Hung from above and still unstable

Every stability result in this collection so far has had a critical load in it — a value of some force at which a member stops being indifferent to being bent. A beam hanging from slings has none. There is no load to increase: the load is the beam’s own weight and it was there before the lift started.

What it has instead is a tilt, and a question about which of two moments grows faster.

Which free body produced the number #

The whole beam, seen end on, taken about the line joining its two lifting points — the roll axis.

Tilt it by an angle $\theta$. The centre of gravity, which sat $y_r$ below the roll axis, swings out to one side by $y_r\theta$, and the weight acting through it produces a restoring moment $W y_r \theta$. So far this is a pendulum and the answer is that it swings back.

Now allow the beam to bend. At a tilt $\theta$ a component $W\theta$ of its own weight acts sideways, about its weak axis, along its whole length. The beam bows, and its centre of gravity moves further out — by $\bar{z}\theta$, where $\bar{z}$ is the lateral deflection of the centre of gravity under a full sideways gravity. There is also an initial offset $e_i$, because no beam leaves the casting bed straight. Both of those are on the overturning side:

$$\underbrace{W y_r \theta}_{\text{restoring}} \quad \text{against} \quad \underbrace{W(e_i + \bar{z}\theta)}_{\text{overturning}}$$

The weight is on both sides and cancels. The heaviest beam and the lightest one of the same section and length are equally stable — which puts this among the small set of checks on this site, with overturning and friction, where the magnitude of the load is not in the answer, which is the first surprise and follows from the fact that everything here is self weight.

Setting them equal gives the equilibrium tilt

$$\theta_{eq} = \frac{e_i}{y_r - \bar{z}}$$

and the factor of safety against reaching some limiting tilt is $\theta_{max}/\theta_{eq}$.

The denominator, which is the whole of it #

$y_r - \bar{z}$. Two lengths, one from the rigging and one from the beam.

$y_r$ is the height of the roll axis above the centre of gravity. For a girder lifted on near-vertical strands from its top flange, that is roughly half the depth: 0.9 m on a 1.6 m beam. It is what a lift plan can most easily change and it is the number everybody looks at.

$\bar{z}$ is the beam’s own contribution and it is the one that decides. It is the lateral deflection of the beam’s centre of gravity when its whole weight is applied sideways — for a simple span with small overhangs, of the order of $5wL^4/384EI_y$, weighted over the length rather than taken at mid-span. It carries $L^4$ and $1/I_y$, and there is nothing in it about the load.

If $\bar{z} \geq y_r$ the denominator is zero or negative and there is no equilibrium at any tilt. Both moments grow with $\theta$, the overturning one grows faster, and the beam goes over from any disturbance whatever. It is not a large deflection; it is the absence of a stable state.

The fourth power, and the cliff #

For the section in these figures — 11 kN/m, weak-axis $EI_y$ of 255,000 kNm² — the numbers run like this:

Twenty-four metres to forty-four is a factor of 1.83 in length and 11.3 in $\bar{z}$ — which is $1.83^4$, exactly. The factor of safety does not follow it, because it depends on the difference $y_r - \bar{z}$ rather than on $\bar{z}$: while $\bar{z}$ is small the difference is nearly $y_r$ and the margin is huge; as $\bar{z}$ approaches $y_r$ the difference collapses and the margin goes with it.

That is why the last column falls off a cliff rather than declining. A thirty-metre girder hangs with a factor of thirteen and a forty-four-metre one of the same section cannot be hung at all, and there is nothing in between that reads as a warning.

Why the intuition fails, precisely #

The rigid-body answer for the same beam and the same hook height is a factor of 18.1 against the real 13.3 at thirty metres. At forty metres it is 13.6 against 2.3, and at forty-four it is 12.4 against none at all.

So the intuition is not merely optimistic; it is increasingly optimistic exactly where the answer is getting dangerous, and it never reports a problem. A rigid body hung from any point above its centre of gravity is stable, at every length, forever. The whole of the risk lives in the term the intuition does not have.

This is the same shape of error as the column that was never straight, where the idealised member reaches a load the real one never does. The difference is that a column’s imperfection sensitivity is well known and taught, and a lifted beam’s is a specialist topic that appears in no general text on statics.

The initial eccentricity, which is a tolerance rather than a property #

$e_i$ is where the beam’s imperfection enters, and it deserves care because it is the only number in the calculation that is not a stiffness or a length.

A precast girder leaves the bed with a sweep — a lateral bow — and the tolerance for it is of the order of $L/960$: 31 mm on a thirty-metre beam. The centre of gravity’s offset is not the sweep itself but the centroid of the bowed arc measured off the chord between the lifting points, which for a half-sine shape is $2/\pi$ of it — about 20 mm.

Two things about that are worth noticing.

Taking $e_i$ equal to the sweep overstates the overturning by 57%, which is a large error in a conservative direction and is a common simplification.

$e_i$ grows with the length too, linearly, since the tolerance is a fraction of the span. So the numerator of $\theta_{eq}$ is growing while the denominator is collapsing, and both are pushing the same way.

Why this is a modern problem #

Girders were lifted for a century before anybody wrote this equation down, and the reason is a scale argument rather than an oversight.

A 15 m steel plate girder has a $\bar{z}$ of a few millimetres against a roll height of half a metre. The denominator is essentially $y_r$, the factor of safety is enormous, and nothing in the lift needs thinking about. The problem does not exist at that size and no amount of care would have found it.

What created it was prestressing. A post-tensioned concrete girder can span forty metres at a depth that a steel one would need a truss for, and it does so because prestress solves the strong-axis problem outright. It does nothing whatever for the weak axis: the section is a bulb-tee, its $I_y$ is a small fraction of its $I_x$, and its weight per metre is three times a steel girder’s. Every term in $\bar{z}$ moved the wrong way at once.

