Radius of Specimen from Flexural Strength

[1question]

Homework Statement

A circular specimen of glass is loaded using the three-point bending method. Compute the minimum possible radius of the specimen without fracture, given that the applied load is 4000 N the flexural strength is 65 MPa, and the separation between load points is 35mm.

Homework Equations

Flexural Strength = (FL)/(pi*R³) where F is the load at fracture, L is the distance between support points (in mm), and R is the radius of the specimen.

The Attempt at a Solution

Flexural Strength = (FL)/(pi*R³)

so

R = cube root of ((4000N)(0.035m)/(pi*6.5*10^7))
R = 8.818 mm

This is not one of the answers given. What am I doing wrong? The only thing I can think of is that the load given is NOT the fracture load, however I don't know how/from where to get that load.

 

[KataKoniK]

your first step should be to double check your equations and make sure they are correct. In this case, your equation for flexural strength is incorrect. The correct equation is:

Flexural Strength = (3FL)/(2pi*R3)

You also need to convert the load from N to MPa, as the flexural strength is given in MPa. So the correct equation becomes:

Flexural Strength = (3FL)/(2pi*R3)

F = 4000N = 4N

Flexural Strength = (3*4N*0.035m)/(2pi*R3)

= 0.0857 MPa

Solving for R:

R = cube root of ((3*4000N*0.035m)/(2pi*0.0857MPa))

R = 10.24 mm

This is the minimum possible radius of the specimen without fracture.