7.6 The derivation of influence functions

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Influence functions or influence lines are based on Betti’s theorem, B(w1, w2) =

0. Because the idea behind influence functions is central for the understanding

of the distribution of the internal actions and also the support reactions in a

structure we start with a short repetition of classical structural analysis.

Influence function for V (x)

To obtain the influence function for the shear force V (x) we introduce a shear

hinge at x and to keep the balance with the applied load the prior internal

actions Vl(x) and Vr(x) now act as external forces. In a second load case two

opposite forces spread the two faces of the shear hinge by one unit length,

G3(x−, x)−G3(x+, x) = 1, apart. According to Betti’s theorem the reciprocal

external work of the two systems must be the same

W1,2 = −V (x) · 1 +

_ l

0

G3(y, x) p(y) dy = 0 = W2,1 (7.188)

530 7 Theoretical details

Fig. 7.7. The punctured domain: a) a point inside Ω, b) a point on the edge

or

V (x) =

_ l

0

G3(y, x) p(y) dy . (7.189)

This is the standard procedure for influence functions. The work W1,2 = 0

because the two opposite forces that spread the shear hinge apart are of the

same size so when they act through the same deflection w(x) their effort is

nil.

To see how influence functions in 2-D are derived, we consider the solution

u of the Poisson equation

− Δu = p on Ω = unit disk, u= 0 on Γ = unit circle, (7.190)

which can be identified with the deflection of a circular prestressed membrane

which bears a pressure p.

The Green’s function for the deflection at the center x = 0 of the unit

disk

G0(y, x) = − 1

2 π

ln r (7.191)

is a homogeneous solution of the Laplace equation at all points y          = x,

− ΔG0(y, x) = 0 y        = x . (7.192)

Next a small circular hole Nε with radius ε is punched in the unit disk. The

center of the hole is the center of the disk, x = 0, and ΓNε is the edge of the

hole. At a point y on this circle the normal vector n = n(y) points to the

center x = 0, and it has components n1 = −cos ϕ and n2 = −sin ϕ if polar

coordinates (r, ϕ) centered at x = 0 are used. For all points y on the circle,

the distance r = |y − x| from the center x is the same. The gradient ∇yr of

the distance r points in the direction opposite the normal vector n, because

this is the direction into which the point y must be pushed if the distance

from x = 0 is to increase at the fastest rate possible. Hence

− 1

2 π

∂

∂n

ln r = − 1

2 π

1

ε

∇yr •n =

1

2 π

1

ε

n•n =

1

2 π

1

ε

. (7.193)

7.6 The derivation of influence functions 531

Next Betti’s theorem is formulated on the domain Ωε = Ω − Nε(x). On the

outer edge Γ both solutions are zero, u = G0 = 0, so that:

B(G0, u)Ωε =

_

Ωε

−ΔG0 udΩy −

_

ΓNε

1

2 π

∂

∂n

ln r udsy

+

_

ΓNε

∂u

∂n

G0 dsy −

_

Ωε

G0 pdΩy

(7.194)

The first integral is zero because −ΔG0 = 0 in Ωε and the third integral is of

order O(ε)

∂u

∂n

· ln ε · ε · dϕ = O(1) · ln ε · ε = O(ε) (7.195)

so that

B(G0, u)Ωε =

_

ΓNε

1

2 π

∂

∂n

ln r udsy + O(ε) −

_

Ωε

G0 pdΩy

=

_ 2 π

0

1

2π ε

u(x + εn(ϕ)

_ _ _

y(ϕ)

) ε dϕ + O(ε) −

_

Ωε

G0 pdΩy

(7.196)

which in the limit becomes

lim

ε→0

B(G0, u)Ωε = u(x) −

_

Ω

G0(y, x) p(y) dΩy . (7.197)

The important observation is that in this approach u(x) is not the limit of a

domain integral, as the notation

−ΔG0(y, x) = δ0(y − x) (7.198)

seems to suggest, and Betti’s theorem evidently seems to confirm

B(G0, u) =

_

Ω

−ΔG0 udΩ −

_

Ω

pG0 dΩy

=

_

Ω

δ0(y − x) u(y) dΩy −

_

Ω

G0(y, x) p(y) dΩy

= u(x) −

_

Ω

G0(y, x) p(y) dΩy . (7.199)

But juggling with Dirac’s delta is not mathematics, rather it is an application

of symbolic algebra. The Dirac delta is a handy symbol to express the algebraic

properties of the Green’s function, but the real properties can only be clarified

by doing mathematics as in (7.194).

532 7 Theoretical details

So all the single terms, the point values ∂0u ≡ u(x), ∂1u ≡ σxx(x), in the

influence functions are the limits of certain boundary integrals over the edge

of the hole

lim

ε→0

_

ΓNε

∂i u∂j ˆu ds = 1· ∂iu(x) i + j = 2m − 1 (7.200)

where the conjugate kernel ∂j ˆu that makes ∂iu(x) emerge has the property

lim

ε→0

_

ΓNε

∂j ˆuds = 1. (7.201)

In 1-D the neighborhood ΓNε consists of two points, in 2-D it is a circle

lim

ε→0

_

ΓNε

Vn ds = 1 Vn = Kirchhoff shear (7.202)

and in 3-D it is a sphere — two points become a circle become a sphere.

