Structural Engineering & Statics
where the load goes
Connections

The joint that has to be as good as the member

A splice exists because members come in lengths and structures do not. It has to deliver the same force, at the same stiffness, in the same distribution across the section, through a discontinuity — and each of those three requirements is met by a different feature of the detail, with the third one usually left to look after itself.

Assumes The connection is not a point, and every diagram on this site says it is, The bolts that do not share and The tear that goes diagonally, and the correction that has no derivation.

17 min read 10 figures The load must go somewhereWhich failure arrives firstOne support too many

A structure is continuous and its members arrive on lorries. Somewhere along every long member there is a place where two pieces were joined, and the job of that place is to be invisible: to deliver the same force, in the same distribution, at the same stiffness, as the material it replaces.

Three requirements, and they are met by three different features of a detail — a division that a connection which is not a point makes for joints in general and that a splice makes unusually sharply, because a splice has no other job. The first is what gets designed. The second is what gets assumed. The third is what nobody writes down, and the order in which they are usually discovered is the reverse of the order in which they matter.

The end bolts do the work and the middle ones very nearly nothingA lap of 10 bolts at 75 mm pitch transferring 800 kN between two plates, with the force each bolt actually carries drawn above it and the flat line a division by the bolt count would have given drawn behind. The end bolts carry 1.16 of their nominal share and the middle ones 0.89. The reason is not in the bolts: at the leading end the first plate is carrying everything and the second nothing, so the two strain at different rates and the slip between them is largest there. In the middle they strain alike, there is almost no slip, and a bolt with no slip across it transfers almost no force. The mean over the worst is 0.861, and the end bolt has to slip 1.16 mm before the rest catch up.800 kN1.161.050.970.920.890.890.920.971.051.16an equal share675 mm = 33.8 dworst bolt 93 kN against a nominal 80 kN
Fig. 1 A lap of ten bolts, with the force each one actually carries drawn above it and the flat line a division by the bolt count would have given drawn behind.

Which free body produced the number

Cut across the splice and take one plate. The force in it has to be handed to the other plate through the fasteners, and how much each fastener hands over is decided by compatibility rather than by equilibrium — equilibrium only says the total.

Follow the two plates along the lap. At the leading end, the first plate carries the whole force and the second carries nothing, so the first is stretched and the second is not, and the relative slip between them is at its largest. In the middle both plates carry about half, they strain alike, and the slip between them is nearly zero. A fastener transfers force in proportion to the slip across it, so the end fasteners work hard and the middle ones barely work at all.

That is the bolts that do not share and it is the single most important fact about a splice. It makes the connection’s efficiency a function of its length, which is a quantity no member calculation contains.

The longer the joint, the smaller the share the worst bolt is doingHow much of a bolted lap is working, against its length in bolt diameters. The elastic answer is tanh(βL/2)/(βL/2), the mean bolt force over the worst one, and it falls without limit — at 90 diameters the end bolt is carrying 2.19 times its nominal share. The code's own reduction is the flat-then-sloping line, and it is milder, because a bolt in bearing is ductile: the end bolt yields, stops taking more, and passes its share inwards. The gap between the two curves is exactly the ductility the connection is being asked for, which is why the reduction starts at fifteen diameters rather than where the elastic distribution first becomes uneven. The same function governs a reinforcing bar's bond, where 1/α is 471 mm, and a cooling fin.0204060800.40.50.60.70.80.91bolted length ÷ bolt diametershare of the joint that is workingthe code's reductionelastic: tanh(u)/u15 d
Fig. 2 How much of a bolted lap is working, against its length in bolt diameters. The elastic answer is tanh(βL/2) over βL/2, and it falls without limit.

At ninety diameters of joint length the end bolt carries 2.19 times its nominal share. The reduction that standards actually apply is milder than the elastic curve, and the gap between the two is exactly the ductility the connection is being asked for: the end bolt yields in bearing, stops taking more, and passes its share inwards. A splice made with fasteners that cannot do that — a fitted bolt, a bonded joint, a brittle weld — has to live on the elastic curve.

The section that is left

The second requirement a splice has to meet is that the pieces it is made of are weakened by the holes it needs.

The net section, and the path the tear takesA 200 mm plate with two holes staggered by 40 mm at a gauge of 60 mm. The straight path through one hole leaves 178 mm; the diagonal path through both leaves 162.67 mm after the s²/4g correction adds 6.67 mm back. The shorter of the two decides, at 81.33% of the gross section.s = 40g = 60net width 162.67 mm of 200the critical path crosses two holes, with s²/4g = 6.67 mm added back
Fig. 3 A plate with two staggered holes, and the two paths a tear can take. The straight path leaves 178 mm and the diagonal leaves 162.67 after the correction, so the diagonal decides at 81% of the gross.

