The crack that never reached forty-five degrees

The shear strength of a reinforced concrete member without links is three variables raised to fitted powers with a size term in front. There is no free body anywhere in it, because the object it describes — a member already crossed by flexural cracks, carrying shear across them by four mechanisms in unknown proportions — does not admit one.

A prestressed web near a support is a different object. It has no cracks in it, it is a continuum, and a continuum under a known state of stress can simply be asked what its worst tension is.

One point, and everything true of it #

Take an element at the centroid of the web, half way up a 1,200 mm section 150 mm thick, near a support. Two things act on it. The prestress puts an axial compression $\sigma_{cp} = P/A = 3{,}000{,}000/420{,}000 = 7.14$ N/mm² along the member. The shear force puts a shear stress $\tau = VS/Ib$ across it, and at the centroid $S$ is at its maximum so $\tau$ is too. Nothing else: the bending stress at the centroid is zero by definition, which is why this element and no other is the one to interrogate.

The algebra is one line. The principal tension is

$$\sigma_1 = -\frac{\sigma_{cp}}{2} + \sqrt{\left(\frac{\sigma_{cp}}{2}\right)^2 + \tau^2},$$

and setting $\sigma_1 = f_{ct}$ and solving for $\tau$ gives

$$\tau_{\lim} = \sqrt{f_{ct}^2 + \sigma_{cp} f_{ct}}.$$

That expression is in every code’s uncracked web-shear clause, usually with an $\alpha_l$ in front of $\sigma_{cp}$ to allow for the prestress not being fully transmitted near the end. It is not calibrated against anything. It is Mohr’s circle rearranged, and it is the only shear rule in concrete design of which that is true.

Why a square root and not a sum #

The shape of the expression is worth more than its value, because it is the reason the second half of a prestress is worth so much less than the first.

At $\sigma_{cp} = 0$ the limit is $f_{ct}$ itself. That is not an accident of the algebra: pure shear is a state whose Mohr’s circle is centred on the origin, so its principal tension equals its shear stress, and a web in pure shear cracks the moment the shear reaches the tensile strength. Concrete’s tensile strength is small, so this is a very low number — 283.5 kN on the section drawn, against an applied shear of 480.

Add compression and the circle’s centre moves left by $\sigma_{cp}/2$ while its radius grows by less, because the radius is a Pythagorean sum in which $\tau$ is the term that matters. The tension shrinks. But it shrinks as a square root of a product:

Doubling the prestress from the value drawn buys 36 per cent more shear, which is the same diminishing-return shape a deeper truss shows against its own chord force and for the same reason: the quantity being bought sits under a root. A prestress chosen for bending is very nearly the right prestress for shear as well, because the marginal return has flattened by the time the bending requirement is met — which is a happier arrangement than it sounds, since the two requirements peak at opposite ends of the span.

The crack turns #

The circle does not only shrink. It rotates, and the rotation is worth as much as the shrinkage.

A crack forms perpendicular to the principal tension, so a flatter crack is one that crosses more of the member’s length. That matters for what happens next. Once the web has cracked, the member behaves as a truss with the cracks as its diagonals, and the shear reinforcement crossing a crack of inclination $\theta$ has to be counted over a length $z\cot\theta$.

Flattening the strut multiplies the number of links a diagonal crack crosses by $\cot\theta$, so the same links carry more shear for nothing. The price is paid by the strut: a flatter strut carries a larger compression for the same shear, and the two effects cross at the angle where the web crushes rather than yields its links. Codes cap $\cot\theta$ at 2.5, which is 21.8°, and the crack angle the prestress produces at the centroid of this web is 21.7°.

That coincidence is not one. The 2.5 cap exists because prestressed webs crack at about that angle and reinforced ones do not, and the cap is a way of letting a designer use the flatter truss where the prestress has earned it. So the prestress is paid twice for a single purchase: once by raising the shear at which the web cracks at all, and again by flattening the truss that carries the shear afterwards.

Which free body produced the number #

The uncracked check has one element and no free body in the usual sense — the element is the free body, and equilibrium of it is what Mohr’s circle encodes. The cracked check has a genuine one.

It is a cut taken along the diagonal crack rather than across the member, severing every link the crack crosses, with the shear on one face carried by the vertical components of those links and nothing else. The number of links in that cut is $z\cot\theta/s$, which is the whole of why the angle multiplies the capacity, and it is also why the cut has to follow the crack: a vertical cut through the same member severs a different number of links and balances a different force.

The two calculations are about different states of the same member, and the honest reading is that the member passes through both. It is uncracked until the shear reaches 711 kN or the moment reaches the cracking moment; after that it is a truss. Which one governs at any station is a question about position along the span.

