Issue 29

S. Terravecchia et al., Frattura ed Integrità Strutturale, 29 (2014) 61-73; DOI: 10.3221/IGF-ESIS.29.07



u

, u c1 

x          y n u n

g

0

x

x

(19a,b,c,d)

u c2 

g

,

0

y

y

with c1 and 2 c being the wideness of the displacements; - for the case of rigid motion of rotation having width f (Fig.2b) f  

(20a)



u

 

x           y n u n

u

f y

g

f n

x

x

y

(21b,c,d,e)

u f x 

g

f n

y

y

x

y

y

x u c1 =

f= f

y u c2 =

x

x

a) b) Figure 2 : Rigid motions: a) translation, b) rotation.

Observations about the load vector as a consequence of the rigid motion 1) If the solid is subject to a rigid motion of translation (Fig.2a) having assigned values of the displacements c1 and 2 c , one has  U 0 and  G 0 but the total load vector has to be u g    L L L 0 with

      

   

*     

)   A C a

(

    0 0

ut

ut

, u tc



)     U

(

(22)

L

u

*

     

 A a

gt

, g tc

  

A

      0 0

ur



 

( ) G

(23)



L

 A C

g

gr

gr

and as a consequence the solution vector X has to be null. 2) If the solid is subject to a rigid motion of rotation (Fig.2a) having assigned rotation vector

f  φ , in the generic node

i one has

i i      G φ n φ s 0 i

i i    U r φ 0

(24a,b)

with s being respectively the unit normal and tangent vectors to the boundary elements on which the node i lies, but the total load vector has to be i r vector distance between the instantaneous center of rotation and the node i , i n e i

u g    L L L 0 with

67