Issue 26

A. Tridello et alii, Frattura ed Integrità Strutturale, 26 (2013) 49-56; DOI: 10.3221/IGF-ESIS.26.06

Therefore, according to Eq. 4, the specimen cross-section that leads to a uniform stress distribution entails the typical Gaussian shape. The total volume of the Gaussian specimen part (i.e., the theoretical risk volume theo V ) can be computed by integrating the cross-section of specimen part 3 with respect to z from 0 to 3 L and multiplying it by two:

2

kL

  

  

L

3/2

2

2       

kL

  D V s z dz   3 2 theo

3

  

  

0 

2

3

2  





e

(5)

erf

k

2

x

2

2      ). Eq. 5 allows to compute the length 3 t e dt

  x

   erf  denotes the Error Function (i.e.,

L for the

erf

where

0

desired risk volume, specimen material (i.e., for a chosen value of k ) and for the diameter 2

D . Part 3 of the specimen is

3 L and k uniquely define the Gaussian specimen part.

2 D ,

thus completely designed, since

In order to determine specimen lengths 1 2 L , equations for wave propagation along a straight and catenoidal specimen profile [1], respectively part 1 and part 2 of the specimen (Fig. 1), are solved. The boundary conditions require to have maximum displacement amplitude equal to in U at the interface between the horn and the specimen (i.e., at   1 2 z L L    ), continuity of displacement and strain amplitude at the interface between part 1 and part 2 (i.e., at 0 z  ) of the specimen. A further boundary condition concerning the required stress amplitude in the risk volume is taken into account. Let define the stress amplification factor of the specimen, M  , as the ratio between the constant stress amplitude in the Gaussian specimen part,  , and the maximum stress amplitude in part 1 of the specimen [1], 1  (i.e.,   1 / / d in M E kU       ). According to the assumption of linear elasticity and introducing the boundary conditions, the stress amplification factor can be expressed as: L and 2 z L   ) and at the interface between part 2 and part 3 (i.e., at







  sin



   

   









L

L

L

cos

tan



k

/

2

2

2

M N 

(6)

  k kL





  







L

tan

/





2





1 tan 

kL

2

3

1





  acosh / 2

2

1 2 / N D D  , being

1 D the diameter of the cylindrical part (part 1 in Fig. 1),

 



kL

N L

and:

where

2

2

  k kL

  tan





  N

  

   

   

   







L

/

1 acosh

3

2



2

atan   





kL

N

(7)

1

  k kL kL  



 

1



/ / 

L

tan

2

3

2

kL , both M  and

1 kL depend on the diameter ratio N and on

According to Eq. 6 and Eq. 7 and for a given value of 3

2 kL . Therefore for a chosen resonance frequency, specimen material, diameter ratio N 1 L giving a stress amplitude equal to  in the Gaussian specimen part are obtained 2 L and

the adimensionalized variable

in U value, the lengths

and

and specimen geometry is thus completely defined.

F INITE ELEMENT ANALYSIS : ACTUAL RISK VOLUME AND STRESS CONCENTRATION EVALUATION og-bone and Gaussian specimens with different theoretical risk volumes are tested through Finite Element Analyses (FEA) by using the commercial finite element program ANSYS. Half of the specimen geometrical model is considered in each analysis due to its symmetry and eight-node quadrilateral elements (plane 82) with the axisymmetric option are used for the finite element models. The numerical models count for a number of elements ranging from 21200 to 53700 elements. A suitable fillet radius between specimen parts 2 and 3 is considered for the Gaussian specimen model. D

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