6.4 Circular arches

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The problem of shear locking in shell elements is best explained by studying

the modeling of arches with finite elements.

Fig. 6.6. Circular

arch

The displacement of a point on the neutral axis can be split into a tangential

movement u and a movement w orthogonal to the axis; see Fig. 6.6.

To first order, the strains in a fiber at a distance z from the axis are [70]

εs = εm + z κ, where εm = u,s +w

R

κ = u,s

R

− w,ss . (6.27)

492 6 Shells

Taking the integral of the strain energy density E ε2s

across the arch thickness

t, the strain energy product becomes

a(u,u) =

_ l

0

EAε2

m ds +

_ l

0

EI κ2 ds u = {u,w} , (6.28)

where E is the modulus of elasticity, A = b t is the cross-sectional area and

I = b t3/12 is the moment of inertia of the cross section of the arch.

In the case of rigid-body motion, the strains and curvatures are zero,

εm = κ = 0, so that

u = b1 cos ϕ + b2 cos sin ϕ + b3, w= b1 sin ϕ − b2 cos ϕ , ϕ = s

R

.

(6.29)

The constants b1 and b2 represent displacements in two orthogonal directions

and b3 is a rotation about the center of the circular arch. Upon such a rotation

of the arch w = 0, and all points move in a tangential direction, u = b3.

In a thin arch the strain εm of the neutral axis is essentially zero, and all

variations in u are mainly attributable to the deflection w:

εm = 0 → u,s +w

R

= 0. (6.30)

Employing linear polynomials for u, and cubic for w,

u = a0 + a1s w= b0 + b1 s + b2 s2 + b3 s3 , (6.31)

we have

εm = (a1 + b0

R

) + b1

R

s + b2

R

s2 + b3

R

s3 . (6.32)

Next, if the thickness t of the arch tends to zero, then so must the strains

in the neutral axis of the arch, εm = 0,

a1 + b0

R

= b1 = b2 = b3 = 0, (6.33)

implying that the flexibility of the arch must tend to zero, because only the

term w = b0 is left to model the deflections. All derivatives of this deflection

curve are zero (w,s = w,ss = w,sss= 0). As t → 0 the element stiffens. This is

the so-called membrane locking. The termEA tends to dominate the term EI

in the strain energy product and any attempt to achieve εm = 0 by increasing

the ratio EA/EI →∞ ensures that the deflection becomes much too small.

These effects can be minimized by reduced integration. The curvature terms

in (6.28) are integrated with a two-point formula, but for the membrane part

only a one-point formula is used, which means that the integrand is only

evaluated at the center point, s = 0. At this point

a1 + b0

R

= 0, (6.34)

and therefore only one degree of freedom must be sacrificed to comply with

the constraint.

6.5 Flat elements 493

Fig. 6.7. Analysis of a water tank with flat plate elements

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