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Chapter 12: Problem 42

A circular specimen of \(\mathrm{MgO}\) is loaded using a three-point bending mode. Compute the minimum possible radius of the specimen without fracture, given that the applied load is \(5560 \mathrm{N}\left(1250 \mathrm{lb}_{\mathrm{f}}\right),\) the flexural strength is \(105 \mathrm{MPa}(15,000 \mathrm{psi}),\) and the separation between load points is \(45 \mathrm{mm}(1.75 \text { in. })\).

Short Answer

Expert verified
Answer: The minimum possible radius of the MgO specimen is approximately 11.4 mm.

Step by step solution

01

Write down the formula for flexural stress

The formula for flexural stress (\(σ\)) in a three-point bending test is given by: \(σ = \frac{3FL}{2π R^3}\) where \(F\) is the applied load, \(L\) is the separation between load points, and \(R\) is the radius of the specimen.
02

Plug the known values into the formula

We are given \(F = 5560\,\text{N}\), the flexural strength \(σ = 105\,\text{MPa} = 105 × 10^6\,\text{N/m}^2\), and the separation between load points \(L = 45\,\text{mm} = 0.045\,\text{m}\). Plugging these values into the formula for flexural stress, we get: \(105 × 10^6 = \frac{3(5560)(0.045)}{2π R^3}\)
03

Solve for the radius R

Now, we need to solve for the minimum radius \(R\) that would prevent the specimen from fracturing. Rearrange the formula to isolate \(R\): \(R^3 = \frac{3(5560)(0.045)}{2π (105 × 10^6)}\) Calculate the value of \(R^3\): \(R^3 = \frac{3(5560)(0.045)}{2π (105 × 10^6)} ≈ 1.4907 × 10^{-6}\) Take the cube root of both sides: \(R = (1.4907 × 10^{-6})^{\frac{1}{3}} ≈ 0.0114\,\text{m}\) Convert the radius to millimeters: \(R ≈ 0.0114 × 1000 = 11.4\,\text{mm}\)
04

State the result

The minimum possible radius of the MgO specimen to prevent it from fracturing under the given load and flexural strength is approximately 11.4 mm.

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