Structural Engineering & Statics
where the load goes
Depth

Series — page 13

A field says what an essay is about. A series follows one idea essay by essay — from the question that introduces it to the one that assumes all the others.
Where plane sections stop staying plane. Strain across a cut face at four span-to-depth ratios, with the straight line the theory assumes drawn faintly behind. For a slender beam the two coincide; for a beam as deep as its span the real distribution is nothing like a straight line, and beam theory has no claim on it.

Plane sections

  1. 1 Plane sections stay plane, and what the assumption costs
1 essay · sections The collapse mechanism of a propped cantilever. A collapse mechanism, with the hinge position found by searching rather than quoted. Every position gives an upper bound on the collapse load; the lowest is 7.29, at a hinge 58.6 per cent along, which is a coefficient of 11.657 times Mp over the square of the span.

Plastic hinge

  1. 1 After the first yield, which is not the end
1 essay · internal-forces The corner is where the plate twists, and it has to be held down. The twisting moment M_xy over a 6 × 6 m simply-supported panel under 10 kN/m², from Navier's double series. It is zero along both centrelines and largest at the corners, which is the opposite of the bending moments and is why no strip reading contains it: 31% of the load crosses this panel in twist, and a strip can only bend. At a free corner the twisting moment is statically equivalent to a downward point force of 2M_xy — 26.7 kN here, 7.4% of the whole load on the panel, applied at a point — so the corner has to be held down. A panel whose corners are free lifts there, deflects more, and cracks diagonally across them. The deflection at the centre is 2.53 mm, against 4.05 mm for the strip that ignores all this — 60% more, which is the size of what the twist is carrying.

Plate torsion

  1. 1 A third of the load crosses sideways
1 essay · deflection Held up by a pressure nobody can feel. An air-supported roof of 60 m span and 9 m rise. The membrane has no bending stiffness whatever, so the only thing that can hold it in tension is a pressure difference, and the pressure has to exceed the load per unit plan area and nothing else: 0.25 kN/m² of fabric plus 0.6 of snow is 0.85 kN/m², so 1.19 kN/m² does it — 1190 pascals, which is 1.17 per cent of an atmosphere and 121 millimetres of water. Ears do not notice it. A door does: at 2.1 kN on an ordinary leaf, the building needs an airlock rather than a handle. And the whole of it arrives at the foundation as 3365 kN of uplift — 17.9 kN on every metre of perimeter — which is the bill the pressure's smallness conceals.

Pneumatic structure

  1. 1 Held up by the air inside
1 essay · structures A portal frame swaying under 20 kN. A portal frame pushed sideways, solved by the stiffness method because statics cannot divide the load between two columns. The base shears come out at 10.0 and 10.0 and add to the applied 20; the peak moment is 22.2. The sway is drawn hugely exaggerated, and the moment diagram is plotted on each member's tension face.

Portal frame

  1. 1 The frame that leans, and what stops it
1 essay · structures The hinge is at mid-height in exactly no storey. The height of the point of contraflexure in each column of a 4-storey, 3-bay frame under lateral load, as a fraction of the storey height, against the portal method's assumption that it is at the middle. The exact solution is a plane-frame stiffness analysis of the same frame. In the bottom storey the zero sits at 0.609 of the height, because a fixed base is stiffer than the joint above it and takes more of the column's moment; in the top storey it sits at 0.359, because there is no column above to share that joint. The average over the whole frame is 0.475, which is why the assumption survives — it is right on average and wrong everywhere. The worst error in the storey shear share is 25%, and the column shears still add to the storey shear to 0e+0 of it, because the method is exact statics applied to an assumed structure.

Portal method

  1. 1 The analysis that assumes the answer
1 essay · internal-forces Two triangles that cross zero, and a block that does not. Stress across a 300 × 700 mm section at each stage, compression positive. The prestress alone gives -6.33 MPa at the top and 20.61 at the bottom; at transfer, with only self-weight on it, the top is at -2.47 MPa and in service the section runs from 7.61 to 3.82 MPa — compression everywhere. The same beam with no prestress reaches -12.67 MPa at the bottom fibre, which is 4.2 times what the concrete can hold.

