5.3 Elements

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The natural choice for a plate element would be a triangular element with

degrees of freedom w, w,x , w,y at each of the three corner nodes, and a cubic

polynomial to interpolate w between. This would result in a linear variation

of the bending moments and constant shear forces. But because a complete

cubic polynomial has 10 terms instead of the 3 × 3 = 9 terms, the element

would be nonconforming, that is, the first derivatives would not be continuous

across interelement boundaries.

A conforming rectangular element can be derived from the unit deflections

ϕi(x) of a beam (see Fig. 5.11)

5.3 Elements 423

Fig. 5.10. Slab: a) uniform load p = 6.5 kN/m2 and the equivalent FE load case ph,

b) element loads, c) vertical forces, and d) moments along interelement boundaries

ϕ1(x) = 1 − 3x2

l2 +

2x3

l3

ϕ2(x) = −x +

2x2

l

− x3

l2

ϕ3(x) =

3x2

l2

− 2x3

l3

ϕ4(x) = x2

l

− x3

l2 ,

(5.16)

using a product approach

ϕe

..(x, y) = ϕi(x) ϕj(y) i, j = 1, 2, 3, 4 , (5.17)

so that each node has four degrees of freedom w, w,x , w,y , w,xy. Unfortunately

this approach is limited to rectangular slabs. Also the degree of freedom w,xy

is not easily accounted for if the element is coupled to other elements.

A truly isoparametric conforming quadrilateral element also requires that

the mapping of the master element onto the single elements be C1. Because

424 5 Slabs

Fig. 5.11. Shape functions based on Hermite polynomials

of the chain rule

w,x = w,ξ ξ,x +w,η η,x (5.18)

the derivatives ξ,x and η,x, etc., of the mapping functions x(ξ, η) and y(ξ, η)

and the inverse functions ξ(x, y) and η(x, y) must be continuous. This means

that the coordinate lines ξ = const. and η = const. cannot change direction

abruptly (no kinks!) upon crossing the interelement boundaries. It is very

costly to establish this property numerically.

six degrees of freedom at the vertices,

w, w,x , w,y , w,xx, w,xy , w,yy (5.19)

and the normal derivatives at the mid-side nodes as additional degrees of

freedom. This conforming element has 21 degrees of freedom.

Stiffness matrices

The strain energy product of two deflection surfaces w and ˆ w is the scalar

product of the bending moment tensor M = [mij] of w and the curvature

tensor K = [κij] of ˆw,

Theoretically the simplest method of deriving a conforming triangular element

is to choose a complete fifth-order polynomial [215] with the following

5.4 Hybrid elements 425

a(w, ˆ w) =

_

Ω

M• ˆK dΩ =

_

Ω

mij ˆκij dΩ (5.20)

=

_

Ω

[mxxˆκxx + 2mxyˆκxy + myyˆκyy] dΩ , (5.21)

or in equivalent notation the scalar product of the vectorm=[mxx,myy,mxy]T

and the vector ˆκ = [κxx, κyy, 2 κxy]T ,

a(w, ˆ w) =

_

Ω

m•ˆκ dΩ =

_

Ω

mi ˆ κi dΩ

=

_

Ω

[mxxˆκxx + 2mxyˆκxy + myyˆκyy] dΩ , (5.22)

where

⎡

⎣

mxx

myy

mxy

⎤

⎦

_ _ _

m

=

⎡

⎣

K νK 0

ν K K 0

0 0 (1− ν)K/2

⎤

⎦

_ _ _

D

⎡

⎣

κxx

κyy

2 κxy

⎤

⎦

_ _ _

κ

. (5.23)

Hence if mi and κi denote the “bending moment” and “ curvature” vector of

the element shape function ϕei

, an individual element of the stiffness matrix

is

ke

ij =

_

Ωe

mi •κj dΩ =

_

Ωe

mTi

κj dΩ =

_

Ωe

(Dκi)T κj dΩ

=

_

Ωe

κTi

Dκj dΩ (5.24)

and the stiffness matrix of an element with n degrees of freedom is

Ke(

n×n) =

_

Ωe

BT(

n×3)D(3×3) B(3×n) dΩ , (5.25)

where column i of matrix B contains the curvatures κxx, κyy, 2 κxy of the

deflection surface ϕei

.

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