Question

A steel rod 5m long and 4cm dia. is used as a column with one
end fixed and the other free. Determine the crippling load by
Euler’s formulae. Take E=2×106kg/cm2

Answer 1

Answer:

N = 2.48 kN

Explanation:

From the above question,

They have given a steel rod 5m long and 4cm dia. is used as a column with one end fixed and the other free.

Determine the crippling load by Euler’s formulae.

             Let's take E=2×106kg/cm2

Length (l) = 5

m = 5 × 103 mm

Diameter of column (d) = 40 mm and

modulus of elasticity (E) = 200 GPa = 200 × 103 N/mm2 .

We know that moment of iertia of the column section,

       l =  $\frac{\pi }{64}$ x $d^{4}$

         =  $\frac{\pi }{64}$ x $40^{4}$

         = 40,000$\pi$m$m^{4}$

Since the column is fixed at one end and free at the other, therefore equivalent length of the column

     $L_{e}$ = 2 l

          = 2 * ( 5 * $10^{3}$ )

          = 10 * $10^{3}$ mm

Euler’s crippling load,

      $P_{E}$  = $\frac{\pi ^{2}EI }{L_{e} ^{2} }$

             = $\frac{\pi ^{2}(200 * 10^{3} * (40000\pi )) }{((10 * 10)^{3}) ^{2} } }$

       $P_{E}$  = 2480 N

N = 2.48 kN

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