2.1 Influence functions or Betti’s theorem Beams

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To calculate the deflection w(x) at the center of a beam with the principle

of virtual forces the scalar product of the two bending moments M(x) and

￣M

(x) is formed (see Fig. 2.3):

1 · w(x) =

_ l

0

M￣ M

EI

dx Mohr’s integral . (2.1)

But according to Betti’s theorem the scalar product of the distributed load p

and the deflection curve G0(y, x) (= Green’s function) which belongs to the

load case ￣ P = 1 would provide the same result:1

W1,2 = 1· w(x) =

_ l

0

G0(y, x) p(y) dy = W2,1 . (2.2)

Betti’s theorem remains valid if the span of the second beam is larger than

the span of the first beam. It is only necessary to cut off the protruding

parts before Betti’s theorem is formulated, and to compensate for this loss

the previous internal actions must be applied as external forces.

Those external forces then contribute external work too, so that Betti’s

theorem becomes rather lengthy, as is seen if the reciprocal statement W1,2 =

W2,1 is solved for w(x):

1 One could also simply integrate (2.1) by parts: (M￣ ,M/EI) = (G0, p ).

2.1 Influence functions or Betti’s theorem 241

Fig. 2.3. Calculation of the deflection δ a) with the principle of virtual forces, b)

with Betti’s theorem (the same beam), c) with Betti’s theorem (a different beam)

w(x) =

_ l

0

g0(y, x) p(y) dy

+ ￣M (0)w

_(0) − ￣M (l)w

_(l) + ￣ V (l)w(l) − ￣ V (0)w(0)

_ _ _

−work on the boundary W1,2

+M(0) g

_

0(x, 0) −M(l) g

_

0(x, l) + V (0) g0(x, 0) − V (l) g0(x, l)

_ _ _

work on the boundary W2,1

.

(2.3)

The deflection curve w = g0(y, x) of the second beam (see Fig. 2.3 c) is

called a fundamental solution, because unlike the Green’s function G0 it does

not satisfy the support conditions of the original beam. But both curves g0

and G0 share the property that the shear force jumps at the source point,

V (x+)−V (x−) = 1. To construct influence functions—to have w(x) emerge—

only the jump in the shear force V is needed.

Kirchhoff plates

The procedure is virtually the same as before: to calculate the deflection at a

point x a single force ￣ P = 1 is applied at this point and the scalar product of

the deflection surface G0(y, x) and the load p is formed:

242 2 What are boundary elements?

Fig. 2.4. Circular plate serving as auxiliary problem

W1,2 = 1· w(x) =

_

Ω

G0(y, x) p(y) dΩy = W2,1 . (2.4)

The problem with this approach is that in general the Green’s function G0 is

unknown. Only if a single force ￣ P is applied directly at the center of a circular

w(r) =

￣ P

16K π

_

3 + ν

1 + ν

R2 (1 − r2

R2 ) − 2 r2 ln R

r

_

. (2.5)

But this suffices. First the radius R of the plate is extended to infinity, R = ∞,

so that all plates that will eventually be considered fit into the interior of the

circular plate. That is, (2.5) is simply stripped of all unnecessary parts and

only the fundamental solution is retained:

g0(y, x) =

1

8πK

ln r . (2.6)

Next the source point x—the center of the infinite circle—is moved to the

point at which the deflection w(x) is to be calculated. Then that part of the

circular plate which coincides in shape and position with the original plate is

cut out of the circular plate, the previous internal actions are applied at the

new edge as external forces, and finally the reciprocal work of the pair {w, g0}

is calculated. The result is an expression of the form

w(x) =

_

Ω

g0(y, x) p(y) dΩy − work on the boundary W1,2

+work on the boundary W2,1 (2.7)

or

plate of radius R (see Fig. 2.4) is the deflection surface known:

2.1 Influence functions or Betti’s theorem 243

c(x)w(x) =

_

Γ

[ g0(y, x) vν(y) − g0ν(y, x)mν(y)

−vν(g0)(y, x)w(y) + mν(g0)(y, x)wν(y)] dsy

+

_

Ω

g0(y, x) p(y) dΩy

+

_

c

[ g0(yc, x) F(w)(yc) − w(yc) F(g0)(yc, x)] , (2.8)

where the underlined terms are the boundary values of the plate:

vν = support reaction mν = bending moment

wν = rotation w = deflection

and the F are the corner forces. The influence coefficients or kernel functions

↓ g0(y, x) _ g0ν(y, x) _ mν(g0)(y, x) ↓ vν(g0)(y, x)

(2.9)

are, in this sequence, the deflection, slope, bending moment and Kirchhoff

shear of the fundamental solution g0 on the boundary. The so-called characteristic

function c(x) on the left-hand side has the value 1 at all interior

points, and the value c(x) = Δϕ/2 π at boundary points where Δϕ is the

angle of the boundary point.