Prestress

  1. 1 The load put on backwards
1 essay · internal-forces Prestress buys shear as a square root, not as a sum. The shear stress an uncracked web can take before the principal tension reaches the concrete's tensile strength, against the axial compression the prestress put there. With no prestress it is 1.35 N/mm², the tensile strength itself, because pure shear has a principal tension of exactly its own magnitude at forty-five degrees. Adding compression gives √(f_ct² + σ_cp·f_ct), which is a square root and therefore flattens: the first newton of prestress is worth far more than the last. At the 7.14 N/mm² drawn the limit is 3.39 N/mm², a gain of 2.51, and doubling the prestress from there takes it only to 4.59. The straight line is what a rule that simply added the two strengths would have promised.

Prestressed shear

  1. 1 The crack that never reached forty-five degrees
1 essay · internal-forces Three materials pulled until they stop. Three stress-strain curves — mild steel, high-strength steel, aluminium alloy — plotted to a strain of 2.0%. One of them has a plateau, so the stress at which yielding starts is something the specimen does rather than something anyone chooses. The 0.2% offset construction is drawn on the high-strength steel: a line of slope E from a strain of 0.002, cutting the curve at 460 N/mm².

Proof stress

  1. 1 The stress at which nothing in particular happens
1 essay · materials The prop load is decided by the digging, not by the hole. Prop forces in a 12 m excavation propped at 3 levels, with the force each prop reaches at any stage of the sequence drawn thick and the force the finished arrangement gives it drawn thin. The pale dots are the individual stages. The middle prop reaches 214 kN while the dig is at 9.5 m and finishes at 88 — a factor of 2.44 between the two, and the larger one is not in the final analysis anywhere. Terzaghi and Peck's apparent pressure diagram, a rectangle of 49.4 kN/m², reproduces the total of the staged maxima to 2% — which is what it is: an envelope of measured prop loads, back-figured into a pressure, and a shape nothing on a wall is ever loaded with.

Propped excavation

  1. 1 Every prop has its own worst day
1 essay · structures The one length a section carries into a column. Four profiles of equal area, with the radius of gyration r = √(I/A) drawn as the distance it is — a pair of lines either side of the centroid, at the depth the whole area would have to sit at to give the section the second moment it has. As a 4 m pin-ended column the same 3000 mm² of material carries between 7 and 3146 kN, in the ratio of the squares of those radii and of nothing else.

Radius of gyration

  1. 1 The one length a section takes into a column
1 essay · sections The stress that leaks away. A restrained shrinkage strain of 300 microstrain in concrete of modulus 32000 N/mm². Ignoring creep it produces 9.60 N/mm², which is above the tensile strength of 3.5 and predicts that every restrained concrete member ever cast has cracked. Counting creep by the superposition integral leaves 2.32 N/mm² after 27 years, and the one-line age-adjusted shortcut at the usually quoted ageing coefficient of 0.8 leaves 3.31. The two disagree — this creep function implies an ageing coefficient of 1.32, not 0.8 — and both are below the tensile strength, so the conclusion turns on counting creep at all rather than on how it is counted.

Relaxation

  1. 1 The strain that was imposed, and the stress that leaked away
1 essay · materials A I-section at 80% of its plastic moment. The same I-section drawn three ways: the shape, the strain across its depth, and the stress that strain produces in mild steel. The strain diagram is a straight line, because plane sections stay plane whatever the material is doing. The stress diagram is not: 0% of the area has yielded, working inward from both faces, and the neutral axis sits at 70.9 mm against a centroid at 100.0 mm. The compression resultant is 322.5 kN and the tension resultant 322.5 kN, on a lever arm of 180.1 mm, which multiplies back to the 58.1 kNm the section is carrying. A rolled residual stress pattern of ±30% of yield is locked in before any load arrives.

Residual stress

  1. 1 The stress that was there before the load
1 essay · materials What a point on a response spectrum is: three structures, three integrations, three points. Three oscillators of periods 0.3, 0.8, 1.8 s, each integrated through the whole of the same ground motion, and the peak of each one plotted against its own period on the curve at the right. The peaks are 12.54, 46.2, 108.95 mm. The complete spectrum is that done 44 times. Nothing in the curve is a property of the earthquake alone: every point on it carries a period and a damping ratio that belong to a structure.

