Physics, 21.02.2020 22:45 annafellows
(a) A 10.0-mm-diameter Brinell hardness indenter produced an indentation 2.4 mm in diameter in a steel alloy when a load of 1000 kg was used. Compute the HB of this material. (b) What will be the diameter of an indentation to yield a hardness of 300 HB when a 500-kg load is used?
Answers: 3
![d=\sqrt{D^2-[D-\dfrac{2P}{(HB)\pi D}]^2}](/tpl/images/0519/7311/67641.png)
![d=\sqrt{10^2-[10-\dfrac{2\times500}{300\pi\times10}]^2}](/tpl/images/0519/7311/f98f0.png)
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(a) A 10.0-mm-diameter Brinell hardness indenter produced an indentation 2.4 mm in diameter in a ste...
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