Find the Euler's crippling load for a hollow cylindrical steel column of 3.8 mm external diameter and 2.5 mm thick. Take length of the column as 2.3 m and hinged at its both ends crippling load by Rankine's formula. Using constants as 335 Km/mm² and 1/7500.

Answer 1

External diameter D=38 mm
Inner diameter d=38 -2•2.5 = 33 mm
E=2.05•10⁵ N/mm²
A=π(38²-33²)/4 =278.8 mm²
I = π(38⁴-33⁴)/64 = 44140 mm⁴
L=2300 mm
P(cr) = π²•E•I/L² =
= π²2.05•10⁵•44140/2300² =
=16.8•10³ N = 16.8 kN
k=sqrt(I/A) = sqrt(44140/278.8) =
=12.6 mm
α=1/7500
σ = 335 N/mm²
P = σ•A/{1 + α(L/k)²} =
=335•278.8/{1 + (2300/12.6)²/7500} =
=17160 N =17.16 kN
Check your data
http://books.google.com.ua/books?id=-3RLtkqAV5AC&pg=PA244&lpg=PA244&dq
page 244, #9.4