1.14 Concentrated forces

Back

A Kirchhoff plate (which ignores transverse shear strains) sustains the attack

of a concentrated force, while a Reissner–Mindlin plate does not: the force

simply cuts through the plate. The same happens if a plate is put on a point

support (an infinitely sharp needle). The plate simply ignores the support.

Why this happens and why structures react differently to point forces will be

discussed in the following.

Assume a concentrated force at the middle of a plate. If we draw a circle Γ

with radius r around the force, the horizontal stresses tx on the circle Γ must

tend to infinity as 1/(2πr) (Fig. 1.32 a), because only this behavior guarantees

that in the limit as r tends to zero, the stresses balance the horizontal force

P = 1:

lim

r→0

_

Γr

tx ds = lim

r→0

_ 2π

0

tx r dϕ = lim

r→0

_ 2π

0

1

2πr

↑

↓

r dϕ = 1. (1.149)

The more the circles close in on the force, the tighter the lines of force are

packed, the more lines pass through each square inch of the cross section of

the plate, and the more singular the stresses become.

In a Kirchhoff plate (Fig. 1.32 b) the Kirchhoff shear vn exhibits the same

behavior, and for the same reason:

1.14 Concentrated forces 45

Fig. 1.32. The edge

tain the balance with

plate; b) slab

lim

r→0

_

Γr

vn ds = lim

r→0

_ 2π

0

vn r dϕ = lim

r→0

_ 2π

0

1

2π r

r dϕ = 1. (1.150)

If this experiment is done in a 3-D elastic solid, the stresses must tend to

infinity as 1/r2, because the integration is carried out over a sphere and the

surface S of a sphere shrinks as S = 4π r2 as r tends to zero.

The spatial dimension

The rate at which stresses tend to infinity thus depends on the dimension

of the continuum. In physics, everything that tends to a point source, the

electrical forces that converge on a point charge e, the gravitational forces

that converge on a point mass m, the stresses that converge on a point load

P must be consistent with the dimension n of the continuum, or rather the

S = 2π r (circle), S= 4π r2 (sphere) , (1.151)

and therefore must counterbalance the rate at which the sphere shrinks, i.e.,

the fields must behave as 1/(2π r) or 1/(4π r2), respectively to reach the

target6 [242]. (The factors 2 π and 4 π are the magnitude of the unit sphere

in R2 and R3 respectively).

But if the strains εxx = σxx/E in a plate (we take ν = 0) behave as 1/r,

then the horizontal displacement u behaves as ln r, the anti-derivative of 1/r.

Hence the displacement at the foot of the concentrated force also becomes

infinite, the point disappears from the screen. But if the force P = 1 has

infinite range then the work done is also infinite, We = (1/2) P ×∞, and

because of Wi = We the strain energy is infinite as well. Hence in the load

case P = 1, the stress and strain field must have infinite energy.

The reason why a Kirchhoff plate sustains the impact of a concentrated

force though the Kirchhoff shear vn (the third derivatives) also tends to infinity

as 1/r is that the deflection w is the triple indefinite integral of vn, and if 1/r

is integrated three times then the result is the function w = 0.5 r2 ln r−3/4 r2

6 Only the 1/r2-law of gravitation makes it possible to concentrate the mass of the

Earth at its center. Given any other law say 1/r or 1/r3 the center of gravity

would not lie at the center of the Earth, [232].

stresses must mainthe

point load: a)

size of the sphere that surrounds the target, see Fig. 1.33,

46 1 What are finite elements?

Fig. 1.33. Near a single point source the line of forces are packed so tightly that a

plastic zone develops. If the load is spread over a short distance then the singularity

is weaker and no plastic zone will develop

which is zero at r = 0 (in the limit as r → 0), that is, the deflection is bounded.

Note that the total deflection is not zero, because the plate deflection is the

sum of this singular function and a regular function; see Equ. (2.5) in Sect.

2.1, p. 242.

Everything hinges on three numbers

i = order of the singularity, the point source

n = dimension of the continuum (1.152)

m = order of the strain energy

The force, or in more general terms the singularity, that deflects the plate

contributes external work which is just the product of the action and its conjugate

quantity (we may neglect the factor 1/2 typical of eigenwork, because

it is irrelevant in this context)

We = force × deflection We = moment × rotation

We = rotation × moment We = dislocation × force .

(1.153)

Because of the principle of conservation of energy, We = Wi, the internal

energy Wi is bounded if and only if the external work We = action × conjugate

quantity is bounded. Because the action, the force P, the moment M, etc., is

always finite—for simplicity it can be assumed that the source has magnitude

1.0—the question of whether We is infinite or not depends on the magnitude

of response of the structure, i.e., the magnitude of the conjugate quantity.

This comprises the subject of the following section.

