Find Euler’s crippling load for a hollow cylindrical column of 80 mm external diameter and 10 mm thick. Both ends of the column are fixed and length of column is 5.5 m. Take Es=2x10^5 N/mm^2. Also determine Rankine’s crippling load for the same column. Take fc=350 MPa and Rankine’s constant α= 1/7500.

Data: L = 5.5 m = 5500 mm.

External dia. (do) = 80 mm

Thickness t = 10 mm

Both ends of the column are fixed.

$Es = 2 \times 10^5 \ N/mm^2$

Fc = 350 Mpa

Rankine constant $\alpha$ = 1/7500

To Find:

[1] Euler’s crippling load.

[2] Rankine’s crippling load.

1] Internal diameter (di) :

$t = \frac{do – di}{2} $

i.e. $di = do - 2t = 80 - 2 \times 10 = 60 \ mm$

$Moment \ of \ Inertia = I = \frac{\pi}{64} (do^4 – di^4) = 1.374 \times 10^6 \ mm^4$

2] Effective length of column:

Both ends of column are fixed.

$\therefore$ $Le = \frac{1}{2} \times L = \frac{5500}{2} = 2750 \ mm$

A] Crippling load by Euler’s formula:

$\therefore$ $P = \frac{\pi^2 \ E \ I}{L^2e} = \frac{\pi^2 \times 2 \times 10^5 \times 1.374 \times 10^6}{2750^2} = 358.63 \ KN$

$(\because 1KN = 10^3 \ N)$

B] Crippling load by Rankine’s formula.

$P = \frac{fc \times A}{1 + (\frac{fc}{\pi^2 \ E}) \times (\frac{Le}{k})^2} = \frac{f_c \ A}{1 + \alpha (\frac{Le}{k})^2}$

$Fc \times A \rightarrow \ Crushing \ Load.$

$\frac{Fc}{\pi^2 \ E} \rightarrow \ \alpha \ (Rankine’s \ Constant).$

$K^2 = \frac{I}{A} = \frac{1.374 \times 10^6}{\frac{\pi}{4} \times \ (do^2 – di^2)} = 624.79 \ (i.e. K = \sqrt{624.79}$

$\therefore$ $P = \frac{350 \times \frac{\pi}{4} \times (do^2 – di^2)}{1 + \frac{1}{7500} \times (\frac{2750}{\sqrt{624.79}})^2}$

$= \frac{350 \times \frac{\pi}{4} \times (80^2 – 60^2)}{1 + \frac{1}{7500} \times \frac{2750^2} {624.79}} = \frac{769690.2}{2.613}$

$P = 294.56 \times 10^3 \ N$