Answer to Question #268139 in Mechanical Engineering for Mbulazi

Question

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A 1.5m long column has a circular cross section of 5 cm in diameter. One of the ends of the columns is fixed in direction and position and the other end is free. Given that the factor is safety is 3. Calculate the safe load through the use of

(i) Rankine's formula, take that the yield stress as 560 N/mm² and α = 1/600 for pinned ends

Answer 1

Radius of gyration $= r=\sqrt{\frac{I}{A}} =\sqrt{\frac{\frac{\pi}{64}d^4}{\frac{\pi}{4}d^2}} =\frac{d}{4}=\frac{50}{4}=12.5mm$

One of the ends of the columns is fixed in direction and position and the other end is free

$l =2L =3=3000mm$

$\lambda= \frac{l}{r_min}=\frac{3000}{12.5} =240$

Rankine Load

$P_r = \frac{f_c A}{1+\alpha\lambda^2}=\frac{560 \frac{\pi}{4}\dot50^2}{1+\frac{1}{600}240^2}=11335.64N$

Safe load

$F_{safe} =\frac{P_r}{FOS}=\frac{11335.64}{3}=3778.55N=3.88kN$