The overstrain the release gives back

A section that has yielded and been unloaded does not return to nothing. It returns to a self-equilibrating field of residual stress, with a permanent set and an elastic range in the direction it was loaded that is wider than the one it started with. Load it again the same way and it stays elastic up to the load it was taken to before; load it the other way and it yields sooner, because the residual field that helps one direction hinders the other. The yield condition is an interval of fixed width, twice the yield stress, and overloading a member moves where the interval sits.

Autofrettage is that fact used on purpose. A thick-walled tube — a gun barrel, a high-pressure hydraulic cylinder, a fuel-injection rail, a waterjet cutter’s pump — is pressurised past yield once, at the factory. The inner part of the wall yields and stretches; the outer part stays elastic and stretched less; and when the pressure is released the outer part squeezes the inner part back, leaving the bore in hoop compression. In service the working pressure has first to undo that compression before the bore feels any tension at all.

The first essay on unloading named autofrettage as one of five places the effect is used and left the arithmetic open. The arithmetic turns out to contain a limit that is easy to miss, because it is decided on the way down rather than on the way up.

The tube, and what the overstrain does to its bore #

The tube here is 2.5 times as wide outside as its bore, in a steel that yields at 900 N/mm², with Tresca’s criterion: the material yields where the difference between the largest and smallest principal stress reaches the yield strength — a shear-stress criterion, and the simplest one with the right shape. In a pressurised tube those are the hoop stress $\sigma_\theta$, which is tensile and largest at the bore, and the radial stress $\sigma_r$, which is minus the pressure at the bore and zero outside. The bore is where yield starts, at the first-yield pressure

$$p_y = \frac{\sigma_y}{2}\left(1 - \frac{a^2}{b^2}\right),$$

378 N/mm² for this tube. Raise the pressure past that and a plastic zone spreads outward from the bore. Inside it $\sigma_\theta - \sigma_r = \sigma_y$ everywhere, equilibrium makes the radial stress grow logarithmically with radius, and the pressure that puts the plastic boundary at radius $c$ is

$$p_A = \sigma_y\left[\frac{1}{2}\left(1 - \frac{c^2}{b^2}\right) + \ln\frac{c}{a}\right],$$

which reaches $\sigma_y \ln(b/a)$ — 825 N/mm² here — when the whole wall has yielded.

The figure follows a single point of material at the bore round its whole history — the same kind of loop the strain path of a yielding specimen draws. Up the first line to yield; along the yield strength as the overstrain grows; then back down when the pressure is released, along a line whose slope is Lamé’s elastic solution for a pressure change at the bore, $2k^2/(k^2-1)$ for a wall ratio $k$.

Two releases are drawn. From 756 N/mm² the bore comes down to exactly −900, the yield strength in reverse, and stops there with nothing to spare. From 825 — the overstrain that yields the whole wall — it reaches −900 while there is still 69 N/mm² of pressure to remove, and from then on it yields backwards. The extra plastic strain it accumulates on the way up is reversed on the way down.

The ceiling the release sets #

What autofrettage buys is the pressure the tube can then take elastically. Reloading from zero, the bore travels back up its release line and stays elastic until its stress difference reaches the forward yield strength again — which, if the release was elastic, is exactly where it was released from. So the elastic pressure after autofrettage is the overstrain pressure, provided the release did not reverse the bore.

If the release did reverse it, the bore starts the reloading from the reverse yield strength rather than from the residual stress it would otherwise have had, and the reload climbs the full interval — forward yield minus reverse yield — before it yields again. With no Bauschinger effect that interval is $2\sigma_y$, which in pressure is

$$p_w = 2 p_y = \sigma_y\left(1 - \frac{a^2}{b^2}\right),$$

756 N/mm² for this tube. That is the ceiling: overstrain past $2p_y$ and the release reverses the bore, the reload starts from the bottom of the interval, and the working range is $2p_y$ whatever the overstrain was. It is the factor of two from the beam, arriving at the bore of a tube.

Each line follows the diagonal — a newton of overstrain is a newton of elastic range — and then either runs out of wall or runs into the ceiling. The 1.5 and 2.0 tubes run out of wall first: the overstrain that yields the whole wall, $\sigma_y \ln k$, is less than $2p_y$, so they can be overstrained right through and the release never reverses the bore. The 2.5 and 3.0 tubes hit the ceiling first and then go flat. For them the overstrain beyond $2p_y$ buys nothing, and it is worse than nothing in a way the figure cannot show: it puts reversed plastic strain into the bore at every production cycle, which is low-cycle fatigue spent before the tube has done any work.

The pressure no wall reaches as made #

The ceiling has a second reading, and it explains why autofrettage exists at all rather than thicker walls.

As made, a tube’s first-yield pressure is $\tfrac12\sigma_y(1 - 1/k^2)$, and that rises with the wall ratio only towards a limit: $\sigma_y/2$, however thick the wall. A tube in a 900 N/mm² steel can never hold more than 450 N/mm² elastically as made. The reason is that the bore carries the whole of the pressure as radial stress and at least as much again as hoop stress, whatever the wall outside it is doing, so the stress difference at the bore is always more than twice the pressure applied. Adding wall moves the hoop stress at the bore down towards the pressure and no further. No thickness buys an elastic pressure above half the yield strength, and the tube drawn, at 2.5 to 1, already has 84 per cent of everything a thicker one could give.

