Determine the maximum permissible eccentricity of the load, Mechanical Engineering

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  ± fb  = 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.

Posted Date: 1/21/2013 7:04:25 AM | Location : United States







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