Answer:
[tex]\sigma_x = 1705 MPa[/tex]
Explanation:
Given data:
Internal pressure = 136.4 MPa
thickness is given as 4 mm
diameter is not given in the question hence we are assuming the diameter of given cylinder to be 100 mm
circumferential stress is given as
[tex]\sigma_x = \frac{P\times r}{t} [/tex]
[tex]\sigma_x = \frac{136.4 \times \frac{100}{2}}{4}[/tex]
[tex]\sigma_x = 1705 MPa[/tex]
