Question

A stainless steel tube with an outside diameter of 54 mm and a wall thickness of 5 mm is used as a compression member. If the axial normal stress in the member must be limited to 190 MPa, determine the maximum load P that the member can support.

Answer 1

Answer:146.26 kN

Explanation:

Given

Outside diameter of tube $d_o=54\ mm$

thickness of tube $t=5\ mm$

therefore inner diameter $d_i=54-2\times 5$

$d_i=44\ mm$

Cross-section of tube $A=\frac{\pi }{4}(D_o^2-d_i^2)$

$A=\frac{\pi }{4}\times 980$

$A=245\pi mm^2$

Stress developed must be less than the limited value

thus $\frac{P}{A}\leq \sigma$

$P\leq \sigma A$

Maximum value of $P=\sigma \times A$

$P=190\times 245\times \pi$

$P=146.26\ kN$