written 3.9 years ago by
teamques10
★ 57k

•
modified 3.9 years ago

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