By replacing the kernel g0 (single force) with the kernel g0n = ∂g0/∂n

(moment), an influence function for the slope ∂w/∂n can be derived in the

same way.

If the observation point x is located on the boundary, these two influence

functions become—in a somewhat simplified notation—a system of two

integral equations:

_

Γ

H(y, x)u(y) dsy =

_

Γ

G(y, x) t(y) dsy

+

_

Ω

U(y, x) p(y) dΩy + R(yc, x) rc x ∈ Γ, (2.10)

for the boundary values of a slab

u = {w(x), wn(x)} t = {mn(x), vn(x)} rc = {wc, Fc} . (2.11)

The subscript c in the vector rc indicates that these terms are the corner

deflections wc and the corner forces Fc.

To solve this system approximately, the boundary functions are interpolated

with polynomial shape functions, and the unknown nodal values are

determined by a collocation procedure, so that the two coupled integral equations

are equivalent to a linear system of equations

244 2 What are boundary elements?

Fig. 2.5. BE analysis of a shear wall

2.1 Influence functions or Betti’s theorem 245

Hu = Gt + d (2.12)

for the nodal values u and t.

If the equation is multiplied from the left first by G

−1 and then by the

mass matrix F,

F G

−1Hu =F G

−1t +F G

−1d (2.13)

and terms are collected, we obtain the system

Ku = f + p (2.14)

which is formally identical to the structural equation in FE analysis, even

though the matrix K is a slightly unsymmetric stiffness matrix.

Linear elasticity

The influence functions for the displacement field u(x) of a plate as in Fig.

2.5 is

C(x)u(x) =

_

Γ

[U(y, x) t(y) − T(y, x)u(y)] dsy

+

_

Ω

U(y, x) p(y) dΩy (2.15)

where t = t(y) is the traction vector Sν = t at the integration point y on the

boundary with the normal vector ν = [ν1, ν2]T . The fundamental solutions

are

Uij(y, x) =

1

8πμ(1 − ν)

[(3 − 4ν) ln

1

r

δ ij + r,i r,j] (2-D), (2.16)

Uij(y, x) =

1

8πμ(1 − ν)r

[(3 − 4ν) δ ij + r,i r,j] (3-D), (2.17)

and the tractions of these solutions on the boundary are

Tij(y, x) = − 1

4απ(1ν)rα [ ∂r

∂ν

((1 − 2ν) δ ij + β r,i r,j )

−(1 − 2ν){r,i νj(y) − r,j νi(y)}] , (2.18)

where α = 1, β = 2 in 2-D, α = 2, β = 3 in 3-D, and

r,i := ∂r

∂yi

= − ∂r

∂xi

= yi − xi

r

. (2.19)

The columns i (or rows) of the symmetric matrix U(y, x) are the displacement

fields of the elastic continuum if a concentrated force P = ei acts at a point

x, and the columns of the matrix T(y, x) are the associated tractions at a

point y with the normal vector ν.

246 2 What are boundary elements?

Fig. 2.6. Shear wall: a) system and loads, b) principal stresses, c) boundary elements

6.5 kN/m Eigengewicht 7.5 kN/m2 9.0 kN/m

4.4 kN/m 7.0 kN/m

23.7 kN/m

23.7 kN/m

130.1 kN/m 141.8 kN/m

x

y

25.0

7.1

6.5 kN/m 9.0 kN/m

4.4 kN/m 7.0 kN/m

23.7 kN/m

23.7 kN/m

130.1 kN/m 141.8 kN/m

2.2 Structural analysis with boundary elements 247

Technical details

The boundary discretization poses no difficulties. The elements should form a

piecewise smooth approximation and the boundary functions should be C0 or

where necessary C1 functions (the deflection w in plate bending). Essentially

the whole technology can be carried over from the FE literature.

More important is an efficient integration of the influence integrals. If the

elements are straight, then in general the integrals can be evaluated analytically.

Analytic integration will not eliminate all problems because in a very

small region near the boundary nodes the solution will still deteriorate, but

these effects can be attributed to the kinks (higher-oder discontinuities) in

the boundary displacements or tractions.

Another problem is that not all the stresses on the boundary can be determined

directly from the integral equations. The traction vector t has two

components, while the stress tensor S has three. In plate bending problems

the moment mt (which requires reinforcement in the tangential direction) and

in some sense also the twisting moment mnt must be recovered artificially from

the boundary values w and w,n by finite differences.

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