Maximum Unsupported Span (Barton)

This is the other half of Barton's chart, and the half people remember: given the rock mass quality and the excavation support ratio, how wide a hole will stand up with nothing in it? The relation is $B_{max} = 2\,ESR\,Q^{0.4}$, and like everything else in the Q-system it is a line drawn through case records rather than a mechanics result.

The exponent is gentle and the ESR is not. Because $Q$ enters to the power 0.4, ground ten times better buys only about 2.5 times the span. ESR, by contrast, enters linearly and multiplies the answer directly. Run the same $Q = 10$ ground twice: as a permanent mine opening at ESR 1.6 you may leave 8.0 m unsupported, and as a rail tunnel at ESR 1.0 you may leave 5.0 m. Identical rock. The difference is sixty percent of the span, and every bit of it is a decision about what a collapse would cost. If there is one lesson in this shard, it is that one, and it is why the anchor tests carry the same pair worked by hand.

The leading 2 is 2 metres. It carries a length, not a pure number, so the equation does not survive a change of units on its own. An imperial reader's answer here has been converted from a metric one and will very rarely be a round number of feet — the $Q = 10$, ESR 1.0 case above comes out at 16.48 ft. That is a property of the equation and not of the ground, and it is worth knowing before somebody assumes a decimal has slipped.

What "unsupported" means is narrower than it sounds. It means the chart indicates no systematic support for an opening this size in ground this good. It does not mean nothing needs doing. Scaling loose material off the roof, spot-bolting a wedge that somebody identifies at the face, mapping every advance and revising the assessment when the mapping disagrees with the assessment — these are ordinary work in any tunnel, they are not systematic support, and none of them appears anywhere in this equation. A self-supporting span is a span that does not need a pattern, not a span that does not need attention.

It also assumes the $Q$ you entered describes the rock over this span, here. That is the assumption that fails most often and most expensively. A $Q$ logged in a competent stretch and carried forward across a shear zone is the classic route from a self-supporting span to a roof fall, and the arithmetic gives no warning at all: the equation is perfectly happy to tell you that eight metres will stand in ground it has never been shown.

On running the relation backwards. Solving for $Q$ inverts the exponent to 2.5, which makes the direction steep: a 20% error in the span becomes a 60% error in the implied quality. Asking what quality a span that has already stood for forty years implies is a fair and useful question, particularly for old openings nobody logged. Working back from the span you would like to the $Q$ you will then claim is not, and the steepness is exactly what makes it comfortable — a small stretch of the span buys a large stretch of the index. Log the face instead.