Structural Engineering & Statics
where the load goes
Connections

The tear that goes diagonally, and the correction that has no derivation

Stagger the holes so that no straight line crosses more than one and the plate does not get its strength back. The tear runs at an angle instead, and the arithmetic that makes it come out right is a century-old piece of curve-fitting nobody has improved on.

Assumes The metal between the holes, which comes out as a block and The connection is not a point, and every diagram on this site says it is.

17 min read 8 figures Drawing as calculationGeometry beats material

A plate carries tension through a bolted connection. Making the connection required drilling holes, and the holes removed material.

The obvious accounting is a subtraction: gross width, less the diameter of each hole on the line, times the thickness. For a 200 mm plate with two 22 mm holes side by side, that is 156 mm of net width and 78% of the gross section.

And the obvious response to that accounting is to stagger the holes, so that no straight line across the plate crosses more than one of them. Which should give 178 mm and 89%, for free, by moving one hole along a bit.

It does not, and the reason is that the tear is not obliged to be straight.

The net section, and the path the tear takesA 200 mm plate with two holes staggered by 50 mm at a gauge of 60 mm. The straight path through one hole leaves 178 mm; the diagonal path through both leaves 166.42 mm after the s²/4g correction adds 10.42 mm back. The shorter of the two decides, at 83.21% of the gross section.s = 50g = 60net width 166.42 mm of 200the critical path crosses two holes, with s²/4g = 10.42 mm added back
Fig. 1 The same two holes staggered by 50 mm. Every candidate tear path is drawn, and the one that governs is picked out: it goes diagonally through both holes, leaving 166.4 mm rather than the 178 mm the straight path would leave. The stagger bought 10.4 mm of the 22 mm it appeared to be worth.

Net section is a search, not a subtraction

The first thing to fix is the shape of the calculation. Net section is not “gross minus holes”. It is a minimum over paths.

A tear path starts at one edge, ends at the other, and passes through some subset of the holes in whatever order takes it across. There are as many candidates as there are subsets, and the plate’s net width is the smallest of them. For two holes there are three candidates — through the first, through the second, through both — and for three holes there are seven.

That framing matters because it makes the answer depend on the arrangement rather than on the count. Two holes can cost 22 mm or 44 mm of width depending on where they are relative to one another, with no change in how many there are or how big they are.

The net section, and the path the tear takesA 200 mm plate with two holes staggered by 0 mm at a gauge of 60 mm. The straight path through one hole leaves 156 mm; the diagonal path through both leaves 156 mm after the s²/4g correction adds 0 mm back. The shorter of the two decides, at 78% of the gross section.g = 60net width 156 mm of 200the critical path crosses two holes, with s²/4g = 0 mm added back
Fig. 2 The unstaggered case, for comparison. Now the only path worth taking is the straight one through both holes, and it leaves 156 mm: the full 44 mm gone. Every candidate path is still drawn, and the diagonal ones are simply longer with nothing to gain.

The correction, and what it is standing in for

Along a diagonal leg between two holes, the rule is to add back

s24g\frac{s^2}{4g}

where ss is the stagger along the member and gg the gauge across it. For the connection at the top of this page, s=50s = 50 and g=60g = 60 give 2500/240=10.42500/240 = 10.4 mm.

Two things about that term are usually misunderstood, and both are worth getting right.

It is not a correction for the diagonal being longer. The obvious thought is that the tear has further to go, so more material has to fail, so add back the extra length. A diagonal leg of stagger 50 across a gauge of 60 is 502+602=78.1\sqrt{50^2+60^2} = 78.1 mm long against a straight 60, which is 18.1 mm of extra length where the rule adds back 10.4.

The two numbers being different in size is the smaller objection. The larger one is that the extra-length account is a correction pointing the wrong way for the right reason: a longer failure surface is a stronger path, and if length were all that mattered the diagonal would never govern at all — a path through two holes would be preferred over a path through one only when the extra length was worth less than the extra hole, which is a different and much weaker condition. Something has to make the diagonal leg worse per millimetre, and the rule is that something, compressed into two symbols.

It is standing in for a change of failure mode. A straight tear across the plate is pure tension: the material separates by pulling apart. A diagonal leg is not. Its normal is inclined to the load, so the material on it is carrying a mixture of tension and shear, and shear reaches its limit at about 0.6 of the direct strength. So the diagonal leg is both longer and weaker per unit length, the two effects partly cancel, and s2/4gs^2/4g is what is left.

That is the physical account. What it is not is a derivation. The term was proposed by Cochrane in 1922, fitted to test results, and has been in codes ever since — and repeated attempts to replace it with something derived have produced expressions that are more complicated and no better against data. It is one of the very few pieces of pure empiricism in structural steel design, and the honest way to hold it is as a fit that works rather than as a result.

