The allowable tensile stress for a V2 belt of thread length 7} 32 inches is 30000 psi. The allowable shear
stress of the material is 15000 psi. Where and how such a bolt most likely to fail, will it fail in the root diameter
or the in the thread.

Failure in thread occurs first due to shear in thread because maximum shear strength (6135.92 lb) is less than maximum tensile strength (8348.19 lb) for given conditions. 

Step-by-step explanation

AAAAAAMAAMAAK 9- din AAAASMMAMA L = 2 in - 32 Given Nominal diameter of bolt (4) - 2 in-- 0.5 ;3 80 noot diameter ( da) = = 0-595 17 0. 8 4 thread length ( L ) - 2 in permissible / allowable tensile stress ( oy+ ) - 30 0 0 0 PS, Allowable shear stress ( Is )allow = 15000 PS , A Now maximum strength in tension ( F )A = (By+ )allow x I (A) 2 = 30060 7 5 x/0 595 ) 1( F ) + = 8348. 19 db) maximum strength in shear ( F )s = ( C )allow x T ( dor ) XL 150 00 x x 0 595 x 2 to)s - 6/ 35 - 92 lb Answer. As F)s< (F)+ $80 failure in thread Toccurs first, due to shear in thread.

Please see attached file