Find the Euler crushing load for a hollow cylindrical, cast iron column of 200mm external diameter, and 25mm thick, if it is 6m long and hinged at both ends. - Take $E = 1.2 x 10^5 N/mm^2$ .Taking $f_c = 550 N/mm^2 and α = 1/1600$. -

Mumbai university > MECH > SEM 3 > Strength Of Materials

Marks: 10M

Year: Dec 2013

End Conditions of the Column: Both sides hinged

$$L_e = L = 6000mm$$

The moment of Inertia:

$$I_{x - x} = I_{y - y} = \frac{\pi}{64}(200^4 - 150^4) = 53.689 × 10^6 mm^4$$

Using Euler’s equation:

$$P_{Euler} = \frac{{\pi}^2EI}{L_e^2}$$

$$P_{Euler} = \frac{{\pi}^2 × (1.2 × 10^5) × 53.689 × 10^6 }{6000^2}$$

$$P_{Euler} = 1766.297 kN$$

The area of cross section is given by:

$$A = \frac{π}{4}(200^2 - 150^2)$$

$$A = 13744.47 mm^2$$

$$k^2 = \frac{I}{A} = \frac{53.689 × 10^6}{13744.47} = 3906.23mm^2$$

$$k = 62.5mm$$

Using Rankine’s Formula:

$$P_{Rankine} = \frac{f_c × A}{1 + α(\frac{L_e}{k})^2}$$

$$P_{Rankine} = \frac{550 × 13744.47}{1 + \frac{1}{1600}(\frac{6000}{62.5})^2}$$

$$P_{Rankine} = 1118.263 kN$$