Question

A 50 x 100 mm timber is used as a column with fixed ends. Determine the minimum length required at which Euler’s formula can be used if E= 10 GPa and critical stress up to proportional limit is 30 MPa. What safe load can be carried with a FS of 2 if the length is 2.5 m

Answer 1

The Euler buckling load is

$P_{cr}=\frac{\pi^2EI}{(KL)^2}\to \sigma_{cr}A=\frac{\pi^2EI}{(KL)^2}$

$L=\sqrt{\frac{\pi^2EI}{K^2\sigma_{cr}A}}$

$I=\frac{100\cdot50^3}{12}=1.04167\cdot10^6(mm^4)=1.04167\cdot10^{-6}(m^4)$

$A=100\cdot50=5000(mm^2)=5\cdot10^{-3}(m^2)$

$L=\sqrt{\frac{\pi^2EI}{K^2\sigma_{cr}A}}=\sqrt{\frac{3.14^2\cdot 10\cdot 10^9\cdot 1.04167\cdot10^{-6}}{0.5^2\cdot 30\cdot10^6\cdot 5\cdot10^{-3}}}=1.65(m)$ . Answer

$\sigma_{cr}=\frac{\pi^2EI}{A(KL)^2}=\frac{3.14^2\cdot 10\cdot 10^9\cdot 1.04167\cdot10^{-6}}{5\cdot10^{-3}(0.5\cdot2.5)^2}=13.1(MPa)$

$P_{cr}=\sigma_{cr}A=65500(N)=65.5(kN)$

safe load $65.5/2=32.75(kN)$ . Answer