Two criteria and one span #

The applied shear is largest at the support and the applied moment is largest at mid-span, so the two things that can crack a prestressed beam are at opposite ends of it.

On a simply supported span under a uniform load the shear is a straight line largest at the support and zero at mid-span, and the moment is a parabola largest at mid-span and zero at the support. The two shear checks inherit that shape exactly: near the support the web is uncracked and the principal tension governs; past the station where the applied moment exceeds the cracking moment the section is cracked in flexure, the diagonal crack starts from a flexural one, and the truss model governs instead.

For the 24 m beam drawn, carrying 40 kN/m, the cracking moment is 2,329 kNm and the applied moment reaches it at 0.287 of the span — 6.89 m from the support. Everything inboard of that station is uncracked and checked by Mohr’s circle; everything beyond it is cracked and checked as a truss. Raising the prestress moves the transition outward and lowering it moves the transition in, so the shape of the check changes with a design variable, not just its value.

This is why prestressed shear design is set out as two clauses with a rule for which applies, and why the rule is written in terms of the moment rather than the shear. A member is uncracked or it is not, and the shear check follows the flexural state rather than the other way round.

There is a third region nobody draws, which is the one where neither expression is honest. Immediately over the bearing the member is a disturbed region: the reaction is applied over a finite width, the strut-and-tie arrangement rather than beam theory is what describes it, and plane sections have stopped staying plane. Codes handle it by permitting the shear to be checked at a distance $d$ from the face of the support rather than at the face, which is a way of saying do not use this model where it does not apply without having to say what does. For the beam here that distance is 1,080 mm, and the shear taken off by moving the check station is 43 kN — a tenth of the demand, from a clause that looks like an administrative detail.

The shear the concrete is never asked about #

Before any of this, the tendon has already removed some of the problem.

For a tendon drooping 0.8 of its eccentricity over a 24 m span, the slope at the support is $4 \times 0.8 \times 380/24{,}000 = 0.0507$, and $V_p = 3{,}000 \times 0.0507 = 152$ kN. The applied shear at the support is 480 kN. Nearly a third of it is cancelled by the geometry of the cable before the concrete is consulted, and the 863 kN of capacity at the support is therefore against a demand of 328 rather than 480.

Two consequences follow that are easy to miss. The first is that a straight tendon buys none of this, so a member with straight strands has both a lower cracking moment profile and no shear relief, and the difference between a draped and a straight arrangement is much larger in shear than in bending. The second is that $V_p$ is a load, so it is subject to the losses that reduce $P$ over time — a beam checked at transfer has more of it than the same beam at fifty years, and the shear check gets worse as the beam ages while the bending check gets better.

The centroid is not always the worst point #

Every number above is computed at the centroid, on the argument that the shear stress is largest there and the bending stress is zero. That is the right point for a beam whose prestress is axial. It is often the wrong point for one whose prestress is eccentric, and the reason is that the prestress has a bending stress of its own.

The compression at a height $y$ above the centroid is $P/A - Pey/I$, so the eccentricity that put 7.14 N/mm² at the middle has put much less near the top. For the section here, $Pe = 1.14\times10^9$ N·mm and $I = 84\times10^9$ mm⁴, so the prestress compression falls by 0.0136 N/mm² per millimetre of height: at the web’s upper end, 450 mm up, it is down to 1.03 N/mm² against 7.14 at the centroid.

The shear stress there has not fallen nearly as fast. $S$ at the top of the web is perhaps 70 per cent of its centroidal value, so $\tau$ is 2.37 against 3.39. Put those two into the same circle:

$$\sigma_1 = -\frac{1.03}{2} + \sqrt{0.27 + 5.62} = 1.91\ \text{N/mm}^2$$

against 1.35 at the centroid. The top of the web cracks first, at a shear well below the one the centroidal check permits, and no amount of care with the centroidal arithmetic will find it.

The first figure on this page answers that directly if the marker is moved to where the top of the web actually sits.

Near the support that is the ordinary condition rather than an exception, because the applied moment there is small and supplies no compression at the top to help. Further along the span the applied moment grows, the top of the web gains compression from it, and the centroid becomes the critical point again — so the worst height migrates down the web as the check station moves inboard.

Which is one of the reasons tendons are lifted toward the centroid at the ends of a member. It is usually explained by the end block and the transfer stresses, and both of those are real; the shear check is the third reason, and it is the one that would still require the lift if the other two were solved. Reducing $e$ at the support flattens the prestress profile through the depth, brings the top of the web back up toward $P/A$, and puts the critical point back where the formula assumes it is.

