7.8 Influence functions for other quantities

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The reach of Green’s second identity B(u, ˆu) = 0 extends well beyond the

calculation of point values.

• To derive an influence function for the strain energy in a single element Ωe

we argue as follows: According to Green’s first identity the strain energy

in a single element Ωe = [xa, xb] of the beam in Fig. 7.9 a is

a(w,w)Ωe =

_ xb

xa

EI (w

__)2 dx =

_ xb

xa

pwdx + [V w−Mw

_]xb

xa

= : p(w)Ωe + [R,w]Ωe (7.239)

where in the abbreviated notation R is short for the internal actions V

and M at the ends of the element Ωe which balance the applied load p.

Next let GE the deflection of the beam if the distributed load p acts on

Ωe alone and when the forces R are applied; see Fig. 7.9 c. Evidently then

a(w,w)Ωe = p(w)Ωe + [R,w]Ωe

_ _ _

W1,2

= a(GE, w) =

_ l

0

GE pdx

_ _ _

W2,1

. (7.240)

Hence the solution GE is the Green’s function for the strain energy in the

element Ωe. The Green’s function for the strain energy in the whole beam

is the solution w itself because a(w,w) = (p,w). Note that GE is load case

dependent.

The implication for the FE solution is that

a(wh, wh)Ωe =

_ l

0

Gh

E pdx (7.241)

where Gh

E is the FE solution of the auxiliary load case pΩe,R. This result

is based on integration by parts and (7.408)

a(wh, wh)Ωe = (ph, wh) + [Rh, wh] =

_ l

0

Gh

E ph dx

=

_ l

0

Gh

E pdx . (7.242)

540 7 Theoretical details

• To derive an influence function for the L2-norm squared of a displacement

field

||u||20

=

_

Ω

u•u dΩ (7.243)

we apply the displacement field u as volume forces. Let uu the solution of

this problem, that is

a(uu, v) = (u, v) v ∈ V , (7.244)

and with u ∈ V it follows

W1,2 =

_

Ω

uu • p dΩ = a(uu,u) = (u,u) = W2,1 , (7.245)

where p is the right-hand side (the volume forces) which belongs to the

original displacement field u. In this particular case the technique is also

known as the Aubin–Nitsche trick .

The extension of (7.245) to the FE solution is evident

_

Ω

uh

u

• p dΩ = (uh,uh) . (7.246)

Here uh

u is the FE solution if the volume forces uh are applied.

• The influence function for the integral value of the displacement in a bar

under the action of a distributed load p is the solution of −EAG__

I = 1

because

W1,2 =

_ l

0

1 × udx =

_ l

0

GI × pdx = W2,1 . (7.247)

That is GI is a quadratic function. Now this is interesting. The influence

function for the sum of the horizontal forces is GΣ = 1, that is a constant

function. The influence function for the integral value of the stress σx =

N/A is a linear function because two opposite forces ±1 pull at the ends

of the bar to generate the influence function Gσ.

const =⇒

_

H = 0 =

_

Hh = Nh(l) − Nh(0) +

_ l

0

ph dx

linear =⇒

_ l

0

σx(x) dx =

_ l

0

σh

x(x) dx

quadratic =⇒

_ l

0

u(x) dx =

_ l

0

uh(x) dx .

That is, if the element shape functions can model constant displacements

then the equilibrium condition is satisfied. If they can model linear displacements

then the integral values of the stresses coincide, (σx, 1) =

7.9 Shifted Green’s functions 541

(σh

x, 1), and if they even can solve the equation −EAu__ = 1 exactly then

it is guaranteed that the integral value of the displacement is the same

_ l

0

udx =

_ l

0

uh dx . (7.248)

(Note that by dividing with the length l we could establish the same results

for the average values). The higher the degree of the shape functions the

higher the moments that will agree (−EAϕ__

i = xk)

_ l

0

uxk dx =

_ l

0

uh xk dx (7.249)

where k = p−2 and p ≥ 2 is the degree of the polynomial shape functions.

Hence it seems that for any quantity we are interested in, there is an influence

function and the important point is that the FE program replaces the exact

Green’s functions in these formulas by approximate solutions or as we say by

shifted Green’s functions.

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