A column of timber section 15 cm × 20 cm is 10-meter-long with it

SOLUTION

Concept:

Crippling load (P) for a column based on Euler’s theory is given as:

\({\rm{P}} = \frac{{{{\rm{\pi }}^2}{\rm{EI}}}}{{{\rm{L}}e^2}}\)

Where

E = Elastic modulus of the material

I = Least value of the moment of inertia

L_e = Effective length of the column

Given:

Dimension of section = 15 cm × 20 cm, Actual length (L) = 10 m = 10000 mm, Young’s modulus (E) = 17.5 kN/mm²

Solution:

Effective length (L_e) = L/2 (when both the ends are fixed)

\(\therefore {{\rm{L}}_{\rm{e}}} = \frac{{10000}}{2} = 5000{\rm{\;mm\;}}\left( {{\rm{\because\;L\;}} = 10000{\rm{\;mm}}} \right)\)

For the given cross-section:

Moment of inertia of the section about X-X axis:

\({{\rm{I}}_{{\rm{XX}}}} = \frac{{15 \times {{20}^3}}}{{12}} = 10000{\rm{\;c}}{{\rm{m}}^4} = 10000 \times {10^4}{\rm{\;m}}{{\rm{m}}^4}\)

Moment of inertia of the section about Y-Y axis:

\({{\rm{I}}_{{\rm{YY}}}} = \frac{{20 \times {{15}^3}}}{{12}} = 5625{\rm{\;c}}{{\rm{m}}^4} = 5625 \times {10^4}{\rm{\;m}}{{\rm{m}}^4}\)

Since I_YY is less than I_XX, the column will tend to buckle in Y-Y direction.

\({\rm{P}} = \frac{{{{\rm{\pi }}^2} \times 17.5 \times 5625 \times {{10}^4}}}{{{{5000}^2}}} = 388.62{\rm{\;kN}}\)

Safe load for the column:

\(\therefore {\rm{\;Safe\;load\;}} = \frac{{{\rm{Crippling\;load}}}}{{{\rm{Factor\;of\;safety}}}} = \frac{{388.62}}{3} = 129.54{\rm{\;kN}}\)