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Chapter 12: Problem 45

(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?

Short Answer

Expert verified
a) Using the given flexural strength, specimen radius, support point separation distance, and applied load, calculate the stress experienced by the cylindrical specimen of aluminum oxide, and determine whether the specimen will fracture under these conditions. Justify your prediction. The calculated flexural stress (\(\sigma\)) is \(575\times10^6\,\text{Pa}\) and the given flexural strength (\(S\)) is \(390\times10^6\,\text{Pa}\). Since \(\sigma > S\), we predict that the specimen will fracture under the given load. This prediction is justified by comparing the calculated flexural stress to the flexural strength of the material, as the stress experienced under the given conditions is greater than the reported flexural strength. b) Discuss the certainty of your prediction. The certainty of our prediction is not 100% as it is based on average experimental results for flexural strength and idealized conditions. Individual specimens may vary due to factors such as manufacturing defects, non-uniform material properties, and slight differences in dimensions. Additionally, the applied load and support point separation distances could also vary slightly during testing, contributing to uncertainty in the actual stress experienced by the specimen.

Step by step solution

01

Calculate the Flexural Stress

To calculate the flexural stress, we can use the formula for the three-point bending test: \(\sigma = \dfrac{3PL}{2\pi R^3}\) Where: \(\sigma\) = flexural stress \(P\) = applied load \(L\) = support point separation distance \(R\) = specimen radius Given: Flexural strength (\(S\)) = \(390\,\text{MPa} = 390\times10^6\,\text{Pa}\) Load (\(P\)) = \(620\,\text{N}\) Separation distance (\(L\)) = \(30\,\text{mm} = 0.03\,\text{m}\) Radius (\(R\)) = \(2.5\,\text{mm} = 0.0025\,\text{m}\) Plugging in these values, we get: \(\sigma = \dfrac{3(620)(0.03)}{2\pi(0.0025)^3} = \dfrac{56.46}{9.82\times10^{-8}} = 575\times10^6\,\text{Pa}\)
02

Compare Flexural Stress to Flexural Strength

Now, we need to compare the calculated flexural stress (\(\sigma\)) to the given flexural strength (\(S\)). If the flexural stress is higher than the flexural strength, the specimen will fracture. \(\sigma = 575\times10^6\,\text{Pa}\) \(S = 390\times10^6\,\text{Pa}\) Since \(\sigma > S\), we can predict that the specimen will fracture under the given load.
03

Justify the Prediction (Part a)

We can justify our prediction by comparing the calculated flexural stress to the flexural strength of the material. Since we found that the flexural stress experienced under the given conditions is greater than the reported flexural strength, the specimen is expected to fracture when subjected to a load of \(620 N\).
04

Discuss the Certainty of the Prediction (Part b)

While our prediction is based on the comparison between the calculated flexural stress and the given flexural strength, we cannot be 100% certain of the prediction. This is because the reported flexural strength value is based on average experimental results, and individual specimens may vary due to factors such as manufacturing defects, non-uniform material properties, and slight differences in dimensions. Additionally, the applied load and support point separation distances could also vary slightly during testing, contributing to uncertainty in the actual stress experienced by the specimen. Therefore, we cannot be absolutely certain of our prediction, as it is based on averages and idealized conditions.

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Key Concepts

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

Three-Point Bending Test
Imagine you're trying to gently snap a pencil by pushing down on its middle while your hands support it at the ends. This is effectively what a three-point bending test does, but in a much more controlled way. It's a popular method to measure the flexural strength of materials by applying a concentrated load to a specimen supported at two points. Engineers use this test to determine how much a material can bend before it breaks.

