Question

A column of timber section $10\times 20$ cm is 5 meter long both ends being fixed. If the youngs modulus for timber is $18.0kN/m{m}^2$. Then safe load for the column is____kN. (Take FOS = 3)

• A. 158

Answer 1

The correct option is A 158
Effective length, $l_e=\frac{l}{2}=\frac{5000}{2}=2500mm$

Moment of Inertia

$I_{xx}=\frac{10\times {20}^3}{12}=6666.67c{m}^4$

$I_{yy}=\frac{20\times {10}^3}{12}1666.6c{m}^4$

$=1666.67\times {10}^{4}m{m}^4$

$I=I_{min}=1666.67\times {10}^{4}m{m}^4$

Cripling load

$P_{cr}=\frac{{\pi }^{2}EI}{l_e^2}$

$=\frac{{\pi }^2\times 18\times 1666.67\times {10}^4}{(2500{)}^2}=473.7kN$

Safe load = $\frac{P_a}{FOS}=157.9\approx 158kN$