The tear that goes diagonally is the essay about that construction. Its consequence for a splice is a familiar tension: staggering the holes recovers net area and lengthens the joint, which costs efficiency by the argument above. The two effects pull in opposite directions and the optimum is short and staggered — which is also the arrangement hardest to fit a spanner into.

An angle bolted through one legA 150 × 90 × 10 angle connected through its 150 mm leg with three bolts at 75 mm pitch. The centroid sits 20.65 mm from the connected face over a connection 150 mm long, so U = 1 − 20.65/150 = 0.86 and 13.77% of the net area is not working.x̄ = 20.65connected legoutstanding legLc = 150net areaU = 0.86 of it worksU = 1 − x̄ / Lc = 0.86both halves are geometry — where the centroid sits, and how long the connection is
Fig. 4 An angle connected through one leg. The centroid sits 20.65 mm from the connected face over a 150 mm connection, so 13.77% of the net area is not working.

The third reduction is the one that is really about distribution. Connect an angle through one leg and the force arrives on one side of the section; the other leg only picks up its share after the force has spread into it, and over a short connection it never fully does. The remedy is a longer connection, which costs efficiency again — the same trade an angle that uses half of itself is about.

Three reductions, all geometric, all interacting, and none of them a property of the member being spliced. Together they mean a splice made of the same material and the same thickness as the member it joins is not as strong as it, and the shortfall is between a tenth and a third depending on nothing but how the holes were arranged.

Matching the section, not the force

The requirement that gets least attention is that a splice should deliver the force to the same parts of the section it came from.

A beam’s moment is carried mostly by its flanges. A splice made with a single deep web plate carrying the whole moment delivers it to the web instead, and the force then has to travel from the web into the flanges through a length of member — a redistribution over about a member depth, with shear stresses in the web that the member calculation never contained.

The usual detail therefore splices flange to flange and web to web, with the flange plates sized for the flange’s own share of the moment (MIf/IM I_f/I, roughly MM times 0.85 on a rolled beam) and the web plate for the shear plus the web’s small share. That looks like a decomposition made for convenience and it is a requirement: a splice reproduces a stress distribution, not just a resultant.

Throat stress round a fillet weld groupA c shape weld group carrying 250 kN at 200 mm from its centroid. The peak throat stress is 2.78 kN per mm of throat, at (79.5, -100); the worst point at maximum radius from the centroid carries 2.78. Checking by radius is right here, and points at identical radius differ by a factor of 1.250 kNpeak 2.78Two points at maximum radiusthe radius rule finds the peak hereequal radius, stresses differ ×1
Fig. 5 Throat stress round a fillet weld group under an eccentric load. The peak is at the greatest radius from the centroid, and the check is a distribution rather than a total.

The same argument decides how a welded splice is made. A full-penetration butt weld reproduces the parent section exactly and needs no argument at all; a fillet-welded lap does not, and its stress distribution has to be checked round the group rather than divided by a length.

Slip, which is the stiffness requirement

The second of the three requirements is that the splice should not change how the member behaves, and for a bolted joint that is a question about slip.

A preloaded joint, before and after it slipsEight preloaded bolts at 172 kN each, on one friction face at μ = 0.5. The joint carries 688 kN by friction with the bolts in tension and not in shear at all; past that it slips into bearing and carries 250 kN with the bolts now in shear. Two different mechanisms, one joint.00.511.522.533.544.550200400600800displacement, mmload, kNfriction 688 kNbearing 250 kNslipthe rising branch is drawn, not solved: it is elastic shear of the plates
Fig. 6 A preloaded joint before and after it slips. It carries 688 kN by friction with the bolts in tension and not in shear at all; past that it slips into bearing and carries 250 kN with the bolts now in shear.

A joint in clearance holes has to travel the clearance before its bolts bear. Two millimetres per joint sounds negligible and is not, because a bracing system may contain several in series and the slip is a rigid-body movement that no stiffness calculation contains. A joint that carries nothing until it slips is the essay about the mechanism.

The two mechanisms in that figure are worth reading as a pair. Preloaded, the joint carries 688 kN by friction with the bolts in tension and not in shear at all. Slipped, it carries 250 kN with the bolts now in shear — a different mechanism, a different capacity, and a different stiffness. One joint, two structures, with the transition at a load that depends on a coefficient of friction and a bolt tension nobody will ever measure again. Which of the two the structure is in at any moment is not recorded anywhere, and the only way to find out is that it has already happened.

What a splice costs, counted properly

The three requirements have three prices, and it is worth putting them beside each other because they are paid to different people.

