Question

Chapter 4, Reserve Problem 018 Three bolts are used in the connection shown in the figure. The thickness of plate (1) is 20 mm. The ultimate shear strength of the bolts is 385 MPa, and the ultimate bearing strength of plate (1) is 385 MPa. Determine the minimum bolt diameter required to support an applied load of P = 180 KN if a minimum factor of safety of 2.5 is required with respect to both bolt shear and plate bearing failure. ela Answer: dbolt min = mm

Answer 1

Given date, Thickness of the plete CH) - 20 mm Ultimate sheer staserath of (+5) - 385 mpa the bolts Ultimate beasing stsength of the (B) = 385 mpa plote is Lood (P) = 180 KN Foctor of sofety (f) = 2.5 No. of bolte (n) - 3 . We know that, allowable Pallowable - the = 385 = 154 mpa Due to duble sheor, the Shear force oching on the top surface is given by 2.5 Vs = 1 20 = 180 2x3 = 30 KN We bo know that, Tolowable = where As = I do I ds = 7 Talloweble

the values of " vs and Pollowable" on we Substituting get I d = 30x10²N 1 1 mpe - N/mm 154 N/mm² >) ds = 15.7491 mm – 15.75 mm - Booking Failure Callowable. = 385 = 25 =154 MPa Shex korice acting on a single boot . Vo = 2 = 180 = 60 KN Allowable shes stress Ponowable - en :-) Ab= Ub . Tarrowoble It do = Vb Terowoble above crustion on we substituting the values in the get ... 60x10²N (20) db = = 154 -

.-). do = 19:4806 mm 19.481 mm since, db>ds i diameter of bolt is fodb: 19:481 mmil