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/mm² in compression & 25 N/mm² in tension.

Solution

External diameter, D = 200 mm

Internal diameter, d = 120 mm

Area of the section,

A = (π/4) (D²  - d ² ) = (π/4) (200²  - 120² ) = 2.01 × 10⁴  mm²

Applied load, P = 400 kN = 4 × 10⁵  N

Direct stress, f0= P /  A=        4 × 105 /2.01 × 104=  19.9 N/mm²  (compressive)

Let the eccentricity of the load = e mm.

Bending moment, M = P × e = (400 × 10⁵ × e) N-mm

Section modulus,

z =  (π /64) (D4  - d 4 ) × (  2/d)

=  (π/32) ((200 ⁴  - 120⁴ )/200) = 68.36 × 10⁴   mm

Bending stress, f _b   =± M/ Z

=± (4× 10⁵ × e) /(68.36 × 10 ⁴)

= ± 0.585 × e

Resultant stress at extreme fibres = f₀  ± 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/mm² (allowable compressive stress)

                              ∴ e ≤ 68.55 mm ---------------(c)

                 (0.585 e - 19.90) ≤ 25 N/mm² (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.