1.18 Proof

Back

It is now time to prove that the FE solution is indeed the scalar product of

the approximate Green’s function Gh0

(which is never calculated explicitly)

and the applied load p. The proof fits on one line8.

Tottenham’s equation

wh(x) = (δ0, wh)

_ _ _

δWe(δ0,wh)

= a(Gh0

, wh)

_ _ _

δWi(Gh0

,wh)

= (p,Gh0

)

_ _ _

δWe(Gh0

,p)

, (1.210)

8 A paper published by Tottenham in 1970 [245] is the earliest reference to this

equation known to the authors.

1.18 Proof 65

0.52

1.04

1.56

2.08

2.59

3 .11

3.63

4 .15

4.67

5.19

5.71

6.23

6 .75

7.26

7.78

8.30

9.34

9.86

9.86

Fig. 1.43. Influence function for the force in the column in the middle of the plate,

and the truck

or in a more conservative notation

wh(x) =

_

Ω

δ0(y − x)wh(y) dΩy = a(Gh0

, wh) =

_

Ω

Gh0

(y, x) p(y) dΩy .

(1.211)

First wh is considered to be the FE solution of the load case p and Gh0

∈ Vh

assumes the role of a virtual displacement:

a(Gh0

, wh) =

_

Ω

Gh0

(y, x) p(y) dΩy (1.212)

then Gh0

is considered to be the FE solution of the load case δ0, and wh

assumes the role of a virtual displacement:

wh(x) =

_

Ω

δ0(y − x)wh(y) dΩy = a(Gh0

, wh) (1.213)

which explains the left-hand side. The symmetric strain energy a(Gh0

, wh)

plays the role of a turnstile.

Because (1.210) contains so much structural analysis, it is perhaps best to

repeat the proof in single steps.

Assume a truck is parked in the middle of a bridge; see Fig. 1.42. The

truck constitutes the load case p. Next two FE solutions are calculated: a) the

FE solution wh of the load case p; b) the FE solution Gh0

, which simulates a

single force δ0 in the middle of the bridge. Both solutions, wh and Gh0

, lie in

Vh. Because Gh0lies in Vh, it follows that

a(Gh0

, wh) = p(Gh0

) Gh0

is a virtual displacement in the load case p

(1.214)

66 1 What are finite elements?

Fig. 1.44. The truck causes the force A in the column while the substitute load ph

causes the force Ah. The force Ah is obtained if the truck is placed on the influence

function Gh0

but it must also be true that

a(Gh0

, wh) = (δ0, wh) wh is a virtual displacement in the load case δ0

(1.215)

and because (δ0, wh) = wh(x), the proof is complete.

Next assume the truck traverses a plate supported in the middle by a

column; see Fig. 1.43. The influence function for the force A in the column

is the deflection P · G0 of the plate if the column is removed and instead a

concentrated force P is applied (for simplicity in the following it is assumed

that P = 1) which pushes the plate down by one unit of deflection. If p denotes

the truck, the force in the column is

A =

_

Ω

G0(y, xA) p(y) dΩy . (1.216)

Concentrated forces are out of reach for an FE program, so the program

replaces the concentrated force with an equivalent load δh

0 that is an aggregate

of surface and line loads, which in the sense of the principle of virtual work is

equivalent to the concentrated force P acting at xA (see Fig. 1.44):

δh

0 (ϕi)

_ _ _

δWe(ph,ϕi)

= fi = (δ0, ϕi)

_ _ _

δWe(p,ϕi)

= P · ϕi (xA) i = 1, 2, . . . n . (1.217)

The deflection surface Gh0

of this equivalent load case δh

0 is an approximate

influence function, and if now the truck is placed on a contour plot of this

deflection surface and the contour lines covered by the wheels of the truck are

traced with a planimeter, the result is exactly the support reaction Ah of the

FE program:

Ah =

_

Ω

Gh0

(y, xA) p(y) dΩy . (1.218)

1.18 Proof 67

0.25 0.12 0.12 0.25

0.06 0.28 0.28 0.06

0.04 0.28 0.28 0.04

0.05 0.03 0.03 0.05

x

y

4.00

1kN

Fig. 1.45. Plate: a) point load δ0; b) FE approximation δh0

with bilinear elements.

