Determine the outside diameter of a hollow steel tube that will carry a tensile load f 500kN at stress of 140 Mpa. Assume the wall thickness to be one-tenth of the outside diameter.

Given:

Tensile Force = 500kN

Stress = 140 Mpa

Find Outside diameter

*Condition: thickness is one-tenth of the outside diameter.*

Solution:

Formula:

Stress = Force / Area

Area = (pi/4)(D^2-d^2)

for hollow steel;

2t = (D-d)

where;

D = Outside diameter

d = inside diameter

t = thickness

if t = D/10

2(D/10) = (D-d)

D/5 = D-d

d= D - (D/5)

d = (4/5)D

find D;

140,000kpa = 500kN/[(pi/4)(D^2 – ((4/5)D)^2]

140,000 kpa = 500kN / (pi/4)(.36)( D^2)

*note: kpa = kN/m^2*

D^2 = 500kN / (pi/4)(.36)(140,000)

D = (0.01263134469m^2)^0.5

D = 0.1123892552 m

D = 112.39 mm (answer)