Answer:146.26 kN
Explanation:
Given
Outside diameter of tube [tex]d_o=54\ mm[/tex]
thickness of tube [tex]t=5\ mm[/tex]
therefore inner diameter [tex]d_i=54-2\times 5[/tex]
[tex]d_i=44\ mm[/tex]
Cross-section of tube [tex]A=\frac{\pi }{4}(D_o^2-d_i^2)[/tex]
[tex]A=\frac{\pi }{4}\times 980[/tex]
[tex]A=245\pi mm^2[/tex]
Stress developed must be less than the limited value
thus [tex]\frac{P}{A}\leq \sigma [/tex]
[tex]P\leq \sigma A[/tex]
Maximum value of [tex]P=\sigma \times A[/tex]
[tex]P=190\times 245\times \pi [/tex]
[tex]P=146.26\ kN[/tex]
