1.33 Singularities

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Stresses are infinite where the strains are infinite, ε = du/dx = ∞, i.e., where

the displacements change “on the spot”. Why this happens is best explained

by the problem of the brachistochrone, the problem to find the fastest route

between two points A and B. The solution of this famous problem is a cycloid;

see Fig. 1.121.

“It is better to start out vertically and pick up speed early, even if the path

is longer” [231]. This is also the tendency we observe in structures. The material

tries to escape as fast as possible from the dangerous zones by starting

with an infinite slope u_(0) = ∞. Such an abrupt growth where the displacements

change stante pede, on the spot, (see Fig. 1.122 a) is described by a

function as

u = rα α < 1 ⇒ σ = E

r1−α , (1.495)

whose derivative du/dr for values of α < 1 is infinite at the start. If the

displacement decays in a soft slope as in Fig. 1.122 b, then α is greater than

one and the stresses remain bounded.

The best known example for abruptly changing deformations of type b is

the rigid punch (Fig. 1.123). Outside the compression zone the displacement

of the soil shoots straight up to taper off very rapidly. This abrupt decrease in

the settlement is the reason for the infinite stresses at the edge of the punch.

In traffic accident research it is said if the braking distance is zero then the

force is infinite. The same holds in structural mechanics. What for a speeding

Fig.

pending on how the

displacements tend

to zero, the stresses

are either infinite or

become infinite because the

Fig. 1.123. Rigid punch on a

half-space. At the edge of the

176 1 What are finite elements?

1.124.

zero and they must assume

car is the acceleration a = dv/dt15 is the strain ε = du/dx or the curvature

κ = d 2w/dx2 for a structure. If a plate cracks, then the strain is infinite,

because in the uncracked concrete the two faces of the crack had the distance

dx = 0 and even an infinitely small crack opening du will result in infinite

strains, du/dx = du/0 = ∞. The same holds for a slab. At a sharp bend the

radius R is zero, and therefore the curvature κ = 1/R is infinite.

Stress singularities occur primarily at the edge, at reentrant corners, or at

points where the boundary conditions change. Some singularities are simply

the result of contradictory boundary conditions. Above the point where the

cantilever beam intersects the wall (Fig. 1.124), the horizontal stress σxx must

be zero, while directly below that point the bending stress σ = M/W attains

its maximum value.

This conflict is not the result of a “discretization error”, which could be

circumvented with a simple trick, but the treatment of the problem is not

adequate. Each abrupt change in the boundary conditions is not in agreement

with the fact that partial differential equations are to be solved.

All abrupt changes in the boundary conditions should theoretically be

replaced by more “blurred” formulations, were it not that an FE program has

its own interpretation of boundary conditions: geometric boundary conditions

are satisfied exactly, but static boundary conditions only in the L2-sense.

In the vicinity of a singularity, the displacement field of a plate consists of

a “non-polynomial” singular part uS and a regular “polynomial” part uR,

u(x, y) = k rα

_

u(ϕ)

v(ϕ)

_

+ uR(x, y) = uS(x, y) + uR(x, y) . (1.496)

The factor k is the so-called stress intensity factor , and the exponent α < 1

depends on the angle of the corner point and the boundary conditions. Because

α < 1 the stresses become singular:

σij = k

1 √

r

. . . (for α = 0.5) . (1.497)

15 If a car hits the wall with a speed v = 100 km/h and is brought to a halt in 0

seconds, the negative acceleration is a = Δv/Δt = −100/0 = −∞.

Fig.

the maximum value

point the stresses must be

At the same

1.34 Actio = reactio? 177

Fig. 1.125. Column and shear wall

The idea to handle only the regular part uR with finite elements implies

that the exact shape of the singular part is known, because when, say, instead

of the exact function r0.5 the function r0.4 is subtracted, not much is gained,

as the FE program still must approximate the missing singular part r0.1 (actually

things are a bit more complicated we cannot simply add and subtract

exponents).

If the solution cannot be split into these two parts, the FE program must

also approximate the singular part, and then one must be careful. One can

then make snapshots of the stress state, which are “correct” for one mesh but

which—in the neighborhood of the singularity—bear no resemblance to the

subsequent stress states as soon as the mesh is refined adaptively.

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