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Now and i.e. greater than Ixx.

105 mm

Iyy = [213 + 8.4(210 - 105)2]10~6 + [9.6 + 8.4 x 1052]10-6

2 222.6 x 10"6 , least k = -——-— = 13.25 x 10-3

Now Euler load for fixed-free ends

_ jt2ei _ tc2 x 208 x 109 x 222.6 x 10~6 ~ 1ZT" ~ 4 x 82

Therefore actual load applied to the column

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