Numerical - Safe load on long column, given modulus of elasticity & FOS

Q.  What is the safe load acting on a long column of 2 m having diameter of 40 mm. The column is fixed at both the ends and modulus of elasticity is 2 x 105 N/mm2? (F.O.S = 2)
- Published on 23 Sep 15

a. 120 kN
b. 124 kN
c. 130 kN
d. 150 kN

ANSWER: 124 kN

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    Discussion

  • Umayal   -Posted on 05 Feb 20
    Thank you...very helpful...
  • Umayal   -Posted on 05 Feb 20
    Thank you...very helpful...
  • Sravanthi   -Posted on 25 Nov 15
    Given: length of column = 2 m = 2000 mm, diameter of column = 40 mm, modulus of elasticity = 2 x 105 N/mm2 ,F.O.S = 2

    Formula: 1) Safe load = Crippling load / FOS

    2) Crippling load = π2 EI / Le2

    here, Le is the effective length, E is the modulus of rigidity, I is the moment of inertia

    3) Moment of inertia (I) = (π / 64) d4


    Solution:

    1) Moment of inertia (I) = (π / 64) d4

    = (π / 64) (40)4 = 125663.7 mm4

    Column is fixed at both the ends, therefore effective length = length of column / 2

    = 2000 / 2 = 1000 mm

    2) Crippling load = π2 EI / Le2

    = π2 (2 x 105) (125663.7) / (1000)2

    = 248050.20 N = 248.05 kN

    3) Safe load = Crippling load / FOS

    = 248.05 / 2

    = 124. 025 kN

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