7.7 Weak form of influence functions

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With Mohr’s integral we can calculate the deflection of a beam

w(x) =

_ l

0

M0M

EI

dy M(x)           =

_ l

0

M2M

EI

dy = 0 (7.218)

but not the bending moment M(x). But if we do finite elements then everything

fits perfectly

Mh(x) =

_ l

0

Mh

2 Mh

EI

dy . (7.219)

This surprising result is the topic of this section.

The classical influence functions are “strong” formulations, are based on

Betti’s theorem or Green’s second identity. Here we study “weak” formulations

which are based on the principle of virtual forces/displacements or else, on

Green’s first identity. By weak we mean that the output value

J(w) = a(Gi, w) ≡ w(x) =

_ l

0

M0M

EI

dy (7.220)

is calculated by forming the strain-energy product between the Green’s function

Gi and the solution w; an engineer would say: with Mohr’s integral. This

is only possible if J(w) is a displacement or a deflection or rotation (Euler-

Bernoulli beams and Kirchhoff) because the strain energy product of higher

order Green’s functions (for stresses and alike) is zero a(Gi, w) = 0 and so

J(w) = a(Gi, w) makes no sense.

The strange thing is that if we approximate the Green’s function z (= Gi)

with finite elements by solving the problem

zh ∈ Vh : a(zh, ϕi) = J(ϕi) ϕi ∈ Vh (7.221)

then the FE solution zh is a reasonable approximation of z and we have

J(wh) =

_ l

0

zh(y, x) p(y) dy = a(zh, wh) . (7.222)

This is just Tottenham’s equation (1.210), p. 64. In simpler terms the Green’s

function zh just does what it is supposed to do

536 7 Theoretical details

a(MA, M) = 0 is

zero

J(wh) =

_

i

J(ϕi) ui =

_

i

a(zh, ϕi) ui = a(zh, wh) (7.223)

because if zh solves (7.221) and wh =

_

i ui ϕi then (7.223) follows naturally.

So in FE analysis we can use both formulas, Betti and a(zh, wh), regardless

of what we calculate. When z is exact then only Betti will do for force terms.

The solve this riddle let us compare the three variants of influence functions:

the exact formulations

∂i w =

_ l

0

Gi(y, x) p(y) dy = a(Gi, w) =

_ l

0

δi(y − x)w(y)dy (7.224)

(1) (2) (3)

and the approximate formulations

∂i wh =

_ l

0

Ghi

(y, x) p(y) dy = a(Ghi

, w) =

_ l

0

δh

i (y − x)w(y)dy (7.225)

(1h) (2h) (3h)

where ∂i w is any of the four values w,w_,M, V .

Let us start at the end: the formula (3) with the Dirac delta is not an

integral in the ordinary sense which can be looked up in an integral table

because the Dirac delta is not a proper function and so (3) is a symbol for

the point value ∂i w(x).

But the formula (3h) is not a symbol. It is an integral—though in a somewhat

abbreviated notation. To see this compare Fig. 1.45 a and 1.45 b on page

67 where the horizontal displacement ux(x) and uhx

(x) of a plate is calculated

with two Dirac deltas δ0 and δh0

respectively.

The point load δ0 in Fig. 1.45 a and its action, the integral

ux(x) =

_

Ω

δ0(y − x) •u(y) dΩy , (7.226)

must be interpreted symbolically. But the action of the approximate Dirac

delta δh0, the integral

Fig. 7.8. The strain

energy product

7.7 Weak form of influence functions 537

uhx

(x) =

_

Ω

δh0

(y − x) •u(y) dΩy =

_

e

_

Ωe

ph0

,e

•udΩ +

_

k

_

Γk

jh0

•uds

(7.227)

can be calculated. It is the work done by the true displacement field u on

acting through the residual volume forces ph0

,e and jumps jh0

in the stress

vectors along the interelement boundaries—these forces try to imitate the

Dirac delta δ0. The forces jh0

are the shaded triangles in Fig. 1.45 b. The

volume forces ph0

,e are given only as element resultants re = ((phx

)2+(phy

)2)1/2

Ωe

.

