A three-point bending test was performed on an aluminum oxide specimen having a circular cross section of radius \(5.0 \mathrm{mm}\) \((0.20 \text { in. }) ;\) the specimen fractured at a load of \(3000 \mathrm{N}\left(675 \mathrm{lb}_{\mathrm{f}}\right)\) when the distance between the support points was \(40 \mathrm{mm}\) (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 \(15 \mathrm{mm}\) ( 0.6 in.) length on each edge. At what load would you expect this specimen to fracture if the support point separation is maintained at \(40 \mathrm{mm}(1.6 \mathrm{in.}) ?\)

The stress formula in a three-point bend test is given by: $$\sigma=\frac{3FL}{2\pi r^3}$$ where \(\sigma\) is the stress, \(F\) is the applied load, \(L\) is the distance between the support points, and \(r\) is the radius of the cross-section.

For the given circular cross-section specimen, we know its radius \((r=5\,\text{mm})\), applied load \((F_1=3000\,\text{N})\) and the separation distance \((L=40\,\text{mm})\). We can now calculate the stress for this specimen: $$\sigma_1=\frac{3F_1L}{2\pi r_1^3} = \frac{3(3000\,\text{N})(40\,\text{mm})}{2\pi (5\,\text{mm})^3}$$ After calculating, we get: $$\sigma_1=36.91\,\text{MPa}$$

Since the material is the same, we can assume a similar stress for the square cross-section specimen. The formula for stress in a three-point bend test for a square cross-section is: $$\sigma_2 = \frac{12F_2L}{s^2w}$$ where \(s\) is the side length of the square cross-section and \(w\) is the distance along the width (which is also equal to \(s\))

For the square cross-section specimen, we know its side length \((s=15\,\text{mm})\), the applied load \((F_2)\), and the separation distance \((L=40\,\text{mm})\). We also know that the stress in the square cross-section specimen should be the same as in the circular cross-section specimen, so \(\sigma_2 = \sigma_1\). Therefore, we can substitute the values in the equation: $$36.91\,\text{MPa}=\frac{12F_2(40\,\text{mm})}{(15\,\text{mm})^2(15\,\text{mm})}$$ Now, solve for \(F_2\): $$F_2=\frac{36.91\,\text{MPa}(15\,\text{mm})^3(15\,\text{mm})}{12(40\,\text{mm})}$$ After calculating, we get: $$F_2=10470\,\text{N}$$ So, we would expect the specimen with a square cross-section of \(15\,\text{mm}\) (0.6 in.) length on each edge to fracture at a load of approximately \(10470\,\text{N}\).