After autofrettage the ceiling is $(1+\beta)p_y$, which tends to $(1+\beta)\sigma_y/2$ — the whole yield strength, for a steel without a Bauschinger effect. The overstrain is therefore not an improvement on a thicker wall; it is the only way past a limit that wall thickness cannot move. The pressure that a tube can carry elastically, with any wall and any overstrain, is bounded by the yield strength itself, and the factor of two is what separates the as-made bound from that one.

The same limit has a name in another part of the subject. A structure under a load that cycles between zero and a peak shakes down — settles into a purely elastic response after its first cycles — if a residual field exists that keeps every point inside its yield interval throughout the cycle. For a thick tube cycled between zero and a pressure, the largest such pressure is the smaller of the whole-wall overstrain and twice the first-yield pressure: the autofrettage ceiling exactly. Autofrettage is shakedown done once, deliberately and in the factory, so that the service cycles never have to do it.

The wall ratio where the ceiling arrives #

The two limits are $\sigma_y\ln k$, from the wall running out, and $2p_y = \sigma_y(1 - 1/k^2)$, from the release. They are equal where

$$\ln k = 1 - \frac{1}{k^2},$$

whose root is $k = 2.218$.

The finding is a number with no material in it. Every tube thicker than 2.218 to one, in any steel that yields equally in both directions, gains exactly a factor of two from autofrettage and no more. A designer who makes a tube thicker to take more pressure gains from the thickness itself — $p_y$ rises with $k$, slowly — but not from the overstrain, whose contribution has topped out. Below 2.218 the gain is $\ln k/(\tfrac12(1 - 1/k^2))$, falling to 1.46 at a wall ratio of 1.5 and towards 1 as the tube becomes thin, where there is no wall for an elastic outer layer to squeeze an inner one with.

Real steels do not yield equally in both directions after a forward plastic strain. The Bauschinger effect — named for the engineer who measured it in Munich in the 1880s — lowers the yield strength in reverse after a forward yield, and in the high-strength steels autofrettage is applied to the reverse yield can be well under the forward one. With a reverse yield strength $\beta\sigma_y$, the interval the bore can travel is $(1 + \beta)\sigma_y$ and the ceiling is $(1+\beta)p_y$. At $\beta = 0.7$ it is 1.70 and arrives at a wall ratio of 1.80; at 0.5, it is 1.50 and arrives at 1.55. The Bauschinger effect costs autofrettage directly in proportion, and it costs it at wall ratios where, without it, a whole-wall overstrain would have been the right answer.

How deep to yield #

The design question that follows is how far into the wall the overstrain should reach, and the answer runs against the intuition that more is better.

Past the crossover the best overstrain is partial, and the thicker the wall the less of it should be yielded: just over half of a 2.5 to 1 wall, 56 per cent; 37 per cent of a 3 to 1 wall; under a fifth of a 5 to 1 wall. The reason is that the residual compression at the bore grows with the depth of overstrain, and the best overstrain is the one that leaves the bore at exactly the reverse yield strength after release. Yield any deeper and the release would have to take the bore further than the reverse yield strength allows, so it yields backwards and the extra depth is bought and given back.

With a Bauschinger effect the best depth is shallower still, and that has a practical edge. Autofrettage is done either hydraulically, with a pressure, or by swaging — forcing an oversize mandrel through the bore — and in either case the depth of overstrain is controlled by how far the bore is stretched. A specification that says “yield the wall through” for a thick tube in a steel with a pronounced Bauschinger effect is asking for an overstrain that the release will partly undo.

Measuring the depth from outside #

An overstrain that is meant to stop part-way through the wall has to be controlled, and the plastic boundary is inside the metal where nobody can see it. The outside surface is where it is measured, and the elastic outer layer makes the measurement exact.

Outside the plastic boundary the wall is a thick elastic tube loaded by the pressure the plastic zone passes on, $\tfrac12\sigma_y(1 - c^2/b^2)$ at radius $c$. At the free outer surface that tube’s hoop stress is $\sigma_y c^2/b^2$ and its radial stress is zero, so the hoop strain a gauge on the outside reads at the peak of the overstrain is

$$\varepsilon_\theta(b) \approx \frac{\sigma_y}{E}\cdot\frac{c^2}{b^2}.$$

The plastic radius is in that strain, and nothing else is. For the 2.5 to 1 tube overstrained to its best, $c/b = 1.84/2.5 = 0.74$, and the outside should read $900/205{,}000 \times 0.54 = 2{,}380$ microstrain — a little over half the strain at which the outside itself would begin to yield. A production line that stops the pump when the gauge reads that figure has put the boundary where the calculation wanted it, without knowing the pressure; one that stops at the pressure has to trust that the steel’s yield strength is the one the calculation assumed. The outside strain is the better control, because it measures the quantity the design depends on.