So the equation is Robert Mast’s and it is from 1989, which is late for a piece of statics this elementary. It arrived when the members did, and the sequence — a technology solves one limit, the next limit turns out to be somewhere nobody was looking — is the ordinary way structural knowledge advances. The lateral-torsional check has the same history a century earlier, and for the same reason: deeper beams became available before the failure mode they introduced was understood.

What a lift plan can actually change #

Four things, in descending order of how much they buy.

Move the lifting points inboard. This is by far the most powerful and it is the standard remedy. The span that generates $\bar{z}$ is the distance between the picks, not the length of the beam, so moving them 10% in from each end reduces the effective span by 20% and $\bar{z}$ by a factor of $0.8^4 = 0.41$. It costs a check on the cantilever moment at the overhangs, which is a strength check and is usually comfortable. It is also the one move that makes the transport case better at the same time.

Raise the roll axis. Converging slings meeting at a single hook put the roll axis at their intersection, which for picks 28.8 m apart at 60° from the horizontal is twenty-five metres above them — an enormous $y_r$, and an enormous headroom requirement. It also puts the beam into axial compression, which softens it laterally and raises $\bar{z}$ slightly: a smaller effect, in the wrong direction, that a spreader beam avoids entirely by raising nothing and compressing nothing.

Brace the beam. Temporary lateral bracing, or a strongback bolted along the top flange, raises $EI_y$ directly. Doubling it halves $\bar{z}$.

Reduce the sweep. Rejecting beams outside a tighter tolerance reduces $e_i$ proportionally. It is the least effective of the four because $e_i$ is in a numerator with a first power while $\bar{z}$ is in a denominator with a fourth.

Reading the factor of safety, which is not a strength factor #

A factor of safety of 2.3 on a strength check means the member could carry 2.3 times the load before something broke. The number in the table above means nothing of the sort, and reading it as though it did is the commonest way to misuse it.

What it is, is a ratio of angles: the tilt considered acceptable, divided by the tilt the beam settles at. So a factor of 2.3 says the beam hangs at 0.17 radians — about ten degrees — against a limit of 0.4. It is already visibly leaning. Nothing has failed and nothing is close to failing in the strength sense, and a rigger looking at it would stop the lift.

That matters because the two failure modes have completely different characters. A strength failure at a factor below one is a break. A stability failure here is a progressive lean that ends in a roll, and the beam gives every appearance of being in trouble long before it goes. Which is a mercy, and is the reason the historical record has more near-misses than losses.

The practice consequence is that a working factor of 1.5 on this check is not the same kind of margin as 1.5 on a bending check, and treating the two as comparable understates how uncomfortable the first one is.

Where this model stops #

The limiting tilt is a judgement. $\theta_{max}$ is taken as 0.4 radians here, which is roughly where a girder cracks under the combined weak-axis bending and the lateral component of its weight. It is not a sharp boundary and different practices use different values, so the factor of safety is a comparison against a convention rather than against a physical limit. What is not a convention is the asymptote: at $\bar{z} = y_r$ the structure has no equilibrium regardless of what tilt anyone thinks is acceptable.

The beam is treated as prismatic and elastic. A cracked precast girder is softer about its weak axis than the gross section says, and it cracks under exactly the lateral moment this problem generates — which is a feedback the linear calculation does not have.

Hanging is not the worst case. The same beam sitting on truck bolsters during transport has its roll axis at the springs, which are below the deck and much softer than a sling, and the road is not level. Transport factors of safety come out well below lifting ones for the same girder, which is why the girders that fall over usually do so on the way rather than in the air.

What the picture cannot show #

The figures draw a static tilt. A real lift is dynamic: the beam is swung, it is stopped, the crane’s boom deflects, wind acts on a very large sail area. All of that is a disturbance, and the factor of safety above is a measure of how large a disturbance the beam can absorb and return from — which is the correct way to read it and is not how a factor of safety on a strength check is read.

Nor do they show that the roll is coupled to a twist. A beam that rolls also warps and twists along its length, so the shape it takes is not the plane rotation drawn; the full treatment is a lateral-torsional problem with the load applied through a moving point. The plane model here is the standard one, it is conservative, and it is what every lift plan uses.

The check nobody makes, and the one everybody does #

There is an instructive asymmetry in how this member is treated across its life.

In service, the girder sits on bearings with a deck cast on top of it, restrained continuously along its compression flange, and every code in the world requires a lateral-torsional check on it anyway. That check almost never governs, because the restraint is real and it is generous.

During the lift, the same girder has no restraint of any kind, is hanging from two points, and is at its most vulnerable — and there is no code check at all. Lifting is a contractor’s temporary works, it falls outside the permanent works design, and whether the equation in this essay gets solved depends on whether somebody in the erection team knows it exists.

That inversion — the checks are strictest where the risk is lowest — is not particular to girders. It is the erection-stage argument in its sharpest form, and the reason it persists is organisational rather than technical: the person who designs the member and the person who lifts it are different people working for different firms under different contracts, and the load case that kills people belongs to neither of them by default.

The generalisation #

The habit worth carrying is about where a stability problem’s variable lives.

Almost everywhere else on this site, stability is a question about a load: raise it until the structure stops being indifferent. Here the load is fixed and the question is about a geometry — a hook height against a deflection — and the load cancels out of the comparison entirely.

That happens whenever the destabilising action and the restoring action are both the structure’s own weight, which is a wider class than lifting. A self-weight buckling column is one. A hanging cable is another. A block sitting on a slope is a third. In every case the answer is a length compared with a length, the material has dropped out, and the useful question is not “how strong is it” but “which of these two lengths is bigger”.