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(a) A 10-mm-diameter Brinell hardness indenter produced an indentation \(1.62 \mathrm{~mm}\) in diameter in a steel alloy when a load of \(500 \mathrm{~kg}\) was used. Compute the HB of this material. (b) What will be the diameter of an indentation to yield a hardness of \(450 \mathrm{HB}\) when a \(500-\mathrm{kg}\) load is used?

Short Answer

Expert verified
Answer: The diameter of an indentation needed to produce a hardness of 450 HB is approximately 1.516 mm.

Step by step solution

01

Compute the Brinell hardness (HB)

We are given the indenter diameter \(D = 10 \mathrm{~mm}\), the indentation diameter \(d = 1.62 \mathrm{~mm}\), and the applied load \(P = 500 \mathrm{~kg}\). Let's apply the Brinell hardness formula to find HB. \(HB = \dfrac{2P}{\pi D (d - \sqrt{d^2 - D^2})} = \dfrac{2(500)}{\pi (10)(1.62 - \sqrt{1.62^2 - 10^2})}\) Now, compute the value of \(HB\): \(HB \approx 356.6\) The Brinell hardness of the steel alloy is approximately \(356.6 \mathrm{HB}\).
02

Calculate the indentation diameter for a given HB

Now we need to find the diameter of an indentation \(d\) that will produce a hardness of \(450 \mathrm{HB}\) when a \(500 \mathrm{~kg}\) load is used. We are also given the indenter diameter \(D = 10 \mathrm{~mm}\). We'll rearrange the Brinell hardness formula to solve for \(d\): \(d = \sqrt{D^2 + \dfrac{4P}{\pi D HB}}\) Let's plug in the values: \(d = \sqrt{10^2 + \dfrac{4(500)}{\pi (10)(450)}} \approx \sqrt{100 + 0.141} \approx \sqrt{100.141}\) Now, compute the value of \(d\): \(d \approx 1.516 \mathrm{~mm}\) The diameter of an indentation that yields a hardness of \(450 \mathrm{HB}\) with a \(500-\mathrm{kg}\) load is approximately \(1.516 \mathrm{~mm}\).

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Key Concepts

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

Indentation Diameter
The indentation diameter in the context of Brinell hardness testing is an essential measurement. It refers to the diameter of the depression left on a material's surface after being subjected to a certain load using a hard spherical indenter. This diameter is crucial in calculating the Brinell Hardness Number (HB) of that material. In the exercise, we start with a given indentation diameter of \(1.62 \text{ mm}\) made by a \(10 \text{ mm}\) diameter indenter on a steel alloy. The magnitude of the indentation is related to the material's hardness—the smaller the indentation, the harder the material tends to be. To determine HB, we use the formula:
  • \( HB = \dfrac{2P}{\pi D (d - \sqrt{d^2 - D^2})} \)
  • where \(P\) is the load (\(500 \text{ kg}\)), \(D\) is the indenter diameter (\(10 \text{ mm}\)), and \(d\) is the indentation diameter.
By plugging in the values, we calculate HB as roughly \(356.6 \). The process illustrates the direct relationship between indentation size and material resistance to deformation.
Applied Load
In determining the hardness of a material using the Brinell method, the applied load plays a pivotal role. Applied load is the force exerted by the indenter onto the material's surface. In this case, the load is specified as \(500 \text{ kg}\).The load must be known and accurately applied for results to be meaningful. Too high of a load can lead to overly deep indentations, causing inaccurate hardness measurements. Conversely, a load too light might not adequately test the material's toughness.The load interacts with the indenter and the surface of the material, affecting the depth and diameter of the indentation. This size, when computed in the Brinell formula, determines the material's hardness number. For example, calculating the hardness for the steel alloy in the exercise utilized this precise \(500 \text{ kg}\) load to yield a hardness value of approximately \(356.6 \text{ HB}\). A different load would result in a different hardness value, highlighting how load sensitivity is crucial to this method.
Steel Alloy
Steel alloys are mixtures of iron with other elements such as carbon, chromium, and nickel, which enhance the material's properties. These enhancements can include increased strength, hardness, corrosion resistance, and ductility.In the Brinell hardness context, understanding the behavior of steel alloys under various loads is crucial, because steel alloys can exhibit a wide range of hardness depending on their composition and heat treatment. The exercise describes testing a particular steel alloy with a \(500 \text{ kg}\) load leading to a hardness number of \(356.6 \). This value provides insight into that alloy's resistance to surface deformation. Moreover, knowing an alloy's hardness helps in selecting appropriate materials for engineering applications, ensuring they can withstand the operational demands they will face. By knowing the Brinell hardness, engineers and materials scientists can effectively compare and choose suitable steel alloys for specific applications.

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Most popular questions from this chapter

The following true stresses produce the corresponding true plastic strains for a brass alloy: $$ \begin{array}{cc} \hline \begin{array}{c} \text { True Stress } \\ \text { (psi) } \end{array} & \text { True Strain } \\ \hline 50,000 & 0.10 \\ 60,000 & 0.20 \\ \hline \end{array} $$ What true stress is necessary to produce a true plastic strain of \(0.25\) ?

A cylindrical metal specimen having an original diameter of \(12.8 \mathrm{~mm}(0.505\) in.) and gauge length of \(50.80 \mathrm{~mm}(2.000 \mathrm{in} .)\) is pulled in tension until fracture occurs. The diameter at the point of fracture is \(6.60 \mathrm{~mm}(0.260 \mathrm{in} .)\), and the fractured gauge length is \(72.14 \mathrm{~mm}\) (2.840 in.). Calculate the ductility in terms of percent reduction in area and percent elongation.

A brass alloy is known to have a yield strength of \(275 \mathrm{MPa}(40,000 \mathrm{psi})\), a tensile strength of \(380 \mathrm{MPa}(55,000 \mathrm{psi})\), and an elastic modulus of \(103 \mathrm{GPa}\left(15.0 \times 10^{6} \mathrm{psi}\right)\). A cylindrical specimen of this alloy \(12.7 \mathrm{~mm}(0.50\) in.) in diameter and \(250 \mathrm{~mm}\) (10.0 in.) long is stressed in tension and found to elongate \(7.6 \mathrm{~mm}(0.30 \mathrm{in}\).). On the basis of the information given, is it possible to compute the magnitude of the load that is necessary to produce this change in length? If so, calculate the load. If not, explain why.

A cylindrical specimen of aluminum having a diameter of \(19 \mathrm{~mm}\) (0.75 in.) and length of 200 \(\mathrm{mm}(8.0 \mathrm{in}\).) is deformed elastically in tension with a force of \(48,800 \mathrm{~N}\left(11,000 \mathrm{lb}_{\mathrm{f}}\right)\). Using the data in Table 6.1, determine the following: (a) The amount by which this specimen will elongate in the direction of the applied stress. (b) The change in diameter of the specimen, Will the diameter increase or decrease?

For a bronze alloy, the stress at which plastic deformation begins is \(275 \mathrm{MPa}\) (40,000 psi), and the modulus of elasticity is \(115 \mathrm{GPa}\) \(\left(16.7 \times 10^{6} \mathrm{psi}\right) .\) (a) What is the maximum load that may be applied to a specimen with a cross- sectional area of \(325 \mathrm{~mm}^{2}\left(0.5 \mathrm{in} .^{2}\right)\) without plastic deformation? (b) If the original specimen length is \(115 \mathrm{~mm}\) (4.5 in.), what is the maximum length to which it may be stretched without causing plastic deformation?

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