Sobolev’s Embedding Theorem

If Ω is a bounded domain in Rn with a smooth boundary and if 2m > n, then

Hi+m(Ω) ⊂ Ci(￣Ω) (1.154)

and there exist constants ci < ∞ such that for all u ∈ Hi+m(Ω)

1.14 Concentrated forces 47

Table 1.1. Admissible (ok) and inadmissible (no) loads

n = 1 n = 2 n = 3

singularity rope, bar, plate,

m = 1 Timoshenko beam Reissner–Mindlin 3-D

i = 0 : ↓ ok no no

i = 1 :

_

_ no no no

singularity

m = 2 Euler–Bernoulli beam Kirchhoff plate

i = 0 : ↓ ok ok

i = 1 : _ ok no

i = 2 : __ no no

i = 3 :

_

_ no no

||u||

Ci( ЇΩ)

≤ ci ||u||

Hi+m(Ω) . (1.155)

The norm of a function u

||u||

Ci( ЇΩ) := max

0≤|j|≤i

____

∂|j|u(x)

∂xj

____(

1.156)

￣Ω

.

This theorem implies that the strain energy due to a point load is bounded

and the conjugate quantity is finite (and continuous) if the three numbers in

(1.152) satisfy the inequality [115]

m − i >

n

2 . (1.157)

The order of the energy is

m = 1 Timoshenko beams, Reissner–Mindlin plates, Elasticity theory

m = 2 Euler–Bernoulli beams, Kirchhoff plates

and the index of the singularity for second-order equations (2m = 2)

i = 0 force i = 1 dislocation

is the maximum absolute value of u and its derivatives up to the order i on

48 1 What are finite elements?

Fig. 1.34. The four singularities of a beam,

_

1

_ ≡ tan ϕl + tan ϕr = 1

and fourth-order equations (2m = 4)

i = 0 force i = 1 moment

i = 2 rotation i = 3 dislocation

The spatial dimensions are n = 1 for ropes, bars, beams, n = 2 for plates, and

n = 3 for elastic solids. Table 1.1 summarizes the inequality (1.157).

If the action is a force, then the shear forces (the third derivatives of a

slab) must behave as 1/r, the bending moments (the second derivatives) as

ln r, the rotations wn (first derivative) as r ln r, and the deflection w finally

as 1/2 r2 ln r − 3/4 r2.

If this is done systematically for all possible point loads (actions), the

following list with the derivatives and antiderivatives respectively of the two

singular functions 1/r (2D) and 1/r2 (3D) is obtained:

→ differentiate

r2 ln r _ r ln r _ ln r _ r

−1 _ r

−2 _ r

−3 _ r

−4

← integrate

To better concentrate on the essential parts, we have dropped all constant

factors and all non-essential parts.

Hence the characteristic singularities in a Kirchhoff plate are the following:

force moment rotation dislocation

w r2 lnr rln r ln r r−1 ←

w,i r ln r ln r r−1 ← r−2

mij ln r r−1 ← r−2 r−3

qi r−1 ← r−2 r−3 r−4

and we learn that for example in the neighborhood of a single moment the

shear forces qi behave as r−2, the moments mij as r−1, the rotations w,i as

1.14 Concentrated forces 49

ln r (the moment rotates infinitely often, a Kirchhoff plate does not sustain

the attack of a moment), while the deflection, w _ r ln r, is bounded.

Traversing the table from left to right, the characteristic 2-D singularity

1/r rises one level higher with each step to the right. The same tendency can

be observed in a beam (Fig. 1.34). In a beam the 1/r discontinuity wanders

from the lower left (influence function for w) to the upper right (influence

function for V ).

In a Reissner–Mindlin plate the deformations are w, θx, θy, and the shear

forces are defined to be

qx = K

1 − ν

2

￣λ

2 (θx + w,x ), qy = K

1 − ν

2

￣λ

2 (θy + w,y ) , (1.158)

and because these shear forces must behave as 1/r in the neighborhood of a

concentrated force, the deflection will behave as ln r, that is w will be infinite

at r = 0, and the rotations θx, θy will be infinite too.

Remark 1.4. Sobolev’s Embedding Theorem deserves some remarks. Let i = 0,

then this theorem states that

Hm(Ω) ⊂ C0(￣Ω ) ||u||

C0( ЇΩ)

≤ c0 ||u||

Hm(Ω) , (1.159)

which means that functions u with the property ||u||m < ∞ are continuous,

and so is the embedding of the space Hm(Ω) into C0(￣Ω)—this is the meaning

of the second part of (1.159).

Hence for each spatial dimension n there is a certain index m beyond which

all functions in Hm(Ω) are continuous, Hm(Ω) ⊂ C0(￣Ω), and this index m

only depends on the dimension n of the space, namely m must be greater than

n/2.

