The modulus of elasticity for spinel \(\left(\mathrm{MgAl}_{2} \mathrm{O}_{4}\right)\) having 5 vol\% porosity is \(240 \mathrm{GPa}\) \(\left(35 \times 10^{6} \mathrm{psi}\right)\) (a) Compute the modulus of elasticity for the nonporous material. (b) Compute the modulus of elasticity for 15 vol\% porosity.

Short Answer

Expert verified
Answer: The modulus of elasticity for the nonporous material is approximately 266.67 GPa and for the material with 15% porosity is approximately 191.99 GPa.

Step by step solution

01

Find the modulus of elasticity for the nonporous material (E_0).

We are given E (modulus of elasticity for 5% porosity) as 240 GPa. Using the equation \(E = E_0 (1 - p)^n\), we can write: $$240\,\text{GPa}=E_0(1-0.05)^n$$ To find E_0, we first need to determine the value of n. We can use the empirical expression for n, which states that \(n=2\) for low porosities (generally less than 15%). Given that the problem is dealing with low porosities (5% and 15%), we can assume that \(n=2\). Now we can solve for E_0: $$240\,GPa = E_0(0.95)^2$$ $$E_0 = \frac{240\,GPa}{(0.95)^2}$$ $$E_0 \approx 266.67\,GPa$$ So the modulus of elasticity for the nonporous material is approximately 266.67 GPa.
02

Compute the modulus of elasticity for 15 vol\% porosity.

Now that we have found the modulus of elasticity for the nonporous material (E_0), we can use the same equation with p=0.15 to find the modulus of elasticity with 15 vol\% porosity: $$E = 266.67\,GPa (1 - 0.15)^2$$ $$E \approx 266.67\,GPa(0.85)^2$$ $$E \approx 191.986\,GPa$$ The modulus of elasticity for a material with 15 vol\% porosity is approximately 191.99 GPa.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Material Porosity
Material porosity refers to the presence of tiny, often microscopic, holes or voids within a solid material. These voids can be connected, forming channels, or isolated, affecting the material's overall density and properties. In materials science, understanding porosity is essential because it can significantly impact physical properties such as strength, durability, and elasticity.

For example, in the exercise where we calculate the modulus of elasticity for spinel that contains 5% porosity, the presence of voids decreases the material's ability to withstand deformation under stress. This concept is particularly important to grasp as it influences how materials behave under load and the design of structures and components in engineering fields.
Spinel Material Properties
Spinel, with the formula \(\mathrm{MgAl}_{2}\mathrm{O}_{4}\), is a mineral with a cubic crystal structure. It is prized for its optical properties and hardness, making it a preferred material for various applications, including abrasives and jewelry. The physical properties of spinel include a high melting point and good chemical resistance. Nevertheless, its mechanical properties, such as modulus of elasticity, are influenced by its porosity.

When discussing spinel in an educational context, emphasizing the relationship between its structure and properties helps students understand why and how materials are chosen for specific applications. For instance, non-porous spinel exhibits higher modulus of elasticity, which translates to greater resistance to deformation, a desirable trait in applications where rigidity and structural integrity are paramount.
Empirical Expressions in Materials Science
Empirical expressions in materials science are formulas derived from experimental data that predict how certain properties of materials change under various conditions. These expressions are crucial tools for engineers and scientists, allowing them to anticipate the behavior of materials without performing exhaustive testing for every scenario. In the exercise, we used an empirical formula to calculate the modulus of elasticity for spinel with different porosity levels.

The formula \(E = E_0 (1 - p)^n\) is one such expression where \(E_0\) is the modulus of elasticity for the non-porous material, \(p\) is the porosity volume percentage, and \(n\) is an empirical constant derived from experimental observation. The value of \(n\) generally equals 2 for materials with low porosity, explaining its usage in the step-by-step solution. Understanding these expressions helps students in predicting the performance of materials in practical applications accurately.

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Most popular questions from this chapter

A hypothetical AX type of ceramic material is known to have a density of \(2.10 \mathrm{g} / \mathrm{cm}^{3}\) and a unit cell of cubic symmetry with a cell edge length of 0.57 nm. The atomic weights of the A and \(X\) elements are 28.5 and \(30.0 \mathrm{g} / \mathrm{mol}\) respectively. On the basis of this information, which of the following crystal structures is (are) possible for this material: sodium chloride, cesium chloride, or zinc blende? Justify your choice(s).

For a ceramic compound, what are the two characteristics of the component ions that determine the crystal structure?

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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 \(40 \mathrm{mm}(1.6 \mathrm{in.}) ?\)

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