Response spectrum

  1. 1 The spectrum is not a load
1 essay · dynamics At thirty degrees each leg carries the whole load. A 100 kN lift on two legs at 60 degrees to the horizontal. Each leg is a two-force member, so its force is the vertical share it carries divided by the sine of its angle: T = W/(2 sin β) = 57.7 kN, which is 0.58 times the whole load in each. The curve on the right is that division, and it is the reason the rule about sling angles exists rather than a convention: at sixty degrees a leg carries 0.58 W, at forty-five 0.71, at thirty exactly 1.00, and at fifteen 1.93. And the horizontal components do not disappear — they run through the thing being lifted, which here carries 57.7 kN of compression between the two pick points. That is what a spreader beam is for: it takes the compression as a designed strut so the legs above it can stand up.

Rigging

  1. 1 The angle that doubles the force
1 essay · equilibrium The cheapest way out of being round. A ring under uniform external pressure, drawn in its first four buckling modes with the pressure each one needs underneath it, in N/mm². The pressure has no direction: it stays normal to the wall wherever the wall goes, so it does work on any change of shape that reduces the enclosed area, and the ring buckles into whichever shape is cheapest. Bare, that is the oval — n = 2 at 3EI/R³ — and the modes rise as n² − 1, so three lobes cost 2.67 times as much. Nothing in the drawing prefers any orientation, which is the point — a column has an axis to buckle about and a ring has none.

Ring buckling

  1. 1 The pressure that needs no direction
1 essay · stability Three ways to apply the same force, and one depth to forget the difference. Three end loads on a member 400 mm deep, all with the same resultant and the same moment: a point load, the same force spread over a fifth of the depth, and the same force split in two. What is plotted is the difference between each of them and the beam-theory answer — the self-equilibrating remainder — as a fraction of the mean stress. The point load starts at 20 times it and is under a tenth of it by 0.77 depths; all three are under one per cent by about 1.18. That distance is the licence every figure in this collection is drawn under, and the exact strip eigenvalue agrees with it: 2.106 + 1.125i, whose real part puts one per cent at 1.09 depths and whose imaginary part means the remainder changes sign on the way out, which no statement of the principle mentions.

Saint-Venant's principle

  1. 1 How far a wrong load reaches
1 essay · internal-forces Where the class limits come from. The width-to-thickness ratio at which two kinds of plate reaches its own elastic critical stress at the yield stress, for three steel grades. A flange outstand (buckling coefficient 0.43) derives to 18.6, 15.2, 13.3 at 235, 355, 460 N/mm², against quoted limits of 14.0, 11.4, 10.0; A web, in bending (buckling coefficient 4) derives to 56.8, 46.2, 40.6 at 235, 355, 460 N/mm², against quoted limits of 42.0, 34.2, 30.0. The derived number is the larger every time, and by the same factor at every grade — flange outstand 1.33, web, in bending 1.35 — because both the derivation and the quoted limit go as one over the root of the yield stress. A constant ratio is what a fixed knockdown looks like: the derivation is for a perfect plate and the quoted limit is for a rolled one, carrying residual stress and not quite flat.

Section classification

  1. 1 The section that cannot reach its own strength
1 essay · materials The same material, four ways. Four cross-sections of identical area, so identical weight and cost, with the second moment of area computed from each profile's own geometry. Only the arrangement differs, and the stiffest is many times the flattest.

Section shape

  1. 1 The same steel in a different shape, and a factor of forty
1 essay · sections Two null spaces of one matrix, and the count is their difference. Two pin-jointed frames, each with the forces it can carry with nothing applied to it drawn on its bars — tension one colour, compression the other, thickness in proportion. That force set is the null space of the equilibrium matrix; a mechanism is the null space of its transpose; and Maxwell's count b + r − 2j is the difference of their dimensions and knows neither of them separately. A square with both diagonals has s = 1 and m = 0. Two bars in a straight line has s = 1 and m = 1 with a count of 0, so the count is satisfied by a frame that both folds and can be prestressed — and the prestress stiffness is positive, which is why a tensioned pair of collinear bars is stiff at all.

Self-stress

  1. 1 The forces that are there with nothing applied
1 essay · structures

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