A hollow cylindrical column, with both ends hinged, is 6m long and has an outer diameter of 120mm and an inner diameter of 80mm. Compare the crippling loads obtained by Euler’s and Rankine’s approach E=80,000 MPa and crushing strength =550 MPa. The Rankine’s constant =1/1600. What is the length of the column if both crippling loads are equal?

Data: Both ends hinged.

L = 6m = 600mm. i.e. Le = L

Do = 120 mm, di = 80 mm.

E = 80,000 Mpa.

Crushing strength Fe = 550 Mpa.

Rankine’s constant = $\frac{1}{1600}$

a] P by Euler's formula.

b] P by Rankine's formula.

c] {L} Length of column when both crippling loads are equal.

[1] Area, moment of Inertia, Radius of Gyration (k) :

$A= \frac{\pi}{4} \times (120^2 - 80^2) = 6283.18 \ mm^2$

$I = \frac{\pi}{64} \times (120^4 - 80^4) = \ 8.168 \times 10^6 \ mm^4$

$K^2 = \frac{I}{A} = \frac{8.168 \times 10^6}{6283.18}$

$\therefore \ k^2 = 1300$

[A] [P] Crippling load by Euler’s Formula:

$P_e = \frac{\pi^2 \ E \ I}{L^2e} = \frac{\pi^2 \times 80 \times 10^3 \times 8.168 \times 10^6}{6000^2}$

$P_e = 179.144 \times 10^3 \ N$

[B] P crippling load by Rankine’s FORMULA:

$P_r = \frac{f6 \times A}{1 + \alpha (\frac{Le}{k})^2} = \frac{550 \times 6283.18}{1 + \frac{1}{1600} \times (\frac{6000^2}{1300})}$

$P_r = 188.76 \times 10^3 \ N$

[C] Length of column when both crippling loads are equal.

$\therefore \ P_e = P_r$

$\therefore$ $\frac{\pi^2 \ E \ I}{L^2e} = \frac{fc \times A}{1 + \alpha (\frac{Le^2}{k^2})}$

$\therefore$ $\frac{\pi^2 \times 80 \times 10^3 \times 8.168 \times 10^6}{L^2e} = \frac{550 \times 6283.18}{1 + \frac{1}{1600} \times (\frac{Le}{1300})}$

i.e. $\frac{L^2e}{1 + \frac{L^2e}{(1600\times 1300)}} = \frac{\pi^2 \times 80 \times 10^3 \times 8.168 \times 10^6}{550 \times 6283.18} = 18.66 \times 10^6$

i.e. $L^2e = 18.66 \times 10^6 + \frac{18.66 \times 10^6}{1600 \times 1300} \times L^2e$

i.e. $L^2e = 18.66 \times 10^6 + 8.97 \ L^2e $

i.e. $7.97 \ L2^e = 18.66 \times 10^6$

i.e. $Le^2 = 2.34 \times 10^6$

$\therefore$ $Le = \pm \ 1529.70$

$i.e. Le = 1529.70 \ mm$

Now, Both ends of column are hinged,

$\therefore$ Le = L

$\therefore$ Length of column L is 1529.70 mm.