Question #268139





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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




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Expert's answer
2021-11-19T00:07:24-0500


Radius of gyration =r=IA=π64d4π4d2=d4=504=12.5mm= 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=3000mml =2L =3=3000mm

λ=lrmin=300012.5=240\lambda= \frac{l}{r_min}=\frac{3000}{12.5} =240

Rankine Load

Pr=fcA1+αλ2=560π45˙021+16002402=11335.64NP_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

Fsafe=PrFOS=11335.643=3778.55N=3.88kNF_{safe} =\frac{P_r}{FOS}=\frac{11335.64}{3}=3778.55N=3.88kN


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