The fuse that makes the span work harder

The weight that lands somewhere else drove a falling weight at a fixed-ended beam, moved it around from one drop to the next, and followed where the plastic rotation went. Half of every drop’s rotation, at any point it landed, turned up at the supports. Every support there was taken to be as strong as the beam. A rock-fall canopy’s beam usually hangs on two bolted connections that are weaker than the beam and much less ductile, so the same half of every rotation lands on the least suitable parts of the structure.

That suggests a design decision that looks attractive at first sight. If the connections are going to go first anyway, let them. Size them as partial-strength fuses: they yield at a known moment, they cap what the supports receive, and a damaged end plate can be unbolted and replaced after a season of drops. The alternative is to make the connections stronger than the beam, so as to move the rotation into the span, where no one can replace it. The usual reasoning prefers the fuse for its maintenance, and it rests on a picture in which the weakest link does the work. For a static load that picture is often right. For a falling weight it is backwards, in a way that follows from counting energy.

A drop is an energy, and a mechanism is a ratio #

The beam is the earlier essay’s: 6 m long, with $EI = 42{,}000\ \text{kN·m}^2$ and a plastic moment of 200 kN·m, a small hardening, struck at mid-span by a weight delivering 4.9 kJ. Each drop is taken as an energy load: the beam is pushed at the point of impact until the work it has absorbed equals the weight’s energy, then let go, keeping its plastic rotations and the residual moments they lock in. The ends now have a resistance of their own, a share $r$ of the beam’s plastic moment.

The statics of a plastic mechanism say two things about this, and they are the whole of the argument. Under a load at mid-span the beam folds at three hinges, and by geometry each end hinge turns through $\theta$ and the hinge under the load through $2\theta$. That ratio has nothing to do with how strong anything is. And the work done in a mechanism is each moment times its rotation:

$$W = 2\,r M_p\,\theta + M_p\,2\theta = 2M_p\theta\,(1 + r).$$

The energy is what the weight delivered. So the rotation each drop adds is

$$\theta = \frac{E_{\text{plastic}}}{2M_p(1 + r)}.$$

As the joint weakens, $r$ falls, and $\theta$ rises, everywhere. The joint has less moment to work against, so it must turn further to dissipate the same energy, and the geometry drags the span’s hinge along at twice the rate.

The computed rotations bear this out. After ten drops at full strength the joints have rotated 45 mrad and the beam under the weight 89. At half strength those are 66 and 125. At a quarter, 78 and 146. The picture to compare is what the phrase “the fuse takes the blow” leads one to expect, which is a tall spike at each end and a clean span. The actual distribution has grown everywhere, and the span’s spike is the taller at every strength.

What each drop adds #

Once the beam has settled into its mechanism the ratio of two is exact, drop after drop, across almost the whole range. A full-strength joint adds 3.2 mrad per drop by the twelfth drop, and the span 6.4. At half strength, 5.6 and 11.3. The figures are lower than the 4.9 and 9.7 of the first drop only because a small hardening has built up in those sections, and hardening is the reason the curve flattens towards the left of the figure: a joint with a tiny resistance has less to lose. Taking $r$ from 1 to 0.5 should raise $\theta$ by $2/1.5$, a third; the ten-drop rotations above go from 45 to 66, nearer a half, because the first few drops also spend elastic energy differently. The direction is not in doubt. Halving the joint’s strength has raised the span’s rotation by 40 per cent.

The first drop is different, and it is the one that misled the fuse’s advocates, or would if it stood alone. On a new beam, under the first blow, the ends reach their low resistance first and turn while the span is still elastic, so the end rotation is proportionately larger: at half strength a first drop gives the ends 9.2 mrad and the span 11.3, a ratio of 1.2. For one rock the joint does appear to have absorbed the blow. Residual moments from that drop then change what comes next, and by the second the ratio is 1.98: 6.9 mrad and 13.7. A test of one drop on one connection would support the fuse, and a season of them would not.

Who does the work #

The work-share is the clearest statement of what went wrong with the idea. From the equation above, the joints do a share $r/(1 + r)$ of the plastic work and the span does the rest. A full-strength joint does half of it, and the computed share in the figure is 0.50. A joint at half strength does a third. A joint at a fifth of the beam’s strength does a sixth. The weaker the fuse, the smaller the fraction of the energy it dissipates. It is the span, not the fuse, that takes up the slack.

