Answer
$σ = 286.48 MPa$
Since the computed value is less than the given ($σ_{fs} = 300 MPa$), the fracture is not predicted.
Work Step by Step
Given:
A three-point transverse bending test is conducted on a cylindrical specimen of aluminum oxide having a reported flexural strength of 300 MPa (43,500 psi). The specimen radius is 5.0 mm (0.20 in.), and the support point separation distance is 15.0 mm (0.61 in.).
Required:
Possibility of fracture when a load of 7500 N is applied.
Solution:
Using Equation 12.7b:
$σ = \frac{F L}{πR^{3}} = \frac{(7500 N)( 15 \times 10^{-3} m)}{(π)(5.0 \times 10^{-3} m)^{3}} = 286.48 \times 10^{6} N/m^{2} = 286.48 MPa$
Since the computed value is less than the given ($σ_{fs} = 300 MPa$), the fracture is not predicted.
