Question

Disk Disk 40 mm Çubuk F=20KN

The d-diameter rod and t-thick round disk in the figure are prepared in one piece.
The place where the rod meets the disk is 40 mm in diameter. When the axial tensile load of 20 kN is applied to the end of the rod, what should be the disc thickness (t) and rod diameter (d) to prevent damage to the disc and the rod? (Damage to the disc and rod begins at 35 MPa shear stress and 60 MPa normal stress, respectively)

Answer 1

disc Normal stress in rod = F A * all = 60MPa elevation 6oN mm² = 20x103 А Plan A = 333.33 mm² А I d ² - 333.33 → 40 mm ald=20.6mm Номер Fro shear stress, area will circumferential area of on which force f will att inner circle (ITA) Area of circumference = Now Tau=35 MPa = F Tdt 2 35 N 20X103 N T(40)t L mm 7 t= 4.547 mm