Find the Euler’s crippling load for the hollow cylindrical column of 50 mm external diameter and 5 mm thick. Both ends of column are hinged and length of column is 2.5 m. Take E=2x10^5 MPa. Also determine Rankine’s crippling load for the same column. Take fc= 350 MPa and α=1/7500.

For hollow cylindrical column:

do = 50 mm

t = 5 mm

$t = \frac{do \ - \ di}{2}$ $\rightarrow$ $di = do - 2t$

$di = 50 - 2 \times 5$

$di = 40 \ mm$

Both ends of column is hinged.

Length of column = 2.5 m.

$Es = 2 \times 10^5 \ Mpa$

$Fc = 350 \ Mpa$

$\alpha = 1/7500$

1] Moment of Inertia = $I = \frac{\pi}{64} (do^4 – di^4) = 181.132 \times 10^3 \ mm^4$

$A = \frac{\pi}{4} \times (d^2_o – d^2_i) = \frac{\pi}{4} \times (50^2 – 40^2) = 706.85 \ mm^2$

2] Effective length of column =

$Le = L$ $(\because$ Both ends are hinged).

$Le = 2.5 \ m = 2500 \ mm$

A] Crippling load by Euler’s formula:

$P = \frac{\pi^2 \ E \ I}{L^2e} = \frac{\pi^2 \times 2 \times 10^5 \times 181.32 \times 10^3}{2500^2} = 57265.81 \ N$

B] Crippling Load by Rankine’s formula:

$P = \frac{fc \times A}{1 + \alpha (\frac{Le}{k})^2} = \frac{350 \times 706.85}{1 + \frac{1} {7500} \times (\frac{2500^2}{\frac{181.32 \times 10^3}{706.85}})} = 58229.94 \ N$

$(\because \ k^2 = \frac{I}{A} = \frac{181.32 \times 10^3}{706.85})$