in a single line loses 19, because the bolt furthest from the centroid is at the top, the most pulled is at the top too, but the torque’s shear there is horizontal while the direct share is vertical, and the tension’s worst direction is pure pull-off rather than a diagonal. Every layout loses something, and none loses nothing: the floor is exact only for a group whose worst shear and worst tension come from the same bolt in the same direction, and no real group is that.

The bolt group has no neutral axis found that the edge model of tension is itself a simplification — a plate pivots about a line that moves with the load, not a fixed edge — and the eigenvalue result inherits it: it is exact for the model, not for the plate. But within any model in which shear and tension are linear in the load’s direction quadrant by quadrant, the ellipse’s worst case is an eigenvalue.

By hand, for the governing bolt #

The group’s polar moment is $J = 4 \times 75^2 + 6 \times 37.5^2 = 30{,}938$ mm². For the top bolt on the offset side, at $(37.5, 75)$ mm, the shear matrix is

$$A = \begin{bmatrix} 0.1667 & -0.3636 \\ 0 & 0.3485 \end{bmatrix},$$

from $1/6$, $150 \times 75/30{,}938$ and $1/6 + 150 \times 37.5/30{,}938$. In the quadrant of directions pushing down and towards the offset side, the plate pivots on its bottom edge, 225 mm from this bolt, and on the side edge the load pushes towards, 62.5 mm from it; with sums of squares of 157,500 and 68,438 mm², and the load 150 mm off the column, the tension is $c \cdot u$ with $c = (0.137, -0.214)$.

With $F_v = 94$ kN and $F_t = 136$ kN, $M = A^{\mathsf T}A/F_v^2 + cc^{\mathsf T}/F_t^2$ has entries $4.16$, $-8.45$ and $31.19 \times 10^{-6}$. Its larger eigenvalue is $33.61 \times 10^{-6}$, so the bolt’s capacity is $1/\sqrt{33.61 \times 10^{-6}} = 172.5$ kN, and the eigenvector $(2.42, -8.45)$, normalised to $(0.276, -0.961)$, points 16.0° off straight down towards the offset side. No other bolt and quadrant gives a larger eigenvalue, so 172.5 kN at 16.0° is the bracket’s least capacity.

What the closed form rests on #

Both force distributions are linear. The elastic method for shear and the edge model for tension each make a bolt’s force proportional to the load, quadrant by quadrant, and that is what makes the demand a quadratic form. The instantaneous centre method is not linear in the load — the centre moves as the direction turns, as the better method flatters the worse layout traced — and for it the worst direction is a search again.

The ellipse is one interaction among several. Eurocode 3 prints the straight line; tests on high-strength bolts under combined shear and tension lie close to an ellipse, which is why straight and kinked code lines are usually described as approximations to one. A designer free to use the ellipse gets both a higher capacity and a closed form for its worst case; one bound to the straight line gets neither, and should search.

Prying is left out. A plate thin enough to bend between its bolts adds prying forces to the tension, which depend on the plate’s bending and are not linear in the load once the plate’s toes lift off the column. The bracket with no worst bolt set prying aside for the same reason; the eigenvalue result holds for a plate stiff enough that the toes stay down.

Still open: the bracket loaded by a set of directions #

Every check here asks for the worst of all directions, which is what a bracket for a load of unknown direction needs. Most brackets are loaded by a known set — a crane rail’s vertical and lateral loads, a façade bracket’s dead load and the wind from either side — whose resultant directions span a sector rather than the whole circle. Under the ellipse the worst direction in a sector is still an eigenvector if it lies inside the sector and an edge of it if not, so a sector is no harder than a quadrant. But the loads in a set also differ in size, and a demand that is a quadratic form in the direction is not one in a vector of loads of different magnitudes combined by a code’s factors. Whether the worst combination of a set of factored loads, each from its own direction, is still the eigenvector of something — a matrix built from the load set rather than from one load — or whether the combination rules break the quadratic form that made one load’s worst case easy, is the question a load table puts to a single arrow.