Structural Engineering & Statics
where the load goes
Concept

Timber — where it appears

Wood used as a structural material: anisotropic, variable, and graded rather than specified. Its members are usually adequate and its connections rarely are, so timber structures are pin-jointed, because a joint able to develop a member's own capacity is larger than the member.

Named by 11 essays across 3 fields — each of them below, with the objects they name alongside it.

Strength at an angle, and the straight line that is not it. Compressive strength against the angle between the load and the grain. Hankinson's formula — f₀f₉₀ ÷ (f₀sin²α + f₉₀cos²α) — is an interpolation rather than a failure theory, and what makes it worth having is how far it sits from the straight line anyone would otherwise draw between 21 and 2.5 N/mm². At forty-five degrees it gives 4.5 N/mm² against the line's 11.8: 38 per cent of it, and 21 per cent of the strength along the grain. The curve drops away in the first twenty degrees because the weak direction starts governing as soon as it has any component at all, which is the same arithmetic as a section's weak axis and the reason a skewed bearing detail is a real loss rather than a small one.

The material that has a direction

Every material in this collection so far has had one modulus and one strength. Timber has three of each, differing by more than an order of magnitude, and the consequence is not a correction to steel design — it is a different set of checks with a different one governing.

materials · Anisotropy Six ways for one dowel to fail, and the capacity is the smallest. Johansen's single-shear mechanisms for a 12 mm dowel through 40 and 40 mm members, each drawn as the shape it is: the dowel straight and the timber crushing, the dowel rotating rigidly, one plastic hinge, then two. The capacity under each is that mechanism's own, and the joint's strength is the smallest — 5.02 kN by mode c, which is the dowel rotates rigidly and both members crush. That is the kinematic theorem of plasticity: every mechanism gives an upper bound and the true collapse is the lowest of them. The crushed timber is shaded, and the circles are plastic hinges in the steel.

The smallest of six failures

Everything else in this collection that fails does so in one way at a time. A dowel through timber does not — the wood can crush while the steel stays straight, or one plastic hinge can form in it, or two — and the capacity is the smallest of the six, which is the kinematic theorem of plasticity applied to a joint rather than to a frame.

connections · Dowel yield Three lines through one point, and three different winners. Modulus against density on logarithmic axes, with a guide line for each of three indices drawn through mild steel. A performance index E^(1/n)/ρ is a straight line of slope n on these axes, so ranking materials by it means sliding the line up and to the left and seeing what it leaves behind. The three lines have three different orders, which is why a table of properties cannot answer the question on its own: for a tie the answer is carbon fibre, for a beam it is timber at 4.28 times steel, and for a plate timber wins by more still. The construction is Ashby's; the arithmetic on it is this site's.

The ranking belongs to the load case

Every table of material properties ever printed ranks by strength, and every one of them answers a question nobody asked. What a structure wants is the least mass for a stated performance, and the combination of properties that gives it changes with the shape of the member — so timber beats steel four to one as a beam and loses to it as a tie, without either material changing.

materials · Material index Ten decades of time for forty per cent of the strength. Strength as a fraction of the five-minute test value, against the length of time the load is held, over 12 decades of seconds. The relation is a straight line on this axis, which is the reason a single duration factor works at all: every decade of time costs about the same amount of strength. A load held for 60 years leaves 59% of the short-term strength, and the member need not have moved — this is not creep, and it is not fatigue, because nothing about the load varies. It is that a static stress slowly breaks fibres. The curve has an asymptote at 18% that nobody quotes: below about a fifth of its short-term strength a member has no time to failure at all, so there is a stress under which duration stops being a question rather than merely becoming a small one.

The load that was left on too long

A timber beam that carries a load for fifty years fails at about fifty-nine per cent of the stress the same beam carries for five minutes in a testing machine. Nothing about the load varies, the member need not have deflected, and the relation between the two is a straight line on a logarithmic time axis over ten decades.

materials · Duration of load A short timber beam is a shear problem, and a steel one never is. Utilisation of the bending and shear checks on one beam, against span-to-depth. The two cross where the ratio equals f_m ÷ f_v exactly — no load, no width and no span survives the cancellation — which for this timber is 6.7 and for steel is 1.73. So a timber beam shallower than about six times its depth is governed by shear parallel to the grain, and a steel beam would have to be shorter than twice its own depth before the same thing happened, which is not a beam. The third check is bearing across the grain, which does not move with the span at all: on the beam drawn it is at 0.40, and it is the one that governs.

