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Chapter 6: Problem 51

(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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Most popular questions from this chapter

For some metal alloy, a true stress of \(415 \mathrm{MPa}\) (60,175 psi) produces a plastic true strain of \(0.475\). How much will a specimen of this material elongate when a true stress of \(325 \mathrm{MPa}\) \((46,125 \mathrm{psi})\) is applied if the original length is \(300 \mathrm{~mm}\) (11.8 in.)? Assume a value of \(0.25\) for the strain-hardening exponent \(n\).

A cylindrical specimen of a hypothetical metal alloy is stressed in compression. If its original and final diameters are \(20.000\) and \(20.025 \mathrm{~mm}\), respectively, and its final length is \(74.96 \mathrm{~mm}\), compute its original length if the deformation is totally elastic. The elastic and shear moduli for this alloy are \(105 \mathrm{GPa}\) and \(39.7 \mathrm{GPa}\), respectively.

Demonstrate that Equation \(6.16\), the expression defining true strain, may also be represented by $$ \epsilon_{T}=\ln \left(\frac{A_{0}}{A_{i}}\right) $$ when specimen volume remains constant during deformation. Which of these two expressions is more valid during necking? Why?

Estimate the Brinell and Rockwell hardnesses for the following: (a) The naval brass for which the stressstrain behavior is shown in Figure \(6.12\). (b The steel alloy for which the stress-strain behavior is shown in Figure \(6.21\).

A bar of a steel alloy that exhibits the stress-strain behavior shown in Figure \(6.21\) is subjected to a tensile load; the specimen is 300 mm (12 in.) long and has a square cross section \(4.5 \mathrm{~mm}(0.175 \mathrm{in}\).) on a side. (a) Compute the magnitude of the load necessary to produce an elongation of \(0.45 \mathrm{~mm}\) \((0.018\) in.). (b) What will be the deformation after the load has been released?
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