Where the stagger stops mattering

If the diagonal path governs at 50 mm of stagger and the straight path governs at very large stagger, there is a crossover, and it is computable.

The two paths are equal when the diagonal’s correction has recovered exactly one hole’s worth of width:

s24g=d⟹s=2gd\frac{s^2}{4g} = d \qquad\Longrightarrow\qquad s = 2\sqrt{gd}

For g=60g = 60 and d=22d = 22, that is 72.7 mm — larger than the gauge itself. Below it the diagonal governs; above it the straight path does and further staggering buys nothing at all.

The net section, and the path the tear takesA 200 mm plate with two holes staggered by 80 mm at a gauge of 60 mm. The straight path through one hole leaves 178 mm; the diagonal path through both leaves 178 mm after the s²/4g correction adds 26.67 mm back. The shorter of the two decides, at 89% of the gross section.s = 80g = 60net width 178 mm of 200the critical path crosses one hole
Fig. 3 Stagger 80 mm, past the crossover. The critical path now passes through one hole only, and the plate has recovered the full 178 mm the naive account promised. Getting there took a stagger longer than the gauge — which is a great deal of connection length for 22 mm of plate width.

The crossover formula is more useful than any individual answer, because it says what a stagger has to be to be worth anything:

gauge stagger needed
40 mm 59.3 mm
60 mm 72.7 mm
90 mm 89.0 mm
120 mm 102.8 mm

Notice which way the table runs. A wider gauge needs a longer stagger to be worth anything, which is the opposite of the intuition that spreading the holes out generally helps. The reason is visible in the formula: the correction is s2/4gs^2/4g, so the gauge is in the denominator and widening it makes each diagonal leg less able to recover width. Spreading holes apart across the plate is good for the straight path and bad for the diagonal one, and the diagonal one is the one that governs.

Every one of those staggers is larger than a bolt pitch would ordinarily be. Which means that in most real connections the stagger is nowhere near enough to make the diagonal path irrelevant, and detailing bolts in a zigzag “so the holes do not line up” is a habit whose benefit is partial and computable rather than automatic.

The net section, and the path the tear takesA 200 mm plate with two holes staggered by 25 mm at a gauge of 60 mm. The straight path through one hole leaves 178 mm; the diagonal path through both leaves 158.6 mm after the s²/4g correction adds 2.6 mm back. The shorter of the two decides, at 79.3% of the gross section.s = 25g = 60net width 158.6 mm of 200the critical path crosses two holes, with s²/4g = 2.6 mm added back
Fig. 4 Half the stagger, at 25 mm. The correction is s²/4g, so halving s quarters it: 2.6 mm added back rather than 10.4, and a net width of 158.6 mm against 156 for no stagger at all. A small stagger is very nearly worthless, which is the quadratic saying so.

Why a rule with no derivation survives

It is worth asking directly why a fitted constant from 1922 is still in every steel code, when almost everything else in this subject has a derivation behind it.

Part of the answer is that the thing it is approximating is genuinely hard. The stress field around a staggered pattern of holes is three-dimensional and disturbed, the failure involves both a fracture and a shear slip, and the sequence in which the two happen depends on the material’s ductility. A closed-form treatment of that has never been produced and is not obviously available.

The larger part of the answer is that the rule is used inside a search, and a search is forgiving of its terms. The net section is the minimum over many paths, and near the minimum the paths are close together — which means that getting one path’s correction slightly wrong usually changes which path wins rather than what the answer is, and the two answers are similar because they were close enough to compete.

That is a real property of minimum-of-many problems and it is worth carrying beyond this rule. A quantity computed as the smallest of a family is systematically less sensitive to errors in the family than a quantity computed directly. It is the same reason the thrust line’s exact position does not matter as long as one fits, and the same reason a lower-bound plastic mechanism does not have to be the real one.

Three holes, and why the search is not optional

With two holes the answer can be reasoned about. With three it is easier to enumerate, and enumerating shows something the two-hole case hides.

Take a 220 mm plate with holes at y=−70y = -70, y=0y = 0 and y=+70y = +70, the middle one offset 45 mm along. Seven candidate paths:

  • through all three, zigzagging: 168.5 mm
  • through the top two: 183.2 mm
  • through the bottom two: 176.0 mm
  • through the outer two, straight: 183.2 mm
  • through any single hole: 198.0 mm

The critical path takes all three, and it is 14.7 mm worse than the best two-hole path. There is no rule of thumb that finds it — “check the straight line and the obvious diagonal” would have returned 176.0 and been 4% optimistic.

That is the argument for treating this as a search. The number of paths grows as 2n2^n, each is a two-line calculation, and a computer does not mind. Reasoning about which path is likely to govern is exactly the sort of shortcut that works for regular patterns and fails silently for the irregular ones that arise when a connection has to fit round something.

What the hole is, and why it is bigger than the bolt

One detail that hides in the arithmetic: the width subtracted is the hole, not the bolt, and the two differ by more than seems reasonable.

A 20 mm bolt goes in a 22 mm hole, which is 2 mm of clearance for a fastener whose whole job is to be a tight fit. And for calculation the deduction is often taken larger still, because a punched hole leaves a rim of damaged material around it that has already used up some of its ductility. Where holes are punched rather than drilled, the deduction is commonly the hole plus 2 mm.

So a 20 mm bolt can cost 24 mm of plate width — 20% more than its own diameter, in a calculation where 22 mm out of 200 mm is already the difference between passing and failing. The clearance exists so that the steelwork can be erected, and erection tolerance is not a structural quantity, which makes this one of the clearer instances of a fabrication decision arriving in a strength calculation with no label on it.

There is one connection type where the deduction vanishes entirely, and it is worth knowing why: a preloaded, slip-resistant joint checked at serviceability transfers its load by friction across the whole interface rather than by bearing at the holes, so no force is passing through the net section at that stage at all. The holes still have to be deducted for the ultimate check, because the joint is expected to slip into bearing before it fails. But the two limit states look at different areas of the same plate, which is a good illustration of why “the section” is not a single well-defined thing in a connection.

The net section, and the path the tear takesA 200 mm plate with two holes staggered by 50 mm at a gauge of 90 mm. The straight path through one hole leaves 178 mm; the diagonal path through both leaves 162.94 mm after the s²/4g correction adds 6.94 mm back. The shorter of the two decides, at 81.47% of the gross section.s = 50g = 90net width 162.94 mm of 200the critical path crosses two holes, with s²/4g = 6.94 mm added back
Fig. 5 The same 50 mm stagger across a wider gauge of 90 mm. The correction falls to 6.94 mm from 10.42, because the gauge is in the denominator — so spreading the holes further apart across the plate makes the diagonal path worse, not better. It is the counterintuitive direction, and it follows directly from where g sits in the formula.
Bearing and tear-out against end distanceA 20 mm bolt in a 10 mm plate. Below 165 mm of end distance the bolt tears a channel out to the end and the capacity is proportional to that distance; above it the plate crushes in front of the bolt and the end distance stops mattering. At 40 mm the capacity is 52.12 kN and the mode is tear-out.020406080100120140160180200050100150200end distance, mmbearing capacity, kN40 mm → 52.12 kNcorner at 165 mmplate crushesbolt tears out
Fig. 6 A third check on the same holes, answering a third question. Net section asks what is left across the plate; block shear asks whether a piece can come out along it; bearing asks what happens in front of one bolt. The same drawing produces three unrelated numbers, and the connection has whichever is smallest.

What this does to the member

The net section is not just an arithmetic quantity; it is a capacity, and the capacity it produces competes with a different one.

A tension member has two tensile limits. It can yield over its gross section, at fyAgf_y A_g, which is a serviceability-scale event — the member stretches, permanently, along its whole length. Or it can rupture at the net section, at fuAnf_u A_n, which is a fracture and is not.

Because the two use different strengths on different areas, which governs depends on the hole ratio. For the plate above at 83% efficiency and S275 steel, gross yield gives 275×2000=550275 \times 2000 = 550 kN and net rupture gives 430×1664=716430 \times 1664 = 716 kN, so yielding governs and the holes are not the problem.

And that is the general case, which is worth saying because it undercuts the anxiety the whole calculation produces. Ordinary hole ratios in ordinary steel leave gross yielding governing, and a member designed for gross yield has net section capacity to spare. The net section becomes critical when the efficiency is poor — many holes, or a wide hole relative to the plate — or when the steel has a high yield-to-ultimate ratio, which is what a higher grade buys and is one of the few places where a stronger steel makes a check harder rather than easier.