That is if the strain energy of the structure is bounded, ||u||m < ∞, then

the displacements are continuous—no cracks. But for this conclusion to be

true it must be m > n/2, i.e., it is true for Kirchhoff plates, 2 > 2/2, but not

for Reissner–Mindlin plates, 1 > 2/2 or elastic solids, 1           > 3/2.

Let R3 be all vectors x = [x1, x2, x3]T. The embedding of R3 into R2—

simply the vertical projection of the vectors x onto the plane—is continuous

because

||x||2 =

 

x21

+ x22

≤ ||x||3 =

 

x21

+ x22

+ x23

. (1.160)

So if two vectors x and ˆxare close in R3 then they are also close in R2. That

is a continuous embedding preserves the topological structure of the original

space.

For additional remarks about Sobolev’s Embedding Theorem and its consequences

for structural mechanics, see Sect. 7.10, p. 552.

50 1 What are finite elements?

Working with point forces

As structural engineers we are well versed in the art of extracting information

from a structure by applying point loads P = 1 or similar singularities. Formally

this information gathering is an application of Green’s first or second

identity

G(G0, u) = 0 B(G0, u) = 0 etc. (1.161)

or stated differently, an application of the principle of virtual forces or Betti’s

theorem.

In light of Sobolev’s Embedding Theorem, it might seem that care must be

taken if these techniques are applied to 2-D and 3-D solids. But the situation

is not so dramatic. One must distinguish between the formulation of Green’s

first identity on the diagonal, G(u, u) = 0, and formulations G(u, ˆu) = 0where

u          = ˆu that are on the secondary diagonal.

If u = G0(y, x) is the displacement field due to a point load in an elastic 3-

D solid the stresses behave as 1/r2. When G(u,u) is formulated with this field

u then the strain energy density σij εijdΩ at x has double that singularity,

and therefore the strain energy in a ball with radius R = 1 is infinite

_

Ω

σij εijdΩ ≡

_ 1

0

O(

1

r2 )O(

1

r2 )O(r2) dr =

_ 1

0

1

r2 dr = ∞. (1.162)

But if ˆu            = u and the field ˆu has bounded stresses, the strain energy density at

x is of the order O(1/r2)O(1)O(r2), and therefore the strain energy product

between the field G0 and the field ˆu is finite.

Similar considerations hold in the case of Betti’s theorem. It is possible

to apply a point load P = 1 to extract information about a regular displacement

field ˆu as in B(G0,ˆu) = 0, but this would fail if we try to formulate

B(G0,G0) = 0, because then the singularities would cancel each other and we

would be left with two meaningless boundary integrals (τ 0 = traction vector

of the field G0)

lim

ε→0

B(G0,G0)Ωε =

_

Γ

τ 0 •G0 ds −

_

Γ

G0 • τ 0 ds = 0. (1.163)

Hence every situation is different, and the presence of singularities requires a

careful study of the limit [115]

lim

ε→0

G(G0,ˆu)Ωε = 0 lim

ε→0

B(G0,ˆu)Ωε = 0. (1.164)

Remark 1.5. Not all is well with point forces. There is one prominent victim

of Sobolev’s Embedding Theorem: Castigliano’s Theorem, which states that

the derivative of the strain energy is the displacement in the direction of the

point load Pi, makes no sense in elastic solids,

1.15 Green’s functions 51

∂

∂Pi

a(u,u) ?=

ui (1.165)

because both the strain energy and the displacement are infinite [115].

Energy estimates involving Green’s functions

In this book we will operate freely with energy estimates involving Green’s

functions with infinite energy. One such inequality is for example

|ux(x) − uhx

(x)| ≤ ||G0[x] − Gh0

[x]||E ||u − uh||E (1.166)

which is an estimate for the (horizontal) displacement error of the FE solution

in a plate. (The [x] is to denote that G0 is the Green’s function for ux(x) at

the point x). Theoretically this equation makes no sense because the strain

energy of the exact Green’s function G0 is infinite (we drop the [x] in the

following because it is not essential here)

||G0||2

E := a(G0,G0) =

_

Ω

σij · εij dΩ (1.167)

and so the distance of the FE Green’s function Gh0

from G0 in terms of the

strain energy

||G0 − Gh0

||2

E := a(G0 − Gh0

,G0 − Gh0

) =

_

Ω

(σij − σh

ij) · (εij − εh

ij) dΩ

(1.168)

would be infinite as well—regardless of how close Gh0is to G0. But if we

read ux(x) as the average value of the horizontal displacement over a small

disk centered at x the corresponding Green’s function G0 would have finite

energy and then (1.166) would make sense. So in this book whenever we use

an expression such as (1.166) we understand this as a statement about the

average value of u(x) (or other terms) over a small disk Ωε with a radius ε

very close to zero.

We should not be deterred too much by the fact that most Green’s functions

have infinite energy. The FE method is surprisingly good at approximating

Green’s functions. Most output we see on the screen is based on the

solution of ill posed problems...

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