The shear that decides a timber beam

A steel beam is never governed by shear, because its bending strength is only 1.73 times its shear strength and no beam is that short. Timber's ratio is 6.7 along the grain and 23 across it, so shear governs at proportions people build every day.

sections · Shear flow The modes that tilt, and the ones that do not. Johansen's mechanisms for a 12 mm bolt with washers in single shear through 40 and 40 mm of timber of density 350 kg/m³, each with the rope term the fastener's axial resistance of 6.64 kN adds to it. Modes a and b, where the fastener only translates, gain nothing; the modes in which it tilts gain a quarter of its axial resistance, capped at a quarter of their own Johansen value. The joint's capacity rises from 5.02 kN (mode c) to 6.28 kN (mode c), 25 per cent.

The pull that Johansen left out

Johansen's mechanisms treat a timber fastener as a beam in a bed of crushing wood, bent and pushed sideways and nothing else. A real fastener that tilts across the joint is also pulled along its own axis, and if a washer or a thread resists the pull, it clamps the two members together and adds to what the joint can carry. The addition is a quarter of the axial resistance, it goes only to the modes in which the fastener tilts, and it is capped by fastener type — nothing for a dowel, a quarter for a bolt, all of it for a screw — which is how a screw keeps gaining strength past the thickness at which Johansen's capacity stops.

connections · Dowel yield Lean a screw and its bending stops mattering. The peak load on an 8 mm fully threaded screw (yield moment 20 N·m, tensile capacity 20 kN) through two 100 mm softwood members of density 350 kg/m³, against the angle it is driven at to the joint: the joint, its two resistances added slip by slip (solid); the thread's alone, resolved along the joint with its friction (dashed); the bending and embedment's alone (dotted); and the square screw with the rope effect, 5.9 kN (thin). At 45 degrees the joint carries 13.4 kN, 2.3 times the square screw, of which the bending gives 1.8; at 30, 19.1. The bending is the larger share only steeper than 85.7 degrees, within 4.3 of square. Withdrawal is taken to peak at 1.5 mm of axial movement and soften after it; bending follows Foschi's curve from EN 1995-1-1's slip modulus — shapes, not measurements.

The screw that is pulled instead of bent

A screw square to a timber joint resists slip by bending, and its thread is engaged only when it has tilted far enough to be pulled. Lean the same screw to 45 degrees and the slip pulls it from the first millimetre. An 8 mm screw through two 100 mm members goes from 5.9 kN to 13.4, reached after 2.2 mm of slip instead of more than twelve, and the bending that Johansen's model is about becomes an eighth of the joint. The price is a joint that holds one way better than the other.

connections · Dowel yield One ratio decides how much of a row works. The share of n single screws' peaks a row reaches, for rows of 10, 20 and 30 8 mm screws at 45° to the joint, 56 mm apart along the grain, in two 100 mm softwood members of modulus 11,500 N/mm² with 40, 80 and 160 mm of member depth each, and threads peaking at 0.5, 1, 1.5 and 2.5 mm, against one ratio: the slip difference the members' stretch puts between the end screw and the middle one at the row's peak, PL(1/EA₁ + 1/EA₂)/8, over the width of a single screw's curve above nine tenths of its peak. Thirty-six rows fall on one curve: every row with the ratio below 0.3 reaches at least 1.00 of its screws' sum; at a ratio of 1 they reach about 0.96; and the most stretched, at 3.5, 0.72. The colour is the thread's peak slip; a brittler thread has a narrower plateau and lands further right for the same row.

