Issue 51

G. Cocchetti et alii, Frattura ed Integrità Strutturale, 51 (2020) 356-375; DOI: 10.3221/IGF-ESIS.51.26

which provides transition mark  =  rm

= h r and then fixes the horizontal thrust within the arch, h = h m (  ),

at h = h rm

under mixed sliding-rotational mode, as linearly decreasing with friction coefficient  :



h

2

r



/ 2  = = = r h 



=

21.5952 ), 

= =

( ) 

( )  = = h



0.395832 (

rm r h h

0.621772;

h

(11)



rm

rm

m

2

Notice that generally h  ( 

) is a non-linear function of  , at variable  . For the complete semi-circular arch (  =  /2;

cos  = 0, sin  = 1), it just becomes a linear function of  , namely h  =  /2  . On the other hand, in analytically seeking the stationary points marked in Fig. 4a, one then searches for the additional joint where further sliding may occur. Thereby, the corresponding stationary condition leads to:



t n tn  −





− −

  +



(1 )cos h

sin

d t

t

1

t

t

t

           

  

  





= =

= − =  = =

t

n

0

(12)

2



d n n 

(1 ) sin h −



+





n

n

n n

cos

n

Thus, since such sliding joint is activated when, there, t / n = –  , while h = h  , one has to solve the following system of three governing equations (independently from thickness parameter  ):



        

t

t

2

2

=

h h  − + = 

0



n n t = −



 +

(sin cos ) h   

−



  −

sin ) 0  =

(cos

(13)

n

  +

 

cos

sin

= =



 −

(sin cos ) h   

−



  +

sin ) 0  =

h h

(cos



sin cos   −

states the map of stationary points of t / n as  2 = h – h 2 = h (1– h ) or  2 + ( h – 1/2) 2 = (1/2) 2

Remark that Eqn. (13) a

(see the circle insert in Fig. 4a). This is also obtained either by eliminating couple (  , 

) from system (13), or variable 

from Eqns. (13) a and (13) b . Instead, by eliminating couple (  , h

) from system (13), and couple (  , h

) from system (13) or variable h

from Eqns. (13) a

and (13) b

, one also gets, respectively:

2

2

2

[2 sin 2(  +

)]    +

− − −

[1 4 cos 2( 

)]     + +

[2 sin 2(  −

 

+ =

)] 0

(14)

2

(2 sin 2 )  +

 

−

 

+ −

2 sin 2 0   =

2cos 2

(15)

the latter equation leading to following symbolic solution  =  ± (  ):

2

cos 2 1 4 2 sin 2     − +





 

=

( )

(16)

where higher root  + holds true for  ≥ cos1/(1+sin1) = 0.293408, with  ≤ 1/2. Thus, from Eqn. (14), with  =  /2, one may numerically solve for position  =  s

of the new sliding joint. Then, in

cascade, at that value of  =  s

, Eqn. (15), or Eqn. (16), leads to transition friction coefficient  ms

(higher root);

gives, at  =  s

finally, Eqn. (13) a

, thrust h ms

(lower root, corresponding to h ms

= h  ms

).

In doing so, or either by directly numerically solving system (13), one gets:

s 

=

=

28.6362 , 



=



=

17.1824 ), 

=

0.499796

rad

0.309215 (

h

0.485714

(17)

ms

ms

ms

364