What is the magnitude of the maximum stress that exists at the tip of an internal crack having a radius of curvature of 3.5 × 10-4 mm (1.378 × 10-5 in.) and a crack length of 4.5 × 10-2 mm (1.772 × 10-3 in.) when a tensile stress of 170 mpa (24660 psi) is applied?

Given:

applied tensile stress, = 170 MPa

radius of curvature of crack tip,   =   mm

crack length = mm

half of internal crack length, a =

a =  

Formula Used:

Solution:

Using the given formula:

= 2726 MPa (395372.9 psi)

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft