Moment-area — the series · Structural Engineering & Statics

The area is the rotation, and its first moment is the movement. A 6 m cantilever under a tip load of 10, with the M/EI diagram beneath it. The shaded area is 180.00, which by the first theorem is the change of slope along the whole member. Its centroid is at 2.000 m, and the first moment about the tip is 720.00 — which by the second theorem is the deviation from the tangent, and for a cantilever that tangent is horizontal, so it is the deflection itself. Integrating the curvature twice instead gives 720.00.

The area of a diagram is a rotation

A deflection is the double integral of a bending moment, and the two constants of integration are the whole difficulty. Mohr's theorems replace them with two pictures — an area, and where that area's centre of gravity sits.

part 1 · deflection

A fixed-ended member's end moments are the edge stresses of a column. A prismatic member, fixed at both ends under a point load 0.333 of the span from its left end, drawn three ways. At the top, the member. In the middle, the column the analogy puts in its place: a strip as long as the member and as wide at each point as 1/EI there, with its centroid, the elastic centre, 0.500 of the span from the left. At the bottom, the simply supported moment diagram, dashed, which is the load on that column; the straight line, which is the stress that load produces, P/A + M(x − x̄)/I; and the member's own moment diagram, which is the difference. The end moments are the column's edge stresses: 0.148 PL at the left and 0.074 at the right, which are Pab²/L² and Pa²b/L², and the largest sagging moment is 0.099 PL.

The fixed-end moment is a column stress

A fixed end forbids a change of slope and a deviation, so on a member fixed at both ends Mohr's two theorems both have the answer nothing. Written with 1/EI as a width, those two conditions are the P/A + My/I of a short column under an eccentric load — the end moments are its edge stresses, and a haunched member's are read from where its elastic centre has moved.

part 2 · deflection

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.

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.

part 3 · deflection

A fixed arch, and the section it is a drawing of. A parabolic arch of 30.0 m span and 6.0 m rise, fixed at both springings, carrying 20.0 kN/m over the span. On the right the analogous column: the arch's own centreline drawn as a section of width ds/EI, so it is narrow where the arch is stiff. Its area is 32.99 and its elastic centre sits 3.86 m above the springings — 2.14 m below the crown and on no part of the arch at all, which is the point of it. The released moments loaded onto that section give a direct stress of 3054.0 kN·m and a bending stress whose gradient is the horizontal thrust, 375.0 kN.

The centre that hangs in the air

Cross's analogy turns a closed frame into a short column and reads its redundant moments off as that column's stresses. Nothing in it requires the frame to be straight. Run round a fixed arch, the analogous column's centroid is a point two thirds of the way up the rise, on no part of the arch at all, and the bending stress about it is the arch's horizontal thrust. For a parabola under a uniform load the answer it returns is a diagram of nothing, which is exactly right.

part 4 · deflection

An unsymmetric frame, and a section with a product of inertia. A portal of 8.0 m span whose columns are 5.0 m and 3.0 m, carrying 10.0 kN/m on its beam, and beside it the analogous column — the frame's own centreline at a width of 1/EI. The section is no longer symmetric about a vertical line, so it has a product of inertia of −18.4 against Ix = 31.6 and Iy = 168.0, and its principal axes are tilted 7.54° from the horizontal — the dashed pair through the elastic centre. A section with a product of inertia does not bend about the axis it is loaded about, so the thrust and the redundant shear come out of a two-by-two rather than out of two divisions.

The axes that have to be turned first

Make one column of a portal shorter than the other and the analogous column stops being symmetric about a vertical line. It acquires a product of inertia, its principal axes tilt seven and a half degrees, and the thrust and the redundant shear stop being two separate divisions. Using the elastic centre and nothing else — which is what the symmetric construction looks like from outside — reports 17.0 kN·m at the left foot where the frame carries 27.8.

part 5 · deflection

A frame bent by nothing at all. A portal of 8.0 m span and 5.0 m columns at EI = 20000.0 kN·m², with its right foot settled 10.0 mm and no load on it anywhere. The moment diagram is drawn on the members: −3.95 kN·m at the left foot, −3.95 at the left knee, −0.00 at the crown, 3.95 and 3.95 on the right. The settlement is drawn hugely magnified; at true scale it is 10.0 mm on an 8.0 m frame. The diagram is antisymmetric, the vertical force the settlement develops is 0.99 kN, and every one of those numbers is proportional to EI.

The load that is not a load

Settle one foot of a portal frame by ten millimetres and the frame develops moments with nothing applied to it anywhere. In the analogous column the case is simpler than a load case, not harder — the section carries no direct stress at all, and the whole answer is one bending stress. And it scales the wrong way: the moments are proportional to EI, so the stiffer the frame, the more a settlement costs it.

part 6 · deflection

Every shape the analogy reaches, which is every single ring. Four frames of 10.0 kN/m on an 8 m span, each solved by the column analogy and each checked against a stiffness solution of the same frame: a portal, largest moment 40.6 kN·m, agreeing to 0.24 per cent; a pitched portal, largest moment 35.6 kN·m, agreeing to 0.15 per cent; a stepped frame, largest moment 46.6 kN·m, agreeing to 0.16 per cent; splayed legs, largest moment 16.9 kN·m, agreeing to 0.09 per cent. The diagrams are drawn normal to each member. Nothing in the method asks what shape the frame is: it needs an area, a centroid and three second moments of the centreline, and a polyline has all five however it bends.

Three, and what three is a property of

The column analogy works because a closed ring cut once has three redundants and a plane section has three stress resultants. That match is the whole method, and it is topological rather than geometric — a portal, a pitch, a step, a splay and a polygonised arch are all one ring and all exact. Add a second bay and the frame has six redundants with nothing to be a drawing of, and the outer ring on its own is out by 190 per cent.

part 7 · deflection