With regard to the second equation, (2), there holds:

a(Gi, w) =

_

∂iw ∂i w = displacement

0 ∂i w = force term (7.228)

that is force terms, M(x) and V (x), cannot be calculated with Mohr’s integral

or in more general terms: stresses σij are out of reach for any weak formulation

σij        = a(G1,u) = 0. (7.229)

This is familiar from beam analysis: the scalar product between the bending

moment M in a beam and the bending moment MA of a redundant XA (≡ G2)

(see Fig. 7.8) is zero which simply means that the slope w_ of the deflection

is continuous at the support.

So to summarize the results: in FE analysis—where we operate with substitute

Green’s functions Ghi

—all three equations (1h), (2h) and (3h) are valid

formulations. If the Green’s function Gi is exact (3) is not computable, it is

a symbol, (2) is only applicable if ∂iu is a displacement, and only (1) will do

in all cases.

Remark 7.3. We trust that the reader is now familiar with the technique and so

we can quickly study the full range of possible weak formulations for influence

functions. Let −u__ = p with boundary values u(0) = u(l) = 0 then we obtain

by formulating Green’s first identity and taking the limit

lim

ε→0

G(Gi, u)Ωε = 0 (7.230)

the results

a(G0, u) =

_

Ωε

−G

__

0 udy + [G

_

0 u]Ωε = u(x) (7.231)

a(u,G0) =

_

Ωε

−u

__

G0 dy + [u

_

G0]Ωε =

_ l

0

pG0 dy (7.232)

a(u,G1) =

_

Ωε

−u

__

G1 dy + [u

_

G1]Ωε =

_ l

0

pG1 dy − u

_(x) (7.233)

a(G1, u) =

_

Ωε

−G

__

1 udy + [G

_

1 u]Ωε = 0 + lim

ε→0

[. . .] (7.234)

538 7 Theoretical details

Fig. 7.9. The Green’s function GE for the strain energy in the element Ωe

where the notation means

lim

ε→0

a(., .)Ωε = lim

ε→0

. . . = result (7.235)

and

Ωε = [0, x − ε] ∪ [x + ε, l] . (7.236)

Because (7.233) is Betti, u_(x) = (G1, p), we conclude that a(u,G1) = 0 and

therefore the limit of the jump terms in the last equation

lim

ε→0

{−G

_

1(x + ε) u(x + ε) + G

_

1(x − ε) u(x − ε)} = 0 (7.237)

must be zero as well. The two opposite horizontal forces G_

l and G_

r press the

cut by one unit apart—in the force method this pair would be a pair X1 = 1—

and in the force method a(G1, u) = 0 would be a test that u is continuous at

x: if both sides of the cut move by the same amount u(x) then the work done

by the two opposite but equal forces, the pair G_

l,G_

r or 1, 1 respectively, is

zero.

So in this light a(w,Gi) = 0 must be zero if Gi is the Green’s function for

a force term—N, M or V —because with a(w,Gi) we test whether u, w_ or w

is continuous at x. A value a(w,Gi)       = 0 would signal a discontinuity at x.

Because a(G0, u) = a(u,G0) also the first two results are the same and the

combined result is Betti for u(x). Note that G__

0 = 0 in Ωe and that the limit

of [G_

0, u] (replace G1 in (7.237) by G0) is u(x) because the normal force G_

0

jumps by one unit at x.

In textbooks a(G0, u) = the principle of virtual forces and a(u,G0) = the

principle of virtual displacements. So both principles, a(u,Gi) = a(Gi, u),

return zero for the strain energy product if δu = Gi is a Green’s function for

7.8 Influence functions for other quantities 539

a force term while lower order Green’s functions, a(u,Gi) = a(Gi, u) = ∂i u,

return displacements

a(Gi, u) =

_

J(u) J(u) = displacement

0 J(u) = force term.

(7.238)

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