Question

Using allowable stress design determine the allowable centric load for a column of 10-m effective length that is made from the following rolled-steel shape: W250x101 Use oy = 250 MPa and E = 200 GPa A=12,900x10-6 m2 ry=65..8x10-m

Answer 1

Given data

$\small \\\\\sigma_y=250 \ MPa \\\\E=200*10^3 \ MPa \\\\A=0.0129 \ m^2 \\\\r=0.0658 \ m$

L=10 m

The ration L/r

$\small L/r=10/0.0658=91.18$

slendreness ration

200000 E SR= 4.71 Oy SR= 133.2189 4.71 V 250

Since SR>L/r so ,

$\small \\\\\sigma_{Cr}=0.658^{{\sigma_y}/{\sigma_C}} * \sigma_y \\\\where ,\sigma_{C}=\frac{\pi^2E}{(L/r)^2}=\frac{\pi^2*200*10^3}{91.185^2} \\\\\sigma_C=237.399 \ MPa \\\\\sigma_{Cr}=0.6589^{(250/237.399)}*250=161.1175 \ MPa$

Allowable Load is

$\small \\P=\frac{\sigma_{CR}*A}{FOS} \\\\P=\frac{237.3991*10^6*0.0129}{FOS}=\frac{2075*10^3}{FOS} \ N$

For

FOS =1

P=2075 KN

but normally FOR this type of cases FOS=1.667 is taken

so,

$\small \\P=\frac{2075}{1.667} \\\\P=1245 \ KN$

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