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)