Issue 48

A. Fesenko et alii, Frattura ed Integrità Strutturale, 48 (2019) 768-792; DOI: 10.3221/IGF-ESIS.48.70 789 We note that when  0 m ,  0 n or  0 n ,  0 m the formula (A8) is not contradictory, it is also possible  0 n (  0 m ). To calculate with this formula, an essential condition is the requirement  m n , i.e     2 1 k k A B . If  / 1 B A , then, to be definite, it is necessary that the condition      2 1 0 , or based on the replacement    sin :    cos sin , i.е    / 4 or    1/ 2 . Let’s consider the case  . A B Taking into account formula (A9), we get                        2 2 2 2 1 1 ( / ) 1 (0, 0, ) ln 2 1 1 ( / ) N k k k k k k k B A C T h N B A . (A10) If N – is an even number, then proceeding from the properties of the logarithm, the expression for the temperature (A10) takes the form                        2 2 2 2 2 1 1 ( / ) 1 (0, 0, ) ln 1 1 ( / ) N k k k k k k k B A C T h N B A . Here we use the property of zeros of the Chebyshev polynomial:    1 N ,     1 2 N ,…, i.е. there are exactly N roots, arranged symmetrically in the gap  ( 1,1) . It can be noted that when N – is odd, there will be a root    ( 1)/2 0 N , which leads to a singularity in formula (A10). For this case we consider the passage to the limit   0          , 0, ( 1) / 2 k k k k N . Using the Lopital rule, we have                                     2 2 2 2 1 ( 1)/2 1 ( / ) 1 1 (0, 0, ) 2 ln 2 1 1 ( / ) N k k k k k k k k N B A C B T h N A B A . Let us consider three cases of local temperature distribution over sections of different sizes: 1)  / 1/ 2 B A ; 2)  / 2 B A ; 3)  / 1/ 4 B A . For the case when the number of nodes N in the formula (A10) was chosen equal to N = 70 аnd the constant of the temperature С = 1, for three cases, the temperature at the corner point of the layer was equal to: 1) (0, 0, ) T h = 0.999445, 2) (0, 0, ) T h = 0.999444, 3) (0, 0, ) T h = 0.998545. The reliability of the results obtained was verified on the basis of formula (3.364(2), [41]). We can justify the result directly, considering the boundary condition (9) of the problem. Proceeding from the applied integral transformations, one can find the transformation of the given function              0 0 sin ( ) ( , )cos cos A A f y f x y xdx C xdx C ,                                  2 sin sin sin sin ( ) 2 . 2 i B i B B i y i y B e e A A A B f f y e dy C e dy C C i We substitute the resulting transform into formula (12), and find the value at the point (0, 0, ) h                      2 2 0 2 sin sin 4 (0, 0, ) . 2 2 C A B C T h d d C