That is backwards from the static case, and it is worth saying why. Static load: a weak connection limits the force that can arrive, and whatever is beyond the connection is protected, because a force cannot pass a limit. Energy: nothing limits the energy that arrives, and a weak connection reduces the rate at which it can be taken out. The fuse that protects the strand and not the wall found the corresponding distinction at the foot of a rocking wall, where a yielding bar caps the strand’s force and leaves the wall’s drift to get on with it. A fuse protects what lies behind it against a force. A falling weight isn’t one.

How many drops #

The practical reading comes with the rotation capacities. A bolted end plate can give perhaps 30 mrad when it fails by plate bending, and a good deal less if a bolt goes first; the joint that has to keep turning sets out the range. A compact beam can give several times that. Taking those two as 30 and 120 mrad, a full-strength joint would be spent in seven drops, and the span in fifteen. At half strength, five and ten. In this comparison weakening the joint has made both worse. The joint runs out two drops sooner, despite doing a smaller share of the work, because it is rotation that uses up a joint’s capacity rather than work. The span loses a third of its life.

Neither of these is the design it might be. The far more interesting comparison is with a third option on the right of the figure: a joint stronger than the beam. The plastic rotation then forms in the beam beside the joint, and the joint itself stays elastic. Fifteen drops, and the limit reached by both of the end sections is the span’s 120 mrad at twice the rotation, so it is the middle that decides. A strong connection does not move the problem into the span. It moves the end rotations out of a detail with 30 mrad to give and into a section with four times that. The beam is then being asked for rotation at three hinges, but it has rotation.

What replacing it buys #

The point of a fuse, it may be said, is not that it makes the beam last longer. It is that it can be replaced, and the replacement gives the joint a new 30 mrad. That is true, and on the count of drops above it removes the joint’s limit: a canopy whose joints are renewed at every five drops never runs out of joint. It is then the span that decides, and the span is the same 125 mrad in ten drops, with its limit at ten rather than fifteen.

The set makes that visible. A new fuse goes back onto a beam in its bent position. The kink it held stays in the geometry, the locked-in moments ease a little, and nothing about the span has been undone. After twenty drops the full-strength beam has sagged 229 mm, and the partial-strength beam 342, or 356 with its fuses changed, a new fuse having none of the old one’s small hardening. The same weight dropped again measured the same kind of vee for a beam in one material, and the shape does not care about the bolts. For a canopy whose job is to stay clear of a road beneath it, the limit may well be this sag rather than either rotation, and the fuse has brought that limit forward by half.

It would make sense to replace the beam with the fuse at a certain point, but the fuse’s whole selling point is that the beam need not be touched. The honest maintenance plan for a partial-strength canopy is to inspect the span for a kink at each change of fuse. If the span is going to be inspected and replaced anyway, then what the fuse saved was the labour of unbolting an end plate.

Where the weight lands #

Mid-span is the kind case for the joints. A mechanism’s hinge rotations under a load at distance $a$ from one end are $\delta/a$ at that end, $\delta/(L - a)$ at the other, and their sum under the load. So a weight landing near one end turns the near joint far more than the far one: at 0.15 of the span, the near full-strength joint adds 7.6 mrad a drop against 3.2 at mid-span. At half strength it is 9.4.

The weight that lands somewhere else found the rotation spreading over the span when the drops wander, with a beam that was glad of the spreading. The joints don’t get that. A joint is at a fixed place and every drop reaches it, nearer or farther. A rock-fall canopy gets its blows where the slope above sends them, usually with some preference for one side, and it is the joint on that side which will run out first. The essay’s hinge count of fifteen sections holds for the span, and the joints are still two sections that collect from every drop.

The kinematics also say something about the shape of the risk. A connection is sized once, and in a partial-strength design it is sized on the assumption that it will yield in a controlled way, usually from a calculation at mid-span. Drops near an end raise the rotation by a factor of two and more, and the capacity that was comfortable for the mid-span case may be spent by a few of those.

