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Chapter 12: Problem 44

A three-point bending test was performed on an aluminum oxide specimen having a circular cross section of radius \(3.5 \mathrm{~mm}\) (0.14 in.); the specimen fractured at a load of \(950 \mathrm{~N}\left(215 \mathrm{lb}_{\mathrm{i}}\right)\) when the distance between the support points was \(50 \mathrm{~mm}\) (2.0 in.). Another test is to be performed on a specimen of this same material, but one that has a square cross section of \(12 \mathrm{~mm}\) ( \(0.47\) in.) length on each edge. At what load would you expect this specimen to fracture if the support point separation is \(40 \mathrm{~mm}\) (1.6 in.)?

Short Answer

Expert verified
Based on the given information and calculations, the expected load at which a specimen of the same material with a square cross-section of 12 mm length on each edge will fracture if the support point separation is 40 mm is approximately 11483.15 N.

Step by step solution

01

Calculate the Moment of Inertia (I) for the Circular Cross Section

For the circular cross section, the moment of inertia (I) can be calculated using the formula: $$ I = \frac{\pi r^4}{4} $$ where \(r\) is the radius of the circle. Here, the radius is given as \(3.5 \mathrm{~mm}\). Plugging this value into the formula, we get: $$ I = \frac{\pi (3.5 \mathrm{~mm})^4}{4} \approx 178.47 \mathrm{~mm^4} $$
02

Find the Bending Stress (f) on the Circular Specimen

The bending stress (f) is calculated using the following formula: $$ f = \frac{M*S}{I} $$ where \(M\) is the moment, \(S\) is the distance between support points, and \(I\) is the moment of inertia. In this case, \(M = P * L/4\), where \(P\) is the load and \(L\) (the distance between support points) is given as \(50 \mathrm{~mm}\) and the load, \(P\) is given as \(950 \mathrm{~N}\). Calculating the moment, M: $$ M = \frac{950 \mathrm{~N} * 50 \mathrm{~mm}}{4} = 11875 \mathrm{~N \cdot mm} $$ Now we can calculate the bending stress, f: $$ f = \frac{11875 \mathrm{~N \cdot mm} * 50 \mathrm{~mm}}{178.47 \mathrm{~mm^4}} \approx 3315.53 \mathrm{~N/mm^2} $$
03

Calculate the Moment of Inertia (I) for the Square Cross Section

For the square cross section, the moment of inertia (I) can be calculated using the formula: $$ I = \frac{a^4}{12} $$ where \(a\) is the length of one edge of the square cross section. Here, the edge length is given as \(12\mathrm{~mm}\). Plugging this value into the formula, we get: $$ I = \frac{(12\mathrm{~mm})^4}{12} = 20736 \mathrm{~mm^4} $$
04

Find the Expected Fracture Load (P) for the Square Specimen

We will use the bending stress from the circular specimen to find the fracture load for the square specimen. Rearranging the stress formula for load (P): $$ P = \frac{4*f*I}{L*S} $$ WithValue of \(L = 40\,\mathrm{mm}\), the fracture load for the square specimen can be calculated as: $$ P = \frac{4 * 3315.53\,\mathrm{N/mm^2} * 20736\,\mathrm{mm^4}}{40\,\mathrm{mm} * 12\,\mathrm{mm}} \approx 11483.15\,\mathrm{N} $$ The expected load at which the square specimen will fracture is approximately \(11483.15\,\mathrm{N}\).

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Most popular questions from this chapter

(a) A three-point transverse bending test is conducted on a cylindrical specimen of aluminum oxide having a reported flexural strength of \(390 \mathrm{MPa}(56,600 \mathrm{psi})\). If the specimen radius is \(2.5 \mathrm{~mm}\) (0.10 in.) and the support point separation distance is 30 \(\mathrm{mm}\) (1.2 in.), predict whether you would expect the specimen to fracture when a load of \(620 \mathrm{~N}\left(140 \mathrm{lb}_{\mathrm{f}}\right)\) is applied. Justify your prediction. (b) Would you be \(100 \%\) certain of the prediction in part (a)? Why or why not?

What point defects are possible for \(\mathrm{Al}_{2} \mathrm{O}_{3}\) as an impurity in \(\mathrm{MgO}\) ? How many \(\mathrm{Al}^{3+}\) ions must be added to form each of these defects?

Magnesium oxide has the rock salt crystal structure and a density of \(3.58 \mathrm{~g} / \mathrm{cm}^{3}\). (a) Determine the unit cell edge length. (b) How does this result compare with the edge length as determined from the radii in Table \(12.3\), assuming that the \(\mathrm{Mg}^{2+}\) and \(\mathrm{O}^{2-}\) ions just touch each other along the edges?

Show that the minimum cation-to-anion radius ratio for a coordination number of 6 is 0.414. [Hint : use the \(\mathrm{NaCl}\) crystal structure (Figure \(12.2\) ), and assume that anions and cations are just touching along cube edges and across face diagonals.]

In your own words, briefly define the term stoichiometric.
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