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This post was last edited by kareale88 on 2019-3-22 14:28 bg4.png [Discussion Post 28] Can the rounding value part of the nominal thickness be ≥0? Nominal thickness = calculated thickness + corrosion allowance + negative thickness deviation + rounded value Design thickness = calculated thickness + corrosion allowance Effective thickness = nominal thickness - corrosion allowance - negative thickness deviation Effective thickness = calculated thickness + rounded value In the question: designThickness() effective thickness. A. Less than ; B. Less than or equal to ; C. Greater than ; D. Greater than or equal to ; Is answer B reasonable? Calculate thickness (≤) effective thickness. More appropriate! If the calculated thickness is 5.7mm, the negative deviation is 0.3mm, and the corrosion allowance is 0mm. It can be equal at this time, do you agree?
Rounding values are usually rounded up to two decimal places.
This post was last edited by yechao25038 on 2019-3-22 13:24 Is there such a special situation, the calculated thickness is an integer, or it is an integer after adding the negative deviation and corrosion allowance, so can the rounded value be equal to 0? Yesterday’s daily question, right? I was a little unsure when I answered the question: lol
The difference is that it is rounded to market specifications, which is generally an integer, and even numbers increase within a certain range.
If the calculated thickness is 5.7mm, the negative deviation is 0.3mm, and the corrosion allowance is 0mm. This can be equal. Is it this?
The root cause is the difference between the thickness increased by rounding and the corrosion allowance. Everything is wrong.
I agree with you, the relationship between design thickness and effective thickness is uncertain.
This analysis makes sense. Design thickness = calculated thickness + corrosion allowance. Effective thickness = calculated thickness + rounding value.
CalculateThickness() effective thickness. A. Less than ; B. Less than or equal to ; C. Greater than ; D. Greater than or equal to ; This seems to be fine