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Chapter 12: Problem 49

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

Short Answer

Expert verified
In summary, for a spinel material with 5% porosity, the modulus of elasticity (E) is given as 240 GPa. Using the formula relating porous and nonporous materials, we find that the modulus of elasticity for the nonporous material (E0) is 240 GPa divided by (1 - 0.05)^n. For a material with 15% porosity, the modulus of elasticity is found to be 204 GPa, accounting for the same material-specific constant (n) as the 5% porosity case.

Step by step solution

01

a) Compute the modulus of elasticity for the nonporous material.

First, let's solve the equation for E0: \(E_0 = \frac{E}{\left(1 - P\right)^n}\) Now, we are given E as 240 GPa, and the porosity, P, as 5 vol%, which is equal to 0.05 when expressed as a fraction. We don't have the value of n - first let's calculate a general formula for E0 that includes n: \(E_0 = \frac{240}{\left(1 - 0.05\right)^n}\) Since n is a material-specific constant, we can assume that it will be the same for both cases.
02

b) Compute the modulus of elasticity for 15 vol% porosity.

Now, we are going to use the same equation for E, but with the value of porosity, P, as 0.15: \(E = \left(1 - 0.15\right)^n E_0\) Substitute the value of E0 from part a: \(E = \left(1 - 0.15\right)^n \cdot \frac{240}{\left(1 - 0.05\right)^n}\) As previously discussed, n should be the same for both cases, and thus, cancels out: \(E = \frac{\left(1 - 0.15\right)^n}{\left(1 - 0.05\right)^n} \cdot 240\) Finally, calculate the value of the modulus of elasticity for 15% porosity: \(E = \frac{\left(1 - 0.15\right)}{\left(1 - 0.05\right)} \cdot 240 = 0.85 \cdot 240 = 204 \mathrm{GPa}\) So, the modulus of elasticity for the material with a porosity of 15% is 204 GPa.

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Key Concepts

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

Materials Science
Materials science is a multidisciplinary field that studies the relationship between the structure of materials at atomic or molecular scales and their macroscopic properties. Understanding the modulus of elasticity is fundamental in this domain. The modulus of elasticity, often referred to as Young's modulus, is a measure of the stiffness of a solid material. It defines the relationship between stress (force per unit area) and strain (proportional deformation) in a material. In the classroom scenario, considering spinel with different levels of porosity helps students explore how microscopic changes influence the overall mechanical characteristics.
Mechanical Properties
The mechanical properties of materials define how they react to physical forces. These properties include elasticity, ductility, hardness, and toughness, among others. The modulus of elasticity is a prime example of such a property, indicating how easily a material can deform elastically when a load is applied. It is crucial for engineers and scientists to understand these properties in order to select the right materials for the job. In materials with porosity, the mechanical strength is affected because the presence of voids can change how the material responds to stress.
Porosity Effects
Porosity refers to the volume fraction of void spaces within a material and is a critical factor in determining a material's mechanical properties. High porosity generally leads to a decrease in the modulus of elasticity, as seen in the exercise where the material's stiffness decreases with higher porosity. This is due to the fact that the voids provide less resistance to deformation under load, making the material 'softer'. When teaching students, it's essential to stress that while some porosity may be detrimental to certain properties, it can also be beneficial for applications requiring lightweight materials with some level of compressibility.
Engineering Calculations
Engineering calculations are methods of applying mathematical and physical laws to solve complex engineering problems. These include computing the modulus of elasticity for materials with various levels of porosity, as demonstrated in the solved problem. By understanding the theory behind these calculations, students can predict how materials will behave in different environments and under various loads. This predictive power is particularly useful when designing new products or systems. If students grasp these calculations well, they will be better equipped to approach material selection and structural design with confidence in their academic and future professional careers.

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Most popular questions from this chapter

The modulus of elasticity for titanium carbide (TiC) having 5 vol\% porosity is 310 GPa (45 \(\times\) \(\left.10^{6} \mathrm{psi}\right)\) (a) Compute the modulus of elasticity for the nonporous material. (b) At what volume percent porosity will the modulus of elasticity be 240 GPa ( \(35 \times 10^{6}\) psi)?

If cupric oxide (CuO) is exposed to reducing at- mospheres at elevated temperatures, some of the Cu2- ions will become Cu. (a) Under these conditions, name one crystalline defect that you would expect to form in order to maintain charge neutrality. (b) How many Cu- ions are required for the crea- tion of each defect? (c) How would you express the chemical formula for this nonstoichiometric material?

A hypothetical AX type of ceramic material is known to have a density of 2.10 g/cm3 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 g/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).

A three-point bending test is performed on a spinel \(\left(\mathrm{MgAl}_{2} \mathrm{O}_{4}\right)\) specimen having a rectangular cross section of height \(d=3.8 \mathrm{~mm}(0.15 \mathrm{in}\).) and width \(b=9 \mathrm{~mm}(0.35 \mathrm{in}\).); the distance between support points is \(25 \mathrm{~mm}(1.0 \mathrm{in}\).). (a) Compute the flexural strength if the load at fracture is \(350 \mathrm{~N}\left(80 \mathrm{lb}_{\mathrm{f}}\right)\). (b) The point of maximum deflection \(\Delta y\) occurs at the center of the specimen and is described by $$ \Delta y=\frac{F L^{3}}{48 E I} $$ where \(E\) is the modulus of elasticity and \(I\) is the cross-sectional moment of inertia. Compute \(\Delta y\) at a load of \(310 \mathrm{~N}\left(70 \mathrm{lb}_{\mathrm{}}\right)\).

Cite one reason why ceramic materials are, in general, harder yet more brittle than metals.
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