Twenty screws that peak together

A leaning screw reaches its peak at two millimetres of slip and then lets go, so a row of twenty along the grain looks like the long bolted joint again — the end screws past their peak and falling while the middle ones have barely started. Follow the row slip by slip and it is not: in a member of ordinary depth twenty leaning screws carry 98 per cent of twenty times one, against the 74 per cent the effective-number rule allows. What decides it is one ratio — the slip the timber's stretch spreads along the row against the width of a screw's peak — and the rule is right only for a brittle thread in a thin member.

connections · Dowel yield Below a limit the row settles; above it, it ratchets. The end screw's slip at the top of each cycle, against the number of cycles, for 20 leaning screws 56 mm apart with 40 mm of member depth per row loaded again and again from new to four levels, each screw, reloaded, reaching the slip it had been to with 85% of the load it had there. At 0.94 it settles at 2.50 mm after 3 cycles; at 0.96 it settles at 2.83 mm after 7 cycles; at 0.97 of the peak the end slip grows every cycle until the row cannot reach the load, at cycle 21; at 0.99 of the peak the end slip grows every cycle until the row cannot reach the load, at cycle 8. The row's first peak is 263 kN; every level drawn is below it.

The row that remembers its overload

A row of twenty leaning screws reaches 98% of twenty screws' peak, once. Loaded again, a screw that has been past its peak comes back along a softer path and reaches the slip it had been to with less load than it carried there. So a row loaded again and again neither simply evens out nor simply ratchets down: below 0.965 of its first peak it settles after a few cycles with the middle screws catching up, and above it the end screws are pushed back onto their curve every cycle, lose a little more each time, and the row fails at a load it carried the first time. One large overload costs it less than many smaller ones near the peak.

connections · Dowel yield A crack starts at the corner and runs along the grain. The end of a timber beam 400 mm deep and 140 mm wide, notched on its underside so that 200 mm stands over the support, with the notch's corner 50 mm from the line of the reaction, drawn to scale. A crack starts at the corner and runs along the grain towards mid-span. With a shear strength of 3.5 N/mm² the un-notched end would carry 130.7 kN; Eurocode 5's factor kᵥ of 0.40 on the reduced depth leaves 25.8 kN, 20% of it.

The notch that costs more than its depth

Cut a timber beam's underside away at the support, so that half its depth sits on the bearing, and a stress check on what is left says the end has lost half its shear strength. The beam says otherwise. A crack starts at the notch's corner and runs along the grain, and whether it runs is a question of energy, not stress — so the answer has a size in it. A 400 mm glulam beam notched to half its depth keeps a quarter of its shear resistance, an 800 mm one keeps 18 per cent, and the deeper the beam the less a notch leaves. Sloping the notch buys much of it back, and moving the corner away from the support costs it again.

sections · Shear flow The more of the load that comes back, the less of it the row can carry for ever. The largest load 20 screws leaning at 45° in a row 56 mm apart carries for ever, as a share of its forward peak, against the share of it that comes back the other way each cycle (0 the same load repeated, 1 fully reversed), for each screw carrying none, half and all of its thread's first loss into the other way. Carrying 0 per cent: 0.96 repeated, 0.96 half reversed, 0.82 fully reversed; carrying 50 per cent: 0.96 repeated, 0.94 half reversed, 0.79 fully reversed; carrying 100 per cent: 0.96 repeated, 0.92 half reversed, 0.76 fully reversed.

The row that is pushed as often as it is pulled

A row of leaning screws in a shear wall is loaded one way by the wind and then the other, and a screw that leans is pulled along its axis one way and pushed the other. Pulled, its axial force clamps the joint and friction helps it; pushed, it prises the joint open and the friction goes. So the row has a weaker side, and under a reversing load it settles exactly as it settles under a repeated one — the reversal does not start the ratchet again — but at a ceiling the pushed side sets, 0.79 of the pulled peak against 0.97. And it settles with a gap in the middle that every reversal crosses carrying nothing.

connections · Dowel yield

Named alongside it

The objects these essays reach for when they reach for this one.

Embedment strengthAnisotropyDuctilityBearingJohansen modelFrictionGrainPlastic hingeClamping forceElastic modulusFree bodyLoad path

All concepts