#### Answer

$F_{f}= 8,800.70 N$

#### Work Step by Step

Given:
A three-point bending test was performed on an aluminum oxide specimen having a circular cross section of radius 5.0 mm (0.20 in.); the specimen fractured at a load of 3000 N (675 lbf) when the distance between the support points was 40 mm (1.6 in.). Another test is to be performed on a specimen of this same material, but one that has a square cross section of 15 mm (0.6 in.) length on each edge
Required:
Load to which the specimen is expected to fracture if the support point separation is maintained at 40 mm (1.6 in.)
Solution:
Using Equation 12.7b to compute the flexural strength:
$σ_{fs} = \frac{F_{f} L}{πR^{3}} = \frac{(3000 N) (40 \times 10^{-3} m)}{(π)(5.0 \times 10^{-3} m)^{3}} 305.58 \times 10^{6} N/m^{2} = 305.58 MPa$
Using Equation 12.7a and solving for $F_{f}$, considering that $b = d =15mm$:
$F_{f} = \frac{2σ_{fs}d^{3}}{3L} = \frac{(2)(305.58 \times 10^{6} N/m^{2})(12 \times 10^{-3} m)^{3}}{(3)(40 \times 10^{-3} m)} = 8,800.70 N$