Determine the maximum permissible eccentricity of the load:
A small hollow cylindrical column carries a compressive force of 400 kN. The external diameter of the column is 200 mm & the internal diameter is 120 mm. Determine the maximum permissible eccentricity of the load, if the permissible stresses are 60 N/mm^{2} in compression & 25 N/mm^{2} in tension.
Solution
External diameter, D = 200 mm
Internal diameter, d = 120 mm
Area of the section,
A = (π/4) (D^{2} - d^{ 2} ) = (π/4) (200^{2} - 120^{2} ) = 2.01 × 10^{4} mm^{2}
Applied load, P = 400 kN = 4 × 10^{5} N
Direct stress, f0= P / A= 4 × 105 /2.01 × 104= 19.9 N/mm^{2 } (compressive)
Let the eccentricity of the load = e mm.
Bending moment, M = P × e = (400 × 10^{5} × e) N-mm
Section modulus,
z = (π /64) (D4 - d 4 ) × ( 2/d)
= (π/32) ((200^{ 4} - 120^{4} )/200) = 68.36 × 10^{4} mm
Bending stress, f _{b } =± M/ Z
=± (4× 10^{5} × e) /(68.36 × 10 ^{4})
= ± 0.585 × e
Resultant stress at extreme fibres = f_{0} ± f_{b} = 19.90 ± 0.585 e
∴ Maximum compressive stress = (19.90 + 0.585 e) ----------- (a)
Minimum compressive stress = (19.90 - 0.585 e)
or Maximum tensile stress = (0.585 e - 19.90) ------- (b)
Therefore, (19.90 + 0.585 e) ≤ 60 N/mm^{2} (allowable compressive stress)
∴ e ≤ 68.55 mm ---------------(c)
(0.585 e - 19.90) ≤ 25 N/mm^{2} (allowable tensile stress)
∴ e ≤ 8.72 mm ---------- (d)
From these two conditions, the allowable maximum eccentricity = 8.72 mm from the centre of the section.