6.5 Flat elements

Back

Flat elements are plate elements in which membrane and bending action are

both included. The coupling between these two actions occurs at the nodes,

and is due to the varying orientation of the elements.

Most shells are probably analyzed with such flat plate elements (see Fig.

6.7, 6.12 and 6.13), because the modeling is easy and the accuracy in most

cases is sufficient. It is guaranteed that flat elements can represent rigid body

motions and because the membrane and bending stresses within an element

are decoupled it easy to understand and control the behavior of such elements.

The first idea is to use triangular elements; see Fig. 6.8. If each node of

the triangle has three degrees of displacement and three degrees of rotation,

then such an element has 18 degrees of freedom,

u = [ui, vi,ϑzi,wi,ϑxi,ϑyi]T , (6.35)

where the individual vectors ui = [u1, u2, u3]T and ϑzi = [ϑz1, ϑz2, ϑz3]T, etc.

are respectively the displacements and rotations of the nodes.

Let the matrix KM be the associated 9 × 9 stiffness matrix accounting

for the membrane stresses of the element. For simplicity, a DKT element is

chosen for the bending stresses. If KB denotes the associated stiffness matrix,

then membrane and bending stresses are indeed decoupled:

Ke u =

_

KM

(9×9) 0(9×9)

0(9×9) KB

(9×9)

_

⎡

⎢⎢⎢⎢⎢⎢⎣

ui

vi

ϑi

wi

ϑxi

ϑyi

⎤

⎥⎥⎥⎥⎥⎥⎦

= f . (6.36)

Only if the midsurfaces of the neighboring elements do not lie in the same

plane will the two stress states (in general) become coupled.

494 6 Shells

Fig. 6.8. Flat elements

Fig. 6.9. An origplane

[160]

Fig. 6.10. Twisted

beam problem [160]

If the CST element is used for membrane stresses, there are no rotational

degrees of freedom about the vertical axes,

KM u =

_

KCST

(6×6) 0(6×3)

0(3×6) 0(3×3)

_⎡

⎣

ui

vi

ϑzi

⎤

⎦ , (6.37)

which means that a “flat” node causes the global stiffness matrix to become

singular. To avoid such unconstrained modes, artificial rotational degrees of

where the four nodes

no longer lie in a

inally flat element

6.5 Flat elements 495

freedom are built in:

αE V

⎡

⎣

1.0 −0.5 −0.5

−0.5 1.0 −0.5

−0.5 −0.5 1.0

⎤

⎦

⎡

⎣

ϑz1

ϑz2

ϑz3

⎤

⎦ =

⎡

⎣

Mz1

Mz2

Mz3

⎤

⎦ . (6.38)

Here E and V are the modulus of elasticity and the volume of the element,

and α is a scaling factor (< 0.5) [258]. In other words the null matrix on

the diagonal in (6.37) is replaced by this matrix. One easily recognizes that

rigid-body motions such as ϑz1 = ϑz2 = ϑz3 will not give rise to couples at

the nodes.

6.11.

structure under gravity load

A good choice for flat elements is a combination of the Wilson Q4+2 element

and a four-node Reissner–Mindlin element. But one must be careful:

while the nodes of a triangular element always lie in a plane this is not guaranteed

in quadrilaterals. Therefore one must modify the stiffness matrix of

the element to account for the chance that the nodes do not lie in a plane; see

Fig. 6.9. One idea is to write

ˆK

= ST KS, (6.39)

where the matrix S represents the coupling between the degrees of freedom

of the displaced nodes and the nodes in the plane of the element,

Fig. Analysis of a

curved staircase with flat elements:

view of the deformed

496 6 Shells

Fig. 6.12. Clarifier, analysis with flat elements: a) system, b) deformations due to

nonuniform temperature changes

u = S ˆu , (6.40)

and the transposed matrix ST transforms the equivalent nodal forces f of the

element into the equivalent nodal forces ˆf of the displaced nodes:

ˆf = ST f , fi = [N(i)

x ,N(i)

y , P(i)

z ,M(i)

x ,M(i)

y , 0]T . (6.41)

Here it is assumed that the displaced nodes are connected to the element via

small rigid bars of length h, and that the equilibrium conditions can thus be

employed to formulate a relation between the equivalent nodal forces f and

ˆf, i.e., to establish the matrix ST .

The lever h will give rise to moments

6.5 Flat elements 497

Fig. 6.13. Intersection of two tubes, analysis with flat elements: a) system,

b) deformations

ˆM

x = ±hNy , ˆMy = ±hNx (6.42)

at the nodes. But as explained in [160], this technique is not to be recommended,

because the nodal moments tend to disturb the pure membrane stress

state. Therefore it is more appropriate to account for the moments due to the

displaced nodes by means of vertical couples. In this way, for example, the

moment due to the two forces F14 and F41 (see Fig. 6.9) is accounted for by

two opposing forces:

F1z = −F4z = h

l14

(F14 − F41) . (6.43)

498 6 Shells

If the twist of a plate strip is to be modeled (twisted beam problem), then

by transferring the moment to the nodes, a moment Mz will also be generated

about the vertical axis, which because there is no corresponding stiffness will

lead to a failure of the model. To avoid this behavior, one can balance the

z-component of the moment M1 (see Fig. 6.10) with a pair of vertical forces:

FA = −FB =

sin α

lAB

M1 . (6.44)

Flat elements model a shell as a faceted surface, and these models are more

sensitive to singularities than a curved shell. This can be seen in the model

of a steel staircase in Fig. 6.11, where under gravity load, stress singularities

develop at those points where the curvature of the structure is large; see Fig.

6.11.

Навигация