What is left in the wall #

The overstrain’s whole value is the residual field it leaves, and that field has the shape every residual field has: a balance of compression and tension that adds to nothing.

The hoop stress is −900 N/mm² at the bore — the reverse yield strength exactly, which is what makes this overstrain the best one — rising through zero inside the formerly plastic zone to a tensile peak at its boundary and falling gently to 202 N/mm² at the outside. Its integral across the wall is zero, as it must be for a stress that was there before the load: nothing is applied to the tube, so its hoop forces can only balance one another. The outer part of the wall is in tension so that the bore can be in compression, and the outer surface of an autofrettaged tube is therefore more highly stressed at rest than the outer surface of one that was never overstrained — which matters for a tube whose outside is corroding or scratched.

The range that does not change #

The point of the residual compression is what it does to the bore in service.

The two lines are parallel, and that is the whole of the second finding. A pressure cycle from zero to the working pressure moves the bore’s hoop stress by Lamé’s elastic amount, $p(k^2+1)/(k^2-1)$, whether or not the tube was overstrained. Autofrettage does not reduce the stress range at the bore; it moves it, from a cycle that runs from zero to a large tension to one that runs from a large compression to a small tension. For a cylinder working at 600 N/mm² — a pressure the tube as made could not take elastically at all, since its bore yields at 378 — the overstrained bore cycles from −900 to −71 N/mm² and never goes into tension at all.

That is why autofrettage is a fatigue treatment as much as a strength one. A fatigue crack grows on the part of the stress cycle that opens it, and a crack at the bore of an autofrettaged tube is held shut for most or all of every pressure cycle. A rule that judged fatigue by stress range alone would give autofrettage no credit, and the fatigue lives of autofrettaged tubes are many times those of tubes that were not — the effect lives entirely in the mean stress, which the range rule throws away.

The whole of it, by hand #

For the 2.5 to 1 tube in a 900 N/mm² steel, every number above comes from four lines.

The first-yield pressure is $p_y = 450(1 - 1/6.25) = 378$ N/mm². The overstrain that yields the whole wall is $900\ln 2.5 = 825$. The release ceiling is $2p_y = 756$, smaller, so the best overstrain is 756 and the tube is past the crossover. The plastic radius for 756 solves $900[\tfrac12(1 - c^2/6.25) + \ln c] = 756$, which gives $c = 1.84$ bore radii, 56 per cent of the way through. The bore’s residual hoop stress is the yield strength less the release’s elastic change, $\sigma_y - 2p_A k^2/(k^2-1) = 900 - 2 \times 756 \times 6.25/5.25 = -900$ N/mm² — minus the yield strength, as the ceiling requires.

Tresca, perfect plasticity and a tube with no ends #

The calculation rests on four idealisations, and each pulls in a known direction.

Tresca rather than von Mises. The two criteria differ by up to 15 per cent in the pressure at which a thick tube yields, depending on whether its ends are closed and how the axial stress is carried; von Mises with closed ends gives higher pressures throughout, by the factor $2/\sqrt3$ in the limiting case. The ratios above — the factor of two, the crossover, the depth — are Tresca’s, and the same argument with von Mises shifts the numbers without changing their shape.

Perfect plasticity. A steel that hardens as it yields gives a higher overstrain pressure for the same plastic depth, and a higher forward yield strength for the reload, so the ceiling moves up a little. A steel that softens does the opposite.

A single reverse yield strength. The Bauschinger effect is not one number: the reverse yield strength falls with the amount of forward plastic strain, so the deeper and harder the overstrain the larger the loss. Treating β as fixed overstates what a deep overstrain keeps.

A uniform tube. Real autofrettaged parts have cross-bores, threads and changes of section, each a stress concentration whose local yielding and residual field is not the tube’s. The cross-bore of a fuel rail is where the fatigue cracks start, and the residual field at it is a three-dimensional problem the tube’s solution only approximates.

Heat, and a crack already started #

The figures cannot show time. A residual stress held at the reverse yield strength in a steel at elevated temperature relaxes by creep, and a gun barrel heated by firing loses part of its autofrettage over its life for that reason. A tube in a hot process line is in the same position: the residual field drawn here is the day it left the factory.

They also cannot show a crack already present. A bore that yields backwards on release at every production cycle — the overstrain beyond the ceiling — has spent some of its low-cycle fatigue life before service, and a crack started by that reversed yielding is a crack the residual compression then has to hold shut.

Still open: the tube made of two #

The other way to put a bore in compression is older than autofrettage: shrink an outer tube on to an inner one, so that the outer is in hoop tension and the inner in compression with no plastic strain at all. A compound tube can be designed so that both its layers reach yield together at the working pressure, and it has no Bauschinger effect to lose, because nothing has yielded. Whether a shrink fit beats an overstrain for a given wall — and whether the two combined, an autofrettaged inner tube with a shrunk-on jacket, can carry the interval past the factor of two that limits either alone — is a question about how many places a designer may put the stress before the load arrives.