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 A thin-walled tube with external radius of 100 mm and wall thickness of 2 mm, is fixed at one end. It is subjected to a compressive force of 1 N acting at a point on the circumference parallel to its length. The maximum normal stress (in kPa) experienced by the structure is _____________ (accurate to two decimal places). 
Correct answer is '‐2.50 to ‐2.20'. Can you explain this answer?
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A thin-walled tube with external radius of 100 mm and wall thickness o...
Given data:
External radius of the tube, rext = 100 mm = 0.1 m
Wall thickness, t = 2 mm = 0.002 m
Force applied, F = 1 N

Formula:
The maximum normal stress, σmax, can be calculated using the formula:

σmax = F / A

Where,
A = π[(rext)² - (rint)²]
rint = rext - t

Calculation:
Given that the external radius of the tube is 0.1 m and the wall thickness is 0.002 m. Therefore, the internal radius of the tube is:

rint = rext - t
= 0.1 - 0.002
= 0.098 m

The cross-sectional area of the tube is:

A = π[(rext)² - (rint)²]
= π[(0.1)² - (0.098)²]
= π[0.01 - 0.009604]
= π(0.000396)
= 0.001243 m²

Now, substituting the values into the formula for maximum normal stress:

σmax = F / A
= 1 / 0.001243
= 803.7 Pa

Converting from Pascal to kilopascal:

σmax = 803.7 / 1000
= 0.8037 kPa

Therefore, the maximum normal stress experienced by the structure is 0.8037 kPa.

Answer:
The maximum normal stress (in kPa) experienced by the structure is 0.80 kPa (approx).
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A thin-walled tube with external radius of 100 mm and wall thickness of 2 mm, is fixed at one end. It is subjected to a compressive force of 1 N acting at a point on the circumference parallel to its length. The maximum normal stress (in kPa) experienced by the structure is _____________ (accurate to two decimal places).Correct answer is '‐2.50 to ‐2.20'. Can you explain this answer?
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