The shear that runs backwards at transfer #

There is one load case in which all the arithmetic above is being applied to a member whose shear is the wrong way round.

At transfer the beam carries its own weight and nothing else, and the prestress is at its highest. For this section — 0.42 m² at 25 kN/m³, so 10.5 kN/m — the self-weight shear at the support is 126 kN. The tendon’s vertical component is 152. The net shear is 26 kN upward, and the principal tension in the web leans the other way.

Nothing about that is dangerous: the magnitudes are small and the web has more capacity than it will ever need at that stage. What it changes is the direction of the crack that would form, and therefore which links would be crossing it. A member whose shear reverses over its life has diagonal tension in both directions at different times, and links, being vertical, cross both — which is one of several reasons links are vertical rather than inclined to suit the crack of the moment.

The comparison that makes the case #

It is worth putting the same web beside itself with the prestress removed, since that is the design decision the arithmetic is for.

The principal tension through the depth of a web peaks at neither of the places its two components do. At the centroid, where the shear stress is largest, the bending stress is zero; at the flange, where the bending stress is largest, the shear stress is nearly nothing. What decides whether the crack starts in the web or from the soffit is where the two profiles overlap worst, and in a prestressed member the prestress profile is the third curve deciding it.

The same 150 mm web, 1,080 mm effective depth, C45 concrete:

• Uncracked, with prestress: 711 kN, from Mohr’s circle.
• Uncracked, no prestress: 283.5 kN, the same circle centred on the origin.
• Cracked, no prestress, no links: 113 kN, from the fitted expression.
• Web crushing, with prestress and links: 919 kN, the ceiling nothing gets past.

The factor between the second and third entries — 2.5 — is what cracking costs, and it is larger than what the prestress buys. It is also the factor that decides whether links are needed at all, which on a precast beam is a fabrication question rather than a strength one: links are the reinforcement that costs most to fix per kilogram, and a web that does not crack does not need them. That ordering is the argument for prestressing a member that is shear-critical: the prestress is not primarily worth 2.5 times the tensile strength, it is worth keeping the member in the state where the tensile strength is the relevant quantity at all.

Where the model stops #

The prestress is fully transmitted. Near the end of a pretensioned member it is not: the force builds up over a transmission length of 60 to 80 diameters through bond, exactly as a reinforcing bar develops its force, and the $\alpha_l$ factor in the code expression is the fraction that has arrived. The support, where the shear is worst, is the one place the prestress is weakest, and the two profiles fight along the same 800 mm.

The tensile strength is a characteristic value of a scattered population. Concrete’s tension is the property with the widest scatter of any it has, and this check depends on it linearly and inside a square root — so the calculated capacity carries the scatter more directly than a compression-governed check does. Choosing which fractile to design to matters more here than almost anywhere.

The section is uncracked because the moment says so. But the moment that cracks it is not only the applied one: a support settlement, a restrained shrinkage, a thermal gradient through the depth all add curvature and can crack a section the applied load would not have. The transition at 0.287 of the span is a calculation about one load case.

The stress state is plane and the web is thin. The transverse direction is assumed unstressed, which is right for an isolated I-beam and wrong for a box girder web that also carries transverse bending from the deck, and wrong again where a diaphragm or an anchorage is nearby and the disturbance from it has not died out.

The links are assumed to yield. The truss model counts every link the crack crosses at its yield force, which requires the crack to open enough to strain all of them past yield and the links to have the ductility to allow it — an assumption that appears in no formula and underwrites the whole method. In a heavily prestressed web the crack opens very little, and the links nearest the flanges may never reach yield at all.

And the pictures show a state, not a history. The circle drawn is the circle at the instant of cracking. Before it the web is uncracked and stiffer than the truss model assumes; after it the load redistributes along the member toward the parts still uncracked, so the first crack does not appear where the first calculation says the demand is highest — it appears where demand and capacity first meet, which is a different station.

The ladder from here #

Later rungs on this anchor: the transmission length properly resolved, with the prestress profile and the shear profile plotted on the same axis and the governing station found where they cross. Shear in a member with inclined flanges, where the flange forces have vertical components of their own and the web sees the residual. The combination of shear, bending and torsion in a box girder web, which is three actions on one element and one circle. Segmental construction, where the joint has no reinforcement across it at all and the shear is carried by friction on a match-cast key under the prestress — a shear-friction problem wearing prestressed clothes. Anchorage-zone bursting, where the same prestress that helps everywhere else is the load that splits the end block. And the reverse case worth knowing: an externally post-tensioned member, where the tendon is outside the section, its eccentricity changes as the beam deflects, and the shear relief it provides is a function of the deflection rather than of the geometry drawn.