Hello,
I'm writing here because I don't know how to solve this question that appeared on an exam (undergraduate, mechanical engineering).
The Vickers hardness test was done on a steel plate twice, once with a smaller force and once with a larger force. The first average length of the diagonal is d=0.181 mm. The second indent (larger force was used this time) left the average length of the diagonal d=0.572 mm. What is the minimum distance between indentations?
The equation we are using is B>3*d , where B is the distance between indentations, d is the average length of the diagonal.
Do I just take the average value for d of 0.181 mm and 0.572 mm = 0.3765 mm? Then B>3*0.3765= 1.13 mm?
I would be grateful if someone could help me with this.
Hi @salveamigos123!
I think this reply is very late already, but I will post here for the sake of future mechanical engineering students encountering the same problem.
I found the above picture on a certain website. From this, the answer to your question is that $d$ is taken as the larger of the two diagonals, which is $d=0.572$ mm. Therefore, $$B>3d=3(0.572)=1.716\text{ mm}$$
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