In Vh the two deltas yield the same result uh(x) = (δ0, uh) = (δh0

, uh)

A converse statement

Consider a plate as in Fig. 1.45 which is subjected to a volume force p (not

shown). In a second load case a horizontal force P = 1 is applied at a point

x; see Fig. 1.45 a. Let ph0

and jh0

be the element residual forces and jump

terms, respectively, of the FE solution Gh0

of this second load case (Fig. 1.45

b). According to Betti’s theorem

W1,2 =

_

Ω

Gh0

• p dΩy =

_

Ω

ph0

•udΩy +

_

k

_

Γk

jh0

•udsy = W2,1

(1.219)

and because of

_

Ω

Gh0

• p dΩy = uhx

(x) (1.220)

we have as well

uhx

(x) =

_

Ω

ph0

•udΩy +

_

k

_

Γk

jh0

•udsy =: (δh0

,u) (1.221)

which means that the horizontal displacement uhx

(x) is equal to the work

done by the “approximate Dirac delta” δh0

= {ph0

, jh0

} acting through the true

displacement field u. In the limit the volume forces more and more resemble

a true Dirac delta, ph0

→ δ0, and the stress jumps vanish, jh0

→ 0, so that

ux(x) =

_

Ω

δ0(y − x) •u(y) dΩy . (1.222)

68 1 What are finite elements?

Fig. 1.46. Effect of

a small perturbation

Δh of the fulcrum on

cient G0 = l1/l2 for

the leverage K

Hence the residual forces {δ0−ph0

, jh0

} are to be minimized. Note that (1.221)

holds for any FE solution. The statement

uh(x) = (δh0

,u) (1.223)

where δh0

stands for {ph0

, jh0

} is just the converse of

uh(x) = p(Gh0

) . (1.224)

Proxies

More will be said about this subject in Chap. 7. For now it suffices to state

that

• the approximate Green’s function Gh0

is a proxy for the exact function G0

on Ph = the set of all load cases based on the unit load cases pi;

• the approximate Dirac delta δh

0 is a proxy for δ0 on Vh.

In other words the Galerkin orthogonality a(u−uh, ϕi) = 0 also holds for the

Green’s functions a(G0 − Gh0

, ϕi) = 0, or after integration by parts (Green’s

first identity)

a(G0 − Gh0

, ϕi) = pi(G0) − pi(Gh0

) = ϕi(x) − pi(Gh0

) = 0 (1.225)

which implies that Gh0on Ph is a perfect replacement for the kernel G0, and

because

(δ0, ϕi) = (δh

0, ϕi) = ϕi(x) (1.226)

the same holds with regard to δh

0 and δ0 on Vh. This means that to calculate

the horizontal displacement uhx

(x) of any displacement field uh ∈ Vh, instead

the influence coeffi1.19

Influence functions 69

of the exact Dirac delta in Fig. 1.45 a, the substitute Dirac delta in Fig. 1.45

b may be invoked. This is a remarkable—but obvious—result. It is obvious

because in the FE method the original Dirac delta δ0 (the load case p) is

replaced by a substitute Dirac delta δh

0 (a load case ph) that is equivalent

with respect to all ϕi ∈ Vh and in a load case such as p = δ0 work equivalence

just means that

ϕi(x) = (δ0, ϕi) = (δh

0, ϕi) = ϕi(x) . (1.227)

Summarizing all these results, we can state that the FE solution uh(x) ∈ Vh

can be written in six different ways

uh(x) = ph(G0) = ph(Gh0

) = p(Gh0

) = (δ0, uh) = (δh

0 , uh) = (δh

0 , u) .

(1.228)

To better pick up the pattern in these equations we use the short hand notation

(ph,G0) instead of ph(G0) for the virtual external work so that

uh(x) = (ph,G0) = (Luh,G0) = (uh, L

∗

G0) = (uh, δ0) (1.229)

uh(x) = (ph,Gh0

) = (Luh,Gh0

) = (uh, L

∗

Gh0

) = (uh, δh

0 ) (1.230)

where L is the differential operator and L∗ = L is the adjoint operator which

is the same operator as L because in structural analysis L is (most often)

self-adjoint.

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