Strength is paid in bolts and plate — the visible cost, and usually the smallest. A flange splice on a rolled beam is a few kilograms of plate and a dozen bolts.

Stiffness is paid in preload and in fit. Preloading eight bolts means torque control or turn-of-nut, an inspection regime, and faying surfaces prepared to a specified condition — which is site labour rather than material, and which is where most of a splice’s real cost sits.

Distribution is paid in the arrangement, and it is nearly free. Splicing flange to flange rather than through a single deep plate costs nothing but the decision to do it.

The ordering explains a persistent pattern in construction. Splices get value-engineered by removing preload, because preload is the expensive part and the strength check passes without it. What is lost is the stiffness requirement, which appears in no calculation that was checked and shows up later as a drift that is larger than predicted or a joint that is heard slipping.

The same reasoning, run the other way, is why a welded splice is often cheaper in the end for a member that has to be stiff. A butt weld satisfies all three requirements at once, by being the member — and it costs more per joint and nothing at all in inspection of a mechanism that might not have been achieved.

Where to put it

Splices are placed where the member’s force is smallest, and the usual instruction is to put them near a point of contraflexure.

2 continuous spans against 2 simple onesThe bending moment in a continuous beam, solved by the stiffness method, drawn over the moment in the same spans made simply supported. The peak sagging moment falls from 182.3 to 102.5, and a hogging moment of 182.3 appears over the supports where there was none.moment102.5 sagging182.3 hogging182.3 if the spans were simplereactions 60.8 202.5 60.8 — the inner supports carry far more than a sharethe continuous case needed stiffness; the comparison did not
Fig. 7 Two continuous spans against two simple ones. The contraflexure points are where the continuous diagram crosses the axis, and they are properties of a load case rather than of the beam.

Three cautions come with that. The moment is zero for one arrangement of load, and it moves as soon as the load is patterned — which is guaranteed, because imposed load can be absent. The shear is not small there; on a continuous beam it is near its largest. And the stiffness requirement does not care where the splice is at all: a joint that rotates changes the member’s behaviour wherever it sits.

What the joint does to the beamEnd moment as a fraction of the fixed-end value wL²/12, against the joint's rotational stiffness, for a beam of EI/L = 17142.86. At the rigid boundary of 137142.86 kN·m/rad the joint delivers 80% of it and at the pinned boundary 20%. Everything between the two lines is a redistribution nobody chose and every analysis assumed away.02000040000600008000010000012000014000016000018000020000022000024000000.20.40.60.81joint rotational stiffness, kN·m/radend moment ÷ wL²/124.99%25.93%57.3%rigid boundarysemi-rigidfixed ended
Fig. 8 What a joint’s rotational stiffness does to the beam it is in. Between the rigid and pinned boundaries the end moment runs from 80% of the fixed-end value to 20%, and everything between is a redistribution nobody chose.

That figure is usually read as being about a beam-to-column connection, and it applies to a splice with more force. A splice in the middle of a member is a rotational spring inserted in it, and if it is softer than the member the moment redistributes away from it — which is safe for the splice and unsafe for wherever the moment went.

Moment against rotation, for three real jointsThree connections on one plot, with the classification boundaries for a beam of EI/L = 20000 drawn as rays through the origin. web cleats is pinned, flush end plate is semi-rigid, extended end plate is semi-rigid. The boundaries are multiples of EI/L, so the same joint is rigid on a short stiff beam and semi-rigid on a long slender one.00.0050.010.0150.020.0250.030.0350.040.0450.05050100150200rotation, radiansmoment, kN·mrigid abovepinned belowweb cleats — pinnedflush end plate — semi-rigidextended end plate — semi-rigid
Fig. 9 Three real connections against the classification boundaries. What counts as rigid is a multiple of EI/L, so the same detail is rigid in one member and semi-rigid in another.

The column splice, which is a different problem

Everything above is about a member in tension or bending. A column splice is governed by something else entirely and is worth separating, because the reasoning inverts.

A column carries compression, and compression can cross a joint by bearing — the two ends in contact, force passing directly from one to the other with the splice plates carrying almost nothing. Where the ends are prepared for contact, a column splice is nominal for the compression it was ostensibly made for.

What it is not nominal for is everything else. It has to carry the tension that arises under wind uplift or a reversed moment; it has to carry the shear at the splice level; and it has to hold the two lengths in line, because a splice that allows a kink has introduced an eccentricity into a member whose whole design is an eccentricity check. That last requirement is a stiffness one and it is why a column splice’s plates are sized for a notional force rather than for a computed one.

