Answer:
The value of minimum possible radius of the specimen without fracture
[tex]r[/tex] = 4.23 mm
Explanation:
Given data
Applied load F = 500 N
Flexural strength = 105 M Pa
Separation between load points = 50 mm
The minimum possible radius of the specimen without fracture is given by
[tex]r_} = \sqrt[3]{\frac{(500) (50)}{(105) (3.14) } }[/tex]
[tex]r[/tex] = 4.23 mm
This is the value of minimum possible radius of the specimen without fracture.