Block shear: the metal between the holesThree bolts in a 10 mm plate end connection. The shaded block tears out along a shear plane 180 mm long and a tension plane 40 mm long. Shear yields first, and the capacity is the sum of two different strengths on two different planes: 421.7 kN, of which the shear plane carries 70.43%.pullshear plane, 180 mmtension plane, 40 mmcapacity 421.7 kN0.6 fu Anv = 322.5 kN · 0.6 fy Agv = 297 kN · fu Ant = 124.7 kNthe yield value governs the shear plane
Fig. 7 And a third reduction on the same plate, which is not an area reduction at all. Block shear does not reduce the section; it proposes a different failure surface entirely, running partly along the member rather than across it. Net section asks how much plate is left on a transverse cut. Block shear asks whether a piece of it can come out sideways. The same holes appear in both and the two numbers are unrelated.
An angle bolted through one legA 100 × 75 × 10 angle connected through its 100 mm leg with three bolts at 75 mm pitch. The centroid sits 19.77 mm from the connected face over a connection 150 mm long, so U = 1 − 19.77/150 = 0.87 and 13.18% of the net area is not working.x̄ = 19.77connected legoutstanding legLc = 150net areaU = 0.87 of it worksU = 1 − x̄ / Lc = 0.87both halves are geometry — where the centroid sits, and how long the connection is
Fig. 8 The other reduction to the same area, and the one that is easy to forget is separate. Shear lag reduces the net area because the connection cannot reach all of the section; net section reduces it because the holes removed material. Both apply, they multiply rather than compete, and a 0.81 shear lag factor on an 83% net section leaves 67% of the gross.

The two reductions multiply

That last point deserves to be stated on its own, because it is a common error and the direction of the error is unsafe.

Net section and shear lag are both written as reductions to an area, they both come out around 0.85, and they both apply to the same member. It is tempting to treat them as two estimates of one effect and take the worse.

They are not. Net section is about material that is not there — the holes. Shear lag is about material that is there and is not fully stressed, because the connection is short and the force cannot spread. Different mechanisms, different geometry, and both true at once. The effective area is UAnU A_n, and 0.81×0.83=0.670.81 \times 0.83 = 0.67.

What to take from it

Net section is a minimum over paths, and the paths include diagonals. Staggering holes does not stop a tear; it makes the tear turn.

The s2/4gs^2/4g term is a fit with no derivation, and it is not about extra length. It stands in for the diagonal leg carrying a mixture of tension and shear. Knowing that is what stops it being applied to geometries it was never fitted to.

The stagger that makes a hole free is 2gd2\sqrt{gd}, and it is larger than most bolt pitches. Zigzagging the holes helps partially and quantifiably, and almost never completely.

Net section usually does not govern — until the steel gets stronger. Gross yielding and net rupture are a race between fyAgf_y A_g and fuAnf_u A_n, and raising the grade moves the finish line towards the net section rather than away from it.

And the three reductions on one plate are three mechanisms, not three estimates. Holes remove material, shear lag leaves material unstressed, and block shear proposes a surface that runs the other way. It is the same lesson the last essay ended on, arriving from the opposite direction: a connection has as many capacities as it has mechanisms, and its capacity is the smallest of them rather than the average of the plausible ones.

Where this goes next

pullshear plane, 180 mmtension plane, 40 mmcapacity 421.7 kN0.6 fu Anv = 322.5 kN · 0.6 fy Agv = 297 kN · fu Ant = 124.7 kNthe yield value governs the shear plane The metal between the holes, which comes out as a block A bolted end connection can fail without a single bolt breaking and without the plate reaching its tensile strength anywhere. A block of metal simply comes out, bounded by two surfaces with two different strengths on them. pulled at 100 N/mm², left and right300-100 — compressionhoop stress, tinted3.0× at the edgewithin 5% by 3.5 radiithe applied stressdistance from the centre, in hole radii12345 The hole that multiplies the stress by three The stress at the side of a hole is three times the applied stress whatever the hole's size, and at the top and bottom of the same hole it is minus one times it — compression in a plate that nothing is pushing. x̄ = 19.77connected legoutstanding legLc = 150net areaU = 0.87 of it worksU = 1 − x̄ / Lc = 0.87both halves are geometry — where the centroid sits, and how long the connection is The angle that uses half of itself Bolt an angle through one leg and the other leg is not fully working. The correction is one over a length — both halves of it are geometry, neither involves a material, and a two-bolt connection throws away a quarter of the section. 100 kNe = 150centroidworst bolt 50.37 kNSix bolts · direct shear 16.67 kN eachelastic vector method The connection is not a point, and every diagram on this site says it is Every free body drawn here has joined its members at points. Real structures fail at the joints far more often than in the members, and the reason is that a joint is exactly the region the theory behind every other page explicitly excludes. 020406080100120140160180200050100150200end distance, mmbearing capacity, kN40 mm → 52.12 kNcorner at 165 mmplate crushesbolt tears out The hole that goes oval, and the one that tears to the edge A bolt pressing on the side of its hole either crushes the plate in front of it or shoves a channel of metal out to the end. Which one happens is decided entirely by a distance that is usually set by a minimum in a table.

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Bolt groupConnectionEfficiencyEmpirical ruleNet sectionShear planeStaggerUltimate strength