The other consequence is about erection. Between the moment the upper length is landed and the moment its splice is bolted up and its bracing connected, the column is a cantilever from the splice, carrying its own weight and whatever wind is blowing — which is the most dangerous day and a load case in which the splice is the only structure there is.

Where the model stops

The bolt distribution above is elastic. The tanh expression assumes every fastener is a linear spring, which is right until the end one yields. Past that the connection redistributes and the ultimate capacity is very nearly the sum of the individual capacities — so the length effect is a serviceability and fatigue problem, and only a strength problem for fasteners that cannot yield.

Fatigue does not forgive. A splice is a discontinuity in a stress field, so it is a fatigue detail — several classes below the parent member — and in a member that sees repeated load the splice may be the whole design.

Nothing here is about erection. A splice exists to allow a member to be delivered and lifted, so its position is decided by lorry lengths, crane capacity and where a temporary support can go, and the structural argument is applied to a location chosen for other reasons.

And the member may be spliced in more than one plane. A beam spliced in its web and its flanges has three bolt groups with three different eccentricities from the section’s own centroid, and the small residual moments they leave on one another are the reason a splice is checked as a group rather than as a set of parts. The same is true of the moment an eccentric reaction leaves behind: a force delivered off a centreline is a force plus a couple, and a splice has several centrelines to miss.

A bolt group under an eccentric loadA 3 by 2 bolt group carrying 400 kN at 150 mm from its centroid, with the resultant force on each bolt drawn to scale, by the elastic vector method. The load is shared equally and the torque is not, so the worst bolt carries 201.46 kN against 66.67 kN of direct shear alone — 3.02 times as much.400 kNe = 150the dashed ring is the group's centroidworst bolt 201.46 kNSix bolts · direct shear 66.67 kN eachelastic vector method
Fig. 10 The other way a group is loaded unevenly. An eccentric load shares its direct component equally and its torque not at all, so the worst bolt carries 3.02 times the direct shear alone.

What the picture cannot show

A splice drawing shows plates and bolts and says nothing about which of the three requirements each feature is meeting. The flange plates are there for the distribution, the bolt count for the strength, the preload for the stiffness, and the arrangement for the fatigue class — and a detail simplified by removing one feature usually loses one requirement silently.

Nor does any drawing show the fit. A splice with a gap between the member ends is a splice whose plates carry the compression too; a splice with the ends in contact is one where they may not. Which of those was built is decided on site, by a saw cut, and it changes which mechanism carries the compression flange.

The generalisation

The habit worth carrying is that continuity has three separate meanings, and a detail can supply one without the others.

Continuity of force is a strength requirement and is what gets designed. Continuity of stiffness is what keeps the analysis true, and it is met or not by features — preload, fit, plate thickness — that the strength calculation does not mention. Continuity of distribution is what keeps the member’s own behaviour intact, and it is met by which parts of the section are connected to which.

The same triple appears elsewhere. A moment connection can be strong and soft, which changes the frame’s drift. A repair can restore strength and not stiffness, which redistributes load away from it. A bearing can carry a reaction and not allow a rotation, which puts a moment into a member designed as simply supported.

In every case the calculation that was done answers the first question, the analysis quietly assumed the second, and the third is a matter of where the pieces were attached. A splice is the clearest example because it has no other purpose: it exists solely to be as good as the thing it interrupts, which makes every way of failing to be a design decision rather than an accident.

Where this goes next

800 kN1.091.010.960.940.940.961.011.09an equal share490 mm = 24.5 dworst bolt 109 kN against a nominal 100 kN The bolts that do not share Every bolted connection in this collection has divided a force by a number of bolts. That is right for a short joint and wrong for a long one, and the reason has nothing to do with the bolts — it is that the plates they join are elastic, and stretch by different amounts at different points along the lap. s = 50g = 60net width 166.42 mm of 200the critical path crosses two holes, with s²/4g = 10.42 mm added back 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. 100 kNe = 150the dashed ring is the group's centroidworst 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. 00.511.522.533.544.55050100150200displacement, mmload, kNfriction 137 kNbearing 188 kNslipthe rising branch is drawn, not solved: it is elastic shear of the plates The joint that carries nothing until it slips Tighten the bolts hard enough and the plates are clamped together with a force nothing applied. The joint then carries shear by friction, the bolts are in tension and not in shear at all, and the load path has nothing in common with the joint it looks identical to.

Named alongside this one

Essays reaching for the same objects. Nobody chose these; they are what the concept index makes visible.

The objects this essay names

Each one links to every other essay that touches it.

Bolt groupConnectionContinuityContraflexureDuctilityJoint stiffnessLoad pathLoad sharingMoment rotationNet sectionPreloadShear lagSlipSpliceWeld group