The safe stress, for a Hollow steel column which carries an axial load of 2.1 x 10^3 kN is 125MN/sq(m). If the external diameter of the column is 30cm, determine the internal diameter?

Detailed Solution:

given;- force = 2.1 x 10^3 kN = 2.1 x 10^6 N,,,,stress = 125 x 10^6 N/sq(m) =125 N/Sq(mm),,, now,, stress = force/Area,,, hence Area = Force/Stress = 2100000/125 =16800 Sq(mm),,, Also Area of Hollow cylinder = 0.785 x (external diameter^2 -internal diameter^2)= 0.785(30x 30 - Internal diameter^2),,,,hence by comparing Area = 16800= 0.785(300x300 - internal diameter^2),, hence,,, 90000- internal diameter^2 = 16800/0.785 =21401.274,,hence Internal diameter^2 = 90000-21401.274=68598.726 ,,hence internal diameter=(68598.726)^0.5 =261.91 mm = 26.2 cm