Crack Paths 2012

º η +β η −− ξ ξ β 2 s i n t a n a 2 c o s 2 c o s h 2 s i n h t a n b b ba 2tip,zy 2 zx = ττ − ª

¨ ¨ © §

¸ ¸ ¹ ·

« ¬

» ¼

η − ξ

2 c o s 2 c o s h

(2π≠β) (8)

η β η ξ −+ ξ

ª 1 2cos2cosh 2 s i n t a n b 2 c o s 2 c o s h 2 inh a b ba 2tip,zy 2 zy = ττ −

º

¨ ¨ © §

¸ ¸ ¹ ·

− η − ξ

« ¬

» ¼

where curvilinear coordinates η and ξ can be given as a function of x and y [11].

0.61

Solid lines: Eq. (8)

τ / τ zy,tip

τ

/ τ

zy

zy,tip

τ

/ τ

0.2

zx

zy,tip

r loc

y loc

ϕ

x loc

ρ/2

-0.62

a

Symbols: F E M

-60

-30

30

60

-90

0

90

ϕ [°]

Figure 4. Plot of the stress components of an inclined elliptical hole (a=1 mm,b=0.1

m mand β=45°) along a circular path of radius rloc= 0.01 m mand comparison with Eq.

(8). Plate thickness t= 6mm.Distance from the mid plane z=2.5 mm.

It is evident that, due the presence of an inclination angle, β, the out-of-plane shear

stress components have both a symmetrical and an antimetrical part. Eq. (8) allows also

to determine the shear stress resultant,

2zy 2 z x τ + τ = τ .

In order to validate this theoretical result some finite element analyses have been

carried out on a uiniaxially loaded wide plate, of finite thickness, containing an inclined

elliptical hole with a=1 mm, b=0.1 m mand β=45°. 20 node brick elements have been

used to carry out the numerical investigations, with a very fine mesh pattern, in order to

get the desired degree of accuracy of the results. Figures 2-4 show a comparison

between the theorical prediction, Eq. (8), and the numerical results. It is evident that in

all the cases the agreement is very satisfactory.

A T H R E E - D I M E N S I OSNHAOLU L D E RPELDA T E

Another interesting example of application is represented by a thick shouldered plate

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