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Drawing as calculation — page 6

Essays 121 to 125 of 125 on this thread, in the same order.

The pole, the strings, and the resultant of any number of forces. Five downward loads on a span of 10 m — 30 kN at 1.5 m, 20 kN at 3.5 m, 45 kN at 5.0 m, 25 kN at 7.0 m, 35 kN at 8.5 m — adding to 155.0 kN. On the right, the loads laid end to end down one line, with a pole 60.0 kN to the left of it and a ray drawn to every division between them. On the left, the funicular polygon: each segment parallel to the ray of the loads it has passed, so the shape is the one a string carrying these loads would hang in. The first and last strings are extended until they cross, at 5.24 m, and that crossing is where the 155.0 kN resultant acts — the same station the moment sum Σ P x / Σ P gives, 5.24 m, reached with no pole in it at all.

The pole decides the drawing, not the answer

Five forces will not pair off the way four do. They need a point that is nowhere on the structure — chosen freely, by whoever is holding the pencil — and the string of lines it generates. Every choice draws a different polygon and finds the same resultant, and the shape it draws turns out to be the beam's bending moment diagram.

$A pin's force does not pass through its centre. Left, a pin of radius 0.15 m in its hole, with a coefficient of friction of 0.15. The reaction at the contact is inclined by φ = 8.5° to the radius through it, because the friction it can develop is that fraction of the force pressing the surfaces together, and the perpendicular distance from the pin's centre to that inclined line is R sin φ = 0.022 m. Every position the contact can take gives a line tangent to the same circle, shaded. Right, a link 0.48 m long pinned at both ends, at the same scale — 3.2 pin radii, which is a stubby linkage rather than a structural tie, drawn that way because at the thirty radii an ordinary tie has, the two circles are smaller than the pencil: its force is a common tangent to the two circles — the two solid lines for the two senses of rotation, the two dashed ones for the senses in which its ends turn oppositely — and the dashed centre line every construction in this collection draws is none of them. A pin of this size carrying 5.0 MN delivers a couple of 111.3 kN·m to whatever it is pinned to, which is the same offset read as a moment rather than as a distance.$ Equilibrium

The pin that is not a point

Every line of action drawn so far passes exactly through a pin's centre, which is true of a frictionless pin and of nothing else. A real one carries its force tangent to a small circle instead, so a link's line is a band, a construction's answer is a range, and a support drawn as a hinge hands a couple of a hundred kilonewton-metres to whatever it is pinned to.

The column under the first interior support is given a quarter more. Two equal spans of 8.0 m carrying 5.0 kN/m, continuous over three supports. The tributary rule gives each interior support one span's worth of load, 40.0 kN, and each end support half of that. The continuous beam gives 15.0 kN, 50.0 kN, 15.0 kN — ratios of 0.75, 1.25, 0.75 to what the areas say. The reactions still add to the whole load, because they must; what has moved is which support gets it. The end supports are relieved because the span next to them hogs over the first interior support, lifting their end of it.

Equilibrium

The column given more than its rectangle

The tributary rule draws a rectangle round each column and hands it whatever stands inside. A floor is continuous over its columns, and a continuous beam does not give each support the load above it — the first interior one takes a quarter more and the end ones a quarter less. On a grid the two directions multiply, and two columns on the same floor differ by a factor of nearly three.

One bracket, three neutral axes, three sets of bolt forces. A bracket 300.0 mm deep with three rows of two bolts at 75.0 mm pitch, carrying 15.0 kN·m about an axis in the plane of the bolts. Taking the axis at the group's centroid puts the outer rows at 50.0 kN of tension and 50.0 of compression, with the middle row idle. Taking it at the plate's compression edge puts every row in tension — 7.1, 14.3, 21.4 kN from the bottom up — with the top row at 21.4. Solving for it instead, with the plate bearing over 200.0 mm of width and the bolts as areas, puts it 40.2 mm above the edge and the top row at 24.6 kN. All three make the applied moment exactly. The top row differs between them by a factor of 2.33.

Connections

The bolt group has no neutral axis

A bolt group carrying a moment about an axis in its own plane has some bolts in tension and something in compression somewhere else. Where that somewhere else is decides the answer, nothing in the group decides it, and the two defensible choices give the worst bolt 50 kilonewtons and 21.

A portal frame, and the section it is a drawing of. A single-bay portal of 6.0 m span and 5.0 m height with fixed feet, carrying 10.0 kN/m on its beam, with the beam's second moment equal to the columns'. On the right, the analogous column: the frame's own centreline drawn as a section as wide as 1/EI at every point, so it is narrow where the frame is stiff. Its area is 16.000 and its elastic centre sits 1.56 m below the beam — inside the frame, on no member at all. The released moments loaded onto that section give a direct stress of 33.8 kN·m and a bending stress whose gradient is the horizontal thrust, 6.4 kN. Together they give 10.6 kN·m at the feet, −21.2 at the knees and 23.8 at the crown, against a free moment of 45.0. The same frame solved by stiffness gives 10.6, −21.2 and 23.8.

Deflection

The elastic centre is not on the frame

Cross's analogy turns a member fixed at both ends into a short column and its end moments into edge stresses. A closed frame's analogous column is the frame's own outline, its elastic centre is a point hanging in mid-air inside it, and the horizontal thrust a gravity load produces is that section's bending stress about the axis through that point.

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