Athree-point bending test was performed on an aluminum oxide specimen having a circular cross section of radius 5.0mm (0.20 in.); the specimen fractured at a load of 3000 n (675 ibf) when the distance between the support points was 40mm (1.6 another test is to be performed on a specimen of this same material, but one that has a square cross sectionof 15 mm (0.6 in.) length on each edge. at what load would you expect this specimen to fracture ifthe support pointseparation is maintained at 40 mm (16 in)?

Load to fracture = 17212.5 N

Explanation:

Firstly we need to find the flexural strength of the material. We do this by using the date obtained by the test of the circular sample and the equation shown below

σ (flexural strength) = F x L/ (π x r³), where F is the load at fracture, L is the distance between the two load points, r is the radius of the circular cross section. Substituting the values we get

σ (flexural strength) = 3000 x 40 x 10⁻³/ (π x 5 x 10⁻³)³ = 306 x 10⁶ N/m²

Now, the flexural stress for a square sample is written as

σ (flexural strength) = 3 x F x L/ 2 x b x d² , and since in a square sample breadth = width the equation becomes

σ (flexural strength) = 3 x F x L/ 2 x d³

Solving for F, we get

F = σ (flexural strength) x 2 x d³ /3 x L

For same material we will use the σ (flexural strength) as calculated above, furthermore L remains the same and d = 15 x 10⁻³. Solving for F

F = 306 x 10⁶ x 2 x (15 x 10⁻³)³ /3 x 40 x 10⁻³ = 17212.5 N

The load required for this specimen to fracture would be 17212.5 N.

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