Hence one cannot monitor a dislocation in a slab by checking, say, the

deflection only at two points, w(x + ε) − w(x − ε)        = 1. Rather one must

complete a full circle to sense a bend δ2 or a dislocation δ3 in a slab

δ2 : lim

ε→0

_

ΓNε

∂G2

∂n

ds = 1 δ3 : lim

ε→0

_

ΓNε

G3 ds = 1. (7.203)

Remark 7.2. For a detailed analysis of the derivation of the hyper-singular influence

function for the slope in a slab (Kirchhoff plate, biharmonic equation)

∂w

∂n

= lim

ε→0

B(G1, w) = 0 (7.204)

see [116] p. 375.

Influence functions for crack tip singularities - don’t exist

If there were an influence function for the singular stresses at a crack tip

σyy(x) =

_

Ω

Gyy

1 (y, x) • p(y) dΩy = ∞, (7.205)

then because the load p is finite, the stress could only become infinite if the

kernel Gyy

1 is infinite, but infinite displacements |Gyy

1

| = ∞ (each kernel is a

displacement field) in almost any patch Ωp of a plate make no sense, as they

would tear the plate apart.

Hence there cannot exist an influence function for the stresses at a crack

tip, only for the stress intensity factor [47].

7.6 The derivation of influence functions 533

The mathematical reason is the following: To derive an influence function

for σyy(x) at a point x, first a small circular region of the point x must be

excluded, and only then are we allowed to let the radius ε tend to zero. In

standard situations, a single term σyy(x) ·1 will be recovered in the limit, but

if the stress field is infinite at x, the effect of the infinite stress is canceled

in Green’s second identity (Betti’s theorem) by a second singular term of the

opposite sign, so that in the limit—which is guaranteed to be zero,

lim

ε→0

B(u,Gyy

1 )Ωε =∞−∞+ finite terms = 0 (7.206)

we are left with meaningless terms, “the ashes”, which convey no real information.

Hence each influence function is the limit of an integral identity. Concepts

as Cauchy principal value or Hadamard’s partie fini integral are just other

names for the limit

lim

ε→0

B(u,Gyy

1 )Ωε = 0. (7.207)

By this process singular or hypersingular integrals are automatically regularized,

because the critical terms drop out.

In Sect. 1.20, p. 80, we argued that in the presence of stress singularities

it is more reasonable to work with resultant stresses. Now we can be more

precise. If the stress σyy becomes singular at the crack tip but if the integral

Ny =

_ l

0

σyy dx < ∞ (7.208)

across the cut is bounded, there exists an influence function for the resultant

stress Ny—in the sense of (7.207)—and the FE program has a chance to

approximate this influence function.

Equivalent nodal forces

The equivalent nodal forces fG

i for the numerical Green’s functions are the

displacements, the stresses, etc. of the shape functions at the point x, see

Sect. 1.19 p. 69,

fG

i = ϕi(x) fG

i = σxx(ϕi)(x) fG

i (x) = qx(ϕi)(x) . (7.209)

But the dimension of each fG

i is force × displacements

fG

i =

_

Ω

δ0(y − x) ϕi(y) dΩy ≡ kN × m (7.210)

To extract the stress from a field we apply a dislocation and calculate the

work done by the stress on acting through the dislocation, [kN/m2] × [m],

534 7 Theoretical details

and to extract a bending moment we apply a dimensionless unit rotation,

so that the work done is M × w_ = [kNm] × [ ], etc.. (A unit slope means

dw/dx = 1× [m]/[m] = 1× [ ] or tan ϕ = 1).

This result is in agreement with the fact that the influence functions are

energy expressions

u(x) [m] × 1 [kN] = . . . σxx(x) [kN/m2] × 1 [m] = . . . (7.211)

So when we calculate a point value by summing over the nodes

uh(x) [m] × 1 [kN] =

_

i

fG

i [kNm] · ui (7.212)

we must divide by the unit, here 1 kN, which extracts the displacement or the

stress, etc., from the field u via (7.210), so that

uh(x) [m] =

1

kN

_

i

fG

i [kNm] · ui =

_

i

ϕi(x) [m] · ui . (7.213)

Recall that the fG

i of the Green’s function for uh(x) are the nodal values of

the shape functions ϕi at the point x.

The nodal displacements ui play the role of weights, that is pure numbers.

The dimension [m] of the displacement u(x) =

_

i ui ϕi(x) =

_

i ui ×

(ϕi(x)[m]) is attached, so to speak, to the ϕi and has already been consumed

in the definition of the fG

i in (7.210).

The net result is that nothing needs to be done: the equivalent nodal forces

fG

i of the Green’s functions are the displacement, stresses, etc. of the shape

functions at x —times the physical dimension of the Dirac delta, [m], [kN],

etc., but because we later divide again by these terms we may ignore them

from the start—and so when we multiply the fG

i with the nodal values ui of

the FE solution we obtain—quite naturally—the pertinent values of the FE

solution at the point x

σh

xx(x) =

_

i

fG

i

· ui =

_

i

σxx(ϕi)(x) · ui . (7.214)

The same can be expressed as

σh

xx(x) =

_

i

fi · uGi

(7.215)

where the uGi

are the nodal values of the Green’s function and the fi are the

equivalent nodal forces of the load case p, see Sect. 1.19 p. 79.

The first formula corresponds to

σh

xx(x) =

_

Ω

δ1(y − x)uh(y) dΩy =

_

i

_

Ω

δ1(y − x)ϕi(y) dΩy

_ _ _

σxx(ϕi)(x)

· ui

(7.216)

7.7 Weak form of influence functions 535

and the second formula corresponds to

σh

xx(x) =

_

Ω

Gh1

(y, x) p(y) dΩy =

_

i

Gh1

(yi, x)

_ _ _

uGi

·

_

Ω

ϕi(y) p(y) dΩy

_ _ _

fi

(7.217)

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