The rock the fuse is for #

All of this has been about a 4.9 kJ rock, well past any elastic limit, and for a smaller one the argument changes. The beam needs a moment of $M_p$ under the weight before the span can yield at all. With joints at half strength a small weight first yields those joints, at about 0.25 kJ, while the moment at mid-span is still well below $M_p$. The joints turn a little, a residual hogging moment settles in, and later drops of the same size repeat elastically: the frame has shaken down, as the structure that settles down would predict, with the ends having taken 3.5 mrad and the beam only a set of 5 mm. Up to 1.31 kJ the span never yields. A full-strength beam reaches its first hinge at 0.95 kJ.

There the fuse does exactly what it was hoped to do. It takes small blows in a part designed to accept them, and it keeps the span clean for rocks about a third more energetic than a full-strength joint would have. But its range is narrow: the half-strength joint is the best on the figure, and at a quarter strength the clean range falls to 0.61 kJ, below the full-strength beam’s. A joint weak enough to yield early is a joint so weak that it has lost the moment which was holding the middle of the beam down.

So a designer choosing between the two has a question to ask about the slope, not the beam. What is the distribution of energies coming down it? For a slope that sheds many small stones and has never sent anything over a kilojoule, a half-strength joint is a sound and cheap choice, and the set under each stone is small. For one that sends a few large blocks, it is the worse of the two for each block, and each of those is what the canopy exists for. Rock-fall records are often of frequency and volume rather than energy, and a mass at a fall height is the number the design needs; the load that is never all there at once is a warning about how quickly such distributions get thin in the tail where the design lives.

The hand check #

The plastic energy of a drop is the weight’s 4.9 kJ, less the elastic energy the beam gives back. For the full-strength case, the collapse load at mid-span is $P_c = 8M_p/L = 267$ kN and the elastic stiffness at mid-span is $192EI/L^3 = 37{,}300$ kN/m, so the energy recoverable at collapse is $P_c^2/2k = 0.95$ kJ. Plastic energy 3.95 kJ. Then $\theta = 3.95/(2 \times 200 \times 2) = 0.0049$ rad at each joint and 0.0099 under the load, matching the computed first drop of 0.0049 and 0.0097.

At half strength, on later drops, $\theta = E_{\text{plastic}}/(2 \times 200 \times 1.5) = 3.95/600 = 0.0066$ rad a drop, and ten of them, 66 mrad, with the span at 132. The computed rotations are 66 and 125.

What the falling weight is not #

A slow push is not an impact. Taking a drop as an energy load is exact for a weight much heavier than the beam, as the earlier essays set out. A rock that is light against the beam spends energy on exciting it, the impact force is higher, and the hinges form in a different order. The essay on the load that is over before it has moved describes that other regime. The argument that the mechanism’s ratio decides everything applies after the plastic hinge pattern has formed, and for a light rock that takes longer.

A plateau for the joint. A real end plate does not have a flat plastic plateau and an unchanged stiffness. It opens up, its bolts stretch, and it pinches on the way back. Under repeated blows it loses preload, and the capacity read off a monotonic test may be well above what a repeated-load path delivers. All these favour the full-strength choice.

Joints that fail by force. A connection’s shear resistance is a separate check, and a weak joint at a fixed peak load can still be overloaded in shear by a rock: the peak force of the first drop is 205 kN at half strength, 270 at full, and half of that reaches each end. This is the one place where the fuse does do what it promises, and a connection whose shear strength is short might well be protected by a lower bending strength. A designer who needs that has a real reason for a fuse, although it is not the reason usually given.

No axial restraint. At these sags the beam begins to act as a tie if its ends are held horizontally. That would add resistance and it would be a separate question about what the joint can do in tension. The cable that pays the factor back follows the same transition for a beam over a lost column. With a vee of 300 mm in 6 m it has started here as well.

Still open: the beam that becomes a cable #

The set in these figures is taken as a shape and nothing more, and the moments are worked out on the original geometry. A 340 mm vee across 6 m is not small: if the ends are held against moving inward, the beam has begun to stretch. Its tension then offers a resistance that grows with the vee, so each later drop would dissipate less in bending and more in stretch, and the set would grow more slowly. It is also a tension the end connections have to carry, in the direction they are poorest at. Whether a canopy with weak joints is saved by catenary action in the later drops, because the deeper vee gets there first, or whether those same joints, already short of rotation, part at the first tension they are asked for, is the question that a set this large brings to a detail that was only ever being asked about bending.