Engineering, 24.03.2020 23:22 Lamr
(a) A 10.0-mm-diameter Brinell hardness indenter produced an indentation 2.3 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 270 HB when a 500-kg load is used
Answers: 3
![=2*1000hg/\pi (10mm)[10mm-\sqrt{(1000^2-(2.3mm)^2} ]](/tpl/images/0561/8846/50f16.png)
![d=\sqrt{D^2-[D-\frac{2P}{(HB)\pi D} } ]^2\\=\sqrt{(10mm)^2-[10mm-\frac{2*500}{450( \pi10mm)} } ]^2](/tpl/images/0561/8846/22bb5.png)
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(a) A 10.0-mm-diameter Brinell hardness indenter produced an indentation 2.3 mm in diameter in a ste...
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