The test setup includes a rigid support at a fixed distance (the span) with the test specimen placed horizontally between the supports. A load is then applied to the middle of the specimen until fracture occurs. The data collected from this point, like the maximum load the material withstands, is crucial for calculating stresses and predicting material behavior under bending. This kind of testing helps in understanding how a material would hold up in real-world applications where bending stresses are common, such as beams in construction or bone pins in medical devices.
Flexural Stress Calculation
Flexural stress is like a silent ticking time bomb inside materials – it's the level of tension or compression a material feels right before it bends or breaks. To calculate it, especially from a three-point bending test, there's a specific formula that must be used:

\( \text{Flexural stress} (\text{\(\sigma\)\})) = \frac{3PL}{2\pi R^3} \
\), where 'P' is the central load applied, 'L' refers to the support span, and 'R' is the radius of the specimen.

Using the formula, we can understand the stress distribution within the material. It helps predict the point at which a material will no longer bend but instead break. The magic of this formula is that it boils down complex material behaviors into a simple number, which is crucial for designers and engineers when choosing materials for applications that involve bending.
Material Fracture Prediction
Predicting when a material will fracture is like forecasting the weather – it involves a lot of data and a bit of educated guesswork. Scientists and engineers do this by comparing the calculated flexural stress to a material's known flexural strength. If our mathematical weather report – flexural stress – predicts stormier conditions than what the material's strength can handle, we'd expect a fracture. This is imperative for safety and design efficiency.

For example, if a beam in a structure is loaded beyond its flexural strength, it will likely fail, potentially causing a collapse. The ability to anticipate this through calculations is a powerful tool but requires a deep understanding of the material properties. Just like a weather prediction, it's not perfect because we base it on average values and the assumption that every testing condition is uniform.
Engineering Material Properties
Materials are as unique as fingerprints – each has a set of properties that define how it reacts to forces, heat, electricity, and other external factors. In engineering, understanding these material properties is akin to knowing the capabilities and limitations of your building blocks.

  • Flexural strength: How much a material can bend without breaking.
  • Elasticity: A material's ability to return to its original shape after bending.
  • Toughness: The total energy a material can absorb before fracturing.
  • Hardness: Resistance to deformation or indentation.

These properties are crucial when making choices for various applications, as they directly affect the performance and durability of the final product. Any design begins by considering these material traits to ensure that each component will behave as expected under expected service conditions.

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Most popular questions from this chapter

Briefly explain (a) why there may be significant scatter in the fracture strength for some given ceramic material, and (b) why fracture strength increases with decreasing specimen size.

The corundum crystal structure, found for \(\mathrm{Al}_{2} \mathrm{O}_{3}\), consists of an HCP arrangement of \(\mathrm{O}^{2-}\) ions; the \(\mathrm{Al}^{3+}\) ions occupy octahedral positions. (a) What fraction of the available octahedral positions are filled with \(\mathrm{Al}^{3+}\) ions? (b) Sketch two close-packed \(\mathrm{O}^{2-}\) planes stacked in an \(A B\) sequence, and note octahedral positions that will be filled with the \(\mathrm{Al}^{3+}\) ions.

A hypothetical AX type of ceramic material is known to have a density of \(2.65 \mathrm{~g} / \mathrm{cm}^{3}\) and a unit cell of cubic symmetry with a cell edge length of \(0.43 \mathrm{~nm}\). The atomic weights of the A and \(X\) elements are \(86.6\) and \(40.3 \mathrm{~g} / \mathrm{mol}\), respectively. On the basis of this information, which of the following crystal structures is (are) possible for this material: rock salt, cesium chloride, or zinc blende? Justify your choice(s).

Compute the atomic packing factor for the diamond cubic crystal structure (Figure 12.15). Assume that bonding atoms touch one another, that the angle between adjacent bonds is \(109.5^{\circ}\), and that each atom internal to the unit cell is positioned \(a / 4\) of the distance away from the two nearest cell faces ( \(a\) is the unit cell edge length).

The modulus of elasticity for beryllium oxide (BeO) having 5 vol\% porosity is 310 GPa \(\left(45 \times 10^{6} \mathrm{psi}\right)\) (a) Compute the modulus of elasticity for the nonporous material. (b) Compute the modulus of elasticity for 10 vol \% porosity.
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